Engineering
Mechanics
Statics and Dynamics
Irving H. Shames
Professor
Dept. of Civil, Mechanical and Eirvirorrmenrul En,qirierring
The George Washington Uiiiver.yiQ
Prentice Hall, Upper Saddle River, New Jersey 07458
Acqui\iiionr Editor: William Stenquiit
Editor in Chic1 Marcia Hoiton
I’mduclion Editor: Ro\e Krrnao
Text I l e s p c r : Christme Wull
Covcr Dc\igsei: Amy Roam
Editorial Ashiavant: Meg Wci.1
Manufacturing Buyer: Lhnna Sullivan
K&Jj
0 1907. 19811. 1Yhh. 1959. 10% hy Prsnlice~Hall,Inc
Simon & SchusterlA Viacom Company
Upprr Saddle River. Ncw Jcrsey (17458
The author and puhlisher ol Ihis hook have u ~ r i ihcir
l
k s t eflurl\ in preparing this hook.
These cfforti include Ihe development, rcsearch, anti crrling ofthe theurieq and progrilms
10 deteiminc their sffeclirenen. ‘The mthur arid puhlirher shall not he liable m any cveiit for
incidcniel o r cunrequeiitial d;mngrs wilh. or arising w l OS, lhc furnishing. pcrformiincc. or
usc of their p n i p m s .
All rights lsscrvcd. N o pact ofrhis h w k may be repruduscd o r
rrmsmillcd in m y form or hy m y mrilns. without written peimi\sion
in writing from Ihc puhliahei.
Printetl in the Unilml States of America
IIJ
0
8
7
6
5
4 3
2
I
I S B N 0-L3-35b924-l
Prcnlice-Hall lntcriiilli~niil(UKI Limited. Lundon
Prcnlicc-Hall of Australia Ply Limmd. SyJocy
Prriilice~HnllCan.tda Inc.. Toconlu
Prcnlicc~HallHispanuamcricima, S . A . . Mexico
Prcnlicc-Hall of India Privillc I.imiled. New D c l h i
Prentics~Hilll01 Japan. Inc.. Tokyo
Simon & Schuhlcr Ask& PIC. Lld., S i n p p o r r
Edilorn Prenlice Hall do Rmsil, I.tda., Rio d,: J;meilo
Contents
Preface
$2.3 Addition and Subtraction
of Vectors 24
2.4 Resolution of Vectors; Scalar
Components 30
t2.5 Unit Vectors 33
ix
1 Fundamentals of Mechanics
Review I 3
tl.1
t1.2
Introduction 3
Basic Dimensions and Units
of Mechanics, 4
71.3 Secondary Dimensional Quantities
71.4 Law of Dimensional Homogeneity
71.5 Dimensional Relation between
Force and Mass 9
1.6 Units of Mass 10
1.7 Idealizations of Mechanics 12
11.8 Vector and Scalar Quantities 14
1.9 Equality and Equivalence
of Vectors 17
t1.10 Laws of Mechanics 19
1.11 Closure 22
2 Elements of vector Algebra
Review II 23
t2.1
72.2
Introduction 23
Magnitude and Multiplication
of a Vector by a Scalar 23
7
8
2.6
Useful Ways of Representing
Vectors 35
2.7
Scalar or Dot Product of Two
Vectors 41
2.8
Cross Product of Two Vectors 47
2.9 Scalar Triple Product 5 1
2.10 A Note on Vector Notation 54
2.11 Closure 56
3 Important vector Quantities
3.1
3.2
3.3
3.4
3.5
3.6
61
Position Vector 61
Moment of a Force about a Point 62
Moment of a Force about an Axis 69
The Couple and Couple Moment 77
The Couple Moment as a
Free Vector 79
Addition and Subtraction of
Couples 80
IV
CONTENTS
3.7
3.8
Moment of a Couple About a Line
Closure 89
4 Equivalent Force systems
4.1
4.2
4.3
4.4
4.5
4.6
82
93
Introduction 93
Translation of a Force to
a Parallel Position 94
Resultant of a Force System 102
Simplest Resultants of Special
Force Systems 106
Distributed Force Systems 107
Closure 143
5 Equations of Equilibrium
151
lntroduction 15 I
The Free-body Diagram 152
Free Bodies Involving
Interior Sections 154
*5.4 Looking Ahead-Control
Volumes 158
5.5 General Equations of Equilibrium 162
5.6 Problems of Equilibrium I 164
5.7 Problems of Equilibrium 11 183
5.8 Two Point Equivalent Loading 199
5.9 Problems Arising from Structures 200
5.10 Static Indeterminacy 204
5.11 Closure 210
5.1
5.2
5.3
6 introduction to structural
Mechanics 221
Part A: Trusses 221
6.1
6.2
6.3
6.4
6.5
The Structural Model 221
The Simple Truss 224
Solution of Simple Trusses 225
Method of Joints 225
Method of Sections 238
6.6
Looking Ahead-Deflection of
a Simple, Linearly Elastic Truss 242
Part B: Section Forces in Beams 247
6.7
6.8
6.9
Introduction 247
Shear Force, Axial Force,
and Bending Moment 247
Differential Relations
for Equilibrium 25Y
Part C: Chains and Cables 266
6.10 Introduction 266
6.11 Coplanar Cables; Loading is a Function
ofx 266
6.12 Coplanar Cables: Loading is the Weight
of the Cable Itself 270
6.13 Closure 277
7 Friction FOrCeS
7.1
7.2
*7.3
7.4
7.5
7.6
7.7
*7.8
7.9
281
Introduction 281
Laws of Coulomb Friction 282
A Comment Concerning the
Use of Coulomb’.: Law 284
Simple Contact Friction Problems 284
Complex Surface Contact Friction
Problems 299
Belt Friction 301
The Square Screw Thread 3 17
Rolling Resistance 319
Closure 323
8 Properties of surfaces
8.1
8.2
8.3
8.4
331
Introduction 331
First Moment of an Area and
the Centroid 331
Other Centers 342
Theorems of Pappus-Guldinus
347
CONTENTS
8.5
Second Moments and the Product
of Area of a Plane Area 355
8.6 Tranfer Theorems 356
8.7 Computations Involving Second Moments
and Products of Area 357
8.8 Relation Between Second Moments and
Products of Area 366
8.9 Polar Moment of Area 369
8.10 Principal Axes 370
8.11 Closure 375
9
Moments and Products
oflnertia 379
9.1
9.2
9.3
9.4
“9.5
*9.6
*9.7
9.8
Introduction 379
Formal Definition of
Inertia Quantities 379
Relation Between Mass-Inertia Terms and
Area-Inertia Terms 386
Translation of Coordinate Axes 392
Transformation Properties
of the Inertia Terms 395
Looking Ahead-Tensors 400
The Inertia Ellipsoid and Principal
Moments of Inertia 407
Closure 410
V
10.5 Looking Ahead-Deformable
Solids 424
Part B: Method of Total Potential
Energy 432
10.6
10.7
Conservative Systems 432
Condition of Equilibrium
for a Conservative System 434
10.8 Stability 441
10.9 Looking Ahead-More on Total
Potential Energy 443
10.10 Closure 446
11 Kinematics of a Particle-simple
Relative Motion 451
11.1 Introduction
45 1
Part A: General Notions 452
11.2
Differentiation of a Vector with Respect
toTime 452
Part B: Velocity and Acceleration
Calculations 454
11.3
10 *Methods of virtual work and
stationary Potential Energy 413
10.1 Introduction 413
Part A: Method of Virtual Work 414
10.2 Principle of Virtual Work
for a Particle 414
10.3 Principle of Virtual Work
for Rigid Bodies 415
10.4 Degrees of Freedom and the
Solution of Problems 418
Introductory Remark 454
11.4 Rectangular Components 455
11.5 Velocity and Acceleration in
Terms of Path Variables 465
11.6 Cylindrical Coordinates 480
Part C: Simple Kinematical Relations
and Applications 492
11.7
11.8
Simple Relative Motion 492
Motion of a Particle Relative to a Pair of
Translating Axes 494
11.9
Closure 504
Vi
12
CONTENTS
13.3
13.4
Particle Dynamics 511
12.1
Introduction
51I
Conservative Force Field 594
Conservation of Mechanical
Energy 598
Alternative Form of Work-Energy
Equation 603
Part A: Rectangular Coordinates;
Rectilinear Translation 512
13.5
12.2
Part B: Systems of Particles 609
12.3
12.4
Newton's Law for Rectangular
Coordinates 5 12
Rectilinear Translation 5 12
A Comment 528
13.6
13.7
13.8
Part B: Cylindrical Coordinates;
Central Force Motion 536
13.9
Work-Energy Equations 609
Kinetic Energy Expression Based on
Center of Mass 614
Work-Kinetic Energy Expressions Based
on Center of Mass 619
Closure 631
12.5
Newton's Law for Cylindrical
Coordinates 536
12.6 Central Force MotionAn Introduction 538
*12.7 Gravitational Central Force
Motion 539
"12.8 Applications to Space
Mechanics 544
Part C: Path Variables 561
12.9
Newton's Law for Path
Variables 561
Part D: A System of Particles 564
12.10 The General Motion of a System
of Particles 564
12.11 Closure 571
13
Energy Methods for Particles 579
Part A: Analysis for a
Single Particle 579
13.1
13.2
Introduction 579
Power Considerations
14
Methods of Momentum for
Particles 637
Part A: Linear Momentum 637
14.1
Impulse and Momentum Relations for a
Particle 637
14.2 Linear-Momentum Considerations for a
System of Particles 643
14.3 Impulsive Forces 648
14.4 Impact 659
"14.5 Collision of a Particle with a
Massive Rigid Body 665
Part B: Moment of Momentum 675
14.6
14.7
14.8
*14.9
585
14.10
Moment-of-Momentum Equation
for a Single Particle 675
More on Space Mechanics 678
Moment-of-Momentum Equation
for a System of Particles 686
Looking Ahead-Basic Laws
of Continua 694
Closure 700
CONTENTS
15 Kinematics of Rigid Bodies:
*16.9 Balancing 838
16.10 Closure 846
Relative Motion 707
15.1 Introduction 707
15.2 Translation and Rotation of
Rigid Bodies 707
15.3 Chasles’ Theorem 709
15.4 Derivative of a Vector Fixed
in a Moving Reference 71 1
15.5 Applications of the Fixed-Vector
Concept 723
15.6 General Relationship Between
Time Derivatives of a Vector
for Different References 743
15.7 The Relationship Between
Velocities of a Particle for
Different References 744
15.8 Acceleration of a Particle for Different
References 755
15.9 A New Look at Newton’s Law 773
15.10 The Coriolis Force 776
15.11 Closure 781
16 Kinetics of Plane Motion of Rigid
Bodies 787
16.1
16.2
16.3
16.4
16.5
16.6
16.7
16.8
Introduction 787
Moment-of-Momentum
Equations 788
Pure Rotation of a Body of Revolution
About its Axis of Revolution 791
Pure Rotation of a Body with Two
Orthogonal Planes of Symmetry 797
Pure Rotation of Slablike Bodies 800
Rolling Slablike Bodies 810
General Plane Motion of a
Slablike Body 816
Pure Rotation of an Arbitrary
Rigid Body 834
vii
17
Energy and Impulse-Momentum
Methods for Rigid Bodies 853
17.1
Introduction
853
Part A: Energy Methods 853
17.2
17.3
Kinetic Energy of a Rigid Body
Work-Energy Relations 860
853
Part B: Impulse-Momentum
Methods 878
17.4
17.5
17.6
17.7
Angular Momentum of a
Rigid Body About Any Point
in the Body 878
Impulse-Momentum Equations 882
Impulsive Forces and Torques: Eccentric
Impact 895
Closure 907
18 *Dynamics of General Rigid-Body
Motion 911
18.1
18.2
18.3
18.4
18.5
18.6
18.7
18.8
Introduction 91 1
Euler’s Equations of Motion 914
Application of Euler’s Equations 916
Necessary and Sufficient
Conditions for Equilibrium of a
Rigid Body 930
Three-Dimensional Motion About
a Fixed Point; Euler Angles 930
Equations of Motion Using Euler
Angles 934
Torque-Free Motion 945
Closure 958
...
VI11
CON'lt.NlS
I9 Vibrations
19.1
19.2
19.3
*19.4
*IY.S
19.6
19.7
19.8
* 19.9
19.10
961
Introduction 961
Frce Vihratioii 961
Torsional Vibration 973
Examples of Other Free-Oscillating
Motions 9x2
Energy Methods 984
Linear Restoring Force and a Force
Varying Sinusoidally with Time 900
Linear Restoring Force with Viscous
Damping 999
I.inelrr Rehtoring Force, Viscous
Damping, and a Harmonic
Disturhance IO07
Oscillatory Systems with Multi-Degrees
of Freedom IO I4
Closurc 1022
APPENDIX I
Integration Formulas xvii
APPENDIX II
Computation of Principal
Moments of Inertia xix
APPENDIX 111
Additional Data For
the Ellipse xxi
APPENDIX IV
Proof that Infinitesimal
Rotations Are Vectors xxiii
Projects xxv
Index Ixxvii
Preface
With the publication of the fourth edition, this text moves into the
fourth decade of its existence. In the spirit of the times, the first edition
introduced a number of “firsts” in an introductory engineering
mechanics textbook. These “firsts” included
a) the first treatment of space mechanics
b) the first use of the control volume for linear momentum considerations of fluids
c ) the first introduction to the concept of the tensor
Users of the earlier editions will be glad to know that the 4th edition continues with the same approach to engineering mechanics. The
goal has always been to aim toward working problems as soon as possible from first principles. Thus, examples are carefully chosen during
the development of a series of related areas to instill continuity in the
evolving theory and then, after these areas have been carefully discussed with rigor, come the problems. Furthermore at the ends of each
chapter, there are many problems that have not been arranged by text
section. The instructor is encouraged as soon as hekhe is well along in
the chapter to use these problems. The instructors manual will indicate
the nature of each of these problems as well as the degree of difficulty.
The text is not chopped up into many methodologies each with an
abbreviated discussion followed by many examples for using the specific methodology and finally a set of problems carefully tailored for
the methodology. The nature of the format in this and preceding editions is more than ever first to discourage excessive mapping of homework problems from the examples. And second, it is to lessen the
memorization of specific, specialized methodologies in lieu of absorbing basic principles.
ix
X
PREFACE
A new feature in the fourth edition is a series of starred sections
called “Looking Ahead . . . .” These are simplified discussions of topics that appear in later engineering courses and tie in directly or indirectly to the topic under study. For instance, after discussing free
body diagrams, there is a short “Looking Ahead” section in which the
concept and use of the control volume is presented as well as the system concepts that appear in tluid mechanics and thermodynamics. In
the chapter on virtual work for particles and rigid bodies, there is a
simplified discussion of the displacement methods and force methods
for deformable bodies that will show up later in solids courses. After
finding the forces for simple trusses, there is a “Looking Ahead” section discussing brietly what has to be done to get displacements.
There are quite a few others in the text. It has been found that many
students find these interesting and later when they come across these
topics in other courses or work, they report that the connections so
formed coming out of their sophomore mechanics courses have been
most valuable.
Over 400 new problems have been added to the fourth edition
equally divided between the statics and dynamics books. A complete
word-processed solutions manual accompanies the text. The illustrations needed for problem statement and solution are taken as enlargements from the text. Generally, each problem is on a separate page.
The instructor will be able conveniently to select problems in order to
post solutions or to form transparencies as desired. Also, there are 30
computer projects in which, for a number of cases, the student prepares hidher own software or engages in design. As an added bonus,
the student will be able to maintain hidher proficiency in programming. Carefully prepared computer programs as well as computer outputs will be included in the manual. I normally assign one or two such
projects during a semester over and above the usual course material.
Also included in the manual is a disk that has the aforementioned programs for each of the computer projects. The computer programs for
these projects generally run about 30 lines of FORTRAN and run on a
personal computer. The programming required involves skills developed in the freshman course on FORTRAN.
Another important new feature of the fourth edition is an organization that allows one to go directly to the three dimensional chapter
on dynamics of rigid bodies (Chapter 18) and then to easily return to
plane motion (Chapter 16). Or one can go the opposite way. Footnotes
indicate how this can be done, and complimentary problems are noted
in the Solutions Manual.
PREFACE
Another change is Chapter 16 on plane motion. It has been
reworked with the aim of attaining greater rigor and clarity particularly in the solving of problems.
There has also been an increase in the coverage and problems for
hydrostatics as well as examples and problems that will preview problems coming in the solids course that utilize principles from statics.
It should also be noted that the notation used has been chosen to
correspond to that which will be used in more advanced courses in
order to improve continuity with upper division courses. Thus, for moments and products of inertia I use I, I,, lxzetc. rather than ly l,, ,P
etc. The same notation is used for second moments and products of
area to emphasize the direct relation between these and the preceding
quantities. Experience indicates that there need be. no difficulty on the
student’s part in distinguishing between these quantities; the context of
the discussion suffices for this purpose. The concept of the tensor is
presented in a way that for years we have found to be readily understood by sophomores even when presented in large classes. This saves
time and makes for continuity in all mechanics courses, particularly in
the solid mechanics course. For bending moment, shear force, and
stress use is made of a common convention for the sign-namely the
convention involving the normal to the area element and the direction
of the quantity involved be it bending moment, shear force or stress
component. All this and indeed other steps taken in the book will make
for smooth transition to upper division course work.
In overall summary, two main goals have been pursued in this
edition. They are
1. To encourage working problems from first principles and thus to
minimize excessive mapping from examples and to discourage rote
learning of specific methodologies for solving various and sundry
kinds of specific problems.
2. To “open-end” the material to later course work in other engineering sciences with the view toward making smoother transitions and
to provide for greater continuity. Also, the purpose is to engage the
interest and curiosity of students for further study of mechanics.
During the 13 years after the third edition, I have been teaching
sophomore mechanics to very large classes at SUNY, Buffalo, and,
after that, to regular sections of students at The George Washington
University, the latter involving an international student body with very
diverse backgrounds. During this time, I have been working on improving the clarity and strength of this book under classroom conditions
XI
xii
PREFACE
giving it the most severe test as a text. I believe the fourth edition as a
result will be a distinct improvement over the previous editions and
will offer a real choice for schools desiring a more mature treatment of
engineering mechanics.
I believe sophomore mechanics is probably the most important
course taken by engineers in that much of the later curriculum depends
heavily on this course. And for all engineering programs, this is usually the first real engineering course where students can and must be
creative and inventive in solving problems. Their old habits of mapping and rote learning of specific problem methodologies will not suffice and they must learn to see mechanics as an integral science. The
student must “bite the bullet” and work in the way he/she will have to
work later in the curriculum and even later when getting out of school
altogether. No other subject so richly involves mathematics, physics,
computers, and down to earth common sense simultaneously in such
an interesting and challenging way. We should take maximum advantage of the students exposure to this beautiful subject to get h i d h e r on
the right track now so as to be ready for upper division work.
At this stage of my career, I will risk impropriety by presenting
now an extended section of acknowledgments. I want to give thanks to
SUNY at Buffalo where I spent 31 happy years and where I wrote
many of my hooks. And I want to salute the thousands (about 5000) of
fine students who took my courses during this long stretch. I wish to
thank my eminent friend and colleague Professor Shahid Ahmad who
among other things taught the sophomore mechanics sequence with
me and who continues to teach it. He gave me a very thorough review
of the fourth edition with many valuable suggestions. I thank him profusely. I want particularly to thank Professor Michael Symans, from
Washington State University, Pullman for his superb contributions to
the entire manuscript. I came to The George Washington University at
the invitation of my longtime friend and former Buffalo colleague
Dean Gideon Frieder and the faculty in the Civil, Mechanical and
Environmental Engineering Department. Here, I came back into contact with two well-known scholars that I knew from the early days of
my career, namely Professor Hal Liebowitz (president-elect of the
National Academy of Engineering) and Professor Ali Cambel (author
of recent well-received book on chaos). 1 must give profound thanks to
the chairman of my new department at G.W., Professor Sharam
Sarkani. He has allowed me to play a vital role in the academic program of the department. I will be able to continue my writing at full
speed as a result. 1 shall always be grateful to him. Let me not forget
PREFACE
the two dear ladies in the front office of the department. Mrs. Zephra
Coles in her decisive efficient way took care of all my needs even
before I was aware of them. And Ms. Joyce Jeffress was no less helpful and always had a humorous comment to make.
I was extremely fortunate in having the following professors as
reviewers.
Professor Shahid Ahmad, SUNY at Buffalo
Professor Ravinder Chona, Texas A&M University
Professor Bruce H. Kamopp. University of Michigan
Professor Richard E Keltie, North Carolina State University
Professor Stephen Malkin, University of Massachusetts
Professor Sudhakar Nair, Illinois Institute of Technology
Professor Jonathan Wickert, Camegie Mellon University
I wish to thank these gentlemen for their valuable assistance and
encouragement.
I have two people left. One is my good friend Professor Bob
Jones from V.P.I. who assisted me in the third edition with several
hundred excellent statics problems and who went over the entire manuscript with me with able assistance and advice. I continue to benefit
in the new edition from his input of the third edition. And now, finally,
the most important person of all, my dear wife Sheila. She has put up
all these years with the author of this book, an absent-minded, hopeless workaholic. Whatever I have accomplished of any value in a long
and ongoing career, I owe to her.
To my Dear, Wondeijiil Wife Sheila
...
XI11
About the Author
Irving Shames presently serves as a Professor in the Department of
Civil, Mechanical, and Environmental Engineering at The George
Washington University. Prior to this appointment Professor Shames
was a Distinguished Teaching Professor and Faculty Professor at The
State University of New York-Buffalo, where he spent 31 years.
Professor Shames has written up to this point in time 10 textbooks. His first book Engineering Mechanics, Statics and Dynamics
was originally published in 1958, and it is going into its fourth edition
in 1996. All of the books written by Professor Shames have been characterized by innovations that have become mainstays of how engineering principles are taught to students. Engineering Mechanics, Statics
and Dynamics was the first widely used Mechanics book based on
vector principles. It ushered in the almost universal use of vector principles in teaching engineering mechanics courses today.
Other textbooks written by Professor Shames include:
Mechanics of Deformable Solids, Prentice-Hall, Inc.
Mechanics of Fluids, McGraw-Hill
* Introduction to Solid Mechanics, Prentice-Hall, Inc.
* Introduction to Statics, Rentice-Hall, Inc.
* Solid Mechanics-A Variational Approach (with C.L. Dym),
McCraw-Hill
Energy and Finite Element Methods in Structural Mechanics, (with
C.L. Dym), Hemisphere Corp., of Taylor and Francis
Elastic and Inelastic Stress Analysis (with F. Cozzarelli), PrenticeHall, Inc.
-
XVI
ABOIJTTHF AI'THOK
In recent ycars, I'rofesor Shalne\ has expanded his teaching
xtivitics and t i a h held two suiiiiner fiicully workshops in mechanics
\ponsored by the State (if Ncw York, and one national workshop sponsorcd by the National Science Foundation. The programs involved the
iiitegr:ition both conceptually and pedagogically 0 1 mechanics from
the sophomore year on through gt-aduate school.
Statics
REVIEW I*
Fundamentals
of Mechanics
+l.l
Introduction
Mechanics is the physical science concerned with the dynamical behavior (as
opposed to chemical and thermal behavior) of bodies that are acted on by
mechanical disturbances. Since such behavior is involved in virtually all the
situations that confront an engineer, mechanics lies at the core of much engineering analysis. In fact, no physical science plays a greater role in engineering than does mechanics, and it is the oldest of all the physical sciences. The
writings of Archimedes covering buoyancy and the lever were recorded
before 200 B.C. Our modem knowledge of gravity and motion was established
by Isaac Newton (1642-1727), whose laws founded Newtonian mechanics,
the subject matter of this text.
In 1905, Einstein placed limitations on Newton's formulations with his
theory of relativity and thus set the stage for the development of relativistic
mechanics. The newer theories, however, give results that depart from those
of Newton's formulations only when the speed of a body approaches the
speed of light ( I 86,000 mileslsec). These speeds are encountered in the largescale phenomena of dynamical astronomy. Furthermore for small-scale
phenomena involving subatomic particles, quantum mechanics must be used
rather than Newtonian mechanics. Despite these limitations, it remains nevertheless true that, in the great bulk of engineering problems, Newtonian
mechanics still applies.
*The reader is urged 10 be sure that Section 1.9 is thoroughly understood since this Section
is vital for a goad understanding of statics in panicular and mechanics in general.
Also, the nutation t before the titles of certain sections indicates thal specific queslions
concerning the contents of these sections requiring verbal answers are presented at the end of the
chapler. The instructor may wish to assign these sections as a reading asignment along with the
requirement to answer the aforestated asssiated questions as the author routinely daes himself.
3
4
CHAPTER I
FUNDAMENTALS OFMECHANICS
t1.2
Basic Dimensions and Units of
Mechanics
To study mechanics, we must establish abstractions to describe those characteristics of a body that interest us. These abstractions are called dimensions.
The dimensions that we pick, which are independent of all other dimensions,
are termed primary or basic dimensions, and the ones that are then developed
in terms of the basic dimensions we call secondary dimensions. Of the many
possible sets of basic dimensions that we could use, we will confine ourselves
at present to the set that includes the dimensions of length, time, and mass.
Another convenient set will he examined later.
Length-A Concept for Describing Size Quantitatively. In order to determine the size of an object, we must place a second object of known size next
to it. Thus, in pictures of machinery, a man often appears standing disinterestedly beside the apparatus. Without him, it would be difficult to gage the
size of the unfamiliar machine. Although the man has served as some sort of
standard measure, we can, of course, only get an approximate idea of the
machine's size. Men's heights vary, and, what is even worse, the shape of a
man is too complicated to be of much help in acquiring a precise measurement of the machine's size. What we need, obviously, is an object that is
constant in shape and, moreover, simple in concept. Thus, instead of a threedimensional object, we choose a one-dimensional object.' Then, we can use
the known mathematical concepts of geometry to extend the measure of size
in one dimension to the three dimensions necessary to characterize a general
body. A straight line scratched on a metal bar that is kept at uniform thermal
and physical conditions (as, e.g., the meter bar kept at Skvres, France) serves
as this simple invariant standard in one dimension. We can now readily calculate and communicate the distance along a cettain direction of an object by
counting the number of standards and fractions thereof that can be marked off
along this direction. We commonly refer to this distance as length, although
the term " length could also apply to the more general concept of size. Other
aspects of size, such as volume and area, can then be formulated in terms of
the standard by the methods of plane, spherical, and solid geometry.
A unit is the name we give an accepted measure of a dimension. Many
systems of units are actually employed around the world, but we shall only
use the two major systems, the American system and the SI system. The basic
unit of length in the American system is the foot, whereas the basic unit of
length in the SI system is the meter.
Time-A Concept for Ordering the Flow of Events. In observing the picture of the machine with the man standing close by, we can sometimes tell
approximately when the picture was taken by the style of clothes the man is
'We are using the word "dimensional" here in its everyday sense and not as defined above.
SECTION 1.2 BASIC DIM!3”SIONS AND UNITS OF MECHANICS
wearing. But how do we determine this? We may say to ourselves: “During
the thirties, people wore the type of straw hat that the fellow in the picture is
wearing.” In other words, the “when” is tied to certain events that are experienced by, or otherwise known to, the observer. For a more accurate description of “when,” we must find an action that appears to he completely
repeatable. Then, we can order the events under study by counting the numher of these repeatable actions and fractions thereof that occur while the
events transpire. The rotation of the earth gives rise to an event that serves as
a good measure of time-the day. But we need smaller units in most of our
work in engineering, and thus, generally, we tie events to the second, which is
an interval repeatable 86,400 times a day.
Mass-A Property of Matter. The student ordinarily has no trouble understanding the concepts of length and time because helshe is constantly aware
of the size of things through hisher senses of sight and touch, and is always
conscious of time by observing the flow of events in hisher daily life. The
concept of mass, however, is not as easily grasped since it does not impinge
as directly on our daily experience.
Mass is a property of matter that can be determined from two different
actions on bodies. To study the first action, suppose that we consider two
hard bodies of entirely different composition, size, shape, color, and so on. If
we attach the bodies to identical springs, as shown in Fig. 1.1, each spring
will extend some distance as a result of the attraction of gravity for the hodies. By grinding off some of the material on the body that causes the greater
extension, we can make the deflections that are induced on both springs
equal. Even if we raise the springs to a new height above the earth’s surface,
thus lessening the deformation of the springs, the extensions induced by the
pull of gravity will he the same for both bodies. And since they are, we can
conclude that the bodies have an equivalent innate property. This property of
each body that manifests itself in the amount of gravitational attraction we
call man.
The equivalence of these bodies, after the aforementioned grinding operation, can be indicated in yet a second action. If we move both bodies an
equal distance downward, by stretching each spring, and then release them at
the same time, they will begin to move in an identical manner (except for
small variations due to differences in wind friction and local deformations of
the bodies). We have imposed, in effect, the same mechanical disturbance on
each body and we have elicited the same dynamical response. Hence, despite
many obvious differences, the two bodies again show an equivalence.
mpcs, thn, Chomcrcrke8 a body both in the action of
na1 a n r a c k and in tlu response IO a mekhnnicd
The pcoperry of
To communicate this property quantitatively, we may choose some
convenient body and compare other bodies to it in either of the two above-
Body A
Body B
Figure 1.1. Bodies restrained by identical
springs.
5
6
CHAPTER I
FUNDAMENTALS OF MECHANICS
mentioned actions. The two basic units commonly used in much American
engineering practice to measure mass are the pound mass, which is defined in
terms of the attraction of gravity for a standard body at a standard location,
and the slug, which is defined in terms of the dynamical response of a standard body to a standard mechanical disturbance. A similar duality of mass
units does not exist in the SI system. There only the kilugmm is used as the
basic measure of mass. The kilogram is measured in terms of response of a
body to a mechanical disturbance. Both systems of units will he discussed
further in a subsequent section.
We have now established three basic independent dimensions to
describe certain physical phenomena. It is convenient to identify these dimensions in the following manner:
length
[L]
time
mass
[tl
[MI
These formal expressions of identification for basic dimensions and the more
complicated groupings to he presented in Section 1.3 for secondary dimensions are called “dimensional representations.”
Often, there are occasions when we want to change units during computations. For instance, we may wish to change feet into inches or millimeters. In such a case, we must replace the unit in question by a physically
equivalent number of new units. Thus, a foot is replaced by 12 inches or 30.5
millimeters. A listing of common systems of units is given in Table 1.1, and
a table of equivalences hetween these and other units is given on the inside
covers. Such relations between units will he expressed in this way:
1 ft
12 in.
= 305 mm
The three horizontal bars are not used to denote algebraic equivalence;
instead, they are used to indicate physical equivalence. Here is another way
of expressing the relations above:
Table 1.1 common systems of
units
SI
c!P
Mass
Length
Time
FOKC
Gram
Centimeter
Second
Dyne
English
Mass
Length
Time
Force
Pound mass
Foot
Second
Poundal
Mass
Kilogram
Length
Meter
Time
Second
Force
Newton
American Practice
Mass
Length
Time
Force
Slug or pound mass
Foot
Second
Puund force
SECTION 1.3 SECONDARY DIMENSIONAL QUANTITIES
The unity on the right side of these relations indicates that the numerator and
denominator on the left side are physically equivalent, and thus have a 1:l
relation. This notation will prove convenient when we consider the change of
units for secondary dimensions in the next section.
t1.3
Secondary Dimensional Quantities
When physical characteristics are described in terms of basic dimensions by
the use of suitable definitions (e.g., velocity is defined2 as a distance divided
by a time interval), such quantities are called secondary dimensional quantities. In Section 1.4, we will see that these quantities may also be established as
a consequence of natural laws. The dimensional representation of secondary
quantities is given in terms of the basic dimensions that enter into the formulation of the concept. For example, the dimensional representation of velocity is
[velocity] = [Ll
[/I
That is, the dimensional representation of velocity is the dimension length
divided by the dimension time. The units for a secondary quantity are then
given in terms of the units of the constituent basic dimensions. Thus,
[velocity units] = [ftl
[secl
A chunge of units from one system into another usually involves a
change in the scale of measure of the secondary quantities involved in the
problem. Thus, one scale unit of velocity in the American system is 1 foot per
second, while in the SI system it is I meter per second. How may these scale
units he correctly related for complicated secondary quantities? That is, for
our simple case, how many meters per second are equivalent to 1 foot per
second? The formal expressions of dimensional representation may he put to
good use for such an evaluation. The procedure is as follows. Express the
dependent quantity dimensionally; substitute existing units for the basic
dimensions; and finally, change these units to the equivalent numbers of units
in the new system. The result gives the number of scale units of the quantity
in the new system of units that is equivalent to 1 scale unit of the quantity in
the old system. Performing these operations for velocity, we would thus have
l(&)
I(*)
>A more precise definilion will be given
= .305(2)
in the chapters on dynamics.
7
8
CHAPTER I
FUNDAMENTALS OF MECHANICS
which means that ,305scale unit of velocity in the SI system is equivalent to
I scale unit in the American system.
Another way of changing units when secondary dimensions are present
is to make use of the formalism illustrated in relations 1.1. To change a unit
in an expression, multiply this unit by a ratio physically equivalent to unity,
as we discussed earlier, so that the old unit is canceled out, leaving the
desired unit with the proper numerical coefficient. In the example of velocity
used above, we may replace ft/sec by mlsec in the following manner:
It should he clear that, when we multiply by such ratios to accomplish a
change of units as shown above, we do not alter the magnitude of the actual
physical quantity represented by the expression. Students are strongly urged
to employ the above technique in their work, for the use of less formal methods is generally an invitation to error.
t1.4
l a w of Dimensional Homogeneity
Now that we can describe certain aspects of nature in a quantitative manner
through basic and secondary dimensions, we can by careful observation and
experimentation leam to relate certain of the quantities in the form of equations. In this regard, there is an important law, the law of dimen.siona1 homogeneity, which imposes a restriction on the formulation of such equations. This
law states that. because natural phenomena proceed with no regard for manmade units, basic equations representing physical phenomena must be valid
f o r all systems of units. Thus, the equation for the period of a pendulum,
7
t = 2 x , / ~ / g , must be valid for all systems of units, and is accordingly said to
be dimensionally homogeneous. It then follows that the fundamental equations
of physics are dimensionally homogeneous; and all equations derived analytically from these fundamental laws must also be dimensionally homogeneous.
What restriction does this condition place on an equation? To answer
this, let us examine the following arbitrary equation:
x=ygd+k
For this equation to be dimensionally homogeneous, the numerical equality
between both sides of the equation must he maintained for all systems of
units. To accomplish this, the change in the scale of measure of each group of
terms must be the same when there is a change of units. That is, if the numerical measure of one group such as ygd is doubled for a new system 0 1 units,
so must that of the quantities x and k. For 1hi.r to occur under all systems
of units, it is necessary that everj grouping in the eyuution have the .same
dimensirmal representation.
In this regard, consider the dimensional representation of the above
equation expressed in the following manner:
SECTION 1.5 DIMENSIONAL. RELATION BETWEEN FORCE AND MASS
[XI
= b g 4 + [kl
From the previous conclusion for dimensional homogeneity, we require that
[XI
= [yg4 = [kl
As a further illustration, consider the dimensional representation of an
equation that is not dimensionally homogeneous:
[LI = [fl’ + [rl
When we change units from the American to the SI system, the units of feet
give way to units of meters, but there is no change in the unit of time, and it
becomes clear that the numerical value of the left side of the equation
changes while that of the right side does not. The equation, then, becomes
invalid in the new system of units and hence is not derived from the basic
laws of physics. Throughout this book, we shall invariably be concerned with
dimensionally homogeneous equations. Therefore, we should dimensionally
analyze our equations to help spot errors.
Dimensional Relation Between
Force and Mass
t1.5
We shall now employ the law of dimensional homogeneity to establish a new
secondary dimension-namely force. A superficial use of Newton’s law will
be employed for this purpose. In a later section, this law will be presented in
greater detail, but it will suffice at this time to state that the acceleration of a
particle3 is inversely proportional to its mass for a given disturbance. Mathematically, this becomes
a = -1
(1.2)
m
where is the proportionality symbol. Inserting the constant of proportionality, F, we have, on rearranging the equation,
-
F=ma
(1.3)
The mechanical disturbance, represented by F and calledforce, must have the
following dimensional representation, according to the law of dimensional
homogeneity:
[ F ] = [ M I -[Ll
[fIZ
(1.4)
The type of disturbance for which relation 1.2 is valid is usually the action of
one body on another by direct contact. However, other actions, such as magnetic, electrostatic, and gravitational actions of one body on another involving
no contact, also create mechanical effects that are valid in Newton’s equation.
‘We shall define panicles in Section 1.7.
9
10
CHAPTER I
FLNDAMENTALS OF MECHANICS
We could have initiated the study of mechanics by consideringfiirce as
a basic dimension, the manifestation of which can he measured by the elongation of a standard spring at a prescribed temperature. Experiment would
then indicate that for a given body the acceleration is directly proportional to
the applied force. Mathematically,
F
m
a; therefore, F = mu
from which we see that the proportionality constant now represents the property of mass. Here, mass is now a secondary quantity whose dimensional representation is determined from Newton's law:
As was mentioned earlier, we now have a choice between two systems
of basic dimensions-the MLt or the FLr system of basic dimensions. Physicists prefer the former, whereas engineers usually prefer the latter.
1.6
Units of Mass
As we have already seen, the concept of mass arose from two types of actions
-those of motion and gravitational attraction. In American engineering practice, units of mass are based on hoth actions, and this sometimes leads to confusion. Let us consider the FLt system of basic dimensions tor the following
discussion. The unit of force may he taken to be the pound-force (Ihf), which
is defined as a force that extends a standard spring a certain distance at
a given temperature. Using Newton's law, we then define the slug as the
amount of mass that a I-pound force will cause to accelerate at the rate of
I foot per second per second.
On the other hand, another unit of mass can he stipulated if we use the
gravitational effect as a criterion. Herc. the pound muxs (Ihm) is defined as
the amount of matter that is drawn by gravity toward the earth by a force of
I pound-force (Ihf) at a specified position on the earth's surface.
We have formulated two units of mass by two different actions, and to
relate these units we must subject them to the sumt. action. Thus, we can take
1 pound mass and see what fraction or multiplc of it will be accelerated
1 ft/sec2 under the action of I pound afforce. This fraction or multiple will then
represent the number of units of pound mass that are equivalent to I slug.
It turns out that this coefficient is go, where g, has the value corresponding to
the acceleration of gravity at a position on the earth's surface where the
pound mass was standardized. To three significant figures, the value of R~ is
32.2. We may then make the statement of equivalence that
I slug
= 32.2 pounds mass
SECTION 1.6 UNITS OF MASS
To use the pound-mass unit in Newton’s law, it is necessary to divide by
go to form units of mass, that have been derived from Newton’s law. Thus,
where m has the units of pound mass and &go has units of slugs. Having
properly introduced into Newton’s law the pound-mass unit from the viewpoint of physical equivalence, let us now consider the dimensional homogeneity of the resulting equation. The right side of &. 1.6 must have the
dimensional representation of F and, since the unit here for F is the pound
force, the right side must then have this unit. Examination of the units on the
right side of the equation then indicates that the units of go must be
(1.7)
How does weight tit into this picture? Weight is defined as the force of
gravity on a body. Its value will depend on the position of the body relative to
the earth‘s surface. At a location on the earth’s surface where the pound mass is
standardized, a mass of 1 pound (Ibm) has the weight of 1 pound (Ibf), but with
increasing altitude the weight will become smaller than 1 pound (Ibf). The
mass, however, remains at all times a I-pound mass (Ibm). If the altitude is not
exceedingly large, the measure of weight, in Ibf, will practically equal the measure of mass, in Ibm. Therefore, it is unfortunately the practice in engineering to
think erroneously of weight at positions other than on the earth‘s surface as the
measure of mass, and consequently to use the symbol W to represent either Ibm
or Ibf. In this age of rockets and missiles, it behooves us to be careful about the
proper usage of units of mass and weight throughout the entire text.
If we know the weight of a body at some point, we can determine its
mass in slugs very easily, provided that we know the acceleration of gravity,
g, at that point. Thus, according to Newton’s law,
W(lbf) = m(s1ugs) x g(ft/sec*)
Therefore,
(1 3 )
Up to this point, we have only considered the American system of
units. In the SI system of units, a kilogram is the amount of mass that will
accelerate 1 m/sec2 under the action of a force of 1 newton. Here we do not
have the problem of 2 units of mass; the kilogram is the basic unit of mass.
However, we do have another kind of problem-that the kilogram is unfortunately also used as a measure of force, as is the newton. One kilogram of
force is the weight of 1 kilogram of mass at the earth‘s surface, where the
acceleration of gravity (Le., the acceleration due to the force of gravity) is
11
12
CHAPTER I
FUNDAMENTALS OF MECHANICS
9.81 m/sec2. A newton, on the other hand, is the force that causes I kilogram
of mass to have an acceleration of 1 m/sec2. Hence, Y.8 1 newtons are equivalent to I kilogram of force. That is,
9.81 newtons
1 kilogram(force)
= 2.205 Ibf
Note from the above that the newton is a comparatively small force, equaling
approximately one-fifth of a pound. A kilonewton (1000 newtons), which
will be used often, is about 200 Ib. In this text, we shall nor use the kilogram
as a unit of force. However, you should he aware that many people do."
Note that at the earth's surface the weight W o1a mass M is:
(1.9)
W(newtons) = [M(kilograms)](Y.81)(m/s2)
Hence:
M(kilograms) =
W(newtons)
_ _ _
9.81 (rnls')
_
~
~
~
Away from the earth's surfxe, use the acceleration of gravity
9.81 in the above equations.
1.7
~
x
(1.10)
rather than
Idealizations of Mechanics
As we have pointed out, basic and secondary dimensions may sometimes be
related in equations to represent a physical action that we are interested in.
We want to represent an action using the known laws of physics, and also to
be able to form equations simple enough to he susceptible to mathematical
computational techniques. Invariably in our deliberations, we must replace
the actual physical action and the participating bodies with hypothetical,
highly simplified substitutes. We must he sure, of course, that the results of
our substitutions have some reasonable correlation with reality. All analytical
physical sciences must resort to this technique, and. consequently, their coniputations are not cut-and-dried but involve a considerable amount of imagination, ingenuity, and insight into physical behavior. We shall, at this time,
set forth the most fundamental idealizations of mechanics and a hit of the philosophy involved in scientific analysis.
Continuum. Even the simpliI"ica1ion of matter into molecules, atoms, electrons, and so on, is too complex a picture for many problems of engineering
mechanics. In most problems, we are interested only in the average measurable manifestations of these elementary bodies. Pressure, density, and temperature are actually the gross effects of the actions of the many molecules
and atoms, and they can be conveniently assumed to arise from a hypothetically continuous distribution of matter, which we shall call the continuum,
instead of from a conglomeration of discrete, tiny hodies. Without such an
'This is particularly true in the marketplace where the word "kilos" is often heard
SECTION 1.7 IDEALIZATIONS OF MECHANICS
artifice, we would have to consider the action of each of these elementary
bodies-a virtual impossibility for most problems.
Rigid Body. In many cases involving the action on a body by a force, we
simplify the continuum concept even further. The most elemental case is that
of a rigid body, which is a continuum that undergoes theoretically no deformation whatever. Actually, every body must deform to a certain degree under
the actions of forces, hut in many cases the deformation is ton small to affect
the desired analysis. It is then preferable to consider the body as rigid, and
proceed with simplified computations. For example, assume that we are to
determine the forces transmitted by a beam to the earth as the result of a load
P (Fig. 1.2). If P is small enough, the beam will undergo little deflection, and
we can carry out a straightforward simple analysis using the undefomed
geometry as if the body were indeed rigid. If we were to attempt a more accurate analysis-even though a slight increase in accuracy is not required-we
would then need to know the exact position that the load assumes relative to
the beam afrer the beam has ceased to deform, as shown in an exaggerated
manner in Fig. 1.3. To do this accurately is a hopelessly difficult task, especially when we consider that the support must also “give” in a certain way.
Although the alternative to a rigid-body analysis here leads us to a virtually
impossible calculation, situations do arise in which more realistic models
must be employed to yield the required accuracy. For example, when determining the internal force distribution in a body, we must often take the deformation into account, however small it might be. Other cases will be presented
later. The guiding principle is to make such simplifications as are consistent
with the required accuracy ojthe results.
Point Force. A finite force exerted on one body by another must cause a
finite amount of local deformation, and always creates a finite area of contact
between the bodies through which the force is transmitted. However, since
we have formulated the concept of the figid body, we should also be able to
imagine a finite force to be transmitted through an infinitesimal area or point.
This simplification of a force distribution is called a point force. In many
cases where the actual area of contact io a problem is very small but is not
known exactly, the use of the concept of the point force results in little sacrifice in accuracy. In Figs. 1.2 and 1.3, we actually employed the graphical representation of the point force.
Particle. The particle is defined as an object that has no size but that has a
mass. Perhaps this does not sound like a very helpful definition for engineers
to employ, but it is actually one of the most useful in mechanics. For the trajectory of a planet, for example, it is the mass of the planet and not its size
that is significant. Hence, we can consider planets as particles for such computations. On the other hand, take a figure skater spinning on the ice. Her revolutions are controlled beautifully by the orientation of the body. In this
motion, the size and distribution of the body are significant, and since a
Figure 1.2. Rigid-body assumption-use
original geometry.
Figure 1.3. Deformable body.
13
14
CHAPTER 1 FUNDAMENTALS OF MECHANICS
particle, by definition. can have no distribution. i t i s patently clear that a particle cannot represent the skater in this case. If; however, the skater should he
hilled as the “human cannonball on skates” and he shot out of a large air gun.
i t would be possible to consider her as a single particle i n ascertaining her Lrajectory, since arm and leg movements that werc significant while she was
spinning on the ice would have l i t t l e effect o n the arc traversed by the main
portion of her body.
You will learn later that the wiitri- ofnirrss 01-muss w i i f r r i s a hypiithelical point at which one can concentrate thc mass ot the body for ccrliiin
dynamics calculations. Actually i n the previous cxamplcs of thc planet and
the “human cannonball on skates,” the particle wc reler to i s actually the
mass center whose motion i s sufficient for the desired infiirmation. Thus,
when the motion of the mass center o f a body suffices for thc information
desired, we can replace the body by a particle. n m e l y the mass center.
Many other simplifications pervade mechanics. The perfectly elastic
body, the frictionless fluid, and so on. will become familiar as you study various phases nf mechanics.
11.8
Vector and Scalar Quantities
We have now proposed sets of basic dimensions and secondary dimensions to
describe certain aspects uf nature. However. more than just the dimensional
identification and the number (if units arc often necdcd to convey adequately
the desired information. For instance, to specify fully the motion o f a car,
which we may represent as ii particle at this tiine. we must answer the lollowing questions:
1. How fast?
2. Which way?
The concept o f velocity entails the information desired in questions I and 2.
The first question, “How fast?”, i s answered hy the speednmeter reading,
which gives the value o f the velncity in miles per hour or kilometers per hour.
The second question, “Which w a y ~ y i,s more complicated. hecause two separete factors arc involved. First, we must specify the angular orientation of the
velocity relative to a reference Srame. Second, we [nust speciSy the sense (if
the velocity, which tells us whether we are moving r o w i r d o r uw’ay.from ii
given point. The concepts o f angular orientation of the velocity and sense o f
the velocity are often collectively denoted as the dire&irr of the velocity.
Graphically, we may use a directed linr, .rrgmr’nt (an arrow) to describe the
velocity of the car. The length o f the directed line segment gives information
as to “how fast” and i s the magnitude o f the velocity. The angular orientation
of the directed line segment and the position of’the arrowhead give inturmation as to “which way”-that is, as tii the direction o f the velocity. The
SECTION 1.8 VECTOR AND SCALAR QUANTlTlES
15
directed line segment itself is called the velocity, whereas the length of the
directed line segment-that is, the magnitude-is called the speed.
There are many physical quantities that are represented by a directed
line segment and thus are describable by specifying a magnitude and a direction. The most common example is force, where the magnitude is a measure
of the intensity of the force and the direction is evident from how the force is
applied. Another example is the displacement vecior between two points on
the path of a particle. The magnitude of the displacement vector corresponds
to the distance moved along a straighr line between two points, and the direction is defined by the orientation of this line relative to a reference, with the
sense corresponding to which point is being approached. Thus, pae (see Fig.
1.4) is the displacement vector from A to B (while p,, goes from B to A).
7
Path of
a particle
*-(. \-
I
1.4
\
Figure 1.4. Displacement vector pAB.
Certain quantities having magnitude and direction combine their effects
in a special way. Thus, the combined effect of two forces acting on a particle,
as shown in Fig. 1.5, corresponds to a single force that may be shown by
experiment to be equal to the diagonal of a parallelogram formed by the
graphical representation of the forces. That is, the quantities add according to
the parallelogram law. All quantities that have magnitude and direction and
that add according to the parallelogram law are called vector quaniities. Other
quantities that have only magnitude, such as temperature and work, are called
scalar quantities. A vector quantity will be denoted with a boldface italic letter, which in the case of force becomes F.5
The reader may ask Don’t all quantities having magnitude and direction combine according to the parallelogram law and, therefore, become
.iYour inslmclm on the blackboard and you in your homework will not be able lo use bld;
face notation lor vcctors. Accordingly, you may choose IO
use a superscript arrow or bar, e.&. F
or F (E or E are other possibilities).
e
F, + F2
Figure 1.5. Parallelogram law.
16
CHAPTER I
FUNDAMENTALS OF MECHANICS
vector quantities? No, not all of them do. One very important example will he
pointed out after we reconsider Fig. 1.5. In the construction of the parallelogram it matters not which force is laid out first. In other words, “ F , combined
with F,” gives the same result as “F, combined with F,.” In short, the combination is commutative. If a combination is not commutative, it cannot in
general he represented by a parallelogram operation and is thus not a vector.
With this in mind, consider the finite angle of rotation of a body about an
axis. We can associate a magnitude (degrees or radians) and a direction (the
axis and a stipulation of clockwise or counterclockwise) with this quantity.
However, the finite angle of rotation cannot he considered a vector because in
general two finite rotations about different axes cannot he replaced by a single
~
I
90”
Figure 1.6. Successive rotations are not
commutative.
SECTION I .Y
finite rotation consistent with the parallelogram law. The easiest way to show
this is to demonstrate that the combination of such rotations is not commutative. In Fig. I .6(a) a book is to he given two rotations-a 90" counterclockwise rotation about the x axis and a 90" clockwise rotation about the i axis,
both looking in toward the origin. This is carried out in Figs. 1.6(b) and (c).
In Fig. 1.6(c), the sequence of combination is reversed from that in Fig.
1.6(b), and you can see how it alters the final orientation of the hook. Finite
angular rotation, therefore, is not a vector quantity, since the parallelogram
law is not valid for such a ~ o m b i n a t i o n . ~
You may now wonder why we tacked on the parallelogram law for the
definition o f a vector and thereby excluded finite rotations from this category.
The answer to this query is as follows. In the next chapter, we will present
cemin sets of very useful operations termed w c t u r algebra. These operations
are valid in general only if the parallelogram law is satisfied as you will see
when we get to Chapter 2. Therefore, we had to restrict the definition of a
vector in order to he able to use this kind of algebra for these quantities. Also,
it is to he pointed out that later in the text we will present yet a third definition consistent with our latest definition. This next definition will have certain
advantages as we will see later.
Before closing the section, we will set forth one more definition. The
/ine (,faction of a vector is a hypothetical infinite straight line collinear with
the vector (see Fig. 1.7). Thus, the velocities of two cars moving on different
lanes of a straight highway have different lines of action. Keep in mind that
the line of action involves no connotation as to sense. Thus, a vector V'
cnllinear with V in Fig. 1.7 and with opposite sense would nevertheless have
the same line of action.
1.9
17
EQUALITY AND EQUIVALENCE OF VECTORS
+==
/
-m
Figure 1.7. Line of action of B vector.
Equality and Equivalence of Vectors
We shall avoid many pitfalls in the study of mechanics if we clearly make a
distinction between the equality and the equivalence of vectors.
Twjo L'ecfors are equal if they have the .same dimcmsions, rnugnirudc,,
and direction. In Fig. 1.8, the velocity vectors of three particles have equal
length, are identically inclined toward the reference xyr, and havc the samc
sense. Although they have different lines of actinn, they are nevertheless
equal according to the definition.
Two vectorr are equivalent in a certain c a p a c i y if each prodnces the
vev ,same e f t k t in this capacity. If the criterion in Fig. I .8is change of elevation of the particles or total distance traveled by the particles, all three
vectors give the same result. They are, in addition to being equal, alsu
"However. wriishingly rniull rotations can be considered a i YCE~UIS since thc commutative
law applies for the combiniltiun of such rotations. A proof of this assertion is presenlcd in Appcndin IV. The tbct that infinitesimal rotations are vectors i n accordance with our definition w i l l be
an irnpoltant consideration when we discuss angular velocity in Chapter 15.
Figure 1.8. Equal-velocity vectors.
18
CHAPTER I
~ U N U A M L N T A L Sor MECHANICS
equivalent fur thcsc capacities. I f the absolute height u l the parlicles above
the .cy plane i s the quesliiin i n piiint, these vectors w i l l no1 he equivalenl
despite their equality. Thus, i t must be cinphasizcd that cquol 1wtor.s need
t i u f uIwri?,s bP ryuivulent: i f deprid.s cririrelv oii / h e situuriori ut hund. Furthermore, vectors that itre n o t equal may s t i l l hc cquivalcnt i n soine capacity. l'hus, i n the beam i n Fig. 1.9, forces F , and
are unequal, since their
magnitudes are IO Ih and 20 Ih, respectively. However, it i s clear from elementary physics cliat their mnments ahiiul the hase 01 the heani are equal,
and su the forces liave the same "turning" action at the fixed end of the
heain. I n that capacity. the forces are equivalenl. If. however, we are interested i n the dellcction of the free end of the heam resulting from each force,
there i s no longer ;in equivalcncc hclwcen thc force.;, since each w i l l give a
different dcllcctiun.
C;
,
I_
,]'
~~~~~
Figure 1.9. F , and I.?equivalent Tor iiioriient
ahw1 A.
To sum up. the ryrcnli~ynf two vecturs i s determined by the vectors
themselves. and thc equivuleurp hctwecn two vccturs i s dctcrniined hy the
task involving the vectors.
In probleins o f mech;mics. we can prufitehly delineate three classes of
situatiuns cunccrning equivalence of veckirs:
1. 5'irirution.s it[ M h i d ~vw/o,-.v miry he p . s i r i o w d unywherr in .spuce wirlwur
1o.u or (.huuKr r,/meuriinp providrd thuc mu,ynilurlr und dir(,<.timu w k e p
intu(.t. Under such circ~iinstaiicesthe vectors are c:nlled free ve('tor.r. For
example. the velocity v c c t i m i n Fig. I.X are lrce vectors a s far as total disLance traveled i h concerned.
2. .Si/iiution.~in w/iir./~Lvt'lor.v mriy 1)r mmw/ u l o q /li<,ir1iur.so f w t i m wirlion/ c'hungr o/ ,nrui,iin,y. Under such circunislainces the vectors are called
truri.smi.s.sible vi't'tutx For cxample. i n towing the object i n Fig. I.IO, we
may apply the lorce anywhere alung t l i t rupc AH or inay piish at point C.
The resulting motion i s lhe same in all cases. s i i lhe Snrce i s a transinissihle
vector for this purpose.
3. Situurions in w/iir.h fhP I ' ~ ~ ' I oi ~n us t br rippli~4NI r1c:finite I1oinl.s. The
point may he represented as the tail or head of the arrow iii thc graphical
representation. For this case. n o other positioii of application leads tu
SECTION 1.10
equivalence. Under such circumstances. the vector is called a bound vector. For example, if we are interested in the deformation induced by forces
in the body in Fig. 1.10, we must be more selective in our actions than we
were when all we wanted to know was the motion of the body. Clearly,
force F will cause a different deformation when applied at point C than
it will when applied at point A . The force is thus a bound vector for this
problem.
We shall be concerned throughout this text with considerations of equivalence.
tl.10
laws of Mechanics
The enure structure of mechanics rests on relatively few basic laws. Nevertheless, for the student to comprehend these laws sufticiently to undertake
novel and varied problems, much study will be required.
We shall now discuss briefly the following laws, which are considered
to be the foundation of mechanics:
1.
2.
3.
4.
Newton’s first and second laws of motion.
Newton’s third law.
The gravitational law of attraction.
The parallelogram law.
Newton’s First and Second Laws of Motion. These laws were first stated
by Newton as
Every particle continues in a state of rest or uniform motion in a straight
line upless it is compelled to change that state by forces imposed on it.
The c b g c of motion is proportional to the naturn1;ferCe impressed and
is made in a direction of the straight line in which the force is impressed.
Notice that the words “rest,” “uniform motion,” and “change of motion’’
appear in the statements above. For such information to he meaningful, we
must have some frame of reference relative to which these states of motion can
be described. We may then ask: relative to what reference in space does every
particle remain at “rest” or “move uniformly along a straight line’’ in the
absence of any forces? Or, in the case of a force acting on the particle, relative
to what reference in space is the “change in motion proportional to the force”?
Experiment indicates that the “fixed stars act as a reference for which the first
and second laws of Newton are highly accurate. Later, we will see that any
other system that moves uniformly and without rotation relative to the fixed
stars may be used as a reference with equal accuracy. All such references are
called inertial references. The earth’s surface is usually employed as a reference in engineering work. Because of the rotation of the earth and the varia-
LAWS OFMECHANICS
19
Figure 1.10. F i s transmissible for towing.
20
CHAPIEK 1
FUNl1AMENTAI.S OF MECHANICS
ticins i n its miition around the sun, i t i s iiot, strictly speaking. iui inertial rcScrence. However, the departure i s xi small Sor m o s t situiitiiins (cxccptions arc
the motion iif guided missile!, and spacccralt) that the trior incurred i s very
slight. We shall, therefore, usually consider the earth's s u r l x c as an inertial
reference, but w i l l keep i n mind the somewhat appr(iximatc iiaturu of this stcp.
As a result n t the preceding discus~ion.we may define equilihriuni as
thuc .slate ($'I hoc/y in which ull its c~instiru~vrt
purtid<,s u m ut r('.st or n i o h g
irn~/?wmlyulon(: u straighl line w l u t i v e to 11ii i i i e ~ ~ i ir&wiiw.
il
The coiivcrse
nf Newton's first law, then, stipulates Ibr the equilibrium stale that there [must
be nu force (os equivalent action of no force) acting on the body. Many situiirions f a l l into this category. The study of bodies in equilibrium i s called S I U I i c s . and i t w i l l be an important consideration in this text.
In addition tn the reference limitations explained above. ii serious limittition was brought to light at tlic turn ill this century. As pointed out carlicr. the
piiineering work 11f Einstein revealed that the laws 01 Newtun become increasingly more approximate a!, the spccd ul' a body incrcii.
Ncar the spccd of
light, they are untenable. hi the vast majority of ciigincc
ciimpututions. the
speed iif a body i s so small compared to the speed
light that these departures
from Newtonian mechanics. called r<dutivi.stic.e[ 1.5, may be entirely disrcfarded with little sacrifice in accuracy. In ciinhidci-ing the motion of highenergy elementary particles occurring i n nuclear phenomena, however, we
cannot ignore relativistic effects. Finally, when we get down t o very small distances. such as those between the protons and neutrons in the iiucleus o f iui
atom. we find that Newtonian mechanics cannot explain many observed pherionieiia. In this case, we must rexiit to quantum mechanics. arid then Newt u n ' s laws give way to the Schrddingcr e i p l i i i n a s the key equation.
Newton's Third Law. Newton stated in his third law:
To every action rhere is always opposed an equul rcucrion, or the mutual
actiuns of mu bodies upon euch other are ulwuys equal arid directed to
contrary points.
This i s illustrated graphically i n Fig. I. I I , where the action and reaction
between two bodies arise Srom direct contact. Other imporrant actions i n
which Newton's third law holds arc gravitationdl attractions (to be discussed
next) and electrostatic forces between charged pat-ticks. I t should he pointed
out that there are actiiiiis that dii nut fiillow this law. nutably the electrinnagnetic fiirces between charged moving bodies.'
Law of Gravitational Attraction. It has alrcady been piiintcd out that these
i s an attraction between the earth and the bodies at its surface, such as A and B
'Llectmmagnclic fi,rcrs b ~ t w r c~hnr g c d m w i n g ~ ~ I I I C I C I
~IDI "dircclcd LO contray poiair."
iiut wllincilr and ~ C I I C L arc
.
iiir
cqu'ti and ,q'pubiiu hui
iiiu
SECTION 1.10 LAWS OF MECHANICS
in Fig. 1.1 I . This attraction is mutual and Newton’s third law applies. There
is also an attraction between the two bodies A and B themselves, but this
force because of the small size of both bodies is extremely weak. However,
the mechanism for the mutual attraction between the earth and each body is
the same as that for the mutual attraction between the bodies. These forces of
attraction may be given by the law of gravitational attractiun:
Figure 1.11. Newton’s third law
Two anicles will be attracted toward each other along their connecting
line ith a force whose magnitude is directly proponional to the product
of the masses and inversely proportional to the distMce squared between
the pqnicles.
~
!
Avoiding vector notation for now, we may thus say that
F = G -m1mz
(1.1 I)
I2
where G is called the universal gravitational constant. In the actions involving the earth and the bodies discussed above, we may consider each body as
a particle, with its entire mass concentrated at its center of gravity.* Hence, if
we know the various constants in formula I , I 1, we can compute the weight of
a given mass at different altitudes above the earth.
Parallelogram Law. Stevinius (1 548-1 620) was the first to demonstrate
that forces could be combined by representing them by arrows to some
suitable scale, and then forming a parallelogram in which the diagonal represents the sum of the two forces. As we pointed out, all vectors must combine
in this manner.
*To be studied in detail in Chapter 4.
21
22
CHAPTER I
FIINDAMENTA1.S OF MECHANICS
1.11 Closure
In this chapter, we havc introduccd the basic dimensions by which we can
describc in a quanlikttivc manner certain aspccw 01 nalurc. These hasic, and
from them secondary, dimensions may be related by dimensionally homogencous equations which, with suitable idcalirations, can represent certain
actions i n nature. The baric laws of mechanics were thus introduccd. Since
the equations of these laws relate vector quantities, we shall introduce a useful and highly dercriplive set of vector operations in Chapter 2 in order to
learn to handle these laws effectively and to gain more insight into mechanics
in general. These operations are generally c:illcd
Check-Out for Sections with 'i
1.1. What are two kinds of limitations on Newtonian mechanics'?
1.2. What are the two phenomena wherein mass plays a key role?
1.3. If a pound force is defined by the extension of a standard spring,
define the pound mass and the slug.
1.4. Express mass density dimensionally. How many scale units of
mass density (mass per unit volume) in the SI units are equivalent
to I scale unit in the American system using (a) slugs, ft, sec and
(b) Ibm, ft, sec?
1.5. (a) What is a necessary condition for dimensionul lumogeneily in
an equation'?
(b) In the Newtonian viscosity law, the frictional resistance T
(force per unit area) in a fluid is proportional to the distance
rate of change of velocity dV/dy. The proportionality constant
pis called the coefficient of viscosity. What is its dimensional
representation?
1.6. Define a vector and a scalar.
1.7. What is meant by line of action o f a vector?
1.8. What is a di.splucement vector?
1.9. What is an inertial reference?
REVIEW I1*
Vector Algebra
72.1
Introduction
In Chapter I , we saw that a scalar quantity is adequately given by a magnitude,
while a vector quantity requires the additional specification of a direction. The
basic algebraic operations for the handling of scalar quantities are those familiar ones studied in grade school, so familiar that you now wonder even that
you had to be “introduced” to them. For vector quantities, these methods may
be cumbersome since the directional aspects must be taken into account.
Therefore, an algebra has evolved that clearly and concisely allows for certain
vely useful manipulations of vectors. It is not merely for elegance or sophistication that we employ vector algebra. Indeed, we can achieve greater insight
into the subject matter-particularly into dynamics-by employing the more
powerful and descriptive methods introduced in this chapter.
t2.2
Magnitude and Multiplication
of a Vector by a Scalar
The magnitude of a quantity, in strict mathematical parlance, is always aposifive number of units whose value corresponds to the numerical measure of
the quantity. Thus, the magnitude of a quantity of measure -50 units is +50
units. Note that the magnitude of a quantity is its absolute value. The mathematical symbol for indicating the magnitude of a quantity is a set of vertical
lines enclosing the quantity. That is,
1-50 units1 = absolute value (-50 units) = +50 units
*The reader is urged to pay particular attention to Section 2.4 on Resolution of Vectors
and Section 2.6 on Useful Ways of Representing Vectors.
-tAgain, as in Chapter I,we have used the symbol t for cenain section headings to indicate
that at the end of the chapter there are questions to be answered in writing pertaining to these sections. The instmcter may wish tu assign the reading of these seclians along with the aforemen-
tioned questions.
23
24
CHAlTEK 2
t L L M t N I ' S 01: VC("I0R
ALGCRRA
Similarly. the miignitude of a vector quantity i s a positive riumhcr 01unit5
corresponding to the length of the vector i n those units. Using our vector
symh~ils.we ciin say that
magnitude u i wctor A = A
1
A
positivc sciiliir qu;inlity. We m;iy iiow di\ciiis the iiiiil1iplic;ition
by ii sciiliii~.
The definitinn (71 the product o i vectoi A h y wiliir iii,written simply iis
m A , i s given i n the following IiiiinncI:
Thus, A i s
_.
oi.
'I b ~ .t
C
i
Thc vector -A iiiay he ciiii\idcred a\ 1he p~iiclucto i thc sciiliir - I ;ind
the vector A . 'lhu\. ironi the 5lateiiicnt ahove u e see that -A d i i f w lrom A i n
that i t h a s an opposite w i s e . I'urtlierniore. [ h a c npcriitims havc nothing to
do with the line ofactinii < > f abector. .;oA and -A may 1 h : i ~ dilfcrcnt lines 01'
x t i o n . This w i l l he lhc ciise n1tlic couple lo he \tudied i n Ch;ipter 3.
.:2.3 Addition and Subtraction
of Vectors
*&<
n
(ill
A tI
C'
ii
0 1
n-ci
.~
-.
,4-C
- &
H
~~~~~~~
fhl
In ;idding a number 01w c t o ~ - \ \re
. miiy rcpeiitcdly cmpluy Ihc parallclogram con\t~-uction.Wc ciiii dci this graphically hy sciiliiig the Icngllis d t h c iirrou's according to the niiigiiiludc~n l the \'ccti)r q i i ~ i i i t i l i ethey
~ rcprcvnt. The magnitude (it.
lhc final iirrou' ciiii then he iiiterpretcd in teiniir o i i t s length by cinployinf tlic
chosen scale f k t o r . A.; xi ea:iinple, ciinsider thi' coplaniir' \cctors A. R. and C
shown in Fig. 2. I(a). .I'he addition of the \cctors A . H. and C h a hccn iicconiplishcd i n two ways. 111 t i g . 2.11hl we lirst add 11 and C and thcn iidd the rcsulting vector (showii cla\lied) lo A . This cnmhin;iti(in ciin he represented hy the
i i o t i i t i ~ nA + IH + CI. 111Fig. 2.llc). \\ idd A iind B. and then add the resultiiig vectoi- (shoum (Iadied) to C. The reprc\eiilatioii 01 this combination is givcii
iis IA
R)
C'. Nolc that thc f i n a ~ectoi-is identical fiIr hotli procedures. Thus.
+
[A -111
~.
-<
'
/--;,
_,
'
C
,.*
+
A
--.-
+
(R
+
CI
=
(A
+
Rl t C
Whcii the quiiiititics itivol\cd iii xi algchraic opcriition
-11
IC1
out rcstriclinii. ilic ~ i ~ ~ c ~ r i i sl i o
s;iiil
i i to
hc
ciin
hc froupcd witli-
n r s i r i o t i i ~ e .Thus. the ;idditinn o i
t o r s i s hnth coninii~tiitivc.iis caplained enrliei-, uid a\sociati\'c,
'To determine ii siin~iiiatiiiii(11. Ict u s h a y . two vectors uritliout recoursc
we need nnly inlakc ii siinplc \ketch o S the vectors approximatcly
sciilc. By tising hiiiiiliw trigonnineti-ic relation\. we can get a dircct cviiluation o i t h c result. This i s illiistratcd iii the f i ~ l l ~ i w i nca;iinples.
g
to graphics.
IO
Figure 2.1. Addition hy pmillclograiii iau.
(2.1)
l('q>law.imcmio: " w n c
p l m . " t \ ,t
ii~biil
SECTION 2.3 A DD ITI O N AN D SUBTRACTION OF VECTORS
Example 2.1
Add the forces acting on a particle situated at the origin of a two-dimensional reference frame (Fig. 2.2). Onc force has a magnitude of 10 Ib acting in the positive x direction, whereas the other has a magnitude of 5 Ib
acting at an angle of 135" with a sense directed away from the origin.
y
I
I
0
R
IOlb
A-
li
Figure 2.2. Find F and a using trigonometry
To get the sum (shown as F ) , we may use the law of cosines' for one of
the triangular poriions of the sketched paiallelogr~m.Thus, using triangle OBA,
+ 5 2 - (2)(10)(5)cos45"]1~2
= (100 + 25 - 70.7)"2 = \'54.3 = 7.37 Ib
IF1 = [ I 0 2
The direction of the vector may he described by giving the angle and the sense.
The angle is determined by employing the law of sines for triangle OBA.?
sin a =
(5)(0.707)
= 0.480
7.37
Therefore.
F = 7.31 Ih
a = 28.6"
The sense is shown using the directed line segment
25
26
('HAPTFR 2
EIXMENTS 01' Vk.CI'OK AlMitHKA
Example 2.2
A simple slingshot (hec Fig. 2.3) i s about to be "fired." I f the entire rubber
band requires 3 Ib per inch o f elongation, what force does the hand exert
on the hand'! Thc total unstretched length of the rubber band i s 5 in.
The top view 01 the slingshot i s shown in Fig. 2.4. The change in
overall length of the ruhher hand A L from i t s unstretchcd length i s
A L = 2(1.5'
+~
2 ) "- ~5
= I 1.28 in.
The teiisiiin i n the entire extended rubber hand i s then ( II .28)(3)Ib. Consequently, the fiircc F transmitted by rorh / c y of the slingshot i s
Figure 2.3. Simplc dingshut.
F = ( I I .2X)(1) =
and the value of
33.84 lb
H
0
=
tan-'
I.5
~
8
= 10.62°
I n Fig. 2.5, we show a paralleliigram involving the Ibrccs Farid their suni
R where R i s the force that the hand exerts on the hand. We ciin use the law
of cosines on either of the triangles to get K . Thus
842
Noting that
cy =
+
-y.1
-. .
13"
Figure 2.4. Top \ i e u 01the slingchor.
33.84' - (2)(33.84)(33.84)cos cx
180" - (?)(10.62") = 158.8" we have
F = 13.X4 Ih
K = [(2)(33.84)'(1 - c o ~ 1 5 X . X ' ) ] ~ ~=~ 66.52
A more direct calculation can be used by considering two right triangles
within the chosen triangle. Then using elementary trigon<iinctry we have
R = (2K33.84) COS (10.62)" = 66.52 Ib
Figure 2.5. Parsllclogram 011orcrs
It must be emphatically pointed out that thc additiim (if vectorsA and R
only involves the vectors themselves and iiot thcir lines of actions or thcir positions along their respective lincs 0faction. That is. we can change their lines 01
action and miive them along their rchpcctive lines of action 50 a s to form two
sides 01a parallelogram. For thc additional vector algebra that we will devclop
in this chapter, we can tnkc siiniliir liheilics with the ~ e c ~ oinvolved.
rs
We inay also add the vectors by moving them successively to parallel
positions so that the head of one v c c t ~ rconnects to the lail of the iiext vector,
and s o (111. The sum o f the \'cctors will then he ii vcctor w l i ~ s etail connects to
the tnil 01 the first vector and whose head connects to the head of the last vector. This last step will lorin a polygon from the v e c t ~ r s .and wc say that the
vector sum then "closcs the polygon." Thus. adding thc IO-lh vector to the
SECTION 2.3
ADDITION AND SUBTRACTION OF VECTORS
27
5-lb vector in Fig. 2.2, we would form the sides OA and AB of a triangle. The
sum F then closes the triangle and is OB. Also, in Fig. 2.6(a), we have shown
three coplanar vectors F,, F2,and Fi. The vectors are connected in Fig. 2.6(h)
as described. The sum of the vectors then is the dashed vector that closes the
polygon. In Fig. 2.6(c), we have laid off the vectors F,, F2.and Fiin a different sequence. Nevertheless, it is seen that the sum is the same vector as in Fig.
2.6(b). Clearly, the order of laying off the vectors is not significant.
Figure 2.7. Subtraction of vectors.
Figure 2.6. Addition by “closing the polygon.”
A simple physical interpretation of the above vector sum can he formed
for vectors each of which represents a movement of a certain distance and
, direction (i.e,, a displacement vector). Then, traveling along the system of
given vectors you start from one point (the tail of the first vector) and end at
another point (the head of the last vector). The vector sum that closes the
polygon is equivalent to the system of given vectors, in that it takes you from
the same initial to the same final point.
The polygon summation process, like the parallelogram of addition, can
be used as a graphical process, or, still better, can be used to generate analytical computations with the aid of trigonometry. The extension of this procedure to any number of vectors is obvious.
The process of subtraction of vectors is defined in the following manner: to subtract vector B from vector A , we reverse the direction of B (i.e.,
multiply by - 1) and then add this new vector toA (Fig. 2.7).
This process may also be used in the polygon construction. Thus, consider
coplanar vectors A , B, C , and D in Fig. 2.8(a). To form A + B - C - D ,
we proceed a s shown in Fig. 2.8(b). Again, the order of the process is not significant, as can be seen in Fig. 2.8(c).
c\P
(h)
B+A-D-C
k
-D
”xD
-c
(c)
Figure 2.8. Addition and
subtraction using polygon
construction.
2.1. Add ii 20-N force pointing in the positive r direction to a
50-N forcc at an nnple 45" to the .r axis in the first quadrant and
dirccted away from the origin.
2.2.
magnitude of force 8 and the direction of forcc C?(For the Eimplcst rcsulti, usc the force polygon, which for this c a w is ;I right
triangle. and pcrform analytical computations.1
Subtract the 20-N force in Prohlem 2.1 from the SO-N force.
Add thc \,ectors in the .x? plane. Do this first graphically.
using the force polygon. and then do it aialytically.
2.3.
(7.S j
2 0 Ih
A
c
)
w
10 N
Fieure P.2.h
Figure P.2.3
A lightweight homemade plane i s bcinf ohserved LIS i t flies
at constant altitude hut in a wries of scparatc comtant directions.
2.4.
2.7. A light cahlc from a Jccp i\ tied to the peak of an A-fi-amr
m d c x c m it lorcc of450 N along thc cahlc. A I,OOO-kg log i s su\pcnded from a .;ecimd cahlc, which i s fastened to the peak. Whal
i, the t ~ , l a l
fri,,n
c.,hle\
,," rhr A-fr;i,,,c'!
At the outset, i t goes rluc east for 5 km, then due north for 7 kin.
thcn southcast fur 4 km. and finally, southwmt for R km. (iraphically determine the shortest dislancc from thc starting poinl to the
end point of the previous ohscrvationi. See Fig. P.2.4.
Figure P.2.7
2.8. Find thr toLal force and i t \ direction from the cahlr acting
on each of the three pulleys. each of which i s free tu turn. The
IOO-N weigh1 i\ stationary.
A
Figure P.2.4
A homing pigcon i s released at point A and is observed. I t
flier I O krn due south, then gocs duc east for IS km. Next i t goes
southcast for I O k m and finally gocs due south 5 km to reach i t s
destination H. Graphically drtrmminc thc \hwtcst distance betweeii
A and B . Ncglcct the earth'\ cuwature.
2.5.
add up to a forcc C that has n magnitude of 20 N. What is the
Figure P.2.8
2.9. If the difference between forces B and A in Fig. P.2.6 is a
force D having a magnitude of 25 N, what i s the magnitude of B
and the direction o f D ?
2.15. A man pulls with force Won a rope through a simple frictionless pulley to raise a weight W. What total force is exerted on
the pulley?
.. ,
2.10. What is the sum ofthe forces transmitted by the structural
rods to the pin at A?
I
f400 N
Figure P.2.10
2.11. Suppose in Problem 2.10 we require that the total force
transmitted by the members to pin A be inclined 12" to the horizontal. If we do not change the force transmitted by the horizontal
member, what must be the new force for the other member whose
direction remains at 40"? What is the total force?
2.12. Using the parallelogram law, find the tensile force in
cable AC, T,,., and the angle a.(We will do this problem differently in Example S.4.)
Figure P.2.15
2.16. Add the three vectors using the parallelogram law twice.
The 100-N force is in the xz plane, while the other two forces are
parallel to the yz plane and do not intersect. Give the magnitude of
the sum and the angle if forms with f h e x axis.
n = so*
W = 1,000 N
T~w=hOflN
x
Figure P.2.16
Figure P.2.12
2.13. In the preceding problem, what should the angle 6 be so
that the sum of the forces from cable DE and cable EA is colinear
with the boom CE? Verify that S = 55".
2.14. Three forces act on the block. The 500-N and the 600-N
forms act, respectively, on the upper and lower faces of the block,
while the 1,000-N force acts along the edge. Give the magnitude
of the sum nf these forces using the parallelogram law twice.
/
2.17. A mass M is supported by cables (I)
and (2). The tension
in cable ( I ) is 200 N, whereas the tension in ( 2 )is such as to maintain the configuration shown. what is the mass of M in kilograms?
(You will leam very shortly that the weight of M must be equal
and opposite to the vector sum of the supporting forces for
equilibrium.)
0
30"
Figure P.2.14
Figure P.2.17
I
29
2.18. Two foothall player, are pushing a hlocking dummy.
Playcr A pushcs with Ill(l-lh forcc whilc player R pushes with
ISO-lh force toward how C of thc dummy. What i s the total furce
cxertcd on the dummy hy the players'!
a
Figure F.2.19
*2.20. Dn prohlcin 2.19 tind then lorm an intciactivc computer
prnpram \o that l l i e wer at ii prumpl i s askcd 10 insert an anglc n
in radian.: fbr which the program w i l l deliver the onricct values of
H
F;, F,. :Ind /7
Twu soccer player5 approach a stntir,nary hall I O ft away
t r i m the gnal. Simitltaner~udy.a player on triini 0 inffensc) kicks
rhc hall with frrcc 100 Ih Snr a split sccnnd while a player on team
rl (defense) kick.: with force 70 Ih durinp the samc timc intcrval.
Does the offense score (asuming lhiit the g d i c i s asleep)'!
2.21.
Figure P.2.18
2.19. What ire the f o ~ m x s and F; and the angle !3fw any given
a to relieve the force of gravity W from the horiiomal \uppart of the hlock at A'! Thc rollers on thc side u l the hlock tlo not
crmtrihute to thc vertical support nf the hlock. The wire5 cnnnect
til the gcrrmerric center of the hlock C.The weight W i s 5 0 0 N .
Form three independent equations for any given a involving the
unknowns f ; . f:, and b.
mglr
Stipulated
directions
Figure 2.9. Two-dimensional
rewlution of vcctor C.
Figure 2.10. Vector C is
replaced hy i t \ components
and is m) longer opcrativc.
2.4
70 Ih
-
l[l'-/
Figure P.2.21
Resolution of Vectors;
Scalar Components
The opposite action (11 addition nf vectors i s c;illed rc~.solrrriori.Thus, liir a
given vector C , we may find 8 pair nf vectors in any two stipulated directions
coplanar with C such that thc t w n \'cctnrs. callcd ~'onii~onrnt.s,
sum ti) the
original vector. This i s a tn.o~ifin~i~n.vionir1
resolution invnlving two component vectors i,oplwzor with the original vector. We shall discuss threedimensional resolution involving three noncoplanar component vectors later
in the section. The two-dimensional resolution citn he accomplished by
graphical construction 0 1 the parallelogram. or by using simple helpful
sketches and then emplnying trigonometric relations. An example nf ~ W O dimensional rcsiilution is shown in Fig. 2.9. Thc two vectors C , and C ,
formed in this way are the compomnt vectors. Wc olten replace a vector b y
its components siticc the cnniponents are alway:, cquivalcnt i n rigid-body
nicchanics to the original vector. When this i s done. it i s ofteii helplul to indicate that the original vcctor i s no longer operative by drawing it wavy line
through the original veclor as shown i n Fig. 2.10.
SECTION 2.4 RESOLUTION OF VECTORS: SCALAR COMPONENTS
Example 2.3
A sailboat cannot go directly into the wind, but must tack from side
to side as shown in Fig. 2. I I wherein a sailboat is going from marker A to
marker I3 5,000 meters apart. What is the additional distance AL beyond
5,OOO m that the sailboat must travel to get from A to B ?
Clearly the displacement vectofl pARis equivalent to the vector sum
of displacement vectors pAc plus pcR in that the same starting points A ,
and the same destination points B,are involved in each case. Thus, vectors
pAc and pcR are two-dimensional components of vector pas Accordingly,
we can show a parallelogram for those vectors for which triangle ABC
forms half of the parallelogram (see Fig. 2.12). We leave it for you to justify the various angles indicated in the diagram. Now we first use the law
(if sines.
~~
AC
sinp
~
5,000
sinn
Marker A
Figure 2.11.
Sailboat
tacking.
And
BC - s,oon
sin y
sinn
L
Hence the increase in distance AL is
AL = (2,418.4
+
2,988.4) - 5,000 =
I-ZO.
406.8 m
'A di.splac?menf vector. we remind you, connects two points A and B in spacc and i s
often denoted as p,, The order of the subacnpts gives the sense of the vector-here going
from A to 8.
Figure 2.12. Enlarged parallelogram.
It is also readily possible to find three components not in the .same plane
as C that add up to C. This is the aforementioned three-dimensional resolution.
Consider the specification of three orthopnal directions' for the resolution of
C positioned in the first quadrant, as is shown in Fig. 2.13. The resolution may
be accomplished in two steps. Resolve C along the z direction, and along the
/
sAlthough the vector can be resolved along three .skew directions (hence nononhogonal),
the orthogonal directions are used most often in engineering practice.
c4
Figure 2.13. Orthogonal or
rectangular components.
31
32
CHAPTER 2
E L E M E N K OF V E r I O K Al.(iEHRA
intersection o f Ihc x? planc and the planc formed by C and the :axis. This is ii
twii-ilimcnsi~inal recolulioii with tlie par;illelogr;im hccmiing a rectangle
because o l the niwinalcy of the :axis to IIic .x? plane. This gives nrthogonal
\'cctoi-s C , and C, t l i i i ~replace original
toi- C . Next take vector C,, and
r c s o h e it along axes . x ~and ? hy ii hecond t u ~ o ~ d i n i c n i ~ ires(iluti~in
nal
involving
a 1-ectangle oncc iigain thus forming 1 ogoiial vectors C , and C , tliat may
replace veckir C,. Clearly (irthogonal
01s C , , C , , illid
add up 10 C and
accordingly can replace C undr~any and all circuni
ces. Hence C , . C2. and
C , w e called orfliiip,wil or ~ ~ ~ ~ ( i uwii/iiJiwii/
~i~iikir
The direction of a rector C relative ttr iiii or~li~iyon;il
rcference i s given
by the cosines 01 the angles fol-med by the \ecror arid tlic rcspcclive c(iordinille axes. These are called dir~wlioii~r)sine.s
and arc dcnoted as
c,
cos ( C . .1) = cos n z i
cos ( C , ?) = cos
F Ill
cos ( C . :) = 1'11s y = ,I
p
(2.2)
p. and yare associ:ited with tlie .c. ?. and :iixcs. rcspectively. Now
le1 us consider the right triangle. whose sides iire C and the comlxmi'iil vector.
C , . \howti sliiided i n Fig. 2. 13. I t tlicii l ~ c c o ~ iclciir.
i e ~ froiii trig~iii~iiiietric
coiisidcriilioiis rif the right triangle. thal lor thc fir71 qiiadrant
uhcrc (x,
,C,I = iC1 co\ y = Cl I1
(2.3)
If wc h:id decided
10 resolve C first i n the direction instcad of the :direction. we would havc pl-duccd ii gc(iiiictry froni which we could ciincludc that
lC,l = ~ C ~ ISiniilwly.
II.
u'c ciiii say tliiit lC,l x ~ C l iWc
. ciin thcn exprey\ IC i n
terms 01' it\ orthogon;il components i n the fi,llowing iminner. using the
Pythagorean theorem."
IC1 =
[( CIIf + (lCin)2 + (~C!ll)~]l
From this equation we cim dchnc tlie I J ~ - I / I O ~ O ~ I Oor
/
rwlm~iiliir
suilur
thc vcctor C having w i ? orientatinn a s
(2.4)
mni~
p m 7 ~ ' i i tof
.~
=
1,
/
(.:~
C, = (C(nr.
C': = ( C , N
(2.51
Note tlial (;. (;, and
may hc ncgative. depending on Lhc sign of the dii-ecfioii cosincs. l-'in:illy. i t i i i i i s t he poinled out tliiil ~ i l l l i o i ~ , C;
y. l ~ C;. mil C~(ire
cmo<.io/edw i i h wr-lui,i ii,rc,v liiid l r e i r w wr/<,i,z i I i w < ~ t i m vtly?
,
i t o i . r hec,i
Iiqml
.si,tiliir.s(irrd iiiii.sr he 11iimIl~~d
( i s .sm/iir,s. l'lius. an cquatiori such
as IOV =
cos i s 1101 correct. bcciiusc (lie lest side i s ii veclor and the right
side i s a scalar. This sh(iuld spui- you lo obscrvc care i n your notation
Sotnctimcs only o w of the sciiliir iirthiigiiiiiil coniporients o f a
ngular sciilar coniponciit) i h desired. Then. .just one direction i s prescribed. as shown i n Fig. 2.14. Thus. h e scalar rectangular coinponent C', i s C 8cos6. Note u e have shii\vn a pair of other recliingular
a
Figure 2.14. Rcctaogular component OF C.
(C(1.
SECTION 2.5
components as dashed vectors in Fig. 2.14. However, it is only the single
component C, that we often use, disregarding other rectangular components.
It is always the case that the triangle formed by the vector and its scalar rectangular component is a right triangle. In establishing C, we therefore speak
of “dropping a perpendicular from C to s ” or of “projecting along s.”
The scalar rectangular component Cycould also be the result of a fwodimensionul orthogonal resolution wherein the other component is in the
plane of C and Cyand is normal to C,. It is important to remember, however,
that a component of a nonorthogonal two-dimensional resolution is nof a rectangular component.
As a final consideration, let us examine vectors A and B , which, along
with directions, form a plane as is shown in Fig. 2.15. The sum of the vectors
A and B is found by the parallelogram law to be C. We shall now show that
the projection of‘C along s is the same as the sum of the projections of the
two-dimensional components of A and B , taken along s . That is,
Cy = Ay
+ E,
On the diagram, then, the following relation must be verified:
ac = ad
+
ub
(a)
But
uc =
ab
+ bc
(b)
Also, it is clear that
ad = bc
(C)
By substituting from Eqs. (b) and (c) into Eq. (a), we reduce Eq. (a) to an
identity which shows that the projection of the sum of two vectors is the same
as the sum of the projections of the two vectors.
2.5
Unit Vectors
It is sometimes convenient to express a vector C as the product of its magnitude and a vector a of unit magnitude and having direction corresponding to
the vector C. The vector a is called a unit vector. The unit vector is also at
times denoted ash. (You will write it as 6.) It has no dimensions. We formulate this vector as follows:
a(unit vector in direction C) =
C
~
IC1
(2.6)
Clearly, this development fulfills the requirements that have been set forth for
this vector. We can then express thc vcctor C in the form
C
=
lCla
(2.7)
The unit vector, once established, does not have, per se, an inherent line of
action. This will be determined entirely by its use. In the preceding equation,
UNIT VECTORS
33
q
c
Figure 2.15. C,= A,
+ B,.
s
34
CHAPTER 2
ELEMENTS OF VECTOR ALGEBRA
the unit vector a might be considered collinear with the vector C. However,
we can represent the vector D , shown in Fig. 2.16 parallel to C, hy using the
unit vector a a s follows:
I ) = lD!a
Figure 2.16. Unit vector a.
(2.7a)
It thus acts as a free vector. Occasionally. it is useful to lahel a unit vector
meant to have the line of action of a certain vector with the lowercase letter of
the capital letter associated with that vector. Thus. i n Eqs. 2.7 and 2.7(a) for
this purpose we might have employed in the place ofa the letters c and d (in
your case i. and &, respectively. Next, if a given vector is represented using a
lowercase letter, such as the vector r , then we oflen make use of the circuniflex mark to indicate the .ociated unit vector. Thus,
r = lrli
(2.7h)
Unit vectors that are of particular use are those dircctcd along the directions of coordinate axes of a rectangular reference, where i, j . and k (your
and &) comespond to the x. y. and
instructor will prohahly use the notation
i directions, as shown in Fig. 2.17.'
Since the sum of a set of concurrent vectors is equivalent in all situations to the original vector, wc can always replace the vector C by its rcctungular scalar components in the following manncr:
?,,r,
Figure 2.17. Unil vec~orslor .r?iaxcs
C = C:i
+
C,/
+
C.k
(2.8)
In Chapter I , we saw that vectors lhal are q u a l havc the same m a p lude and direction. Hence, if A = B , we can say that
A,i
+ A , j + A;k
= 0.i
+
B,j
+ Brk
(2.9)
Then, since the unit vectors have mutually different dircctions. we conclude
€rom ahove that
A,i
=
R,i
A , j = B,j
A,k = B:k
It then follows that
A , = R,
A,
=
B,.
A ; = B:
'Curvilinear caardinalc rystems iiilvc associated sei* (11 u r ~ YCCLOIE
l
just 8s do the rectangular coordinatc ayslcrns. As will be seen Iaitcr. however. culain or Ihew unit vectors do not a11
have tined directions in space for a given rclcrcncc as tlu rhc vectois i . j . arid k .
SECTION 2.6 USEFUL WAYS OF REPRESENTING VECTORS
Hence, the vector equation, A = B, has resulted in three scalar equations that
in totality are equivalent in every way to the vector statement of equality.
Thus, in Newton's law we would have
(2.1021)
F = m
as the vector equation, and
F,
F, = ma,,
= ma,,
F,
= maI
(2. IOb)
as the corresponding scalar equations.
2.6
Useful Ways of Representing Vectors
Quite often, we show a rectangular parallelepiped with sides oriented parallel
to the coordinate axes and positioned somewhere along the line of action of a
vector (see Fig. 2.18) such that this line of action coincides with an inside
diagonal of the rectangular parallelepiped. The purpose of this rectangular
parallelepiped and diagonal is to allow for the easy determination of the orientation of the line of action and hence the orientation of a vector. AB in the
diagram is such a diagonal used for the determination of the line of action of
vector F . Numbers for this purpose are shown along the sides of the rectangular parallelepiped without units. Any set of numbers can be used as long as
the ratios of these numbers remain the ones required for the proper determination of the orientation of the vector. That determination proceeds by first
replacing the displacement vector p,,, from comer A to comer B by a set of
three vector displacements going from A to B along the sides of the rectangular parallelepiped. We thereby can replace the vector p,, by the sum of its
rectangular components. Thus, for the case shown in Fig. 2.18 we can say8
PA,
l O j - 4i
=
+ 6k
=
z
4
i+ lOj
+ 6k
B
6
I
X
A
IO
/
Figure 2.18. Rectangular parallelepiped used for specifying the direction of a vector.
XImagineyou are "walking" from A to B hut restricting your movements to he along the
coordinate directions. This movement i s equivalent to going directly from A to B in that the Same
endmints result.
35
36
CHAPTER 2
ELEMENTS OF VECTOR AI.OE~RA
Now using the Pythagorean theorem, divide pAn by i t s magnitude, namely:
+ 6',
We thus form the unit vectorbA8. That is.
,42 +
bAt{
-
PA#
-
7
lP,,Hl
-4;
\42
+ I O j + 6k
= -.3244i
+ IO' + 6'
+ .81 I l j + .4867k
As a final step we can give vector F as fMows:
F = F(-.3?44i
+
.81 I1.j + .4867k)
If F = 100 N we then can say:
F
,
h
= -32.441'
+
81.1 1.j
+
48.67k N
Note that the rectangular parallelepiped can he anywhere along the line
of action of F including cases where F i s not inside of the parallelepiped or
extends heyond the parallelepiped (see Fig, 2.19). In the two-dimensional
case. a right triangle serves the same purpose as the rectangular parallelepiped i n three dimensions. This i s shown i n Fig. 2.20 where vector V i s i n
the x? plane. Here we can say.
I
v=v
9
i + ,
.;2?
\.2' + 9 2
= V(.9762i
+ 97
+ .2169j)
Figure 2.19. Other ways to use the reclangular
parallelepiped.
-~
L-
Figure 2.20. Right triangle used fur sppecifying
the direction ot it vector m two dirnensionq.
There are times when the rectangular parallelpiped i s not shown explicitly. However. thc Icngth o f the sides o f one having the proper diagonal may
hc availahle so that the replacement of the diagonal displacement vector into
rectangular components can he readily achieved. The simplest procedure i s IO
mcntally move from the beginning point o f the diagonal to the final point
always moving along coordinatc directions, or, i n other words, always moving along the sides 0 1 the hypothetical rectangular parallelepiped. Thus. i n
Fig. 2.21, for the v e c t ~ rF, we can consider A B to he the diagonal and in
p i n g from A to B we could first move i n the minus x direction by an amount
-1. then move i n the plus ?' direction by the amounl 1.5, and finally i n the I
dircclion hy an amount 3 . This would take us from initial point A to final
point IT. The corresponding displacement veclor would then he
p,, = -1;
+
I.Sj
+ 3j
The following example w i l l illustrate the use of orthogonal resolution
as well a s the use of rectangular components of vect~rs.
SECTION 2.6 USEFUL WAYS OF REPRESENTING VECTORS
Example 2.4
A crane (not shown) is supporting a 2,000-N crate (see Fig. 2.21) through
three cables: AB, CB, and DB. Note that D is at the center of the outer edge
of the crate; C is 1.6 m from the comer of this edge; and B is directly
above the center of the crate. What are the forces Fl, F2, and Fi transmitted
by the cables?
We will soon l e a n formally what our common sense tells us, namely
that the vector sum of force F,, force 5. and force Fi must equal 2,OWk N.
We first express these three forces in terms of rectangular components. Thus,
-li + 1.5j + 3k
= .;(-.2857i
1
+ ,42861 + .8571k) N
= F,(-.1761i - ,44021
+ .8805k) N
We now sum the three forces to equal 2,OOOk N.
F,(-.2857i
+ .4286j + .8571k) + F2(.3162i + .9487k)
+ F3(-.I761i - .4402j + .8805k) = 2,OOOk
We have three scalar equations from the previous equation.
-2857 Fl + ,3162 F2 - ,1761 F3 = 0
,4286 Fl + 0 - ,4402 F3 = 0
,8571 Fl
+
,9487 F2
+
,8805 F7 = 2,000
Solving simultaneously, we get the following results:
3m
Figure 2.21. A crate is supported by three forces.
37
2.22. Resolve thc 100-1h force into a set of components along
the slot shown and in the vertical direction.
F i g i r e P.2.2.5
Two tughants are maneuvering an w e a n liiw The desired
iota1 inrcc i s 3,000 Ih at an angle r r f I S " a h cliown i n the diagram.
If thc tughaat frrrces have dircctim\ a s shown, what inwt the
forces I.', and F- he''
2.26.
Figure P.2.22
A lamer needc to build a fence from the corner of his ham
to the corner of hic chicken house 10 m away in the NE dircction.
However, he wants to enclose ils much of the harnyard as possihie. Thus, he nins thc fcncc cmt, from thc corner of his ham to thc
prnpeny line and then NNE to thc corncr of his chickcn housc.
How long i s the fencc?
2.23.
* /
2.24. Resolve the force F into a component perpendicular to AR
and a component parallel to NC.
Figure P.2.26
I n the previous pruhlem, if F2 = I.000 Ih and = 40",
what should I.; and n he so that F, + F? yiclds the indicated
2.27.
vmn-ih ~ C K C C ' !
2.28. A I.000-N force is resolvcd into component.; along AH and
Ac. If the component along AH i s 700 N.determine the angle n
and the value of the component along AC.
Figure P.2.24
2.25. A simple truss (to be studied later in detail) supports two
forces. If the forces in the members are colinear with the memhcrs, what arc the forces in thc mcmbcrs? Him; The lorccs in thc
mcmherq must have a vector w m q u a l and opposite to the vcctoi
sum of E; and F,. The entire Eystem i s coplanar.
Figure P.2.28
2.29. Two men are trying to pull a crate which will not move
until a 150-lb total force is applied in any one direction. Man A
can pull only at 45' to the desired direction of crate motion,
whereas man B can pull only at 60" to the desired motion. What
force must each man exert to start the box moving as shown?
y
Desired
motion
+
A
C'
1
5001b
Figure P.2.31
2.32. The orthogonal components of a force are:
x comvonent 10 Ib in oositive x direction
y component 20 Ib in positive y direction
z component 30 Ib in negative z direction
(a) What is the magnitude of the force itself?
(b) What are the direction cosines of the force?
2.33. What are the rectangular components of the 100-lb force?
What are the direction cosines for this force?
Figure P.2.29
z
\
2.30. What IS the sum of the three forces? The 2,000-N force IS
in the y z plane.
I
A
4
,
,
\
/ '45"
I
'\\
II
Figure P.2.33
2.34. The 1,000-N force is parallel to the displacement vector
<A while the 2,000-N force is parallel to the displacement vector
CB. What is the vector sum of these forces?
Y
x
Figure P.2.30
2.31. The 500-N force is to be resolved into components along
the AC and AB directions in the xy plane measured by the angles a
and p. If the component along AC is to be 1,000 N and the component along AB is to be 800 N, compute a and p.
x
Figure P.2.34
39
2.35. A S(l-m-long diagrinal inemher Ot i n ii space iiallir I\
30" t o the I and Y BXCS. respeclixly.
inclined a1 a = 70" and
What is y? How long mist rncmhel-c OA. A ( ' . ON. HC, and ('E hi.
to suppafi
b: <)i
01;'
7:
I
t:
/'
Figure P.2.35
What is thc orthogonal total f h x cwnponcnt in tlic .I
direction 01 the ioice tiansmittcd to pin A of a roo1 t n h i h i tlic
four rncrnher,'! What is the total cirmpi,nent in the ? dbrcction"
2.36.
Cahlcc :KC i n
1
,
Figure P.2.37
40
Figure P.2.43
2.44. Express the 100-N force in terms of the unit vectors i, j , 2.45. Express the unit vectors i,j , and k in terms of unit vectors
and k . What is the unit vector in the direction of the 100-N force'! e;. eo, and e7.(These are unit vectors for cvlindricul coordinates.)
Express the 1,000-lb force going through the origin and through
The force lies along diagonal AB.
point (2, 4, 4) in terms of thc unit vectors i, j , k and E?, eti, eZwith
0 = 60". (See the footnote o n p. 34.)
Figure P.2.44
2.7
Figure P.2.45
Scalar or Dot Product of
m o Vectors
In elementary physics, work was defined as the product of the force component, in the direction of a displacement, times the displacement. In effect, two
vectors, force and displacement, are employed to give a scalar, work. In other
physical problems, vectors are associated in this same manner so as to result
in a scalar quantity. A vector operation that represents such operations concisely is the scalar product (or dot product), which, for the vectorA and B in
Fig. 2.22, is defined as'
A . B = IAI IBl cos a
(2. I 1 )
where a is the smaller angle between the two vectors. Note that the dot product may involve vectors of different dimensional representation, and may be
positive or negative, depending on whether the smaller included angle a is
less than or greater than 90". Note also that A * B is equivalent to first projecting vector A onto the line of action of vector B (this gives us IAI cos a),
and then multiplying by the magnitude of vector B (or vice versa). The appropriate sign must, of course, be assigned positive if the projected component of
vector A and vector B point in the same direction; negative, if nol.
The work concept for a force F acting on a particle moving along a
path described by s can now be given as
W =JF.ds
where ds is a displacement on the path along which the particle is moved
''TOensure that there is 11" c u n l u h n between the dot product of two vectors and the ordiR = C as "A
nary product or two scalars that you have used up to now, we urge you to read A
dotted intoR l i e l d s C."
<
Figure 2.22. a is smallest
angle between A and B .
42
CHAPTER 2
ELEMENTS OF VECTOR ALGEBRA
As with addition and subtraction of vectors, the dot product operation
involves only the vectors themselves and not their rcspective lines of action.
Accordingly, for a dot product of two vectors. we can move the vectors so a s
Lo intersect at their tails as in Fig. 2.22. Remember in so doing we must not
alter the magnitudcs and directions of the vectors.
Let us next consider the scalar product of mA and n u . If we carry it out
according to our definitions:
( n A ) * ( n B ) = Id/
jrrR1 cos (d.
nBJ
= (inn) IAI IRI cos (A. R ) = (mnJ (A
B)
(2.131
Hence, the scalar coefficients in the dot product of two vectors multiply in the
ordinary way, while only the vectors themselves undergo the vectorial operation as we have defined it.
From the definition, clearly thc dot product is commutnrivr. since the
numher /AI lB1 cos ( A , BJ is independent of the order of multiplication of its
terms. Thus,
(2.14)
A - R = R - A
Let us now consider A * ( B + C).By definition, we may project the
vector ( B + C ) onto the line of action of A and then, assigning the appropriate sign, multiply the magnitude of A times the projection of B + C . However, in Section 2.4 we showed that the projection of the sum of two vectors
is the same as the sun1 of thc prtijections of the vectors, which means that
A.(R +C) = A * B+ A .C
(2.15)
An operation on a sum of quantities that is the same as the sum of the operations on the quantities is called a di.rtrihurive n p r n r i o n . Thus, the dot product
is distributive.
The scalar product between unit vectors will now be carried out. The
product i * j is 0, since the angle a in Eq. 2. I I is 90". which makes cos a = 0.
On the other hand. i * i = I . We can thus conclude that the dot product of
equal orthogonal unit vectors for a given reference is unity and that of unequal orthogonal unit vectors is zero.
If we express the vectors A and B in Cartesian components whcn taking
the dot product, we get
A * B = (A,i t A ,j
=
+ Ark) *
(B,i
+ H,j +
B:k)
(2.16)
A t B z t A y B y t A.B.
. .
Thus, wc see that B scalar producl of two vectors is the sum of the ordinary
products of the respective components."'
"'Thus the ordinary griide school producr of two numhers, i.e. (o)(h), is n special
the dot product n * b where the the vectors have thc w n c d i r e c l i m Thus
<ti
hi
=
(ii)(h)
C ~ S C0
1
SECTION 2.1 SCALAR OK DOT PRODUCT OF TWO VECTORS
43
If a vector is multiplied by itself as a dot product, the result is the
square of the magnitude of the vector. That is,
A
A = /AI IAl = A’
(2.17)
Conversely. the square of a number may be considered to be the dot product
of two equal vectors having a magnitude equal to the number. Note also that
A
+ A;. + A:
A = A:
= A’
(2.18)
We can conclude from Eq. 2.18 that
~~~
+ A: + A:
A = ./A:
which checks with the Pythagorean theorem.
The dot product may be of immediate use in expressing the scalar rectangular component of a vector along a given direction as discussed in Section 2.4. If you refer back to Fig. 2.14, you will recall that the component of
C along the direction s is given as
c, = IC/ cos 6
Now let us consider a unit vector s along the direction of the line s. If we
carry out the dot product of C and s according to our fundamental definition,
the result is
c
s = IC1 Is1 cos 6
Similarly, the following useful relations are valid:
C,=C*i,
C.=C.k
C,=C.j,
Finally, express the unit vector i directed out from the origin (see Fig.
2.23) in terms of the orthogonal scalar components:
-
3 = (i i ) i
+
-
(3 j ) j
+
(i * k)k
/
L
0
,
1-
But
Similarly, 3
k
-
i * i
j = m and 3
=
Figure 2.23. Unit vector idirected from 0.
131 lil cos (3, n) = /
k = n. Hence, we can say that
3 = li
+
mj
+
nk
(2.19)
Thus, the orthogunal scalar components of a unit vectur are the direction
cosines of the direction uf the unit vector. Now, computing the square of the
magnitude of i , we have
li12= 1 =
/2
+
mz
+
nz
(2.20)
We thus arrive at the familiar geometrical relation that the sum of the squares
of the direction cosines of a vector is unity.
Example 2.5
Cables GA and GB (see Fig. 2.24) are par1 (11 it guy-wire system supporting two radio transmission towers. What are the length:, o f GA and GLI and
- -
the angle a hetween them?
We may directly set up thc vectorsGA and GB hy inspecting the diagmni. l h u s on moving along the coordinate directions, i t is easy to scc that
ZA
Gii
+ 5OOk rn
100i + S O O ~m
= 3Wlj - 400i
=
3Oij
+
-
Using the Pythagorean theorcm, we can ray Sur the length:, ofGA and&
<A
;
= (300’
G H = (300’
+
+
:
400L + 5002)1/L= 707 in
100’ + 5 0 0 2 ) ” 2 = 592 111
Now we usc the dot product definition to find thc angle.
-
(;A *
6
3=
(GA)iGLI)cos
(Y
Thcrclbrz.
Hence,
a = 44.18”
A
Figure 2.24. Kvdio transmission tiiwers.
2.46. Given the vectors
2.50. Show that
A = 10i + 20j + 3k
B = -l0j + 1%
cos (A, E ) = 11'
what is A * B ? What is cos (A, E)? What is the projection of A
along E ?
+
mm'
+
nn'
where I, m, n and 1', m', n' are direction cosines of A and E ,
respectively, with respect to the given xyz reference.
2.47. Given the vectors
A
= 16i
+
E
3j.
=
10k - 6i,
C = 4j
(a) (A * E )
(b) (A E )
compute
(a) C(A * C )
(b) -C + [ E
-+
2.51. Explain why the following operations are meaningless:
-+
C
C
B
(-A)IC
2.52. A block A is constrained to move along a 20' incline in
2.48. Given the vectors
A=6i+3j+10k
B=Z-Sj+Sk
C = 5i - 2j + l k
the yz plane. How far does the block have to move if the force F
is to do I O ft-lb of work?
z
1
what vector D gives the following results'?
F
=
10i + 20i
D - A = 2 0
D . B = 5
D.i=IO
2.49. A sailboat is tacking into a 20-knot wind. The boat has a
velocity component along its axis of 6 kn but because of side slip
and water currents, it has a speed a speed of .2 kn at right angles
to its axis. What are the x and y components of the wind velocity
and the boat velocity? What is the angle between the wind d o c ity and the sailboat velocity'?
+ 15k Ib
Y
X
Figure P.2.52
2.53. An electrostatic field E exerts a force on a charged panicle of qE, where q is the charge of the particle. If we have for E :
E = 6i
+
3j
+
2k dyneslcoulomb
what work is done by the field on a panicle with a unit charge
moving along B straight line from the origin to position x = 20
mm, y = 40 mm, z = -40 mm?
2.54. A force vector of magnitude 100 N has a line of action
Figure P.2.49
with direction cosines 1 = .7, m = .2, n = .59 relative to a reference xyz. The vector points away from the origin. What is the
component of the force vector along a direction a having direction cosines 1 = -.3, m = . I , and n = 9 5 for the xyz reference'?
(Hint: Whenever simply a component is asked for, it is virtually
always the rvctrrngulrrr component that is desired.)
4:
2.55. What is the anglc bctwccn the I ,0011-N force and the axis
A B ! The force is i n thc diagonal plane GCDE.
2.60. What is the rectangular component UC the 500-N forcc
alirrig the diagonal Iron, R to A'!
I
Fieure P.2.60
~
/I I\,l,lllllli.
A radio tiiwci is held by guy wires. It A H were t<>he
moved to inleriect CII whilc rcmaining pardkl ID i t s original
2.61.
Figure P.2.55
(iiven a force F = Ioi + sj
Ak N. If this force is
have a rectangular component of X N along a line having a unit
vectur i= .hi + .Xk, what shuuld A hc'! What is the angle
2.56,
+
position, what is the angle brtwccn AH and U T !
hetween F arid i?
2.57. Given a twce Ai + Bj + 2llk N, what must A and H he to
give a rectaiigular ctimpr~nent01 Ill N iii the direction
i , = .?i
+ .6j + .742k
as wcll as il cornpuncot of 18 N i n the dircctiun
+
i, = .4i
.Yj
+
.1732k!
2.58. Find the dot prtiduct ul the vectors represented hy thz
diagonal5 from A to I" arid trmn 1) to G. Whal is thc angle hetween
them?
(?
I
c
l(1'
I
,, -----,,'
XI'
*--_
Figure P.2.61
2.62. What is the angle bctweeii the 1.000-N force and the position vector r'!
Y
1)
Figure P.2.58
2.59.
A force F is givcn as
F = 8OOi
+
hOOj - I ,000k N
What is the recfrmyulur cornporrenr along an axis A ~ Aequally
inclincd tu thc positive x, s, and :axes?
46
x'
Figure P.2.62
SECTION 2.8 CROSS PRODUCT OF TWO VECTORS
2.8
Cross Product of N o Vectors
There are interactions between vector quantities that result in vector quantities. One such interaction is the moment of a force, which involves a special
product of the force and a position vector (to he studied in Chapter 3). To set
up a convenient operation for these situations, the vecfor cross product has
been established. For the two vectors (having possibly different dimensions)
shown in Fig. 2.25 as A and B , the operation" is defined as
(2.21)
A X B = C
where C has a magnitude that is given as
IC1 = IAI lB1 sin a
(2.22)
Figure 2.25. A x B = C.
The angle a is the smaller of the two angles between the vectors, thus making
sin a always positive. The vector C has an orientation normal to the plane of
the vectors A and B . The sense, furthermore, corresponds to the advance of a
right-hand screw rotated about C as an axis while turning from A to B
through a-that is, from the first stated vector to the second stated vector
through the smaller angle between them. In Fig. 2.25, the screw would
advance upward in rotating from A to B, whether the procedure is viewed
from above or below the plane formed by A and B. The reader can easily verify this. The description of vector C is now complete, since the magnitude
and direction are fully established. The line of action of C is not determined
by the cross product; it depends on the use of the vector C.
Again we remind you that the cross product, like the other vector algebraic operations, does not involve lines of action, so in taking a cross product
we can move the vectors so as to come together at their tails as in Fig. 2.25.
As in the previous case, the coefficients of the vectors will multiply as
ordinary scalars. This may he deduced from the nature of the definition.
However, the commutative law breaks down for this product. We can verify,
by carefully considering the definition of the cross product, that
(A X B ) = -(B
"Again, we urge you to read A
X
X
(2.23)
A)
B = Cas "A crossed
into B
yields C.'
47
48
CHAPTER 2
EI.EMENTS OF VECTOR AI.GERRA
P
,
We can readily show that the cross product, like the dot product. is a
distributive operation. To do this, consider in Fig. 2.26 il prism mniipqr with
edges coinciding with the vectors A , H , C, and (A + B ) . We can represent
the area of each face of the prism as a vector whose inngnitude equals the
area o l the face and whose direction is norm;~lto the face with a sense pointing out (by convention) from thc body. It will be left to the student to justil'y
the given formulation for each of the vectors in Fig. 2.27. Since the prism is a
closed surface, the net projected area in any direclion must be zero, and this.
in turn, means that the total arm \sector must he zero. We then gct
(A
it
+ B) X C + $A
X
B
+ 4B X A + C X A + C X H = 0
Figure 2.26. Prim using A , B, and C.
Figure 2.27. Area vzctorc tor prism f i c m
Noting that the second and third expressions cancel each olher. we get.
rearranging the terms.
C
X
(A+B)=C
X
A + C
X
B
011
(2.24)
We have thus demonstrated the di,stributiw property of the cross product.
Next. consider the cross product of rccliinpular unit vectors. Here, the
product of equal vector5 is zero hecause a and. consequently. sin CY are iero.
The product i X j is unity i n magnitude. and hccnusc (if h e right-hand-screw
rule must he parallel 111 the :axis. If the :a x i s has been erected in a seiise conIstt-hand triad
/
Ki@t~handtriad
(ill
ixj=-k
(hi
Figure 2.28. Diffcrent kinds of rcfcrcnccs.
AA A v
A:
By
4
j
k
B.\
i
(2.26)
(2.27)
For the products along the dashed diagonals, we must remember in this
method tu multiply by -I. We then add all six products as follows:
A , B > k + H,A,j
+ A,B:i
-
A ; B , i - B,A,j
-
A,B,k
= ( A , B - A$, ) i + (A,B, - A,B;)j + (A,B, - A,B,)k
Clearly, this is the same result as in Eq. 2.25. It must be cautioned that this
method of evaluating a determinant is correct only for 3 x 3 determinants. If
the cross product of two vectors involves less than six nonzero components,
such as in the cross product
( 6 i + lOj) X ( 5 j - 3 k )
then it is advisable to multiply the components directly and collect terms, as
in Eq. 2.25.
Example 2.6
A pyramid i s shown i n Fig. 2.30. If the height of the pyramid i s 300 I t .
find the angle brtweeii the outward n(irnia1s ti1 planes A D B and HIlC."
We sliall first find the unit normals to the aforesteted planes. Then.
using lhc dot product between these normals. we can easily find the
desircd angle.
To get thc unit normal n I lo plane A B D , we Sirst coinpulc the area
vectorA for this plane. Thus. from simple trigiinomelry and the delinition
o f the cross priiduct.
,
A,=;A%xA%
/ A
N11k next that
100'
H
Figure 2.30. Pyramid.
"G =
-
l00j ft
Furthcrmiire. we cai cxprcss A D i n rcctangular components by moving
lrom A tii I)along coordiiiatc directions as f ~ i l l i i w s :
A% = S O j
-
5Oi
+ 300k f t
Hcnce.
A, =
=
i (I 00j ) x (-5Oi + soj + 3nok )
15,oooi + 2.S00k ft?
Accordingly
A I - 1 ~ , 0 0 0+i 2,sook
15,0002 + 2.5002
IA,
= .c)X64i t . lh44k
~~~
"I
=
1 ,
(a)
~~
A h for unit n(iriiiiil ti2 corresponding Lo plane UIlC, whohe arm vectiir we
denote as A ,. w e havc
A,. = .~ 7x
.n7,
Niitc chat
2
=
-I OOi I t
-
And once again. moving along coordinate directions. we have for B D
871
=
-SOj -- SOi
+ 300k ft
SECTION 2.9 SCALAR TRIPLE PRODUCT
Example 2.6 (Continued)
Hence,
A, = &(-1OOi) x (-50; - 5 0 j
= I5,OOOj
+ 300k)
+ 2,500k ft2
Accordingly,
n ----4 - I5,000j+2,500k
2 lA21 115,0002+ 2,5W2
= .9864j + .1644k
Now, we use the dot product of n, and n2. Thus,
n,
n2 = cos /3
(C)
where p is the angle between the normals to the planes. Substituting from
Eqs. (a) and (b) into (c), we get
cos
p
=
,0270
Therefore,
We see from this example that a plane surface can be represented as a vector,
and if that plane surface i s part of a closed surface, by convention the area
vector is in the direction of the outward normal.
2.9
Scalar miple Product
A very useful quantity is the scalar triple product, which for a set of vectors
A , B , and C is defined as
(A x B) * C
(2.28)
This clearly is a scalar quantity.
A simple geometric meaning can be associated with this operation. In
Fig. 2.3 I , we have shown A . E , and C as an arbitrary set of concurrent vectors. We have set up an xyz reference such that the A and B vectors are in the
xy plime. Further, a parallelogram abcd in the xy p k n e is shown in the diagram. We can say that
IA
X
Bl = IAI lB1 sin a = area of ahcd
51
,
/'
_;11111-1
Figure 2.31. A and R in p planc.
Using thi.; gemictrical iiitzrpretati(in (if the \ c a l x triplc pr(iduct. the
reader can e x i l y conclude that
(A x R ) * C
=
~
(A
X
C)
B =
~
(C
X
B) * A
(2.29)
Thc computation 01 ihe scalar lriple pmduct i s a very s m i ~ h t h r w i ~ r d
pi-ocess. It will he left iis an cxercise (Prohleni 2.72) for you to demonstrate that
I
,A,
( A X H ) - C = 13~
C.)
As
A~
19>
n~
(2.30)
C> C<
I n liiler chapters. we shall einploy the svalar triple product. although wc chall not
alwayh want 10 associate ihc preceding gcomclric inteq~reltationof this prodticl.
Another opcration involvinz thrcc vectors i s ~ l l cI
defined Ihr vectors A . B. and C a s A x ( B x C).Thc vector triple product
i s a vector quantity and w i l l appear quitc o l t c n i n sttidier oidynamics. It w i l l
be left h r you lo demonstrate that
A
X
(B
X
C ) = R(A * C ) - C(A * B )
(2.11)
Notice here that the vector triple product can he carried oi11by uhing only dot
products.
SECTION 2.9 SCALAR TRIPLE PRODUCT
Example 2.7
In Example 2.6, what is the area projected by plane ADE onto an infinite
plane that is inclined equally to the x, y , and z axes?
The normal n to the infinite plane must have three equal direction
cosines. Hence, noting Eq. 2.20 for the sum of the squares of a set of direction cosines, we can say that
Therefore
Hence,
The projected area then is given as
A,, =
(;
A ~ Ix
-
AZ) n
= [ $ ( - 5 O i + 5 0 j + 3 0 0 k ) x ( - I O O i ) ] *. 3- L ( i +j + k )
The preceding result is a scalar triple product that can readily be solved as
follows (disregarding the final sign):
53
2.10
A Note on Vector Notation
When expressing ryirations. we must at all times clearly denote sciilar and
vectoi- qiiaiititics ;ind hmdlc them ;iccordingIy. When
ire simply identifying quantities i n il di.scu.s.sioii or i n a rliugi-(irrm. however, instead of using the
vector reprcscntation. I;. we can .just LISZ I.. On the other hand. / ~w
' i l l be
undcrstiiiid to rcprcscnt i n ;in equatioii the magnitude o l the vector F . Thu,
using f i n thc itnit \'cctor in the direcIion of I>.w e can then say:
F = /'f
= b'lcos (I;. r)i
+
cos tF. ! ) j
+
co\ ( F , : ) k ]
A \ another exaniplc. wc might wiiiit to employ the lorcc F, which i s shown in
the coplanar diagr;im o f Fig. 2.32ia) al a k n ~ i w i i i i c l i n a l i ~ nand acting a1 a
point (1. A correct representation o l this Iiirce i n a vector equation would he
I.(-cos ai
sin aj).
As for \ c a l x components 01 a n y vecfor 1:. we shall adopt the following
understanding. The Iiotiltim t';,h i , or t'; labeling s o m e vector component i n a
r/iu,y,-rinrw i l l hc undcrsloiid to rcprcscnt the niu,yi~ii~ide
o f that particular c o n ponent. Thus, in Fig. 2.32(b) the two ctimponcnLs shown arc cqual in magnitude hut opposite i n sense. Ncvcrtheless. thcy xrc both Iahclcd F,. H11wever.
i n an cquetion inviilving these quantities. thc scnsc must propcrly he
accounted for by the appropriatc usc o f signs.
+
2.63. If A = I O i + hj - 3k and B = 6i,find A x B and B
x A . What is the magnitude of the resulting vector'! What are its
direction cosines relative to the xyz reference in which A and B
2.69. If the coordinates of vertex E of the inclined pyramid are
( 5 , SO, 80) m, what is the angle between Outward normals to faces
A L E and BCE!
are expressed?
2.64. What are the cross and dot products for the vectors A and
B given as:
A = 6 i + 3 j + 4k
B = Ri 3j + 2k?
E
~
2.65. If vectors A and B in the xy plane have a dot product of SO
units, and if the magnitudes of these vectors are I O units and R
units, respectively. what is A x B !
2.66. (a) If A * B = A B'. does B necessarily equal E'?
Explain.
(b) If A X B = A
Explain.
-
X
Y
E , does B necessarily equal B"!
2.67. What is the cross product of the displacement vector from
A to B times the displacement vector from C to D'!
X
I
Figure P.2.69
2.70. In Prvblem 2.69, what is the area of face ADE of the p y m
mid? What is the projection of the area of face ADE onto a p l a ~
whose normal is along the direction 6 where:
C
X'
~ = O . h i- 0.Rj
2.71.
(a) Compute the product
/
(A
Figure P.2.67
2.68. Making use of the cross product, give the unit vector n
normal to thc inclined surface ABC.
I
B)
X
-
C
in terms of onhogonal components.
(b) Compute (C X A ) B and compare with the result in
part (a).
2.72. Compute the determinant
Ar
Ay A:
gv B:
c1 cy c:
B,
where each row represents, respectively, the scalar components of
A , B , and C. Compare the result with the computation of ( A x B )
C by using the dot-product and crowproduct operations.
2.73. In Example 2.5, what is the area vector for GAB assuming
a straight line connects points A and E ! Give the results in kilo-
/
Figure P.2.68
meters squared.
Fixure P.2.71
2.11
Closure
Check-Oulfor Sections with t
2.1 What i s meant by thc n i u p i t ~ o~ fka vector? What sign iiwst i t hauc'?
2.2 Can you multiply a vector C by ii sciilar s ? If so, dcscribc the rcsull.
2.3 What are the Im, o f u m i w s and the law qfsiiirs'?
2.4 What i s nicaiil by the <rs.w(.iariwlaw of addition'?
2.5 Describe two ways to add ;my three vectors graphically.
2.6 How do you subtract vector D Srom vector F'?
2.7 Given a vector D , how would you forni a m i t vector collinear with D'!
2.8 What are the cciiliir equations 01 thc following vector equation'?
Di
+
E,j - Ihk = 20i
+
iI S
+
G)k
2.78. Flight 304 from Dallas is tlying NE to Chicago XJU miles
away. To avoid a niilssive storm Srvnt, the pilot decides instead to
Ily due north to Topcka, Kansas, and then ENE (see Fig. P.2.4 f i x
compass settings) to Chicago. What are the distances that he must
travel from Dallas to Topeka and from Topeka ti) Chicago'!
N
Chicago
Figure P.2.80
2.81. Cuntractori encountered an irnpasiahle swnmp while buildinp a road from town T to city C SO kin SE. To avoid the swamp,
they built the road SSW from T and thcrr ENE to C. How long is
the road? ( H i m : See the compass-settings diagram, Fig. P.2.4.)
45"
E
Dallas
Figure P.2.78
2.79. What is thc cross product between the 1,000-N force and
the diiplaccrnent vector p,,,"'?
2.82. Sum all forces acting on the block. Plane A is parallel to
the .r.v plane. We will later m d y the special properties of two parallel forces (called a c o u p k ) that are opposile in direction atid
in magnitude'
.
//"
I,11011 N
-y
(3.4, - 2 )
Figure P.2.79
Figure P.2.82
2.80. Four member, of a space frame are loaded as shown. What
are the orthogunill scalar components of the forces on the ball
joint at O?l h e 1,000-N force goes thmugh points D and E of the
rectangular p;irallelepiped.
2.83. The r and I compunsnti of the force F arc known tu be
100 Ih and -311 Ib, respectively. What i q thc f k c F and what are
its direction cosincc"
SI
,
with a speed of 100 d s e c . What are the force components on the
coulombs.
electron'? The charge of the electron is l.hOlX x
500 Ib
400 Ib
s
L.
Figure P.2.89
2.90. A skeet shooter is aiming his gun at point A . What is the
heirht
iof
point A'?
7
Figure P.2.92
.
2.93. For the line segment A X , determine
A(2,15,z)m
zx
and direction
cosines m and n.
Figure P.2.90
2.91.
n
?--/
If F, and the 5 W N force sum vectorially to FT, determine
/
F, and F,
Figure P.2.93
2.94. Using the scalar triple product, find the area projected onto
the d a n e N from the surface ABC. Plane N is infinite and is normal to the vector
r = 50i
+ 40j +
30k
ft
500 N
Figure P.2.91
2.92. The force on a charge moving through a magnetic field B
is given as
F = ~ V X B
where q = inagnitude of the charge, coulombs
F = force on the hody, newtons
V = velocity vector of the pafiicle. meters per second
B = magnetic flux density. webers per meter2
Suppose that an electron moves through a uniform magnetic field
of IO" Wh/mz in a direction inclined 30' to the field, as shown,
A
Y
x
Figure P.2.94
2.95. A 500-lh crate is held up by three forces. Clearly the three
forces should add up to a force of 500 Ib going upward. What
should forces F , and F, he for this condition'? All forces are
coplanar (in the same plane).
59
I
!hl Il a hze-bee gun i s to sliont down thc ballowi, a s s u n ing i t is morncnlarily slali<mary ill C. what arc thc
direction corinrs ol the prupcr lint. r i f sight?
Figure P.2.96
60
2.100.
"CCtOl
What i s thc anglr S hclwccn
p,,,,.!
Figire P.2.100
r
and the displaccmcnt
Important
Vector
Quantities
3.1
z
Position Vector
In this chapter, we shall discuss a number of useful vector quantities. Consider first the path of motion of a particle shown dashed in Fig. 3.1. As indicated in Chapter I, the di,splacemenr vecfor p is a directed line segment
connecting any two points on the path of motion, such as points I and 2 in
Fig. 3.1. The displacement vector thus represents the shortest movement of
the panicle to get from one position on the path of motion to another. The
purpose of the rectangular parallelepiped shown in the diagram is to convey
the magnitude and direction of p as explained earlier. We can readily express
p between points I and 2 in terms of rectangular components by noting
the distance in the coordinate directions needed to go from I to 2. Thus, in
Fig. 3.1, p l r = - 2i + 6 j + 3k m.
The directed line segment r from the origin of a coordinate system to a
point P in space (Fig. 3.2) is called the position vector. The notations R and p
are also used for position vectors. You can conclude from Chapter 2 that the
magnitude of the position vector is the distance between lhe origin 0 and
point P. The scalar components of a position vector are simply the coordinates of the point P. To express r in Cartesian components, we then have
r = xi
+ y j + zk
(3.1)
We can obviously express a displacement vector p between points 1
and 2 (see Fig. 3.3) in terms of position vectors for points 1 and 2 (Le., r , and
r2)as follows:
p = r? - r , = (x2
~
x,)i
+ cy2
~
y,)j
+
(z, - z,jk
(3.2)
Figure 3.1. Displacement vector p
hetween points 1 and 2.
)/;
y"
. o
j
Y
x
x
Figure 3.2. Position vector.
__---
__
Figure 3.3. Relation hetween a
displacement vector and position
Yectors.
61
62
("AFTER
3 IMI'ORTANT VECTOR QUANTITIES
Example 3.1
T w o bets ul' refercnces. .YK and XYZ. arc shown in Fig. 3.4. The position
vector o l the origin 0 o l r y : rclative to XYZ is given as
+
R = IOi
6j
+
The position vector. r'. i f a point P relative
r' = 3i
+
Sk m
til
(a)
XY7 is
2.1 - hk m
What is the position vector r of point 'f relative
dinates .I,y. and iof I"?
From Fig. 3.4. i l is clear that
(h)
ti1 .v?i?
What arc the coor-
r ' = K + r
IC1
Therefore.
r = r' -- R = (3i
+
Zj
-
r =-7iL4j-
6k)
-
(IOi
Ilkm
+
hJ
+
Sk)
I4
We can then conclude that
' X
x = -7m
y = -4m
z = -11 m
t"
3.2
(c)
Figure 3.4. Refercnccs .q;
and X ) %
scpamtcd hy position vector it.
Moment of a Force About a Point
Case A. For Simple Cases. The moment of a force about a point 0 (see
Fig. 3.5). you will recall from physics, is a vector M whose magnitude equals
the product of the force magnitude times the perpendicular distance d from 0 to
the line of action of the force. And the direction of this vector is perpendicular to
the plane of the point and the force, with a sense determined from the familiar
right-hand-screw rule.' The line of action of M is determined by the prohlem at
hand. In Fig. 3.5, the line of action of M i s taken for simplicity through point 0.
Case B. For Complex Cases.
Figure 3.5 Moment 0 1 force F nhoul 0 is F d
Another apprwach is to employ a position
vcctcir r from point 0 to nnypoinf P along the line ol'action of force F as shown
in Fig. 3.6. The nioment M of F about point 0 will he shown to he given as1
M = r x F
(3.3)
SECTION 3.2
MOMENT OF A FORCE ABOUT A POINT
63
For the purpose of forming the cross product, the vectors in Fig. 3.6 can he
moved to the configuration shown in Fig. 3.7. Then the cross product between
r and F obviously has the magnitude
lr X FI = /rl IF\ sin a = IF1 lrl sin p = IF1 r sin fi = Fd
where r sin fi = d, the perpendicular distance from 0 to the line of action of
F , as can readily he seen in Fig. 3.7. Thus, we get the same magnitude of M
as with the elementary definition. Also, note that the direction of M here is
identical to that of the elementary definition. Thus we have the same result as
for the elementary definition in all pertinent respects. We shall use either of
these formulations depending on the situation at hand.
The first of these formulations will he used generally for cases where
the force and point are in a convenient plane, and where the perpendicular
distance between the point and the line of action of the force is easily measured. As an example, we have shown in Fig. 3.8 a system of coplanar forces
acting on a beam. The moment of the forces about point A is thenZ
MA = -(5)(1,000).4 - (4)(600)k + (11)R,k ft-lb
Figure 3.6. Put r from 0 to any point along
the line of action of F .
= ( I IR, - 7,400)k ft-lb
x
Figure 3.7. Move vector r end F
Figure 3.8. Coplanar forces on a beam
For a coplanar force system such as this, we may simply give the scalar form
of the equation above, as follows:
MA = IIR, - 7,400ft-lh
The second formulation of the moment ahout a point, namely r X F ,
is used for complicated coplanar cases and for three-dimensional cases. We
shall illustrate such a case in Example 3.2 after we discuss the rectangular
components of M.
Consider next a system of n concurrent forces in Fig. 3.9 whose total
moment ahout point 0 (where we have established reference xyz) is desired.
We can say that
M=M, +M,+M,+...+Mn
= r x F , + r x F , + r X F ,
+ . . . + r x F,
(3.5)
‘Please nole that we still use lhe right-hand-screw rule in determining Ihe signs of the
respective moment%
Figure 3.9. Concurrent forces
Now. hecause (if the distrihutive pixipei-ty of the c r w s product. Eq. 3.5 can
he bvrittcn
M = r X
(6+ F, + F3 + . . . + I;,,)
(3.61
Wc ciin conclude iriiiii the preceding eqiiatioiis that thc s u m of the minimits
ahout it point of a system of concurrent fixcss i s the same iis ihc ~iionicnt
ahout tlic point of the \uni 01 rhc force^. l h i r i-esul~i s kiiinvn iis \'uri,qmm'v
I/i('ofwi~.
which y>um a y UCIIreciill Srofii physics.
Ax ii hpccial ca\c of Varigniin's thcorcni, we inay find i t convenient io
deconipose a iorcc F inti1 i h rcclangular ciimpiincnts (Vi;. 1. IO). and tticn t o
iise these coniponents ior taking iiioiiieiits ahwt ;I point. Wc cim Ihrn s;ly that
M = r x F = r x (P;i +
/I./ +
Fk)
(3.7)
SECTION 3.2 MOMENTS OF FORCE ABOUT A POINT
Example 3.2
Determine the moment of the 100-lb force F , shown in Fig. 3.11, about
points A and E , respectively.
As a first step, let us express force F vectorially. Note that the force
is collinear with the vector pljEfrom D,, to E, where
pllK= Xi
+ 4 j - 4k
To get a unit vectorp in the direction of P , , ~ .we proceed as follows:
=
P,], - Xi + 4 j - 4k
lPDE/ is2 + 4 2 + 4 2
~
(b)
= .X16i + .408j - .408k
We can then express the force F in the following manner:
F=I.'pf,,=(100)(.816i+.408j-.40Xk)
=81.6;+40.Xj-40.8k
(c)
To get the moment MAabout point A , we choose a position vector
from point A to point D which is on the line of action of force F. Thus, we
have, for cn,
G~~= IOi + 4 j - Xk ft
(d)
and for MA,we than get
=I%
MA = r A , , X F = ( 1 0 i + 4 j - 8 k ) X
,
m,
=
(81.6it40.8j-40.8k)
-
W.6 40.8 - .
(10)(40.8)k + (Xl.6)(-8)j + (4)(-40.8)i
-(-8)(40.8)i - ( 4 0 . 8 ) ( 1 0 ) j- (4)(81.6)k
Therefore.
MA = 163.21 - 245j + X1.6k ft-lh
(e)
As tor the moment about reference point E , we employ the position
vector rBI, from B to position D,again on the line of action of force F.
Thus, we have
rHfj= 4j - Xk ft
Accordingly,
M,
F = (4j - Xk) x (81.6i + 40.Xj - 40.8k)
= (4)(81.6)(-k) + (4)(-40.8)(i) + (-X)(XI,6)(j) + (-X)(40.8)(-;)
= rHIl x
MB = 163.23 - 653j - 326k fklb
(t)
Figure 3.11. Find moments at A and 8.
65
3.1.
What is thc position v e c t ~ r from the origin (0, 0, 01 to the
poinl (3, 4, 5 ) ft'! What are its magnitude and direction cosines?
3.2. What is the displacement vector from position 16. 13, 71 ft
to position ( I O , --3, 4) ft'?
A surveyor determines that the top 0 1 a radir, transmission
tower is at position r, = (1,OOOi + I,OOOj + 1,000k) m relillivc
:(I her position. Similarly. the top of a sccond tower is locatcd hy
5 = (2,000i+ 500j + 700k) m. What is the distance hetwcen
3.3.
3.7. A particle mmes along ii circular path in the xy plane. What
is the position vector r of this p;irticle as ii function of the wordi.
nxte r"
I
I
.he two tower tops?
3.4. Reference XJZ is rotated 30" counterclockwise about its ~i
rxis to form reference XYZ. What is the position vector r for refer:nce xyz of a point having a position V C C ~ O Ir' h r rcfcrencc X Y 7
:iven as
r'
=
hi'
+ 107 + 3k'
Figure P.3.7.
m?
3sc i , j , and k (no primes) lor unit vectorb auaciatcd with refer:nce 1y7.
1.5. Find the momcnt of thc SO-lb forcc ahout the support at A
ind a h w l support A of the simply supported hzam.
3.8. A particle moves along a paraholic path i n thc i;plane. If the
particle has a1 ow point n pmition vrctiii r = 4 j + Zk, give the
piisition ~ c c t oa1
i ~any point on Ihe path i i i ii fimotion of Ihr :c o w
din;ilc.
v
-
61)'100'----t
)
-
Fieure P.3.S.
1.6. Find the moment of the two lorces first ahout point A and
hen ahout point R.
(a) Do not u s e r X F fbrmat-only scalar prodncts.
(h) lJsc vector approach.
I,OOO N
x
-2m---,
Figure P.3.h.
Figure P.3.8.
3.9. An nnillery spotter on Hill 350 (350 rn high) eqimiltes the
posirioii o i an cncmy tnnk as 3.000 m NE of him at a n C I c v a t i m
200 m helow hi? position. A 105-nim howmer unil with n range
of I1.(100 m i s 10,1100 m doc south oltht. spotter. and il 155-mm
h o w i t m unit with a rangu of 1.000 m is 11,001) m SSE o l t h r
spottcr (rce Fig. P.2.41. Both gun units are located at an elevation
OS 150 111. Can eillirr o r hoth gun unit, hit thc tank, o r r m w t an air
mike be callcd in?
3.10. Find the moment of the forces about points A and B .
(a) Use scalar approach.
(b) Use vector approach.
3.13. The total equivalent forces from water and gravity are
shown on the dam. (We will soon be able to compute such equivalents.) Compute the mnment of these forces about the toe of the
dam in the right-hand comer.
(Asrume a11 forces
+ L -kI
8’
‘Toe
Figure P.3.13.
Figure P.3.10.
3.11. The crew of a submarine patrol plane, with three-dimensional radar, sights a surfaced submarine 10,000 yards north and
5,000 yards east while flying at an elevation of 3,000 ft above sea
level. Where should the pilot insmct a second patrol plane flying
at an elevation of 4.000 ft at a position 40,000 yards east of the
first plane to look for confirmation of the sighting?
3.12. A power company lineman can comfortably trim branches
I m from his waist at an angle of 45” above the horizontal. His
waist coincides with the pivot of the work capsule. How high a
branch can he trim if the maximum elevation angle of the arm is
75“ and the maximum extended length is 12 m?
3.14. In an underwater “village” for research. an American flag
is in place as shown. It is of plastic material and can rotate so as to
be oriented parallel to the flow of water. A uniform friction force
distribution from the flow is present on both faces of the flag having the value of 10 N per square meter. Also the flagpole has a
uniform force from the flow of 20 N per meter of length of the
flagpole. Finally there is an upward buoyant force on the flag of
30 N and on the flagpole of 8 N. What is the moment vector of
these forces at the base of the flagpole?
Figure P.3.14.
3.15. Three transmission lines are placed unsymmetrically on a
power-line pole. For each pole, the weight of a single line when covered with ice is 2,ooO N. What is the moment at the base of a pole?
1.5 m
Figure P.3.12.
Figure P.3.15.
67
3.17.
A truck-mounted crane has a 20-m hrxnn inclincd a1 60" to
the horizontal. What i \ the moment ahout the honm pi\ot due t o a
liftcd wcight of 30 k N ? Do hy vector and hy scilliii~methodi.
3.22. What i\ the moment of a IO-lh I w c e F directed ;ilonf lhe
di:igmaI of a cuhr ahout thc c ~ m c nof the cuhc'! 'The sidc 01 thc
w h c i s (i I t .
Figure P.3.22.
3.23.
Three guy wires are used in the suppon syslem lor a trle\ i s i o n transmicrion towcr that i\ h0O m tall. Wircb A and R arc
tiphtcnrd t o a tzmion o f hO LN. whereit\ w,irc C has iinly 30 kN 01
t c m i o n What i s thr nomrnt 01 h e wire I w c v a h w l the hax 0
o f lhe tower'i Thc y :,xi\ i': ciillincar with A O .
Figure P.3.17.
3.18. A small hlirnp i s lempr,l-arily mrxrred
as shown in the d i i i gram wherein / I C and the centerline of AH arc coplanar. A
force F from wind, weight. and buoyancy i s shown ;sling ill Ihc
centerline of the hlirnp. It
F
=
I
5i + l l j + IXk kN
what are the moment w c t m from F ahwt A . N. and
c?
I
hi1 111
Figure P.3.18.
68
F i p r e P.3.23.
3.24. Cables CD and AB help support member ED and the
I ,000-lb load at D. At E there is a ball-and-socket joint which also
supports the member. Denoting the forces from the cables as Fer,
and FAB. respectively, compute moments of the three forces about
point E . Plane EGD is perpendicular to the wall. Get results in
terms of Fro and T,r
Figure P.3.24.
3.3
Moment of a Force About an Axis
Case A. For Simple Cases. By means of a simple situation, we shall
set forth a definition of the moment of a force about an axis. Suppose that a
disc I s mounted on a shaft that is free to rotate in a set of bearings, as shown in
Fig. 1.12. A force F, inclined to the plane A of the disc, acts on the disc. We
decompose the force into two coplanar rectangular components, one normal to
plane A of the disc and one tangent to plane A of the disc, that is, into forces FB
and
respectively, so as to form a plane shown tinted, normal to plane A.
5,
Figure 3.12.
5 turns disc
We low )m experience that Fs does not cause : disc rota1
know from physics and intuition that the rotational motion of the disc is
69
70
CHAPTER 3
IMPORTANT VEC'TOR Ol~\Wl'l'llliS
delermincd by the product (if F;1 and the perpendicular distance d I m m the
centerline (ifthe shaft to the line of action of F4. You w i l l remember from
physics, this product i s nothing more than tlie mmncnt 01. force F ahout the
axis (if the shaft. We sliall next generalire from this simple case tn the general
case o f taking Ihc moment d'un\. force F ahout ow? axis.
T o compute the moment (or tnrque) of a force F i n a planc perpendicular
to plane A ahout an axis B-B (Fig. 3.13). we pass any plane A perpendicular to
the axis. This plane cuts i3-R a1 (I and thc line of action of forcc Fat some poini
P. 'The fnrce F i s then pro.jectcd to fomi a reclmgular component
along ii
line at P normal 10 plane A and tliiis parallel tu R-B. a s shown in the diagram.
'The intersection of plane A with lhe plane of Snrceh F,$ and F (the latter plane i h
shown shaded and i h a plane throush F and pcrpeiidicular to plane A ) sives ;I
direction C-C along which the other rcclangular component of I.: denoted :is ti.
can he projected." The moment of F about the line 8 - ~ Bi s then defined as the
scalar reprcsentation of the mnment of lj ahout point ( I with a inagnitude equal
to <</--a
prcihlcni discussed a1 the heginning OS the previous section (Case A).
Thus in accordance wilh the definition, the component kjr which i s parallel to
the axis 11-8. contributes 110 moment about the iixis, and wc may say:
r;,
'
H
Figure 3.13. F(1rmula% the
ahout an axis R -8.
mOll1ellt
Mnment ahout a x i s B-H = l<)ldl =
(cos a ) ( d )
with nn appropriate sign. The moment about an a x i s clrarly i s a scalar. evcn
though this niomcnt i s associated with a particulx axis that has ii distinct
direction. 'The situation i s the same as it i s with the scalar components V,, V,.
etc.. which arc associated with certain directinns but which are scalar\.
The reader will he quick to noLe that Fig. 3.13 reprchents ii generalization (it
Fig. 3.12 having an axis H-B. a plane A normal to this axis and fillally an
arhitrary force E To explain further. we have redrawn Fig. 3.13 Isec Fig.
3.14Ia)l showing only plane A , axes R-R and C-C, and the force I;. In Fig.
3.14(h), we have also included the moment vector M . This latter diagram then
takes us back Lo Fig. 3.5 where wc first defined the moment vector o f '.I 1.nrce
ahout a point i n a most simple manner. Accordingly, we note, nn the cine
hand. for the moment ahout point (I [scc Fig. 3.14(b)], we can get the vector
M , whereas 011 the other hand i n Fig. 3.I4la) we can get the scalar iiiomciit M
ahout an axis B-8 at point n and perpendicular to planc A . Thus, by taking the
scalar value o f M in Fig. 3.14(b), wc get the moment about the a x i s at puint ( I
normal t o plane A as formulated in the development of Fig. 3. 13.
v,
Figure 3.14. Comparison of Figs. 3.13 and 3.5.
SECTION 3.3 MOMENT OF A FORCE ABOLIT AN AXIS
We can thus conclude, on considering Fig. 3.15. that the moment of F
about point C can be considered in two ways as follows:
Figure 3.15. Consideration of the moment
of F about point A .
Before continuing, we wish to point out that 4 in Fig. 3.13 can be
decomposed into pairs of components in plane A . From Varignon's theorem
we can employ these components instead of 6 in computing the moment
about the B--B axis. For each force component, we multiply the force times
the perpendicular distance from a to the line of action of the force component
using the right-hand-screw rule to determine the sense and thus the sign.
Case B. For Complex Cases. When we discussed the moment of a
force about apoint, we presented a formulation useful for simple cases (i.e., a
vector of magnitude Fd) as well as a more powerful formulation that would
be needed fix more complex situations (Le., r x F ) . Thus far, for moments
about an axis, we have presented a formulation Fd that is useful for simple
cases? and now we shall present a formulation that is needed for more complex cases. For this purpose, we have redrawn Fig. 13.13 as Fig. 13.16(d). In
Fig. 13.16(b), we have shown axis B-B of Fig. 13.16(a) as a n n axis and have
set up coordinate axes y and z at any point 0 anywhere on axis B-B. The
coordinate distances x, y . and z for the point P are shown for this reference.
The position vector r to P is also shown. The force component FH of Fig.
3.16(a) now becomes force component F,. And, instead of using
we shall
decompose it into components and 4 in plane A as shown in Fig. 3.16(b).
We now compute the moment about the x axis for force F using this
new arrangement which does not require F to be in a plane perpendicular
to plane A . Clearly, F, contributes no moment, as before. The force components
and 5 are in plane A that is perpendicular to the axis of interest and so,
as before in the case of 5,we multiply each of these forces by the perpendicular distance of point a to the respective lines of action of these forces.
<.
FA,
<.
'That is. for cases where the force and point in question are in a plane easily seen to he nor^
mal to the ru-is in question. thus allowing for an easy determination of the perpendicular distance
between the point and the line of action of the force.
71
72
CHAPTER 3
IMPORTANT VECTOR amNri-riEs
For Sorce t;. this pcrpcndicular distance i s clc;irly x. a h can rcadily be
the diagmni, and, lor force E ; , this perpendiculx distance i s :. Using the righthand-scrcw rule fiir :isccrt;iining llic sense oleach o i the momenls. wc: can say:
~ii~iiieiit
ahout~raxis=
IxF
-
(3.10)
:.f< j
Were we to hikc m o m e n t s OFF ahout the origio 0, wc \vould get (scc Eq. 1.8)
M=M,i+M,j+M:k=rx F
= IyF
-
:F:)i
+
-
+
xF;),j
(.r/,;
-
yF\;,)h
(3.1 I )
Comparing Eqs. 3.10 arid 3.1 I. wc ciin conclude that tlie iiioiiieiil about the .I
axis i s simply M , . 1he.v coiiilionent o l M about 0.We ciin thus concludi: that the
moment about the .v axis 01the Iilrce F is ihc component i n the .\ direction of the
moment 0 1 F about a point 0 positioned m ~ w l w r calong
,
the Iaxis, T ~is. I
+~/
,
"
moment ;ihout
/'
Figure 3.17. M,,= ir X F ) *
iC1.i
12.
1
axis
7
M, = M,,
-
i = (r
X
F) * i
(1.12)
We may generali/e tlie preceding discussion iis Iollows. Consider iiii arbitral-y
axis ii-ii to which we have ;is\igned a unit vector 11 (Fig. 3.17). An uhitrary
force I; i\ a l m shown. 'lo gut the iiioiiicnl M,, of I'orcc F ahout axis 11-PI. \e
clioosc any point 0 along 11-11. Then draw ii p o s i t h i vector r Iron1 point 0 to
any point dong Ihc line 01; d u n of F . Thih h a been shown i n [he diagraiii.
We can then hay. fironi our prcvious discussion.
M,,= (r
X
I;)
-
n
(3.13)
SECTION 3.3 MOMENT OF A FORCE ABOUT AN AXIS
(Notice from Eqs. 3.12 and 3.13 that the moment of a force about an axis
involves a scalar triple product.) Equation 3.13 stipulates in words that:
The moment of a force about an axis equak the scalar component in the
direction qf the axis of the m m e n f vector taken about any point along
the axis.
This is lhe more powerful formulation that can he used for complex cases.
Note that the unit vector n c d n have two opposite senses along the axis n,
in contrast to the usual unit vectors i, j, and k associated with the coordinate
axes. A moment M,, about the n axis determined from M * n has a sense consistent with the sense chosen for n. Thdt is, a positive moment M,, has a sense
corresponding to that of n, and a negative moment M,, has a sense opposite to
that o f n . If the opposite sense had been chosen for n, the sign of M n would
be opposite to that found in the first case. However, the same physical moment
is obtained i n both cases.
If we specify the moments of a force about three orthogonal concurrent
axes, we then single out one possible point in space for 0 along the axes.
Point 0,of course. is the origin of the axes. These three moments about
orthogonal axes then become the orthogonal scalar components of the
moment of I; about point 0, and we can say:
M = (moment about the x axis)i
+
(moment about they axislj +
(moment about the z axis)k = M,i
+ MJ + M:k
(3.14)
From this relation, we can conclude that:
The three orthogonal components ojthe moment of a force about a point
e
are the muments of this force about the three o
the point as a n origin.
You may now ask what the physical differences are in applications of
monients about an axis and moments about a point. The simplest example is
in the dynamics of rigid bodies. If a body is constrained so it can only spin
about its axis, as in Fig. 3.12, the rotary motion will depend on the moment of
the lorces about the axis of rotation. as related by a scalar equation. The less
familiar concept of moment about a point is illustrated in the motion of bodies that have n o constrhints, such as missiles and rockets. In these cases, the
rotationiil motion of the body is related by a vector equation to the moment of
forces acting on the body about a point called the center ofmass. (The center
of mass will he defined completely later.)
13
14
CHAPTER 3 IMPORTANT VECTOR QUANTITI~S
Example 3.3
+
Coinpiitc the moment of a force F = 1Oi
hj N. which gocs through
po5ition
= 2 i + 6.j in (see Fig. 3.181, about a line going through points
I and 2 having the respective position vectors
cz
i!
I
rl = hi
+
I O j - i k in
r, = -3i - 1 2 j
+ 6k in
T u compute this moment. we can take the moment o f F about either
point I or point 2, and then rind the component or this vector along the
/'
r,'
direction of the displacement vector hetween I and 2 or between 2 and I.
Mathematically, wc have, using a displacement vector from point I tu
point u. namely (c, - r , ) .
Figure 3.1X. Find momenl d F
M I'
=
Iir,,
-
r , ) x FI *
fi
(3)
where i j i s the unit vector along thc line chosen to have a sense going from
point 2 to point I. The formulation ahove i s the s c d a r triple product exaniincd in Chapter 2 and we can usc the determinant approach for the calculation m c e the components 01 the vectors (r,, - rl). F , and i j have been
determined. Thus. we have
iI
r(! - rI =
( 2 i + 6 j )- (hi
+ lOj - i k )
= -4i - 4.j + 3k m
F = IOi
+6 j N
= .354i
+ X66j
-
.354k
We then have, for M,,:
!
i
#
Mp
i
=I
-4
-4
3~
10
h
01 =
3 5 4 ,866 - . 1 d
13.94 N-m
ib)
Because M,, i s pobilive. wc have a clockwisc inoiiieiil about thc line as we
look from point 2 to point I . I 1 we had chosen p lo have an opposite sense,
then M,, would have been computed a s -13.94 N-in. Then. we would con:
i cludc that MI, i s a coiintcrclockwix rniimcnt about the line a s one looks
! from point I to point 2. Note that the same physical inoinent i s determined
i
in both cases.
-.
..
ahout line.
SECTION 3.3 MOMENT OF FORCE ABOUT AN A X I S
Example 3.4
A deep submergence vessel is connected to its mother ship by a cable
(Fig. 3.19). The vessel becomes snagged on some rocks and the mother
ship steams ahead in a forward direction in an attempt to free the submerged vessel. The connecting cable is suspended from a crane directed
up over the wdter 20 m above the center of mass of the mother ship and 15
m out from the longitudinal axis of the mother shim The cable transmits a
force of 200 kN. It is inclined 50" from the vertical in a vertical plane
which. in turn, is oriented 20" from the longitudinal axis of the ship. What
is the moment tending to cause the mother ship to roll about its longitudinal axis (].e., the x-axis)?
The position vector from the center of mass C to point A is
I
r
=
-1Sj
0
-144
M, =
1
+ 20km
-15
20
-52.4 -128.6 kN-m
0
0
75
3.25. Disc A has a radius of 600 mm.What i s the moment of the
%rces ahout the center of thc disc? What i s the torque of thew
brces ahout the axis of the shaft'!
I
/2
3.29. A blimp i\ moored to a tmvcr at A . A forcc on A Irmr t h i b
blimp i s
F
=
Si
+
3j
+
I.Xk k N
What i s the moment ahaut tixis C ' c m the gmund'! Knowledge 01
this m o m c ~ and
~ t othcr ni~nirnt\at the b a s t is nreded to pnqxrly
dchign the fciundation of the tower.
I I\N
kN
Figure P.3.25.
L2h.
A fmce F acts at position ( 3 , 2, 0) ft. It I S in thc ,Q planc
md is inclined at 30" from the I axis with a s c n x directed away
rom the origin. What is the momcnt of this force ahout an axis
ping through thc paints (6, 2. 5 ) ft and (0, -2. - 3) ft?
1.27. A force F = IOi + C r j N goes through the origin of the
.uordinate system. What is the nioment 01this fircr F about an
!xis going through points I ;md 2 with position vcctors?
+
r , = 6i
3k m
r2 = 16j - 4 k m
c,
Figure P.3.29.
3.30. Compute thc thrurt 01 the applicd tnrcci shown aluog
the axis nf the rhair and the torque of the firrcr? about the a x ~ sof
1.28. Given a forcc F = IOi + 3j N acting at posirion
= S j + I0k ni, what i s the turquc ahout the diagonal showti iii
he diagram? What i?the t n ~ m e n tabout point E!
the shaft.
100
Ih
. ...
.
L)
3m
,...t...-;h,--i
2m
Figure P.3.28.
'6
A/
3.31. What
Figure P.3.30.
i s the milximuni load W that the craiic c u i lift witliout tipping about A'! Hirrr: when tipping is irnpcnding, what i s Lhc
suppoiring force tit thc wheel at IT!
L
.x
3.34. The base of a Sire truck extension ladder is rotated 75”
counter-clockwise. The 25-m ladder j s elevated 60‘ from the horizontal. The ladder weight is 20 kN and is regarded as concentrated at a point I O m up from the base (the lower part of the
ladder weighs much more than the upper part). A 9 0 - N fireman
and the 500-N young lady he is rescuing are at the top of the ladder. (a) What is the mument at the hase of the ladder tending to tip
over the fire truck’! (b) What is the moment about the horizontal
axis having unit vector 6 shown in the diagram?
Figure P.3.31.
3.32. Find the moment of the I ,000-lh force about an axis going
hetween points D and C.
View A-A
Figure P.3.34.
Figure P.3.32.
3.33. In Problem 3.24, what is the mumenr of the three indicated
forces about axis CD?
3.4
I A\,-F
The Couple and Couple Moment
A special arrangement of forces that is of great importance is the cuuple. The
cuuple is formed by any two equal parallel forces that have opposite senses
(Fig. 3.20). O n a rigid body, a couple has only one effect, a “turning” action.
Individual forces or combinations of forces that do not constitute couples
may “push” o r “pull” as well as “turn” a body. The turning action is given
quantitatively by the moment of forces about a point or an axis. We shall,
accordingly, be most concerned with the moment of a couple, or what we
shall call the couple moment.
y
1
Figure 3.20. A couple.
I1
78
CHAI'PdK 3
IMI'OKrANT VECTOR QUANTITIES
e
=
1rl-r:)
-F
Let us now evaluate the moment of the couple about thc vrigin. Position vectors have heen drawn in Fig. 3.21 to points I and 2 anywhere along
the respective line OS action of each force. Adding the monient of each force
about 0, we have f i r the couple moment M
M = rl x F
+
rL
x (-F)
(3.15)
= (rl - r.) x F
We can see that (rl - r , ) i s a displacement vector between points 2
and I, and if we call this vector-e, the formulation above hecomes
Figure 3.21. Cornpule ~ i i i m c nof~ couplc
abuut 0.
M = e
X
13.16)
F
Since e i s i n thc planc o f the couple. i t i h clear froin thc dcfinition of a cross
product that M i h in an i r i e n t a t i ~ nnorinal to the plane o S the couple. The
sense in this case may he seen in Fig. 3.22 to he directed downward. in accordance with the right-hand-screw rule. Notc thc use ofthe double arrow to represent the ciiuple miiment. Note also that the riitatiiin of e to F , a s stipulated
in the cross-product lormulation, i s i n the siiiiie direction as the "turning"
action u t the two force vectors, and from now vn wc shall use the latter criterim for determining the sense of r ~ t a t i o nto he used with the right-handscrew rule.
M
Figure 3.22. I he cuuplc moment M
Now that the direction of couplc ~nornentM has hem established for
the couple, wc need only compute the niagnitude for a complete description.
Points I and 2 may be chosen anywhere along the lines o f action of the forces
without changing the resulting moment. since the forces arc transmissible for
taking moments. Therefore, to compute Ihc magnitude 01 the ciiuple moment
vector it will be simplest to choose positions I and 2 s o that e i s p r p e n d i r u lar to the lines of action of the forces ( e i s then dcnvted as eL). From the definition of thc cross product, we can then say:
IMI = lell IF! sin '10"
=
led
=
IFId
(3.17)
where the more familiar notation. (1. has been used in place v1 lell as the perpendicular distance hetween the lines of action 01the forces.
SECTION 3.5
THE COUPLE MOMENT AS A FREE VECTOR
Note that in the computation of the moment of the couple about origin
0, the final result in no way involved the position of point 0. Thus, we can
assume immediately that the couple has the .same moment about every point
in space. More about this in the next section.
3.5
The Couple Moment as a Free Vector
Had we chosen any other position in space as the origin, and had we computed the moment of the couple about it, we would have formed the same
moment vector. To understand this, note that although the position vectors to
points 1 and 2 will change for a new origin O‘, the difference between these
vectors (which has been termed e ) does not change, as can readily be
observed in Fig. 3.23. Since M = e X F, we can conclude that the couple
hus the sume momenr about every point in space. The particular line of action
of the vector representation of the couple moment that is illustrated in Fig.
3.24 is then of little significance and can be moved anywhere. In short, the
couple moment is n free vecror. That is, we may move this vector anywhere in
space without changing its meaning, provided that we keep the direction and
magnitude intact. Consequently, for the purpose of taking moments, we may
move the couple itself anywhere in its own or a parallel plane, provided that
the direction of turning is not altered-i.e., we cannot “flip” the couple over. In
any of these possible planes, we can also change the magnitude of the forces
of the couple to other equal values, provided that the distance d is simultaneously changed so that the product IFld remains the same. Since none of these
steps changes the direction or magnitude of the couple moment, all of them
are permissible.
Figure 3.23. Vector e is the same for both references.
19
80
CHAPILR 1
IMI'OKTANT VFCTOR OI'ANTITIFS
As we pointcd out ciirlicr. the only effect of a couple on ii rigid hody i s
represented quantitatively by lhc miimcnt of the
couple-i.c., thc ciiuplc iniimcni. Since this i s its sole effect. it i s only naturiil
to represent the couple hy specifications of i t s moment; its magnitude. then.
becomes IFId and its direction that of i t s monient. Thi\ i h tlic siliiic a s identilying a person by Iicrlhih joh (i.e., its :I teacher. plumber, etc.). Thus, in Fig.
1.24. the cmiple niiiment C inay he used to seprcmit the indicated couplc.
i t s turning action. which i s
3.6
loo lh-ft
25 Ib-I1
Addition and Subtraction of Couples
Since couples thcinscl\,cs liavc x r o iiet lorccs. addition per se of coiiplcs
alw;iys yields 7ei-c lorcc. Fur this rciis1111. the ;iddition and suhtsactioii of
ciiuples i s interpreted to iiieaii ;idditiiin and subtsaction 01the t?zotnwii.s of thc
couples. Since ciiuple nioniciits arc lrec vector\. we ciiii tilways arrange to
have a concurrent system 01 vector\. We r h l l iiow tahe the iipportuiiity ki
illustrate inany 01the ciirlies remarks about couples by adding the two couples
\tiown on the face 111 the cuhe iii I"g. 3.25. Noticc that the couple miiment
vectors 01thc couples have heeii drawn. Sincc lliese vectors arc frcc. lhey may
be moved ttr ii convenient position and then added. The total couplc inoiiienl
llirii hcciimcs 103.2 lh-ft iit an anglc 111 16- with the Iiorimiitill. as shown in
Fig. 3.26. The couple that crcatcs [hi\ lusning action i s in ;I plane at right
;inglcs ti) thih oriciitation with ii cli>chwise senw iis (ihscrvcd froin below.
100 lh-It
4 103.? lh-li
Figure 3.25. Add c r ~ i p l e s .
25 Ih-ft
SECTION 3.6 ADDITION AND SUBTRACTION OF COUPLES
This addition may be shown to he valid by the following more elementary procedure. The couples of the cube are moved in their respective planes
to the positions shown in Fig. 3.27, which does not alter the moment o f the
couples, as pointed out in Section 3.5. If the couple on plane B is adjusted to
have a force magnitude of 20 Ib and if the separating distance is decreased to
514 ft, the couple moment is not changed (Fig. 3.28). We thus form a system
o f forces in which two of the forces are equal, opposite, and collinear and,
since these two forces together cannot contribute moment. they m a y he deleted.
leaving a single couple on a plane inclined to the original planes (Fig. 3.29).
The distance between the remaining forces is
Figure 3.27. Movc couplcs.
425
+
ft = 5.16ft
and so the magnitude of the couple moment may then be computed to be
103.2 Ib-ft. The orientation of the normal to the plane of the couple is readily
evaluated as 16" with the horizontal, making the tntal couple moment identical to our preceding result.
Figure 3.28. Change values o l two
furcea.
Figure 3.29. Eliminate collincnr 20-lb
f0VXS.
A common notation for couples in a plane is shown in Fig. 3.30. The
values given will be that o f the couple moments.
Figure 3.30. Repressentiltiun of couple moments in
a plene.
81
82
C H A P T ~ Ki IMP O RT A N T VECTOR Q U A N T I T I ~ S
Example 3.5
Replace the system of forces and couple shown in Fig. 1.31 by ii single
couple moment. Note that the 1.000-N-in couple m ~ m c i i ti s in thc diagonal plane ABCD. As a first step in the prohlem. identify ii second couple
moment i n addition to the couple moment i n the diagonal plane.
Examine the vertical forces. There i s an upward sun) of 1.700 N
clearly not colinear with the downward 1,700-N force. These forces form
the second couple. To gel the couple moment for these forces. we take
m~imcntsol these forces about the origin as follows:
C,
= 3k
x 80Oj
+
(3k
+
2;) x (700
-
17O(l)j + 2; x 200j
= 600; - 1.600k N-m
As for the I ,000-N-ni couple moment. wc look in
axis toward the origin as shown in Fig. 1.32.
The angle a in Fig. 3.32 i s given as follows:
tan
a = 314
it
direction along lhe .\
:_
a = 36.87''
Hence we have fix the second couplc moment C,.
C , = -1,000 cos 36.87"k + I.000 sin 16.87'j
= -800k + 6OOj N-m
N o w we can add the two couple moments to get C,,,,,.
C,,,,
= C,
+ C2 =
CmAL = 6001
(600;
-
1,600k)
+
(-800k
+
600j)
- 2,400kN-m
3.7
Figure 3.32. View along the .I a i \ .
Moment of a Couple About a line
I n section 3.3. we pointed out that the moment of a force I: about a line A-A
(see Fig. 3.33) is fiiund by first taking the moment of F about or!? point P on A A arid then dotting this ifector into a. the unit bector along the l i i Thai
~ is,
Figure 3.33. To tind moment of F
ahout A-A.
M,,,, =
(r
X
(3.lX)
Consider now the moment i i f a couple about a liiic. For this purpose, we show
a couple niomeiit C and line A-A in Fig. 3.13. As belirrc, wc Irrst want the
nioineiit of the couple about any point P along &,l. Hut the inomcnt of C
ahout ever? point in space i s simply C itself. Therelore, tu gel the moment
aboul thc line &A all we nccd to do i s dot C into a . Tho\,
h ~ , ~=, ,C
Figure 3.34. 1 0 find moment of couple
about A-A.
F) * a
-
a
(3.10)
Since C i s a free vector, the moments o l C about all lines parallel to A-A m u s t
have the same value.
SECTION 3.1 MOMENT OF A COUPLE ABOUT A LINE
Example 3.6
Consider the steering mechanism for a go-cart in Fig. 3.35. The linkages
are all in a plane oriented at 45" to the horizontal. This plane is perpendicular to the steering column. In a hard turn, the driver exerts oppositely
directed forces of 30 Ib with each hand in order to turn the 12-in. diameter
steering wheel clockwise as the driver looks at the steering wheel. What is
the moment applied to each wheel about an axis normal to the ground?
Assume half the transmitted torque goes to each wheel
Steerine
X
+
I-
View looking down
steering column
Figure 3.35. Steering mechanism of a go-cart.
The couple moment applied to the steering wheel is readily determined as
C = (30)(12)(.7071' - .707j) = 254.51
~
254.5jin-lb
The torque about the vertical axis for each wheel is now easily evaluated.
Thus
Torque =
-
I ( 2 5 4 . 5 - 254.53) j =
2
~
,
.
-127.3 in-lb
83
84
('HAPIER 3
IMPORTANT VECTOR QUANTITIES
Example 3.7
I n Pis. 3.36, find
(a) the sum 01 the lbrces
(b) the sun1 of thc couplcs
(c) the torquc of the entire system about axis C-C having direction
cosines I = .46 and rn = .61 mil going through point A .
Figure 3.36. Force system in spisr.
=
700i
+ 267.3; + 534.5j + 8OI.Xk
X F = 967.3i
(h)
c
C = 400k
+ 800
+ 534.5j +
+ 500
4j
4'
-
21
801.8kN
Si-8/'+7k
,5'
+Ik
+ 2' + I'
+ 82 + 72
WIT1
SECTION 3.1 MOMENT OF A COUPLE ABOUT A LINE
Example 3.7 (Continued)
ZC = 400k + 212.8i
z
i
- 340.53
+
- 297.1i
+
297.9k
445.7k
+
594.2j
+ 253.1j t 1 , I W
(c) To find M,, we proceed by first finding the unit vector along C-C,
which we denote as P. Thus from geometry
/* + m2 + n2 = I
:_
.462 .63* n2 = I
+
n =
+
,6257
Hence,
c?
= .46i
+
.63j
+
.6257k
We now get M<.(..
M,,
+ (-3i - S j - 16k)
(267.3 + 534.51 + 801.8k) + (212.81' - 140.5j + 297.9k)
= {(6i - 3i - S j - 16k) x (700i)
x
+ (594.2j - 297.11' + 445.7k) + 400k)
(.46i + 6 3 j + .6257k)
Carrying out calculations in the large bracket, we get:
M,, = {(5,60Ok - II,20Oj)+(2,138i-l,872j +534.9k)
+ (212.8i - 340.5j + 297.9k) + (594.2j - 297.1i + 445.7k)
+ 400k) (.46i + .63j + .6257k)
Mc.c= 944 7 - 8,075 + 4,554 =
-2,516
+-
85
i
Figure P.3.37.
i
Vigilre l'..UIl
3.41. A couple is shown in the yz plane. What is the moment of
this couple about the origin'! About point (6, 3, 4) m'! What is the
moment of the couple about a line through the origin with direction cosines I = 0. rn = .8, n = -.6? If this line is shifted to a
parallel position so that it goes through point (6, 3, 4) m, what is
the moment o f the couple about this line'?
3.44. What is the moment of the forces shown about point A and
about a point P having
position vector
r,, = IOi
+
7j
+
ISkm'!
I
E
I
ii
10N
3m
200 N
y+v.
Figure P.3.41.
x
Figure P.3.44.
- 3.42.
Given the indicated forces, what is the moment of these
forces about points A and E?
3.45. Find the torque about axis A-A developed by the 100-lb
force and the 3,000-ft-lb couple moment. The position vector r, is:
r , = l O i + 8 j + 12kft
Figure P.3.42.
3.43. An eight-bladed windmill used for power generation and
pumping water stops turning because a bearing on the blade shaft
has "frozen up." However, the wind still blows, so each blade is
subjected to a 25-lb force perpendicular to the (flat) blade surface.
The force effectively acts at 2 ft from the centerline of the shaft to
which the blades are attached. The blades are inclined at 60" to the
axis of rotation. What is the total thrust of all the blade forces on
the windmill shaft'? What is the moment on the stalled shaft?
/
Figure P.3.45.
3.46. ~i,,d M~~ N~~~that the 400.ft.]b couple
the diagonal from point A point 8,
is along
I
of blade
u
Figure P.3.43.
Figure P.3.46.
87
Figure P.3.49.
Figure P.3.53.
3.54. An oil-field pump has two valves, one on top and one on
the side, that must be closed simultaneously. The valve wheels are
each 21 in. in diameter and are turned with both hands by workers
who can exen hetween 50 Ib and 125 Ib with each hand. I f a weak
worker tums the side wheel and a strong worker turns the tap
wheel, whet ic the total twisting moment (couple mument) on the
pump'?
3.55. What is the mal moment about the origin of the force system shown'!
3.56. Add the couples whose forces act along diagonals of the
sides of the rectangular parallelepiped.
Sm
SN
,
,
,
10 m
Figure P.3.56.
Y
/
Figure P.3.55.
3.8
closure
In this chapter, we have considered several important vector quantities and
their properties. In particular, for rigid bodies we found we could take certain
liberties with a couple without invalidating the results.
Note in particular that in the chapter on vector algebra, the line of
action was of no significance. However, it should now be abundantly clear
that in taking moments we cannot change the line of action of the force. On/?
the line of action of a couple momerrf can be changed to any parallel position
for rigid bodies. Moreover, i t is important to remember that we can always
move a force along its line of action any time we are computing moments.
We are now ready to pursue in greater detail the impottant subject of
equivalence of force systems for rigid body considerations. We will see that in
equivalence considerations of rigid bodies, we again must he careful about
what to do with lines of action. They will play a vital role in our deliberations.
89
3.57. A n A-framc fbr hoiyting and dragging equipment i? hcld in
the position shown by a cahle C. To determine thc cahlc force. the
mnment of the applicd force about axib 8-8 must bc known. What
is that momcnt whcn n 1,000-N lbrcr is applied a i shown'!
What i s thc moment ahout A of thc 500-N force and the
3.110(1-N-m c i ~ u p l cncting 011 the cantilcvcr hr:rm?
3.60.
000 N-ni
Figure P.3.60.
t I.nno I\:
Figure P.3.57.
3.58.
A plumber places his hands I8 in. apart on a pipe threader
3.61. A furcr F = I h i + l O j - 3k Ih goes through point i t
havinz ii position w u t i i i r,, = I h i - j + 12k ft. What is the
innnienl ahout an :,xis p i n g Ihl-oush prrintc I arid 2 having rcspcct i w pixitm, vectors given :s\
r , = hi
r~ = 3;
+
3j
-
-
4.j
+ IZk
2h fi
li'!
md can push (and pull) with 80 Ih of force. What couple moment
joes he exert'.' How much could he exert i f he moved his hand5 t u
he ends s o that his hands are 24 in. apart'! Whet force musi hc
ipply at the ends to achieve the came couple nwnxnt :IF when he
ield his hands 18 in. apart?
0
-I
I X"+
Figure P.3.58.
1.59.
Find the torque ahnut e linr gving fi-im point I to point 2.
/
P
=
Figure P.3.59.
1.OOOi t 6OOjN
3.63. What i i the tulal c w p l c imoment of thc three couples
shown? What IS the i n ~ ~ n e n~f
t thir force systcm ahout point
(3. 4. 1) It'! What i s Ihc rnomcnt nf this f k c system nhout the
pwition vector r = 3; + 4 j + 2k fl takcn ab an axis'! What i s
thr total fkroc of thi, \\,stun'!
3.?il
I
3.66. Compute the moment of the XX)-lb force about points P,and
p,.
/ XI0 Ib
300 Ih
Figure P.3.63.
3.64.
Find the torque about axis AB.
I
Figure P.3.66.
3.67. A tow tNck is inclined at 45" to the edge A-A of a ravine
with sides sloping at 45" to the vertical. The operator attaches a
cable to a wrecked car in the ravine and starts the winch. The cable
is oriented normal to A-A and develops a force of IS kN. What are
the moments tending to tip over the tow truck about the led1
wheels (rocking backward). (Hint; Use the position vector from C
to B in Fig. P.3.67(c).)Notc that view 0-0 is n m n d l to A-A and
parallel to the incline.
/
Figure P.3.64.
3.65. A force is developed by a liquid nn a pipe any time the
pipe changes the direction or the speed of flow as a result of an
elbow or a nozzle. Such fbrces can he of considerable magnitude
and must be taken into account in building design. We h a w
shown three such forces. What moment stemming from thece
forces must be counteracted by the support at O?
I
I.',
Figure P.3.67.
./
Figure P.3.65.
I
3.68. A surveyor o n a IIlO-m-high hill dctcrrnines that Ihr cnrnrl~ 3.71. Find the t m p e of lhr f w c c sy.;tcm ahout
of a huilding at thc hair of the hill is 600 m east :ind 1.500 m
north of her pmition. What is tho posilian of the huildiog c o m r i
relati\-e to another surveyor on top of a 5,~100-m-highmountain
that is 10,000 m wcst and 3,000 ni south of the hill'? What i\ ihc
distance from thc second \ u i n r y w to the huilding c m ~ r ' !
3.69. ComQUIr the rnomcnt of the 1.00ll-lh furcr :ih<,ut support^
ing points A and H .
1
I,000ih
IN
..
,
~i
Figure P.3.71.
Figure P.3.69.
3.70. What is the turning action OS the forcsc shown about the
diagonal A-n?
Figure P.3.72
Figure P.3.70.
92
iixi5,4-/<.
~
Equivalent
Force
Systems
4.1
Introduction
In Chapter I , we defined equivalent vectors as those that have the same
capacity in some given situation. We shall now investigate an important class
of situations, namely those in which a rigid-body model can be employed.
Specifically, we will he concerned with equivalence requirements for force
systems acting on a rigid body. Parenthetically, we will begin to see that the
line of action plays a vital role in the mechanics of rigid bodies.
The effect that forces have on a rigid body is only manifested in the
motion (or lack of motion) of the body induced by the forces. Two force systems, then, are equivalent if they are capable of initiating the same motion of
the rigid body. The conditions required to give two force systems this equal
capacity are:
1. Each force system must exert an equal “ p u s h or “pull” on the body in any
direction. For two systems, this requirement is satisfied if the addition of
the forces in each system results in equal force vectors.
2. Each force system must exert an equal “turning” action about any point in
space. This means that the moment vectors of the force systems for any
chosen point must be equal.
Although these conditions will most likely be intuitively acceptable to the
reader, we shall later prove them to be necessary and, for certain situations,
sufficient fbr equivalence when we study dynamics.
As a beginning here, we shall reiterate several hasic force equivalences
for rigid bodies that will serve as a foundation for more complex cases. You
should subject them to the tests listed above.
1. The sun1 o l a set of concurrent lorces i s a singlc liirce that i s equivalent to
the original system. Convei-sely, a single f171ce i s equivalent to any coinplctc set of i t s cc~niponcnts.
2. A Sorce may he moved along its linc oiaction (i.c., forces are transmissihlc vectors).
3. The only elSect tliat ii couplc develops on ii rigid hody i s cmhodied i n the
couplc momen[. Since the ciiople monicni i s a l w a y it lree vector. for our
purposcs at present the couplc may he altcl-cd i n m y way as long as thc
c1)uplc i i i o m e i i l i\ tnot changcd.
Note for ( I ) and ( 2 ) ;ihiive. we ciiiinot change thc linc OS action alone while
maintaining equivalence.
In succccding sectioiis. k e shall present other equivalence relatioils for
rigid bodies and then examine peifectly general h r c e \yslcnis with a view to
rcplacing Lhcm with iiiiii-e convcnicnt and simplcr cquivaleiit force systems.
Thcsc simple i-epl;tccnicnts air oftcn called nmilruiir.r iif the more general
YYStems.
4.2 Translation of a Force
to a Parallel Position
I n Pig. 4.1. let us c o n d c r [he pos<ihility of miiving ii force F (solid arrow)
acting [in a rigid hody to a parallel position at point (I wliilc maint;iining rigidhody equivalence. If at posilioii (I we apply equal and iipposite forces, one of
which i s F and the othcr - F, ii syctcm of thiee Iirces i s fortricd that i s clearly
equivalent 111 the original single forcc F. Note tliiit Ihc original liircc F and
thc new Sorcc i n the oppositc sense form a couple (thc pair i s identified hy a
wavy connecting line). As usual. we represent the couple hy its moment C.as
shown i n Fig. 4.2. iiorniiil to the plane A of point a and llic original h c c F .
Thc magnitude of thc couple imonic~itC i s 1I;d. where d i s (he perpendicular
distance hetween point (I and (he original line iiiaction o f the fol-cc. The couplc imonient may hc moved 10 any palallel liositioii. i n c l u d i n ~the orib'
Tin. as
indicated i n Fig. 4.2.
SECTION 4.2 TRANSLATION OF A FORCE TO A PARALLEL POSITION
Thus, we see that a force may be moved tu any parallel position, provided rhat a couple moment of rhe mrrecf orientation and magnitude is
.simnltaneously provided. There are, then, an infinite number of arrangements
possible to get the equivalent effects of a single force on a rigid body.'
We now present a simple method for computing the couple moment
developed on moving a force to a parallel position. Return to Fig. 4. I and
compute the moment M of the original force F about point a. We can express
this as [see Fig. 4.3 (a)]
,
x
1
(b)
(a)
Figure 4.3. Couple moment on moving F is p x F .
M = p X F
(4.1)
where p is a position vector from a to any point along the line of action of F .
Now the equivalent force system, shown in Fig. 4.3 (h), must have the same
moment, M, about point a as the original system. Clearly, the moment about
point a in Fig 4.3 (b) is due only to the couple moment C. That is,
M = C
(4.2)
Accordingly, we conclude, on comparing the previous two equations, that
C = p X F
Thus, in shifting a force to pass through some new point, we introduce a
couple whose couple moment equals the moment of the force about this
new point.
We illustrate this in the following examples.
' A moment of thought will give credence to the ahove procedure of maintaining rigid hody
equivalence while moving B force to a different line of action. By moving F IO go rhruugh point
a, you are eliminating the moment ahout a that existed before the move. The couple moment
inserted at 1ht: time of the move realores this lost moment.
95
'Thcrclore.
C = -12k - 1 2 j
I
C = -4Xi
-
54; -t
+
12j
(ii
+ 24; +
+ 42k ft-lb
51k
SECTION 4.2
T RA N SL A TIO N O F A FO RC E TO A PARALLEL POSITION
Example 4.2
What is the equivalent force system at position A for the 100-N force
shown in Fig. 4.5?
Figure 4.5. Find equivalent force system at A .
The 100-N force can he expressed vectorially as follows:
F = -48.Oi
- 68.55'
+ 54.8k'N
We then have the force given above at A . In addition, we have a c ~ u p l e
moment, C, found using a position vector, r, from A to any point along the
line of action of the 100-N force. Thus, choosing point B for r, we have
C = (I0i - S j
+ 8 k ) x ( - 4 8 . 0 i - 68.5j + 54.8k)
= (10)(-68 5 ) k
-
(8)(-68.5)i
+ (-48)(8)j + (-8)(54 8 ) i
-
(54.8)(10)j- (-E)(-48 O)k
Therefole,
C = 109.6i
-
- 9321 - 1,069191-m
(h)
_
I
91
98
CHAPTER 4
E Q U I VA LE N T FORCE SYSTEMS
The reverse of the proccdurc just presented may be instituted in reducing a force and a c(iuplc in rhe S N ~ pP l m e to a .sitigle cquivalcnt force. This
is illustrated in Fig. 4.6 (a) where ii couple composed 01 lorces R and -8 a
distance d , apart and :I force A are shown i n planc N . The moment repreuentation of thc couple is shown with force A i n Fig. 4.6 ( h ) .
(1,) Fulrs A and couple moment
Figure 4.6. A coplanar frircr and couple.
Equal and opposite forces A and -A may i i r x t he added ti) the systcm at
a specific position e (see Fig. 4.7). The purpose olthis step is to form another
couple moment with a magnitude lAld2 equal to IHId,and with a directon of
turning opposite to the original couple moment (see Fig. 4.8) and this dictates
the position of P . The couple ninrnents then cancel each other out and we are
left with only a single force A going through point r . Therelbre. we can
alwdys reduce a rwcc and a couple in the s3me plane to a single force which
clearly must have a .spwjfk line ,!f nction f i x the case at hand.
Figure 4.7. Equal and opposite ~orccs
placed at e
I
_/ ”
‘
Figure 4.8. Adjiisr d2 so that couple
moments cancel.
We now exemplify the ahovc procedure in the following example
SECTION 4.2 TRANSLATlON OF A FORCE TO A PARALLEL POSITION
Example 4.3
In Fig. 4.9(a), we have shown a cantilever beam supporting a single force
and a couple in the xy plane. We wish to reduce this system to a single
force equivalent to the given system for purposes of rigid-body mechanics.
In Fig. 4.9(b), we have shown the couple moment and a point P to
which we shall shift the I ,000-N force. It should be clear on inspection that
the couple moment accompanying this shift will have a sign opposite to the
original couple moment. Our task now is to get the correct distance d so as
to effect a cancellation of the couple moments. Thus we require that
di x (1,000) (.707i
:.
-(707d)k
+
~
,707j)
550k = 0
+ SSOk
tf s
= 0
J78m
Note we could have reached the above result more simply if we had resolved
the force into rectangular components first. Only the vertical component has
nonzero moment about e and so dispensing with vectors, we can directly say
-707d
,
+
550 = 0
d = ,778 m
I
I
(a) Coplanar loading
(h) Move force to point e
x
i
( c ) Single force at e
Figure 4.9. Reduction of a coplanar force and couple to a single force with a specific line of action.
99
4.1.
Replace the 100-lb force hy an equivalent \ystcm, fmm n
rigid-body point of view, at A . Do the same lor point H . Do thk
problem by the technique 01 adding equal and opposite collinear
forces and also by using (he cross prriduct.
r
-
100 Ih
3
111
~-
*
Figure P.4.4.
+-3(l,+
Figure P.4.1.
5. A plumber c x c n s a vertical 60-lh fol-cc on il pipe wren< 1
inclined at 10" to thc hnrimnlal. What f w c r and couple nomcnt
on thr pipe are equivalent tn thc plumber'& action?
4.2.
Tri hack an airplane away from the boarding gate. it tractor
pushes with a force of 15 kN on [he nose wheels. What i s rh?
equivalent farce system on the landing-gear pivot point. which i s
2 m above the point where the tractor pushes?
~~~~~~
/
/A3
'
PivotPoint
Figure P.4.5.
Figure P.4.2.
1.3. Replace the 1,000-1b force by equivalent systcms at pointr
4 and 6. Do so by using the addition of cqud and opposite
iollinear force components and hy using the c r o s ~product.
4.6.
A tractor pera at or i s attzmpting to lift a IO-kN bouldcr.
What arc thc cquiialent lwuc systims :it A a n d at H l r m thr
hnuldcr'!
+
n
cy,
Figure P.4.3.
Figure P.4.6.
4.7. A small hoist has a lifting capacity of 20 kN. What are the
largest and smallest equivalent force systems at A for the rated
maximum capacity?
r
4.10. A carpenter presses down on a brace-and-bit with a 150-N
force while turning the brace with a 200-N force oriented for
maximum twist. What is the equivalent force system on the end of
the bit at A?
I
x
Figure P.4.7.
4.8. Replace the forces by a \ingle equivalent force
1001b
1200lb
Figure P.4.10.
+ 4k lb goes through point (6, 3,
2) ft. Replace this force by an equivalent system where the force
goes through point (2, - 5, IO) ft.
4.11. A force F = 3 i - 6 j
A force F = 20i - 60j + 30k N goes through a point
(10, - 5, 4) m. What is the equivalent system at point A having
position vector rA = 20i + 3j - 15k m?
412.
20' +20,
+20
Figure P.4.8.
4.9.
Replace the forces and torques shown acting on the apparatus by a single force. Carefully give the line of action of this force.
Figure P.4.9.
4.13. Find the equivalent force system at the base of the
cantilever pipe system stemming from force F = 1,000 Ib.
Figure P.4.13.
101
4.14. Replace the h.000-N force and the I0,OOtI-N-m ciiupli
moment hy ii single Iorcu. Whcrc docs this force cross tht
x axis'!
4.17.
A wpplementary mppoiling guy-wirc systcm for a 00-rn~
Ld1 tower IS lightened. The cables ;ire fiistcncd t o thc gmond at
[101t11\ 120" apart mtl 100 m irom thc t o w u hasc. Whnt i\ the
cquivalent i ~ c hyslent
e
acting on thc tower hasc whcrt the tensioii
i s SO kN in cahlc AT, 75 kN i n 81: a n d 25 k N in ('7'1
7
6,000 N
Figure P.4.14.
In Prohlrrn 4.13. the pipe weighs 20 Ih/ft. What i\ thc
rquivalcnt force system at A froin the weight of thc pipc'! [Hirir:
Concentrate the weight, of the pipe sections at thc rwpective c c w
ters of grsvity (gcomstric centcis i n thi, ca\c).I
4.15.
'The operator o i a sinall hoam-type crane i s rrying t u drag a
chunk of cnncrete. The hoorn i s IO" ahove the hwimntal and
rotated 30" clackwisc as seen l r o r ahovr. Thc cahlc i s directed a s
shown in thc diagmn arid has 60 kN of tension. What is thc cquivnlent force systcm at thc hoom pivot point!
4.16.
4.3
Figure P.4.17
Resultant of a Force System
As defined at the beginning of the chapter. :I resultrint o f r r . f i ~.s?srrm
r~
is n
simpler equivalent [ w e systcm. I n many compulatims it i s desirable first to
establish a resultant before cntcring into other comptrtations.
For a general arrangcmcnt of forces, no niattcr h o w complex. we can
always move a11 forces and couplc imimcnts. the latter including hoth those
given and those lorined from the movement 01 fwces. to proceed through any
single point. The result i s then a systcm (if concurrenl forccs at the point and
a system (if concurrent couple moments. These systems may then he coin-
I02
SECTTON4.3 RESULTANT OF A FORCE SYSTEM
bined into a single force and a single couple moment. Thus, in Fig. 4.10 we
have shown some arbitrary system of forces and couples using full lines. The
resultant force and couple moment combination at the origin of a rectangular
reference is shown as dashed lines.
z
Figure 4.10. Resultant of general force system
Thus, any force system can be replaced at any point by equivalents no
more complex than a single force and a single couple moment. In special
cases, which we shall examine shortly, we may have simpler equivalents such
as a single force or a single couple moment. Finally, for equilibrium of a
body, it is necessary that at any chosen point the simplest resultant system of
force and couple moment acting on the body he zero vectors-a fact that will
he discussed in dynamics2
The methods of finding a resultant of forces involve nothing new. In moving to any new point, you will recall, there is no change in the force itself other
than a shift of line of action; thus, any component of the resultant force, such as
the x component, can simply be taken as the sum of the respective x components
of all the forces in the system. We may then say tor the resultant force
(4.3)
The couple moment accompanying FR for a chosen point a may then be
given as
C, = [rl X Fl + r2 X F2 + . . . I + [C, +C,+.. . I
(4.4)
where the first bracketed quantities result from moving the noncouple forces
to a, and the second are simply the sum of the given couple moments. The
vectors r are from a to arbitrary points along the lines of action of the forces.
In more compact form, the equation above becomes
(4.5)
The following example is an illustration of the procedure
‘When we refer hereafter to a resultant, we yhall mean the simplest resultant
103
104
r
CIIAPTER 1 CQUIVALCNI' I:ORCI< S Y S TEM S
Example 4.4
T w o forces and a couplc arc shown in Fig. 1.I I, the couplc hciiig pmiiioncd
in the I? plnne. We shall find thc resultant (ifthe
teiir at thc w-igin 0.
Figure 4.11. Find ircwltiim
A t 0 we will have
added to give 1,;:
ii
at
0
set of two concurrcnt fcirce5. which nlay hc
+ hi; + ( 3 + 3 ) j + ( 6
4 = 16; + 6 j + 4 k N
= (10
-
2)k
The resultant couple niornent ill piiint 0 i s the vector ~ U I I I or ihc ciiuplcmnmetit vector? devclopcd by mii\'ing the two forces. plu\ Ihc couple
rnornent o f the couplc in thc :
!plane. Thus,
C, = r l
X
PI
+ r, x
P, - 30; N-in
NOW
r l x Fl = (10;
=
rr
X
F, =
2I i
(10;
= -6;
+
-
Sj
3Oj
+
-
3k)
X 1 IOi
20k N-111
+
+
3ji
X
(hi
+
2O.j
+
IZk N-til
+
3.j - ? k )
Hence,
C, = - 1 5 - l O j - SkN-m
The rehultanl i h shown in Fir_.3.12.
I
3.j
+
hkl
SECTION 4.3 RESULTANT OF A FORCE SYSTEM
* Example 4.5
What i s the resultant at A of the applied loads acting in Fig. 4.13’! The
forces are directed to intersect the centerline of the shaft along which we
Figure 4.13. Find resultant at A ; F , and F , are concurrent.
have placed the z axis. We first express the loads vectorially. Thus,
= -40.2i
+ 53.7j - 134.lk Ib
= 94.8i - 176k Ib
F3 = -100jlb
C
= -50k ft-lb
We can now readily find the resultant force system at A. Thus,?
F, = (-40.2
+
94.8)i
+ (53.7 -
FR = 54.62
C, = (-Ilk)
= -Ilk
X
X
X
F; +
(-40.2i
-
l0O)j
46.31’
(-8k)
X
+
(-134.1 - 176.0)k
- 3IOklb
(Fz + FJ
+ 53.7j -
+ (-50k)
134.lkJ + (-8k)
(94% - 17h.Ok - IOOj) - 50k
CR = -u)9i
- 316j - 50k ft-lb
‘Remember that for C rcauliing lrvin il muvement of a force, the posirion vector gora
from the point (in this CBSC point A) to the line of action of the force.
105
106
CHAPTER 4 EQLIJVALENTPVKCE S Y S TE M S
Simplest Resultants
of Special Force Systems
4.4
We shall now consider special but important force systems and will establish
the sirnplrst resultants possible for each case. Examples will serve to illustrate
the method of procedure.
Case A. Coplanar Force Systems. In Fig. 4.14, a system of forces
and couples is shown in plane A. By moving the forces to a common point a
in plane A, we will form additional couples in the plane. The force portion of
the equivalent system at such a point will he given as
FK = [ T ( < , ) t ] i + [ T ( < , ) v ] j
x
(4.6)
Figure 4.14. Coplanar force system.
The couple moment portion of the equivalent system can be given as
(using the right hand rule h r proper signs):
c,
= (Fd,
+
F*d2
+
.
. .)k + (C,
+
C,
+..
.
)k
(4.7)
where d,, d,, etc., are perpendicular distances from point a to the lines of
action of the noncouple forces, and C,, C,, etc.. are the values of the given
couple moments. The resultant at o is shown in Fig. 4.15.
If FR# 0, (i.e.. if
Fr t 0 andor
P
x
Figure 4.15. Resultant at point u.
b; f 0 ) we can move the force
I'
from a to yet a new parallel position so as to introduce a second couple
moment to cancel C, nf Fig. 4.15 in the manner described earlier in Section
4.2. Since the x and y directions used are arhitrary, except for the condition
that they be in thc plane of the forces, we can make the following conclusion.
~ t h e . f i i r ccornpnents
(~
in any direction in rht. plane add to uther rhun zero,
we may replace the entire coplunar .system by ( I singlr,forcr x'ifh a specific
line flcti,lrl.
What happens if
F, = 0 and
F, = O ? Without a force at point
(?r
c
P
P
a, we can no longer eliminate a couple in plane A . Thus, our second conclu-
7, are
Ft arid
sion is that if
P
v r o , the resiiltont m u f he u couple
I'
mament ur be z r o .
In thc coplanar case, therefore, the simplest equivalent force system
must be a single force along i) specific line of action, a single couple moment,
or a null vector. The following example is used to illustrate the method of
determining such a resultant directly without the intermediate steps followed
in this discussion.
SECTION 4.4
107
SIMPLEST RESULTANTS O F SPECIAL FORCE SYSTEMS
Example 4.6
Consider a coplanar force system shown in Fig. 4.16. The simplest resulF c and
tant is to be fnund. Since
I'
F~ are not zero, we know that we
P
can replace the system by a single force, which is
We now need to find the line of action in the plane that will make this single force equivalent to the given system. To be equivalent for rigid-body
mechanics, this force without a couple moment must have the same tuming actinn about any point or axis in space as that of the given system.
Now the simplest resultant force must intercept the x axis at some point X4
We can determine X by equating the moment of the resultant force without
a couple moment about the origin with that of the original system of forces
and couples. Using the vector Ti as a position vector from the origin to the
line of action of FR (see Fig. 4.17), we accordingly have
xi
X
(6i + 13j)
x (6i +
(Si + 2 j )
3 j ) + (Si + 3 j ) x (IOj) - 30k
Figure 4.16. Find simplest resultant
L
=
fb)
Carrying out the cross products,
24k
~
12k
+ S0k
Figure 4.17. Simplest resultant.
- 30k = I 3 2
(C)
Hence.
By specifying the x intercept, X, we fully determine the line of action of the
simplest resultant force. We could have also used the intercept with t h e y
axis, y, for this purpose. In that case, the position vector from the origin
out to the line of action is y j , and we have, on equating moments about 0
of the resultant without a couple moment with that of the original system:
j$ X
(6i + 13j)
=
(Si + 2 j )
X
(6i + 3 j )
+ (Si + 3 j )
X
(IOj) - 30k
Y=
'If the resultant force is parallel to the x axis, the intercept will be at infinity.
108
CHAPTER 4
EQUIVALENT FORCE SYSTEMS
Example 4.7
Compute the .simplest resultant for the loads shown acting on the heam i n
Fig. 4. IX(a). Givc thc intercept with the x axis.
/+ifJ-
...
75 N
IL)
Figure 4.18. Find siinplcst resultant.
I t i s immediately apparent on inspection of the diagram that
F , = loOi - 75j N
(a)
Let x he the intercept with the x axis of the line o f action of I$ when this
line o f action corresponds to zero couple moment C, [see Fig. 4. IS(b)l. I n
Fig. 4.18(c). we have decomposed F, along this line of action into rcctatigular components so as to permit simple calculations of moments about
the origin 0 (here we mean moments about the z axis). Accordingly,
equating moments about the z axis of F, without a couple moment, with
that 0 1 the original system o f loads, we get,
-(75)(i;)
= 50
-
(2.5)(75) - (.4)(100)
X = 2.37 m
I
Thus, the simplest resultant i s a force lOOi - 7 5 j N intercepting the bearrl
axis at a position I = 2.37 ni.
As pointed out earlier, i n the instance wherein F, = 0. we then possihly have as the simplest resultant a couple moment normal to the plane o f the
coplanar force system. There i s also the possibility that there i s zero couple
moment, i n which cahe the lorces o l the coplanar force system i~ornplulrly
cancel each other’s effects on a rigid body. To find the couple moment for the
case where FK = 0, we simply take inutnents o f the coplanar force system
ahout m y p i n t i n space. This moment, if i t i s not equal to zero, i s clearly the
couple-moment vector sought.
SECTION 4.4 SIMPLEST RESULTANTS OF SPECIAL FORCE SYSTEMS
Example 4.8
What is the simplest resultant for the forces shown acting on beam AB in
Fig. 4 . 1 9 '
L
Jt.6
m
6
1.2 "1
-
Figure 4.19. Cuplanar loading un a simply aupported beam
Our first step will he to compute the resultant force by adding up the
force vectors. Thus
F, = 1,500j - 666.2 - (I,SXS.X)(.S)j
+ (1,585.8)(.866)i- 7 0 7 . l i 7 0 7 . l j
~
Collecting terms, we have
F, = (-666.2 + 1,373.3 - 707.l)i + (IS00 - 792.9 - 701.1)j = 0
The simplest resultant clearly must he either a couple moment or be a null
vector. For this information, we shall take moments about point A.
C,
C,
= ( [ . 3 - (.2)(.707)]i - (.2)(.707)jJ
x (-666.2)
+
(.3)(1,5OO)k - (.6)(1,SX5.8)(.5)k
+
(li
= -94.20k
+
.2j) x (-707.li - 7 0 7 . l j )
+ 4SQk -
- 707.lk
+
47S.7k
141.4k = '-685.6ki-J-m
We have a couple moment in the minus z direction having any arbitrary
line of action as the simplest resultant.
It is important thet the nature of the equivalence just instituted be
clearly understood. Thus, for finding the supporting force system, we can use
the undeformed geometry and hence the single force replacement. However,
for finding the deflection of the heam, it should be obvious that the replacement is invalid. Note, finally, that there is only one point on the beam that
will allow for a single force to he equivalent to the original system for purposes of rigid-body considerations.
109
110
CHAPTt3 4
EQUIVALENI FOK('E SYSTEMS
Case B. Parallel Force Systems in Space.
Nuw. consider the syst e m 01I I p a ~ t l l e forces
l
i n Fig. 4.20. wherc the :direction has been selected
parallel to the forces. We also include m couples whose planes are parallel to
the :dircction bec;ruse such couples can be considered to he composed of
equal and opposite forces parallel t o the :direction. We can i n o x the forces
s o that they all pass through the origin uf the .q:axes; the force portion of the
equivalent hysteni i s then
F,? =
Figure 4.20. Parallel system rrf force\.
1
E'<,)k
(4.8)
l h c couple inomenl ponion of the equivalent system i s found by applying Eq.
4.5 to this case:
wlicrc
F represents tlic noncouple Ibrces. Cairyiiig out the crocs product, we get
Froin this. we sec that the couple mi)nicnt must always be parallel tu the .I?
plane (i.e., perpendicular to the directioii of the forces). We then have at the
origin il single force and a single couple moment at right angles to each uther
[see Fig. 4.2I'a)l. llFM# 0, wc can move
again to anothcr line of action
in a plane A perpendicular t o C# /see Fig. 4.21(h)Jand. choosing the proper
value of d , ensure that /$I = lC,l with ii sense opposite to CMSsuch that we
eliminate the couple ~noinent.We thuh eiitl up wirh a single force having a
particular line 01 action specified hy the intercept T,? of the line of action of
the f'urcc with the .ky planc. I f the suninration of forces should happen tu he
Lero. the equivalent system tnust thzn he a couple moment or a inull vector.
Thus, ltw sirnplrst rrrulrunl syrfewi o ( o parrillt.1 fiJWC sy,sfvrri i.5 either
u , f . r wifh
~ ( I .sp(witic line o f w t i o n , u .siiiRle wupl<,niommt, or (1 null i'ri'tor.
The following cxample will illustrate how we can directly dctcrmine lhe sinplest resulrant.
Ft
Figure 4.21. Simplcst rcsultnnt f o r
parallel furce system.
SECTION 4.4 SIMPLEST RESULTANTS OF SPECIAL FORCE SYSTEMS
Example 4.9
Find the simplest resultant of the parallel force system in Fig. 4.22(a)
i
I’
X’
(b)
Figure 4.22. Find simplest resultant.
The sum of the forces is 30 Ib in the negative z direction. Hence, a
position can he found in which a single force is equivalent to the original
system. Assume that this resultant force without a couple moment proceeds through the point X, 7 [Fig. 4.22(b)]. We can equate the moment of
this resultant force about the x and y axes with the corresponding moments
of the original system and thus form the scalar equations that yield the
proper value of Y and 7. Equating moments about the x axis, we get
(30)(2) - (20)(2) - (4O)(lO) = -3Oj;
Therefore,
7=
12.7
Equating moments about they axis, we have
-(30)(2)
+
(20)(4)
+
(40)(4) = 30T
Therefore.
You can also show, as an exercise, that the same result can he reached for
x, j by equating moments of the resultant force without a couple moment
~
about the origin with that of the original system about the origin.
111
i 12
CHAPTER 1
FQIIIVA1,CNl I,OR('I-: SYSTEMS
Example 4.10
Consider the parallel Iorcc system i n f:ig. 4.23(a). What i s tlic 4inplcxt
resultant?
Here we have a c:ise whcrc Ihc sum (ifthe forces is x i - c mil \(I
FR = 0. Therefore, 1hc simplest resultant inusl be a couple i n o n i c n ~or he a
null vector. To gel Lhih ciiuplc iiiiinienf, Cr we c m lake inioiiieti1s 0 1 the
fkrces about rmy poi111 in space. Thii tnonient bectiir then c q i d s tlic
desircd couplc niwneiit C,?.
One proccdurc i s ti) use llic origin nl'thc referrncc as the point ahout whicli 10 lakc iniiiiients. Then we ciin \ay (ha1
CK = (4i + 2 j ) x (-.iOk)
= -2Oi + 20j N-iii
+ ( 3 ; + 2.1')
X
(4llk)
+ C?i +
411 N
(3.2)m
,
/'( 4 . 2 )
(2.4Jm
m
,ill
4.1') X (-1Ok)
(a!
The rectangular conipincnls (if C, along the .v and y iixes are the iniiimcnl~
of the brcc system ahout these axes. Thus.
(C,), = -20 N-iri
(C,), = ?ON-Ill
(h!
We can get the riiciments o f t h e foi-ce\ about [lie x aiid y a x e \ directly
and thus generate [he ciimpoiieiits or [lie dcsircd ciiuple iiioiiienl <it.
Accordingly, using rhc clt'meiitary definilion 01 l l i e iiioiiiciil 01 :I force
about a line as presented earlier. we have
(e,)\ = -(10)(4j + (40K2) - ( 3 0 ! ( 2 )= -20 N-ni
(C,)? = (10)(2) - (40)(1) + (10)(4) = 2 0 N-iri
Thus, the moment of lhe force syslcin abmit the origin, ; ~ n dhcncc ahout
any point. i s 1lien [he desired couple niii~ncnt(Fig. 4.23Ih))
C, = -2Qi
.
.
I
"
C,<= X l i
4
2Ilj \ ~ w
117,
I;igure 4,23, P;,I-allcl
f,lrce sr,tr,ll,
+ 20jN-m
.. .
.
, .,
..
,.,...
N(iw rhal v e l i i i \ ~ cciriisidered the coiiccpl (if the siinplcsl rcsultant for
coplanaI and parallel hi-ce systems. wc wish to pi1 hach to tlic ,qewrdf;ww systems for a irioiiieiil. We Iciirned eai-lier ltial ~ ' caii
c nlw;iys replace such ii systenr
iii r i g i d - h d y inechanich hy a single fhrcc F<and a single couple moinent C , iil
IC
this ;ilw;iys tlie very simplest systeiii 1o1-rigid-body iiiech;rii\how this. deciini~nisctlie couple iniinicnt C,, iiito t w o rectangular c(iiiipoiicn1s (; arid
peq~cndicularto the forcc and collinear with [he
liirce. respectively. We ciiii iiow concciu;ibly 1111)vethe force 10 ii specilic p a i d
kl piisiti(iii iuid ciin cliiiiiniile C,. thc c ~ i ~ i i p o n ciilciiuple
nl
iiioiiiciil iioriiial 10 the
force. Hiiwever. there i s i i d i i n g that u'e can d(1 ahout thc C;, conipiinent of couple inoiiient collincx (or parallrl) to the force. The rc;ixin Ibr this i s that any
inoveiiieiil OS thc forcc 10 a pafiillcl 1nisitioii i,/wiys introduces a couplc iiioiiienl
/~',rp~~it[l;(,i,fli~r
to thc liirce. Thus the ciinipiineiit
canniil he affected. By eliiiiirialing C, we end up with the force
;iid C, collinear with F,. This systcni i s tlii'
siniplcst iii tlie g e i i c r i case aiid il i\ c;illcd a w r ~ m d 7iscc Fig. 4.24). H(iwcver.
%'e shall nlit generally use the \ ~ r e n c tcoiicept
i
in this text and will work instead
with llic rcsultiint hrcc F,<;uid tlie couplc iniimeiit C, iil m y chosen point.
any chosen point.
ics? No. i t i s iiiit. '1.0
,cy1*
%
'
Figure 4.24. Enamplci of thc v - c a l l c d
wrench. This is ,he \jmplcst leprearnt;ltj~,ll
01 a general force \ystern.
c3
4.20. Find the resultant of thc fvrce systcm at point A . The 100N. 200-N, and 900-N Iruds are at the centers r i f the pipe sections.
?'
I
500 Ih
c-d
i
YO0 N
j
400 N
Figure P.4.18.
sin N
Figure P.4.20.
4.19. Compute the iesultnnt force bystem at A stemming from
the indicnted 50-lb force. What is the twist developed about the
axis of the shaft at A'! The 50-lh force is normal to the wrench.
4.21. A 20-kN car and an XO-kN truck are stoppcd (in a bridge.
What is the resultant force system of these vehicles at the cenler
of the bridge'? At the center of the left end of the bridge? The distanceq given to truck and car are to respective centers of gravity
where we can concentrate the weights.
Ac-
/I
k l X +
Figure P.4.19.
Figure P.4.21.
4.22. Two heavy machinery criltes (A weighs 20 kN and B
weighs 3 1 kN) are placed on a truck. Whilt is the resultant force
syblem at Ihc center of the rear axle'! The ~ e n t c r sof gravity 01 the
crates. where we can conccntriite the weights, are at the geometric
4.25. Find thr simplcw resuliilnt of the lul-ces shown acting o n
the beam. Give thc intel-ccpt with the axis of the beam.
I.5I)O It,
centers.
500 Ib
Figure P.4.25.
4.26. Find the . s i m p l ~ wredfarit of thc forces \hewn acting on
the pulley. Givr the intwcrpt with thc .t axib.
Figure P.4.22.
4.23.
Replace the system of forces hy a iesultilnt at A
Figure P.4.26.
4.27. A inail raises a 50-lb bucket of water to the top of a bricklayer's scaffold. Also. a Jeep winch i s used 10 m i x a 2In-lh load
of bricks. What i c the ,sinrple.sr resultant fi~rcciyrtem on the scaffold'? Givr the iintercept. Consider the pulleys to bc frictionless
so that the 50-1b force and the 200-lb force are trtinimitted respectivcly to the man and tu the Jeep.
Figure P.4.23.
4.24. Evaluate Farces F ; . F2, and F;, so that thr resultant of the
forces and torque acting on the plate is /ern in both force and couple moment. (Hint; If the resultant is zero for one point. w i l l it not
be zero for any point? Explain why.)
I
S O Ih
4.2% Find the resullanl x 1 A .
I.ooo s
Figure P.4.24.
I14
Figure P.4.27.
Figure P.4.28.
4.29. Compute the simplest resultant for the loads acting o n the
beam. Give the intercept with the axis of the beam.
v
25'*
S'
4.32. A parallel system of forces is such that: a 20-N force acts
at position x = 10 m, y = 3 m: a 30-N force acts at position x =
5 m, y = -3m; a SO-N force acts at position x = -2 m, y = 5 m.
(a) If all forces point in the negative z direction, give the
simplest resultant force and its line of action.
(b) If the 50-N force points in the plus idirection and the
others in the negative idirection. what is the simplest
resultant?
2 0 ' 4
Figure P.4.29.
4.33. What is the .simplrst resultant of the three forces and
couple shown acting on the shaft and disc'! The disc radius is 5 ft.
4.30. Find the .simpIm resultant for the fiirces. Give the location
of this resultant clearly.
/
SO0 Ib It
Figure P.4.33.
Figure P.4.30.
4.31. ~~~l~~~ the system of forces acting on the rivets of the
plate by the simplest resultant. Give the intercept of this resultant
with the x axis.
4.34. What is the simplest resultant for the system of forces?
Each square is 10 mm o n edge.
y
I
It
I l +
.--I I .-I--
U
Figure P.4.31.
Figure P.4.34.
1.35. What i s the simplc>t rculvant? Where docs
icriori crash lhc .t ani\'!
i t s line 11f
1+-22'-I
Figure P.4.3X.
Figure P.4.35,
1.36. What i s thc simplest r e d t a n t 01 the loadings shown? Be
rure 1c, give its line d a u t i o n .
1110 N
uesullilllt
/
Figure P . 4 3
Figure P.4.36.
1.37. T w o hoists arc operaled UII the w n c ovcrhzad track. Hoist
4 has a 1.000bkN I<iilil.arid hoisl N has a 4,000-kN load. Whal is
ihe r e d m l Cwcr syslcm at the left cnd 0 of the track'! Where
due\ the ,simpip.rr reiiiltiint f w c e act'.'
Figure P.4.37.
4.38. A I<,-hoy trailer weighs l h . 0 0 0 Ih atid i\ Ihded with :I
I5.000-1h hulldozcr A arid il I2.0011-1b front-rnd l o ~ d e M.
r What is
the siinplcst rcsultanl 1orce and where does i t act'! The weiphts ot
the michincs and trililer act at lhcir rchpeilivc centcis of gve\ity
(C.G.I.
II 6
20 k N
Figure P.4.40
SECTION 4.5 DISTRIBUTED FORCE SYSTEMS
4.5
Distributed Force Systems
Our discussions up to now have been restricted to discrete vectors-in particular, to point forces. Vectors as well as scalars, may also be continuously distributed throughout a finite volume. Such distributions are called vecror and
,sr.ulur,field,s,respectively. A simple example of a scalar field is the temperature distribution, expressed as T(x,y , ,II), where the variable t indicates that
the field may be changing with time. Thus, if a position x,,,yo. zl) and a time
fl) are specified, we can determine the temperature at this position and time
provided that we know the temperature distribution function (i.e., how T
depends on the independent variables x, y , z, and t). A vector field is sometimes expressed in the form F(x, y. z, t). A common example of a vector field
is the gravitational force field of the earth-a field that i s known to vary with
elevation above sea level, among other factors. Note, however, that the gravitational field is virtually constant with time.
In place of the vector field, it is more convenient at times to employ
three scalar fields that represent the orthogonal scalar components of a vector
field at all points. Thus, for a force field we can say:
fcxce component in .r direction = g(x, J’, i,
I)
force component in y direction = h(x, y, z,t)
force component in z direction = /(x. .y. i,
f)
where g, h. and 1 represent functions of the coordinates and time. If we substitute coordinates of a special position and the time into these functions, we
get the force components F,. <., and F for that position and time. The force
field and its component scalar fields are then related in this way:
F(x, y .
i,
f) = n(x. y, ,It)i
+
h(x, y. z, t)j
+
l(x, y.
i,
t)k
More often, the notation for the equation above is written
F(x. y. z, t ) = FJx. Y , z, i)i + F,.(x. y, z, t)j + F(x, y, z, t)k (4.1 1 )
Vector fields are not restricted to forces but include other quantities such as
velocity fields and heat-flow fields.
Force distributions, such as gravitational force, that exert influence
directly on the elements of mass distributed throughout the body are termed
body fiirce distributions and are usually given per unit of mass that they
directly influence. Thus, if B(r, y, z, t) is such a body force distribution, the
force on an element dm would be B(x. y . 7,t)dm.
Force distributions over a surface are called surfhceforce distributionsh
T(x. y , z, t) and are given per unit area of the surface directly influenced. A
simple example is the force distribution on the surface of a body submerged
in a fluid. In the case of a static fluid or of a frictionless fluid, the force from
the fluid on an area element is always normal to the area element and directed
“Surface forces are often called suriuce rrucrions in solid mechanica.
I 17
1 18
CHAPTER 4
EQIIIVALtNl FOR(.'I! SYSTEMS
i n toward thc hiidy. The force per unit area stemming Sriim such fluid action
i s c a l l e d p r ~ s s u r rand i s denoted a s p Pressure i s a scalar quantity. The direction of thc force resulting lroni a pressure on a surface is given by the uricntatimi of the surfiicc. [You w i l l recall f r m Chapter 2 that an arm clement can
be considered as a vector which i s normal to thc arc3 clement and directed
outward from the enclosed body (Fig. 4.25).1 The infinitesimal force on the
area clement i s then h'
'Ivell 21s
Figure 4.25. Area
vector
df = -1' &
A more specialiml. but nevertheleqs common, force distribution i s that
of a continuous load on a beam. This i s oftcn a parallel loading distribution
that i s synimctrical about the center planc .x?. of a beam, as illustrated i n Fig.
4.26. Various heights of hricks stacked on ii beam would he an example of
this kind of Iwading. We can replace such a Ii~adinghy an equivalent coplanar
distributioii that acts a l the cciilcr plane. The loading i s given per unit length
and i s denoted as M', the iiireiisif?. 01 loiidblg. Thc lorce on an element dx of
the heam, then. i s IV d r .
We have thus presenlcd force systeriis distributed throughout volumes
(biidy lorcesj, over surfaces (surlace forces or trxtionsj. and over lines. The
conclusions about resultants that were reached carlicr for general. parallcl,
and coplanar point force systems are also valid for these distrihuted force systenis. These concIusii~iisare true hecause each distributed force system can he
considered as an infinite nurnher US infinitesimal point lurces of the type used
heretofore. We shall illustrate the handling of force distributions i n the following cxmiples.
Case A. Parallel Body Force System-Center of Gravity.
Figure 4.27. G r w i t y hody
force distrihution.
Consider a rigid body (Fig. 4.27) whose density (masslunit volume) i s given
as p(x, y, zj. I t i s acted 011 by gravity, which. for a small body, may be considered to result i n a distributcd parallel lorce field.
Since we have here ii parallel system O S lorces i n space with the same
sense. we know that a single lorcc without a couple moment along a certain
line of action w i l l he equivalent to thc distribution. Thc gravity body furce
SECTION 4.5 DISTRIBUTED FORCE SYSTEMS
B(x, v , 2 ) given per unit mass is -gk. The infinitesimal force on a differential
mass element dm, then, is -g(p dv)k, where d v i s the volume of the element?
We find the resultant force on the system by replacing the summation in Eq.
4.8 with an integration, Thus,
FR = -J,g(pdv)k
=
-gk[ V pdv = -gMk
where, with g as a constant, the second integral becomes simply the entire
mass of the body M.
Next, we must find the line of action of this single equivalent force
without a couple moment. Let us denote the intercept of this line of action
with the xy plane as X, j ; (see Fig. 4.27). The resultant at this position must
have the same moments as the distribution about the x and y axes:
- FRY- --- gJ" YPdlI
FRx = -gJV x p dc',
Hence, we have
x=-
x p dv
~
M
y p dt.
y=-
M
Thus, we have fully established the simplest resultant. Now, the body is reoriented in space, keeping with it the line of action of the resultant as shown in
Fig. 4.28. A new computation of the line of action of the simplest resultant
for the second orientation yields a line that intersects the original line at a
point C. It can be shown that lines of action for simplest resultants for all
other orientations of the body must intersect at the same point C . We call this
point the center ( f g r u v i t y . Effectively, we can say for rigid-body considerations that all the weight of the body can be assumed to be concentrated at the
center of gravity.
I
Line of action
om firpt onentation
Y
X
Line of action from
second orientation
Figure 4.28. Location of center of gravity
'Note that gp is the weight per unit volume. which is often given as y. the so-called specific
weight.
119
Find Ihc cciitcr 01 gravity of the tii;ingiilar block having a uniloriii dcnxily
p shown i n Fig. 4.20.
The totiil weight ~ S l t i chody i s easily c v i i l i ~ i i t c diis
=
:'p(d,<.'2)
(21)
To find 7.we w i l l equiitc the in11111c1iIo i I,;? ahoui the .I i i h i s with ihiit
of the weight distrihution O S Ihc hlock.
kicilit:~ic lhc lalter. we shall
choo\c within the hlock h!/iuilc.~brrol
elements whosc weights are ciisily
coinputed. A h Ihc inoriiciit 01 the weight o S ciicli clciiicnl ahout the I
axis i s 111 he likewise easily coniptiteil. Inlinilcsimal slices < i t thickncss (I?
paiallcl to the .r; plant! l u l f i l l (1111 requiicliiciils nicely. The weight 0 1 w c h
a slice i \ simply !:h cly),i,y. whci~c: i s Ihc height 01tlie \lice (scc Fig. 4.2'1).
because all points (11tht slice i i i c iit tlie siiiiic criiirdin;itc disliiiicc v irom
the .r axis. clc;irly the ~ i i i m c i iof
t tlie weight 01 the \lice i s easily coiiiputed
a s -?(:/J
i1yjp.q. By letting i' 11111 Sroiii l l 10 ( I during iiii i i i t c g ~ i t i ~wi c. ciiii
~ i c c o u ~lor
i t iill the \lice\ iii the hody. T l i i i s . \vc have
,
I
I<.:
=
'1'11
</l'Jp,V
/i:!(:/J
!I11
The tcrni :ciin he cxpl-essed with the ;11d 171 similar lrimglrs in Ici-iii\ ol
lhc inlcgralion v;il-iahli. J
.<'
:=
1'
~~
(i
(;;J
We then have for tk.lh!. on r c p l x i i i g I.,,,using F:q. (a).
(CJ
I
SECTION 1.5 DISTRIBUTED FORCE SYSTEMS
Example 4.12
Find the center of gravity for the body of revolution shown in Fig. 4.31.
The radial distance of the surface from they axis is given as
r =
(a:
& y 2 ft
The k ~ d has
y constant density p , is IO ft long, and has a cylindrical hole at
the right end of length 2 ft and diameter I ft.
‘/?
I- 7 ’ 4
IO’
-I
Figure 4.31. Body of revolution. Find center ofgravity.
We need only compute 7, since i t is clear that 2 = X = 0, owing to
symmetry. We first compute the weight of the body. Using slices of thickness dy, as shown in the diagram, we sum the weight of all slices in the
body assuming it is whole by letting y run from 0 to I O in an integration.
We then subtract the weight of a 2-ft cylindcr of diameter I ft to take into
account the cylindrical cavity inside the body. Thus, we have, noting that
the area of a circle is nrz or nD2/4
Using Eq. (a) to replace
12
in terms of y , we get
121
122
CHARKR 4
EQlIIVALk~NTFORCE SYSTEMS
Example 4.12 (Continued)
To get y , we equate the moment of W about the x axis with that of the weight
distribution. For the latter, wc sum the momcnts ahout thc .v axis of the
weight of a11 slices, assuming first n c inside cavity. Then. we subtract fm111
this the moment about the .r axis of the weight of il cylinder corresponding to
the cavity in the body. Because p is constant, the center of gravity of this latter cylinder is at its geometric center s o that the monient arni froin thc ~Y axis
for the weight of the cylinder is clearly 9 ft. Thus. we have
Suppose as will be the case in Pmblern 4.43 that y (= pg), which
is the .specific fi,eiRht (weight per unit volume). varies with y. That is,
y = y ( y ) . Then for this problem, note that
also
Note here we cannot take the short cuts used in the original problem wherc
y was a constant,
123
SECTION 4.5 DISTRIBUTED FORCE SYSTEMS
Example 4.13
A plate is shown in Fig. 4.32 lying flat on the ground. The plate is 60 mm
thick and has a uniform density. The curved edge is that of a parabola with
zero slope at the origin. Find the coordinates of the center of gravity.
The equation of a parabola oriented like that of the curved edge of
the plate is
y = CX*
(a)
We can determine C by noting that y = 2 m when x = 3 m. Hence,
2 = C * 9
(b)
I
Figure 4.32. Find center nf gravity
of plate.
Therefore,
c=2
9
The desired curve then is
y =
~
‘
2 $
9-
Therefore,
=
-1y”2
42
(C)
We shall consider horizontal strips of the plate of width d y
(see Fig. 4.33). Using the specific weight, which is weight per volume
and is equal to pg. we have for the total weight W of the plate:
where f is the thickness. We replace x using Eq. (c) to get
Figure 4.33. Use of horizontal strips.
Integrating, we get
2
W = r y + (1y 3 / * ) ( z ) (
L’
2
= ry&(2)3”
= 4ry N
311
We next take moments about the x axis in order to get 7.Thus,
(d)
124
C'HAPTER 4
COLIIVAI.ENT FOKCL SYSTEMS
Example 4.13 (Continued)
Using 4ry i o r W from Eq. (d). we get, for
v:
I fl
.y=$m
Tii get i.
we take miiments ahout the !axis, s t i l l utili7iiig the horizontal strips of Fig. 4.33. The center of gravity of a strip is at its center
since y i s constant and s o the miiinent aim 1111- ii strip ahout tlic y a x i s i c .xi?.
W.?
=
1;;
( r .r (l!)?,
(F)
Continuing with the calculations. we liavc
On replacing W according to Eq. Id), we get fir.i:
.r=#m
(,hl
Next, we proceed to get .X using vertical strips as shown i n Fig. 4.34.
Equaling moments of the weights of lhcsc vertical strips ahiiut the y axih
with that of the tntal weight W at its ccnlcr of gravity .I. wc gel oil rioting
that ( 2 - J) i s the length 0 1 Ihe strip
Figure 4.34. Use of vertical strips.
Using W = 4y1, we can now solvc for .r. l h u s , we again gel
7 = <I ni
Finally. i t i h clear that the ? ciiordinate i s zero for rehence
centcr plane of the plate.
~
_-. - "
,,
.I;Y ill
thc
..I~___._
.
111the prcviiios problems, we used sliccs of the body having a thicknesses
rl? or d z . If the spccilic weight were ii fiinction of position. y(.i.
no1 readily uhe such slices, since wc cannot ziisily cxpirss the weight 01' such
iliccs iii a simple iniiniicr. The reason fix this i\ that in the .r and r directions the
dimensions 01 the elemen1 are finite. and s o y would vary in thcse din-ections
thimughout thc clement. If, however. we c h i m e an elenicnt that i s infinitesimal
i n rill r l I r ~ i ~ l i i ~ t such
t . s , as an inrinitesimal rectangular parallelepiped having vi11unie dx cl! 11:. then y ciin he aswmed to he constant throughout the clcincnt. The
wcight of the elcmcnt i s then easily seen to he y(d.r d? d:.), whcre the coordinates
of ycorrespond Lo the position o i the elenienl. We now illuslriite a simple case.
SECTION 4.5
Example 4.14
Consider a block (see Fig. 4.35) wherein the specific weight y at comer A
is 200 Ihf/ft’. The specific weight in the block does not change in the x
direction. However, it decreases linearly by 50 IhWft’ in 10 ft in the y
direction, and increases linearly by 50 Ihf/ft’ in 8 ft in the z direction, as
has been shown in the diagram. What are the coordinates 2,g of the center
of gravity for this block’?
X
Figure 4.35. Block with varying y
We must first express y at any position P(x, y , z ) . Using simple proportions, we can say
v
y = 2 0 0 - ~ ( 5 0 ) + q8S ‘O )
IO
= 200 - 5 y
(a)
+ 6.252 Ibf/ft’
We shall first compute the weight of the block (Le., the resultant
force of gravity), We do not use an infinitesimal slice o r rectangular rod of
the block, as we have done heretofore. With the specific weight varying
with both y and z , it would not he an easy matter to compute the weight
and moment of a slice or a rod. Instead, we shall use an infinitesimal rectangular parallelepiped having volume dx dy dz, located at a position having coordinates x, y , and z as has been shown in Fig, 4.36(a). Because of
the yanishingly small size of this element, the specific weight y can he
considered constant inside the element, and so the weight dW of the element can he given asx
dW = y(dx dy dz) = (200 - 5y
“We are dcleting higher order quantities.
+
6.252) dr dy dz
DISTRIBUTED FORCE S YS TEMS
125
126
CIIAPTER 4
~ I J I J I V A L E N IFORCE SYSTEMS
Example 4.14 (Continued)
T u include thc weight o l u l l such elenicnls in the block. we first let .v “run”
from 0 lo 4 ft while holding J and z fixed. Thr rectangular parallclcpiped
OS Fig. 4.3hla) then heconics a rectangular rod ac shown in Fig. 4.36(hl.
Having ruii its course. x i s no longer a variable i n this summation priicess.
Next. let y ”run” from 0 to IO while holding Iconstant. The rectangular
rod of Fig. 4.36(h) then bcciinies an infinitesimal slice. a s shown i n Fig.
4.361~).The variable y has thus run its course and ir. no longer a variablc.
This lcaves only the variable ;, and now we let r “run” froni 0 to X.
Clcady, we cover the entire hkick hy this process.
We can do this inathcmatically by a process called inid~ipleinrrfimtion. We pcrt~irmthree iiitcgrations, paralleling tlic thrcc steps oullined i n
the previous paragraph. Thus. we can lormulate W a s ffoll~iws:
We first consider integr;ition with reqpecl
That i c
J(:(200 - SJ
ti1
x holding ? and I constant.
+ 6.25;)dr
As iii the firsl step set lorth in the previous paragraph, to go from a rectangular pnrallelepiped to a rectangular rod, wc integrate with respect (o .i
from ,x = 0 tu .t. = 4 n,hile holding? and :constant. ‘Thus,
I
j(,(200 - 5y
+ 6.25:)rh
=
= (2l~lIx
- S?X
+ 6.25:.*)1;
xoo - 20y + 25;
With .v no longer a variable (since i t has run its course), the equation for W
becomes
Now, we hold 3 constant and integrate with respect t o y from 0 to IO. (This
takes us from a rectangular rod to ii slice.) Thuh,
(C)
Figure 4.36. (a) Infinitesimal element
N o w ? has I-UII i t s cw~rse.and wc have
w=J
X
11
(7,000 + 250:)
at
l b ) I runs from 0 IO 4. while i
and v are fixed, to form rcctangular rod:
( c ) ? runs from 0 to I O , while holding:
fixed. 11, frrrm slice.
P ( . y y, :);
if:
SECTION 4.5 DISTRIBUTED FORCE SYSTEMS
Example 4.14 (Continued)
By integrating with respect to z. we sum up all the slices, and we have covered the entire block. Thus,
To get j , we equate the moment ahout the x axis of the resultant
force without a couple moment with the moment of the distribution. Thus,
using multiple integration as described above:
-(64,0OO)p =
-I I /
R
10
n n
4
n'
v(200 - 5 y
+ 6.252) dx dy dz
Therefore, integrating with respect to x, then y. and then z as before,
we have
64,OOO.v = lRji"(200yx- Sy'x
+ 6.25yzx)/,,4 dydz
n o
=
/,"I:"
=
~ox(40,000
- 6,667 + 1,250z)dz
=
[3 3 , 3 3 3 ~+ 1250 5*
(800y - 20)''
+ 25.~2)dy dz
I
= 307,000
and
y = 4.79 a
Because y dots not depend on x, we can directly conclude by
mpection that X = 2 ft.
Next, in the case of a body made up of simple shapes (subbodies) such
as cones, spheres. cylinders, and cubes, we can find the center of gravity of
the body by using the centers of gravity of the known subbody shapes. Thus,
we can say on taking moments about they axis that
y",a[(z)= X.w;(3'
(4.12)
where W is the weight of the ith subbody and where [z), is the x coordinate to
the center of gravity of the ith subbody. Bodies made up of simple subbodies
are called composite bodies.
127
<-
L2is
SECTION 4.5 DISTRIBUTED FORCE SYSTEMS
Case B. Parallel Force Distribution over a Plane SurfaceCenter of Pressure. Let us now consider a normal pressure distribution
over a {dune surface A in the ,ty plane in Fig. 4.38. The vertical ordinate is
taken as a pressure ordinate, s o that over the area A we have a pressure distribution p ( x , y ) represented by the pressure surface. Since in this case lhere is a
parallel force system with one sense of direction, we know that the simplest
resultant is ii single force, which is given as
Figure 4.38. Prcsure distribution
The position x , y can hc computed by equating the moments about they and x
axes of the resultant force without a couple moment with the corresponding
momcnts of thc distribution. Solving f<)rZ and 7,
I
p r [/A
x =
,:=
~~
J
~~
/I
d.4
iv dA
11 </A
Since wc know that 11 is a function of .r and y over the surface, we can carry
out the preceding integrations either analytically or numerically. The point
thus determined is called the center ofpressure.
(In later chapters. we shall consider distributed frictional forces over
plane and curved surfaces. In these cases, the simplest rcsultant i s not neccssarily a single forcc as it was in the special case above.)
I29
130
CHAPTkR 4
EQUIVALENT FORCE SYSTEMS
Example 4.16
A plate ABCD on which both distributed and point force systems act is
shown i n Fig. 4.39. The pressure distribution is given as
/ I = -4?
+ 100 psf
(21)
Find the simplest resultant lor the system.
To get the resultant for-ce, we consider a strip dy along the plate as shown
in Fig. 4.39. The reason fix using buch a strip is that the prcssure 17 is uniform
along this stnp, as ciln he seen lroni the diagram. Hence, the force from the
pressure on the strip is simply / I dA = p[dy)lS). Thus. we can ~ i l ythat
fifl =
-jSp(s,(m., - son
11
-*!
;t-
FK = ( 2 0 ”
-
Figure 4.39. Find simplest m u l l a n t .
500y)’ - 500 = -2,167 lb
3
~rl
To get the position ~F,J of the resultant force FKwithout a couple moment.
wc cquate moments of FK about the I and y axes with that of the original
system. Thus. starting with the x axis, we have using strip d? as before:
-2,1677
j i ) ’ p ( s d y ) - (500)(2)
=
-
=
-j$-4U.’
+ 100)dU.- 1.000
-
Therefore,
i.ooo = -4.125
7 = 1.904 ft
Now, considering they axis, we still use the strips dv because / I is uniform
along such strips. Howcvcr. the force <If=11 dA = p(S)[<fy) may he conyidered acting at the center of the strip, and accordingly has a moment arm
about the ? axis equal to SI2 for each slrip. Hence, we can say that
2.167.r =
-
I’2- pis(/y) +
I1
q - 4 , ’
2
-
500
(sno)12)- :--12
+ IOO)dJ+ 1.000-41.7
= 5.125
~___._....I_...
~
--_ll-...--.......l.-
~
In a statlonay liquid. the pressure at the surface of the liquid is transmitted
uniformly Ihmughout the liquid. In addition, there is superposed a lineal-ly increasing pressure with depth resulting from gravity acting on thc liquid. Thus. i l we
havc [I,,,,~, at the surface (often called the free surface). the prewtre in the liquid is
P = P,,,z,, +
P
where y i s the specific weight of thc liquid and I’ is the depth helow the surh c c . Thus, for a given liquid the presburc is unilorm at any constant distance
below the frcc surface. With this in mind. examine the following example.
SECTION 4.5
DISTRIBUTED FORCE SYSTEMS
Example 4.17
In Fig. 4.40 find the force on the door AB from water whose specific
weight is 9,806 N/mZand on whose free surface there is atmospheric pressure equal to 101,325 Nlmz (= 101,325 Pa).q Also find the center of pressure.
The pressure on the door AB is then
11
~ + ~(y)(r)
, ~ = P , , , ~+ (y)(.s)(sin 45")
= P
where s is the distance from 0 along the inclined wall OR. The resultant
force is then
,
Door A B
Figure 4.40. Door AB is exposed on
one side to water and on the other
+ (9,806)(.707)~](2)ds
= 1sy[101,325
101,325s + (9,806)(.707)%
To get the center of pressure equate moments about 0 with that of the
resultant. Thus using the notation S to locate the center of pressure we have
1.199 x 10hs =
5
9
5
+
s [ ~ = , ~(y)(s)(sin45")](2)ds
= [lO,,32S$
:.
I:
+ (9,806)(.707)%
(2)
3 P 7.061 m
Clearly the total force on the door from inside and outside would
include the contribution of the atmospheric pressure on the outside. We can
e a d y determine this force by deleting p<,<,,,in the preceding calculations.'U
'The unit of pressure in the SI system is the pascal, where I pascal = 1 Pa = I Nlm'.
"We have touched here on the subject of hydrostnrics. For a treatment of this suhjcct
that may he similar to what you will study in your upcoming course in fluid mechanics, see
Shames, I.H.. Mechunics off nu id.^, 3rd Edition, 1992. McCraw-Hill, Inc.,Chapter 3
We finish this series of examples with a multiple integration problem.
side t o air.
13 1
*Example 4.18
What i s thc simplest rehulkint and the ccnier iif pressure l11r the prccsurc
distribution shown i n Fig. 4.41'?
Notice that thc pressure varics linearly i n the x and y directions. Thc
pressure at any poinl x,y i n the distrihution can he givcn :I\ follows with
the aid of similar triangles:
=
2:
I'
,
!
211 N i m ~
/
+ 6.1Pa
1,'igurc 4.41.
We cannot employ a convenient strip here along which the pressure i s u n form, an i n Example 4.16. For this reasnn we consider rectangular :ire3 elcment d-r h. tn work with (scc Fig. 4.41). For such i1 siiiiill area. w ciiii
assume the pressure a s constant s o that p d x i/y i\ Ihc lirrce 1111 lhc cIcnicn1.
To find the resultant lnrce, we mint inlcgnlte over thc IO x 5 r e ~ t i l n g l e .
This integration involvcs twn vnri;ihle\ a n d i s again ii case o l m ~ , t t i p itif<,/~~
Rmtion. Thus, we can say that
N i r n u n i l u r m I~ICSSLIIC
diwihiitioii.
wherein we first integrate with respect lo .x while holding v constant and
then inlcgrate with respcct t o s ( i n this way we cover the cntil-e 10 x 5 I-cctangular arca). Thus. we have
Ill
j,, (II)\.+ 7 5 )(/y
= -
FR =
- 1,250 N
To find ~V for FH without il couple mon1cnt. we equate moments 01 F;?
about the x axis with that of the distrihittioti. Thus.
~I
.,^.___.. __"
",
~
.
.
..
., ,. " , ,
. ".
.
... ..
~~
SECTION 4.5 DISTRIBUTED FORCE SYSTEMS
Example 4.18 (Continued)
Therefore,
v=
i n I 5( 2 ~ + 6 x ) ~ d x d y
-I 1
1,250 n n
~ -I J
-
5
( 2 v 2 r6 x+ 2~ ~ )dyl
'0
1,250 o
=
0
(1oy2 + 7 5 y ) d y
1,250
1In
10
10
0
p=
A5
5.67 m
for 3, we proceed a3 follow9
X(1,250) =
Therefore,
~
I=-
'
5, I,px d x dy
in
5
SIn J5(2y
1,250 o
o
+ 6 x ) x d x dy
+y)l
5
- l,&oJ:0(2~$
_
=
---I
0
I
' 0 (25y + 250)dy
1,250 n
x =,3
The center of pressure is thus at (3.00, 5.67) m.
Case C. Coplanar Parallel Force Distribution. As we pointed
out earlier, this type of loading may be considered for beams loaded symmetrically over the longitudinal midplane of the beam. The loading is represented
by an intensity function w(x) as shown in Fig. 4.26. This coplanar parallel
force distribution can he replaced by a single force given as
FR = -J w ( x ) d x j
We find the position of FRwithout a couple moment by equating moments of
FR and the distribution w about a convenient point of the beam, usually one of
the ends. Solving for .F, we get
133
134
CHAPTER 4
EQUIVALI:NT I'ORCE SYSTEMS
Example 4.19
A simply supported heam i s shown in Fig. 4.42 supporting a 1,000-lh
point h r c e , a S O 0 Ib-ft couple. and a coplanar. parabolic. distrihutcd load
w Ih/St. Find thc simplest i-esulliint of this force systcm.
To exprcss the intensity of loading lor the coordinate system shown
in the diagram. we hegin with thc gcneral iormulation
=
Lt,?
at
+
h
Note from the diagram that when A = 25 we have w = 0, and when
x = 6s. wc h w c IV = 50. Suh.jccting Eq. ( a ) to these conditions, wc can
determine fi and ti. Thus.
+
i,
2.500 = iil6S)
+
0 = 4251
ih)
h
(e)
Subtracting. we can gel ( i as follows:
-2.500 = -4Iki
Therefore,
(i
= 62.5
From Eq. (h). we gcl
h = -(25)(62.i) = -1,562.5
Thus, wc have
I+=
h2.5.1
-
1,562.5
(d)
Summing forces. wc get i c r b;.
hi
F+( = -1,000
-
J25 .,'62.5~t
-
1,502.5 N I X
(el
To integratc this. we may change rari;ihlcs as follows:
11
= 62.S.t - 1.562.5
Thercfim.
N;LI = h 2 . i d t ~
Substituting into the intcgral i n Eq. ( e ) , wc have"
i0
Fig''re 4.42. Find qimQlest
reSU'tan'.
SECTION 4 5
DISTRIBUTED FORCE SYSTEMS
135
ExamPle 4.19 (Continued)
We now computer for the resultant without a couple as follows:
1 x>/62.5x
65
- 2,3337 = -(10)(1,000) -
25
-~
- 1,562.5 dx - 500
(g)
We can evaluate the integral most readily by consulting the mathematical
formulas in Appendix 1. We find the following formula (No 6):
In our case b = 62.5 and a = -1,562.5, so the indefinite integral for our
case is
Putting in limits, we have
= 65,333 - 0 = 65.333
..
Going back to Eq. (g), we can now solve easily for X, Thus,
x
~
=
1_ [-(10)(1.000) - 65,300 - 5001
-_
2,330
Z
= 32.5 ft
Before closing, it will he pointed out that, for a loading function w(x).
the resultant [,'w dx, equals the area under the loading cuwe. This fact is
particularly useful for the case of a triangular loading function such as is
shown in Fig. 4.43. Hence, we can say on inspection that the resultant force
has the value
5
I FR =
(;i(5i(1.000)
Fn = ~(5)(1,000)= 2,500 N
Furthermore, you can readily show that the simplest resultant has a line
of action that is (2/3) x (length of loading) from the toe of the loading.''
Thus, 4 without a couple moment is at a position (2/3)(5) to the right of a
(see Fig. 4.43). You are urged to use this information when needed.
"In Chapter 8, you will learn that the simplest resultant force for a distribution w(x) goes
through the cvnrrnld of the area under w(x). The centroid will be carefully defined at that lime.
Figure 4.43. Triangular loading resultant.
Figure P.4.15.
Figure P.J.4.3.
Figure P.4.46.
Show that thc volume and center 01 gravity of the conical
frustum are, respectively.
4.47.
r;
+ qr2 + r i )
and
In Problem 4.49 find the distance from the ground to the
center of gravity if the total weight is 2.37 x I O x kN.
4.50.
*4.51. A plate of thickness 30 mm has a specific weight ythat
varies lincarly in the x direction from 26 k N / d at A to 36 kN/m3
at R,and varies in they direction a i the square 01) from 26 kN/m’
at A to 40 kN/m‘ at C. Where is the cenler of gravity of the plate?
s
A
crowyection I - h l
Figure P.4.47.
Find the center of gravity of the piate bounded hy a
straight line and a parahola.
4.48.
sm
n
Figure P.4.51.
4.52. You are looking down on n plate with a hole in it as shown.
Thc thickness has a constant value equal t o t and the specific weight
y is constant. Find the coordinates F, 7 o1Ihe e w e r qf,vuviq.
A
Figure P.4.52.
Figure P.4.48.
4.53.
4.49. A massive radio-wave antenna for detcction of signals
from outer space is a body of revdutinn with a parabolic face (see
the diagram). These antennas may be carved from rock in a valley
away from other disturbing signals. What would the antenna
weigh if invde Srom concrete (23.6 kN/m’) for locatiun in a
remrxe desert area’!
I
Find the center of gravity of the plate having uniform
thickness and uniform specific weight. You are looking down
onto the top of thc plate.
?’
-
Figure P.4.49.
Figure P.4.53.
137
4.54. Thc
tup view of a platc is shown. Find center 01 grevit!
cnordinatcs
r, v.
V.57.
Suppose in Pmhlem 4.12 that
.f = ( . O I ~ k
Y i s conslant
and thickness I
?
.2 m
-1
.I r
n
d i s constant
+
2 1
+
. k ) k ot./lhni
Find Ihe .sinrplr.sr resultant for p = 450 Ihmlfl'. Find the plopelline of action.
il fast stop and swerve 10 Ihc left. the load of sand
(specific wcight = I S kNhn') in a dump truck i s i n the positirm
shown. What i s thc simplest resultan1 force on thc truck ti-om
thc sand and where does i t act'! If the truck was full (with a I c v d
tup) hcforc thc stop, how inuch sand spillcd'.' U w the results of
4.58. After
Problem 4.46.
.2 "I
Figure P.4.54.
MI N/m. What i s the >
coodinarc of i t s C C ~ ~ CofI gravity?The rud fumis me-half of a circle
4.55. The thin circular rod has a weight of
Figure P.4.58.
In Problems 4.5Y rhmufiir 4.62
U.IP
rhc k n o w , posiliom o/ <cnfe~.s
of ,q'at,iry of ,simple .)hiipc,.s.
4.59. A n I-heain cantilwered out from a wall weighs 30 Ihlft
Y
Figure P.4.55.
4.56. Find J for the uentcr of gravity of the horizontal plate with
a hole in i t .
and support\ a 300-1h hoist. Steel (4x7 Ihift') cover plates I i n
thick are welded on thc hcam near the wall to increme the carrying capacity o f the heam. What i s the moment at the wall duc t o
the weight ofthe reinforced b c m and the hoisted load of4.000 Ih
at thz outermost position of the hoist? Whilt is thc .sinzplv.st resultant force and its location'?
Y
Im
5m
I
Figure P.4.56.
138
Figure P.4.59.
4.60. The bulk materials trailer weighs 10,ooO Ib and is filled
with cement ( y = 94 Ib/ft’) in the front compartment (sections 1
and Z), and half-filled with water ( y = 62.5 Ib/ft3) in the rear
compartment (sections 3 and 4). What is the simplest resultant
force, and where does it act? What is the resultant when the water
is drained? Use the center of gravity and volume results from
Problem 4.47 (conical frustum).
4.62. Find the center of gravity of the body shown. It has a constant specific weight throughout. Cone and cylinder are on block
surfaces.
Cimeh=.Ilm
Figure P.4.62.
4.63. Find the .simplest resultant of a normal pressure distribution over the rectangular area with aides a and b. Give the coordinates of the center of pressure.
1’
2
Trai1erC.G.
3
4
P
Figure P.4.60.
vf the center of gravity of the
loaded conveyor system. The centers of gravity of the crates C
and D are at their geometric centers. WE is the weight of the frame
whose C.G. is at its geometric center.
4.61. Find the coordinates (x, y),
Figure P.4.63.
4.64. Find the simplest resultant acting on the vertical wall
ABCD. Give the coordinates of the center of pressure. The pressure varies such that p = E/Cv + I ) + F psi, withy in feet, from
IO psi to 50 psi, as indicated in the diagram. E and F are constants.
Y
We= 1001b
Wc=80lb
WD = 300 Ib
WE= 1,000lb
WBELT = 5 Ib/ft
2
Figure P.4.61.
IFigure P.4.64.
139
4.65. One lloor of a warehouse i s divided into lour area?. Area I
i s s t a c k 4 high with TV scts such that thc distrihutcd load i s
p = 120 Ihlft2.A r m 2 has refrigerators with 11 = 65 Ibift'. Area 3
has stcreos stacked s o that 11 = XO Ih/ft2. Area 4 has wa.;hing
machines with 11 = SO Ihlft'. What i q thc simplest resultant force
and where does i t act'!
1-
4.68. (a) Find the torque about axis i\H from thc wrcocl~.
An
(h) Find the torque ahout axih
lroni lhc distrihutcd
loads. ( H i m ; I.ooh dowm lirwn a h w e to hclp vicw
prohlcm.)
t
100'
Figure P.4.65.
4.66. Consider a prcssurc distrihution 1, fbrminp a hemispherical
surpace over a domain 01 radius 5 in. IS the inaximuni pressure i s
5 Pa, what i s the s i m p k s l resultant from this prcs\urc distrihution'!
Ill m
I'
Figure P.4.68.
4.69. For the system 01 lorccs shown. determine the turquc
ahout the a x i s going trorn A t o B. N o i r : 'The 100-N force and thc
triangular load distribution are i n the y: planc and the 100-N-m
couple i s mi the tup facc d thc rcuiimpular hux.
v
s(m)
Figure P.4.66.
-
4.67. A reclangular lank contains watcr. If the tank i s rotated
clockwise IO" ahout an axis normal t o the page. what toque i s
required to maintain the configuration? Width o f tank i s I 11.
1x it
___
2-1,
2-1,+-
Figure P.4.67.
t
Figure Y.4.6Y.
4.70. A manornerrr is a simple pressure measuring device. One
such manometer called B U rube is shown in the diagram. The tank
contains water including the tube to level M where mercury i s present. M and N are at the same Icvel. What is the gage pressure
(ie., the pressure a b w z atmosphere) at point LI in the tank for the
following data:
d , = .2
d, = . 6 m
111
y,,
4.12. (a) Calculate the force on the door from all fluids inside
and outside. The specific gravity of the oil is 0.8. This
means that the oil has a specific weight y which is 0.8
times lhat of water (y,,,
= 62.4 Ihlft'). Note that a
uniform pressure on the surface of a liquid extends
undiminished throughout the depth of the liquid.
(b) Determine the distance from the surface of the oil to
the .rimpIrst rrsultant force on the door from all fluids.
yH:o = 9,806 N/m3
= (13.7)(yHP) Nim'
[ H i n t : M and N are at the same level and are joined by the
same fluid, namely mercury. Hence, the pressures there are
equal.]
t,
ft
2qS
10 i t
ft
Figure P.4.72.
Mercury
Figure P.4.70.
4.71. Imagine a liquid which when stationary stratifies in such a
way that the specific weight is proportional to the square root of
the pressure. At the free surface, the specific weight is known and
has the value & What is the pressure as a function of depth frvm
the free surface? What is the resultant force un one face AH of
a rectangular plate submerged in the liquid! The width of the plate
is b.
4,73,
Ar what height
The door is
will the water
the door to
wide, Neglect friction and the weight
o f the door.
piitm
X
36 kN
L
Figure P.4.71.
Figure P.4.73.
141
4.74. Find the krcc on the door front the inside and uulsiile
prebsureh. Give the poritiuri of the re%dlml lurce ahvvc the
base of the door. The specific grevity .S of the oil i s 0.7, i.c.,
= K.7)Yk, 0.
x>,,
/p\
4.77.
A hlock I ft thick i h suhniergcd i n water. Computr thr
siinplest resultant force and the center of prehsure uti [he huttom
.;urt~icc.Take y = 62.4 Ih/tt'.
,,Top i\ ripen
Figure P.4.74.
4.75. Find the total f k x o n door Ail fri,m fluid\. The spccific
gravity S o l a fluid i h y; .,",J x , ~~.
Take$>,, = 1l.h. Find the pinition
of this force from thc hottuirlof the dour.
Figure P.4.77.
*4.78. Wha1 i s thc r ~ s ~ l t m
force
l from wiler and where does i l
act on thc 40-m-high circular concrctc d a m hetwero two wall, of
a rocky gorge? (Water wcishs Y.XO6 N l m ' . )
4
40
I?tl
J it
H
Figure P.4.75
4.76. What i \ the simplest r5sd13111 flircc t r i m the water and
where due\ i t acl un the h O - m - h ~ ~X00-m-laog
h
rtraight canhiill
dam'! (Water weigh, O.XO6 N/m'.1
Figure P.4.76.
t-
111
Tup "icw
Figure P.4.78.
4.79. The weight of the wire AHC'L) per unit length. w, incrcascs
linearly lr<m4 orlft at ,t ti) 2 0 o,lft nl 11. Where is thc center 01
gravity <,1thc wire?
Figure I'.4.79
4.80. Find the center of gravity of the wire. The weight per unit
length increases as the square of the length of wire from a value of
3 ozlft ai A until it reaches the value of 8 ozlft at C. It then
decreaaes I orJft for every 10 ft of length.
4.82. At what distance ifrom point A can the system of thin
rods be suspended so that it will just balance? Use the formula
developed in the previous problem for the radial distance to the
360r. 9
center of gravity of a thin circular rod, namely E9 s'n-i
Y
I,
W.
= .IR m
= 500 Nim
Figure P.4.80.
What is the center of gravity coordinate y, for a thin circular rod shown in the diagram? It has a weight of w Nlm. The rod is
placed symmetrically about they axis. The angle Q is in degrees.
4.81.
1 4;
=
70'
Figure P.4.82.
Y
Figure P.4.81.
4.6
Closure
We now have the tools that enable us to replace, for purposes of rigid body
mechanics, any system of forces by a resultant consisting of a force and a
couple moment. These tools will prove very .helpful in our computations.
More important at this time, however, is the fact that in considering conditions of equilibrium for rigid bodies we need only concern ourselves with this
resultant to reach conclusions valid for any force system, no matter bow complex. From this viewpoint, we shall develop the fundamental equations of stdtics in Chapter 5 and then employ them to solve a large variety of problems.
143
4.88 A Seep weigh5 I I kN and has hoth a front winch and a r e a
powcr takeoff. The tension in the winch cable is 5 kN. The puwer
take-off develops 300 N-tn of torque T ahout an axis parallel to
the x axis. If the driver wcighs XU0 N, wh;it is the resultant force
y = Ill* (x'
+ 3x) ;:
Parabola
sy\tem at the indicated center of gravity of the Jeep where we can
consider the weight of thc Jcep to he concentrated?
Scencenter
1
c1
I-
x
-
-1.Sm-
ii
+1.2m
3m
1
Figure P.4.XX.
Figure P.4.')0.
4.91.
Find the torque about axis OH from the system of forces.
7
4.89. What is thc .simpk~tr e d t a n t for the forces and couple
F =200i+300kN
actin3 011 the bean'!
311 Ih
4
Figure P.4.91.
100 Ih
Figure P.4.89.
92. A rectangular plate chown as ARC can rotate about hinge
E . What length 1 should RC be so there is zero tarquc about H
from the water, air. and weight o l thc platc'! Take this weight as
1,000 Nlm o l Icngth. Thc width i ? I m.
= 9,806 Nlm'.
4.90. A parabolic body of revolution has cut out of it a second
parabolic body uf revolution starting at A and forming a sharp
edge at H with x r o dopz at A .
(a) What is I' as a function ol .r for the cut-out body of
revolution?
(h) Set up an integral for computing W (weight) and then
thc center of gravity coordinate i.
Notr; y vxic:; with x. Do not solvc the integral.
Figure P.4.92.
x,,~
I45
4.93. An open rectangular bank u t water 1 5 partially filled with
water. The dimensions iirc \hewn.
t i l ) r k t e r n t i n e thc hrcc on ihc h m w n 01 thc t a d l r o n ~
the water.
(hl Detemmine the force on the donr <hmvn at thc Cidc 01
the tank. Indicate the pwition of this firrcc 1r<m l l i c
bottom of the tank.
Note that the atmospheric pressure develops equal and opposite
forces on both sides of the door and hence yizlds n o inut f<,rcc.
,
,i
Figure P.4.93.
4.94. Sandhags are piled on a bean. Each hag i h I 11 witlc and
weighs 100 Ih. What i s the simplcst resultant force imd where doe&
it act’?What linear mathematical function ofthe di\crihuted Ioiiil ciin
he used to represent the sandhap <rvecthe left 1 ft 01the tieiiiii’l
i
Figure P.4.94.
4.95. A cantilcvcr b u m i s subjected t o
a linearly w q i n p I h d
over p d n 01 i t s length. What i s the .sit!rpl<ml rcwlt:int f~rrce.and
where does it act’! What i c the moment :It thc suppomd cnd?
146
4.98. Find the resultant force system at A for the forces on the
bent cantilever beam. BC is parallel to z axis
4.100. The L-shaped concrete past supports an elevated railroad.
The concrete weighs 150 Ib/ft3. What is the simplest resultant
~~
Force from the weight and the load and where does it act? Load
acts at center of top surface.
V
40,000 Ib
15'
1
Figure P.4.100.
Figure P.4.98.
4.99. (a) Find equations describing both parabolas.
(h) Using rjerfical strips (and a composite body approach),
find weight of plate in terms of yt.
(c) Using vaticill strips, find X of C.G.
4.101. Explain why the system shown can he considered a system of parallel forces. Find the simplesr resultant for this system.
The grid is
of
squares,
X
( I O , - 3) ft
Figure P.4.99.
X
Figure P.4.101.
I47
l.102. A plate 01 thickness
f
has as the uppcr edge a parabolic
:uwc with infinite slope at the origin. Find the x, y coordinates 01
hc criitei ut gravity fnr this plate.
Figure P.4.1U4.
*4.105. A block has 8 rectangular portion removed (darkened
region). If ths spccific weight is given a h
2'
Figure P.4.102.
tind x for the
CCIIISI
of gravity
/
7
A rectangular tank conmitih a liquid. At the tup U Ithe
iquid there i s a pressure ol ,1380 Nlmm' absolute. What i s the
.iinplesl resultatit lbicr in the inside surface of the dour AH!
Nhwe is thc ceiitcr ot pressuse relative to the bottom of the door'!
rake y = 8,190 Nlm' for the liquid.
1.103.
AI?m
y
Air
f
5m
/
I2 n,
Figure P.4.105.
Liquid
bigure P.4.103.
l.104. The sprcilic weight ot thr material i n a right circular
m e varies directly :IS the s q u a r ~Crt the distance? from the hase.
f 1;, = S O Ih/ft3 is the cpecific weight at the base. and i f y' = 70
hlft' is the specific wcight at the tip, when, i \ the center u l g r a v ly ,!I the cone'! (Szz hint in Pmhlrm 4.45.)
48
4.106. Compute the sirnl,ics1 resultant fix the loads shown act^
irig un the simply supported beam. t i i v r the line of action
*4.109. The pressure pO at the corner 0 of the plate is SO Pa and
increases linearly in they direction by 5 P d m . In the x direction, it
increases parabolically starting with zero slope s o that in 20 m the
pressure has gone from S O Pa to 500 Pa. What is the simplest
resultant for this distribution'! Give the coordinates of the center
of pressure.
4.107. Fmd the center ot gravity ot the plate.
--v
x
Figure P.4.107.
~/c-'"m+
Figure P.4.109.
4.110. A sluice-gate door in a dam is 3 m wide and 3 m high.
The water level i n the dam is 4 m above the top of the door. The
gate is opened
the
level f a l l s 4.,, What is the
resultant force on the closed door at hoth water levcls? Where du
the forces act (i.e., where i s the "center of pressure" i n each case)?
Water weighs 4,806 Nlm'.
4.108. Find the center of gravity of the W s s . All members have
the same weight per unit length.
v
Figure P.4.108.
Figure P.4.110.
149
__
~
~
~~
~-
A cylindrical tank of watcr is rotated at cmbtilnt angular
speed Wuntil the water ceases t o c h m g c shapc. The result is a Srw
4.111.
surface which, from lluid inrchaiiicr cansidcrations, is that of a
paraboluid. If the prcswl-e varics direclly as the depth hclow Ihc
I r e sutfke. w h a t i \ the w s d ~ m force
t
on a qu:idmnt ~ r fthe hasr
01the cylirrder? Take y = 62.4 Ihlft'. [ H i m : Use circular strip in
quadrant having area l l 4 ~ 2 n - d r ) . ~
Figure P.4.111.
I SO
4.112. Firid lhc 'i arid Y coordinates UT the center of gravity 0 1
the hodich \hewn. Thehe c i i n s i q of:
( a ) A plate AH('whose thicknsss I = SO niin.
(hj A rod I> 01d ~ i m z t e r.? m :md leneth 3 m.
(i.) A hlock F whocc thickness (not \huwn) is 3 in.
'Ihc density of thc three bodich i \ the SJIIIC.
Figure P.4.112.
Equations
of EquiIibrium
5.1
Introduction
You will recall from Section 1.10 that apurficle in equilibrium is one that is
stationary or that moves uniformly relative to an inertial reference. A body is
in equilibrium if all the particles that may be considered to comprise the body
are in equilibiium. It follows, then, that a rigid body in equilibrium cannot be
rotating relative to an inertial reference. In this chapter, we shall consider
bodies in equilibrium for which the rigid-body model is valid. For these bodies, there are certain simple equations that relate all the surface and body
forces, or their equivalents, that act on the body. With these equations, we can
sometimes ascertain the value of a certain number of unknown forces. For
instance, in the beam shown in Fig. 5.1, we know the loads F; and and also
the weight W of the beam, and we want to determine the forces transmitted to
the earth so that we can design a foundation to support the structure properly.
Knowing that the beam is in equilibrium and that the small deflection of the
beam will not appreciably affect the forces transmitted to the earth, we can
write rigid-body equations of equilibrium involving the unknown and known
forces acting on the beam and thus arrive at the desired information.
Note in the beam problem above that a number of steps are implied.
First, there is the singling out of the beam itself for discussion. Then, we
express certain equations of equilibrium for the beam, which we take as a
rigid body. Finally, there is the evaluation of the unknowns and interpretation
of the results. In this chapter, we will carefully examine each of these steps.
Of critical importance is the need to be able to isolate a body or part of
a body for analysis. Such a body is called afree body. We will first carefully
investigate the development of free-body diagrams. We urge you to pay
special heed to this topic, since if is the must important step in the solving of
mechunics problems. An incorrect free-body diagram means that all ensuing
5
W
Figure 5.1. Loaded beam.
151
+
w
I
I<
5.2
The Free-Body Diagram
SECTION 5.2
I
M
THE FRELBODY DIAC;KAM
Rollcra
(ill
(CI
Figure 5.6. Standard connections
In general, to ascertain the nature (11 the fnrcc system that a body M is
capable of transmitting to a second body N through some connector or support, we may proceed in the following manner. Mentally move the bodies relativc to each other in each of three orthogonal directions. In those directions
where relative motion is impedcd or prevented by the connector or support,
there can be a Sorce component at this connector or support in a free-body
diagram ofeither body M or N. Next, mentally rotate bodies M and N relative
to each other about thc orthogonal axes. In each direction about which relative rotation is impeded or prevented by the connector or support, there can
bc a couple-moment component at this connector or suppon in a free-body
diagram of body M or N. Now as a result of equilibrium considerations of
body M or A',certain force and couple-moment components that are capable
of being generated at a support or connector will he zero for the particular
loadings at hand. Indeed. one can often readily recognize this by inspection.
For instance, consider the pin-connected beam loaded in a coplanar manner
shown partially in Fig. 5.7. I f we mentally move the beam relative to the
gnmnd in the I,y. and 7 directions, we get resistmce from the pin for each
direction, and so the ground at A can transmit force components A,, A,, and
Figure 5.7. Pin connectim
153
154
CHAPTER 5
IiQlJATIONS OF EQLIILIRRIIJM
A ~ Howe\~cr.
.
hecause the loading is coplanar iii the .i\' plane, the force coniponcnr A _ niuit bc Lero and can he deleted. Next. inlentally rotale the be;m
rclati\'c to the ground at A about the three orthogonal axes. Because of the
smooth pin conlicetion. thcrc i s t i i ) resistance ahout the : axis and MI
M; = 0.But there i s resistancc ahout t h c ~ and?
r
axcs. Howevcr, the coplanar
loading iii the .r?. plane cannot e x t i t iiiorncnts ahout the ~r arid s axes. and s o
the couple inioiiients M , and M, arc 721-0. All told. then. we jus1 have Icirce
cc)mp~~nenls
A , and A , a1 the pin connection, a\ hiis heeri shown earlicr in
Fig. S.h(bJ.whcrciii w e rclicd on physical reasoning for this result.
5.3
B
\
Figure 5.8. Rigid hod) i n cquilihrium
Free Bodies Involving
Interior Sections
l,tt us consider;^ rigid body i n equilihriuiii as shown i n Fig. 5.8. Clearly, every
portinti ( i t t h i s hocly inul iilso he in equilihriuin. If we consider the hody as
two pirts A and 8.w e can present cithei- part i n a lice-body diagram. To do
this, we niust includc 011 the poition chosen to he the free hody the Ibrces f i m t
flip o t / i ( , t - / ~ u rthat
t x i s c at the coniinon section (Fig. 5.Y). The suilacc between
both seetioils may he any curved or plane surface, and ovcr i t there will hz a
c ~ n t i n u o uforce
~ dislrihution. I n the general c a e . wc know that such ;I distrihutioii can he replaced by a single force and ii single couple nioinent (at any
chosen point) and this has been done in the tiec-hody diagram or parts A aiid
B in Fig. 5.4. Nolice that Newton's third law has heen ohserved.
Figure 5.9. krcc hodics ot parts .1 and H
Figure 5.10. Cantilc\w beam.
IV
Figure 5.1 1. Free-hdy diogr;im
01cmtilever heam.
As a special case. consider a beam with one end cmbeddcd i n a iiiassivc
wall (canlilcver beam) aiid loaded within the .x:v plane (Fig. 5. IO). A free hndy
01 the p o r t i m (if Ihc heam cxteiiding from Ihe w a l l i s shown i n Fig. 5. I I
Recause 01 the geometi-ic symmetry OS the hcam ahout the .(I plane and the
Iact that thc loads arc i n this pliine. thc cxposcd forces i n the cut section can
he considered coplanar. Hence, these exposed I i r c e q ciin he replaced by a
single lorce and ii single couplc inonienl i n the centcr plane. and il is the usuiil
practice to dccomposc the I'orcc into components F, and PI. Allhough a line 111
actiori for the liirce can hc found that would enahlc us to climinate the couple
ninlnent. i t i s dcsirahle in slructural problems to work with an equivalent i y s tcni that has the lorcc p i n g through the center 01 the heam cross section.
arid thus to liave a couple moment. 111 the iicxt section. we will see how F a n d
C. can he ;iscertiiincd.
SECTION 5.3 FREE BODIES INVOLVING INTERIOR SECTIONS
Example 5.1
As a further illustration of a free-body diagram, we shall now consider the
frameZ shown in Fig. 5.12, which consists of members connected by fric-
tionless pins. The force systems acting on the assembly and its parts will
he taken as coplanar. We shall now sketch free-body diagrams of the
assembly and its parts.
Free-body diagmm of the entire assembly. The magnitude and direction of
the force at A from the wall onto the assembly is not known. However, we
know that this force is in the plane of the system. Therefore, two components
are shown at this point (Fig. 5.13). Since the direction of the force C i s known,
there are then three unknown scalar quantities, A,, A,, and C , for the free body.
Free-body diagram of the component parts. When two members are
pinned together, such as members DE and AB or DE and BC, we usually
consider the pin to be part of one of the bodies. However, when more than
two members are connected at a pin, such as members AB, BC, and BF at
B, we often isolate the pin and consider that all members act on the pin
rather than directly on each other, as illustrated in Fig. 5.14. Notice four
sets of forces that form pairs of reactions have been enclosed with dashed
lines. The fifth set is the 1,000-lb force on the pin and on the member BE
Figure 5.12. A frame.
A
1,OOU Ih
Figure 5.13. Free-body diagram
of frame.
I
Figure 5.14. Free-body diagrams of parts.
frame i s it system of connected straight or benr, long. slender members where some
of the connecting pins are not at the ends of the members as is the case for structures that we
will study kter called I T U I S ~ S .
155
Example 5.1 (Continued)
Do niil he cwncerned about the pi-nper ,SPII,SP [if an unknown force
ciimlmient that yrw draw 1111 the free-body diagram, for y o u m a y choose
cithei- a pnsiti\'e or n c g a t i x scnse for tlicsc components. When the \piilues
01 these qi~antities:ire ascertained by methods iil stiltics, the proper sense
(breach coinponcot can then he estahlishcd: hut. liavins chosen a seiisc for
a mniponcnt, you must he sure that the rrw(.tion to this ciimponent has the
oppasir<, s e i i s c ~ ~ ~ - eyiru
l s e aril1 \ irilate Newtiin's third law.
Free-body d i q r n m ofportion of the assemhly to the right o f M - M . I n
making a free body 01 the portion to the right [if section M-M (see
Fig. 5 . ISa). we must rememhcr to piit i n the weight of the portions n l the
inernhers remaining &v the cut liiis hccn made. At the two cut5 made hy
h--M we must replacc coplanar fiirce distrihutions hy resulkints. as i n the
case nf the previously coiisidercd cnntilcvcr heaiii. This is accomplished
by inserting two h r c c co~iipiincnlsnsu;illy normal and tangential to the
cross section and a couple mnnient a h \vas done for the cantilever bcam.
Note i n Pig. 5. I X b ) tliat there arc w v c n unhnown scalar quantities lor this
free-body diagram. They iirc C',, CL. t;.F,. t;?.
and Fd. Apparently. the
numher 01 unknowns varieq widely lor the various free bodies that may
he drawn O r thc system. For this reawn. you nitist choose the free-body
diagram that is suitahle 101- your ncedr wilh snmz discretion in ordcr ti)
eflectively solvc l i r the desii-ed unknowiis.
SECTION 5.3
FREE BODIES INVOLVING INTERIOR SECTIONS
Example 5.2
Draw a free-body diagram of the beam AB and the frictionless pulley in Fig.
5.16 (a). The weight of the pulley is W, and the weight of the beam is K8.
Figure 5.16. (a) Beam AB: (b) Free-body diagram of AB;
(c) Free-body diagram of Pulley D
The free-body diagram of beam AB is shown in Fig. 5.16(b). The
weight of the beam has been shown at its center of gravity. Components U,
and U,, are forces from the pulley D acting on the beam through the pin
at B. The free-body diagram of the pulley is shown in Fig. 5.16(c).
Some students may he tempted to put the weight of the pulley at U in
the free-body diagram of beam A B . The argument given is that this weight
“goes through B.” To put the pulley weight at U on free body AB is strictly
speaking an error! The fact is that the weight of the pulley is a body force
acting throughout the pulley and does not act on the beam UD. It so happens that the simplest resultant of this body force distribution on D goes
through a position corresponding to pin B. This does not alter the fact that
this weight acts on the pulley and nor on the beam. The beam can only feel
forces U, and U, transmitted from the pulley to the beam through pin U .
These forces are related to the pulley weight as well as the tension in the
cord around the pulley through equations of equilibrium for the free body
157
Example 5.3
We finish this series of free-hody diagrams with a three~diineiisional
case. In Fig. 5.17(a) we sliou' a structure having h;ill-joint connections at
A and n, a fixed-end support at U and. finally. at i"
the column R resting
directly on the foundation. Draw the frcc-body diagratn fcir the slruclure.
We show the frcc hody diagr;nn O r this case i n Fig. S.I7(h).
"5.4
looking Ahead-Control Volumes
I n rigid-body irw<./iuiri(,swc use Ihc free hody whcrein wc isolate a hody or a
portion ( f a hody and wc identify all the external f i m e 7 acting on the body so
that we can employ Newtiin's l a b .
In ./7uid mw/wiiii.s. we may either make :I free body of some chosen
chunk ot lluid (here i t i s called a system). but more likely i t w i l l be more protitable to identify some volume i n space involving fluid flow through the volunic. Such a v(iIu~iici s celled ii ( * i t i f r . d inlinfne. Here, as in the case of a free
body, wc must specify crll the cxrrrnol fi)rms such as triictions o n the hounding suifaccs of the conrrol v ~ l u m eand. i n addition, hody forces on the material inside the c o n t r ~ lvoluiiic. Thih identification and force specification i s
needcd to ensure, i n appropriate equations, that Newton's law and other laws
iire satisfied for the fluid and other bodies inside the control voluiiic at any
Lime f.
Thos. you must dcveliip sensitivity at this early stage o f your studies in
depicting cxternal fotres fiir a frcc hody. The ~ a m ecarc w i l l he needed in
your upcoming courses i n lluid nicchanics.
5.1. Draw the free-body diagram of the gas-grill lid when it is
lifted at the handle to a 45" open position.
5.3.
Draw a free-body diagram of the A-frame
Figure P.5.1.
5.2. A large antenna is supported by three guy wires and rests on B
large spherical ball joint. Draw the free-body diagram ofthe antenna.
Figure P.5.3.
5.4. Draw complete free-body diagrams for the member AB and
for cylinder D. Neglect friction at the contact surfaces of the
cylinder. The weights of the cylinder and the member are denoted
as y , and W,,, respectively.
Figure P.5.2.
Figure P.5.4.
5.11. Draw lire-body diagrams of each pan of the me-branch
trimmer.
Branch 2 3
Figure P.5.11.
Figure P.5.14.
5.12. Draw free-body diagrams for the two booms and the body
E of the power shovel. Consider the wcight of each part to act at a
central location. (Regard the .;havcl and payload as concentrated
forces, W, and W,,., respectively.)
5,15, Draw a free-body diagra,n of members CG, AG, and the
disc B. Include as the only weight that of disc B . Label all forces.
(Hitit: Consider the pin at G as a scparate free body.)
P
Figure P.5.12.
Figure P.5.15.
5.13. Draw the free-body diagram for the hulldozer, B , hydraulic
ram, R,and tractor, T. Cunsidcr the weight of cach part,R, X, and T.
5.16. Draw the free-body diagram of the horizontally bent can
tilevered beam. Use only X:I components of a11 vectors drawn.
..
Figure P.5.13.
5.14. Draw a free-body diagram of the whule apparatus and of
each of its parts: AB. AC, BC, and I). Include thc weights of all
bodies. Label forces.
Figure P.5.16.
16
5.5
General Equations of Equilibrium
For every free-hody diagram. we can replace the system of forces and couples
acting on the hody by a single lorce and ii single couple moment at a point ( 1 .
The forcc will have the same magnitude and direction, n o matter where point
a is chohen 10 n i o ~ the
e entire system by methods discussed earlier. However.
the couple-moment vector will depend on the point chosen. We will pmve in
dynamics that:
The necessary cimditiuns,fnr a rigid body IO be in equilibrium are rhat the
resultant force F, and the resultant couple moment C, .for any point a be
7ero vectors.
That is,
FR = 0
= 0
c,
W e shall prove i n dynamics, furthermore, that the conditions above are .s@
c.iriit tn maintain an inifiully .rfationury body in a state nfequilibriutn. These
arc the fundamental equations of statics. You will rememher from Section 4.3
rhat the resultant FR is the sum nf the forces moved to the cotnmon point, and
that the couple moment C, is equal 10 the sum of the moments of all the original forces and couples taken about this point. Hence, the cquations ahove can
hc written
(5.221)
(S.2h)
where Ihe p,'s are displacement vectors from the conirniin point (1 to any
p i n 1 on the lines of action of the respective forces. Frnm this form of the
equations oS statics. we can conclude that for equilibrium tn exist, the w m r
.rum of tlw ~;Jw..s m u s I be zero and the moinewt of rhr .y>'.s/rrn,!f,fi)rw.s mnd
miiiplm about miy point in .ypace musf be zero.
Now that we have summed forces and have taken moments ahout a
point ti. we will demonstrate that we cannot find another independent equation hy taking moments ahout a diffrrent point ti. For the hody in Fig. 5.18.
wc have initially the fdlowing equations of equilibrium using point (I:
x F~ + p, x F ,
+
11,
x F,
+
pd x F~ =
n
(5.41
SECTION 5.5 GENERAL EQUATIONS OF EQUILIBRIUM
Figure 5.18. Consider moments about point h
The new point h is separated from a by the position vector d. The position
vector (shown dashed) from h to the line of action of the force F, can be
given in terms of d and the displacement vector p, as follows. Similarly for
(pJh,which is not shown, and others.
( P , )=
~ (d + P,)
(p2Ih= (d + P&
etc.
The moment equation for point b can then he given as
(P, +
dl
x
4 + (P, + 4
X
F2 + (P, +
4
X
F3 + (p4 + 4
X
F4 = 0
Using the distributive rule for cross products, we can restate this equation as
( P , X F, +
+d
X
(F,
X
F2 + P,
X
F3 + p4
+
F2 + F?
+ F4) = 0
X
F4)
(5.5)
Since the expression in the second set of parentheses is zero, in accordance with Eq. 5.3, the remaining portion degenerates to Eq. 5.4, and thus we
have not introduced a new equation. Therefore, there are only two independent vector equations
of equilibrium for any
single free body.
We shall now show that instead of using Eqs. 5.3 and 5.4 as the equations of equilibrium, we can instead use Eqs. 5.4 and 5.5. That is, instead of
summing forces and then taking moments ahout a point for equilibrium, we
can instead take moments about rwo points. Thus, if Eq. 5.4 is satisfied for
point a, then for point b we end up in Eq. 5.5 with
dX(F,+F,+F,+F,)=O
(5.6)
If point b can be any point in space making d arbitrary, then the above equation indicates that the vector sum of forces is zero. If a point b happens to be
chosen making Eq. (5.6) identically 0 = 0. (and hence useless), choose
another point b. We then have equilibrium since F, = 0 and C, = 0.
Using the vector Eqs. 5.2, we can now express the scalar equations of
equilibrium. Since, as you will recall, the rectangular components of the
moment of a force about a point are the moments of the force about the onhogonal axes at the point, we may state these equations in the following manner:
163
164
CHAPTER 5
EQUATIONS OF t?QUIL.IRRIlJM
From this set o f equations, i t i s clear that !io more than si.^ unknown scularqiiun1ifir.s i l l / / ! e,qmrru/ m s c ~ r hen s o l d hy lnellior/.s of .stuli~~s,fi~i'
u .sin,s/e
. f W C h<J</Y.'
We can easily express on? nrrrnhu oT scalar equations of equilibrium
for a free hody by selecting references that have different axis directions,
along which we can sum forces and ahout which we can take moments. However, i n choosing s i x indepewlmr equations, we w i l l find that the remaining
equations w i l l he dependcnt (111 these six. Thac is, these equations w i l l he
sums, differences, ctc., of the independent set and so w i l l he o f no use i n solvi n g lor desircd unknowns othet- than for purposcs (it' checking calculations.
5.6
Problems of Equilibrium I
We shall now examine prohlems o f equilihrium i n which the rigid-body
assumption i s valid. 'lo solve such prohlems, wc musl find the value o f cerlain unknown fhrces and couple niomcnts. We first draw a rree-body diagram
o f the entire system or portions thcreof to clearly rxpose pertinent unknowns
for analysis. We then write the equilibrium equations i n terms of the
unknowns along with Lhe known forces and geometry. As we have seen. for
any free hody there i s a limited number of independent scalar equations o f
equilibrium. Thus. at times we must employ several free-body diagrams for
portions of the system to produce cnough independem equaiinns to solve all
the unknowns.
For any lrcc hody, wc may proceed hy expressing two hasic vector
equations of statics. After carrying out such vector operations as cross products and additions i n Ihe equations, we form scalar equations. These scalar
equations are then solvcd simultaneously (logcther with scalar equations from
other free-body diagrams that may hc needed) to find the unknown forces and
SECTION 5.6
couple moments. We can also express the scalar equations immediately by
using the alternative scalar equilibrium relations that we formulated in previous sections. In the first case, we start with more compact vector equations
and a m v e a1 the expanded scalar equations by the formal procedures of vector algebra. In the latter case, we evaluate the expanded scalar equations by
carrying out arithmetic operations on the free-body diagram as we write the
equations. Which procedure is more desirable? It all depends on the problem
and the investigator’s skill in vector manipulation. It is true that many statics
problems submit easily to a direct scalar approach, hut the more challenging
problems of statics and dynamics definitely favor an initial vector approach. In
this text, we shall employ the particular procedure that the occasion warrants.
In statics problems, we must assign a sense to each component of an
unknown force or couple moment in order to write the equations. If, on solving the equations. we obtain a negative sign .for a component, then we have
guessed the wrong .sen.sefor that componenr. Nothing need he redone should
this occur. Continue with the remainder of the problem, retaining the minus
sign (or signs). At the end of the problem, report the correct sense of your
force components and couple-moment components.
We shall now solve and discuss a number of problems of equilibrium,
These problems are divided into four classes of force systems:
1. Concurrent.
2. Coplanar.
3. Parallel.
4. General.
The type of simplest resultant for each special system of forces is most
useful in determining the number of scalar equations available in a given
problem. The procedure is to classify the force system, note what simplest
resultant force system is associated with the classification. and then consider
the number of scalar equations necessary and sufficient to guarantee this
resultant to he zero. The following cases exemplify this procedure.
I
[Case A. Concurrent System of Forces. In this case, since the
simplest resultant is a single force at the point of concurrency, the only
requirement for equilibrium is that this force he zero. We can ensure this condition if the orthogonal components of this force are separately equal to zero.
Thus, we have three equations of equilibrium of the form
As was pointed out in the general vector discussion, there are other
ways of ensuring a zero resultant. Suppose that the moments of the concurrent force system are zero about three nonparallel axes: a,b, and y. That is,
C(M,)!
= 0,
c c M p ) i = 0,
i
c ( M y ) i= 0
(5.9)
PROBLEMS OF EQUILIBRIUM I
165
166
CHAFTER
s ~ Q I I A T I O N SOF EQIJILIBRIUM
Any m e of the follnwing three conditions must then hc true:
1. The rcsultant forcc F; i s ~ c r o .
2. F; cuts all three axcs (see, Fig. 5.19).
3. FA cuts two axes and i s parallel 10 the third (see Fig. 5.201
Figure 5.19. F , cuts t h m axes.
Figure 5.20. F,<cuts t
w axes
and i s parallel
10
third.
We can guarantee condition I and thus equilibrium i f we select axes a, 0, and
y so that no straight line can intersect a l l three axes or can cut two axes and
he parallel to the third. Then wc can use Fqs. 5 9 as the equations o f equilihrium under the aforcstdted conditions rather than using Eqs. 5.8. What happens i f an axis used violates these conditions? The resulting equation will
either he an identily 0 = 0 or will be dependent on a previous independent
equation of equilibrium for onc of the axcs. No harm i s done. Onc should usc
nther axes until three independent cquationq are found.
Similarly, m e can sum forcer i n one direction and take moments ahoul
two axes. Setting these equal lo zero can yicld three independent equations nf
equilibrium. If not. use other axes.
The essential conclusion 10 he drawn i s that there nrc three independent
srnlnr f,quntions of equilibrium / i w a coiicizrrmt ,fi,n.c .system. For such
systems i t i s most likely that you will always w n i forces rather than take
moments. However. for olher fnrce systems that we shall undertake, there
w i l l be ample opportunity to profitably cmploy alternate fnrms o f equations
other than those that we shall at first prescribe. The imporvdnt thing to
remember i s that, just as i n (he concurrent force systems, only a definite number of equations for a given free hody will he independent. Simply writing
more equations beyond this number w i l l only lead to identities and equations
that w i l l he of no use for wlving for Ihe desired unknowns.
SECTION 5.6 PROBLEMS OF E Q U I L IBRIUM I
Example 5.4
What are the tensions in cables AC and AB in Fig. 5.21? The system is in
equilibrium. The following data apply:
a
j3 = 50"
W = 1,000 N
= 37"
Figure 5.21. Derrick holding a beam.
Immediately i t should be clear from observation of the diagram that
TAD
= 1,000 N (tension)
A suitable free body that exposes the desired unknowns is the ring A,
which may be considered as a panicle for this computation because of its
comparatively small size (Fig. 5.22). Physical intuition indicates that the
cables are in tension and hence pulling away from A as we have indicated
in the diagram although, as mentioned previously, it is nor necessary to
recognize at the outset the correct sense of an unknown force. The force
system acting on the particle must be a concurrent system. Here it is also
coplanar as well, and therefore we may solve for only two unknowns.
Hence, we can proceed to the scalar equations of equilibrium. Thus,
1,000
CFv=O
~
TAc cos 37" - TARcos 50' = 0
:. .7986TA,
Solving for
c,
and
c,
+ .6428TA,
= 1,000
(a)
from Eqs. (a) and (h), we get
9"
4
l&
= 602.6N
ce
&l
= 767.1 N
Since the signs for TAc and
are positive, we have chosen the correct
senses fix the forces in the free-body diagram.
i
T,,= 1,000N
Figure 5.22. Free-body
diagram of pin A .
167
168
EQUAIIONS OF F<OUII.IIJKIIIM
CHAPTER S
p Example 5.4 (Continued)
I
Anotlicr way (if ai-riving at the ~ ~ ~ u t i o
i stto
l , collsider tile,fivce /J,J/ygon that wils discussed i n Section 2.3. Bccause the forces arc i n equilihrium.
the polygon niiisl close; that is. lhc haid of the final lorcc niiist coincide with
thc tail of the inilial Force. In this cast. we have il triillifde. ils ~ h o w ' I 1ill
Fig. 5.23 ;ipproximately 10 scale. We can nnw use the l i i w 6f sines
I.ONIN
T,,,= 602.6 N
The force polygon m:iy lhu.; be used lo good ad\wita$c whcn threc
concurrcnt coplanar forces arc i n cquilihrium.
As a final allernative. ICI 11s now initiate thc ci)rnputation lor the
unknown tensions from the hasic i'C('fOl' equ;ltionc (11' Stiltics. pirxt. U c
must expresh all forccs i n \'ectur notation.
?<.=
C,,
TI, (-sin 1 7 ' i - c o h 17" j j
= T,,+ (\in
SO" i
-
cos So" . j )
We get the lollowing equation when the \'eclor s u m of the lorccs i s sct
equal lo zero:
%(. (-.6OlXi
- .7YXbj)
+
r;, (.766Oi
-
,612X.j)
+
I,OOOj = 0
Choosing point A. tlie point OF concun-ency, we clearly scc thiit the \uni of
moments 01 thc fiirces about this point i s zero. s o the \econd hasic equation of equilibrium i s intrinsically satisfied. Wc now regroup tlic terms o i
the preceding cquiition i n tlie following niiinncr:
(-.601X?;,.
+
.76hO3,,)i
+
(-.7'>863,
-
.h4?X7;,j
+
l.OO0j.j = 0
To satisfy this equation. each of tlie quantities i n parentheses must he wr(i.
This gives the ccalar equations t i i ) and (h) sliilctl e:irliei-. fi.oiii which the
scalar quantities
!;,j
iintl
Cc.cim he sol\cd.
Thc three dtcrniitive inelhods of solutivn iue apparently 01equal uyclulness in this simple pri)hlcin. Howcvcl-. the lcircc p ~ i l y g o ni h only (if practical use for tlirce c01icur1~111
c ~ i p l ~ u iforces,
ar
where thc trigi)nomctric properties
01;I triangle can he directly uscd. The olhcr method\ can be readily cxtended
to inure complex concurreiit pi~ohlcms.
169
SECTION 5.6 PROBLEMS OF EQUILIBRIUM I
Find the forces in cables D B and CB in Fig. 5.24. The 500-N force is
parallel to the y axis. Consider B to be a hall joint located in the xz plane.
Rod A B is a compression member, with a hall joint at A .
In Fig. 5.25 we have indicated the forces acting on joint B. Clearly
we have a three-dimensional concurrent force system with three unknowns. We cim readily determine the unknown forces here hy simply setting the sum of the forces equal to zero. However, since we only want the
forces in the two cables, we shall proceed by setting the moment of the
forces about point A of rod A B equal to zero, thereby not including
the forcc in member AB.
Denoting the force in BC as
and the force in BD as
we
proceed inow to establish the rectangular components of these forces.
c.
==)
+ 9 j + 5k
-1%
-,
-15
+ 92 + S’
+ S j + 13k
\ 152
+ 5’ + I?’
1152
Y
A
G,
=
Tc(-.824i
+ .495j + ,275.k) N
=
T,(-733i
+
244j
+
63%) N
Figure 5.24. Rod A B and cables
CB and LIB support a 500-lb force.
The poyition vector that we shall use for the moment about point A
i b % H . which is
‘aH = 1Si
+ 5j -
5k rn
We now set the moments ahout point A equal to zero.
E M A= O
(1%
+
S j - Sk) X [T,.(-.624i
7;,(-.733i
+
244j
+
+
,495j + ,275k) +
.63Sk) - SOOJ] = 0
We simplify the calculations further by noting that cable BD has a direction inclined to the plene A C B in which the other three forces lie. This can
only mean that the force 7, must have a zero value. Hence, deleting this
force in the above equation and carrying out the cross products, it is easy
to get the remaining nonzero force. We thus have
q=649N
&=ON
Figure 5.25. Free-body diagram of joint E .
170
CHAPTEK 5
F,QLIAlIONS OF EQUILlt(KII!M
Example 5.5 (Continued)
Finally. as indicated at the outset. we could proceed hy summing
Sorces in the coordinate directions. The resulting scalar equations for all
three unknown forces are
-0.8247;
-
0.7337;,
+ 0.90ST4 = 0
+ 0.2447;, +
(3.2757;. + 0.6157;, 0.495T.
0 . 3 0 2 ~ ;= 500
0.3027; = 0
We may now solve the sitnultaneuus equations using C'rrinwr'r d r . Thus
we calculate first the determinant of the coelficicnts of the unknown\. Thus
I
I
-0.824 -0.733
0.905
0.495
0.244
0.302
0.275
0.635 -0.302
To calculate
T . we pruceed as follows:
=
0.272
0 -0.733 O.YO5
1500 0.244 0.302
Note that the first column of the deterniinant consists of thc right side
terms of the set of simultaneous equations i n placc 01 the coellicients of
the desired unknown. We can solve for the other unknowns similarly. We
then would have the comprcssive h r c e in member AB which i b 541 N .
I Case B. Coplanar Forces System. 1 we h;lve shown that the sinip l e g resultant for a coplanar force system (see F i g 4.14) is a single force or
a single couple monicnt nomial to the plane. Thus. to ensure that the resultant
furcc is 7.ero. we require lor a coplanar system in which all forces are in the
~ v xplane:
(b-<J , = 0 .
( F ; J, = 0
(5.10)
c
To ensure that the resultant couplc moment is zero. we require for mninenls
abuut any axis parallel to the z axis:
We conclude that there are three scalar equations olequilibriuni for a coplanar lurce system. Other combinations, such as two moment equations for twu
axes parallel to the .: axiz and a single force summation. il properly chosen,
may bc employed to give the three independent scalar equations of equilihrium, as was discussed in case A .
SECTION 5.6 PROBLEMS OF EQUILIBRIUM I
Example 5.6
Figure 5.26. A car i s being towed up an incline at a constant speed.
A car shown in Fig. 5.26 is being towed at a steady speed up an incline
having an angle of 159 The car weighs 3,600 Ib. The center of gravity is
located as is shown in the diagram. Calculate the supporting force on each
wheel and force 7:
Y
\
3,6W Ib
N,
Figure 5.27. Free-body diagram of the car
A free-body diagram of the car is shown in Fig. 5.27. The forces N,
and Nz are the total forces, respectively, for the rear wheels and the front
wheels. Note, because the wheels are rotating at cnnstant speed, there are
no friction forces present. We have thus formed on this free body a coplanar
force system involving three unknowns and hence the unknowns are
solvable by rigid-body statics. Using axes tangent and normal to the incline
we have
171
172
CHAPTER 5
EQUATIONS OF liQllll.lRRIUM
Example 5.6 (Continued)
C F ,= o
T - WsinO =
.:
o
T = i3,hOO)(siii 15") = 931.7 Ih
i
1
T = 931.7 1h
+io11 7)(1 5 ) l = 1.785 Ih
From Eq.( I ) . we can now per N , . Thu,
N, = 3,477
-
N2 = 3.477
-
1,785
=
1,692 Ib
Clcarly each rear whccl has acting on it 21 normal fbrcr OF N,l2
and each front wheel has ii normal force o l N , l 2 = XY2.5 Ih.
..
=
846. I Ib
Rear wheel support force = 846.1 Ib
Front wheel support force = 892.5 lb
We may now check this soIuliu~iby using a rcdundant equation uf
equilibrium. Thus
We have here il roundoff error, which wc can scccpt fix thz accuracy ol
the calculations Laken in this pnrblcm.
SECTION 5.6 PROBLEMS OF EQUILIBRIUM I
Example 5.7
A frame is shown in Fig. 5.28 in which the frictionless pulley at D has a
mass of 200 kg. Neglecting the weights of the bars, find the force transmitted from one bar to the other at joint C.
D
Figure 5.28. Loaded frame
we form the free body of bar
To expose force components C, and Cy,
BD. This is shown as F.B.D. I in Fig. 5.29. It is clear that for this free body
we have six unknowns and only three independent equations of equilibr i ~ mThe
. ~ free-body diagram of the bent b a AC is then drawn (F.B.D. I1
in Fig. 5.29). Here, we have three more equations hut we bring in three
Y
FBD 1
F B D I1
-
1.962 N
5.000 N
F.B.D. 111
Figure 5.29. Free-body diagrams of frame parts
41t should be noted that it is possible to have situations wherein there we more
uriknowns than independent equations of equilibrium for a givm free body, but wherein
some of the unknowns--perhaps the dcsircd onea--can be still determined by the equations
available. However, not all the unknowns “ I the free body can be solved. Accordingly, be
alert for such situations, so as to ,minimize the work involved. In this case. we must consider
other free-body diagrams.
113
I14
("AFTER
5
FQLIATIONS OF FQIIII.IHKIUM
Example 5.7 (Continued)
more unknowns. Finally, the tree-body diagram 0 1 the pulley (F.B.D. 111 in
Fig. 5.29) gives threc more equations with no additional unknowns. We
now have nine equalions available and nine unknowns and can proceed
with confidence. Since only two ot. the unknowns are desired, we shall
take select scalar equations from each of the free-hody diagrams to arrive
at the components C , and <.! most quickly
From F.H.D. 111:
Therefore.
Therefore.
-I ,962
-
s.000
+
0,= 0
Therefore.
11, = 6.962 N
From F.B.D. I:
Therefore,
Fmnz F.B.D. 11:
-(1.3)(14) - (7')(3.l)
-
C,(4)
+
C s ( 2 . S )= 0
Thereiore,
C, = 24.300 N
We can give the Sorce at C (transmittcd from bar AC lo bar B D ) as
C =24,W
+
11.313jN
5.17. In a tug of war, when team B pulls with a 400-lb force,
huw much force must team C exert for a draw? With what force
does team A pull'?
5.20. Find the tensions in the three cables connected to B. The
entire system of cables is coplanar. The roller at E is free to turn
without resistance.
A
Ropes tied to a ring
-
Y B
B
C
Figure P.5.17.
Figure P.5.20.
5.18. Find the tensile force in cables AB and CB. The remaining
5.21. A 700-N circus performer causes a .15-m sag in the
middle of a 12-m tightrope with a 5,000-Ninitial tension. What
cables ride over frictionless pulleys E and F.
additional tension is induced in the cable? What is the cable tension when the performer is 1 m from the end and the sag is .I2 m?
Figure P.5.21.
-1
Figure P.5.18.
5.19. Find the force transmitted by wire BC. The pulley E can be
assumed to be frictionless in this problem.
5.22. A 27-lb mirror is held up by a wire fastened to two hooks
on the mirror frame. (a) What is the force on the wall hook and the
tension in the wire? (b) If the wire will break at a tension of 32 Ib,
must the wall hook be moved (i.e., the wire lengthened or sholtened and the 4-in. rise distance changed)? If so, to what point?
Wall hook
Figure P.5.19.
Figure P.5.22.
17.
c(M,),
= 0. Axe\ d and e arc
/'
in,>[
~ ~ i l i t i l l ctol the
A;
plane.
Moreover, the axe, are oricntcd so that the linc 01 action 0 1 the
re\ultanl force cannot inter\rct hoth axcs.
5.24.
Cylinders A and H weigh 500 N ciich mil c)llndc~. ( '
weighs 1,000 N. C m ~ ~ p ~alli l contact
e
force&.
'%
/i
Figure P.5.25.
Figure P.J.2X.
5.29. Find the supporting force systems far the beams shown.
Note that there is a pin connection at C. Neglect the weights of
the beams.
v
. ... .. ...
1.000 N
I00 N/m
n--I
Figure P.5.31.
5.32. What are thc supporting forces for the frame'! Neglect all
wights except the IO-kN weight. Disregard friction.
klTlTL
8m
Figure P.5.29.
5.30. Find the supporting force systems at A and B. The length
of C R is 8 m.
R
Figure P.5.32.
5.33. Find the supporting forces at E and K Pulleys A and B
offer n o rotational resistance from friction at the beerings.
A
Figure P.5.30.
5.31. What are the supporting forces at A and D fbr the frame
shown? What are the forces in members AB, BE. and BC?
Figure P.5.33.
177
Figure P.5.34.
35. An e l a l i c cord AH i \ .just tau1 beliore
1,000
force i \
iplied. I f il takes 5.0 N / m m of elongalion of the card, what is the
r l i i i ) i i 7'in the curd after the I ,000-N force i s applied'! Set up the
Iuation for 7'hut do ~101solve.
Figure P.S.35.
36. A thin hoop o1 radius I 1x1 and weight 500 N rests on an
:line. What friction force/ at A i s needed for this configun~atiun'?
hat i s the tension iii wire CB?
Kigure P.5.36.
8
k ipure P.5.39.
5.40. What is the supporting force system at A for the cantilever
5.43. Find the supporting force aystem at A
beam? Neglect the weight of the beam.
D
I' -kid
Figure P.S.40.
5.41. In Problem 5.40, find the force system transmitted through
the crms section at B.
A
Figure P.S.43.
5.44. A light bent rod AD is pinned to a straight light rod CB at
C. The bent rad supports a uniform load. A spring is stretched to
connect the two rods. The spring has a spring constant of IO4 N/m,
and its unstretched length is .8 m. Find the supporting forces at
5,42. A
beam AB is pinned at to a simply suppone,jA and B. The force in the spring is I O 4 times the elongation
beam RC. For the luads given, find the supporting force systcm at in meters.
A . Determine force components that are normal and tangential to
the cross-section of beam AB. Neglect the weights of the beams.
Y
-
2m
Figure P.S.42.
Figure P.S.44.
175
5.45. Light rods A I ) and HC are pinncd together at C and supw i t a ?OWN and a IOO-N Indd. What arc the supporting forces a1
I and H?
5.47. Sulve fbr the supponing f r ~ e at
s A and (T. AB weighs 100
Ih, and RC weighs 150 Ih.
300 N
Figure P.5.45.
Figure P.5.47.
i.46. A light rod CI) is held in a horiiontal position hy a strung
,lastic hand AH (shock cord), which acts like a spring in that it
akrs IO3 N per meter felong gat ion vfthe band. The upper part of
he band i s connected to a small wheel free to roll on a horizontal
llrfdce. What is the angle a needed tu suppoit a 200-N l w d
.s shown?
Figure P.5.46.
80
5.48. What torque 7 i s ncsded to maintain the configuration shown for the cumprcssor if p , = 5 psig'! The system lies
horizontally.
Figure P.5.48.
5.49. Wurk Problem 5.48 for the system oriented vertically with
BC weighing 3 Ib and CD weighing 5 Ib.
5.52. Find the supporting forces at A and G. The weight of W is
500 N and the weight of C is 200 N. Neglect d l other weights.
The cord connecting C and D is vertical.
5.50. Neglecting friction, find the angle p of line AB for equilibrium in t e n n s of ai,
a, W,, and W,.
Figure P.5.52.
Figure P.5.50.
5.53. What torque T is needed for equilibrium if cylinder R
weighs 500 N and CD weighs 30U N?
5.51. If the rod CD weighs 20 Ib, what torque T i s needed to
maintain equilibrium? The system is in a vertical plane. Cylinder
A weighs 10 Ib and cylinder B weighs 5 Ib. Disregard friction. At
D there is a slot.
30"
Figure P.5.51.
Figure P.5.53.
A har A l l i s pinned tn twn identical planewy g a r s ellch of
jiameler 311 in. Gear 1: is pinned t n har AI3 and meshes with the
wo planctary pears. which i n turn mesh with statimary gear I). II
1 tirrquc 7 ' 0 1 100 N ~ im
s applicd 10 bar AH, ~
1 c't~
~ c m1i l ltorquc
s ncedcd t o he applied t u thc upper planctary g c x 10 iniiintain
5.54.
A
I
I
I.5
111
.
Figure P.5.54.
3.121 m
i.56. A Bucyrus+Eris tianhi1 crilnr i s holding il chimnc) having
I weight 0120 kN. The chirnnq is held by il cahle that g w \ w e r
L pulley at A , thcn gee. i w c i a cccntid pullry at I). and thcn to a
uinch ill K . The position of baom AI! (on to,,) is inaintaincil hy
WII heparate cahles, one from A t o /I. and the other from R to p~kIey C. Find the tensions in cohle5 AH and RC. Nuts that HC is oii,ntcd 30" from [he vertical f o r the setup show!. Consider only ttir
veight of the load and neglect frictim.
82
/
Figure P.5.57.
SECTION 5.7 PROBLEMS OF EQUILIBRIUM 11
5.7
Problems of Equilibrium II
I
lease c.
Parallel Forces in Space. In the case of parallel forces
in space (see Fig. 4.20), we already h o w that the simplest resultant can be
either a single force or a couple moment. If the forces are in the z direction,
then
=0
(5.12)
ensures that the resultant force is zero. Also,
guarantees that the resultant couple moment is zero, where the .r and v axes
may be chosen in any plane perpendicular to the direction of the
Thus, three independent scalar equations are available fur equilibrium of parallel forces in space.
A summary of the special cases discussed thus far is given below.
For even simpler systems such as the concurrent-coplanar and the parallelcoplanar systems, clearly, there is one less equation of equilibrium.
System
Concurrent
(three-dimensional)
Coplanar
single couple moment
Parallel
(three-dimensional)
Simplest Resultant
Single force
Number oJ Equations
for Equilibrium
3
Single fkce or
3
Single force or
single couple moment
3
'For parallel forces in the I direction, a simplest resultant consisring ot a couple moment
only murl have this couple moment parallel to the xy plane (see diagram). Recall from Chapter 3
that the orthogonal q z components of C, equals the torque of the system about these axes.
Hence. by setting x ( M r ) ,=
= ( I , weareensuring thatC, = 0.
C(Mv),
183
184
CHAPTER 5
I~QUATIONS< ) F F.QlllLlUKlllM
Example 5.8
Dcteriniiie ttic ~ W C C S required to suppoi-t (hi' unifi)rm heirin in Fig. 5.30
showii Iwded with ii couple. a point force. iiiid ii downw;ird par;ihi)lic (listrihutim of lo:ld hnring 7er-0 slope :it the origin. Thc weiftit o1 thr, hzam i s
I O 0 Ib.
&~
20'
~~~
~
--
lO(1 Ihiul.1
Figure 5.30. Find \ u p p < r l i n gforce\
Since ii couple ciin hn rotated w i l l i w t afleclins Ihc cquilihriurli of
the hody. we ciiii orieiil lhc couple s o llliil the lorccs arc vertic:il. AccoIdingly. wc have here ii hcam loaded hy ii system 01 parallel coplanar loads.
Clearly. Ihc siippoi-lin$ forces imusi hc wrtical, i i h shown ill F:ig. 5.31.
wlicre %'e havc ii frcc-hirdy di;igl-aiii of ttic hciuii. Siiicc tlicrc ;ire imly IH(I
unknown quantities. UT ciin handle the prohlcm hy \Liiticid considcralion
01this free body.
H,
KI
Figure 5.31. f+cc,-hodj di:igr:wi
+
The cquatinn Ibr thc loading curve must he I V = m'
h. whcrc o
and I, are ttr IRdctcriniiicd l i r ~ nthe
i Ir,;iding ilat,~and tlic c h k c iilrrfereilce.
With an .qrcfcreiicc ;it thc lelt end. tis shown, M'U then havc the conditions:
1. W h c n x = (1.
2. When v
=
bt' =
20,
11'
0.
=
400.
To satisfy these cmiditimih, h niust he /ern ;ind (I niusl hc uiiity: the Itxiding
liinctiim i\ lhus given a\
II'
= ~ x ' lhifi
SECTION 5.1 PROBLEMS OF EQUILIBRIUM I1
Example 5.8 (Continued)
In this problem, we shall agein work directly with the scalar equtitions. By summing moments about the left and right ends of the beam, we
can then solve for the unknowns directly:
-500 - (10)(100) - (15)(500) -
I,
2u
xy dx
+ 20R2
= 0
Replacing y by xz, then integrating and canceling terms, we get
-9,000 -
4'!:'I+
20R, = 0
By inserting limits and solving, we get one of the unknowns:
R, = 2,450lb
Next,
-20Ri
- 500 + (10)(100) + ( 5 ) ( 5 0 0 )+ 1,:(20
- x)ydx
= 0
Replacing y by x2, integrating and then solving for R , , we have
R , = 817lb
As a check on these computations, we can sum forces in the vertical
direction. The result must he zero (or as close to zero as the accuracy of
our calculations permits):
R,
+ K2 - 100 - 500
~
I
20
U
x 2 d,x = 0
Therefore,
2,667
-
2,661 = 0
Always take the opportunity to check a solution in this manner (i.e., by
using a redundant equilibrium equation). In later problems, we shall rely
heavily on calculated reactions (supporting forces); thus, we must make
sure they are correct.
185
186
C H A PT ER 5
E Q UA TI O N S OF FQIIII.IBRIIIM
Example 5.9
I n Fig, 5.32. find the suppnrting I i r c e s at A. D. and I ) . Note thc pin
ncction at C. Also inole that at I<we h a w n weldcd connection
~(111-
Y
Figure 5.32. h l c m h c n . A ( ' and ( ' I ) arc pinncd 31 ('.
The free-hody diagram for thc entire system is shown i n Fig. 5.33.
We have a coplanai s y s k m (ifforces fkr this free body and s o w e hmc
only three independent equations of equilibrium. Howcvcr we have Iiei~r
four unknowns. One of the unknowns, namcly A , , can he s e r i i hy itispection to he /,cro lea\'ing now a coplanar parallel system with three
unknowns but with o n l y two equations ofeqiiilihriuin.
\'
I
Figure 5.33. Rcc-hudy d i a g a n i I.(F.R.I). I)
We shell next consider the frcc hody 01 mernhcr [ I ) . This is shown
in Fig. 5.34 where we have simplified the distributed loading i n order to
i i i t i s t he x r o . Thus.
better facilitate the ensuing computakiiins. Clearly.
for F.B.D. I I in Fig. 5.34. we will then haue only two unknmwis lor which
we havc two equations of equilibrium. By raking moments ahout point C'
in Fig. 5.34, we can directly dererniine [Ir.
7'
<',
Figure 5,34,R-ec-hody diagl~am
(F.H.D. 11)
SECTION 5.7
Example 5.9 (Continued)
C M c = 0:
~
~~~
(D,)(IS) - ( 2 c u ) ( l s ) ( y ) -
(:)(1s)(300)[(~)(15)]
=0
We may now go hack to Fig. 5.33 to solve for the remaining two
unknowns. Thus, simplifying the loading between C and D as we have
done i n Fig. 5.34. we have
c
A> + E )
I
+ 3,000 - (200)(34) - 1(300)(1S)
=0
M, = 0 :
~-
+ (3,000)(21)
-A,(13)
~
(200)(34)(2; - 13)
- ~2( 3 0 0 ) ( 1 5 ) [ h + ( ~ ) ( I S )=
] 0
From Eq. (h), we get
And from Eq. (a), we have
,*
4.- 6,069N
Note that A, was negative indicating that we had inserted the wrong
sense for this force in F.B.D. 1. We did not make any changes in the ensuing calculations while going to Eq. (a) (valid for F.B.D. 1) to calculate B,
(i,e., we used the result from Eq. (c) with the negative sign of A, intact).
Also, if you were tempted in F.B.D. 1 to include the force C, at pin
C in this diagram you were flirting with the prime error in the statics of
rigid bodies-namely, including a force which is infernal to the particular
free body drawn.
general case is a force and a couple moment. Six equations of equilibrium
can be given for each free-body diagram. We now examine two examples
for this case.
PROBLEMS OF E Q UILIBRIUM II
187
Example 5.10
A derrick i s shown in Fig. 5.35 supporting a 1,000-1b load. The verticil1
heam has a ball-and-socket coniiection inti, the ground at r l and i s held hy
guy wircs ( I C ' and tic Neglect the weight of the mcinhers and guy wires.
and find the tensiirns in the guy wires u r ' , hr. and ('e.
Figure 5.35. Loaded rlrrriok.
If we select as ti free hody both membcrs and the interconnecting
guy wire w , we shall cxpose two of the desired unknowns (Fig. 5.36).
Note that this i s a general three-dimensional h r c e system with iinly five
unknowns." Allhough all these unknowns can be solved by statical considerations of this free hody, you will notice that, if we take monicnts ahout
point d , we w i l l involve in a vector equation only thc desired unknowns
Th,.and T,, . Accordingly. all unknown for-ces need riot he computed kir
this free-body di;ipr;im. You should always look for such short cuts in situations such as this.
To determine the unknown tension ?;.<,, we musl employ another
free-body diagram. Either the vertical cor horimntal member will expose
this unknown i n it manner susecptihle ti) solulion. The latter has been
selected and i s shown in Fig. 5.37. Note that we have here a coplanar force
system with ttircc unknowns. Again. you ciln Fee that, by taking moments
about p i n t , / . wc will involve only the desired unknown.
'% \ h i ~ o l dhc clciii on 8sipection ,,I big. 5.35 Ihiit. duc 11) symnclry. forma in the ~ W O
supp<,niog ~ i i h l ~nwst
b
he eqiiill. Usmg i l i i s intoimiitmr1 i\ l i i n t i i r n ~ ~ i nt ol using cine of thc
cquiitiow d c q o i l i h r i i ~ i n Hiwcrcr.
.
ror praclice we w i l l not uhc this inforination and we w i l l
b u l \ c lor h o l h 0 1 ihcxl cnhlcs and d e m ~ ~ n w alhcir
t c cqu"liiy. Also, iiutc that there i s it sixth
cquiiliori d c q u i l i h r i i i r t i llijll i, iklcniicnlly \ulialicd. To hcc this. look a1 rnooients of the forcer
about a x ~ qmi i n E g . 5.36 Why i s Ihc l ~ t i t iiioiiiciit
l
dcnticidly quill i m zero ahuut this axi,,
lhuc denyiiip \ I \ im q u i l t i o n io hclp ~ I V C for unLnrwns'! I \ the derrick complcicly coiictraiocd" F.xplisin
Figure 5.3h. F'er-budY
diagram I.
4
P
6;
I,OOO Ih
Figure 5.31. Free-body diagram 2.
SECTION 5.7
Example 5.10 (Continued)
The vector T,, may then be given as
Similarly, we have for Tbc,
Using the free-body diagram in Fig. 5.36, we now set the sum of
moments about point d equal to zero. Thus, employing the relations above,
we get
13k x __
'' ( S i
m
- l O j - 13k) + 13k
- l O j - 13k)
x - (Tbc
-Si
x'333
+ l O j x (-1,000k) = 0
When we make the substitution of variable
f2 = Tb<
the preceding equation becomes
/m3,
tl =
[130(tl t t2) - lO,OOO]i t [104(tl - t 2 ] j = 0
(c)
and
(d)
The scalar equations,
130(tl
+
10,000 = 0
104(tl - t2) = 0
fz) -
can now be readily solved to give f l = f2 = 38.5. Hence, we get
= 38.5 ~ 6 %
= 702 Ib and qc = 38.5 (333 = 702 lb.'
cc
Turning finally to free-body diagram 2 in Fig. 5.37, we see that, in
summing moments about f,the horizontal component of the tension q,has
a zero-moment arm. Thus,
(10)(0.707)T,,
~
(IO)(l,OOO)
= 0
Hence,
'By taking moments in Fig. 5.36 about the line connecting points a and d (see Fig.
5.35). we could get Tbcdirectly using the scalar triple product. We suggest that you try this.
PROBLEMS OF EQUILIBRIUM II
189
190
CH.\PIt% 5
EQlli\llONS OF CQLIII.IHKIIIM
Example 5.11
A blimp i s \ h o w in Fig. 5.38 fixed at Ihc mooring lower n by a
ball-ioint coiiiieclion. and held by ciihles A B and A<'. The lhlimp has a
ma!.\of l,SOO hg. Thc s i m p l c b l resultirnt forcc P from air prcssure (including the effect\ 01 wind) is
F = I7.500i
+
I .OOOlj
+
I ,500k N
at ii position hhown in the diagram. Compute the tensior in the cables a s
well as the I0rcc tran\initled L o the ball ,j(iint at the top 111 the towx i i t D.
Also, what Sorce syslcni is transinittcd to the ground at G through the
mooring tower? l h e towcr weighs 5.000 N.
Figure 5.38. Telhrl-cd hliiiip.
We shall first consider ;I frec-body diagram of the hlinrp. as shown
i n Fig. 5.34. We have five unknown 1orcez hcrc. and we can d v e all 01
tlieiii by using equation!. ~Sequilihriulnfor this free
As it Sirst step,
we express the cahlc tensions vectorially. That is.
We now gii hack to the basic vector cquation ~ I e q u i l i h r i u n i 'l'hur
.
Figure 5.39. Fize-body diagram OF
hlimp.
SECTION 5.7
I
Example 5.11
(Continued)
The scalar equations are
DL + 2,785
+
D. +
+
.447%, = 0
1.500 - .359%, = 0
1,000
(a)
(b)
(C)
Next take moments about point D
1 3 j x (9.81j(I,SOO)(-i)+ I h j x (17,500i + I,OOOj + 1,500k)
+ 2 9 j x TA,(-.933i - .359k) + 291 x TA,(-.894i + ,4471) = 0
Carrying out the various cross products, we end up nnly with k and i components, thus generating two scalar equations.' They are
-10.41T,,.
25.9%,
+
191
I
- .933%,. - .894T,, = 0
Dy
PROBLEMS OF EQUILIBRIUM I1
+
24,000 = 0
(d)
27.I%,. - 88,700 = 0
(e)
We now have five independent equations for five unknowns. We can thus
solve these equations simultaneously. From Eq.(d), we have
T,, = 2,305 N
From Eq. (e), we have
TAB= 1,012
From Eq. (cj, we have
Dz = -673 N
From Eq. (b), we have
D, = -1,452 N
T h e lhird equation is 0 = 0.That is, there are no moments ahout they axis. because
all forces pass through the Y axis.
Hcncc.
Frim
tliih.
we gcl
Mb= 0
M , = 17.470 N-III
M = -37,xoo N-Ill
!
We ciiii conclude
ground is
tliiit. iil the ceiiler
I' = 5.2701'
i
hilhc.
l,4S2j
~
C = 17.47Uj
i
ol thc
-
-
37.XOOk
the Siircc systcm from the
h72k N
N-iii
Thc Ibrcc system acting on the gr<~und
iit the ccntcr 0 1 tlic hasc i\ tlrc reaction t o Llie sy\te~iiahiibc. Thus.
6
<,,>
,,,,
= -.5,27Oi
C,cc,t,,, = -17,470j
+
1.452j
+
+
612kN
37,800k N-m
5.58. The triplc pulley sheave and the double pulley sheave
weigh I S Ib and 10 Ih, respeclively. What rope fbrce is necehsary
to lift a 350-1b engine'! What is the force o n the ceiling hook!
I
Counter
welght
I
Figure P.5.60.
5.61. A Jcep winch is used to raise itself by a force of 2 kN.
What are the reactions at thz Jeep tires with and without the winch
load'? The drivcr weighs 800 N. and the Jeep weighs I I kN. The
cenler of gravity of the Jeep is shown.
Figure P.5.58.
5.59. A multipurpose pry bar can be used tn pull nails in the
three positions. If a fbrce of 400 Ib is required to remove a nail
and a carpenter can exen SO Ib, which position(s) must he use'?
I-
t ~ . ~ m + )
1.3m
3.1 m
Figure P.5.61.
5.62. A diflerrnriul p d l q is shown. Compute F in terms of W ,
r, . and rz.
~
I .25"
c
F
Figure P.5.59.
At what position must the operator of the cuunterwcight
crane locate the SO-kN counterweight whcn he lifts a IO-kN h a d
of steel?
5.60.
Figure P.5.62.
19
5.63. What is the longest portion of pipe weighing 400 Ib/ft that
can he lifted without tipping the 12,000-lh tractor'! Take the center
of rravity. of the tractor at the L'eonietiic centcr.
5.66. An 1-hcam canrilevcred out Srom a u d l wcighs 30 lhift and
supprim a 700-lh hoist. Steel (4x7 Ihilt'l TIIVCI plates I iii. thick
arc uvlded on thc heam iirx the wiill lo incieiice the mmientCarl-yinp capacity of the heam. What arc the reiictioiis at the u.all
when a 400I-lh In;id is hoisted at thc out~rmostpmition of the
hoist'.'
I,>,\,
>P>p
It-
8'-+
I
Y/
<y+
-5'-l
9
Figure P.S.63.
5.64. The L-shilped concrete post supports an clevated railroad.
The concrete weighs 1501 Ih/W What are the reactions at thr hese
of the post'!
Figure 1'5.66.
5,67,
shr,wn
the
fc,rcc
fill-
ci
~
,,,,
iltlvcI.he;
(.,
40.000 Ih
r
i
5.68. Find the \upporting force sysrcm lor ttir cmtilewr heams
cmncctcd to har A H hy pins.
-8
Figure P.5.64.
5.65. Two hoists arc operated on the same overhead track. Hoist
A has a 3,000-lh load, and hoist R has ii 4.000-lh load. What are
rc.cti~,ns
the ends c,c Lhr trach when the hoi,ts arc i n [he
position shown?
I
Figure P.S.6X.
The tciilcr weighs SO kN and is loadcd w i t h crate\ wigtiiiic thc rriictions at thr ,car wheel ;ind
rm the tractor at ,I..'
5.69.
i n s ')O kN and 4 1 k N . What
Figure P.5.65.
194
I'igure P.S.69.
5.70. What load W will a pull P of IO0 Ib lift in the pulley system? Sheaves A, H, and C weigh 20 Ih, 15 Ih, and 30 Ib, respectively. Assume first that the three sheaves are frictionless and find
W, Then, calculate W that can he raised at constdnt speed for the
case where the resisting torque in each of sheaves A and H is .01
times the total force at the bearing of each of sheaves A and H .
5.73. A 20-kN block is being raised at constant speed. If there is
no friction in the three pulleys, what are forces F,, 4 , and F7
needed for the job! The block i s not rotating in any way. The line
of action of the weight vector passes through point C as shown.
Figure P.5.73.
Figure P.5.70.
5.74. A 10-ton sounding rocket (used for exploring outer space)
5.71. A piece of pop art is being developed. The weight of the
body enclosed by the full lines is 2 k N . What is the smallest distance d that the artist can use for cutting a .5-m-diameter hnle and
still avoid tipping'? The body is uniform in thickness.
4 m *-2
I
m-1
I
- = t
i m
has a center of gravity shown as C.G.,. It is mounted on a
launcher whose weight is SO tons with a center of gravity at C.G.,.
The launcher has three identical legs separated 120" from each
other. Leg At4 is in the same plane as the rocket and suppolting
arms CDE. What are the supporting forces from the ground? What
torque is transmitted from the horizontal arm C D to the ramp ED
by the rack and pinion at hinge D to counteract the weight of
the rocket'?
='I
Figure P.5.71.
5.72. What is the largest weight W that the crane cdn lift without
tipping? What are the supporting forces when the crane lifts this
load? What is the force and couple-moment system transmitted
through section C of the beam? Compute the force and couplemoment system transmitted through section D . The crane weighs
1 0 tons, having a center of gravity as shown in the diagram.
50'
'c
20
e
~
I
Figure P.5.72.
Figure P.5.74.
195
5.75.
A door i s hinged at A and N and contains watcr whosc specific weight y is 62.5 Ihift'. A firrce 1.' nornral to the dnor keeps
the door closed. What are the forccs on the hinges A and II and thr
force F to ctrunteract the water? As noted in Chapter 4, thc preisure in thc water above atmosphere i s givcn as yd. where rl is the
pcrpendicular distance from the k e surface of the wjatcr.
A
7
111
1
t '
Figure P.5.77.
5.78.
Figure P.S.75.
Find the supporting force and couple-moment system for
the cnntile\icr hcam. What i s the force and couple-moment system
Irammitted through a cross section of the heam at R'!
5.76. A row of hooks of length 750 niin and weighing 200 N sits
.m a three-legged tahle as shown. 'The legs are equidistant from
:ach other with one leg N coinciding with the y axis. l'he crthcr
:wo legs lie along a line parallel to the ,t axis. If thc tahle weighs
100 N,will i t tip'! If not, what hie the fhrces on the legs'?
I
500 Ih
Figure P.S.78.
5.79. A
structure i s supported hy a ball-and-socket joint at A , a
pin connection at N offering no resistance in the direction A B , and
a himplr roller support at C. What are the supporting forces fcrr the
lnads shown'!
Figure P.S.76.
i.77.
A small helicopter i s in a hovering m a n c u v c ~ ~The
. lhcliopter rntor hladcs givc a lifting force I ; hut therc result\ from the
ir forces on the hlades B lorquc c', . The rex rotor prevents thc
ielicupter from rotating ahout the i axis hut drvclaps n torquc C , .
:ompute the force F; and couple C, i n terns of the weight @
iow arc F, and (I, related'!
96
+
2(H) kN
Figure P.5.79.
5.80. Compute the value of F to maintain the 200-lb weight
shown. Assume that the bearings are frictionless, and determine
the forces from the bearings on the shaft at A and B.
5.83. Determine the vertical force F that must be applied to the
windlass to maintain the 100-lb weight. Also, determine the supporting forces from the bearings onto the shaft. The handle DE on
which the force is applied is in the indicated XI plane.
F
Figure P.5.80.
Figure P.5.83.
5.81. A bar with two right-angle bends supports a force F given as
F
=
10i
+
3j
+
IOOkN
If the bar has a weight of I O N/m, what is the supporting force
system at A'?
5.84. A transport plane has a gross weight of 70.000 Ib with a
center of gravity as shown. Wheels A and B are locked by the
braking system while an engine is being tested under load prior to
take off. A thrust T of 3,000 Ib is developed by this engine. What
are the supporting forces?
Figure P.5.81.
5.82. What is the resultant of the force system transmitted across
the section at A? The couple is parallel to plane M.
V
i
-1
I.zm+
Figure P.5.82.
Figure P.5.84.
5.85. Two cahles GH and K N support
a rod A R which connCCtC
to a ball-and-socket joint wpporl at A and q q m r l s ii 500-kg hady
C at R . Whal are the tensioni in the cahle and the supporting
forces at A’!
Figure P.S.87.
What force P is needed tu huld thr door i n a hwimntal
pohilion? ‘lhc door weighs SO Ih. Dcterminc the SuppOl-li~lf
l o ~ w c :it
\ A and 8.At A there i s a pin and at /I therc i\ il hall-and-
5.88.
mckrt
,joint.
Figure P.5.85.
5.86.
I
What changc in elentirm fkr thc 100-lh weight will a coiiple of 300 Ih-ft support if wc neglect li.ictian in the hearings :it A
and R? A l w , determine the suppolling fbrce cwnponcnts at the
bearings for this crrnfiguration.
Figure P.S.88.
*S.RY.
A onilbnn has of length / a n d weight 1V i s connecled to
the ground hy a ball-and-wckct iomt, and r a t s on a wnicylindcr
i r o n which i t 15 not allowed to slip down hy n wall at U . If we
crinsidcr the wall and cylindcr 10 he frictiotilcah. dctcmiine the
supporting I r c z c at A for the follou,ing data:
I
<
,’
= I m
= 30 111
_I
.?O
111
h
= ..101n
W = 100N
I
Figure P.5.86.
Dctcrmine thc frircc P requircd to ep the 150-N
or of
an airplane open SO” while in Ilight. Thc Sorce P IS excrfed ill il
dircctinn normal to the fiiselagc. There is a i i ~ prersure
t
increase
o n thc ootcide s u r f x x of .02 N/,nmz. A l w . determine the supporting forces at thc hinge&.Consider thal the top hinge supports any
vertical force on the door.
5.87.
198
Figure P.5.89.
SECTION 5.8 TWO POINT EQUIVALENT LOADINGS
5.8
199
Two Point Equivalent Loadings
We shall now consider a simple case of equilibrium that occurs often and
from which simple useful conclusions may readily be drawn.
Consider a rigid body on which the equivalent systems of two separate
forces are respectively applied at two points a and b as shown in Fig. 5.41. If
the body is in equilibrium, the first basic equation of statics, 5.l(a), stipulates
that F, = -F2;that is, the forces must be equal and opposite. The second fundamental equation of statics, 5.l(b), requires that C = 0, indicating that the
forces be collinear so as not to form a nonzero couple. With points a and b
given as points of application for the two forces in Fig. 5.41, clearly the common line of action for the forces must coincide with the line segment ab. Such
bodies, where there are only two points of loading, are sometimes called
“two-force” members. Such members occur often in structural mechanics
problems. Furthermore, there is considerable saving of time and labor if the
student recognizes them at the outset of any problem.
a
4
Figure 5.42. Compression
and tension members
b )
Figure 5.41. Two-force member.
We often have to deal with pin-connected structural members with
loads applied at the pins. If we neglect friction at the pins and also the weight
of the members, we can conclude that the equivalent of only two forces act on
each member. These forces, then, must be equal and opposite and must have
lines of action that are collinear, with the line joining the points of application
of the forces. If the member is straight (see Fig. 5.42). the common line of
action of the two forces coincides with the centerline of the member. The top
member in Fig. 5.42 is a compression member, the one below a tensile member. Note that the bent member in Fig. 5.43, if weightless, is also a two point
loaded member. The line of action of the forces must coincide with the line
ab connecting the two points of loading. However, the beam in Fig. 5.44 is
not a two point loaded member since at the left end there will be a couple
moment, And, clearly, such a loading must entail two points of loading to
accommodate the two equal and opposite forces comprising the couple. There
are then in effect three points of loading for this member. Accordingly, use
caution here when dealing with a cantilever beam.
Before considering an example, it should be emphasized that the forces
4 and 5 in Fig. 5.41 may be the resultants of systems of concurrent forces at
a and b, respectively. Since concurrent forces are always equivalent to their
resultant at the point of concurrency, the member in Fig. 5.41 is still a twoforce member with the resulting restrictions on the resultants & and F2.
Figure 5.43. Line of action
of F collinear with ab.
Figure 5.44. Cantilever beam is not an
example of a two point loaded member.
+
A device lor crushing rock\ ic chown i n Fig. 5.45. A pistiin I ) having iiii
%in. diameter is aclivated hy il pressure 11 of 5 0 p i g ( ~ ~ i i u n dper
h sqliale
inch above that or the atmosphere). Rod?, A B , 11C'. ;1nd Ill) can he considered wcightlcss for this prohlcni. What is the horimnkil force tranhrnitled
at A tn the trapped rock shown i n the diagram'?
We have he]-e thrcc two-force menibcrh coining kigethci- iit 11.
Accordingly. if we i d a t e pin I1 a s a rrec body. we will h;ir.c Ilircc forces
acting o n thc pin. These forces must he collinciir with the cenlerliiieh 01 the
respectivc mcnihers. a s explained earlier (Fig. 5.42).
Thc force I.;>is easily computed by considcring the iictioii 1 i i Ihc
piston. Thus. we get
Figure 5.45. Rock crusher.
/;] = 150)nx' = 2,51(1 Ih
4
1
i
The lorcc transmitted to the rock i n the horimiitiil dircction ic flieii 4.850
cos I S " = 4.690 Ih.
:. Horizontal force transiniltcd
to
thc rock = 4,690 Ib
5.9
I;,
Figure 5.36.trce-hmly dia;rarn
o i pin H.
Problems Arising From Structures
SECTION 5.9 PROBLEMS ARISING FROM STRUCTURES
Example 5.13
Rod C shown in Fig. 5.47 is welded to a rigid drum A , which is rotating
about its axis at a steady angular speed w o f 500 RPM. This rod has a mass
per unit length w , which varies linearly from the base to the tip starting
with the value of 20 kg/m at the base to 28 kglm at the tip. If normal .strexs
is defined as the normal force at a section divided by the area of the section (similar to pressure except that the force can be pulling away from the
section rather than always pushing against the section), what is the normal
stress at any cross section of the cylinder at a distance r from the centerline
B-B of the drum due only to the motion?
,Rod
Figure 5.47. Rods attached tu
C
il rotating
rigid drum
First we expose a section of the rod at a distance r from B-B in a
,frre-bod? diagram as shown in Fig. 5.48 in which we denote the normal
stress acting on this section as rrr.The force from this section must restrain
the centrifugal force stemming from the angular motion of the portion of
the rod beyond position r. For this purpose, we consider an infinitesimal
slice of the rod (see Fig. 5.48). We shall use the variable q to denote the
distance to the slice from the centerline B-B. Furthermore, the thickness of
the slice shall he d q . We shall use rl to position slices bctween end position r of the free-body diagram and the tip of the rod. We are using this
approech for bookkeeping purposes. Now you will recall from freshman
physics that the centrifugal force on this slice is given by
201
202
CHAPTEK 5
EQUATIONS OF EQlJlLIBRlUM
Example 5.13 (Continued)
,XCentritugal turcr
F.B.D
R
B
Figure 5.48. Free hody exposes T,, at
section r.
Note that q is a dummy wriahle.
Consequently, the total centrifugal force for the material beyond position r
in the free body of Fig. 5.48 will he found by integration to be
f,,,,
7
=
WJJ
dq 0
'
Next, note that the term w varies linearly with
w = 20
+
JJ
(b)
as follows:
'1.2
8 kg/m
Clearly from this equation we see that when q = .2, we get a' = 20
kglm and when JJ = .7, we get M' = 28 kg/m. Also, the variation is
linear. Now, going back to Eq. (b), we have
Integrating
f,,,,, = 2.742 x IO'
=
2.742 x 10'[4.116
+ 1.829 - 8.40r2
~
5.333~~1
We can give the desired stress T , by
~ dividing by the cross-sectional area
nD2/4 = 7.854 x 10-'mZ. We thus have the desired normal stress distribution for sections of the rod as follows:
SECTION 5.9 PROBLEMS ARISING FROM STRUCTURES
Example 5.14
Consider a thin-walled tank containing air at a pressure of 100 psi above
that of the atmosphere [see Fig. 5.49(a)]. The outside diameter D of the
tank is 2 ft and the wall thickness t is 114 in. We consider as a free body
from the tank wall a vanishingly small element such as ABCE in the diagram having the shape of a rectangular parallelepiped. What are the stresses
on the cut surfaces of the element? Neglect the weight of the cylinder.
(4
(b)
(C)
Figure 5.49. Thin-walled tank with inside gauge pressure p .
We can examine face BC by considering a free body of part of the
tank, as shown in Fig. 5.49(h) exposing BC. (Note that we have not
included the pressure on the inside wall). Because there is a net force from
the air pressure only in the aria1 direction of the cylinder, we can expect
normal stress r,, (like pressure except that it is tensile rather than compressive) over the cut section of the cylinder, as has heen indicated in the
diagram. Furthermore, because the wall of the tank is thin compared to the
diameter, we can assume that the stress r,,, is uniform across the thickness.
Finally, for reasons of uxiul symmet? of geometry and loading, we can
expect this stress to he uniform around the entire cross section. Now we
may sily from considerations of equilibrium in the axial direction that
L
.:
7
nl
J
(IOO)(24 - $1'
p-( D - 2 t ) 2
==
D2 - ( D - 2 t ) 2 = -24* - 23.52
2.325 psi
(a)
Hence, on face BC we have a uniform stress of 2,325 psi. Clearly, this
must also be true for face AE. This is called an uxiul stress.
To expose face EC next, we consider a half-cylinder of unit length
such as is shown in Fig. 5.49(c). Because it is j a r from the ends and
because the wall is thin, we can assume as an approximation that the stress
rn2 shown is uniform over the cut section." A pressure p is shown acting
"Nenr the ends of the tank lhe stress distribution varies in value because of 1he prorimily ofthe comp/icoredyrornrrry atid the contributions toward equilibrium afthe end plates.
203
204
('HAPTEII s
EQUATIONS or EQUILIBKIIJM
Example 5.14 (Continued)
:4
normal to the insidc wall surface. (The stress r,,, computed earlier i s licit
; shown, to avoid cluttering the diagram.) Now we consider an end view of
! this body i n Fig. 5.50 lor cquilihriuin i n the vertical directiiin. Wc havc
lrp(;
:. r,,2 = 21I [,,('2'
2[(r,,?) ( / ) ( I ) ] --
-
/)c/Q(l)sinQ =
o
(b)
-- I ) ] ( - cos Q ) ~ n
I1
3
y
t
r,,2 =
I
:
2
Figure 5.50. I k r bridy of pari of
-
1) =
cylinder.
4,700 psi
I
!
i
,J(
We puinl out now that the force in a piirlicular direclion friini a unilorm Dressitre on a curved surfice equals lhc Drcssurc times the .uroiected
.
area of this surface i n the directkin (if thc dcsircd force. (You w i l l leiirii
5,,
r
this i n your studies (if hydrostatics.) Thus for the case at hand tlie projected
area i s that of a rectangle I x (11 - 21). s o that the second expresion of
Eq. (h) becomes p ( D - 2).You inay rcadily verify that this give\ thc
r
"\
sanie result as ahovc.
r,,, i s
called the hoop .srrc.ss; i t i s about twice the u~rirrl
,sfre.s,sr,, . Wc show element A M % with the stresscs present in Fig. 5.51
The stress
......., _"l_l_..
,
.,"
,__x_I
5.10
I{,
Figure 5.52. Skitically drtel-minatc problem
..
. ,.
~
....
,
TI,
Figure 5.51. Free body u l ~ n
clcrncnt of the cylinder.
;_______..
.
,.
Static indeterminacy
Examine the simple beam i n Fig. 5.52, with known external loads and weight.
I S thc defiirniation of the heam i s small, and the final positions o f the external
loads alter deformation differ only slighlly Srom their initial positions, we can
unic the bcem to be rigid and. using the wideformed gcomctry, we can
solve the \upporting forces A, I<<, and B,. This is piissihle since we have three
equationa (if equilibrium available. Suppose, now, that an additional support
i s made availahle to the beam, as indicated in Fig. 5.53. Thc beam can still be
coiisidered a rigid body, since the applicd I m d w i l l shift even less hecause 0 1
de1orm;ition. Thcrclore, the resultant force coming from the ground to countciilct the applied loads and wcight of the beam niust he the sume as before. In
tile first case, i n which two supports wei-e given, howcvcr, a unique set 01values for the Siirces A . B l , and By gsve us the required rcsullant. In other words,
we wcrc able to solve for these liirccs hy stdtics alone, without further cun-
SECTION 5.10 STATIC INDETERMINACY
siderations. In the second case, rigid-body statics will give the required same
resultant supporting force system, but now there are an infinite number of
possible combinations of values of the supporting forces that will give us the
resultant system demanded by equilibrium of rigid bodies. To decide on the
proper combination of supporting forces requires additional computation.
Although the deformation properties of the beam were unimportant up to this
point, they now become the all-important criterion in apportioning the supporting forces. These problems are termed statically indeterminute, in contrast to h e statically determinate type, in which statics and the rigid-body
assumption suffice. For a given system of loads and masses, two models-the
rigid-body model and models taught in other courses involving elastic behavior-are accordingly both employed to achieve a desired end. In summary:
Figure 5.53. Statically indeterminate
problem.
indeterminate problem, we must satisfy both the equaiions of
fur rigidbodies and the equations that stemfrum deformation
tions. In statically determinare problems, we need only satisfy
ons of equilibrium.
In the discussion thus far, we used a beam as the rigid body and discussed the statical determinacy of the supporting system. Clearly, the same
conclusions apply to any structure that, without the aid of the external constraints, can be taken as a rigid body. If, for such a structure as a free body,
there are as many unknown supporting force and couple-moment components
as there are equations of equilibrium, and if these equations can be solved for
these unknowns, we say that the structure is externally statically determinate.
On the other hand, should we desire to know the forces transmitted
between internal members of this kind of structure (i.e., one that does not
depend on the external constraints for rigidity), we then examine free bodies
of these members. When all the unknown force and couple-moment components can be found by the equations of equilibrium for these free bodies, we
then say that the structure is internally statically determinute.
There aTe structures that depend on the external constraints for rigidity
(see the mucture shown in Fig. 5.54). Mathematically speaking, we can say
for such structures that the supporting force system always depends on both
the internal forces and the external loads. (This is in contrast to the previous
case, where the supporting forces could, for the externally statically determinate case, be related directly with the external loads without consideration of
the internal forces.) In this case, we do not distinguish between internal and
external statical determinacy, since the evaluation of supporting forces will
involve free bodies of some or all of the internal members of the structure;
hence, some or all of the internal forces and moments will be involved. For
such cases, we simply state that the structure is statically determinate if, for
all the unknown force and couple-moment components, we have enough
equations of equilibrium that can be solved for these unknowns.
Figure 5.54. Nonrigid structure
205
5.90. Draw tree-hudy diagram\ liir the hoe. aims, and tractor 01
the bdckhuc. Conhidcr thr weight of cach part to act at a central
location. Thc backhoe IS inot digging at thc insmil shown. Neglect
the weights of the hydraulic systcms CE, A H , and F l i ,
Figure P.5.90.
5.Yl. A parking-lot gate ami weighs 150 N. Becauw 01 thc
.itper, thc weight can he regarded a\ conccntl-a~cdat a puint I .?5
11 from the pivot point. What iorce must he exrrtcd hy the soleloid to l i l t the galc? What mlcnoitl force i\ necehsary it a 300-N
:r,unlerwcight i s plised .25 111 t o the left o l t h e pivot point?
i.92. Find thz force dclivcrcd ill c' i n a lhiiiimnval direction
v crush the rock. Presrurc p = 100 p i g arid p , = hO psig
pressures nieebured a h w r atmmpheric prehwrci. The diameter\
)Ithe pistons are h iii. each. Nrglcot thc wciplit d thr rock
Figure P.5.94
5.95. Find the values of F and C so that members A B and CD
fail simultaneously. The maximum load for A B is 15 kN and for
CII is 22 kN.Neglect the weight of the members.
5.98.
(a) Find the supporting forces at B. Neglect the weights of
the members.
(b) What is the force in the member CB?
1 1
100 mm
Figure P.5.95.
5.96. The landing carriage uf a transport plane supports a stationary total vertical load of 200 k N There are two wheels on
each side of shock strut AB. Find the force in member EC, and the
forces transmitted to the fuselage at A, if the brakes are locked and
the engines are tested resulting in a thrust of 5 kN, 40% of which
is resisted by this landing gear.
Figure P.5.98.
5.99. Find the suppotting forces at B and C. Disc A weighs
200 N. Neglect the weights of the members as well as friction.
5.97. Find the magnitudcs (if the supporting forces fur the frame
show You may only use fuv free-body diagrams for this problem. k t forth a complete system of equations for solving the
desired unknowns hut do not carry out the algebra for actually
solving these equations for the desired unknowns.
Figure P.5.97.
Figure P.5.99.
5.100. Find the supporting force system at C. Neglect friction.
Figure P.5.100.
207
c
U
Figure P.5.102
Figure I'.5.I04.
5.105. A trdp door i s kept open hy a rod CD, whose weight we
shall neglect. The door has hinges a1 A and R and has B weight OS
200 Ib. A wind hlowing against the outside surface of the door
5.107. Find the supporting forces at A and C. You must show
and use only m e free-body diagram.
creates a prersure increase o f 2 Iblft'. Find the force in the rod,
assuming that it cannot slip from the position shown. Also determine the forces transmitted to the hinges. Only hinge B can resist
motion along direction A R .
2m
k+
Figure P.5.107.
5.108. A coupling between two shafts transmits an axid load ot
5,000 N. Four bolts ha\ing a diameter each of 13 m m , cmnects
the two units. Befbrc loading of the shafts, thesc bolts have anly
negligible tensile forces. Assuming that each bolt carries the same
Inad, what i s the average normal stress in each bolt stemming
from the 5.(100 N lotid?
Figure P.5.105.
5.106. Find the Ibrce BO. A l l conncctims are ball-and-socket
,ioints. Neglect the wcights of a l l the members. Member AB has
two XI' bends. Member RI) i s in the yz. plane.
CF"
50Nm
Figure P.5.108.
5.109. A circular shaft i s suspended from above. The specific
weight OS the shaft material i s 1.22 x I O * Nlm'. What i s the tensile stress r.. o n cross sections of the shaft a i a function of r?
Figure P.5.109.
Figure P.S.106.
5.110. Do the previous pmblrm k r the czce whex the specific
weight y varies with the syuure of z starting with the value of
6.50 x IO" Nlm' at the top and reaching a value of 7.50 x
IO4 Nlm' at the bottom.
209
Figure 1'.5.1 11.
5.11
Closure
5.113. Determine the tensions in 811 the cables. Block A has a
mass of 600 kg. Note that GH is in the yz plane.
5.116.
Find the forces on the block of ice from the hook.; at
A and F .
Figure P.5.113.
5.114.
Determine the force components at G. E weighs 300 Ih.
A
+I
5'
*I
Figure P.5.116.
5.117. Members AB and BC weighing, respectively, 50 N and
200 N are connected to each other by a pin. BC cunnecls to a disc
K on which a turque T, = 200 N-m is applied. What turque T is
needed on A B tu keep the system in equilibrium at the configuration shown'?
Figure P.5.114.
5.115. A scenic excursion train with cog wheels for steep
inclines weighs 30 tons when fully loaded. If the cog wheels have
a mean radius to the contact points of the teeth of 2 ft, what torque
must he applied to the driver wheels A if wheels B run free? What
force do wheels B lransmit to the ground?
I
I
Figure P.5.115.
/-
Figure P.5.117.
21
A transport j e t planr has ii weight without frirl uf220 kN.
If om wing is I<,aded with SO k N r i f f k l , what are the force? in
eich 01 the three landing gear?
5.1 I X .
H--'
4
111
4
111
Figure P.5.118.
5.119. A rod AB i s connected by a hall-and-wcket joiiit to a
frictionlesc sleeve at A , arid by a ball-and-sochet .joint t o ii fixcd
position at 8.What are the supporting fbrccs at B and at A i l we
ncglect the weight of AB? The 100LIh load i\ connected 10 the
C C I I ~ C I0 1 A l l .
i
8'
Figure P.5.119.
212
Figure P.5.122.
5.123. Light rods RC and AC are pinned together
at C and support a BOO-N load and a 500-N-m couple moment. What arc the
supporting forces at A and B ?
5.125. A bent rod AUCB supports two weights-one at the center of AU and one at the centcr of UC. There are ball-and-socket
joint supports at A and R . With one scalar equation using the triple
scalar product, determine the tension in cable Uc'.
2m
10,
Figure P.5.123.
I
x
Figure P.5.125.
5.124. A rod AB is held by a hall-and-socket joint at A and supports a 100-kg mass C at 8.This rod is in the z? plane and is
inclined to the? axis by an angle of 159 The rod is 16 m long and
F is at its midpoint. Find the forces in cables DF and ER. Crosahatching indicates part of the V E plane.
5.126. Find the tension in cable FH. The disc C weighs 500 Ih.
Use only one free body.
H
x
Figure P.5.124.
Figure P.5.126.
213
Figure I'S.131
5.132. A uniform block weighing 500 Ih is constrained by three
wires. What are the tensions in these wires?
n
A
5.135. A mechanism consists OS two weights W each OS weight
50 N,tour light linkagi: rods each of length (iequal 10 200 mm,
and a spring K whose spring constant i s X Nlmm. The spring i c
unextended when H = 450 I f held vertically, what is the angle B
fix equilihrium'! Neglect friction. The force from the spring equals
K times the compression of the sprins.
Figure P.5.132.
5.133. Find the suppvrting forces for thz frame shown
Figure P.S.135.
,,-
~
k
2,000 Nini
5.136. Find the compressive force in pawl AB. What is the resultant supporting force system at E!
2Figure P.5.133.
5.134. Four cables supporl a black of weight 5,000 N. The edges
of the block are parallel to the coordinate axes. Point B i s at
(7, 7, -15). What are the forces in the cables and the direction
cosines for cable CI)?
,
~, ,,
.,:__
fi:
Figure P.5.136.
5.137. A IO-kN load is lifted in the front loader bucket. Whal
are the furces at the connections tu the bucket and to ann AEI
Hydraulic ram DF i s prrpcndicular to arm AE, and AC is h o r i m n ~
tal. Points A and I. ;ire iit the same height above the friiund.
IS'",
Figure P.5.134.
Figure P.5.137.
n
L - 1
Figure P.5.142.
5.144. What force F do the pliers develop on the pipe section Ll?
Ncglect friction.
Wind load
normal to arc
10 m m
ti
A
Figure P.5.146.
5.147. An arch is formed by uniform plates A and R. Plate A
weighs 5 kN and plate R weighs 2 kN. What are the supporting
forces at C, 1). and E'!
20 kN
'13ON
I
Figure P.5.144.
5.145. What are the supporting forces for the frame'? Neglrct a11
weights except the IO-kN weight.
Figure P.5.147.
5.148. Find the iupponing forces at the ball-and-socket connections A , 0, and C. Mcmbers A 8 and DB are pinned together
through member EC at R.
I.000 Ib
Parallel
n
Figure P.5.145.
5.146. A 20-m circular arch must withstand a wind load given
lor O < B < 7 ~ 1 2as
f = 5,000( I -
n)
e Nlm
where B is measured in radians. Note that for 8 > 7~12,there i?
n o loading. What are the supporting forces? (Hint: What is the
point lor which taking moments is simplest'!)
x
Figure P.5.148.
217
5153. A bar can rotate parallel to plane A about an axis of rotation normal to the plane at 0. A weight W is held by a cord that is
attached to the bar over a small pulley that can rotate freely as the
bar rotiltes. Find the value of C for equilibrium if h = 300 mm,
W = 30 N, @ = 3(Y, I = 700 mm, and d = 500 mm.
x
.4? m
30 m
...
x
5.157. Find the supporting forces at A , A, and C. Neglect the
weight ofthe rod. Use only one free-body diagram.
IW Ib
Figure P.5.153,
5.154. Find the supporting forces at A and B in the frame.
Neglect weights of member?.
I
I O Ih/ft
F
40'
Figure P.5.154.
c
5.155. A holt cutter has a force of 130 N applied at each handle.
What is the force on the bolt from the cutter edge?
30"
Figure P.5.157.
5.158. The 5,000 Ib viin A of an airline food catering truck rises
straight up until its floor is level with the airplane tloor. A hydraulic
ram pulls on the right bottom support of the lift mechanism at which
we have rollers to prevent friction. The two members of the lift
mechanism are pinned at their center. The center of gravity of the
van is its geometric center. What is the ram force for this position?
,Bolt
I-
1 2 ' 4 No friction
130 N
Figure P.5.155.
5.156. A steam locomotive is developing a pressure of .20 Nlmm'
gage. If the train is stationary, what is the total traction force from
the two wheels shown? Neglect the weight of the various connecting rods. Neglect friction in piston system and connecting rod pins.
Figure P.5.158.
219
E
I'
!!
Figure P.S.159.
S.IU1. For lhc structure \houii. dclemiine i h c i m c e i n the
;:ihlc Et'.
Figurc P.S.lh0.
220
Figure P.S.162
Introduction
to Structural
Mechanics
Part A Trusses
6.1
The Structural Model
A trus,? is a system of members that are fastened together at their ends to support stationary and moving loads.' Everyday examples of trusses are shown in
Figs. 6. I and 6.2. Each member of a truss is usually of uniform cross-section
along its length; however. the various members typically have different crosssectional areas because they must transmit different forces. Our purpose in
Part A of this chapter is to set forth methods for determining forces in members
of an elementary class of trusses.
As a first step, we shall divide trusses into two main categories according to geometry. A truss consisting of a coplanar system of members is called
a plane Irus.7. Examples of plane trusses are the sides of a bridge (see Fig.
6.1) and a roof truss (see Fig. 6.2). A three-dimensional system of members,
on the other hand, is called a p a c e truss. A common example of a space truss
is the tower from an electric power transmission system (see Fig. 6.3). Both
plane trusses and space trusses consist of members having cross-sections
resembling the letters H, I, and L. Such members are commonly used in many
structural applications. These members are fastened together to form a truss
by being welded, riveted, or bolted to intermediate structural elements called
Russet plates such as has been shown in Fig. 6.4(a) for the case of a plane
truss. The analysis of forces and moments in such connections is clearly quite
complicated. Fortunately, there is a way of simplifying these connections
' A i i u s , ~is different than a,frame (see the footnote on p. 157) in that the members of a truss
are always connected together at the ends of the members. as will soon become evident,
a frame has some rnemhers with connections not at the ends of the member.
whereas
22 I
222
CHAPTER 6
INI'ROI~IICTIONTO STRUCTURAL MECHANICS
Figure 6.1. Foot hridgc near author's former home. Sidcs oi sirucfurc are plane irusscs
Figure 6.2. Roof trusses that are plane trusses
SECTION 6.1
Figure 6.3. Space tmsscs supporting transmission lines
sending power into the northeast grid of the United
States.
such as to incur very little loss in accuracy in determining forces in the memhers. Specifically, if the centerlines of the members are (nncurrenf at the
connections, such as is shown in Fig. 6.4(a) for the coplanar case, then we
can replace the complex connection at the points of concurrency by a simple
pin connection in the coplanar truss and a simple ball-and-socket connection
for the space truss. Such a replacement is called an iderzliiution of the system.
This is illustrated for a plane truss in Fig. 6.4, where the actual connector or
joint is shown in (a) and the idealization as a pinned joint is shown in (b).
In order to maximize the load-carrying capacity of a truss, the external
loads must he applied at the joints. The prime reason for this rule is the fact
that the members of a truss are long and slender, thus rendering compression
members less able to c a y loads transverse to their centerlines away from the
ends? If the weights of the members are neglected, as is sometimes the case,
rce
and accordit should be apparent that each member is a t ~ ' ~ ~ f i )member,
ingly is either a tensile member or a cornpression member. If the weight is
not negligible, the common practice as an approximation is to apply half the
weight of a member to each of its two joints. Thus, the idealization of a member as a two-force member is still valid.
'You will understand these limitations innre clearly when you study huckling in your
strength 0 1 materials course.
THE STRUCTURAL MODE
223
rriiioviil 01 ;my of i t s menrherc dcstioyc i t s rigidity. Ii rcmo\:ing :I mrniher
does not dcstioy rigidity, the \ti-ucture I \ said IO be (iv?r~ri,qid.Wc shiill he
SIXTION 6.4
METHOD OF JOINTS
225
forces acting on it from the members can always he found. (One such joint is
the last joint formed.) Each unknown force from a member onto this joint
must Iiavc a direction collinear with that member, and hence has a known
direction. There are, then, only three unknown scalars, and since we have a
concurrent forcc system they can he determined by statics alone. We then
find another joint with only three unknowns and so carry on the computations
until the forces in the entire structure have been evaluated. For the simple
plane truss, it similitr procedure can be followed. The free body of at least one
joint has only two unknown forces. We have a concurrent, coplanar force system, and we accordingly can solve the corresponding two equilibrium equations in two unknowns at that joint. Wc then proceed to the other joints,
thereby evaluating all member forces by the use of statics alone.
6.3
Solution of Simple Trusses
Generally, the first step in a truss analysis is to compute the supporting forces
in the ovurall truss. This calculation of the external forces or reactions that
must exist to keep the truss in equilibrium is independent of whether the truss,
internally, is statically detcrminate or statically indeterminate. Simply regard
the truss as a rigid body to which forces are applied, some known (given
applied forces) and some unknown (rcactions),' and solve for the reactions as
we did in Chapter 5. We have shown a simple plane truss in Fig. 6.7(a) and
have shown the features of the truss in Fig. 6.7(b) thal are essential for the calculations of the reactions. Note that members such as CA, DB, and DE need
not he shown in the free body since they provide irirmnirrl forces for the body.
Once the free-body diagram has heen carefully drawn, use three equations of equilibrium to determine the reactions of a plane tmss (six equations
for a space truss). It is highly advisable to then check your results by using
another (dependent) equation of equilibrium. You will be using the computed
reactions for many subsequent calculations involving forces in internal memhers. Accordingly, with much work at stake. it is important to start off with a
correct set of reactions.
We shall present two methods for determining the forces in the memhers of the truss. One is called the rnezhod oj'joinrs and the other is called the
merhod of serliorrs. As will be seen in the following sections, the prime diiference between these method:, lies in the choice of the free bodies to he used.
6.4
Method of Joints
In the method of joints, the free-body diagrams to he used. once the reactions
are determined, are the pins or hall joints and the forces applied to them hy
the attached members and external loads. Note that we havc already alluded
'Suppming Ibrrch are o r h c j l l c d rwcliom i n S I ~ U F I U T . ~mcchanics
~
c
*,
I)
10 m
t
R,
Z.(xx) N
(b)
Figure 6.7. Free-hody analysis of truss.
226
CHAPTER 6
INTRODUCTION TO STRUCrtJKAI. M[.C'HANICS
to this method in Section 6.2. Consider first Ihc trian$ular plane truss shown
i n Fig. h.X(a). Niitice we havc already determined the I'L'ilctions.
'
'
1.000 N
51111 N
I .lIOO N
5110 N
ilil N
(Bl
lhl
Figure 6.8. Mcthod ol.juint\
IC1
joint B.
Next, consider the free hody of pin B [Fig. h.X(b)l. The unknown forces
from the members arc shown cnlliiiear with the centerlines o f Ihese member5
since they are two-liirce nienihers. We can s o l v e fiir these foi-ccs by setting
the sum of forces equal to zern i n the hnrizont;il and vertical directinns. to g c ~
Because both forces arc p m h i r q against pin B. the corresponding members arc
compressive rathcr than tension rnenihers. We can mnst readily see this fact hy
considering Fig. 6.X(c), where members AH and C" have been cut ;it vwinus
places. Notice that AH i s also pushing againsl pin A as dncs ('8against pin <'.
577 N
500 N
(b)
Figure 6.9. Procedure for method of joints
(a) Notation for mcmhers AB and CH:
(b) free-body diagram of A.
Thus, nnce having decided that the incmhers arc. compressive meinhers 21s ii
result of considerations itf a pin at one end o f the member. we can conclude
that the member i s pushing with equal force against the pin at the other end.
To make for specd and accuracy in we go Sviiiii nnr .joint to ;mother. wc recornmend that, once the nature of the Inading i n a mcmher has heen established
hy considerations at a pin. we mark down this viilue using a T for t e n s i ~ nor ii
C for compression after i t on the trusb diagram. as shown i n Fig. h.Y(a). Note
also that apprnpriate arrows arc drawn in the memhcrs. Thcsc imows rcprescnr
forces developcd by the tnrmher,~,,n die phia. Hence. for minprevsior! the
arrnws point lowcirri the pins, ;uid for r c v s i o r i they point c i w < y lrom the pins.
Acciirdingly. if we now comidcr the Iree hody of pin A a s shiiwii in Fig.
h.S)(h). we know the direction and valuc nf the Ihrcc 011 A lrnm nicmher AB.
If a negative valuc i s found fiir a I i r c e at ii pin. the seiise CIS the force
has heen taken incorrectly iit the outset. With thi\ in mind. wc deciclc whether
the memher ~issociatedwith Ihc fnrce is a teii\ion or comprehsiiin inemher
and we label the mcrnher accordingly, a b s h w m in Fig. h.Y(a) for use later i n
examining the pin at the other end 01the incinher as a ltee hody.
We now consider the solutiiin nf ii plane ti-ucs prnhlem hy the method
ofjnints i n prcater detail.
227
SFCTION 6.4 METHOD OF JOINTS
Example 6.1
A simple plane truss is shown in Fig. 6. IO. Two I ,000-lh loads are shown
acting on pins C and E. We are to determine the force transmitted by each
member. Neglect the weight of the members.
In this simple loading, we see by inspection that there are 1,000-lb
vertical forces at each support. We shall begin, then, hy studying pin A , for
which there are only two unknowns.
D
1.000lh
Pin A. The forces on pin A are the known 1,000-lb supporting force and
two unknown forces from the members AB and AC. The orientation of
these forces is known from the geometry of the truss, but the magnitude
and sense must be determined. To help in interpreting the results, put the
forces in the same position as the corresponding members in the truss diagram as is shown in Fig. 6.1 l . That is, avoid the force diagram in Fig.
6.12, which is equivalent to the one in Fig. 6.11 but which may lead to
errors in interpretation. There are two unknowns for the concurrent coplanar
force system in Fig. 6.1 I and thus, if we use the scalar equations of equilibrium, we may evaluate
and FAc:
sR
A
FA(.
I
=0
~
F, = 0
~~~
1,000Ih
Figure 6.10. Plane truss
F~~ - n 707~,,= 0
1,000 Ih
Figure 6.11. Pin
-0 7 0 7 +~ 1,000
~ ~= n
Therefore,
FAB = 1,414B ,
FAc
0
1,mm
Since both results are positive, we have chosen the proper senses for the
forces. We can then conclude on examining Fig. 6.1 I that AB is a compression member, whereas AC is a tension member:? In Fig. 6.13, we have
labeled the members accordingly.
If we next examine pin C , clearly since there are three unknowns
involved for this pin, we cannot solve for the forces by equilibrium equations at this time. However, pin B can be handled, and once Gcis known,
the forces on pin C can be determined.
Pin E. Since AB is a compression member (see Fig. 6.13) we know that it
exerts a Sorce of 1,414 Ib directed against pin B as has been shown in Fig.
6.14. As for members BC and E D , we assign senses as shown.
FAX
Figure 6.12. Pin A--avoid this diagram.
F
A
1,000 Ih T
I
sHad we used Fig. 6.12 as a lree hody. the State of loading in the memkrs (i.e.. tension
or compression) would not he clear..Therefore. we strongly rccommend putting forces reprc~
senting memkrs in positions coinciding with the members.
1,000 Ib
I,(XX)lh
1,000lh
1.000lh
I.OMlIh
Figure 6.13. Notation for members A 8
and AC.
228
CHAPTER 6
INTRODIICTION TO S I K t ~ ( ' T 1 I R A I . MM'HANII'S
Example 6.1 (Continued)
Figure 6.14. Pin li,
Summing forces on pin
A
(Fig. 6.14). wc get
(1.414)(0.707)+ Fk(
=
0
FBc = -1,000 Ib
Hcrc we have ohtained Lwo negativc qumtitic\. indicating t l i i i t wc
have made incorrect clioiceb 0 1 sense. Kccping t h i h i n mind, we can conclude that memhes R D i s a compression membcr, when-ens mcnihcr IK' i\
a tension mcmher. Noticc that wc havc shown thebe liirccs propci-1) iii
i
!
Fig. 6. I S .
We can proceed i n this nianner i r o m .joint to ,joint. A I thc last .joint
a11 the forces w i l l havc hecn computed without using il iis ii frcc hody.
Thus, i t i s available to he used iis ii check on the solution. 'lhnt is. thc win
of the known forces for thc last joint in the iand directions should hc
zero or closc to zero, dcpcnding (in the accuracy of yiiur ciilciilation\. We
urge you to take advantage of this check. The final wlutiim i h ~ I i o In
~ n
Fig. 6.15. Notice that membcr CI) has zero load. This docs not i n e m that
we can get rid of this nieniher. Othes luxlings expccted for thc t r i i s y hill
result in nonzero forcc lor C.'D. Furtherninre. without C I ) thc truss will not
he rigid.
Figure 6.15. Solution lor tius?.
I
SECTION 6.4 METHOD OF JOINTS
Example 6.2
A bridge truss is shown in Fig. 6.16 supporting at its pins half of a roadbed
weighing 1,000 Ib per foot. A truck is shown on the bridge having estimated loads on pins E , G, and I of the truss equaling, respectively, 1 ton,
1.5 tons, and 2.5 tons. The members weigh 45 Ib/ft. Include the weight of
the members by putting half the weight of a member at each of its two supporting pins. Find the supporting forces.
A
H
F
J
L.
Figure 6.16. A bridge truss supporting a roadbed and a truck
We first determine the forces on the pins from the weight of the
members denoting them as (W,)j.Thus we have
I ~ a d 1:
s Weights of members
(W,)*= (W,),
= 2[?(20)(45)]
I
= 900
Ih
I 20 (45) = 1,536.5 Ib
( y ) R= (T), = 2[;(20)(45)] + -2(.707)
~
(W,),. = (W,),= (W,)(=;3[;(20)(45)]+
(W,)”
= (W,))=
H (W,)E = 3[;(20)(45)]
2[1(m)(45)]
I 20
= 2,623 Ib
= 1,350 Ib
Next we get to the roadbed
Loads 2: Weights from the roadbed
1
(Wz),,
= (W2), = 2(500)(20)= 5,000 Ib
(W2)<:= ( W 2 ) E= (Wz)G = (500)(20)= 10,000 Ib
Finally, we list the loads from the truck.
Loads 3: Weightsfrom the truck
(W?), = 2,000 Ih
(W,),; = 3,000 Ib
(WJ, = 5,000 Ib
229
230
CHAPTliR 6
INTROL~lJ('Tl0NTO S'IKII('TC'RAL MIX'IIANICS
ExamtAe 6.2 (Continued)
I
We are now rcady to determine [he supporling fiwcec for the ti-ins. Taking
the entire truss a s a frce hody we firs1 take niulncnts ahout joint I ( w e
Fig. 6.17).
!
1
S.lUl0 Ih
SECTION 6.4 METHOD OF JOINTS
231
Example 6.3
Ascertain the forces transmitted by each member of the three-dimensional
truss [Fig. 6.18(a)].
We can rcddily find the supporting forces for this simple structure
by considering the whole structure as a free body and by making use of
the symmetry of the loading and geometry. The results are shown in
Fig. 6.18(b).
Figure 6.18. (a) Space tmss and (b) free-body diagram
Joint F. It is clear, on an inspection of the forces in the x direction acting
on joint F, that
:FpE = 0 , since all other forces are in a plane
at right
angles to it. These other forces are shown in Fig. 6.19. Summing forces in
F
they and z directions, we get
I,&JOlb
Figure 6.19. Free-body diagram of joint F.
Therefore,
cz
FBD = 2,240lb compression
I
F =0:
-FAF
10
+ 1,000 - 2,240=0
4500
Therefore,
CF =
I.000 - 1,000 = 0
232
CHAPTER h
INTRODUCTION TO STRUCTURAL. MI~CIIANK'S
Example 6.3 (Continued)
Joint B. Going to .join1 B, w e see ]Fig. 6.1X(h)l that
tiE= 0
,
Since thcre are
110
and
uh
= 0
and
olhcr liirce components on pin B i n the
directions of these members. Finally.
Joint A . Let
FnR
FBc = 2.000 Ih tension
ct(.
ncxt consider,joini A (Fig. (i.20). Wc cim express I11l-c~
5,:vectorially. Thuc.
Joint I). We now consider.ioint D (Fig. 6.21). Forces
4.,)and F:,,i ~ r e
expressed 21s lollows:
Figure 6.21. I'rec-hmly diasmrii of jainl IJ.
..
.
~
.
..,.
.
I
"
__-
SECTION 6.4 METHOD OF JOINTS
I
233
Example 6.3 (Continued)
Hence, summing forces, we get
-2,000j
- 1,000k
+ F,,(-.40Xi
-
F,,,i
+
- .816j
2,240(.894j
-
+
.447k)
.408k) = 0
(d)
Thus,
-2,000
-1,000
We see here that
+
+
2.000 - .81bF,,
= 0
(e)
l&
+
.408F,,
= 0
(0
1,000
-
,4085, = 0
FED = 0 and
F,,
= 0
(8)
,
Joint E. The only nonzerc) forces on joint E are the supporting forces and
F;.&, as shown in Fig. 6.22(a). We solve
= Z&Wlbcompression .
I&
Joint C. As a check on our problem, we can examine joint C. The only
nonzero forces are shown on the joint [Fig. 6.22(b)]. The reader may readily verify that the solution checks.
I
E
X
1,000 Ib
1,000 Ib
(a)
(b)
Figure 6.22. (a) Free-body diagram ofjoint E and (b) joint C.
234
('HAPTER 6
INTKOIXJ(:TION TO STKIICIIIRAI.MECHANICS
Belore proceeding with the problems. i t will he well to comment on
the loading 01 plane roo1 trusscs. IJsually there will be a series of \cp;ir;itetl
parallel t t u h z e h supporting the loading from the r o d such a s i h S ~ O W I I i n
Fig. 6.23, wlicre a wind pressurc is shown on a roof as 11. Niiw the Oisiilr tru\h
can he considered to support the loading over a region extending h;lllway tc
each neighboring trus? (shown as distanced). Furthermore. pins A and B suppori thc force exerted on area Ilmk while pins B and C suppol-t the forces
excrtcd on area /IT/,. When dcaling with the entire inside truss a s a free hody,
you can use thc resultant force from pressure ovcr krwn. Howcvcr. when
dcaling with thc pins as ii free hody you must use the forces coming on tu
each pin as described ahove and no1 the totnl resultanl, which was iiscd tor
lhc free hody of the entire internal truss. Clearly. thc outhide trusses support
halt the Iimds descrihed ahove.
Figure 6.23. R o d trussc? supporting 21 wind load
Finally, we wish Lo remind you that a curved meinhcr i n a truss. such as
appears in Prohleins 6.5 and 6.8. is a two-forcc incinher with forces coming
only lrom the pinz. Rccdll that. for such nienihrrz, the force transmitted to the
pins niiist he collinear with the line connecting the pins. such as is shown i n
Fig. 6.14.
6.1. Slate which of the trusses shown are simple trusses and
which are not.
(a) Pratt INSS
+4m+
~-
Figure P.6.3.
(b) Fink
6.4. A rooftop pond is filled with cooling water from an air conditioner and is supported by a series of parallel plane trusses.
What are the forces in each member of an inside truss'! The roof
trusses are spaced a1 10 ft apart Water weights 62.4 IMA.',
~NSS
i:
f
10'
i
(c) Special-purpose t r u s ~
Figure P.6.1
6.2. Find the forces transmitted by each member of the truss
Figure P.6.4.
6.5. Find the forces transmitted by the straight members of the
truss. DC is circular
5,000 Ib
Figure P.6.2.
6.3. The simple country-road bridge has floor beams to carry
vehicle loads tn the tmss joints. Find the forces in all members for
a truck-loaded weight of 160 kN. Floor beams I are supported by
pins A and B, while floor beams 2 are supported by pins B and C.
*
B
Figure P.6.5.
23
I
6.6. Roof tiusscs such 215 the o n r shown arc spaced 6 rn apart in
il long, ncctangular huilding. During thc wintcr. MIOW loads of LIP
to I kN/rn2 (or I kPa) ac~uniulateon the central portion o l the
nnrf. Find the force in each memher for il t r u \ no1 at the cnda of
the huilding.
6.9.
p
Find the lol-crs in thc mcmhrrs of the truss.
I,OOO N
8
IO",
('
10m
F
IO 111
45"
Figure P.h.h.
A
.
Figure P.6.Y.
6.7.
'Ihc bridge supportr a rtradway Imd of I.000 Ihilt lor each
0 1 the twu trusses. Each mcinhcr weighs 30 Ihilt. Crmpute the
fiirces i n the rnernhers, accounting approxim;itcly for the weight
[if
t h e rnrrnhcrs.
I'
6.10.
In Example 6.1. incluilc the weights ut the memhcrs
;tppruxim:itcly. The mmmhrrs each weigh 100 Ih/ft.
H
6.11.
Dctcriiiiiic the lorurs in thc mrmhel-s. T h r pulley5 ill c'and
I'cach weigh 300 N. Ncplecl all iithct vicighth. Bc surc yuu have
il check OI your iiiluIiuii.
Figure P.6.7.
6.8. Find
1.000 N
a
the force, i n the straight incmhcrs u l Ihc tiuhh
/I
IO'
1 0 ,
1)
IO'
+T
5 0 0 Ih
500 Ih
Figure P.h.8
Figure P.h.11
6.12 Roadway and vehicle loads are transmitted to the highway
bridge ~ N S S as the idealized forces shown. Each load is 100 kN.
What are the forces in the members?
D
-1
6 panela at 6 m
2
Figure P.6.12.
Figure P.6.14.
6.15. Find the forces i n the members of the truss. The 1.000-lb
Force is parallel to they axis, and the 500-lb force is parallel to the
:axis.
6.13. A hoist weighing 5 kN lifts railroad cars for truck repair.
The hoist has a 150-kN capacity and hangs from a truss with an Lshaped member to clear boxcars. What are the forces in the
straight members for full capacity of the hoist’?
Figure P.6.15.
6.16. Find the forces in the members and the supporting forces
for the space truss ABCD. Note that BDC is in the XI plane.
Figure P.6.13.
6.14. A 5-kN traveling hoist has a 50-kN capacity and is s u s ~
pended from a beam weighing I kN/m, which, in turn, is fastened
to the roof truss at I and G as shown. In addition, wind pressures
of up to 2 kN/m2 (or 2 kPA) act on the side of the roof. The resulting force is transmitted to pins A and J . If the trusses are spaced S
m apart, what are the forces in each member of the truss when the
hoist is io the middle of the Span?
Figure P.6.16.
231
rJ-
n
6.5
23R
Method of Sections
SECTION 6.5 METHOD OF SECTIONS
well inside a truss and avoid the laborious process of proceeding joint by joint
until reaching a joint on which the desired unknown force acts.
Generally a free body is created by passing a section (or cut) through
the truss such as section A-A or section B-B in Fig 6.25(a). Note that the section can he straight or curved. The corresponding right hand free-body diagrams [see Fig. 6.25(b) for cut A-A and Fig. 6.25(c) for cut B-B] involve
B
A
(c)
Figure 6.25. Section cuts.
coplanar force systems. We have, accordingly, three equations of equilihrium available for each free body. One can also choose to use left-hand freebody diagrams for these cuts. Note that in contrast to the method of joints,
one or more equilibrium equations can most profitably he moment equations.
The choice of the section (or sections) to find the desired unknowns inside a
truss involves ingenuity on the part of the engineer. Helshe will want the
fewest and simplest sections to find desired forces for one or more members
inside the truss. The method of sections is used for efficiently finding limited
information. The method of joints for such problems is by contrast one of
"brute force."
We now illustrate the method of sections in the following examples.
239
SECTION 6.5 METHOD OF SECTIONS
Exa iple 6.5
A pla tru% is shown in Fig. 6.21 for which only the force in member AB
is des rd. The supporting forces have been determined and are shown in
the di :ram.
1,000 N
I
7x9 N
1.077 N
Figure 6.21. Plane truss.
Fig. 6.28(a) we have shown a cut J-J of the truss exposing force
lis is the same force diagram as that which results from the free-body
I of pin A,) We have here three unknown forces for which only two
ns of equilibrium are available. We must use an additional free body.
hus, in Fig. 6.28(b) we have shown a second cut K-K. Note that by
moments about joint B, we can solve for <,c directly. With this
ition we can then return to the first cut to get the desired unknown
cordingly, we have, for free body 11:
I
5w (
diagr;
equat
takin;
infori
CH.
i
C M B=0:
-(10)(500)
+ (30)(789) - (FAc)(sin 30")(30) = 0
(Note ve have transmitted FAc to joint H in evaluating its moment contrbutio ) Solving for FAc we get
FAc = 1,245 N
.
a) Free body I from cut J-J
(a) Free body I1 lrom C u t K-K
Figure 6.28. Free bodies needed for the computation of fiirce FAD.
~~~
241
242
('HAPTLIK 6
INTKOIIIICTION 7'0 STKllCTliKAL. ME('IIANI('S
Example 6.5 (Continued)
Summing forces for free body 1. we have"
F,,,,cos 30"
~
FA<.cos 30"
~
1.000 sin 30" = 0
Thcrcfbre,
Pi,,,
= I .X22N
FAB
= -66IN
Therefore.
We see that member A B i s a Leiision nieinber rathcr than ii compressim
member a i wiis our initial guess i n drawing thc free-hody diagrams.
I n retrospect. you w i l l note that. i n the method of joints, errurs made
early w i l l (if neccssity propagate through the calculations. There is, on the
other hand, much less likelihood of this occurring in the method of. sections.
However. for simple trusses with many members, we may profitahly use the
method of joints i n conjunction with a computer for which the brute-force
approach of the method o f joints i s ideally suited. I1 i s to be pointed out that
there i s computer soltw;rre availahle which makes this kind of computation
routine and quick.
*6.6
Looking Ahead-Deflection
of a Simple, Linearly Elastic Truss
I n the solids or structures courses you will soon hc laking. you w i l l he given
inhtruction for dctcnnining the nioxnient o f the pins of a linearly elastic, s i n ple truss steiriniing from the cxtcrnal loads. It i s true that you can now determine the forces in il simple truss and from this you can determine the change
i n length of each memher. However, using t h i s data as well a s the accompanying changes o f orienllition (if the meinhen, you w i l l find i t next to impossible
to get the movements o f the pins for all hut the most trivial trusses.
!
SECTION 6.6 LOOKING AHEAD-DERECTION OF A SIMPLE, LINEARLY ELASTIC TRUSS
243
~
You will learn later of a neat method of doing this. In this method, each
movable (unconstrained) pin is imagined to be given first a hypothetical
movemeit’ 6, in one of two orthogonal directions, say the x direction here,
while no movement is allowed for that pin in the y direction. All other pins
are held fixed for the preceding action. We now evaluate via geometry the
changes in length stemming from 6, of all members affected by this hypothetical < isplacement. Also, the energy of deformation8 for these members is
computei. The aforementioned displacement is shown in Fig. 6.29 for 6,. In
addition to the energy of deformation nf the affected members, we compute
the worb done by external forces that undergo movement from 6,, keeping
the forcqs constant during this movement. Here the work is F(cos a)6,.This
procedure is canied out for each such x and y displacement for all of the movable pin?. We then add up all the energy terms and all the work terms, multiplying tke latter by - I . We then have a function of all of the n 6’scomprising
the x a n i y components of all movable pins. This function, which is commonly dmoted as n, is then extremized with respect to the n 6’s. We then
form n s multaneous equations from
(%)i
= O
i = l,2, ... n
truss, th- members were deformable and we mentioned that we needed the
energy c f deformation for the use of the total potential energy principle.
’This is called a virruol displucemenl and will be discussed further in Chapter 10 when we
discuss the: method of virtual work.
xYoi will learn to calculate the energy of deformation in your solid mechanics or your
where S is the axial change in length of the member, A
structures EOUTSB. The formula is
2L
is the cross-sectional area. E is the modulus of elasticity. and L is the length of the member.
Figure 6.29. A simple truss has its free pin
H given a hypothetical dnplacement 6,
while keeping all other displacement
components fixed. Note that only members
CH, EH, and H E are involved. Work done is
F(cos a)&,.
6.25.
Find the Iorucs in members C'D. DG, and llC i n the plane
11U\\.
I
.oi)o ih
Figure P.6.22.
6.27. 'The roof is subjected t o :I wind hading of 20 Iblft'. Find
the 1orurs iii rncinhers LK and K.101 an interior tius5 if thc irusse\
arc spaccd 10 11 apdl~l.
Figure P.6.23.
'44
6.31. A pair of trusses supports a roadway weighing 500 Ib/ft.
By method of sections, find the forces in DF and DE. The roadway is supported at pins A , D, F, and H on the two trusses.
15’
I”
F
15’
m
M
IIj0-k;
Guideway
1 “5-kN
t 45-kN
hoist
lOdd
Figure P.6.31.
supported at
J and G only
Figure P.6.28.
6.29. In Example 6.2, determine the forces in members FG and
CE.
I
I
6.30. (a Find the forces in members DG and DF by the method
of .secrio 2. Stare whether the members are tension or compression
members
members AC, AB, CB, and CD by the
diagram in an appropriate manner and
are in tension or compression.
80 kN
I
I
6.32. Find the force in members H E , FH, FE, and E‘C of the
truss.
SOkN
I
SO kN
SK
,
I
Figure P.6.30.
Figure P.6.32.
245
6.33.
Find the force in member JF in the truss.
+ +
Ion1
?,500 N
6.35. A railmad engine i \ stairing to cros\ the deck-type truh,
hridge chawn. If ths wcighc of the engine i s idealiied by Ihr hulSO-kip loads," find che l i ~ e in
\ mrmbcrs AH. HL. Cti. CL, /.ti.
/)ti.K.I. and I N
SOK
2,500 N
2,500 N
2.500 N
I
Figure P.6.33
''A kip
246
ib
a kilopound, 01 1.000 Ib.
XIK
X!K
50K
I
SECTION 6.8 SHEAR FORCE, AXIAL FORCE, AND BENDING MOMENT
Part IB:
Section Forces in Beams'"
6.7
/Introduction
6.8
Shear Force, Axial Force,
and Bending Moment
241
, . I
I
M'
.
fl
1UJ
Figure 6.30. Kesoltant
;it ii
scctiiiii
additional slicilr c i m p i m c n t V~ !ree Fig. 6.3 I). oiie additional hendingmomeiit cnmponenl M > ,
and ii couple inomcnl along the axis of tlic beam M\.
which we shall call the twt.\f;ti,q tnom(w.
Notice i n Fig. 6.3!)(c)that a sectind ll-cc-hody diagram h a s hcen drawn
which exposcs [lie "othei- side" o f thc cross-section at position x. The shear
Force. axial f w c c . and bending momrnl Cor [his section h w e heen prinicd i n
ilic d i a p n i . We know f r o m Ncwtiin's third law that they should he equal and
opposite 10 the ciirresponding unprimcd qunnlities i n part ( b ) of the diag'ani.
We can thus choose for our conipuralions cither ii left-liand or n right-hand
free-body diagram. But this poses somewhat of ii prohlem for 11s whcn u e
m m c ki reporting the s i g i \ o f Ihe transmitted forccs and couple morncnts :I( ii
section. We catin01 use the direction of a lorcc or couple inotnenl a1 thc sccLion Clearly. this would hc inadcquiite since the s e n x o l Ihc force or couplc
t i i ~ m e n tiit :I wctioii would depend on whether a left-hand or a right-hand
lrce-body diagram wiis used. 'Tu associate an unambiguous sign lor shcar
lorcc, axial force. and bending tmoriicnt i i t a secfion. we ;tdiipt the following
invention:
1
SECTION 6.8
SHEAR FORCE, AXIAL FORCE. AND BENDING MOMENT
i
A forc componenffor a section is positive
sectio and theforce component both
in the egative dbctions of any orie
The sam is true for the bending moment.
T s, consider Fig. 6.30. For the left-hand free-body diagram, the area
vector f r section x points in the positive x direction. Note also that H, y,,and
the vect 'a1 representation of M2also point in positive directions of the xyz axes.
Hence, a cording to our convention we have drawn a positive shear force, a positive axi I force, and a positive bending moment at the section at x. For the righthand fre -body diagram, the cross-sectional area vector points in the negative x
direction And, since H', V!', and Mlpoint in negative directions of the x, y , and z
axes, th 'e components are again positive for the section at x according to our
conventi m. Clearly, by employing this convention, we can easily and effectively
specify e force system at a section without the danger of ambiguity.
~
i
i
1
iI
A
rigid-bo
provide
will dep
bending
P
Figure 6.31. Section resultant for three-dimensional loading
pointed out earlier, we can solve for Vy,H, and Mz at section x using
y mechanics for either a left-hand or a right-hand free-body diagram
that we know a11 the external forces. The quantities y,, H , and M2
nd on x, and for this reason, it is the practice to sketch shear-force and
moment diagrams to convey this information for the entire heam.
e now illustrate the computation of V and M.
"S me authors employ the reverse convention for shear force from the one that we have
proposed. O u r convention is consistent with the usual convention used in the theory of elarticity
f k thc si n of stress at a point. and it is fnr this rcilson that we have employed this mnvznlian
rather tha the other one.
249
250
CHAPTER 6
INTRODUCTION TO STRIICTIIRAI. MECHANIC’S
Example 6.6
We shall express the shear-fiirce and hending-m(irnent equations fnr thc
simply supported heam shown in Fig. h.RZ(a). whose weight we shall
neglect. The support forces ohtained frcini equilihrium are S O 0 N each.
To get the shear force at a section x, we isnlate either thc left or ]right
side of the beam at x and employ the equations of. equilibrium 011 thc
resulting free hody. If.r lies between A and C [if the heam. the only noniiiternal force present for a left-hand free hody i s the left supporting fwcc
[see Fig. 6.32(h)l. Notice that we have used directions for Vand M (rherc
i s no need for suhscripts in the simple pniblem) correspiinding to the po.sitiw states frrirn the point o f view [if our convention. Clearly, the (i/,qehroi(.
sign we get for these quantities from equilihrium calculations will then
correspond t o the umvmrion sign. If li i s between C and H for such a free
body. two external forces appear lsee Fig. h.32(cJl. Therefore. i l thc shcar
force i s to be cnprcssed as a function of .r. clearly separate equation5 C O Y ering the two ranges. 0 < .r < 112 and 112 < x < /, are necessary. Summing
forces we then get
o<li__<112:
500 + V = 0: therefore. V = ~ ~ 5 0N 0
112 < x < I:
S O 0 - 1.000 + V = 0; therefore. V = 5Oll N
~~~~
~
(h)
Notice from the above results that therc i s a sudden chaiigc o f the shear
force from -S(X) N to +XK) N as we pas? the position of the concentrated
I ,000-N load. Clearly then, the value l i t shear niiist perforce he in~lr/r,rninrrtr at the position of this concentrated load. I t i s f(ir this very reason that.
in the ranges o l applicability of Eqs. (a) and (hl. we have excluded the
positions o f the three concentrated loads ol the problem. Note further that
ilthere wcrc only a diatrihuted load starting at point C. therc would not he
discontinuity in shear and so we would not have to delete the position of
point C i n the rangc applicahility of the shear equatinns.
Now let us turn ti1 the bending-moment equ;lti~ins. Again, UK inust
consider two discrete regions. Taking moments ahout position 1. we gct
s
+
-S00x
M = 0;
therefore,
M = 5011s N-in
IC1
1
t
//2<r 5 i : ~
-soox +
therefore.
i . n 0 0 ( ~- $1
+
M = (I:
M = S O O ( / - x) N-in
,
._--I_
(11)
. ..
...
.
,
.
_”__
..
SECTION 6.8 SHEAR FORCE, AXIAL FORCE, AND BENDING MOMENT
i
4
Exa pie 6.6 (Continued)
In the present problem there are no point couples
(d), we include the entire beam in the combined
Y
A
V
500 N
y
1.000N
V
500 N
(c)
Figure 6.32. Simply supported beam.
It/is generally the practice to express the shear and bending-moment
equatio s successively under common ranges of applicability. In such cimmstances we shall adopt the practice of excluding from any range of applicability a y points of discontinuity for eirher the shear or the bending-moment
equatiobs.
I
25 1
252
lNTR0I)L'ClION TO STRUCTURAL MK'HANICS
CIIAI"I1IK h
r Example 6.7
r
j
!
:
Dctei-mine llic hhcar-force and hending-moment equations for the simply
supported bean shown i n Fig. 6.33. Neglect the weight of tlie hcam.
W e inus1 first find [he supporting Iorccs for the hcani. Hence. we
liave tising the right-hand irulc
K,(2?)
a
-
soo - (50)(X)(X) - (I.ooo)(x,
=0
Ttrcreliirc.
i
R, = 532 Ih
;
ii
i
In Fig. 6.34(a) we have shown a l r a - b o d y diagram exposing sections between the left suppnrt and the uniform Inad. Summing forces and
taking moments ahout a point i n the seclinn. where we have drawn V aid
M a s positive according IO o u r wiivcntioii. we gct
(l<r~<_
4:
868
-86Xi
I
8
+ V
+M
= 0: thcreliire,
= 0: therefore.
V = -868 Ih
M = X68.x ft-lh
The iicxt i i i t e i - v i
i h hctweeii ttir heginning o l Ihc uniform lond and the
poinl fiirce. 'Thus, ohwrving I'ig. h.?4(h):
4~5
.l.~<&
liir
V,
808
:.
-
V
5O(x - 4 ) + V = 0
= 50.x - 1.068 Ih
and fix M
:.
I
+ I.llh8.~- 400 fl-lh
We now ciinsider the interval hctwccn the poinl force and the end
utiilorin l o x i Thus. ohserving Fig. 6.34(c):
?-Ll--S
I
M = -25r'
lor
trf
the
12:
Figure 6.34.Free-hody diagrams irir
various r a n p a
v.
868
-
so(^, - 4 ) - m o +
1' = S t r ~.68 Ih
:.
v =o
SECTION 6.8 SHEAK FORCE, AXIAL FORCE, AND BENDING MOMENT
ple 6.7 (Continued)
,
and forM,
I
-868x
~
i,..
*
+ 5 0 ( x - 4)2 + I,WO(x - 8 ) + M
:. M = -25x2 1 6 8 1 + 7,600 ft-lb
~~
= 0
The xt interval is between the end of uniform loading and the point couple. e can now replace the uniform loading by its resultant of 400 Ib, as
show in Fig. 6.34(d). Thus,
18:
868 - 400 - 1,000 + V = 0
.: 1' = 532 Ih
I
i
~
andforM,
~
-868x
~
I
+ 1,4OO(x - 8) + M
i8
.: M
~
=0
= -5321 111,200 ft-lb
The I st interval goes from the point couple to the right supp&rt. It is to be
point d out that the point couple does mi contribute directly to the shear
force and we could have used the above formulation for V for interval I8 <
x<
. However, the couple does contribute directly to the bending moment,
thus uiring the additional interval. Accordingly, using Fig. 6.34(e), we get
(el
< x < 22:
i
V = 532 Ih
(as in previous interval)
Figure 6.34. (conrinurdJ Free-body
diagrams for various ranges.
~
Whepas for M we have
I
~
j
-868x
+ 1,4OO(x - 8)
.: M
= -532x
-
500 I M = 0
+ 11,700 ft-lh
a downward force P, as shown in Fig. 6.35(a), in-
it a positive bending moment P t [see
P induces on sections 5 to the right
6.36(b)]. Finally, as can be seen
to the right of it (it does not
253
254
CHAPTER h
INTRODUCTION 1
'
0 STRUCTURAL MECHANICS
require from equilibrium a shear force), whereas a counterclockwise couple
moment C [Fig. 6.37(h)l requires ;I negalive bending moment 4' on sections to the right of it. 111 the following example. we shall show how by this
reasoning we may more directly formulate the shear-force and hendingmoment equations.
I
I
Figure 6.36. Rending mornen1 mduced h) P
Figure 6.37. Bending moinent induced hy C.
SECTION h.8 SHEAR FORCE, AXIAL FORCE, AND BENDING MOMENT
Exi
lDle 6.8
Eva
shoy
te the shear-force and bending-moment equations for the beam
in Fig. 6.38.
4 free-body diagram of the beam is shown in Fig. 6.39. We can imely compute the supporting forces as follows, remembering to use
ht-hand rule.
med
the I
CLL
M -0.
The
i
x
”
J?L
Figure 6.38. Simply supported beam.
+ (500)(21) - 800 + (500)(5)= 0
-R,(26)
255
ore,
R, = 4691b
E M , = 0:
Q(26)
The
-
(500)(21) - 800 - (500)(5)= 0
ore,
R, = 531 Ib
We
vie\
111 now directly give the shear force Vand bending moment
M while
g Fig. 6.39. Thus,
n<x<5:
V = -469 Ib
M = 469xft-lb
5 < x < 13:
V = -469
+ 500
= 31 Ib
M = 469s - 500(s - 5 ) = -31x
+ 2,500 ft-lb
13 < x 5 16:
__
9-
~~
V = 31 Ib (same as previous interval)
M = 469x - 500(x - 5 ) + 800 = -31x
i
-
f,
+ 3,300 ft-lb
16 5 x < 26:
V = -469 + 500 + 50(x - 16) = -769 + 50x Ib
M = 469s - 500(x -- 5 ) + 800
= -25~’
50(.x
-
-
- 16)?
2
+ 769x - 3,100 ft-lb
We all present effective methods of sketching the shear-force and bending- m e n t diagrams in Section 6.9.
-
Figure 6.39. Free-body diagram of beam
256
CHAPTEK 6
INTKiXW(710N TO SIRLICTUKAI. R.IECHj\KICS
Bcliire we proceed further. i f must he carefully pointed uut that the
replacement (if a distributcd load by ilsingle rcsultiiiit lorcc is iinly mcaniiigful
fix the pili-ticular lrec hody on which the i i r c e distrihurion acts. Thus, to conipute the rcacticiiis Sur the entire heaiii taken as n tree hody (Fig. 6.40). we can
replace the wcight di\tribution ivO hy the torill weight at position 1.12 (Fig.
6.41). FoI llic hendiiig iiioiiieiit iit ~ x ,the restillant of the loading lur thc Sree
beam.
x
~
Figure P.6.40.
6.41. Formulate ths shear-force and bending-moment equalions
for the simply supported beam.
300 N
Figure P.6.37.
:
!
6.38. F rmulatz Ihe shear-force and bending-moment equationr
for the c' ntilever beam. Do not include the weight of the beam.
J
Figure P.6.41.
-X
6.42. Compute shear force and bending moments for the bent
beam as functions o f s along the centerline of the beam.
IO lhlfl
Figure P.6.38.
and bending-moment equations
Figure P.6.42.
6.43. A simply suppmted beam is loaded in two planes. Thi:
means there will be shear-force components V, and Vz and bend
ing-moment components M: and My.Compute these as function
of x . The beam is 40 ft in length.
I(1)
I
or the beam shown, what is the shear force and bending~
6.40.
moment at the following positions?
( ) 5 ft from the left end
I2 ft from the left end
&)
5 ft from Ihc right end
Figure P.6.43.
2:
1
I
What are tlic shear force, hcnding moment. and axial force
for tlic three-dimensional cantilever hram! Give your results separately for the three porlians A H , H C . and Cl). Neglrct the weight
of thc memhcr. Usc ,s as the dictairce along the ccnterline from I ) .
6.44.
6.47. A hoist can move along a bcnm whilc supporting a l0,OOIlIh l o x . It the hoist st:~rts31 lhe left and moves liom 4 = 3 to I=
I?.dctcl-mine Ihe shear f k e and hmding momcnt at A in terms
01i.
At what priiition i do we get thc ineninurn shear fbrcc ill A
and thc mnxiiniiin hending mornenl a1 ,I,
What
!
ills
their v;ilues'!
I,noo N
x
Y'
,
"
10.000 Ih
Figurc P.6.44
6.45. Oil flow\ from a tank through il pipc AB. Thc oil weighs
40 Ibift' and. in flowing, dzvelops a drag on the pipe r r f I Ihift.
The pipe has an inside diainctci 013 in. and a length of211 ft. f k w
condition?,arc assumed ID he the same along the entire length of
the pipe. What arc the shear force, bending nrmmt, :md axial
force along the pipe? The pipe weighs IO lhirl.
F i p r e P.6.47.
6.48. A pipe weighs I O lhift and has a n imide diameter of 2 i n
Ifil i s full of wiitcr and the p r e w r c irf the water i y that r i f the
atmrnphei~ciit the ciitratlct. A. coniporc thc \hear force, i i x i i i l force.
ant1 hending mimien1 01the pipe from A to 11. l l s e coordinate i
mcaured f m m A olonp the centcrline 01 the pipe.
I
i,'
~1
.
Figure P.h.45.
Determine the shear iixcc, bending moment, and zixiiil
force a s lunctions u i H for thc circular heam.
6.46.
Figure P.6.48.
6.49.
After finding the supporting iorces, detzrmine fur Pnrhli-m
6.37 the \hsar-l<rcc and hending-moment equations without thc
fiirthei iiid of free-hwiy diagrams.
ll&
A
1
4 5 " ~
Figure P.h.46.
258
6.50.
Deterrninc the <hcac-torcr and hendingmoment equittiona
trri Frohlcm 6.7U without the ;lid of frcc~hodydi.',Pl<U"\.
.'
6.51. I n Priihlrin 6..39. alter delermlning the supporting forces.
tletcimine thr shcar-fbrce and hending-motncnt equation? without
thr aid of li-er-hody di,:'Ail:l",b.
i
6.54. Formulate the shear-force and bending-moment equations
for the simply supported beam. [Suggestion: For the domain 5 <
x < IS, it i s simplest to replace the indicated downward triangular load, going from 400 Nlm to zero, by a uniform 4Ml-NIm uni6.53. Gi e the shear-force and bending-moment equations for form downward load from x = 5 to x = IS plus a triangular
the cantil ver beam. Except for determining the supporting forces, upward load going from zero to 400 Nlrn in the interval.]
do not us free-body diagrams.
6.52. In Problem 6.40, after finding the supporting forces, write
the shear force and bending moment as a function of x for the
beam wit out the aid of free-body diagrams.
i
i
1,000 N
4Ml Nlrn
x
I
-
-l5m------t
~
20m
Figure P.6.54.
Figure P.6.53.
I
11 for
Differential Relations
Equilibrium
6.9
1
In Secti n 6.8, we considered free bodies of finite size comprising variable
portion of a beam in order to ascertain the resultant force system at sections
along t beam. We shall now proceed in a different manner by examining an
infinite 'mal slice of the beam. Equations of equilibrium for this slice will
then yi Id diferenrial equations rather than algebraic equations for the variables V n d M .
C nsider a slice AX of the beam shown in Fig. 6.43. We adopt the conventionl that intensity of loading w in the positive coordinate direction is
of the beam has been included
diagram of the element in Fig. 6.44. Note we have
and bending-moment convention as presented
Figure 6.43. Element Ax of beam.
z
'
'
V+AV
Figure 6.44. Free-body diagrdm of element.
forces:
L
F, = 0:
,
i
iL
-V
+ ( V + A V ) + WAX = 0
Taking moments about corner a of the element, we get
M, = 0:
260
CHAPTER 0
INTRODUCTION TO S T R U C T I K A I , MECIIANICS
p
where
i s some lraction which. when multiplied hy Ax, gives the proper
moment arm OS the force I$’ Ax ahout corner (1. Thcsc equations can he written
i n the following niaiinei- after w e cancel tcrins and divide through hy Ar:
i\v -Ax
-I”
AM = -I/
AI
I n the liniit a s h r
i
+ ivPAi
(1. we get the following differential equations:
dV
dx = -w
(6.321)
We may next integrate Eqs. 6.3(a) and 6.3(b) from position I along the
beam to position 2. Thus, UT ha\c
(6.4)
and
(M)?-(M),
= - p h
I
Thercforc,
( M ) , = ( M ) , - JIVdx
(6.5)
Equation (6.41 incans that the ctmige i n the shear lorce hetwccn two points
on a hcam ecluiils m i n u (lie area under the loading curve hetwccn these
pointq provided that there i s no point force present i n thc interval.? Note that.
if w ( . x ) i s positive i n an inlei-val. the area under this curve i b positive i n this
interval: i l w ( l i i is nepative i n a n interval. Ihc iirea under this curve i s negative
in this interval. Siinilarly, FA]. 6.5 indicates that thc change i n hending
moment hctweeri Iwo points on a beam equals minus the area (if the shearforce diagram hetwccn these points pro\’ided that there arc n o point couplc
moments applied in the inter\al. If Vi.r) i s positivc i n an interval. [hc area
undcr this curvc i s positive i n tliib interval: i l V ( x ) i s negative i n an interval,
the area under thc curve i s ncpativc lor this intcrval. In cketching the diagram.
we shall make use oSEq. 6.4 and 6.5 a s well as thc diffei-entii equations 6.3.
“lbu dilirrenfial c q i i ~ l i o i ib.~i,:i)I C o ~ l ymeenmgful with i t conimuous Ionding p r e m ~ i .
h X h j i \ w>ly \ h I iri the Aihirnce 01 p o i n t ‘m~tplcriiorncnt\.
while t$
j
I
SECTION 6.9 DIFFERENTIAI. RELATIONS FOR EQUILIBRIUM
i.
261
Exa pie 6.9
and bending-moment distributions for the simply
in Fig. 6.45 and label the key points.
x
Figure 6.45. Simply supported beam.
~
IThe supporting forces R, and R, are found by rules of statics. Thus,
(-R,(20)
+ (500)(14) + (50)(10)(10/2) - 100 = 0
I
R, = 470 Ib
=
0:
R2(20) - (500)(6) - (50)(10)(15) - 100 = 0
~
Therd fore.
R, = 5 3 0 l b
530 Ih
sketching the diagrams, we shall employ Eqs. 6.3, 6.4, and 6.5-
of the shear-force and bending-moment equations, evaluating
ia)
on the shear diagram that the 470-lb supporting force
x
-470
(b)
2.820
2.025
2,7u(1
-
M
A
V, and we have a 30-Ib shear force at point D. Again, since w = 0 in
1.925
r
11
iC)
I
B
262
rttAPTCR 6
INTRODUCTION TO STRIJCTUKAL blFCIIANICS
Example 6.9 (Continued)
load at D, there i s ino sudden change iii shear as we crash this point. Ncnt.
the change in shear hetween 1) and I1 i s i i i i i i i i s the area (11 the loading
curvef4 i n this interval i n ;scordance with Eq. 6.4. But this ;rea i s
(-SO)( IO) = -500. Hcnce, from Eq. 6.4 the value (IS
(.iu\t Io (he left lit.
the support) i s
- (-500) = 530 Ih. Also, since N i s incpti
stant hetween I1 and B. the slope of the shear curve should he positive and
constant, i n accorkincc with Eq. h.Xa1. Hence. we can draw it straight line
between V,, = 30 Ih and V, = 530 Ih. As we now cross the ]right support
530 Ih on scclioiis 10 thc
force. we see that it induces a negative h e a r <if
right of thc support, and s o at I3 the shear c u r w conics hack t i l zero.
We now proceed with the bending-moment curve. With iio point coilple moment presenr at A. the value o f M,* must he zcrii. The changc in
moment between A and C i s then minus the area underneath the shear curve
in this interval. We can then say from Eq. 6.5 that M,. = M , - (-470)i6)
= 0 + 2,820 = 2,820 It-lh, and we denote this in the nioineiil diagram. Furthermore the value of V i s a negative constiin1 in the interval and. accordingly [see P:q. 6.3(h)J, the slope (ifthc moment curvc i s positive and
constant. We can then draw a straight line hctweeri MAand M,.. Hetwecn C
and 11 the area for the shear diagram i s 120 Ih-St, and s o we can say thal M,,
= M,. - (120) = 2.820 - 120 = 2,700 St-lh. Again. with Vconslant and
positive in the iiitcrvd, the slope of the moment curve must he negativc and
constant in the i n t e r d and has heen so drawn. Hetweeii I ) anti F the iirea
under the shear curve i s readily sccn to he ( 3 0 ) ( 5 )+ + ( 5 ) ( 2 5 0 )= 775 ll-lh.
Hence, the bending moment goes from 2,700 St-lb at 11 to 1,925 St-lh at F.
Now the shear cuwe i s positive and irrrrv,.sing i n value ;IF we go from D to
F. This means that the slope of the bending-moment curve i s ncgatike atid
hecoming steeper a s we go from D to I:.As wc go by F wc eiicounter lllc
100-ft-lh point couple moment and we can say that this point couple moment
induces a positive 100-St-lh mnmcnt on sections to the right (if point F.
Accordingly, there is a sudden increase i n bending ni(iment o l 100 It-lh at F,
as has heen shown i n the diagram. The area ofthe shear diagram hetwcen
F and L( is readily seen from Fig. 6.46(h) to he (2XO)(S)
;(5)(2SO) =
2,025 ft-lh. We see thcn that the bending miiment gocs to zero ill H. Since
the shear Iorce i s positive and inr.reu.sing hetween F and H. wc conclude
that the slope o f the hcnding-inomcnt curve i s negativc and hecoining
,stef’pcras we approach R. We have thus drawn the shcar-force and bending-moment diagram and have labeled all key points.
Note that to he correct, both the shear-force and hending-nioment
curves must go to zero at the end of the beam to the right iif the right sopport. This serves a s a check o n the correctnes iir the calculations.
vj
yj
+
laNotc tlmt the pnim couplr mnment llas a r.cru nrf frorce and b n nerd no1 he ill concern
in the inlcrvill from D lo B a s far as shear i s conccrned. Howcwr. i f will hc ii p m i l ulicic
auddrn chengr occurs in the hending-momcnr diagram
-
. ~
.
-
.
I
I
SECTION 6.9 DIFFERENTIAL RELATIONS FOR EQUILIBRILM
In xample 6.9, we can get equations and diagrams of shear force and
bending moment independently of each other. With simple loadings such as
point fo ces, point couples, and uniform distributions, this can readily be
done. In eed, this covers many problems that occur in practice. Usually, all
that is n eded are the labeled diagrams of the kind that we set forth in the previous pr blem. In problems with more complex loadings, we usually set forth
the equ ions in the customary manner and then sketch the curves using the
equatio s to give key values of V and M (the areas for the various curves are
no long r the simple familiar ones, thus precluding advantageous use of Eqs.
6.4 and 3;the key points are then connected by curves sketched by making
use o f t e slope relations as in Example 6.9.
It ill be helpful to remember that if a curve has inareusing magnitude
(absolut value), the subsequent curve must have a steepening slope over the
hand, if a curve has decreasing magnitude
curve must have aflatrening slope over the
1
n
Figure 6.48. At V = 0, possible
maximum for M.
and bending-moment curves where there is zero value of
be possible maximum values of shear force and bending
263
10 Ihlit
IO K
n
++I++
Figure P.6.55.
I ,lli)i~
Ih
Figure P.6.58.
Si10
N-111
c
Xi1 N l m
+I+++
-+
Figure P.h.57.
!64
Figure P.6.60
6.61. ! tch the shear-force and bending-moment diagrams for
6.64. A cantilever beam supports a parabolic and a triangular
the sin
mamen
load. What are the shear-force and bending-moment equations'!
Sketch the shear-force and bending-moment diagrams. See the
suggestion in Problem 6.54 regarding the triangular load.
,idally loaded beam. What is the maximum bending
/
w
=
sin i i x / L Ib/ft
-X
Figure P.6.61.
Figure P.6.64.
6.62.
for the
imulate the shear-frxce and bending-moment equations
tm. Sketch the shear and moment diagrams.
6.65. Determine the shear-force and bending-moment equations
for the beam. Then sketch the diagrams using the aforementioned
equations if necessary to ascertain key points in the diagrams,
such as the position between the supports where V = 0. What is
the bending moment there?
1,000 Ih
I 00 I b/ft
througt
at right
'imply supported I-beam is shown. A hole must be cut
e web to allow passage of a pipe that mns horizontally
gles to the heam.
lrilrl af
Where, within the marked 24-ft section, would the hole
t the moment-carrying capacity of the beam?
to least
In the same marked section, where should the hole go
e c t the shear-carrying capacity of the beam'?
6.63.
21'-
r
Figure P.6.65.
24'-4
4,000 Ih
4'
60,000 Ib-ft
8,000 Ih 5.000 Ib
3'
l7'-p5'1
\
111 1 1 1 1
Ul 11 1
80 Ib/fl
t
Figure P.6.63.
265
266
CHAI'TCK 6
INTRODIICTION TO SI'KIJ('TIJRA1. MFCllANlCS
Part C: Chains and Cables
6.10
Introduction
We often cncountcr I-elatively llzxihlc cables or chains that are used to s u p
port Iuads. I n suspension hridgcs, lur cxample, we h i d a coplanar arrangement i n which a cable wpports a large load. The weight of ttic cable itself in
m c h cascs may oliun be considcrcd negligible. 111 traiisniissiixi lincs, oil ttrc
ollicr hand, the principal Iircc i s the weight o l the cahle itscI1. In Part C. wc
shall c\'iiluate the sh;ipc 01 and the tension in the cahles f o r both these cascs.
T u Ihcilitalc computations. thc model OS Llic structural system will he
assunicd 1 0 bc perfectly flexible and inextensihlc. The llexihility assumption
ineiins that at the center of any cross section 01the cahle only a tensile force
i s lransmittcd and there can hc no bending municnt there. The Surcc transmitlcd Iliriiugh tlie cablc inusl. under these conditiiins. he tangent to the cable at
all posilions along thc cahle. The inextcnsibility assumption means that the
length of thc cable i h consveiit.
6.11
;I/-\
Function of x
r~_r/
~
/
~-
Coplanar Cables: Loading is a
I,
We shell iiiiw consider lhc case of a cahle suspended helwecn two irigid S U ~ porls A and 1) under thc action o f a loading function W)I(
given per unit
length a s measured in the hnri:ontnl direction. This Ii,ading w i l l bc considcrcd to he copliinar with the cahle and directed vertically. as shown in Fig.
6.49. Consider an clcnient o f t h e cahlc of length A.7 a s a (tee hody (Fig. 6.50).
~~
~~
~~~~
~
Figure 6.49. Copl;inar cahlr; i v =
Y
~~~~
~
I*'(,)
Summing liirces in the .r and
directions, respectivcly. we get
- 7 . ~ 1 s 6'
-Tsin H
+
(T
+
+
('I'+ A T ) cor($
A / ' ) sin (6'
+
+
A@ = 0
(6.6a)
AH) - w , ~A.l; = 0
(h.6b)
SECllON 6.11 COPLANAR CABLES; LOADING IS A FUNCTION OF X
i
I
$
where
taking t
i
is the average loading over the interval Ax. Dividing by Ax and
limit as A x + 0, we have
~
~
lim
Ax-0
i
i
[( T + AT)cos(BA+x AO) - TcosO
~~
lim [ ( T + AT) sin(O + AO) - T sin B
Ax
41-0
iI
The terd w is now the loading at position x . The left sides of the equations
accordance with elementary calculus, and so we can
(6.7a)
(6.7b)
From Ed. 6.7(a), we conclude that
Tcos O = constant = H
(6.8)
where c/early the constant H represents the horizontal component of the tensile forcb anywhere along the cable. Integrating Eq. 6.7(b), we get
i
TsinO = / w ( x ) d x + ~ ;
(6.9)
~
d
where
is a constant of integration. Solving for Tin Eq. 6.8 and substituting
into Eq. 6.9, we get
~
t
Noting hat sin Olcos O = tan 8 = dyldx, we have, on canying out a second
integrat on:
t
Equatioh 6. 10 is the deflection curve for the cable in terms of H , w(x). and the
constan s of integration. The constants of integration must he determined by
the bou dary conditions at the supports A and B .
261
+
SECTION 6.1 I
i
COPLANAR CABLES; LOADING IS A FUNCTION OF X
Exa ple 6.10 (Continued)
the largest 0 occurs at x = 112 (i.e., at the supports).
we have, for
Cons+pently, we get for T,,:
H
T,,, = cos[ tan-I ( ~ 1 1H2 ) ]
Fromltrigonometric consideration of the denominator,
I
Substituting for H using Eq. (c), we then get, on rearranging the terms,
I
]Finally,
to determine the lrngth ofrhe cuble for the given conditions,
we m st perform the following integration:
P
F
Now he slope, dy/& equals wxlH [see Eq.
H Lse Eq. (c)] becomes 8hxll.z Therefore,
This
(01, which on substituting for
inay be integrated using a formula to be found in Appendix I to give
I
Subs ituting limits, we have
1
r
Rea anging so that the result is given as a function of the sag ratio hll and
the s an 1, we get finally
!
269
,
SECTION 6.12 COPLANAR CABLES: LOADING IS THE WEIGHT OF THE CABLE ITSELF
i
I
Dividinglthrough by As and taking the limit as As
analogo+ to Eqs. 6.7.
+ zero, we get equations
Integratilg, we have
I
i
~
TcosO = H
(6.I I a)
Tsine=J’w(s)ds+~;
(6.1 1b)
Eliminating Tfrom Eqs. 6.1 I , we get, as in the previous development:
9
=
w(s)d.s + C,
dx
equation is a function of s. Thus, we cannot directly inteAccordingly, note that
dy = (d.? - dx2)1’2
~
Hence, flom this equation,
(6.13)
Substituting for dyldx in Eq. 6.12 using the preceding result, we get
Solving lor &/&,
we have
Separatihg variables and integrating, we get
I
~
i
I
~
(6.14)
i
~
t step, if possible determine the constant C , by applying a slopein Eq. 6.14, solve for s as a func-
271
272
CHAFTER 6
INTRODUCTION IO STRUCTURAL M1:CHANICS
Example 6.11
Consider a uniform cablc having a span / ;uid ii sag /I a\ s h i n \ n i n Fig.
6.52. The weight per u i i i l length I I ' the
~ cahle is a constant. Ikteriiiiiic ihc
shape of the cahle when i t i( loaded only by its own weight.
For simplicity. we have placed a relerencc at the ccntcr of the span
where the slope or (he ciiblc i h Lei-o. Acciirdingly. considw Eq. 6. I2 for
rhia case:
When s = 0 we I-equirc Ihal &Idr
sider Eq. 6.14:
= 0.whcrcupiin
i s /.cro. NOM- c o w
Integrating the right side 01 the equation using integralion forniul;i IO lrom
Appendix 1, wc gct
SECTION 6.12
COPLANAR CABLES; LOADING IS THE WEIGHT OF THE CABLE lTSELF
Example 6.11 (Continued)
The constant C, must also he zero, since x = 0 at s = 0. Solving for
from Eq. (cj, we get
s
H .
s
xw
= - sinh -
w
H
Substituting for s in Eq. (a) using the preceding result, we have
dv
rr;r
. w
= sinh - x
H
Integrating, we get
Since y = 0 at x = 0, the constant C, becomes -Hlw. We then have the
deflection curve:
This curve is called a catenmy curve.''
To determine H , we set y = h when x = 112. Thus,
This equation can be solved by trial and error by the student or by a computer. We may then proceed to determine the maximum force in the cable
as well as the length of the cable in the manner followed in Example 6. 10.
'"rhe Latin for chain i s curenu
213
Example 6.12
A watcr ckicr i s shown i n Fig. 6.53 dmpling Iron1 a kite that is towed hy a
powerhoal at a speed i r I 30 inph. 'The hoat devzlops a thi-ust (11 200 Ih. Thz
drag o n the hoar from the water i s estiinated a s 100 lti. At Ihc sopport A.
the ropc has a tangent ol 30". I1 the man weighs I50 Ib. find the height and
the lift of thc kite as well as the maximum tension i n the rope. The kite
weighs 25 Ih. Thc u i i i l i r n i rupc i!. SO I t long ; ~ ~ weighs
id
3 Ih/It. Neglect
aerodynamic eflccts on the rope.
We stail with Eu. 6.12. which heciimcs f i r this ca.e:
lJsing a reference at A as shown i n the diagram, we know that dy1d.t = tall
30- = 377 when i = 0. Thus, we get for C , .
= 577
Equation 6.14 i b c o n d e r e d next. We have
+
Integrating by making a change i n variable to replacc [(w/H).s ,5771 and
using the inlegration f(irmula 10 i n Appendix I. we get
Solving kir s, we have
Substituting Sor s in Eq. (a) using Erl. ( 2 ) . we get
lnlegraling again, we have
We must now zvaluatc thc unknown constants C,, <'~+. and H using
thc boundary conditions and data of thc priihlem. First.iince H is the horizontal comp~inentof fbrce transmittcd hy the rope. we k n o w that H i s the
thrust of Lhe hoat minus the drag of Ihe watcr. 'Thus.
H = 100lh
Also, x = 0 when s = 0 . s o that from Eq. (,e).we get C2 a s iollows:
sinh(-
5
('
100
A
) = ,577
-
SECTION 6.12 COPLANAR CABLES, LOADING IS THE WEIGHT OF THE CABLE ITSELF
Example 6.12 (Continued)
Therefore.
- loo
_
" C
2
=
sinh-' ,577 = ,549
Hence,
C, = -109.8
Finally, note that x = 0 when y = 0. From Eq. (e), we can then get constant C, in the following manner:
C, = -Tcosh[s(-109.8)]
100
100
= -200 cosh ,548 = -23 1
We may now evaluate the position x', y' of point B of the kite. To
get x', we insert for s in Eq. (b) the value of S O ft. Thus,
Now from Eq. (e) we can get y' and consequently the desired height.
+ 109.8)-]-100
5
y' = K c o s h [ ( 4 0 . 9
.s
231
,,.
..-X' = 28.6 fc
(0
The maximum tension in the rope occurs at point B, where 0 is
greatest. To get Omdx, we go back to Eq. (a). Thus,
Therefore,
=
39.6"
Hence, from Eq. 6.1 I(a) we have for :,"I
To get the lifting force of the kite, we draw a free-body diagram of
point B of the kite as shown in Fig. 6.54. Note that F, and F,c are, respectively, the aerodynamic lift and drag forces on the kite. The lift force F y of
the kite then becomes
F, = 175
+
Tmaxsin 39.6" =
'
(i)
Figure 6.54. Free-hody diagram
of kite support.
275
6.66.
Find the length of a cahlc wctched hetween twu suppolts
\pen / = 200 ft and s q I t = 50 fi, if it
i s huh.jccted to a vertical h a d u t 4 lblft uniformly dist~ibutedin thc
h o r i i ~ n t direction.
~l
(Assume that the weight of the cable is cithcr
nrgligihle o r includcd in the 4~lhIftdistrihutiwi.) Find the maxiat thr sainc elevation with
m u m tcnsioii.
The lefl side of a cahle i s mourned at an elcvatim 7 111
hclow the right side. The \ag. msaiorcd from the left iuppon, i s 7
m. Find the niaximum tension if the cahlr has il un(fiwn2 loading i n
the vertical direction of 1500 Nlm. [ B c , y f i c v f i o , z ; Place rcfcrence at
position of zero slope and dctcrmiiie the location 01 this point
fram the houndary conditions.]
6.70.
6.67. A cahlc xqports an X000-kg uniform har. What is the equation d w r i h i n g the shape of the cahle and what is the inaxiinurn
tension i n the c;ihle'l
-
i
?Ktn +-I
1
Figure P.6.70.
Figure P.6.67.
6.6X. A cable supports a uniform londing U I100 Iblft. If the lowest point of the cablc occur&20 ft S r m ? point A as shown, what is
thc n i i i h i i n ~ i nt e i i h i i m i n the cahle and its length? Usc A as the origin of rrtcrrncc.
X(]'.__~
6.71. A blimp is dragging a chain of leiiglh 400 f t and weight 10
IMft. A thrust of 300 Ib is developed hy the hlimp a b i t moves
against an air resistniicc of 2~10Ih. HOWmuch chair, i\ 011 the
ground and how high i ? the hlimp? l h e vertical lift of !tie blimp
on the cable is takzn as 1,000 Ib.
-L
.
R
i
Figure P.6.68.
6.69. A unifixm cahlc ih \ h o w whose weight we shall ncglrct.
If a hading givcn ac 5r N l m is imposed wi thc cahle, what i s the
dctlcctim curve of the cable i t thew i s a zero slope of thc curve at
point A'! What is the maximum tunsian?
v
Figure P.6.69.
Firurr P.6.71.
A Iargc halloon has a hrioymt fixci. of 100 Ib. 11 i s held hy
a l50-ft cahle whose weight is .5 Ih/ft. What i \ !he hcight /z of the
balloan above the ground w h m a steady wind cuuscs i t 111 assun~s
thc position shuw,n? What i s the maximum tension i n thc cahlc'l
6.72.
Figure P.6.72.
6.13. What is the detlection curve for the uniform cable shown
weighing 30 N/m? Find the maximum tension. Compute the
height h of the support B .
6.16. A flexible, inextensible cable is haded by concentrated
forces. If we neglect the weight of the cable, what are the supporting forces at A and B? What are the tensions in the chord AC and
the angle d![Hint: Proceed by using finite free bodies and working from first principles.]
Figure P.6.76.
Figure P.6.73.
6.14. A search boat is dragging the lake floor for stolen merchandise using a 100-m chain weighing 100 N/m. The tension of
the chain at support point B is 5,000 N and the chain makes an
angle of SV there. What is the height of point B above the lake
bed'? Also, what length of chain is dragging along on the bottom'!
Do not consider buoyant effects.
-
6.11. A system of two inextensible, flexible cables is shown
supporting a 2,000-lb platform in a horimntal position. What are
the inclinations of the cable segments AB, B C , and DE to accomplish this and what lengths should they be'! Neglect the weight of
the segments and note the hint in Problzm 6.76.
Figure P.6.71.
Figure P.6.14.
6.15. A cable weighing 3 Ib/ft is stretched between two points
on the same level. If the length of the cable is 450 ft and the tension at the points of support is 1,500 Ib, find the sag and the distance between the points of support. Put reference at left support.
6.13
closure
Essentially what w e have done in this chapter is to apply previously developed material t o situations of singular importance in engineering. Further
information (in structures can be found in hooks on strength of materials and
structural mechanics. We turn again to new material in Chapter I , where w e
will discuss the Coulomb laws of friction.
8LZ
I
6.82. Express the shear-force and bending-moment equations
with the aid of free-body diagrams. Then express Vand M without
the diagrams.
Y
6.85. Sketch the shear-force and bending-moment diagrams
labeling key point.5.
Y
x
x
Figure P.6.82.
m
m
Figure P.6.85.
6.83. Express the sheat-force and bending-moment equations
6.86. Find the shape of a cable stretched between two points on
without the aid of free-body diagrams.
the same level, I units apan with sag h , and subjected to a vertical
loading of
w ( x ) = 5cos
l
500 N/m
n.x
rN/m
distributed in the horizontal direction. The coordinate x is measured from the zero slope position of the cable.
6.87. (a) By inspection, which members in the truss shown have
a zero force for the given loads?
Figure P.6.83.
(b) For the Fink ~ N S S in Fig. P.6.I.with vertical loads on
the bottom pins, which members will carry a zero force?
6.84. Give the shear-force and bending-moment equations for
the beam, and sketch shea-force and bending-moment diagrams.
At what position between supports is the bending moment equal
ti) Lero?
I .ooo Ib
+
10'
t
50 Ib
Figure P.6.84.
6,000 N
Figure P.6.87.
279
30'
A
4,000 N
Figure P.h.88.
Figure P.6.91.
Friction
Forces
7.1
Introduction
Friction is the force distribution at the surface of contact between two bodies
that prevents or impedes sliding motion of one body relative to the other. This
force distribution is tangent to the contact surface and has, for the body under
consideration, a direction at every point in the contact surface that is in opposition to the possible or existing slipping motion of the body at that point.
Frictional effects are associated with energy dissipation and are therefore sometimes considered undesirable. At other times, however, this means
of changing mechanical energy to heat is a beneficial one, as for example in
brakes, where the kinetic energy of a vehicle is dissipated into heat. In statics
applications, frictional forces are often necessary to maintain equilibrium.
Coulomb .friction is that friction which occurs between bodies having
dry contact surfaces, and is not to he confused with the action of one body on
another separated by a film of fluid such as oil. These latter problems are
termed lubrication proh1em.s and are studied in the tluid mechanics courses.
Coulomb, or dry, friction is a complicated phenomenon, and actually not
much is known about its true nature.’ The major cause of dry friction is
believed to be the microscopic roughness of the surfaces of contact. Interlocking microscopic protuberances oppose the relative motion between the
surfaces. When sliding is present between the surfaces, some of these protuberances either are sheared off or are melted by high local temperatures. This
is the reason for the high rate of “wear” for dry-body contact and indicates
why it is desirable to separate the surfaces by a film of fluid.
‘Fur a imme complcle discussion ol friction. rce F. P. Bowdm and D. Tabor, The Friction
und Luhrkution oj‘Solk/.~.Oxfwd Unircraiiy Press, New York. 1950.
28 I
We have previously employed the ternis “smooth’ and “rough” surfaces of contact. A “smooth” surface can only suppnlt a normal force. On the
other hand. ii “rough” surface in addition can support a force tangent to the
contact suiface (i,e,, a frictiiin iorcc). In this chapter. we shall consider certain
situatinns wherehy the friction Snrce can be direclly related to the normal
force at a surface of contact. Olhcr than including this incw relationship, we
use only thc usual static equilibrium equatiwis.
7.2
(‘omlition of imprndine
Figure 7.1. Idealized plot of applied forcc P.
L.r
Fig,ure 7,2,
Laws of Coulomb Friction
Everybody has gnne thniugh the experience nf sliding lurniture along a floor.
We exert a continuously increasing force which i s ciiinpletely resisted hy friction until the nhject heginc to nii)ve-usuall)
with ii lurch. .The lurch occurs
hecausc once the oh.ject hegins to move. there i s ii decrease i n firictional force
from the maximum force attained under static condili(ins. An idealized plot nf
thi? force as a function of tiinc i s shown in Fig. 7. I where the force P applied
to the furniture, idealizcd as ii hloch in Fig. 1.2. i s shown 10 drnp from the
highcst or limiting valiie 10 a Iowcr value which i s constiint with lime. This
latter constant value i s independent iif the velocity o f the nhject. The condition corresponding to the maximum value i s termed thc conditiiin nf irnpfvidirrg rnoriiiii or itnpmdinI: slippqqe.
By carrying out experiments on hlocks tending to i n w e without rotation or actually moving without ~ri~tatioii
on flat surfaces, Couliimh in 1781
presented certain concIusions which ;ire applicable at the c(inditiini of
impmding .slippafie os once clippuge I i a . ~hegirn. These h a w since hecomc
known as Coulomh’s laws of friction For hliick psnhleinr. hc reported that:
1. The total fnrce nf f r i c l i m that can he develiiped i s independent of the
magnitude (if the area nf c ~ n t a c t .
2. For low relative velocities between sliding objects. the lrictional force i s
practically independent n i veliicily. However. the sliding lrictiiinal force i s
lcss than the frictional fnrce corres~oiidingt n inipending slippage.
that can he developed i s prnportional to the tiorinal force transmitled iicrnss the surfice of cnnlacl.
3. The lotal frictional force
Conclusions I and 2 iiiay cnime as a surprise IO inost of you and be coiilrary to your “iiitiiition.” Nebel-theless. they arc iiccuraft’ enough statements
for many engineering applicalions. More precise studies 01 friction, as wds
pointed out earlicr, are complicated and involved. We can rxprcss conclusion
3 mathematically a h :
1 - N
Therefore.
f = }‘A’
where ji i s called the (.mf/i,.irniq:fi-icriori.
(7.
I)
SECTION 1.2 LAWS OF COULOMB FRICTION
Equation 7.1 is valid only a t conditions uf impending slippage or while
body is slipping. Since the limiting static friction force exceeds the
dynamic friction force, we differentiate between coefficients of friction for
those conditions. Thus, we have coefficients of static friction and coefficients
of dynamic friction, pTand p,,, respectively. The accompanying table is a
small list of static coefficients that are commonly used. The corresponding
coefficients of friction for dynamic conditions are about 25% less.
the
Static Coefficients of FrlctionZ
Steel on cast iron
Coppe,.ron steel
Hard steel on hard steel
Mild steel on mild steel
.40
36
.42
Rope on wood
.57
.70
Wood on wood
.20-.75
Let us consider carefully the simple block problem used to develop the
laws 0 1 Coulomb. Note that we have:
1. A plane surface of contact.
2. An impending or actual motion which is in the same direction for all area
elements of the contact surface. Thus, there is no impending or actual rotation between the bodies in contact.
3. The further implication that the properties of the respective bodies are uniform at the contact surface. Thus, the coefficient of friction p is constant
for all area elements of the contact surface.
What do we do if any of these conditions is violated? We can always choose
an infinitesimal part of the area of contact between the bodies. Such an infinitesimal area can he considered plane even though the general surface of contact of which it is an infinitesimal part is not. Furthermore, the relative motion
at this infinitesimal contact surface may he considered as along a straight line
even though the finite surface of which it is a part may not have such a simple straight motion. Finally, for the infinitesimal area of contact, we may consider the materials to he uniform even though the properties of the material
vary over the finite area of contact. In short, when conditions 1 through 3 do
not prevail, we can still use Coulomb’s law in the small (i.e., at infinitesimal
contact areas) and then integrate the results. We shall call such problems
complrx surface contact problems and we shall examine a series of such
problems in Section 1.4.
ZF. P.Bowden and D. Tabor. The Friction and Lubrication ojSolidr, Oxford University
Press. New York, 1950.
283
"7.3 A Comment Concerning
the use of Coulomb's Law
Ac a simple illu\lralion <if why curvature
(11 the surtacc of contilcl is sccond
order and hence negligihlc. consider yourself at a location on the earth where
i t i s perfectly round as ii planet. As you look around you over a small area
compared to ii significant portion o f thc earth':, surSacc. there i s no evidence to
your ohservatiini ur in what yiio normally do that indicares the presence of curvature of the earlh. In Lhc same way, an infinitesimal arca ciin he considered
Slat c v c ~ii f i h part OS II finite curved :,urhcc when considering Couloinh's law.
Furthcrinore. a s~ntion;iry. iiiinrolatinp ohserver in inertial space looking at ii similar sniiill iirzii on the earth's surfice a1 Ihc equator sccs the ~ e l o c ity of the area essentially as given by R o where R i s thc radius o f t h e Earth
;ind w i:, its angular velocity. All parts o f this small arca w i l l have this same
velocity up ti) only s c c o n d - d e r varialio~ias seen hy this observcr. And, so
friim this viewpoint. all points within the area are esscntially translating with
the af~,rcmentioned speed. 'lo explain lurther. we w i l l now say that lhis small
area has a maximum dimension given by thc Icngth r. The rolutioilul specd of
;my point seen by an inertial and herice inonriitating observer at the area and
tran4atinp witli the area must he no greater than v u . Clearly then, this rotational spccd i s iie,qli~yih/ysmoll when compared with Rw. Hence, an infinitesimal arca at a finite dislancc from the axis (iI~r(it~tio11
01ii rotating surface can
he considcrcd as having primarily a translational velocity and so permits the
use of Ciiulnmh's law ''in the small."
7.4
Simple Contact Friction Problems
We iiow examine simple ciintacl problems where Ciiuliimh's laws apply to
the coiitiict surtice ci.s CL ~ ; h o l rwithout requiring integration pri~cedures.We
shall thus consider uniforni blocklike hodics akin to those used by Coulomb.
Also. we shall considcr hodies which however complex havc very smrrll coliL&t
' . surf'accs.
.
such a:, i n Fig. 7.3(al. Clearly. the whole c ~ n t a e tsurface can
then he coiisidcred an inlinitesinial plane area and for rzasons set forth earlier, we shall directly use Coulomh's laws whcn approprialc as has been
rhown i n Fig. 7.3(h).
Beiorc proceeding to the examples. we have one additional point to
make. For ii l'inite simplc surface of contact. such as the block shown i n Fig.
7 . 2 . we iiiust note that wc do nut generally know the line [if action fur the
simplest rehultant supporting force N, since wc do not generally know the
iiorniiil force dislrihution hetween the two hodics. Hencc, we cannot take
moments for such free-body diagrams without intriiducing additional unknown
distanccs i n the equatiiili. Conccquently. for such prohlems we liniit ourselves
to suiiirniiig iorces only. This i h tior true. however, when w e havc a poiiu
SECTION 1.4 SIMPLE CONTACT FRICTION PROBLEMS
contact such as in Fig. 7.3(a). The line of action of the supporting force must
be at the point of contact, and we can thus take moments without introducing
additional unknown distances.
Figure 7.3.(a) Small contact surface;
(b) Coulomb's laws applied.
Two common classes of statics problems involve dry friction. In one
class, we know that motion is impending, or has been established and is uniform, and we desire information about certain forces that are present. We can
then express friction forces at surfaces of contact where there is impending
or actual slippage as pN according to Coulomb's law and, using for other
friction forces, proceed by methods of statics. However, the proper direction
must be given to all friction forces. That is, they must oppose possible,
impmding, or actual relative motion ut the contact surfaces. In the second
class of problems. external loads on a body are given, and we desire to determine whether the friction forces present are sufficient to maintain equilibrium. One way to attack this latter type of problem is to assume that
impending motion exists in the various possible directions, and to solve for
the external forces required for such conditions. By comparing the actual
external forces present with those required for the various impending
motions, we can then deduce whether the body can he restrained by frictional
forces from sliding.
The following examples are used to illustrate the two classes of
problems.
L.
285
286
CII,APTCR7
~ R I C T I O NFORCLS
Example 7.1
A n automnhile i s shown i n Fig. 7.4(;1) o n ii roadway inclined at an anplc H
wilh thc hprizontal. I1 the coefficients of. static and dynamic friction
'
between the (ires and the road are taken RS (1.6 and 0 . 5 . rcspectibely. what
; i s the ni;lxinnum inclination H,><;,%that the car can climb at uiiiS<)rm speed'!
I t has rear-wheel drive and has a total Iuadcd weight 013.600 Ih. The celltcr (11gravity ior this loaded conditinn ha\ been shown in thc diagram
I.ct us assume that the drive wheels do not "spin"; thit
rclative v c l i i ~ i t yhetween the tire sui-fke and the niad sur1.a
Thai, clearly, the maximum friction liirce pixsihlc i s
times the
: c1111t~1ct.
inorinill forcc at lhis contact surldcc, iis ha\ hccn indicated i n Fig. 7.4(hj.'
We cain considcr this to be a u ~ p / ( i n ( i iproblem
with thrcc unknnwns.
1 N,. N2. and Hc,l,th. A c c d i n g l y . sincc the fricticin force i s reslricted to a
point. three equation\ [if equilihriutn arc avnilablc. IJsing llie rcicrcnce .rv
shown in the diagi-an. we have:
~
!
1
F; = 0 :
~~
~
~
.6N, - 3,600 sin B,,,,,
= 0
(a)
CF; = f):
N,
CM:,
ION,
+ N,
3.600 cos H,,,,,, = 0
~
(hi
0:
~
(3,h(10cosH,,,,,)(~)
+ (3.600sinH,,,;,,)(I)
= 0
'l'n solve Sor H,,,:,,. wc eliminate N , Srnm Eqs. (a) and (bl, gctting a s
the equatiiin
N,
= 3.600 cos
e,,,,,
~
0,000 sin
(Ci
B
rcsult
(d)
Now. elimin;iting N, froin Eq. ( c j using Eq. (dl, wc get
I X . 0 0 0 cos B,,,,, - 56.400 sin 0,,,:, = 0
~
'There fire,
tan Hrl,2,x= ,320
(C)
B,,,, = 17.1"
(1)
Hence.
!
!
i
i
i
-
If the drive wheel5 wcre caused to spin. we would have to use p,, in place
iit ii smaller H,,,,,,. which S i x
0 f p 5for this pnihlein. We would then arrive
this problem would bc 14.7".
'You will nolicc lllnl lhcrc I \ IIO iriclicin Imcc o n th? Iront wheel\. 'This IS SI) hecauw
m q u c ciirrririg iron1 rhr autorrwh~ie'sIMmmission onlcl lhrse uhcei,. whilc it1 the
siiiiic liinc llic whceh .IK 1oiii1~ngill c o i i h t i i n i speeil. Note we are ncplroing "nilling" ve\is.
~iinccilsmriiiig lmni the i l c l o ~ ~ ~ m0 i1oihc
n ro.d \urTacc and lhc mc. i j s~iiiillIorcc 10 he c o w
iidered an Sciiion 1 . 8 . a \i;wved hcciion
ihcir
I \ IIW
SECTION 7.4
SIMP1.E CONTACT FRICTION PROBLEMS
Example 7.2
Using the data of Example 7. I , compute the torque needed by the drive
wheels to move the car at a uniform speed up an incline where f3= 15".
Also, assume that the brakes have " locked while the car is in a parked
position on the incline. What force is then needed to tow the car either up
the incline or down the incline with the brakes in this condition? The
diameter of the tire is 25 in.
A free-body diagram for the first part of the problem is shown in Fig.
7.5(a). Note that the friction f0rce.f will now he determined by Newton's
law and not by Coulomb's law, since we do not have impending slippage
between the wheel and the road for this case. Accordingly, we have, f0r.f:
c
F =n:
______
1
.f
-
3,600 sin 15' = 0
Therefore,
,f= 932 Ib
The torque needed is then computed using the rear wheels as a free body
[see Fig. 7.5(a)]. Taking moments about A, we have
torque = ( f ) ( r )= (932)(*$)
971 ft-DJ
=
For the second part of the problem, we have shown the required free
body in Fig. 7.5(b). Note that we have used Coulomb's law for the friction
forces with the dynamic friction coefficient p,, on all wheels. We now
write the equations of equilibrium for this free body.
TuD- S ( N , + N z )- 3,600sin 15'
=0
(N, + N 2 )- 3.6Oocos 15" = 0
(a)
(b)
Solving for N, + N2 from Eq. (b), and substituting into Eq. (a), we can now
solve for Tup.Hence,
Ttzp= (.5)(3,600)( 966) + 932 =
(c)
287
,
-
I,,,,.,
'
SECTION 7.4 SIMPLE CONTACT FRICTION PROBLEMS
Example 7.3
In Fig. 7.6 a strongbox of mass 75 kg rests on a floor. The static coefficient of friction for the contact surface is 20. What is the largest force P
and what is the highest position h for applying this force that will not
allow the strongbox to eithcr slip on thc floor or to tip ?
The free-body diagram for the strongbox is shown in Fig. 7.7. The
condition of impending motion has been recognized by the use of
Coulomb’s law. Furthermore, by concentrating the supporting and friction
forces at the left corner, we are stipulating impending tipping for the prohlem. Thesc two impendin&conditions impose the largest possible values of
P and h that we are seekine.
The pertinent forces constitute a coplanar system of forces at
the midplane of the strongbox. We procecd with the scalar equations of
equilibrium:
-1
.h m
4
Figure 7.6. Strongbox
being pushed.
I
N = 75g = 7 3 6 N
Figure 7.7. Impending tipping and
dlpplng
P = .2N = 147.1.5N
~ ( 7 5 g ) ( . 3 ) +(147.1S)h = 0
Therefore, we get for the largest P and the largest h the following
results:
P,,,sx = 147.15 N
h,,,
= 1.50 m
i
Thus, the height of the applied load must he less than or equal to I .SO m in
order to avoid tipping.
The three examples presented illustrated the,first type of friction prohlem wherein we know the nature of the motion or impending motion present
in the system and we determine certilin roorces or positions of certain forces.
In the last example of this series, we illustrate the .second type of friction
problem set forth carlier-namely the problem of deciding whether bodies
will move or not move under prcscrihed external forces.
289
290
CHA1"Ik;K 1 FRICTlClN bOK('ES
Example 7.4
1 Thc ciicfficienl of static friclicin kir all contiic1 surfaces i n Fig. 7.X
I
1
i s .2.
Does thc 50-lb force move the block A iip. hold i t in equilihrium. or i s il
too small to prcvmt A from coming down and H fr(nn moving w t ' ! The
50-lb fiircc i s exerted at the niidpl;inc iif the hlircks s o llial w e cain consider
this ii coplanar pmblcm.
4
.i
I
200 Ib
We ciin cmipute ii liirce P i n placc 111. the 50-lh fhrcc I(I cause i n pending m o t i i i ~01
i block H to the left, and ii forcc P fb1- impcnding m i l i o n
of hlock R to the right. I n this way, we can judge hy comp;irism the action
that tht: SO-Ih tiirce w i l l C:IUSZ.
The f r o - b o d y &igrains for impending nr(ilion o l h l o c k H to the l e f t
have hccn shown i n Fig. 7.9, which conlains the unknown fwcc I' iiieiitioned abiivc. We necd not be ciincerned ahoul the correct Iocatioii (11the
centers of gfii\'ity n1 thr hloch\, since wc shall only add forces i n the
analysis. ( W c do not know the line of action o f thc inorniiil liirccs iit
lhc cwitact surlxec and thcrefore in not takr moments.) Summing forces
on block A . wc gel
iV2cos I S " j - N , s i n
+ N,i
-
.2NZco\ 15"i
IS''i-~ . 2 N , j - X ( ) j
'. .2N, sin IS':j = 0
The scalar c q i i i i t i i i i i ~arc:
N , .?59N, .1932N,
-
.OhhN, - .?N,- 200
-
-
=0
.051XN, = 0
Solving cimultiineously. wc get
N, = 243 Ih.
N, = 1OY.R Ih
SECTION 7.4 SIMPLE CONTACT FRICTION PROBLEMS
Example 7.4 (Continued)
For the fre.e-body diagram of B. we have, on summing forces,
-N2 cos 157 + Nz sin 15"i - Pi + .2N3i + N , j
- 1 0 0 j + . 2 N 2 c o s 1S"i+.2N2sin ISy=O
This yields the following scalar equations:
+ 62.9 + .2N3 + 46.9 = 0
+ N , - 100 + 12.6 = 0
-P
-235
Solving simultaneously, we have
P = 174 Ib
Clearly, the stipulated force of 50 Ih is insufficient to induce a motion of
block B to the left, so further computation is necessary.
Next, we reverse the direction of force P and compute what its value
must be to move the block B to the right. The frictional forces in Fig. 7.9 are
all reversed, and the vector equation of equilibrium for block A becomes
N2 cos 15y - Nz sin 1S"i + .2N,j - 2OOj
+ N,i + .2N2 cos 1% + .2N2 sin l5y = 0
The scalar equations are:
N , - .259N2 + .1932N2 = 0
.966N2 + .2N, - 200 + .0518N2 = 0
Solving simultaneously, we get
N2 = 194.1 Ib,
N , = 12.80 Ib
For free body B we have, on summing forces,
-N2 cos l 5 . j + N2 sin 15Oi + Pi - .2N3i + N 3 j
- lOOj - .2Nz cos 157 - .2N2 sin 157 = 0
The following are the scalar equations:
P - .2N3 + 50.3 - 37.5 = 0
-100
+ N3 - 187.5 - 10.05 = 0
Solving, we get P = 46.7 Ib. Thus, we would have to pull to the right to get
block B to move in this direction. We can now conclude from this study
that the blocks are in equilibrium.
291
A hlock has a force F applied to it. If this force has a timc
variatioii as shown in the diagram, draw a simple sketch showing
the friction force variation with time. Take p3 = .1 and p,, = .2 for
the problem.
7.1.
Figure P.7.4.
7 . 5 Explain how a violin how. when drawn oyer a string, maintains the vibration oi the string. Put thi.: in leimi 0 1 friction forccs
and the difference i n static and dynamic cocfficients of frictirin.
I
7.6. Whit is thc valuz of thc force F. inclined at 30" to thc hnri~ o i i t a l ,nccded to get thc hlock juct ctartetl up the incline? What is
the forcc F needed t u keep it .just moving up at s constant speed'!
The coefficients of ctntic and dynamic friction arc .3 ;md .?lF,
respectively.
W = 100 Ih
Figure P.7.1.
7.2. Show hy increasing the inclination @ o nan inclined surface
unlil there is impending slippage o f supported hodics. we rrach
the anfile ofrepose 9, so that tan $$= p>
Figure P.7.6.
1.3. Tu what minimum angle must the drivcr elevate thc dump
bed of the truck to cause the wooden cratc of weight W to slide
out? For wood on steel, p, = .6 and p,, = .4.
7.7. Bodies A and L( weigh 500 N and 300 N, respeclivcly. The
plalfol-m nn which they are placed is raised from thc horizontnl
poaition t o an anplc H. What i'i the ~ , i u i n i r o nanplc that can he
reached before the bodies slip down the incline'? Takc p> for body
N and the plane as .2 and fl?for hod! A and the plane as .i.
Figure P.7.3.
7.4. A platform is suspendcd by two ropes which are attached 1~
docks lhat can slide horizontally. At what value of W does the
hitform begin to descend'? Will W stan tipping'?
!92
Figure P.7.7.
7.8. What is the minimum value of p,vthat will allow the rod AB
to remain in place? The rod has a length of 3.3 m and it has a
weight of 200 N.
7.11. A 30-ton tank is moving up a 30" incline. If ps= .6 for the
contact surface between tread and ground, what muximum torque
can be developed at the rear drive sprocket with no slipping?
What maximum towing force F can the tank develop? Take the
mean diameter of the rear sprocket as 2 ft.
Data
W=200N
L = 3.3 m
A
Figure P.7.11.
Figure P.7.8.
7.9. Find the minimum force P to get block A moving
7.12. A SCi-lb crate A rests on a I,OM)-lb crate B. The centers of
gravity of the crates are at the geometric centers. The coefficients of
static friction between contact sufaces are shown in the diagram.
The force Tis increased from zero. What is the fin1 action to occur?
Dofa
W, = 200 Ih
W, = 90 Ih
F'= 3
Figure P.7.9.
7.10. What minimum force F is needed to staR body A moving
to the right if px= .2S for all surfaces? The following weights are
given:
W, = SO
W, = 125 N
N
W,, =
100 N
The length oSAB is 2.5 m.
Figure P.7.12.
7.13. What force F is needed to get the 300-kg block moving to
the right? The coefficient of static friction for all surfaces is .3.
p,
Figure P.7.10.
=
.3
Figure P.7.13.
29:
Y
Figure P.7.19.
7.20. The cylinder shown weighs 200 N and is at rest. What is
the friction force at A ? If there is impending slippage, what is the
static coefficient of friction? The supporting plane is inclined at
60" to the horizontal.
Figure P.7.22.
Figure P.7.20.
7.21. Armature B is stationary while rotorA rotates with angular
speed 0. In armature B there is a braking system. If M~ = .4, what
is the braking torque on A for a force F of 300 N? Note that the
rod on which F 1s applied is pinned at C to the armature B
Neglect friction between B and the brake pads C and H.
1.23. An insect nies to climb out of a hemispherical bowl of
radius 600 mm. If the coefficient of static friction between insect
and bowl is .4, how high up does the insect go? If the bowl is spun
ahout a vertical axis, the bug gets pushed out in a radial direction by
the force m m 2 , as you learned in physics. At what speed o will the
bug just be able to get out of the bowl?
Figure P.7.23.
Figure P.7.21.
7.22. A 200-lb load is placed on the luggage rack of the 4,500-lh
station wagon. Will the station wagon climb the hill more easily
or with greater difficulty with the luggage than without'? Explain.
The static coefficient of friction is .55.
1.24. A block A of mass 500 kg rests on a stationary support B
where the static coefficient of friction ps = .4. On the right side,
support C is on rollers. The dynamic coefficient of friction .udof
the support C with body A is .2. If C is moved at constant speed to
the left, how far does it move before body A begins to move?
Figure P.7.24.
295
Figure V.7.27.
296
Figure P.7.32.
7.33. The block of weight W is tu he moved up an inclined
plane. A rod of length c with negligible weight is attached to the
block and the force F is applied to the tup of this rod. If the coefficient of static friction is
determine in terms o f u , d, and p,rthe
mnximum length c for which the block will begin to slide rather
than tip.
7.36. What is the minimum static coefficient of friction required
to maintain the bracket and its 500-lh load in a SVdtic position?
(Assume point contacts at the horizontal centerlines of the arms.)
The center of gravity is I in. from the shaft centerline. Hint: Note
that there is clearance between the veitical shaft and the horizontal arms.
Figure P.7.33.
Figure P.7.36.
7.34. Determine the range of values of W, for which the block
will either slide up the plane or slide down the plane. At what
value of W, is the friction force zero? W2 = 100 Ih.
W-
=
100 Ih
7.31. If the static coefficient of friction in Problem 7.36 is .2, at
what minimum distance from the centerline of the vertical shaft
can we support the 500-lh load without slipping?
Figure P.7.34.
7.38. A rod is held hy a cord at one end. If the force F = 200 N,
and if the rod weighs 450 N, what is the mruimum angle a that the
rod can he placed for between the rod and the floor equal to .4?
The rod is I m in length.
$, J !
7.35. A 200-kN tractor is to push a 60-kN cuncrete beam up a
15' incline at a construction site. If pc,= .5 between beam and din
and if pris .6 between tractor tread and din, can the tractor do the
job? If so, what torque must be developed on the tractor drive
sprocket which is .8 m in diameter'? What force P is then developed to push the beam'?
Figure P.7.3S.
Figure P.7.38.
297
7.39,
Suppose tliat a i l ice liftcr IS uscd to support a hard block nf
matcrial hy friction only. What i s the ,ni,iimunr coefficient of static
friction, u,, to accomplish this for any weigh1 W a n d frri the g e m etry shown i n the di;ig!ram'!
10"
t
,~
7.41.
A hcam supports loud C' wcighiiip 5 0 0 N. At suppoflh .4
and H. the static coefficient of frictioil i\ .2. At thc Contact surlac~!
helween load C and the bcam. the dynamic coefficient 0 1 friction
i s . l S . It force I: m o v e > C'steadily tu rhc Icit. how far doe\ it IIIII\IC
b e f k thc bcam hegins tu move'! The beam w e i g h 200 N.
Neglect tlic hcighi iol the hriuii in p u r ciilculiltioiii.
A
i
~
111
Figure P.7.41.
7.42.
DO Problsni 7.41 fur thc
imv account. 'lakc I = 120 i n i n
cilx
whcrc [lie height
1
i, takcn
7.43.
A roil is supported h i IWO u,hrrls spinning in oppoiitc
directions. Ifthe wheels were horirmtal, the rod wimld he placed
centrally over thc wheel, for equilihriuni. However, the wheels
have an inclination ol20" a s shown, and lhr rnd m u h t he placed at
LI pusition off center for equilibrium. If thc cocflicienl of fricliuii
is L I , , = .8, how many lcct oll w i t c r iiiuhi the rud he placed?
Figure P.7.39.
I
I O 0 Ih
7.40. A rectangular case is loaded with uniform vertical thin
rods such that when i t i s full, as shown in (a), the case has a total
weight of 1,000 Ih. The case weighr 100 Ib whcn ciripty and hits a
cocfficient of static friction of .3 with tlic tlwx as shown in thc
diagram. A forcc 7'of 200 Ih i s nnintained un the ci\se. It the r d s
are unloaded a\ shown in ( h ) , what i s thc limiting value o f x for
Figure P.7.43
equilibrium to he inaintiiined'!
How much forcc /: !nust hc applied ti) the wedge I n
hepin to iilisc the cratc? Neglcct changes in gcrmetry. What force
must the hiopprr hlock provide to prevent the crate frnm inwing t o
the Icft'! The slatic coefficient o i friction hetwccn all wtiaccs i s 3
P.7.44.
(h)
(a)
Figure P.7.40.
c-..(,'d
Figure P.7.44.
7.45. What is the ninrimum height x of a step bo that the force P
will roll the 50-lb cylinder over the step with no slipping at a?
Take p,, = 3.
1.41. If we neglect friction at rhc mllen, and the coefficient of
static friction is .2 for all surfaces, ascendin whether the 5.000-lb
weight will go up, go down, or stay stationary.
Figure P.7.45.
7.46. The rod AI1 ic pulled at A and it moves to the left. If the
coefficient of dynamic friction for the rod ar A and B is .4, what must
the minimum value of W, be to prevent the block from tipping
when a = ZO"? With this value of W,, determine the minimum
coefficient of static friction between the block and the supporting
plane needed to just prevent the block from sliding. W , is 100 N.
-
4,000 Ib
0 nim
A
60mm
1-
1-
Figure P.1.46.
7.5
Complex Surface Contact
Friction Problems
In the examples undertaken heretofore, the nature of the relative impending
or actual motion between the plane surfaces of contact was quite simple-that
of motion without rotation. We shall now examine more general types of contacts between bodies. In Example 7.5. we have a plane contact surface hut
with varying direction of impending or slipping motion for the area elements
as a result of rotation. In such problems we shall have to apply Coulomb's
laws locally to infinitesimal areas of contact and to integrate the results, for
reasons explained in Section 7.2. To do this, we must ascertain the distribution of the tiormal force at the contact surfdce, an undertaking that is usually
difficult and well beyond the cdpahilities of rigid-body statics, as explained in
Chapter 5. However, we can at times approximately compute frictional
effects by ertimating the manner of distribution of the normal force at the surface of contact. We now illustrate this.
Figure P.7.47.
300
CHAPTFK 7
FKlCrlON FOKC'FS
Example 7.5
Compute the frictional resistancc to rotation of a rotating solid cylinder
with an attached pad A pressing against a flat dry surface with a h r c e P
(see Fig. 7.10). The pad A and the stationary tlnt dry surface constitute a
dry rlirust hrurinR.
The direction of the frictional forces distributed over the contact surface is no longer simple. We therefore takc an infinitesimal area for e x a n illation. This area is shown i n Fig. 7.10, where the clement has been
formed from polar-coordinate differentials so as lo he related simply to the
boundaries. The arca CIAis equal to r d0 dr. We shall ussuirw that the normal force P is uniformly distributed over the entire area of contact. The
normal force on the area element is then
\
A
P
The friction force associated with this lcirce during moliori is
Figure 7.10. Dry thrust bearing.
The direction oE df must opposc the relative motion between the surfaces.
The relative ni(iti~inis rotation 01 concentric circles ahout the ccnterline,
so the direction of a force dfl (Fig. 7. I I) must lie tangent to a circle of
radius r. At 180" from thc position of thc area element for df;, we may
carry out a similar calculation fhr ii force (If,, which for the same rmust be
equal and opposite to df;,thua forming a couple. Since thc cntire area may
he decomposed in this way, wc can conclude that there are only couples in
the plane of contact. If we take moments of all infinitesimal forces ahout
the center, we get the magnitude of the total frictional couple moment. The
direction of the couple moment is along the shaft axis. First, consider area
elements on the ring (if radius r:
Figure 7.11. Friction forces f < r mcouplcs.
SECTION^.^
Example 7.5 (Continued)
i
Taking pd as constant and holding r constant, we have on integration with
respect to 8:
dM = p d -
2m2dr
nDZ/4
We thus account for all area elements on the ring of radius r. To account
for all the rings of the contact surface, we next integrate with respect to r
from zero to 012. Clearly, this gives us the total resisting torque M. Thus,
What we have performed in the last three steps is multiple integration,
which we introduced in Chapter 4 when dealing with rectangular
coordinates.
7.6
Belt Friction
A flexible belt is shown in Fig. 7.12 wrapped around a portion of a drum,
with the amount of wrap indicated by angle p. The angle p i s called the angle
of wrap. Assume that the drum is stationary and tensions T , and T, are such
that motion is impending between the belt and the drum. We shall take the
impending motion of the belt to be clockwise relative to the drum, and therefore the tension T , exceeds tension T2.
ds
I
Figure 7.12. Belt wrapped around drum.
BELTFRICTION
301
302
CHAFTER 7
FRICTION
bowi~s
Consider an infinitesimal segment 01' the bell as a lrce hody. This segment subtends an angle d 8 a1 thc drum cenier as shown in Fig. 7.13. Summing force components i n the radial and transverse directions and equating
them to zcrn as per ryrrilihriurn, we get the following x a l w equations:
Figure 7.13. h e - h o d y diagram of regiirnr (if
helt; impending blippage.
Therefore,
dN
dT cos 7
-
=
fll ON
Therefore.
rl t)
-2T sin dt)
-~
- dT sin -$ f dN = 0
2
-
The sine of a very small angle approximately equals thc angle ilself i n radians. Furthermore, to the same degree of accuracy, the cosine of a sinal1 angle
apprwches unity. (That these relations are true inay bc seen by cxpanding the
sine and cosine i n a powcr series and then retaining only the firs1 tcrms.) Thc
preceding equilibrium equations then become
dT = 0 %
(IN
-7a8 - dT d8
2
+ dN
(7.2a)
=0
(7.2b)
SECTION 7.6 BELT FRICTION
303
In the last equation, we have an expression involving the product of two
infinitesimals. This quantity may he considered negligible compared to the
other terms of the equation involving only one differential. Thus, we have for
this equation:
Td0=dN
(7.3)
From Eqs. 1.2a and 7.3, we may form an equation involving T and 0. Thus,
by eliminating dN from the equations, we have
dT = &T d 0
Hence,
Integrating both sides around the portion of the belt in contact with the drum,
we get
01
We therefore have established a relation between the tensions on each
part of the belt at a condition of impending motion between the belt and the
drum. The same relation can he reached for a rotating drum with impending
slippage between the belt and the drum i f w e neglect centrifugal eflects on the
belt. Furthennore, by using the dynamic coefficient of friction in the formula
above, we have the case of the belt slipping at constant speed over either a
rotating or stationary drum (again neglecting centrifugal effects on the belt).
Thus, for all such cases, we have
(7.5)
where the proper coefficient of friction must he used to suit the problem, and
the angle ,8 must he expressed in radians. Note that the ratio of tensions
depends only on the angle of wrap 0 and the coefficient of friction p. Thus, if
the drum A is forced to the right, as shown in Fig. 7.14, the tensions will
increase, hut if ,8 is not affected by the action, the ratio of T,IT, for impending
or actual constant speed slippage is not affected by this action. However, the
torque developed by the belt on the drum as a result of friction is affected by
the force F. The torque is easily determined by using the drum and the portion
of the belt in contact with the drum as a free body, as is shown in Fig. 7.14.
Tj
w
-c
2’
l?igure 7.14. F
not T,IT,.
Q ~ FCaffects
~
T,
and T~ but
304
C HA PTER 7
wt(-rro[V FORCES
Thus.
SECTION 7.6
Example 7.6
A drum (see Fig. 7.15) requires a torque of 200 N-m to get it to start rotating. If the static coefficient of friction usbetween the belt and the drum is
.35, what is the minimum axial force F on the drum required to create
enough tension in the belt to start the rotation of the drum?
Torque required = 200 N-m
Radius = 250 mm
72
Figure 7.15. Drum is driven by a belt.
The angle of wrap /3 clearly is
n radians. To get the minimum force
F, we use the condition of impending slippage between belt and drum so
that we can say
T - $5,
L
~
=
3 00
(a)
T2
For the condition of starting rotation we have
Mu = torque required = 200 N-m
~~
:.
(T, - T2)(.25) = 200
(b)
It is now a simple matter to solve the previous two equations simultaneously to get
T , = 1,200 N
T2 = 400 N
Finally we can determine the minimum value of F for this problem by
summing forces in the x direction as follows:
Fnli,l= 1,200 + 400 =
1,600 N
BELT FRICTION
305
i
J
SECTlON 7.6 BELT FRICTION
Example 7.7 (Continued)
For the idler pulley (no resisting or applied torques), we have
T3 = T2
(C)
From Eqs. (c) and (d), we conclude that
T2 = T3 = 250 Ib
(e)
From Eq. (b) we now get for the maximum tension T , :
TI = T,+37I =
(f)
We must next check the driving pulley to ensure that there is no slippage occumng. For the condition of impending slippage, we have, using
as T2 the value of 250 Ib and solving for T ,
TI - T *e4 X = (250)(3.51) = 878 Ib
Clearly, since the T , needed is only 621 Ib (see Eq.(f)), we do not have slippage at the dnving pulley, and we conclude that the maximum tension is
indeed 621 Ib.
Driving pulley
Belt on conveyor frame
Torque
SO0 Ib
j = .OS N
I
/
Idler pulley
Figure 7.17. Various free-body diagrams of parts of conveyor.
307
308
3'
1
1
I
( ' I I A P I I I K 1 I ~ K I ( T I O Nl:OU('l?i
Example 7.8
An electi-ic motor (not \hewn) i n Fig. 7.18 drives at constant speed the
pulley B. which c~innecl\to pulley A by ii belt. Pulley 4 is uit~ncctcdto il
compressor (not shown! which rcquircs 700 N-m torque 10 drive i t ill ~1111staiit speed 01,~.
I f ,u, Ibr the belt and either piillcy is .A, what nnininium
value o f t h e indicated force F is required to have ino dipping anywhcrc?
if^-_^-_____
,,,,
I
Figure 7.18. Deli- driver comprrssuu'.
As ii f'ir\t stcp. we determine the m g l e s of wrap p for the respeclive
pulleys. For thi, pui-p(ix, wc first compute a (Fig. 7.19). Note that thc
radii 0 , l l and OljE. k i n g perpendicular 10 lhc same line DE..iire thei-cforc
pariillel to each r1ther. Drawing E<'par;lllel to 0,Ow we t l i m forin ix i n the
shaded triiinglc. Hence. wc can say:
,
Fi'iQure7.19. Find an$lcs of wrap
$
1
1
i
Note i n Fig. 7.1') Ihiil
is pcrpendicular to <:E and [hat (I.,(; is perpendicular 10 Ill;, Thcrclore. the included angle between OA6' aiid O~,Gmost
cqual the included angle hctwccn C1:' and Ill.'. Thih an&!le i s n. Clearly. the
;mgk hctwccn O,j.l ;ind O,,ti is also this angle ix. We cim inow express t h r
angles (if w r q fur holh pulleys a s follows:
p,
p,
..
.,
.
,l_"
.
.
= 1x0"
= 180"
+ 2(5.74) =
191.5'
3 1 . 7 4 ) = 168.5"
"".,,
.
..
.
.. .
..
.
SECTION 7.6 BELT FRICTION
Example 7.8 (Continued)
Now consider pulley A as a free body in Fig. 7.20. Note that the minimum
force F corresponds to the condition of impending slippage. Accordingly,
for this condition at A, we have
Figure 7.20. Free-body diagram of A.
Also, summing moments about the center of the pulley, we have
[(TI), - (T2),I(.50) - 700 = 0
Solving Eqs. (a) and (b) simultaneously, we get
= 1,898
N;
( T J A = 498 N
From equilibrium, we can compute force FA as follows:
(1,898 + 498) cos 5.74"
~
FA = 0
Now go to pulley B to see what minimum force FR is needed so that
the belt does not slip on it during operations. Consider in Fig. 7.21 the freebody diagram of pulley E . For impending slipping on pulley E , we have
Figure 7.21. Free-body diagram of B
309
31 0
CHAlTEK 7
FRIrTION FOKCES
Example 7.8 (continued)
The torque for pulley B needed I o dcvelop 700 N-m on pulley A is next
computed. Thus,"
Therefore,
M,
= 320 N-in
Summing moments in Fig. 7.21 ahout the ccnler of. B. we then have on
using the above resull
-
[(TI),, . - lT2),](.30) + 420
=
0
Therefore,
(TI),$
~
Cr,), =
1,400
(e)
Solving Eqs. (d) and (c) simultaneously, we gct
(TI),, = 2,112.5 N:
(T2),j = 625 N
Hence, the minimum F,, needed for pulley B is
r;, = (2.025
Notc that the ratios
cylinders art'
til'
+ 625) cos
5.74" =
the tcnsions for irnpendiny .sli/qiiiKe f o r the two
1)riven cylindcr.4
Driving cylinder B
ti^
T
T
&
5
= 3.81
FA = 2,384 N
= 1.24
fi;
=
2,637 N
If we use impending slippage for driven cylinder A , we will have slipping
for driving cylinder B, which we cannot tolerate. Thub, if we use the larger
force of 2,617 N. we will hc well under the impending slippage condition
of the driven cylinder A . Clearly. the optimum result to uvoid slip[i~geand
to minimize the belt tension ior greater life of the belt is to have a force
somewhere above 2.617 N . Here is a place for good engineering judgement
and experience.
"Note [hill the ratio oi tranimitlcd torques MJM, between directly connected pulleys
and gears will cquiil r J r , or l ) J l l ,01 the pullcyr o r gear*. Can you vcriiy lhis yourself'
7.48. Compute the frictional resisting torque for the concentric
dry thrwt bearing. The coefficient of friction is taken aspc
Compute the frictional torque needed to rotate the truncated cone relative to the fixed member. The cone has a 20-mmdiameter base and a 60" cone angle and is cut off 3 mm from the
cone tip. The dynamic coefficient of friction is .2.
7.51.
7.49. The support end of a dry thrust bearing is shown. Four
pads form the contact surface. If a shaft creates a 100-N thrust
uniformly distributed over the pads, what is the resisting torque
for a dynamic coefficient of friction of , 1 '?
7.52. A 1,000-Nblock is being lowered down an inclined surface. The block is pinned to the incline al C, and at E a cord is
played out so as to cause the body to rotate at uniform speed about
C. Taking pd to be .3 and assuming the contact pressure is uniform along the base of the block, compute T for the configuration
shown in the diagram.
In Example 7.5, the normal force distribution at the contact surface is not uniform but, as a result of wear, is inversely
proportiunal to the radius r. What, then, is the resisting torque M?
7.50.
7.57. The seaman pulls with 100-N force and wants to stop the
motorboat from moving away from the dock under power. How
few wraps n OS the rope must he make around the post if the
motorboat develops 3,500 N of thrust and the static coefficient o f
friction between the rope and thc post is . 2 ?
Figure P.7.60.
Figure P.7.57.
7.58. A length of belt rests on a flat surfact and runs over a
quarter of the drum. A load Wrests on the horizontal portion of
the belt, which in turn is supported by a table. If the static coefficient of friction for all surfaces is 3, compute the maximum
wcight W that can be moved by rotating the drum.
7.61. A mountain climber of weight W h a y s freely suspended
by one rope that is fatened at one end to his waist, wrapped onehalf turn about a rock with I(, = .2, and held at the othcr end in
his hand. What minimum force in terms of W must he pull with to
maintain his position'! What minimum force must he pull the rope
into himself with to gain altitude?
Figure P.7.58.
7.59. The rope holding the 50-lb weight E passes over the drum
and i s attached ~t A. The weight of C is 60 Ib. What is the mini^
mum static coefficient of friction between the rope and the dNm
to maintain equilibrium of the drum'?
Figure P.7.61.
7.62. Pulley B is turned by a diesel engine and drives pulley A
connected to a generator. If the torque that A must transmit to the
I
I
generator is 500 N-m, what is the minimum static coefficient of
friction between the belt and pulleys for thz case where the force
F is 2,(100 N ?
Figure P.7.59.
7.60. What is the maximum weight that can be supported by the
system in the position shown? Pulley B connor turn. Bar AC is
fined to cylindcr A , which weighs 500 N. The coefficient of static
friction for all contact surfaces is 3.
2m
-l-
Figure P.7.62.
313
A hand hrake ik \ h o w Iip<, = .4, what i s the rcristing
li~l-i]oewhen thc rhait is rotating'! What are the supporting fbrce.;
mi the rod AB.
7.63.
1.66.
( a ) What force P i s necded t o drvelop a resisting torque 01
hS N-m on the rrrtating drum'! 'The dynamic coellicient
of friction / I < ,i s 0.4.
(h) With thc same force P from part (a), %,hat must the
value o f p,, he i n wdcr 11, increilw the resisting t i r q u r
hy IO N-m'.'
Figure P.7.63.
7.64. A conyeyor i s shown with two driving pulleys A and B .
Drivel- 4 has no angle o f W d p of 330". wherca5 B has a wrap of
1x0". If the dynamic cuefficient nf friction between the hell and
the hed o l the wn\,ryr,l- is .I. atid the m a l weight to he Iran\ported is 1(1,000 N. what is the .s,nnlirrf sfatic coefficient of friction between the belt and the driving pulleys'! One-fifth of the
Iuad can he assumed to he hetween thc twn pulleys at dl times.
and the tension in the slack side (underneath) i s 2,000 N. Therc i s
a free-wheelin$ pulley at the left end uf the conveyor. You will
have t o SOIYC
a n cquation hy (rial and ciror.
Figure P.7.64.
7.65. A freely turning idlcr pulley i s used 10 increase the angle
of wrap for the pullcys shown. If the tension in the slack side
a h w e is 200 Ih, find the nrmi~numtorque that a n be transmitted
hy the pulleys for a static coefficient uf friction of .3
Idler pulley
Figure P.7.65.
314
Figure P.7.66.
What is the maximum value r f a x i a l l o a d P 10 maintain a
rritational speed 0 = 5 r;id/s of the dry thrust hearing'! The sum
of the hell forces is 200 N . The cwifact surtaor helween the hclt
and drum and hetween the dry thrust hcaring and base havc the
mmc H , and p,, of .4 ;and 3. r c y v x t i v c l y .
1.67.
7.68. Rod AR weighing 200 N is supported by a cable wrapped
around a semicylinder having a coefficient of friction pTequal to
250 mm
.2. A weight C having a mass of IO kg can slide on rod AB. What
is the maximum range x from the centerline that the center of C
can be placed without causing slippage?
/
7.71. From first principles, show that the normal force per unit
B
A
1.2m
length, w, acting on a drum from a belt is given as
w, =
2r
Use the indicated diagram as an aid. [ H i n o Start with Eq. 7.2(a)
and use Eq. 7.4 for any point a.]
"
7.69. The cable mechanism shown is similar to that used to move
the station indicator on a radio. If the indicator jams, what force is
developed at the indicator base to free the jam when the required
torque applied to the turning nob is IO b i n ? Also. what are the
forces in the various regions of the cable? The static coefficient of
friction is .IS.
J
T2
I P
,
"
I
T,
Idler
Pulley
Figure P.7.69.
Figure P.7.71.
7.72. what minimum force F is needed so that dmm A can transmit
a clockwise torque of 500 N-m without slipping? The coefficient of
friction, pa,between A and the belt is .4. What minimwn caefftcient
of static friction is needed between drum R and belt for no slipping?
~
F
7.70. What are the minimum possible supporting force c o m p nene needed for pulley B as a result of the action of the belt? The
static coefficient of friction between the belt and pulley B is .3 and
between the belt and pulley A is .4. The torque that the belt delivers
to A is 200 N-m.
+lm+
Figure P.7.72.
315
induced hy hod) (. weighing S O 0 N ?The weight (11 A i \ 100 N .
The m t i c c w l l i c ! r n t ,if friction hetween the hclth arid A i s .J. and
between A iuid the w;illi i s . I .Neglect li-ictiun at pulley (;.
IO0 1111
Figure P.7.1
7.14.
A V-hell i \ \huwn. ShnN that
Figure P.7.16.
7.77.
Figure P.7.74.
.A pullcy A i\dri\cn hy an outadc ;ipenl at a spccd w o i 100
I-pm. A bclt weighing 30 Nlm i s driven hy thc pulley. IlT2 = 21x1
N. what i s the ,,ro.rirmm pmhihle tciisiiin 7, computed without considering ccntrilirp;il r f f e a c ? Cornputt: T,accounting lin- ccntrifugal
ellcuts, and g i w the perccntqc cmor incurred hy tnot including CCIItrilupal cffccts. 'The \tittic cncfficient ot triution hetween tho helt
iind thc pullq i\ 3. Suc Prohlerri 7.76 hcfirre doing this pmhlem.
Figure P.7.17.
SECTION 7.7
7.7
THE SQUARE SCREW THREAD
The Square Screw Thread
We shall now consider the action of a nut on a screw that has square threads
(Fig. 7.22). Let us take r as the mean radius from the centerline of the screw
to the thread. The pitch, p. is the distance along the screw between adjacent
threads, and the lead, L, i s the distance that a nut will advance in the direction
of the axis of the screw in one revolution. For screw threads that are singlethreaded, L equals p . For an n-threaded screw, the lead L is np.
Forces are transmitted from screw to nut over several revolutions of
thread, and hence we have a distribution of normal and friction forces. However, because of the narrow width of the thread, we may consider the distribution to be confined at a distance r from the centerline, thus forming a
“loading” strip winding around the centerline of the screw. Figure 7.22 illustrates infinitesimal normal and frictional forces on an infinitesimal pan of the
strip. The local slope tan a as one looks in radially is determined by considering the definition of L, the lead. Thus,
Figure 7.22. Square screw thread
L =np
slope = t a n a = 2nr
2nr
All elements of the proposed distribution have the same inclination
(direction cosine) relative to the z direction. In the summation of forces in this
direction, therefore, we can consider the distribution to be replaced by a single
normal force N a n d a single friction force f at the inclinations shown in Fig.
7.23 at a position anywhere along the thread. And, since the elements of the
distribution have the same moment ann about the centerline in addition to the
common inclination, we may use the concentrated forces mentioned above in
taking moments about the centerline. There is thus a “limited equivalence”
between N and f and the force distribution from the nut onto the screw. The
other forces on the screw will be considered as an axial load P and a torque M7
collinear with P (Fig. 7.23). For equilibrium at a condition of impending
morion to raise the screw, we then have the following scalar equations:’
-P
+ Ncos a - ptNsin a = 0
+,,N cos a r - N sin a r
+
M, = 0
’The equalions also apply to .steady rotation of the n u t on the scccw, in which case one uses
the dynamic coefficient of friction fld in the equations.
Figure 1.23. Free-body diagram.
3 11
3I8
CHAPTER 7
FRIC TIO N FORCES
These cquatioris ]nay he used to eliminate the Scirce N and so get a relation
between P and M. that will hc o S practical significance. This may readily hc
done by solving S& N i n ( a ) and substituting into ih). Thc result is
A n important question arises when we employ the scrcw and nut in the
form of a jack as shown in Fig. 7.24. Once having raised a Imad P by applying the torque M. to the jackscrew, does the dcvice maintain the load at the
raised position when the applied torque i s rcleased. (11 docs the screw unwind
,
,/
’ ,’
_,,\,
,
,,
Figure 7.24. Jackscl-rw.
under the action of the load and thus lower the load? I n other words, i s this a
selflorking device’! To examine this. we go hack to the equalions of equilihrium. Setting hf. = 0 and changing thc direction o f the friction Sorces, we
have the condilion for impending “unwinding” 01 the screw. Eliminating N
l‘rom the equations. we get
/‘!-(-/I<
cos01
c11sa-+rS
+ sina)
= 0
sina
This requires that
-pr cos a
+
\in a = 0
Thereforc,
p . = tan a
(7.9)
We can conclude thal. if the coefficient of friction prequals or excccds tan a.
we w i l l have a self.-locking condition. If M, i s less than tan 01, the screw %,ill
unwind and will not support ii load P without the proper external torque.
SECTION 7.8
Example 7.9
A jackscrew with a double thread of mean diameter 2 in. is shown in Fig.
7.24. The pitch is .2 in. If a force F of 40 I b i s applied to the device, what
load W can be raised'? With this load on the device, what will happen if the
applied force F is released! Take / I , = .3 for the surfaces of contact.
The applied torque M, is clearly:
M... =
12
(4
(40) = 26.7 Ib-ft
The angle a for this screw is given as
Therefore,
a
= 3.64'
Using Eq. 7.8 we can solve lor P . Thus,
p =
-
a
M. (cos a
- @,$ sin a )
+ sin a )
r ( p , cosa
(26.7)[.998 ~ (.3)(.0635
_ _ _ -~
_
,&[(.3)(.998)+ ,06351
~
The load W is 864 Ih. The device is self-locking since @,s exceeds tan
= ,0636.
p, > tan (r
,,:.
8e
ng,
To lower the load requires a reverse torque. We may readily compute this torque by using Eq. 7.8 with the friction forces reversed. Thus,
864(&)[-(.3)(.998) + ,06351
,998 + (.3)(.0635)
( MZ)down = ~-
"7.8
16.71 Ih-tt
(d)
Rolling Resistance
Let us now consider the situation where a hard roller moYes without slipping
along a horizontal surface while supporting a load W at the center. Since we
know from experience that a horizontal force P is required to maintain uniform motion, some sort of resistance must he present. We can understand this
resistance if we examine the deformation shown in an exaggerated manner in
R O I . I . I N CRESISTANCE
;
319
vi
I'
lo;rd
~
~
/>,I U t l C C i
l w c c wquiwtl tor
c,~nit,,,ll
>pccct
Fig. 7.75. If foi-cc P i s ahng the cciiterline a h shown, Ihc cquivalenl liirce systein c w i i i n g 01110 tlie roller from the regioti o f cnntiicI niiist he ihat O S a force
N whosc line O S iiclion i i l s o gocc through t h center
~
OS the rnller since, you
w i l l recall S i r i n i Chapter 5. three notipal-allel forces niiist he c o n ~ i ~ r r cfor
~it
equilibrium, 111ordcr to develop ii r c s i ~ t u i c e10 ~ i i ~ t i oclearly
n.
N i i i u s i he oriented iit iin anglc 4 with the vertical direction. i i s i s shown in Fig. 7.25. The
sciiliir equ;iti,iiis O S q u i l i l i i . i u i i t hccome
6:
W = N cos
Therefore.
P = A' cin 4
;
Figure 7.25. Rolling ie*i\liincc modcl.
=
lilt1
(7.10,
@
Since ilic iireii 01coti1iic1 i s sni:ill, we ticite tliiit q5 i s ii s m a l l angle and that tali
@ = \in $. The h i l i @ i\ sccn to he u i i ~Srnni I:ip. 7.25. Thcrcriirc, we inay sa)
Illill
Solving fort', we gel
,
w<r
-
I -
( 7 .I I h)
The dist;incc ( I i n tliex cqu;itirins i h ciillcd the c , w / f ; ( . i oj'tdIii8,y
~~
rrsisru,i(.e.
Couloinh suggesled that l i i r wiriorhle Ioadc 1V. the riitin PIW is c o n s t m i
for given m;ikrials and ii gibe11 geometry ( r = cm\tant). Liioking at Eq.
7.I la, w e scc tliiit (I must then he ii c ~ n c t i i i i tlcii~given geomc~ryiind inateriiilc. Ciiuloinh added that, Sor given materials and iwrinhic I-adius, the riitin
PIU'varics in\'ei-sely iis r ; that is, a s the radiuh OS the cyliridcr i s increased. the
reci~tancclo uniform iriutioii lor ii given load W decre;ises. Thus. considering
Ikl. 7. I l a iigain. we may cnnclude that. for given nialcriiils, (I i s a l s o c ~ n ~ t m t
for iill s i x s o i rollers atid loads. Ilowevcr. other invcsligiitors have contested
huth ~ t a t e ~ i i c n t s1xirticul;irly
,
the laltei- one. ;ind ihcrc i s :I nccd for further
invcstig;ition i n l l i i s iireii. I.aching hetter datii. we prcwnt the following list of
rolling ~ ~ c l l i c i c iStir
i t ~your iise. hot we tiiust ciiution th;il you ehould iint
expect great ;iccur;icq froiii t h i s general procediir:~.
~ . . ~.
..
..
..
~
~
~
Coefficients of Rolling Resistance
-.
~
~
~
.
..
~
~~
~.
~~~~
~
(in.)
,007 .!II 5
.Oh .IO
~!I?
- SI1
.!I4 - .Oh
ti
S t c d on hieel
Sicel ,111 wood
Pwum;itic tires , , I / viiuiith raid
Pncuni;ilic liw 011 mod ro:td
H;iidrnrd \tee1 on h:il-deneJ \ t c d
~
.0!,02
.0005
SECTION 7.8 ROLLING RESISTANCE
Example 7.10
What is the rolling resistance of a railroad freight car weighing 100 tons?
The wheels have a diameter of 30 in. The coefficient of rolling resistance
between wheel and track is ,001 in. Compare the resistance to that of a
truck and trailer having the same total weight and with tires having a
diameter of 4 ft. The coefficient of rolling resistance a between the truck
tires and road is ,025 in.
We can use Eq. 7. I 1 b directly for the desired results. Thus. for the
railroad freight car, we haveK
For the truck, we get
We see a decided differences between the two vehicles, with clear advantage toward the railroad freight car.
T h e number of wheels n plays no role here since we divide the load by n to get the
load per wheel and then niultiply hy n to get Lhc m a l resirtanfe.
321
7.78.
A simple C-clamp i s used t o hold two pieces of metal
together. The clamp has a single square thread with a pitch 01
12 in. and a mean diameter of .75 in. The static coefficient of friction i s 30.Find the torque required i f a 1,000-lh compressive h a d
i s required on the blocks. If the thread I S a double thread. what i s
the required torquc?
c
Figure P.7.78.
1.79. The mast of n railhoat i s held hy wires called shroud,, iis
rhown in the diagram. Racing sailors are carelul t o get the pmpcr
ension in the shrouds hy adjusting the lurnhucklc at the hattom or
:he shrouds. When wc do this we say wc are .'tuning" thc hoat. II
i tension of I S 0 N exists in the chroud. what torque I S needed l o
itart tightening further by turning the tumhuckle? 'lhc pitch of the
;ingle threaded screw i s 1.5 mni and the mean diameler i s X . 0 mni.
The static coefficient of friction i s .2.
Figure P.7.79.
Forccs F of SO Ih are applied to the jackscrew shown. The
hread diametcr is 2 in. and the pitch i\ in. The %,tic coefficimt
1.80.
4
if friction Tor the thread IS .OS. The wripht I?J and colliir arc nm
ierniitted to rotate and so the collar must rotatc on the shaft o1 the
crew. Ifthe static cocfficient of friction hctween the colliir and
haft is .I.determine the wcight W that can he liftcd hy this system.
122
Figure P.7.82.
7.83. Consider a single-threaded screw where the pitchp = 4.5
mm and the mean radius is 20 mm. For a coefficient of friction p,\
= .3, what torque is needed on the nut for it to turn under a load of
1.(XI0 N ? Compute this for a square thread and then do it for a triangular thread where the angle p is 30". See Problem 7.82 before
doing this problem.
7.84. If the coefficient of rolling resistance of a cylinder on a flat
surface is .05 in., at what inclination of the surface will the cylinder of radius r = I ft roll with uniform velocity?
7.86. In Problem 735, suppose there is only rear-wheel drive
available. What is the minimum static coefficient of friction
needed between tires and ground for the vehicle to move?
7.87. A roller thrust bearing is shown supporting a force P of 2.5
kN. What torque Tis need to turn the shaft A at constant speed if
the only resistance is that from the hall hearings? The coefficient
of rolling resistance for the balls and the bearing surfaces is
,01270 mm. The mean radius from the centerline of the shaft to
the balls is 30 mm.
7.85. A 65-kN vehicle designed for polar expeditions is on a
very slippery ice surface fur which the static coefficient of friction
between tires and ice is ,005. Also, the coefficient of rolling resistance is known to be .8 mm. Will the vehicle be able to move?
The vehicle has four-wheel drive.
~ 1 . 2 m 4 - 1.3 m-.,
Figure P.7.85.
7.9
Figure P.7.87.
Closure
In this chapter, we have examined the results of two independent experiments: that of impending or actual sliding of one body over another and that
of a cylinder or sphere rolling at constant speed over a flat surface. Without
any theoretical basis, the results of such experiments must be used in situations that closely parallel the experiments themselves.
In the case of a rolling cylinder, both rolling resistance and sliding
resistance are present. However. for a cylinder accelerating with any appreciable magnitude, only sliding friction need be accounted for. With no acceleration on a horizontal surface, only rolling resistance need be considered.
Most situations fall into these categories. For very small accelerations, both
effects are present and must be taken into account. We can then expect only a
crude result for such computations.
Before going further. we must carefully define certain properties of
plane surfaces in order to facilitate later computations in mechanics where
such properties are most useful. These plane surface properties and other
related topics will be studied in Chapter 8.
323
7.88. If thc static coefficient of friction for all surV~cesis 3 5 , find
the force F needed to start the 2 0 - N weight moving to the right.
i
,30"
7.91. A lriction drive is shown with A the drivcr dicc and R the
dri\eti disc. If fiwce F pressing R onto A is I S 0 N. what i ? the
ninrimu~rrtorque M , that can he developed'? For this torque. whal
is the lurquc M , needed for thc drive disc A ? The static coefficient
o f friction bctwccn A and H is .7. What mitical force must COLI
G
wilhsrand f o r thc action dcscrihed ahrivr!
Figure P.7.XX.
7.89. A loaded crate is shown. Thc crate weighs 500 Ih wilh il
ccnier of gravity at its geometric center. Thc contact w h c z
between crate and floor has a static coefficient of friction of .2. II
H = 90", show that the crate will slide hefore m c can increasc I
enough fcw tipping to occur. If a stop is to bc inserted in rhc floor
at A to prevent slipping so that the cmte could he tipped. whal
minimum horizontal force will he exerted on thc stop?
Figure P.7.91.
7.92. Detmninc thc weight of hlock A for impending motion to
the right. Thc static coefficient of friction hctwern the vahle and
thc wrlaccs which it conlacts is 0.2. Thc m l i c crrcfficienl 0 1 friction helwrrn hkrch A a i d the burface upon which it rei15 is (1.4.
Tht. two prist\ are circular with ii diameter of0.25 in.
A
Figure P.7.89.
Figure P.7.92.
In Problem 7.K9, compute a value o f H and I whzre dipping and tipping will occur siinultancously. If the actual angle H i \
smaller than this value of 8, is there any further nerd of the slop at
4 to prevent slipping'?
7.90.
324
7.93. lh Pnrhlem 7.92 till- impending slippage of hody A to
the kil.
7.94. A tug is pushing a barge into a berth. After the barge turns
clockwise and touches the sides of the pilings, what thrust must
the tug develop to move it at uniform speed of 2 knots fanher into
the berth? The dynamic Coefficient of friction between the barge
and the sides of the berth is .4. The drag from the water is 3,000 N
along the centerline of the barge.
The drum is driven by a motor with a maximum torque
capability of 500 Ib-ft. The static coefficient of friction between
the drum and the braking strap (belt) is .4. How much force P
must an operator exert to stop the drum if it rotates ( I ) clockwise
and (2) counterclockwise? What are the belt forces in each case'?
7.96.
Side
P
Figure P.7.96.
Figure P.7.94.
7.95. The static and dynamic coefficients of friction for the
A of the
are ps = .4, pd = .3,
upper surface of
and for the lower surface of contact B are p,, = .I and pd = .OS.
What is the minimum force P needed to just get the cylinder
moving'!
7.97. The four drive pulleys shown are used to transmit a torque
from pulley A to pulley D on an electric typewriter. If the static
coefficient of friction between the belts and the pulleys is .3, what
is the torque available at pulley D if 10 b i n . of torque is input to
the shaft of pulley A? What are the belt forces?
R
-
Figure P.7.93.
Figure P.7.97.
325
7.98. A scissors jack i s shown lifting the cnd of a car s o that K =
6.67 kN. What torque T i s needed krr thic operation? Notc that A
i s merely a bearing and at B we linvc a nut. 'I'he \crcu is sin&
threaded with a pitch of 3 mm and a mean diamcter o f 20 mnl.
The static coefficient of frictinn hetween the screw and nut at 13 is
.3. Neglect the weight ofthe members and evaluiltc Tfor B = 45"
and for 0 = 60".
7.100. A hot rcctanpolar mctal i n g u IS to he flatlcncd hy passing
through cylindrical roller\. I I tlic ingm i \ 10 he drawn into the
r d e n hy friction once il toiiches the ~DIICIS,what i\ the minhrrum
thickness I ofthe ingnl that can he auhiwed hy thi5 pmccss on one
p i s 9 'I'hc static cocfficicnl of friction for tlic contact between
inpwt and cylindcr i'l .3. The cylindel~, rotate a i showu with ang'ulal- rpced w.
1"
Pigwe P.7.100,
A ccmc clutch i~bhown. Assuming that uniform pressure\
surlacei, compute thc m ~ x i , m mtorque
that can hc tran\mittcd. Thc \tiitiu c d f i c i e n t of I i ~ i c t i u ni s .30 and
the activating force F i c 100 Ih. IMim Acsurne that the niming
cone transmits i l h 100-1h axial f w c c trr the statinmry c m e hy pressure primarily. That i\. wc will neglect the friction-torcr compo^
nent un thr cone wriace normal ti, thc lrmsverse direction.l
7.101.
exist hetwcen thc w r i t a c t
Figure P.7.98.
1.99. A hlack C weighing 1 0 k N i q being moved on rollerr A
ind B each weighing I kN. What force P i s nceded to m;iintain
steady motion? Take the coefficient o f rolling rcsiitancr hetween
.he mllers and the ground to he .h mm and hetween hlock L' ;MI
hc rollers to be .4 mm.
t
"I
C
+- l[y-
F
Figure P.7.99.
126
~~
Figure P.7.101.
100 Ih
t
7.102. In Fig. P.7.18, delete the external forces and couple at
point G, and replace with a force of 6,000 N at point G at an angle
a with the horizontal going from left to right. For the condition of
impending slippage in the downward direction. what should the
angle a be?
Figure P.7.104.
7.105. Shaft CD rotates at constant angular speed in a set of dry
jouinal bearings (as a result of poor maintenance). Approximately
what torque is needed to maintain the angular motion for the following data'?
M A = 50 kg
M B = 80 kg
Shaft CD = 40 kg
ud =
.2
See Problem 1.104before doing this problem.
60 mm
R
A
Figure P.7.102.
7.103. In Problem 1.39, what is the minimum angle between the
supporting links CD and ED to support any weight W if the static
coefficient of friction is p~ = .4?
7.104. A shaft AB rotates at constant angular speed in a pair of
dry journal bearings. The total weight of the shaft and the cylinder
it is supporting is W. A torque T i s needed to maintain the steady
angular motion. There is a small clearance between the shaft and
the journal, resulting in a point confacf between these bodies at
some position E as shown. Form a two-dimensional force system
acting on the shaft by moving W, the total friction force, the total
normal force on the shaft surface. and the torque to a plane normal
to the centerline of the shaft and at a location at the centei of the
system. View this system along the centerline.
Figure P.7.105.
7.106. Two identical light rods are pinned together at B. End C
of rod BC is pinned while end A of rod AB rests on a rough floor
having a coefficient of friction with the rod of pd = .5. The
spring requires a force of 5 Nlmm of stretch. A load is applied
slowly at B and then maintained constant at F = 300 N. What is
the angle 0 when the system ceases to move'? The spring is unstretched when 0 = 45". (HincYou will have to solve an equation
by trial and error.]
F
(a) From considerations of equilibrium, what and where
is the resultant force vector from the friction force and
the normal force onto the shaft?
(b) If the angle @ in the diagram is very small and if the
coefficient of dynamic friction is p<!,explain how we
can give the following approximate equation
T = Wp<,r
where I is the radius of the shaft.
Figure P.7.106.
327
I70 iiim
Figure P.7.107.
L . . . l
Figare 1'.7.109.
I
A
I
7.112. In Prohlem 7.27, what maximum value should W, be ir
order to S t a n moving the system to the left? All other data art
unchanged.
I
7.113. A block rests on a surface for which there is a coefticienl
of friction p , = .2. Over what range of angle p will there be nc
movement of the block for the 150-N force? (You will have to
solve an equation by trial and error.)
Figure P.7.110.
7.111. A triangular pile of width I ft is being driven into the
ground slowly by a force P of 50,000 Ib. There is pressure an the
lateral surfaces of the pile. This pressure varies linearly from 0 ai
A to /io
at B, as has been shown in the diagram. If the coefficient
of dynamic friction between the pile and the soil is 0.6, what is the
ma.ximum prcssure ,I(,'!
F L , =2
Figure P.7.113.
7.114. What is the largest load that can be suspended without
moving blocks A and E'?The static coefficient of friction for all
plane surfaces of cuntact is 3. Block A weighs 500 N and block B
weighs 700 N. Neglect friction in the pulley system.
Width of
pile = I ft
Figure P.7.111.
Figure P.7.114.
329
What is the minimum force F to hold the cylinders. each
weighing 100 Ih'! Takc IC, = .2 liir all surf.ict.s of contac~.
7.115.
=
I 10 Nlmm'
Figure P.7.116.
7.117.
Find the cord tension ifhlock I attains mnximuni friction
Figure P.7.115.
1.116. A compressor is shown. If the prcsaurc in the cylinder is
1.40 N1mm2 above atmosphere (gage), what mininzum torque 7 i h
iccdcd to initiate mution in rhc systcm? Neglect the weight o f t h c
:rank and connecting rod as far as thcir ctintrihution triward imovng the system. Consider fi-ictioii only hetwecn thc piston and
:ylinder walls where the coefficient 01 Irictiun
= .IS.
330
L.-l
Figure P.7.117
Properties
of Surfaces
8.1
Introduction
If we are buying a tract of land, we certainly want to consider the size and, with
equal interest, the shape and orientation of the earth’s surface, and possibly its
agricultural, geological, or aesthetic potentials. The size of a surface (Le., the
area) is a familiar concept and has been used in the previous section. Certain
aspects of the shape and orientation of a surface will he examined in this c h a p
ter. There are a number of formulations that convey meaning about the shape
and disposition of a surface relative to some reference. To he sure, these formulations are not used by real estate people, hut in engineering work, where a
variety of quantitative descriptions are necessary, these formulations will prove
most useful. In general, we shall restrict our attention to coplanar surfaces.
8.2
First Moment of an Area
and the Centroid
A coplanar surface of area A and a reference xy in the plane of the surface are
shown in Fig. 8.1. We define thefirst moment of area A about the x axis as
Mx =
jAY&
(8.1)
and the first moment about the v axis as
These two quantities convey a certain knowledge of the shape, size, and
orientation of the area, which we can use in many analyses of mechanics.
331
332
CHAPTHK 8
PKOPEKTII<SOF SIJKPACES
\
I
-
:
Figure 8.1. Plane area.
You will no doubt notice the similarity of the preceding integrals 10
those which would occur f(ir computing moments about the x and? axes from
a parallel force distribution oriented normal to the area A in Fig. 8.1. The
moment of such a force distribution has been shown for the purposes of rigidhody calculations to be equivalent to that of a single resultant force located at
a particular point 1.7 . Similarly. we can concentrate lhe entire area A at a
posilion x , , y,, called the w i i / i - o i d ' where, for computations of lint moments.
this new amngement is equi\dent In the original distribution (Fig. 8.2). The
coordinatcs x . and are usually called the wntruidul ~nordiririrr.~.
To conipute these coordinatcs. we simply eqiiatc miimenls 111 the distributed arcii
with that of thc conccntratcd area ahout both axes:
,
Figure 8.2. Centmdal 'oord!n.iie\
'The location of the centriiid of an area can readily bc shown to be indeppn&iit ( i t the r&wiiw
U X ~ S
employed. That is. the centroid is a proprro, only
of the owii ilsclS. We have wked the render to prove this in Problem 8.1
If the axe5 ,ry havc their origin at the centroid, thcn thcsc axes are called
nd clearly the first nioineiits about (hcsc axes must be zero.
Finally. we point out that all axes going through the centroid of an arc3
arc called renrroiilirl ii.ies for that area. Clearly. the .fir.v n i o i ~ i ~ not s/ r m ureu
uhoiit nti.y < ! / i / . v wntroidul ii.ir.s iniisf hr :era. This inust he true since the perpendicular dislmce frnm the cenlroid lo the centroidal axis must he zero.
SECTION 8.2 FIRST MOMENT OF AN AREA AND THE CENTROID
333
Example 8.1
A plane surface is shown in Fig. 8.3 hounded by the x axis, the curve y’=
25x, and a line parallel to they axis. What are the first moments of the area
about the x and y axes and what are the centroidal coordinates?
We shall first compute M, and My for this area. Using vertical infinitesimal area elements of width dx and height y . we have noting that
y = 5x,’
t
”’
pdL
10’
To compute M ,we use horizontal area elements of width dy and length
(IO-x) as shown in the diagram. Thus
v
could alsc we used vertical strips for computing M, as follows u: R
centroids of the vertical strips:
M,
= I -(ydx)=I
lo
Y
0 2
0
725x
d.x
10
=
(l’2.5)($)~
=
625 ft3
n
To compute the position of the centroid (x,., y,), we will need the
area A of the surface. Thus, using vertical strips:
= 105.4 ft2
Figure 8.3. Find centroid,
250’
334
f
CHAPTER 8
PROPERTIES OF SURFACES
Example 8.1 (Continued)
The centroidal coordinates are, accordingly,
To get the moment of the area about an axis y’. which is 15 It Io thc
lelt of t h e y axis. simply proceed as follows:
=
105.4 6.00
+
15 = 2.213 ft’
. ~...
i
Consider now a planc arca with an arI.\ ,!/.sytritiicJrry such a s is \howl1 ill
Fig. X.4, where the J axis is collinear with the axis 01symmetry. I n coinputing x,. for this area. we havc
5ymmctry
Figure 8.4. Arca with one axis of symmelry
In evaluating the inkgrid ahove. we i i i n considcr iircii element\ in syn~metric
pairs such as shown in Fig. 8.4. where we have hhown ii pair of area elcnicnls
which are iiiii-ror images OS cach other ahout the axis 0 1 iymmetry. Clearly,
the first moment of such a pair about the axis of symmetry is ~ e r oAnd.
.
since
the entire area can he considered as composed of such pairs. we can conclude
that ,r? = 0. Thus, the cenlriiid of a11 area with onc axis of symmetry must
therefore lie somewhcrc along thih axis of cyminetry. The axis of symmetry
then is a centroidal axis. which is another indication that the first ninmcnl of
area must he 7.ero about such an axis. With two orthogonal axcs of s y n ~ m e t ~ - y ,
the centroid inust lie at the inlersccti(in O S these axes. T h w . lor such areas 21s
circles and rectangles. the centroid is easily determined by inspcction.
In many problem>, the arca (11 interest ciin he imcidered formed hy [he
addition o r subtracti(in iif ciinple familiar arciis whiihe centroids arc known by
inspection as well as by otlier firmiliar areas. cuch iis triangles and sectors of
circles whosc centroids and areas are givcn i n handhoohh. Wc call areas rnade
up of such simple areas u i r n / m ~ I / earcas. ( A listing of familiar areas is given
fiir your convcnicncc on the inside bach covcr of this text.) For such p r i b
Icms. we can say that
where .< and \;, (with proper signs) are the cenlroickil co(irdinatcs to simple
area Ai,and where A is the totid area.
SECTION 8.2 FIRST MOMENT OF AN AREA AND THE CENTROID
Example 8.2
Find the centroid of the shaded section shown in Fig. 8.5
t
60 rnm
1-
200 mm
d-
X
Figure 8.5. Composite area.
We may consider four separate areas. These are the triangle ( l ) , the
circle (2) and the rectangle (3) all cut from an original rectangular 200 x
140 mm2 area which we denote as area (4). In composite-area problems,
we urge you to set up a format of the kind we shall now illustrate. Using
the positions of the centroid of a right triangle as given in the inside covers
of this text, we have:
~~~
~
A , = -f(30)(80)
A, = - 7 ~ 5 0 ~
A, = -(40)(60)
A, = (20Ol(140)
= -1,200
10
= -7,850
100
= -2,400
180
= 28,ooo
100
A = 16,550 mm2
Therefore.
c
-12,000
-785,000
-432,000
2,800,000
113.3
70
110
-136,000
-549,780
-264,000
70
1,960,000
AjXj =
1.571 x 106mm3
EAjY, =
1.011 x 106mm3
335
336
CHAPTER K
PKOFERTIFS OF SUKI'ACES
We now illustrate ho\+ wc can use the composilc-;i~-c:a approach lor
finding the ceiitniid ol iin m i l c(imp~iscd<iff m i l i a r parts iis just described.
I n clo.;ing. wc would likc to piiint out that the centroid ciincept can he
of use in finding Llic simples1 resultant ol a dislrihulcd loading. Thus. consider thc di\ti-ihuted kiiiding ii.f.rI
shmvn in Fig. X.6.The rcsultant fiirce
01'
t t i i i loading. also shown i n the diagi-ani. i\ given iis
<,
8.1.
Show that the centroid of area A is the same point for axes
and x'y'. Thus, thc pasition of rhc ccntroid of an area is a pmpcrty only of the area.
XJ
v'
-1,-
I
~
Figure P.8.4.
t
8.5. What are the centroidal cormiinates for the shadzd arza?
The curved boundary is that of a parabola. [Hint: 'The general
equation for parabola? of the shape shown is J = cr.r2 + b.1
Figure P.8.1.
8.2. Show that thc centroid of the right triangle is x, = 2a/3.
\ = MD.
v
T
J I..
v
Parabola
20 ni
Figure P.8.2.
8.3. Find t h r ccntroid of thc area under the half-sine wave. Whal
is the first moment of this area about axis A 4 1
ki.
l5m
?
Y =
L,
sin .r
7
A
-__(
Figure P.8.5.
8.6. Show that the centroid of the
area under a semicircle is ab
shown in the diagram.
Figure P.8.3.
8.4. What are the first moments of the area about the x and y
ares? The curved houndary is that of a parabola. [Him: The general equation for parabolas of the shape shown is y 2 = ai + h.1
n
Figure P.8.6.
33:
I*--
/
+-I
Figure P.8.7.
Figure P.8.10.
8.11. Show that thc ccntrrrid of thc trianglc is at x, = ( o +
b]/3. v, = hl?. [Hint: B i n ~ ~thc
h triaoglc inlo two right trianglcr
i w which the uetitmids arc known fioiri Problcni 8.2.1
4'
h
I
I
~~
~~
t-1
20'
Figure ILX.8.
t
1-
--I
li
+I
"
-~
Figure P.8.11.
8.12. What are the cetitroidal cnordinates for the shaded area'?
The outer hoondziry is that ,if it circle having ii radius of I 111.
ti
-I
4
Figure P.X.V.
338
kigure P.8.12.
8.13. What are the coordinates of the centroid of the shaded area?
The parabola is given as y2 = 2x withy and x in millimeters.
Y
y = 7mm
x
Figure P.8.15.
In rhr remaining problems of this secrion, U J Y cenmidal
p0SirilJn.Cof'rimple ur?u.s us jilund in rhe inside co~erx.
8.16. Find the centroid of the end shield of a bulldozer blade.
x
=
10 mm
Figure P.8.13.
8.14. Find the centroid of the shaded area. The equation of the
curve is y = 5xz with x and y in millimeters. What is the first
moment of the area about line AB?
1 - 4 4
Figure P.8.16.
8.17. In Example 2.6, determine ic
for the centroid of the uiangular faces of the pyramid plus the base area ABCE. The height
of the pyramid is 300 ft.
y
I
I
I
8.18. A parallelogram and an ellipse have been cut from a rectangular plate, What are the centroidal coordinates of what is left
of the plate? What i h M,,for this area? Use formulas given on the
inside ofthe hack covers.
c
80 mm
Y'
\
AIY, 7)
+
1mmm
+
Figure P.8.14.
8.15. Find the centroid of the shaded area. What is the first
moment of this area about line A-A. The upper boundary is a
parabola y2 = 501 with x and y in millimeters.
c
Figure P.8.18.
339
8.19. A mrdiu,i axis has the mnic aira o n one side of the axis as
it does on the other side. Find the distance hetween the horizontal
(median l i n e and a parallel centmidal axis.
/(----A
t
50 mm
f
I:I+-
60'
I---.
100 irim
Figure P.8.19.
*I
-
5'
.*I
Figure P.X.22.
8.23.
X.20.
1,
Find thc centroid of the end of rhc buckct of B sinall frrrnt
end lvilder.
Find thc ccntroid of the truss gnwt.1 platc.
Find the centroid of thc indicatcd area.
Figure P.8.23.
v
I
8.24.
Whew is the czntroid of the airplane's vertical stabilizer
(whole area)'!
r
( - ' 0 2 4. l
Figure P.X.21.
8.22. Find the crntruidal cuordin;acs ftrr the shaded area shown
tiiw rhc results in meters. [Hinr: Sec Fig. P.8.f.l
I
340
1-
J 111
-1
Vigure P.8.24.
I
t I 111
What is the first moment ofthe shaded area about the diagonal A-A? [Hinr: Consider symmetry.]
8.25.
8.27. A wide-flange I beam (identified as 14WF202 I beam) is
shown with two reinforcing plates on top. At what height above
the bottom i s the centroid of the beam located'?
.r
Figure P.8.25.
1
Figure P.8.27.
8.26. A built-up beam is shown with four 120-mm by 120-mm
by 20-mm angles. Find the vertical distance above the base for the
centroid of the cross-section.
8.28. Compute the position of the centroid of the shaded u t a .
[ H i m See Fig. P.8.6.1
0 In,"
1-
5
21 mm
1-35
mm+
f
4T-
13 mm
41 mm
A+35 mm
Figure P.8.28.
34 I
Figure P.8.30.
ICL,--rl
Figure P.8.2Y.
8.3
I
Other Centers
We employ the concepts of n i ~ i n c n t sand ccntroids in mechanics for threedimensional bodies as wcll as for plaiie areas. Thus, w e intriiduce now the
first tnoment 01a volunie. V. 01a hiidy (see Fig. X.X) ahout a point 0 where
(8.6)
Figure 8.8. Center of voIume. C.V.. of a
v?.
hody.
=
jjjrdr.
i
Thcrrlore.
r' =
v! Jjjri / r '
(8.7)
1
We see that the center of volume is the poinl wherc we could hypothetically
concentrate thc entire volu~neof il body for purposcs of computing the firhi
moment of the volume lit. the hody about some point 0. The components of
Eq. 8.7 pibe the ( ~ ~ ~ ~ ~ , , i d d i ~ofi ~volttine
i i i ( . ( ,.x,
.~
, );, and z c . Thus, we have
342
343
SECTION 8.3 OTHER CENTERS
x do. it should be noted, gives the first moment of volume
The integral
about the yz plane, etc.
If we replace dv by dm = p du in Eq. 8.6, where p is the mass densiv,
we get thefirst moment o f m a n about 0.That is,
moment vector of mass
=
111r
p dz:
(8.9)
V
The cenfer of mass rc is then given as
(8.10)
where M is the total mass of the body. The center of mass is the point in space
where hypothetically we could concentrate the entire mass for purposes of
computing the first moment of mass about a point 0. Using the components
of Eq. 8.10, we can say that
"=
lls
XP
YP dL'
dv
IljpdZJ '
yc =
JjJpdv
,
2, =
ISf =pdv
IIJ p
dv
In our work in dynamics, we shall consider the center of mass of a
system of n particles (see Fig. 8.9). We will then say:
$/.Me
Therefore,
V
mi';
-
i=l
M
(8.1 I)
where M is the total mass of the system. Clearly, if the particles are of infinitesimal mass and constitute a continuous body, we get hack Eq. 8.10.
Finally, if we replace dv by ydu, where y ( = pg) is the specific weight,
we arrive at the concept of center of gravity discussed in Chapter 4. We have
used the center of gravity of a body in many calculations thus far as a point to
concentrate the entire weight of a body.
You should have no trouble in concluding from Eq. 8.10 that if p is
constant throughout a body, the center of mass coincides with the center of
volume. Furthermore, if y ( = pg) is constant throughout a body, the center of
gravity of the body corresponds to the center of volume of the body. If,
finally, p and g are each constant for a body, all three points coincide for the
body.
We now illustrate the computation of the center of volume. Computation for the center of mass follows similar lines, and we have already computed centers of gravity in Chapter 4.
x
e
m,
Figure 8,9. System of panic,es showing
center of mass, C.M.
c
SECTION 8.3 OTHER CENTERS
Example 8.4
8
!
What is the coordinate s? for the center of volume of the body of revolution shown in Fig. 8.1 I'! Note that a cone has been cut away from the left
end whilc. at the right end, we have a hemispherical region.
-1
4 mm-11
mm
Figure 8.11. Compovte valume
We have a composite body consisting of three simple domains-a
cone (body I ) , a cylinder (body 21, and a hemisphere (body 3). Using
furinulas from the inside covers, we have:
Tliereforc,
We have presented 8 number of three-dimensional problems for detcrmining the center of volume, ccnter of mass, and the center of gravity of composite bodies. We will leave it to the student to work hidher way through
these problems, working from first principles, without the need of examples.
However, we ask that you follow the following format, which clearly is an
extension 01' what we havc been doing up to this point.
345
346
CHAPlEK X
PROP1IRTIES I F SURFACES
.
(p;
I,
...
I n closing. wc wish to point o u t further that curved surfaces and lines
have centroids. Since we shall have occasion in the next section to consider
the ccntrnid of a line, wc simply point nut inow (see Fig. 8.12) that
Figure 8.12. Czntniid frir ciirwd line.
(8.12a)
(X.12hi
wlicre 1. i s the length nf the line. Note that the centroid C will not gcnerally
lie along the line.
Consider next a curve tnadc u p nf simple curves each of whose cem
troids i s known. Such i s the case shnwn i n Fig. 8.13. inade up nf straight
7,.
lines. .The linc segment I.,.has for instance centroid C, with coordinates il,
iis has heen shown i n the diagram. We can then suy for the enlire curve that
Figure 8.13. Ccntroid for coinporite line.
(8.13)
SECTION 8.4 THEOREMS OF PAPPUS-GULDINUS
"8.4
Theorems of Pappus-Guldinus
The theorems of Pappus-Guldinus were first set forth by Pappus about 300
A . D . and then restated by the Swiss mathematician Paul Guldinus about 1640.
These theorems are concerned with the relation of a surface of revolution to
its generating curve, and the relation of a volume of revolution 10 its generaling area.
The first of the theorems may be stated as follows:
Y
Generating curve
Figure 8.14. Coplanar generating curve.
To prove this theorem, consider first an element d/ of the generating curve
shown in Fig. 8.14. For a single revolution of the generating curve about the
x axis, the line segment d/ traces an area
dA = 2Ry dl
For the entire curve this area becomes the surface of revolution given as
A = 2 n I y d l = 2ny,L
(8.14)
where L is the length of the curve and y, is the centroidal coordinate of the
curve. But 2 % is~the~ circumferential length of the circle formed by having the
centroid of the curve rotate about the .r axis. The first theorem is thus proved.
341
348
CHAPTEI< K
PROPERTIES OF St!KFA(TS
An(ithcr way 01intcrpreting Eq. X.14 i h to note that the area ofthe hody
<)Irevolution i s e q u l to 211 times thc .fii:st monwir 0 1 the generating curve
about the axis of rc\'oIulioii. If the genernting c u I \ ~ ci h coinposcd of himplc
curves. I-,. whose ccntroidh iirc known. such a s the cncc shiiwn i n Fig. 8.13.
thcn we can exprcs, A iic l i i l l o w s :
whcl-c 7,i s the centniidsl c~iordinateto the ith line segment L,.
Thc second theoreni may he stated a s liillciws:
Consider a plane su$ace and an axis of revolution coplonar with the sur,face bat oriented such that the axis cui1 intersect the surface only as u tangent at the boundary or havv no intersection ai all. The volume of the
body ufrevolution developed by rotating the plane surface about the axis
uf revolution equals the product O f the area of the suface times tbe circumference of the circle fiirmed by the centroid qf the suface in the
process of generating the body uf revolution.
SECTlON 8.4 THEOREMS OF PAPPUS-GULDINUS
Thus, the volume V equals the area of the generating surface A times the circumferential length of the circle of radius yc. The second theorem is thus also
proved.2
Another way to interpret Eq. 8.16 is to note that Vequals 2a times the
first moment of the generating area A about the axis of revolution. If this area
A is made up of simple areas At, we can say that
(8.17)
where y,, IS the centroidal coordinate to the ith area A , .
We now illustrate the use of the theorems of Pappus and Guldinus. As
we proceed, it will be helpful to remember the theorems by noting that you
multiply a length (or area) of the generator by the distance moved by the centroid of the generator.
'11 is ta be pointed out that the cenVoid of a volume of revolution will not be coincident
with the centroid of a longitudinal cross-section taken along the axis of the volume. Example: a
cone and its triimgular. longitudinal cross-section.
Example 8.5
Determine the surface area and volume of the bulk materials trailer shown
in Fig. 8.16.
Figure 8.16. Bulk materials trailer.
We shall first determine the surface area by considering the first
moment about the centerline A-A (see Fig. 8.17) of the generating curve of
lte"'7-
e
20'-
A
A
Figure 8.17. Generating curve for surface of revolution
349
350
CHAPTEK 8
I'KOPFRTIES 0 1 ' SIJKI:A('FS
Example 8.5 (Continued)
the surface of rwiilutimi. This curvc is a h c t (11' 5 straight line5 e d i of
whose centroids is misily k n o w n by inspection. Accordingly we niay use
Eq. 8.15. For clarity, we use a column forinat for the data as follows:
v, ( f l )
L, ( f l )
I .s
3.S
i
3.5
1.5
I. 1
2. \,,X'
+ I2
L, T,
= I(.Oh
3. 20
3. 8.06
5. 3
If13
4.5
28.21
80
X2l
1.5
L,V,
=
115.43
Thcrcfore.
A = (2rO(135.33) =
914ft'
To get the voliinic. we next show i n Fig. 8. I X the generating arcii lor
Ihe hody of rcvolutioii. Notice il has been dcconiposed into simple compositc areas. We shall ciiiploy Eq, 8.17 arid hencc wc shall iieed the Sirst
nionient ofarca iibiiut the iixis A-A of the composite arcas. Again. we shall
employ a columii formal for the data.
A
I. 24
2. $(8)(1) = 3
3. 80
,A
3+
I .s
16
;= 3.33
4. 1
2
1.137
5. 24
I .S
13.73
I60
13.33
16
1A,T, = 2513.7
Therefore.
V = 2 i i x A , T , = ( 2 1 ~ ) ( 2 5 8 . 7=)
1,625ft3
The theorems or Pappus and Guldinus h a w enahled us to computc the surface area and the volume of the bulk inaterials trailer quickly and easily.
8.31. If r2 = ax in the body af revolution shown. compute the
centroidal distance x, of the body.
Figure P.8.34.
Figure P.8.31.
8.35. Find the center of mass for the paraboloid of revolution
having a uniform density p.
i
8.32. Using venical elements of volume as shown, compute the
centroidal coordinates x, , yCof the body. Then, using horizontal
elements, compute ?<.
i
1
Figure P.8.35.
Figure P.8.32.
8.33. Compute the center of volume of a right circular cylinder
of height h and radius at the base r.
8.36. A small bomb has exploded at position 0.Four pieces of the
bomb move off at high speed. At f = 3 EZC, the following data apply:
m (kgi
r (m)
I.
.2
.I
2.
3.
.IS
4
??
Zi
+
3j + 4k
4i + 4j 6k
-.li + 2j - 3k
2 - 3i + 2k
~
Whdt is the position of the center of mass?
Figure P.8.33.
8.34. Determine the position of the center of mass of the solid
hemisphere having a uniform mass density p and with a radius a.
Figure P.8.36.
75
I
4
8.43. Find the center of gravity of the bent plate. The rectangular
cutout occurs at thc geometric center of the surface in the-ic plane.
I
8.46. Two thin plates are welded tugether. One has a circle of
radius 2llO mm cut out ns shown. If each plate weighs 450 Nlm',
what is the position 01 thc center of mass?
.3 m
IJ
Figure P.8.46.
8.47. Where is the center of mass of the bcnt wire if it weighs
Figure Y.8.43.
Ill Nlm'!
8.44. A hen1 aluminum rod weighing 30 N l m is fitted intu a
plastic cylindrr weighing 200 N, as shown. What are the centers
of volumc, mass. and gravity?
i
I
Figure P.8.47.
8.48. Find the centci of mass of the bent wire shown in the
plane. The wirc weighs IS Nlm.
Figure P.8.44.
z?
8.45. An illuniinum cylinder fit5 snugly into a brass block. The
brass weighs 43.2 kN1nv' and the illuniiiium wzighh XI kN1m'. Find
the center af volume, the center of mass, and the center of gravity.
400 mm
y
T
6OU m m
l?
-I"/
500 m m
-AJ
Figure P.8.45.
\
900 mm
Figure P.8.48.
8.49. In Prohlcm 8.41, involving a wooden cone-cylinder with i
cylindrical hole, find the center of mass for the case where tht
cylinder has a density of4h.C Ibmlft' and the cone has a density oi
30.0 Ibmlft'.
35:
8.50. The volume uf an ellipsoidal body ofrevolution is known
from calculus to be inab2.If the area of an ellipse is linbI4, find
the centroid of the itred for a semiellipse.
l-~
1.51. Find the centroidal coordinate y , of the shaded area shown,
ising the theorems of Pappus and Guldinus.
~u
in m m
I 2 0 mrn 2nn m m
Figure P.8.50.
Figure P.8.54.
8.55.
Find the volumc and r u f i x c nrca n1 the Apollo rpaceship
used Sor lunar cxploralion.
Y
Figure P.8.51.
1.52. The cutting tool of a lathe is programmed to cut along the
lashed line as shown. What are the volume and the area of the
rody of revolution fnrmed on the lathe'?
Figure P.8.55.
8.56.
Find the cenler of volume 5 for thc machine elerncnt shown.
b f ? L20" 4 3 ' J + 8 " 4
Figure P.8.52.
i.53. Find the surSace area and volume of the right conical
r
I
IO
Figure P.8.53.
54. Find the surface area and volume of the Earth entry capile for an unmanned Mars sampling mission. Approximate the
xmded nose with a pvinted nosc as shown with the dashed lines.
54
I
1.-
-11)-
j y
Figure P.8.56.
SECTION 8.5 SECOND MOMENTS AND THE PRODUCT OF AREA OF A PLANE AREA
8.5
Second Moments and the Product
of Area3of a Plane Area
We shall now consider other properties of a plane area relative to a given reference. The second mnrnenfs of the area A about the x and y axes (Fig. 8.19),
denoted as I,, and lyv,respectively, are defined as
j y? dA
Ivy= j,,x z dA
I,, =
I
A
(8.18~1)
(8.18b)
I yFigure 8.19. Plane surface
The second moment of area cannot be negative, in contrast to the first
moment. Furthermore, because the square of the distance from the axis is used,
elements of area that are farthest from the axis contribute most to the second
moment of area.
In an analogy to the centroid, the entire area may be concentrated at a
single point (kx, kJ to give the same second moment of area for a given reference. Thus,
The distances k,, and kv are called the radii ofgyration. This point will have a
position ihui depend.y nui only on the shape (fl the area but also on the position o f t h e reference. This situation is unlike the centroid, whose location is
independent of the reference position.
The product .f area relates an area directly to a set of axes and is
defined as
I,, =
1, x y d A
(8.20)
‘We often usc the expressions moment and product of inenin for second moment and product of area. respectively. However. we shall also use the former expressions in Chapter 9 in connecticin with mass distributions.
355
356
CHAPTER
x PROPERTIES OF S ~ J R F A C E S
?
dA
T
Area synlnlrlric
axis,
This quantity may hc negative.
I I the area under consideration has an axis of symmetry, the product of
area for this axis and any axis orthogonal to this axis must he zero. You can
readily reach this conclusion hy considering the area in Fig. 8.20, which is
symmetrical ahout the axis A-A. Notice that the centroid is somewhere along
this axis. (Why ?) The axis of symmetry has hccn indicated as the y axis. and
an arbitrary x axis coplanar with thc area has hecn shown. Also indicated arc
two elemental areas that are positioned as mirror images ahout the J axis. The
contnhution to the product of area of each elcmciit is .KJ CIA. hut with opposite
signs, and so the net i-esull i h x r o . Since the ciitire iireii can he considered to
he composed OS such pairs;. it hecomes evident that the product of area for
such cases is zero. This should no/ he taken to nican that a nonsymmetric area
cannot have a zero product of area ahout a set of axes. We \hall discuss this
last condition in more deud later.
8.6
Transfer Theorems
We shall now set foith a theiircm that will he 01gre;u use in computing sccond moments and products of are8 for areas that can he decomposed into simple parts (composite areas). With this theorem, wc can find second moments
or products of area ahout any axih in terms of second moments or products of
area ahout a pnmllel set of axes going through the cmtroid of the area in
question.
An .r axis is shown in Fig. 8.21 pal-allel to and at B distance d from an
axis x’ going through the centroid of the area. The latter axis you will recall is
a centi-ooidal axis. Thc second momcnt of area about rlic I axis is
where the distance y has been replaced hy (J’
operation and integrating, le;ids to the rcsiilt
=
+
y” </A + 2dj,, y’dA
:
I
,cl
Figure 8.21. x and x’ are parallzl axes.
d ) . Carrying out the squaring
+ Ad’
The first term on the right-hand side is by definition I r , < , . The second term
involves the first moment of area ahout the .I’axis. But the .x’ axis here is a
centroidal axis, and so the second term is zero. We can inow slate the transfcr
thcorem (frequcntly called the parallel-axis theorem):
u y axis
- ‘about
a par die^
+ Ad2
(8.21)
a i s at eenmoid
where d is the perpendicular distancc hctween Ihc axis for which I is being
ciimputed and the parallel ccntroidal axis.
SECTION 8.7 COMPUTATION INVOLVING SECOND MOMENTS AND PRODUCTS OF AREA
In strength of materials, a course generally following statics, second
moments of area about noncentroidal axes are commonly used. The areas
involved are complicated and not subject to simple integration. Accordingly,
in structural handbooks, the areas and second moments about various centroidal axes are listed for many of the practical configurations with the understanding that designers will use the parallel-axis theorem for axes not at the
centroid.
Let us now examine the oroduct of area in order to establish a Darallelaxis theorem for this quantity. Accordingly, two references are shown in Fig.
8.22, one (x’, y’) at the centroid and the other (x, y ) positioned arbitrarily but
parallel relative to x’y’. Note that c and d are the x and y courdinates, respectively, of the centroid of A as measured from reference xy. These coordinates
accordingly must have the proper signs, dependent on what quadrant the centroid ofA is in relative toxy. The product of area about the noncentmidal axes
xy can then be given as
In =
I,
xy dA =
5,
Y’
Figure 8.22. c and d measured from xy.
(x’
+ c)(y’ + d ) dA
Carrying out the multiplication, we get
Clearly, the first term on the right side by definition is Ix.,.. whereas the next
two terms are zero since x’ and y ’ are centroidal axes. Thus, we arrive at a
parallel-axis theorem for products of area of the form:
It is important to remember that D and d a r e measured from the xy axes
centroid and must have the appropriate sign. This will be carefully
pointed out again in the examples of Section 8.7.
to the
8.7
Y
351
Computations Involving Second
Moments and Products of Area
We shall examine examples for the computation of second moments and
products of an area.
3.58
(‘HAPTER X
PROPERTIES O b S I J K b N ‘ t S
Example 8.6
A rectangle is shown in Fig. 8.23. Compute the second moments and prod.
ucts o l area about the centroidal x’y’axci iis well as ahout thc .q axcs.
iii
i~.
x
L
Figure 8.23. Rectangle: hasc h. height h.
l y , vI,r,,s,. For computing It ,,,, we can use a strip of width d?‘ at a distance s ’ from the x’ axis. The area r l A then hecomes h &‘. Hence, we have
This is a common result and should well he remembered since it occurs so
often. Verhally. fnr such an axis, the second moment 01 area is equal tn %.
the hase h times the height h cubed. The second moment of area for the J’’
axis can immediately he written as
(h)
where the hase and height have simply heen interchanged.
As a result of the previous statement? on symmctry. we iinmediatcly
note that
6.I>\, I,,
Emplnylng the tran\ler thcoreni\, we get
I LI = hbh’ +bhe2
I VY = &hb3
+ bM2
In computing the product of area. we must he careful to cmploy the proper
signs for the transfer distances. In checking the derivation of the transfer
theorem, we see that these distances are measured from the noncentroidal
axes to the centroid C. Therefore, in this prohlem the transfer distances are
( + e )and (H).
Hence, thc computation of It, becomes
I+ = 0
+
and is thus a negative quantity.
:”.,>
(hh)(+r)(-d) =
(0
SECTION 8.7 COMPUTATTON INVOLVING SECOND MOMENTS AND PRODUCT OF AREA
359
Example 8.7
What are 6,
I,. and IX,for the area under the parabolic curve shown in
Fig. 8.24?
To find I, we may use horizontal strips of width dy as shown in
Fig. 8.25. We can then say for IXx:
10
I, =
But
Therefore,
y2[dy(10- XI]
x =
I,
Y
qmiy2
Figure 8.24. Plane area.
10
=
0
yz(IO - &6y1'2)dy
As for I",. we use vertical infinitesimal strips as shown in Fig. 8.26.
We can, accordingly, say:
Figure 8.25. Horizontal strip.
Finally, for LYwe use an infinitesimal area element dn dy shown in
Fig. 8.27. We must now perform multiple integration4 Thus, we have
I0
v=r1/10
s =I0jVX0
*ydyh
Notice by holding x constant and letting y first run from y = 0 to the curve
y = 3/10 we cover the vertical strip of thickness dn at position x such as is
shown in Fig. 8.26. Then by letting x run from zero to IO, we cover the entire
area. Accordingly, we first integrate with respect toy holding x constant. Thus,
k - 1 0 m m 4
Figure 8.26. Vertical strip.
Y
/A+
y=M,
Next, integrating with respect to x, we have
b l O m m 4
"This multiple integration involves boundaries requiring some variable limits, in contrast to previous multiple integrations.
x
Figure 8.27. Element for multiple
integration
360
CHAPTER 8
PKOPERTIES OF SURFACES
..-.,.
Example 8.8
Compute the second niomcnt (if arcil (if a circiiliii- iireii ah(ii11ii di;iiiictcr
(Fig. 8.28).
i
Figure 8.28. Circular area with p d a r ciiwdinarc.;.
Using polar coordinates, we hiivc' lor i l l ;
Completing the integration. uc h a w
:'-
"..".
>..
SECTION 8.7
COMPUTATION INVOLVING SECOND MOMENTS AND PRODUCT OF AREA
Example 8.9
Find the centroid of the area of the unequal-leg Z section shown in Fig.
8.29. Next, determine the second moment of area about the centroidal axes
parallel to the sides of the Z section. Finally, determine the product of area
for the aforementioned centroidal axes.
Figure 8.29. Unequal-leg Z section.
We shall subdivide the Z section into three rectangular areas, as
shown in Fig. 8.30. Also, we shall insert a convenient reference xy,as shown
in the diagram. To find the centroid, we proceed in the following manner:
A, (in.?)
1. (2)(1) = 2
2. (8)(1) = 8
3. (4)(1) =A
EA, = I4
?, (in.)
I
2.50
S
V,. (in.)
A,rj(in.')
7.50
4
2
20
.so
20
A,,?, = 42
Figure 8.30. Composite area.
A;?; (in.')
15
32
2
A,Y, = 49
361
362
CHAPTER x
PRO PERTIE S OF SURFACES
Example 8.9 (Continued)
Therefore,
We h w e shown the centroidal axes .xc);. in Fig. 8.31. We now find
lxcxc
and ly,vt,
using the parallel-axis theorem and the lormulas &hh’ and
&hb3 for the second moments of area about centroidal axes of symmetry
of a rectangle.
+[(&1(4)(13)+ (4)(3*)] =
4
0
I?<>?
. = [(&)(1)(23 + ( 2 ) ( 2 2 ) ]+ [(/2)(8)(13) + (8)(fP]
0
0
+[(&)(I)(49 + (4)(22)] =
i
Figure 8.31. Centroidal axe,
32.
<
a
Finally, we consider the product of area Cc?<,.Here we must be cautious in using the parallel-axis theorem. Remember that xcy, are centroidal
axes for the entire ales of the Z section. In using the parallel-axis theorem
for a subarea, we must note that x<y<are not centroidal axes for the subarea. The centroidal axes to he used in this problem for subareas are the
,,
are simply axes ahout
axes of symmetry of each subarea. In short, xV
which we are computing the product of area of each subarea. Therefore, in
the parallel-axis theorem, the transfer distances c and d are measured.from
the x,y, uxer to the rentroid in each subarea, as noted in the development
of the parallel-axis theorem. The proper sign must be assigned each time
to the transfer distances with this in mind. We have for lrcJt:
XJ
8.57. Find lxx,/vY. and I=\, for the triangle shown. Give the
results in feet.
8.61. Find I,\ for the shaded area. You must first determine the
constant c.
Y
Y
I
Figure P.8.57.
8.58. What are the second moments and products of area of the
ellipse for reference xy'? [Hint: Can you work with one quadrant
and then multiply by 4 for the second moments?]
Y
Figure P.8.61,
8.62. Find I v y for the area between the curves
y = 2 sin x ft
y = sin 2x ft
Figure P.8.58.
8.59. Find I,, and I,, for tbe quarter circle of radius 5 m
fromx = O t o x = nft.
Y
8.63. Find I", for the areas enclosed between curves y = cos x
and y = sin x and the lines x = 0 and x = nI2.
x
5m
Figure P.8.59.
8.64. Show that I_ = bh3/12, 5, = b3k/12, and
for the right triangle.
:i\
1
I,, = b2h2/24
Y
8.60. Find lix,I">, and I., for the shaded area.
Y
I
I+-+-
Figure P.8.60.
X
Figure P.8.64.
363
8.74. In Problem 8.73, show that I,,,, = bh3t36, Iv+ =
(bhI36Kb2 - ab + a'). and lxcjc
= (h2bI72)(2a - b ) for the trianele.
~. IHint: Use the results of Problems 8.1 I and 8.73 and the
parallel-axis theorem.]
8.77. Find In,.,! lxsc,and I,
Disregard all rounded edges.
C~ c
for the stmctural "hat" section
8.75. Find I-. Ivy. and I,, of the extruded section. Disregard all
rounded edges. Do this problem using 4 areas. Check using 2
areas.
Y
Figure P.8.77.
I
8.78. Find In. Ivy. and
<,"of the hexagon.
x
X
Figure P.8.75.
8,76, Find the second moment of area of the rectangle (with a
hole) about the base of the rectangle. Also, determine the product
of area about the base and left side.
k R d /
Figure P.8.78.
8.79. A beam cross-section is made up of an I-shaped section
with an additional thick plate welded on. Find the second
Of area for the
'CY,. Of the
crosssection. What is
Give the results in millimeters.
+9"
i_2:l(t:i
Figure P.S.76.
i
? b6" - t i
Figure P.8.79.
365
I
I
I
Figure P.X.XI.
8.8
Relation Between Second Moments
and Products of Area
We sh:ill iiow show that wc can iisccrliiin second miiments and producl 0 1
area relatiw to a rolalcd rclcrcncc .A'!' i f w e know these quilnlities for referencc r y that has tlic .surne o r i ~ i ~Such
i.
ii reference x'y' tmated an anglc a
froni x? (countcrclockwise :is positive) i s shown i n Fig. X.32. Wc shall
assume that the second miimcnts wid product of area for the unprimed reference are kiiowii.
Hefiire proceeding, wc mu\t know Lhr relation hetween the coordinalcs o1
thc area clcmcnts (/A 101- the Iwo references. Froiii Fig. X.32. yi)u inlay show that
a + y sin n
-isin a + ? c o s n
~ i '= i
s' =
IX.23a)
cos
lX.23h)
With relation 8.23h. wc can cxprcss I t , $i n, the follo\ving manncr:
= j,,,ty'~c/~
= j , ~ t - . ~ s i n a + y c ~ ~ s ~ ~ i (X.24)
~[/~
Figure 8.32. Rimlion of axe\
('arrying ouI thc scj~iiire.we h a b e
/s,t,
=
sin? c x j
n
~ tA
<!?
~
?hili n c o s a j . x v < / ~+ c o s i a j y?'/il
1
A
Thrrefiire.
I>,>,
=
366
sin'
a+
cos2
n
~
21$)sin cxcos a
(8.25)
SECTION 8.8 RELATION BETWEEN SECOND MOMENTS AND PRODUCTS OF AREA
A more common form of the desired relation can he formed by using the following trigonometric identities:
cos2 a = $(I + cos 2a)
sin2 a = $ ( I - c o s 2 a )
2 sin a cos a = sin 2a
(a)
(b)
(C)
We then have6
To determine I;,., we need only replace the a in the preceding result by (a +
ED).Thus,
Note that cos (2a + z) = - cos 2a and sin (2a + n) = -sin 2a. Hence,
the equation above becomes
-€
- I
am 2a + 1,
(8.27~
Next, the product of area /+.can be computed in a similar manner:
/xc.
= j,x’y(d~ = I A ( x c o s a + y s i n a ) ( - x s i n a + y c o s a ) ~
This becomes
1
I
X Y
I
= sin a cos a (5,
- /J
+
(cos’ a - sinZ a)/xY
Utilizing the previously defined trigonometric identities, we get
=
+ I,coSzq i
(828)
Thus, we see that, if we know the quantities 1,. I,, and I,, for some reference
xy at point 0, the second moments and products of area for every set of axes
at point 0 can be computed. And if, in addition, we employ the transfer theorems, we can compute second moments and products of area for any reference in the plane of the area.
“!3quations 8.26. 8.27, and 8.28 are called rrnnsfomarion equations. They will appear in
the next chapter and in your upcoming solid mechanics course for variables other than second
moments and products of area. In the remaining portions of this chapter, you will see that a number of imponant properties of second moments and products of area are deducible direcrly from
these transformation equations. This primarily accounts for the importance of the transformation
equations. Chapter 9 will give you additional insight into this topic.
367
S EC T ION 8.9 POLAR MOMENT OF AREA
8.9
Polar Moment of Area
In the previous section, we saw that the second moments and product of area
Sor an orthogonal reference determined all such quantities for any orthogonal
reference having the same origin. We shall now show that the sum of the
pairs of second moments of area is a constant for all such references at a
point. Thus, in Fig. 8.35 we have a reference xy associated with point a.
Summing I,, and Ivywe have
Figure 8.35. J = I,,
+
I.>.
Since r 2 is independent of the orientation of the coordinate system, the sum
IA, + I _ is independent of the orientation of the reference. Therefore, the
sum of second moments of area ahout orthogonal axes is a function only of
the position of the origin u for the axes. This sum is termed the polur moment
of area, J.x We can lhen consider J to be a scalar field. Mathematically, this
statement is expressed as
J = J(x: y ' )
(8.29)
where x' and y ' are the coordinates as measured from some convcnicnt
reference x'y'for the point of interest.
That the quantity ( I r r + IJ does not change on rotation of axes can
also be deduced by summing transformation equations 8.26 and 8.27 as we
suggest you do. This group of terms is accordingly termed an invuriunt. Parenthetically, we can similarly show that (/&v
- I;,) is also invariant under a
rotation of axes.
XQuite oftcn I,, i s used for the polar moment of are&.
369
370
CHA1”I‘ER X
PROPERTIES OF SURFACES
8.10
Figure 8.36. Principal axes
Principal Axes
Still othcr conclusions may he drawn about second iiiotiients and products of
a r m associated with ii point i n an area. In Fig. 8.36 ail area i s showti with a
reference xy having its urigin at point o . We shall assume that I,.<. I $ , , and I,,
are known for this reference. and shall ask at what angle CY we shall find an
axis having the inuirnuni second mu~iientof :ireti. Since the sum of thc second moments of area i s constant Iur any reference with origin at (1. the nrirtirnurn second niniiicnt of area must then correspond 10 iui iixis at rifihf anglcs
to the axis having the maximimi second tiinmetit. Sincc second nimnen~sof
area have heen expressed i n Kqs. 8.26 and 8.21 a s lunctiotis 0 1 lhc variable a
at a point. these extremcs may readily he determined by setting the partial
derivative nf I \ , & , with rcspcct to uequ;il to (erii. Thus.
II we denote the value nf a that satisfies the equation above iis iU, we h a w
Hence,
This formulation gives us the angle ri. which corresponds tO an extreme value
of I > , , , (i.e.. ti) a maximum or minimum valuc). Actually, there are two possible valucs u f 2 N which are nradians apart that w i l l satisfy the equation above.
Thus,
or
2ix =
p
+i[
This means that we have two values o f ri, given a s
Thus. there are two axes orthogonal to each other having extreme values for
the second monienl of area at (1. On tie of thehe axes i s the n i a x i i n ~ msecond
moment of area and. a s pointed out carlicr. thc minimum second moment nf
area must appear on the other axis. These axes are called the p r i i i ~ ~ i LpLlI P S .
SECTION 8.1o PRINCIPAL AXES
Let us now substitute the angle ti into Eq. 8.28 for Q,,:
I..=xi.
~
2
I,'v
sin 2 8
'~
+ /_, cos 2 8
(8.31)
6)and angle 22
such that Eq. (8.30) is satisfied we can readily express the sine and cosine
expressions needed in the preceding equation. Thus
If we now form a right triangle with legs 2'- and (Ivy -
By substituting these results into Eq. 8.31, we get
Hence,
(,y,
= 0
Thus, we see that the product of urea corresponding to the principul
u e s is zero. If we set I,.,. equal to zero in Eq. 8.28, you can demonstrate the
converse of the preceding statement by solving for a and comparing the
result with Eq. 8.30. That is, if the product of area is zero for a set of axes at
a point, these axes must be the principal axes at that point. Consequently, if
one axis of a set of axes at a point is symmetrical for the area, the axes are
principal axes at that point.
The concept of principal axes will appear again in the following chapter in connection with the inertia tensor. Thus, the concept is not an isolated
occurrence hut is characteristic of a whole family of quantities. We shall,
then, have further occasion to examine some of the topics introduced in this
chapter from a more general viewpoint.
31 1
372
VHAPTER
x
PKOPEKTIES 01.
S U RF A C E S
Example 8.11
Find the principal second moments of area at the centroid of the Z section
of Example X.9.
We have from this example the following results that will be of use
LO Ub:
/ccrc = 113.2
I Y(
V(
= 32.67 in.4
I /'?< = -42.0 in.4
Hence, we have
2 h = 46.21 O, 226.2"
For2ir = 46.21":
I , = 113.2+32.67
2
= 72.9
+
27.9
+
113.2-32.67cos~46,21~~-~-42~~,n46,21~
2
+
131.1 in.4
30.3 =
For 2 0 = 226.2":
l2 = 72.9 - 27.9 - 30.3 =
~~,..',~..
.
&.&%%$
As a check on our work, we note that the sum of the second moments
of area are invariant at a point for a rotation of axes. This means that
I r , I(
+
IVt
V/
= 1, +
'2
113.2+32.7= 131.1+14.75
Therefore,
145.9 = 145.9
We thus have a check on our work.
Before closing, we wish to point out that there is a graphical construction
called Mohr's circle relating second moments and products of area for all
possible axes at a point. However, in this text we shall use the analytical relations thus far presented rather than Mohr's circle. You will see Mohr circle
construction in your strength of materials course where its use in cnnjunction
with the important topics of plane stress and plane strain is very h e l p f ~ l . ~
"See I . H . Shames, Iniroducnon 10 Solid Mechontr..>.Second Edition. Prmrice-Hall. loc..
Englrwuod Clills. N.J.. 1989.
8.82. It is known that area A is IO ft2 and has the following
moments and products of area for the centroidal axes shown:
I= = 40 ft4,
I , = 20 ft‘,
Y
Lx
5, = 4ft4
Find the moments and products of area for the x’y’ reference at
point a.
Figure P.8.85.
v’
X’
8.86. Use the calculus to show that the polar moment of area of
a circular area of radius I is nr41? at the center.
Y
6 - x
Figure P.8.82.
Figure P.8.86.
8.83. The cross-section of a beam is shown. Compute I,;.. ly.,.,
and
in the simplest way without using formulas for second 8.87. Find the direction of the principal axes for the angle secmoments and products of area for a triangle.
tion at point A.
c,,
Y’
Y
. ,
Figure P.8.83.
Figure P.8.87.
8.84. Find I,=,I,\, and I,, for the rectangle. Also, compute the
polar moment of area at points a and b.
8.88. What are the principal second moments of area at the origin for the area of Example 8.7?
I
8.89. Find the principal second moments of area at the centroid
for the area shown.
42
” k
Figure P.8.84.
8.85. Express the polar moment of area of the square as a function of x , y, the coordinates of points about which the pola
moment is taken.
1- IT^ t
Figure P.8.89.
4
7
L
1,1111
I
t
"I-
r
-
L,
3'
Figure P.8.91.
8.YZ. Show that thc axe5 for which the product of area i s
imuiii are rotated from n by an angle a so that
il ~ i i i i x -
Figure P.8.92.
8.93.
What is thc value of the mgle 01 lor thc principal axss at A.
I'
"
-2.-
kigure P.8.93
I
SECTION 8.1 I CLOSURE
8.11
closure
In this chapter, we discussed primarily the first and second moments of plane
areas as well as the product of plane areas. These formulations give certain
kinds of evaluations of the distribution of area relative to a plane reference xy.
You will most certainly make much use of these quantities in your later
courses in strength of materials.
In this chapter, we have touched on subject matter that you will encounter in the next chapter and also, most assuredly, in later courses. Specifically, in Chapter 9 you will he introduced to the so-called second-order
inertia tensor having nine terms which change (or transform) in a certain particular way when we rotate coordinate axes at a point. These particular transformation equations define the inertia tensor. Any other set of nine symmetric
terms that transform via the same form of equations, are symmetric secondorder tensors. You will also learn that the second moments and products of
area form a two-dimensional simplification of the inertia tensor. The transformation equations 8.26 through 8.28 are thus special cases of the threedimensional defining equations of the inertia tensor. Take note that certain
vital results emerged from these simplified two-dimensional transformation
equations. They included
1. Invariant property at a point for the sum (In + ZJ on rotation of axes.
2. Principal axes and principal moments of area at a point.
3. A graphical construction called Mohr’s circle that depicts the transformation equations. This topic has been omitted in this chapter hut will be
described in your solids course where yon will study the two-dimensional
simplifications of the stress and strain tensors. At that time one can make
much use of the Mohr’s circle construction and it is a simple matter then
to describe Mohr’s circle for second moments and products of area.
We will emphasize in the next chapter that there are many symmetric sets
of nine terms that transform exactly like the terms of the inertia tensor and
we classify all of them as second-order tensors. Identifying these sets as
second-order tensors immediately yields vital properties common to all of
them such as those presented above for the two-dimensional moments and
products of area.
There will be more to be said on tensors in a “looking ahead” section of
the next chapter which we invite you to examine.
375
I
r -
8.102. A half body of revnlution is shown with the x: plane as a
plane of symmetry. Determine the centroidal coordinates. The
radius at any section x varies as the square o f x .
8.105. Using the theorems of Pappus and Guldinus, find the centroid of the area of a qualter-circle.
J
Figure P.8.11)5.
Figure P.8.102.
8.103.
(a) Find
Irk, I,,. and 18>fur the .r? axes at position A .
(b) Find the principal recond moment5 of area at point A.
8.106. A tank has a semisphericdl dome at the left end. Using
the theorems af Pappus and Guldinus, compute the surface and
volume of the tank. Give the results i n meters.
Figure P.B.106.
8.107. Find I
and I,,,, for the set of axes at point A fur the
VX
v'
X.104. Whet are the directions
of the principal axes at point A'?
Figure P.8.103.
t
12m
8m
A
10m-!-ll
~i;
(11
Figure P.8.1117.
S'
Figure P.8.104,
8.108. Find the centroid of the aren, and then find the second
moments of thc area about centroidal axes parallel to the sidcs of
the area.
311
8.112. A wide-flanged I heam i\ s h ~ ~ w Far
d " the upper flangr.
what i \ the first imonienl 0 1 t h ~\haded area as a function o h meii\tired froiii .v = 0 at the right rrid t o where s apprrlaches the weh'!
Next pet hl, fur the entirc a m i ahove the position shown a\ i n
t h r wch. Let Y go from (.in the lop of thc wch.
Figure IL8.108.
Find the principal semiid moments of area at a point
where I&>
= 321 in?.
= I I X . 4 in?, and 1 , ) = 1.028 i d .
1.109.
1715-
lc--
Figure P.8.112.
I,
+
2 0 I,,
~~
+
Figure P.X.110.
1.111.
( a ) What are the centmidal coordmatrs o f t h e shaded area'!
(b) What ai(: M&,,and M , , for axes x'v' at A'? 1Il;nr: Ucr
ormulas fur the sector o f a circle given on the imide hack cover.I
,
-
8,"
Figure P.8.113.
Figure P.8.111.
78
and Products
of Inertia'
9.1
Introduction
In this chapter, we shall consider certain measures of mass distribution relative to a reference. These quantities are vital for the study of the dynamics of
rigid bodies. Because these quantities are so closely related to second
moments and products of area, we shall consider them at this early stage
rather than wait for dynamics. We shall also discuss the fact that these measures of mass distribution-the second moments of inertia of mass and the
products of inertia of mass-are components of what we call a second-order
tensor. Recognizing this fact early will make more simple and understandable
your future studies of stress and strain, since these quantities also happen to
be components of second-order tensors.
9.2
Formal Definition of Inertia
Quantities
We shall now formally define a set of quantities that give information about
the distribution 0 1 mass of a body relative to a Cartesian reference. For this
purpose, a body of mass M and a reference xyz are presented in Fig. 9.1. This
reference and the body may have any motion whatever relative to each other.
The ensuing discussion then holds for the instantaneous orientation shown
at time t. We shall consider that the body is composed of a continuum of particles, each of which has a mass given by p do. We now present the following
definitions:
Figure 9.1. Body and reference at time f.
#This chilpter may he covered at a later stage when studying dynamics. In that case, it
should be covered directly after Chapter 15.
379
380
CHAPTER 9
MOMENTS AND PRODUCTS OF INERTIA
(9.1~1)
(9. I b)
(9.1C)
The terms lxr,I+ and in the set above are called the muss momerifs ujinertiu of the hody'ahout the x,y. and 7. axes. respectively.' Note that in e x h such
case we are integrating the mass elements p d7: times the peipmdicular distance squared from the mass elements tn the coordinate axis about which we
are computing the moment of inertia. Thus, if we look along thex axis toward
the origin in Fig. 9.1, we would have the Yiew shown in Fig. 9.2. The quantity y * + z 2 used in Eq. 9.la for /,ct is clearly d i the perpendicular distance
squared from dti to the x axis (now seen as a dot). Each of the terms with
mixed indices is called the mass product of inertio ahout the pair of axes
given by the indices. Clearly. from the definition of the product of inertia, we
could reverse indices and thcrehy form three additional products of inertia f i r
a reference. The additional three quantities formed in this way. however, are
equal to the corrcsponding quantities of the original set. That is,
Figure 9.2. View of hody along x axis.
We now have nine inertia terms at a point for a given reference at this point.
Thc values of the set of six independent quantities will, for a given body.
"Nc use the same clntilticm as wa? used for second inoinents and product\ 01 a n n , which arc
dsu suinetinies called inornents and pruducts of inertia. 778s i i standard practice in mechanics. Thm
nccd he no cunfusion i n using thew qaaniitic\ i f we keep the cmlext ofdiscucsinnh clearly in mind.
SECTION 9.2 FORMAL DEFINITION OF INERTIA QUANTITIES
depend on the posifion and inclination of the reference relative to the body.
You should also understand that the reference may be established anywhere
in space and need nor be situated in the rigid body of interest. Thus there will
be nine inertia terms for reference xyz at point 0 outside the body (Fig. 9.3)
computed using Eqs. 9.1, where the domain of integration is the volume V of
the body. As will be explained later, the nine moments and products of inertia are components of the inertia tensor.
z
Figure 9.3. Origin of xyz outside body.
It will he convenient, when referring to the nine moments and products
of inertia for reference xyz at a point, to list them in a matrix array, as follows:
Notice that the first subscript gives the row and the second subscript gives the
column in the array. Furthermore, the left-to-right downward diagonal in the
array is composed of mass moment of inertia terms while the products of
inertia, oriented at mirror-image positions about this diagonal, are equal. For
this reason we say that the array is symmetric.
We shall now show that the sum of the mass moments of inertia for a
set of orthogonal axes is independent of the orientation of the axes and
depends only on the position of the origin. Examine the sum of such a set of
terms:
Combining the integrals and rearranging, we get
I _ + I v y + Iz,
= IJ/2(x2
V
+ y * + z Z ) p d u = IS/2lr(’pdu
(9.2)
V
But the magnitude of the position vector from the origin to a particle is independent of the inclination of the reference at the origin. Thus, the sum of the
moments of inertia at a point in space for a given body clearly is an invariant
with respect to rotation of axes.
38 1
382
C"A1"I'b:K V
MOMliNTS ANI) PRODLICTS OF INKRTI.4
Clcal-ly. 011 inspecli(in nf thc equations 0. I. i t i s clear that the iiioniciits
(11 inertia must always be positive, while the products o f inertia may he positiYe or negativc. 01intercst i s the case where one 01 the coordinate planes i s
ii plww nf.s\mmrrr\ h r the miss distrihution of the body. Such a plene i s thc
I
,! \ plane shown i n Fig. 9.4 cutting a hody into two parts, which. by definition
"
of symmctry. are niirror iniagcs o f each other. For the computation of I > : ,
1%
,
Figure 9.4. ?: i\ plane 0 1 cymmrlry.
each half w i l l give ii crintrihuti(in ofthe same magnitude hut ol'oppositc sign.
We can most readily hcc that this i s so by loohing along the y axis tiiwai-d thr
origin. The plane of symmetry then appears as a line coinciding with thc z
i i x i c (ECC Fig. 9.5). We can cnnsidcr the body to he composed of pairs of inass
elements diii which are mirror imagcs of each other with respect to pnsition
and shape ahout the planc of symmetry. The product (if inertia
Sor such ii
pair i s then
I:
dm
I<. =
J .x;
~
Y:
dwt = 0
Thus. we can cnncludc that
-- I({in -
1:
(1111
= 0
i---i
"FbI
<I,/ 111.1 I,,
Figure 9.5. V i r % along v axic.
lkh
tl<li,,li,,
This c(inclusion i s a l s o (rue for I \ $ . We can say that I\>,= 1,: = 0. But on
consulting Fig. Y.4. you shmild be ahle to readily decide that the terni I.\ w i l l
h a w a positivc value. Note that those products of inertia having .r as ;in index
are /era and tlial the .v coordinate axis is normal to the plane of symmctry.
Thus, we can conclude that if i w , ~ ~ ~ ~ . s .Nf pi li ur w
~ nof,swnmc,rry f i i r ihr ~no.s.s
di,strihu~iono f u l x d y , tlte p,-o'dum o r i n e r l i a hoi'ing (1,s r m index the cooro'inote thar i,s normal to the plaw o / . s w m e l r ! . w,ill he ZPIO.
Considcr next it body ofri~r.olutio~~.
Take the z axis to coincide with the
axis of symmetry. It i s easy Io conclude for the origin 0 of xy? anywhere
iil(ing the a x i s of symmetry that
I
I:
= I ,: = I > > = 0
I x 5 = I\, = co11stallt
for all possihlc .xy axes iormed hy rotating ahout the z axis at 0. Can you justify these conclusions?
Finally. we define m d i i o f , s y u i i o n i n a manner analognus to that used
for second moments o f iircii i n Chaptei- 8 . Thus:
where k 5 , k v . arid k . i r e lhe radii
(11
gyration and M i s thc total Iiiass.
SECTION 9.2 FORMAL DEFINITION OF INERTIA QUANTITIES
Example 9.1
Find the nine components of the inertia tensor of a rectangular body of
uniform density p about point 0 for a reference xyz coincident with the
edges of the block as shown in Fig. 9.6.
We first compute I., Using volume elements dii = dx dy dz, we get
on using simple multiple integration:
=
( a 9 + a'bc
- )=
p PV ( b 2 + a * )
3
where V i s the volume of the body. Note that thex axi5 about which we are
computing the moment of inertia I,x IS normal to the plane having sides of
length a and h, Le., along the z and y axes. Similarly:
I _ = -PV
-(c2 + a 2 )
..
3
I.,,= --(b
PV
.~
3
2
+e2)
We next compute lxv
Note for f x v , we use the lengths of the sides along the x and y axes
We accordingly have, for the inertia tensor:
i
Figure 9.6. Find /,!
at
0
383
384
8
C",\PTIR U
M O M E N I S A N D I'R0r)UCI'S OF INtRTlh
Example 9.2
I
Computc the components of thc inerlia icnsor iit the ccnicr or
sphere 0 1 uniform dcnsity p iis shown in Fig. 9.7.
il
s111id
SECTION 9.2 FORMAL. DEFINITION OF INERTIA QUANnTIES
Example 9.2 (Continued)
With the aid of integrdtion formulas from Appendix I, we have
PI,n I,,
Iv,,
=
2n
r* cos? $[-
4cos @(sin2e + 2)]/: dqdr
Integrating next with respect to
4, we get
Finally, we get
R' 4
R' 4
I,, = p - - r r + p - - n
5 3
5 3
8
: . I = -prrR'
x'
15
But
M = p'nR3
3
Hence,
I_ =
$ MR2
Because of the point symmetry about point 0, we can also say that
I _ = liZI
= $MR2
Because the coordinate planes are all planes of symmetry for the mass distribution, the products of inertia are zero. Thus, the inertia tensor can be
given as
385
386
CHAPTER Y
MOMENTS AND PRODIJCTS OF INERTIA
9.3
Relation Between Mass-Inertia
Terms and Area-Inertia Terms
We now relate the second inomrnt and product of area studied in Chapter 8
with the inertia tensor. To do this, consider a plate of constant thickncss I and
uniform density p (Fig. 9.9) A rcferencc is selected so that the ~x? plane is in
the midplane of this plate. The components of the inertia tensor are rewritten
for convenience as
Figure 9.9. Platc <fthickness I
Now consider that the thickncss I is sninll comparcd to the lateral dimension\
of the plate. This means that I is restricted to ii range of values having a small
magnitude. As a resull, we can make two simplifications i n the equations
above. First. we shall set i equal to zero whenever it appears on the right side
of the equations abuve. Second. we shall express d i ’ a s
r/z, = I
dA
SECTION 9.3 RELATION BETWEEN MASINERTIA
where dA is an area element on the surface of the plate, as shown in Fig. 9.10
Equations 9.3 then become
A
X
Figure 9.10. Use volume elements r dA
Notice, now, that the integrals on the right sides of the equations above are
moments and products of area as presented in Chapter 8.Denoting mass-moment
and product of inertia terms with a subscript M and second moment and product
of area terms with a subscript A , we can then say for the nonzem expressions:
Thus, r a
ct pt throughout, we can
compute the inertia tensor components for reference xvz (see Fig. 9.9) by
using the second moments and product of area of the surface of the plate relative to axes xy.
I t is important to point out that pr is the mass per unit area of the plate.
Imagine next that f goes to zero and simultaneously p goes to infinity at fates
such that the product pt becomes unity in the limit. One might think of the
resulting body to be a plane area. By this approach, we have thus formed a
plane area from a plate and in this way we can think of a plane area as a special mass. This explains why we use the same notation for mass moments and
products of inertia as we use for second moments and products of area. However the units clearly will be different.We now examine a plate problem.
TERMS AND AREA-INERTIA TERMS
387
388
CHAPTER 9
MOMENTS AN D PRODUCTS O F INERTIA
Example 9.3
Determine the ine
tive to the indicated axes xyz. The weight of the plate is ,002 N/mm.I For
the top edge, y = 2 ~ ' xwith x and y in millimeters.
It is clear that for pt we have, remembering that this product represents mass per unit area:
-
002
pt = - - 000204 kglmm'
9.81 - '
We now examine the second moments and product of area for the surface
of the plate about axes xy. Thus?
1ilil
=I,
I,
1""
=
y=z\*.~
Figure 9.11. Plate of thickness I.
v2 dydx
v="
2.-*
yi
dx
IO0
=
0
u
*
7x3'zdx
.
= 1.067 x IO' mm4
I,,
Ino
=
-~
x 2 y l ~ ' ~dx
' =
IW
n
x2(2\x)dx
i
I?
= 5.71 x IO6 mm4
xy d y dx
=
dure
= 6.67 x
los mm4
'Note we have multiple integration wherc one of [he boundaries is variable. The proce
follow should he evident from the cxample.
to
!
SECTION 9.3 RELATION BETWEEN MASS-INERTIA TERMS AND AREA-INERTIA TERMS
Example 9.3 (Continued)
i
Using Eq. (a), we can then say for the nonzero inertia tensor components:
Y
(I&
= (.000204)(1.067X IO') = 21.
.71 x 106) = I16
= (.
.67 x 10')
= (.
Note that the nonzero inertia tensor components for a reference xyz on a
plate (see Fig. 9.9) are proportional through pt to the corresponding areainertia terms for the plate surface. This means that all the formulations of
Chapter 8 apply to the aforementioned nonzero inertia tensor components.
Thus, on rotating the axes about the z axis we may u s e the transformation
equations of Chapter 8. Consequently, the concept of principal axes in the
midplane of the plate at a point applies. For such axes, the product of inertia
is zero. One such axis then gives the maximum moment of inertia for all axes
in the midplane at the point, the other the minimum moment of inertia. We
have presented such problems at the end of this section.
What about principal axes for the inertia tensor at a point in a general
three-dimensional body? Those students who have time to study Section 9.7
will leam that there are three principalaxes at a point in the general case. These
axes are mutually orthogonal and the products of inertia are all zero for such a
set of axes at a point? Furthermore, one of the axes will have a maximum
moment of inertia, another axis will have a minimum moment of inertia, while
the third axis will have an intermediate value. The sum of these three inertia
terms must have a value that is common for all sets of axes at the point.
If, perchance, a set of axes xyz at a point is such that xy and xz form two
planes of symmefv for the mass distribution of the body, then, as we learned
earlier, since the z axis and they axis are normal to the planes of symmetry,
Ix," = I,, = I,, = 0. Thus, all products of inertia are zero. This would also be
true for any two sets of axes of xyz forming two planes of symmetry. Clearly,
axes forming two planes of symmetry must be principal axes. This information will suffice in most instances when we have to identify principal axes.
On the other hand, consider the case where there is only one plane of symmetry for the mass distribution of a body at some point A. Let the xy plane at A
form this plane of symmetry. Then, clearly, the products of inertia between
the z axis that is normal to the plane of symmetry xy and any axis in the xy
plane at A must be zero, as pointed out earlier. Obviously, the z axis must be
a principal axis. The other two principal axes must be in the plane of symmetry, but generally cannot be located by inspection.
5The third principal axis for a plate at a point in the midplane is the
plate. Note that (/zz)M must always equal
+ (IJW Why?
389
z axis normal
to the
9.1. A uniform hum<igcneous 4cnder rod r i f mass M is shown
Compute I , > and
Figure P.9.1.
Figure P.9.5.
9.2. Find I _ and I , , for the thin rod of Prohleni Y. I for the case
wherc the mass per unit length at the left end is 5 Ihm/ft and
increases linearly so that at the right cnd it is 8 Ihmirt. Thc rod i s
2 0 f t in length.
for the half-cylinder
9.6. Compute the moment of inertia, I,,
5hown. The hody is homogeneous and has a mass hi.
9.3. Compute I,>for h e thin homogcneous hrxrp of mass M.
B
Figure P.9.6.
Figure P.9.0.
compute
I r < , I.,, I:?, and I , , for the ~lomoyrneous
lar parallelepiped.
9.4.
9.7. Find 1 . ~and l l vlor the homueencous right circular cylinder
of maw M.
r
i
Figure P.Y.7.
Figure P.9.4.
3 . 5 A wire having the shape of a parabola is shown. The curve
s in the yz plane. If the mass of the wire is .3 N/m,~whatare I,,,
ind lrz'?
[Hinr: Replace dr along ihe wire by l ( d y 1 & ) 2 + I dz. 1
9.8. For the cylindzr in Prohlem Y.7, the density increases lincarly in the 2 direction from a value o f . 100 gramsimm' at the left
end to a value of ,180 pramsimm' at the right end. Take I = 30
and I.. .
mm and I = IS0 mm. Find
9.9.
+
Show that I, for the homogeneous right circular cone is
Y
I
MR~.
Figure P.9.9.
Figure P.9.13.
9.10. In Problem 9.9, the density increases as the square of z in
the z direction from a value of ,200 gramslmm’ at the left end to a
value of ,400 grams/mm3 at the right end. If r = 20 mm and the
cone is 100 mm in length, find lz,.
9.11. A body of revolution is shown. The radial distance r of the
boundary from the x axis is given as r = Z.?. m. What is I, for a
uniform density of 1,600kg/m3?
9.14. Find the second moment of area about the x axis for the
front surface of a very thin plate. If the weight of the plate is
.02 N/mm2, find the mass moments of inertia about the x and y
axes. What is the mass product of inertia I=??
x
y
I
x
Figure P.9.14.
Figure P.9.11.
9.12. A thick hemispherical shell is shown with an inside radius
of 40 mm and an outside radius of 60 mm. If the density p i s 7,000
kglm,’ what is
*9.15. A uniform tetrahedron is shown having sides of length a,
b, and c, respectively, and a mass M. Show that lyz = &Mac.
(Suggestion: Let z mn from zero to surface ABC. Let x run from
zero to line AB. Finally, Let y run from zero to B. Note that the
equation of a plane surface is z = a* + py + 1: where a,p, and y
are constants. The mass of the tetrahedron is pabc16. It will he
simplest in expanding ( I - xlb - y/c12 to proceed in the form [(I
- ylc) - (x/b)I2, keeping (1 - ylc) intact. In the last integration
replace y by [ - c(1 ylc) + cl, etc.)
~
Figure P.9.12.
9.13. Find the mass moment of inertia I, for a very thin plate
forming a quarter-sector of a circle. The plate weighs .4N. What
is the second moment of area about the x axis? What is the product of inertia? Axes are in the midplane of the plate.
Figure P.9.15.
391
9.18. Can you identify hy inspection any o i i h r principii1 iixe\ 01
inertia iit A'! At R' Explain The dcnsitj oithu matcriiil i\ onii<>l-ni.
392
Figure P.9.19.
SECTION 9.4 TRANSLATION OF COORDINATE AXES
393
Note that the quantities bearing the subscript c are constant for the integration
and can be extracted from under the integral sign. Thus,
where p dz: has been replaced in some terms by dm, and the integration
JIS p
V
in the first integral has been evaluated as M, the total mass of the body. The
origin of the primed reference being at the center of mass requires of the
first moments of mass that I j j x ’ d m = I I J y ’ d m = / / j z ’ d m = 0. The
middle two terms accordingly drop out of the expression above, and we recognize the last expression to be Jz,;,. Thus, the desired relation i s
I..
‘. = I,.
.,.,
+
M(x:
+ y):
(9.7)
By observing the body in Fig. 9.12 along the z and z’ axes (Le., from directly
above), we get a view as is shown in Fig 9.13. From this diagram, we can see
that y: + x,Z = d.’ where d is the perpendicular distance between the z‘ axis
through the center of mass and the z axis about which we are taking moments
of inertia. We may then give the result above as
Izz =
i
Md2
Let us generalize from the previous statement.
The momea of inenia of a body
any a i s
the
inenia of the body about a parallel a i s that goes tkough the center of
mass, plus the total mass times the perpendicular db
a e s squared.
We leave it to you to show that for products o f inertia a similar relation
can he reached. For I t +for example, we have
Here, we must take care to put in the proper signs of xc and y, as measured
.from the xyz reference. Equations 9.8 and 9.9 comprise the well-known parallel-uxir lhc~oremsanalogous to those formed in Chapter 8 for areas. You can
use them lo advantage for bodies composed of simple familiar shapes, as we
now illustrate.
Y’
Figure 9.13. View along L direction
(from above).
394
CHAPTER 0 MOMENTS AND PROOLJCTS OF INERTIA
Example 9.4
Find I,, and I,, for the body shown in Fig. 9.14. Take pas constant for the
body. Use the formulations for moments and products of inertia at the center of mass as given on the inside front cover page.
We shall consider first a solid rectangular prism having the outer
dinlensions given in Fig. 9.14, and we shall then subtract the contribution
of the cylinder and the rectangular block that have been cut away. Thus,
we have, for the ovcriill rectangular block which we consider as body I,
From this, we shall take away the contribution of the cylinder, which we
denote as body 2. Using the formulas from the inside front cover page,
(Ixr)2 =
M(3rZ + h z ) + Md’
&[pn(1)2(1S)][3(l’)
= 5,243~
=
+ IS2] + [ p ~ ( l ) ~ ( l S j ] [ 6+’ 7.5’1
(b)
Also, we shall take away the contribution of the rectangular cutout (bod)
3):
( l x x )=
3 & M ( u 2 + h 2 ) + Md2
= &[P(8)(6)(4)](4*
+ 6 2 ) + [P(8)(6)(4)](2’ + 3’)
(c)
= 3,328~
The quantity lrAfor the body with the rectangular and cylindrical cavities
is then
l a x= (231,200 - 5,243
3,328)~
~
I,,
=
2
(d)
We follow the same procedure to obtain lxy.Thus, for the block as a
whole, we have
(IkV),
= (I,,,,
+ Mx,yc
At the center of mass of the block, both the (x’), and ( y ’ ) ] axes are normal
to planes (if symmetry. Accordingly. (I,,),. = 0. Hence,
C~),
= 0 + [P(20)(8)(1S)1(-4~(-IO)
= 96,nnop
(e)
For the cylinder, we note that both the ( x ‘ ) and
~ (.y’),- axes at the center of
mass are normal to planes of symmetry. Hence, we can say that
(I,>),
= 0 + 1p(~)(I2)(15)1(-8)(~6)
=
2,262~
(f)
IS’
Figure 9.14. Find lu and ilv.
SECTION 9.5 TRANSFORMATION PROPERTIES OF THE INERTIA TERMS
Example 9.4 (Continued)
Finally, for the small cutout rectangular parallelepiped, we note that the
( x ' ) ~and cy'), axes at the center of mass are perpendicular to planes of
symmetry Hence, we have
(I&
The quantity
is then
I
/Iy
= 0 + [~(8)(6)(4)1(-2)(-16)
= 6,144~
(g)
for the body with the rectangular and cylindrical cavities
If p is given in units of lbdft,) the inertia terms have units of Ibm-ft.z
"9.5
Transformation Properties of the
Inertia Terms
Let us assume that the six independent inertia terms are known at the origin of
a given reference. What is the mass moment of inertia for an axis going
through the origin of the reference and having the direction cosines 1, m, and n
relative to the axes of this reference? The axis about which we are interested
in obtaining the mass moment of inertia is designated as kk in Fig. 9.15.
Y
x
Figure 9.15. Find lhh
From previous conclusions, we can say that
(9.10)
395
396
CHAPTER 9 MOMENTS AND PRODUCTS OF INERTIA
where @ is the angle between kk and r . W e shall now put sin2 @ i n t o a more
uscful form by considering the right triangle Sormed by the position vector r
and the axis kk. This triangle is shown enlarged in Fig. 9.16. The side a of the
triangle has a magnitude that can be given hy the dol pniduct of r and the unit
vector E, along kk. Thus.
0 =
r
-
Ek
= (.xi
+ ~f +
zkj
(/i
+
mj
+
nk)
(9. I 1)
Figure 9.16. Right triangle Sormcd by rand kk
Hence
11
= /.r
+
niy
+
IIZ
Using the Pythagorean theorem, we can now givc side b as
h? =
lr12
-
02
= (.r2 + y2 + .?1
-
(/Y
+ m2j2 +
Thc lerm sin2 $may next be givcn
ii2?
+
?hry
+
2lrzxz
+
2mnyz)
iis
Substituting hack into Eq. 9.10. wc get, on canceling kernis.
-(Pr’
+ m’j’
+iiZ;’
+ 2 / n ~ r ?+ 2/nrz + ~ m n y z j l p d i ,
Since I’ + in2 + ti2 = I , we can multiply the tirst bracketed exprcssion in
thc integral by this sum:
SECTION 9.5 TRANSFORMATION PROPERTIES OF THE INERTIA TERMS
397
Carrying out the multiplication and collecting terms, we get the relation
Refemng back to the definitions presented by Eqs. 9.1, we reach the desired
transformation equation:
We next put this in a more useful form of the kind you will see in later
courses in mechanics. Note first that 1 is the direction cosine between the
k axis and the x axis. It is common practice to identify this cosine as ab
instead of 1. Note that the subscripts identify the axes involved. Similarly, m
= ub and n = akz.We can now express Eq.9.13 in a form similar to a matrix
etc.
array as follows on noting that Izy =
6,
This format is easily written by first writing the matrix m a y of I‘s on the
right side and then inserting the a’s remembering to insert minus signs for
off-diagonal terms.
Let us next compute the product of inertia for a pair of mutually perpendicular axes, Ok and Oq, as shown in Fig. 9.17. The direction cosines of
Ok we shall take as I, m, and n, whereas the direction cosines of Oq we shall
take as ,‘l m’, and n’. Since the axes are at right angles to each other, we know
that
Ek.€
Y
k
=o
Y
Therefore,
II’
+ mm’ +
nn’ =
0
(9.15)
Noting that the coordinates of the mass element p dv along the axes Ok and
Oq are r * ek and r * eq,respectively, we have, for Ikq:
Using xyz components of r and the unit vectors, we have
=
jjj[(i+ yj + ~
V
k * )(1i + n ~ +‘ A)]
x
4
398
CHAPTER 9
MOMENTS .
\
i
W
PKODIK'TS OF I N L K I I A
Hcnce.
+ y n l ' + yrmn' + wi/' + -\.nni')p
+xdin' t .x:lii'
(11'
Collecting k i l n s and bringiiig ~ I i ciiirrctii)n cosines outside the inlegratiolis.
wc gel
Noting lhc definitions i n Eil. 9 I . we
li,, =
~
iiitlir/~r
~
,illr/
~
+
+
(Id
ciiii
statc Ihc desired Lransforinalion:
+
(/n1'
In/')/,,
d ! I ,+
~
+
(111li'
+
Ii?d!/),
(9.19)
We can now rewrilc the p r e \ i w s cquarion in ii more u\cIuI mil simple torin
using U ' Y iis direction cosine\. Thus. noting !hat I' = (I,,).ctc.. we proceed as
in Eq. [I. 14 to uhtain
SECTION 9.5 TRANSFORMATION PROPER'IlES OF THE INERTIA TERMS
399
EXaI'rIple 9.5
Find Iz,7, and I,,,, for the solid cylinder shown in Fig. 9.18. The reference
x'y'z' is found by rotating ahout they axis an amount 30", as shown in the
diagram. The mass of the cylinder is 100 kg.
It is simplest to first get the inertia tensor components for reference
xyz. Thus, using formulas from the inside front cover page we have
z
2
Ja = -IM r 2 = L(lOO)(!$)
2
2
=
21.13 kg-mz
-Ew
=
85.56 kg-m2
30"
Noting that the xyz coordinate planes are planes of symmetry, we can conclude that
l x z = IY l = I YZ
=o
Next, evaluate the direction cosines of the z' and the x' axes relative
to nyz. Thus,
For z' axis:
60" = ,500
cos
90" = 0
Z'Y
ai.? = cos 30" = ,866
a:,= =
a
COS
=
For .c' axis:
aXlr = COS
30" = ,866
a*Y = cos90° = 0
a~c,z
= cos 120" = -SO0
First, we employ Eq. 9.14 to get J,,,,.
= (85.56)(.500)z
+ (21.13)(.866)'
= '37.h
Finally, we employ Eq. 9.20 to get I,;,.
-Ix,,, = (85.56)(.500)(.866) + (21.13)(.866)(-,500)
Therefore,
x'
b l . 3 m ~ +
Figure 9.18. Find lz.,,and ldz,
400
CHAPTER 9
MOMWTS AND PROIlI!('TS OF INEXI'IA
SECTION 9.6 LOOKING AHEAD: TENSORS
401
You will leam that because of the common transformation law identifying
certain quantities as tensors, there will be extremely important common chardcteristics for these quantities which set them a p a t from other quantities. Thus, in
order to leam these common characteristics in an efficient way and to understand
them better, we become involved with tensors as an entity in the engineering sciences, physics. and applied mathematics. You will soon he confronted with the
stress and strain tensors in your courses in strength of materials.
To explore this point further, we have shown an infinitesimal rectangular parallelepiped extracted from a solid under load. On three orthogonal
faces we have shown nine force intensities (Le., forces per unit area). Those
with repeated indices are called normal stwsses while those with different
pairs of indices are called shear stresses. You will leam, that knowing nine
such stresses, you can readily find three stresses, one normal and two onhogonal shear stresses, on any interface at any orientation inside the rectangular
parallelepiped. To find such stresses on an interface knowing the stresses
shown in Fig. 9.19, we have the .same fran.~jbnnationeyuutions given by Eqs.
9.21 and 9.22. Thus stress is a second-ordijr tensor.
/
/
-.x
r1.r
Figure 9.19. Nine stresses o n three orthogonal interfxes at a point.
A two-ilimensionalsimplification of r!, involving the quantities T ~ T,,~., ,
and T~~ (= T,J as the only nonzero stresses is called plune .stress. This occurs
in a thin plate loaded in the plane of symmetry as shown in Fig. 9.20. Plane
stress is the direct analog of second moments and products ofurea, which is
a two-dimensional simplification of the inertia tensor. Clearly, plane stress
and second inoments and products of area have the same transformation
5,. 5,. T~;.replacing
equations, which arc Eqs. 8.26 through 8.28 with
I<,,
I+ -IrY,
- - I r ,respectively.
,
In solid mechanics, you will also learn that there are nine terms sjj that
describe deformation at a point. Thus consider the undeformed an infinitesimal
~~
Figure 9.20. The case of plane stress.
402
CHAPTER Y
MOMENTS AND PRODUC'IS OF INERTIA
rectangular parellelepipcd iii Fig. %2l. Whcn there is it dcforniation there are
namml .struiri.s q,, t~.along thr direction 01 the darkened edges which give
the change7 of length per unit original length o i these edpcs. Furthennore, when
there is a dcfi~rmation.therc arc six slwrrr . s ~ , r r b r , s e t $ = E , , , E,: = e, tyzI .
di.
Figure Y.21. An iniinilesimal rectangular
parnllclrpiped with three cdgrh highlighted
~
t:, that give the ch;inge i n angle i n radialis i r m that of the right angles 1)f the
three darkened edges. Knowing these quantities. we can find any othcr strains in
thc rectangular parallelepiped. Thesc other strains ciin be found by using transli)rmation Eqs. 9.2 I atid 9.22 arid so v t n i i r i is illso a .sec~~rzd-orrkr
teiisor.
The two-diniensional simplification 0 1 t involving thc quantities, e l , ,
arid
( = e \ , ) a s the only iii)iizero strains is calledplaiie slrrriir and rcpresents the strains i n a prismatic body constraincd at the ends with loading
noriixil to the ccntcrhne in which the lodding docs not vary with :(see Fig.
4.22). Also, thc prismatic body must not be suhjcct ti) bending. Plane strain is
an analogous mathematically t o plane stress and sccond moments and products of area. 411 three are two-dimensional simplifications of second-ordcr
sylrlnielric tcnsors and have the .\(1111? t r ~ ~ ~ i . ~ : ~ ~r ry ~u r~i rt ir~a~~n21s
i. ~~well
: r ~ as
other matheniatical properties. Finally. in elcctroniagnetic theory and nuclear
physics. you will be inti-oduced ti) the quadruple tensor.'
In thr ,fi,llowing prohlenis. use rhe lbrmulus ,fin’ momen1.s unci
produluctr of inertia ut rhr mo.s,s cenler 10 be ,found in the inside
front cover pugr.
9.20. What are the moments and products o f inertia for the XJZ
and x’y’i’ axes for the cylinder?
x
Figure P.9.22.
Figure P.Y.20.
9.21. For the uniform block, compute the inertia tensor at the
center of mass, at point u, and at point h far axes parallel to the x y i
reference. Take the mass of the body as M kg.
9.23. A thin plate weighing 100 N has the following mass
moments of inertia at mass center 0:
I,, = I5 kg-m2
I\?= 13 kg-m*
i
I,,, = -10 kg-m2
What are the moments of inertia lx,r,,I v,?, and I,,i, at point
P having the position vector:
r = .5i + .2j
+
.hk m
Also determine I,,:, at P.
Y’
?
I
X
1
Figure P.9.21.
9.22. Determine l t r + l v , + 1:: a i a function of x, y, and 7 for
all Doints in mace far the uniform rectangular parallelepiped. Note
that I?: has its origin at the center of mass and is parallel to the
sides.
Figure P.Y.23.
40
A crate with its contents weighs 20 k N and has its center
of ,,\a'.i Ilt
9.24.
r, = 1 . 3
+
?j
+ .Xk in
9.26. A block having a uniform d e n i i y of 5 frmlsicm' ha\ a
hole of diamctcr 41) ntm cut out. Whilt arc the principal m c m e ~ ~ t s
US iiicitia a1 p<lintA ill the ccntroid of thr right lace of the block?
40 m
-e+--
100
,"",
--t
Figure P.Y.26.
/
9.27. Find niilxiiiiuni and minimum miiniciits 01 inertia at point
A. The hlnck wrighc 20 N and thc ccmr weighs 14 N.
8
J 11,
Kigure P.Y.24.
9.25 A cylindrical crate and its cmtents wzigh 500 N. 'The cellter V I IIUS i s at
rc
It
= .hi
+
.7j
+
15
I
2k m
i\ kiiown that at A
Figure P.Y.27.
(I,,),, = X S hg-ni'
( I , ,),,, = -22 kg-m'
Find I s , and I , \ 211 I<.
Figure P.9.25.
9.28. Solid sphcl-cs C arid I ) each weighing 25 N and having
radius 01 S O inm are attilched to a thin solid rod wcighing X1 N.
Also. solid sphcrcs E and G each weighing 20 N and having radii
of 30 rmn are attached to a thin rod weighing 20 N. The rods iiic
attachctl to he unhugunal to each other. What air thc principal
nwmeiits of inertia at point A ?
Figure P.Y.28.
9.29. A cylinder is shown having a conical cavity oriented along
the axis A-A and a cylindrical cavity onented normal to A-A. If
the density of the material is 7,200 kg/m3, what is lap?
A-
-A
Figure P.9.29.
9.30. A flywheel is made of steel having a specific weight of
490 Ih/ft.3 What is the moment of inertia about its geometric axis?
What is the radius of gyration?
Figure P.9.32.
9.33. A disc A is mounted on a shaft such that its normal is oriented IO' from the centerline of the shaft. The disc has a diameter
of 2 ft, is 1 in. in thickness, and weighs 100 Ib. Compute the
moment of inertia of the disc about the centerline of the Shaft.
Figure P.9.33.
Figure P.9.30.
9.34. A gear B having a mass of 25 kg rotates abuut axis C-C. If
the rod A has a mass distribution of 7.5 kgim, cumpute the
moment of inertia ofA and B about the axis C-C.
9.31. Compute Ivy and
for the right i ular cylinder, I
has a mass of 50 kg. and the square rod, which has a mass
kg, when the two are joined together so that the rod is radial I
cylinder. The x axis lies along the bottom of the square rod.
t
C
mss-9ection
5 mm X 25mm
X
I
Figure P.9.31.
9.32. Compute the moments and products of inertia for the
axes. The specific weight is 490 Ib/ft' throughout.
*y
C
Figure P.9.34.
405
Y.35. A hlock wcighing 100 N ih h w r i . Cimipolt. the tmimieiit
, A inertia ahoiit thc diagimiil 11-I).
9.38. A bent rod weighs . I Nlmm. What i s I,,,, lor
c,, = .iOi
+
.4Sj + .X4lk?
Figure P.9.35.
Figure P.9.38.
1.36. A solid qphere A of diameter I ft and weight 1110 Ih i i C O ~ I iccted to the shaft B-B hy a \ d i d rod weighing 2 Ihllt and having
9.39. Evaluate the n u t r i x of direction cmine?, for the primcd
I iliamclei~0 1 I in. Computc I.... for the rod and hall.
axcs rclativc to the unprimcd axrs
.L?'
v 20"
A
B
-
/
$4YY
.
l.37. In Prohlem 9 13. Figure
wc found
P.9.36.
the fdlowiny r r w l t s lor Ihr
'iin platc:
I , l = I , , = .IOl')grams-m~
I , , = ,1164') gram<-rn'
'ind a11 cmponcnts lor I h r incrtia tensor for relerence x'J':'.
LXCS ~s'y'l i e in Ihr midplane ot the plate.
~~
I'
j(P
.1
Figure P.Y.39.
9.40. The hlock i s uniform 111density and weighs ION. Find I,,:,
4
.-
Figure P.9.37.
Figure P.Y.40.
A thin rod of length 300 mm and weight 12 N is oriented
relative 10 x'v'z' such that
*9.41.
c,! = .4i'
+
.3j'
+
.X66k'
Show that the transformation equation for the inertia tencomponents at a point when there is a rotatiun of axes &e.,
Eqs. 9.14 and 9.20) can be given as follows:
V.42.
SOT
where k can be x', y', or z' and y can be x'. y', or 7'. and where i
and j go from .r to y to z. The equation above is a compact definition of s ~ n n d - o r d e rt m m r s . Remember that in the inertia tensor
you must have a minus sign in front of each product of inertia
term (i.e., -Irv, -Iv2, etc.). [Hint: Let i = 5 ; then sum overj; then
let i = y and sum again o w j ; etc.]
/ 1
x'
Figure P.Y.41.
*9.43. In Problem 9.42, express the transformation equation to
get I+:, in terms of the inertia tensor components for reference xyz
having the same origin as x's'z'.
k
"9.7
The Inertia Ellipsoid and Principal
Moments of Inertia
Equation 9.14 gives the moment of inertia of a body about an axis k in terms
of the direction cosines of that axis measured from an orthogonal reference
with an origin 0 on the axis, and in terms of six independent inertia quantities
for this reference. We wish to explore the nature of the variation of Ikk at a
point 0 in space as the direction of k is changed. (The k axis and the body are
shown in Fig. 9.23, which we shall call the physical diagram.) To do this, we
will employ a geometric representation of moment of inertia at a point that is
developed in the following manner. Along the axis k , we lay off as a distance
the quantity OA given by the relation
where d is an abitraly constant that has a dimension of length that will render
OA dimensionless, as the reader can verify. The term d<ilM is the radius
of gyration and was presented earlier. To avoid confusion, this operation is
shown in another diagram, called the inertia diagram (Fig. 9.24), where the
new 5, rl. and <axes are parallel to the x, y , and z axes of the physical diagram. Considering all possible directions of k, we observe that some surface
will be formed about the point O', and this surface is related to the shape of
the body through Eq. 9.14. We can express the equation of this surfdce quite
Y
Figure 9.23. Physical diagram.
6'
f
Figure 9.24. Inertia diagram.
407
N o w replace ttic d i r c c l i o i i cosine\ i n 141. 9.13. using the relations above:
S EC TION ‘3.7 TH E INERTIA ELLIPSOID AN D PRINCIPAL MOMENTS O F INERTIA
therefore imagine that the XK reference (and hence the {qcreference) can be
chosen tu have directions that coincide with the aforementioned symmetric
axes, 0‘1, 0 ’ 2 , and 0 ’ 3 . If we call such references x’y’z’ and
respectively, we know from analytic geometry that Eq. 9.27 becomes
t’q’c,
c
where <’,q’, and
are coordinates of thc ellipsoidal wrtace relative to the
are mass moments of inertia of the body
new reference, and fx,,z., I,.,,, and IZ.?,
about the new axes. We can now draw several important conclusions from
this geometrical construction and the accompanying equations. One of the
symmetrical axes of the ellipsoid above is the longest distance from the origin
to the surface of the ellipsoid, and another axis is the smallest distance from
the origin to the ellipsoidal surface. Examining the definition in Eq. 9.24, we
must conclude that the minimum moment of inertia for the point 0 must correspond to the axis having the maximum length, and the maximum moment
of inertia must correspond to the axis having the minimum length. The third
axis has an intermediate value that makes the sum of the moment of inertia
terms equal to the sum of the moment of inertia terms for all orthogonal axes
at point 0,in accordance with Eq. 9.2. In addition, Eq. 9.28 leads us to conclude that Ix,?, = fV,:, = /,., = 0 . That is, the products of inertia of the mass
about these axes must be zero. Clearly, these axes are the principal uxrs of
inertia at the point 0.
Sincc the preceding operations could be carried o u t at any point in spdce
for the body, we cm conclude that:
AI each point there is a set of principal axes having the extreme values of
moments of inertia for that point a n d having zero products of inertia.’ The
orientation of these axes will vary continuously j%m point to poinr
throughout space for the given body.
All symmetric second-order tensor quantities have the properties discussed above for the inertia tensor. By transforming from the original refercnce to the principal reference, we change the inertia tensor representation
from
In mathemalical parlance, we have “diagonalized” the tensor by the preceding operations.
“A gcncriil prucedure fur coinpuling principal ~~loments
of inertia is set forth in Appendix !I.
409
dl 0
CHAPTER 9
MOMENTS ANLJ PKODtICTS OF INFRTIA
9.8
closure
I n this chapter. we first introduccd the ninc cnnipiincnts comprising the incrlia tensnr. Ncxt. we ciiniidered thc ciise o f llic very thin flat plale i n which llie
.qaxes form the 1nidp1;iiie of the plale. We Ioond that the m a s s n i m i e n t s and
prodocts of inertia terms (I,),,,
Ibi- tlie plate are pmpnrand
tional, rcspcctively. t o ( I , < ) , . ( I > , I,.,. and (/,,I,,. tlie iecond nioinents and product [ifarea 111 the plate surface. As a result. we could set forth the concept of
principal axes for the incrtiii tensor a s an extcnsion of thc work i n Chapter X.
Thus, wc pointed nut that for thesc axes tlic prnducts o f inertia w i l l he zerii.
Furtherinnre. one principal a s k corresponds lo the iiiaximuni moment 01
inertia at the point while another of the pi-incipal axes corrcspoiids to the ininimum moment of incrtia at the point. We pointed out that for hodies with two
nrthogm;il planes of symmetry. llie principal axes al any point o n the line (if
intcrsectioii of the planes of symmetry inust hc alnng this line nf inter~ectiiiii
and nnrinal lo this liiic in the p1;ines of symmetry.
Those readers who studied the starred sections from Section 9.5 onward
w i l l have Ioond proofs 111 the extensinns set lorth earlier ahiiut principal axes
frnm Chaptcr 8. E w n niore imporlant i s the disclnsure that the incrtia tensor
components change their \'slues when the axes are rotated at a point i n
exactly the same way a s inany cithcr physical qiiaiititirs having ninc cornponents. Such quantities ;ire called second-order teninrs. Because OS the c ~ r ninn transformation equation for such quantities, thcy h;we many imporlant
identical properties. such a s principal axcs. In your cnurse i n strength O S
inalerials you shnold learn that stress and strain arc second-order tensors iind
hence h a w principal axes.'" Additionally. y < u w i l l find that a two-dimens i n i i a l stress distributim called I h n e sIres.s i s d a t e d to the stress tensor
csactly a s thc moments and products iif area are related to the incilia tensnr.
The sanie situation exists with strain. Consequcntly. there are similar inntheinaticiil formulatiiins for plane stress and the corrcsponding
(plane itrain). Thus. hy tahing the extra time to coiisider llie mathcniatical
considerations n l Sections 0 3 thi-ough 9.7. yciu w i l l find unity hetween Chnpler 9 and some vcry important aspects of strength 01. materials to he studied
later i n your program.
I n Chapter I O , we shall introducr another apprnxh tn studying eqiiilihriiini heyoiid what we liavc used thus far. This ;ipprn;ich i s valuahlc lor certain importanl classes 01 sltdtics priihlenir and ;it the same time fni-ins the
grnundworh for ii nuinher of advanced lechnique? that many students will
study later i n theii- programs.
9.44. Find for the body of revolution having uniform density
of .2 kg/mm?. The radial distance Out from the Iaxis to the surface is given as
9.47. In Problem 9.46, what are the principal axes and the prin-
where z is in millimeters. [Hint:Make use of the formula for the
moment of inertia about the axis of a disc, M?.]
9.48. What are the principal mass moments of inertia at point O?
7
cipal moments of inertia for the inertia tensor at O!
Block A weighs 15 N. Rod B weighs 6 N and solid sphere C
weighs I O N. The density in each body is uniform. The diameter
of the sphere is 50 mm.
5
x
Figure P.9.48.
Figure P.9.44.
9.45. In Problem 9.44, determine 1,: without using the disc formula but using multiple integration instead.
9.49. The block has a density of 15 k&.?
Find the moment
01
inertia about anis AB.
9.46. What are the inertia tensor components for the thin plate
about axes xyz? The plate weighs 2 N.
E
E
100 mm
50 mm
5
-X
/
I
/O
Figure P.9.46.
Figure P.9.49.
41
9.50. A crate and its contents weigh.: IO kN. The center o f mass
9.52. Find ivr
and ivr.
The diameler of A is 0.3 m. R is thc centel
0 1 thc crate and its contents is at
of the right Cam of the block. Take p = p,, kglm.’
r, = .40i
+
.3Oj
+
.6Ok m
If at A we knuw that
lv,, = 800 kg-m’
1,. = 500 kg-m?
find /?, and 1,: at 8.
Figure P.9.52.
4
9.53. A body is composed of two adjoining hlocks. Both blocks
have a uniform mass density p equal tn I O kg/m.’
la) Find the mass moments of inertia, I < %and lzz
(b) Find the product of inertia iL3.
IC) Is thc product of inertia 1,: = 0 (yes or no)’? Why’!
x
Figure Y.9.50.
1.51. A semicylinder weighs SO N. What are the principal
noments of inertia at o? What is the prnduct of inertia I,,,::? What
onclusirm can you draw about the direction of principal axes at O ?
-’
,/x’
I
-I.-
206 m;n
/+300mm
Figure P.9.51.
12
f
Figure P.9.53.
"Methods of Virtual Work
and Stationary
Potential Energy
10.1 Introduction
In the study of statics thus far, we have followed the procedure of isolating a
body to expose certain unknown forces and then formulating either scalar or
vector equations of equilibrium that include all the forces acting on the body.
At this time, alternative methods of expressing conditions of equilibrium,
called the method of virtual work and, derived from it, the method of sfationary pofential energy, will be presented. These methods will yield equilibrium
equations equivalent to those of preceding sections. Furthermore, these new
equations include only certain forces on a body, and accordingly in some
problems will provide a more simple means of solving for desired unknowns.
Actually, we are making a very modest beginning into a vast field of
endeavor called variational mechanics or energy methods with important
applications to both rigid-body and deformable-body solid mechanics.
Indeed, more advanced studies in these fields will surely center around these
methods. I
A central concept for energy methods is the work of a force. A differentia1 amount of work d d k due to a force F acting on a particle equals the
component of this force in the direction of movement of the particle times the
differential displacement of the particle:
dWk = F - d r
(10.1)
Wi
And the work
on a panicle by force F when the particle moves along
some path (see Fig. 10.1) from point 1 to point 2 is then
'i"lIk =
1;, F * dr
/?
Figure 10.1. Path ofparticle on
which F does work.
( I 0.2)
'For a treatment of energy methods for deformable solids. see 1. H. Shames and C. Dym.
Energy and Finite E l m e m Method?in Structural Mechanim Taylor and Francis Publishers. 1955.
4
Note that the value and direction o f F can vary along tlie path. This fact iiiusl
he taken into account during the intcgratiiin. We shall have inure to say ahoul
the concept of work i n liitcr sections.’
Part A:
10.2
,
1’
Figure 10.2. Paltick (VI
ii
hictionles\ surface
Method of Virtual Work
Principle of Virtual Work
for a Particle
For our intrixluction to the principle of virtual work, we will first consider a particle actcd on hy cxternal loads K,.K,,..., K,,, whose rcsiiltant f i x e pushcs the
particlc ;igainsi ii rigid constraining surface S in space (Fig. ((1.2).This sur(ace S
is assuincd Iu he frictionless and will thus exert a constraining forceNon the parLick which i s iioriniil to .F. The forccs K, arc called r,ctiwfi,n~e5 in cciniicctioii
with the method 01‘ virtual wnrk. while N retiiiiih llie identilication 01 a rnnvrruinirrfi .fiirw a\ iised prcviiiusly. Employing the resultiint actiw forcc K,. we ciin
give the neceswy and sufticicnt’ conditions lor cquilihriuin ioi- tlie particle :is
K, + A’
=
0
(10.3)
We shall iiiiw prove that we caii expre\\ [he iiccescary and sufficient
ciinditions 01 equilibrium i n yet another way. Let us iiniiginc that we givc the
particle an infinilesimiil tiypothctical arhitrary displaceincnt that i s consistent
n i t h the coiihtraints (i.c., along the surlace). while keeping the forces K, and
“Vonstant. Such a displacement i s termed ii i.irtunl ‘i;,~,,l‘,~,~,n,~,,~t,
and w i l l he
denoted by &, i n conlmst to a rciil infiniicsiinal displaceincnt. dr, which
might iictudly iiccur during a lime intcrviil ilr. We can Illen take the dot priiduct oi the vector Sr with the Ibrce wxtors i n thc equation above:
K, * Sr + N * Sr
=
0
(10.4)
Sincc N is normal 10 thc surface and Sr is tangential to the surface. the coriespunding scalar product must he 7.ero. leaving
K, * Sr
=
0
(I0.5)
The uxpres\ion K,<.Sr i s called (tic virniol i n d 01 the \y\tcrn (if tbrces and i s
denoted as 8Tt’v,,~,.Thu\. thc \:iriuiil wnrk by the active forces oil a plirticlc
SECTION 10.3 PRINCIPLE OF VIRTUAL WORK FOR RIGID BODIES
415
with frictionless constraints is necessarily zero for a particle in equilibrium
for any virtual displacement consistent with the frictionless constraints.
We shall now show that this statement is also sufficient to ensure equilibrium for the case of a particle initially at rest (relative to an inertial reference) at the time of application of the active loads. To demonstrate this,
assume that Eq. 10.5 holds but that the particle is nor in equilibrium, If the
particle is not in equilibrium, it must move in a direction that corresponds to
the direction of the resultant of all forces acting on the particle. Consider that
d r represents the initial displacement during the time interval dt. The work
done by the forces must exceed zero for this movement. Since the normal
force N cannot do work for this displacement,
K,.dr>O
(10.6)
However, we can choose a virtual displacement 6r to be used in Eq. 10.5 that
i s exactly equal to the proposed dr stated above, and so we see that, by admitting nonequilibrium, we amve at a result (10.6) that is in contradiction to the
starting known condition (Eq. 10.5). We can then conclude that the conjecture that the particle is not in equilibrium is false. Thus, Eq. 10.3 is not only a
necessary condition of equilibrium, but, for an initially stationary particle, is
in itself sufficient for equilibrium. Thus, Eq. 10.5 is completely equivalent to
the equation of equilibrium, 10.3.
We can now state the principle of virtual work for a panicle.
The case of a particle that is not constrained is a special case of the situation
discussed above. Here N = 0, so that Eq. 10.5 is applicable for all infinitesimal displacements as a criterion for equilibrium.
10.3
Principle of Virtual Work for
Rigid Bodies
We now examine a rigid body in equilibrium acted on by active forces K , and
constrained without the aid of friction (Fig. 10.3). The constraining forces Ni
arise from direct contact with other immovable bodies (in which case the con4This test breaks down for il panirk thar is moving. Consider a panicle constrained 10
move in a circular path in a horirontal plane, as shown in the diagram. The particle is moving
with constant speed. '%her
no active furces. and we consider the constraints as frictionless.
rnent cumistent with the constraints at any time I gives us a m o
The work for a virtual dis
result. Nevenheless, the p
i s not tn equilibrium, since clearly there is at time I an accelera~
tion toward the center of curvature. Thus. we had to restrict the sufficiency condition lo panicles
that are initially stationilry.
~i~~~ 10.3, ~ i ~body
, d
and ideal constraining forces.
forces
straining forces iirr oricnted inormil to Ihc c o n t ~ i csur1iucl
l
01- irom conkict wilh
iniinovahlc bodies thniugh pin and ball-joint ciinnections. We shall considelthe body to he made up o f elementar) particles for the purposes o i discusion
Now consider ii particle of inass m,. Actibc loads. cxterniil coiislraining
iorccs. ;ind forces iron1 other piiiticlcs i i u y possibly he acling on the pal-tick
The forces from uther particles :ire internal forces S,which maintain h c rigidity ofthc body. Using tlic rcsultiinls ot'lhcsc !<ari(ius forces on thc particle. wc
m i y slate iron1 Nen'tiin's liiw that ttic neceswy arid sufficient5 condition 1111equilihi-ium of the ith parriclc i s
(lO.7)
I K , ) , 4. I N , ) , + ( S i < ) ,= 0
Nov.,. u'e give the pal-tick ii virtual displ;iccnient fir, that i s consistent
with the exlcrior coiislriiitit~and with (he condition that thc body i s rigid.
Taking lie dot priiiluct 01 the vectors in the cquatiiin ahovc wilh 6ri. we gcl
(K,);Sr,
t (N,);Sr;
+(SI,), -6r, = I )
1IO.Xl
Cleai-ly,(IV,), * 6r, tiiust he zero. lheciiiise Sr, i s normal ior N j ior constraint
stemming from direct contact with itnniovahlc bodies or hecausc Sr, = 0 for
conslraint stcmniing from pin and Ixill-j~iiiitcoiiiieciioiis with irnmovirtilc
bodies. l.et us then sun1 the resulting cquarions of rhc iol-m 10.8 for all tlic
particles that are considel-cd to mahc up the body. Wc Ixive. for 8 1 panicla.
1x1 us now consider i n niorr detail the internal forces i n order to sho\v
that the second qunnlity on the left-h;ind side of the eqoalioli aho\e i s LCIO.
The force on m, from pxticle iii, w i l l be equal ;ind iippositc IO ~ h cforce (in
liarticle i i i from parliclc iii,.according to Ncwton's third liiw. The interniil
I
iorces on Lhesc particles :Ire sliiium as S and S,,i n Fig. 111.4. The first s u b
script identifies the particle on which ii force acts. while the second subscript
idcntifies the particle exerting this fiircr. Wc ciiii tlirii say that
s
=
-s
( I O . IO)
Any virtual motion wc give t u :my pair of particles m u s t niiiiiitnin a colistiuit
dihliince between the particle\. Thih r e q i i i ~ ~ n i c sitetiis
nl
ironi Ihc rigid-hody
condition and will he truc if:
I. Both parliclcc are given the siiiiie di\placcment SH.
2. The particles ill-e rulatcd 6@relative to eiicli other.''
We liow consider the general ciise where both motions Lire prcscnt: that i\.
hoth I N , untl I W , arc given ii \irliial displaceincnt SH,a n d lurthel-nirirc, HI, i s
' 7 h c d l i c i e n q I~C~IIICIIICIII
iagiiiii ; ~ t > p li~o \an irr~i~atty
hla~ionaryp~ritclc.
"Thc viirlual ~ t i ~ ~ d ~ (Sr,
~ c,)Ic e;wh
~ n 01c ~ill?~ iwt)
~ ~ P:IIIICIC\r m i Ihcn
~
hu Ihc ICIUIL
< u l x r p w i i , m 01 6R iimt bo?.
,,i 11,c
SECTION 10.3 PRINCIPLE OF VIRTUAL WORK FOR RIGID BODIES
rotated through some angle 64 about inj (Fig. 10.4). The work done during
the rotation must he zero, since Sj, is at right angles to the motion of the mass
m.. Also, the work done on each particle during the equal displacement of
I
both masses must he equal in value and opposite in sign since the forces
move through equal displacements and are themselves equal and opposite.
The mutual effect of all particles of the body is of the type described. Thus,
we can conclude that the internal work done for a rigid body during a virtual
displacement is zero. Hence, a nrcessais condition for- equilibrium is
2 (K,),
* 6r, = 6~,~;,,=
o
;,nd
SI, =
Y-
(10.11)
Thus, the virtuol w r k done by nctiveforcrs on a rigii body having frictionless constraints daring virtual iii.spluccments consistent with rhr consfrainfs
is zero ifthe body is in equilibrium.
We can readily prove that Eq. 10.I I is a silfficienr condition for equilibrium of an initially stationary body by reasoning i n the same manner that
we did in the case of the single particle. We shall state,first that E4. 10.11
is validfor a body. If the body is not in equilibrium, it must begin to move.
Let us say that each particle ini moves a distance dr, consistent with the constraints under the action of the forces. The work done on particle mi is
-
-
drj z 0
(10.12)
-
But (A',), * drj is necessarily zero because of the nature of the constraints.
dr,
When we sum the terms in the equation above for all particles, c(SH)j
must also be zero because of the condition of rigidity of the body as previously explained. Theresore, we may state that the supposition of no cquilibrium leads to the following inequality:
i ( K , ) ; 'dr, > 0
(10.13)
,=I
But we can conceive a virtual displacement 6r, equal to dr,for each particle to
he used in E q . IO. I I , thus bringing us to a contradiction between this equation
and Eq. 10.13. Since we have taken Eq. 10.11 to apply, we conclude that the
supposition of nonequilihrium which led to Eq. 10.13 must be invalid, and so
the body must he in equilibrium. This logic proves the sufficiency condition
for the principle ol virtual work in the case of a rigid body with ideal constraints that is initially stationary at the time of application of the active forces.
Consider now severiil movable rigid hodies that are interconnected
by smooth pins and hall joints or that are in direct frictionless contact with each
other (Fig. 10.5). Some of these bodies are also ideally constrained by irnmow
able rigid bodies in the manner described above. Again, we may examine the
system of particles mi making up the various rigid bodies. The only new kind of
force to he considered is a force at the connecting point between bodies. The
force on one such particle on body A will he equal and opposite to the force on
the corresponding particle in body B at the contact point; and so on. Since such
-s,,
$8
Figure 10.4. Two panicles of a body
undergoing displacement 8R and rotation S+.
,=I
(K,II dr, + i N J j * rfri+ (S,);
417
418
CHAPTER 10 METHOIIS OF VIRTUAL WOKK AND STATIONARY POI'ENTIAI. ENERG\
Figure 10.5. System 0 1 ideally co~islrilinetlrigid hodie\ acted on hy forces K ,
pairs ut' contiguous particles have the same virtual displacement, clearly the virtual work al all connecting points hetween hodies i s zero for any virtual displacement [if
the system cnnsisteiit with the cwistrainls. Hence, using thc same
rmsoning ils heforc. we can xiy,(hr- (1
!em o / iniliall? s m r i o n r i n : rigid bodies.
the n r ~ ~ r . s s(2nd
r i ~ .rriffi&nf i.onrliriorr offtquilihiiuin is thor the virtrrul work o/
tlw uctivr f i i r w s be :era f i w t i l l po,vsihle i,irtud displuwmntrs (.onsi.slrirr M , i r h
rlir con.s!ri~int,s.We may then use the fiillowirig equation instead [if equilihriuin:
.
(K,), Sr,
= 6'71 '\,T,
= 1)
(II). 14)
where ( K N ) ;arc the iicti\,c forces [in the system of rigid bodies and Sr, are the
movements of the point of application n1 these forces during a virtual displacement of the system consistent with the constraints.
10.4
Degrees of Freedom
and the Solution of Problems
We have developed equations sufficient lor equilihriuni 01 initially stationary
I
Figure 10.6. Plene pciiduluin.
systems i i l hiidies by using the concept of \'irtual work for \'irtual displacements consistent with the constraints. These equations do lint involve rclactions or connecting hrces. arid when these forces are lint of interest, the
method i s quite useful. Thus, we may snlve for as inany unknown a<.tiw
forces as there arc indqwmkwr cquations stemming from \,irtuaI displacenients. Thcn our prime interest is to know hnw many independent equations
can he written lor a system stemming from virtual displacements.
For this purpose. we definc iIw ruitiihcr ofilq,ur.s ,!/rrr(4,,in o / o
tein l i s tlw itumh(~ro / g~~nenrli:rdc o ~ ~ r d i t t i i r ~n ~k .i ~
d '! i.s required 10 full!
.sprcIfi rhr, wnfijiurotion r$tlre .sy.rtmt. Thus. tni- the pcnduluin i n Fig. 10.6.
which i s restricted to i n w e in ii plane. one iriilqienrlenr cnrirdinatc 0 locates
the pendulum. Hence, this system has hut w e degree of lreedom. We m a y
ask: Can't we specify i and y nfthe hob. and thus aren't there two degrces or
' G m c m l r i ~ dcoonlirratrm are a y \et of indqi"mi<,rii v r i r i d h ~ihal can i u l y q i z c i i y the
coniigsralicr " f a aysiein. Gcnrrnlizcd cmrdioatc* cm iriclutk a n y 01 lhc usual c o o r d i n u x such
:is C a r l w i x C O i ) l t l i n ~ t e b or cylindricvl coordin;~le~.
huI iiecd no^. We shall only cunuder l h o w
CIISEL wlierc lhc iiiiiiil coi,rdmalec SZPC its ~ h r
~ e n c i m i r s dc o o r ~ l i ~ i i i i c s .
SECTION 104 DECR!z.ES OF FREEDOM AND THE SOLUTION 01. PROBLEMS
freedom? The answer is no, because when we specify x or y, the other coordinate is determined since the pendulum support, being inextensihle, must
sweep out a known circle as shown in the diagram. In Fig. 10.7, the piston
and crank arrangement, the four-bar linkage,8 and the balance require only
one coordinate and thus have hut one degree of freedom. On the other hand,
the double pendulum has two degrees of freedom and a particle in space has
three degrees of freedom. The number of degrees of freedom may usually be
readily determined by inspection.
Since each degree of freedom represents an independent coordinate. we
can, for an n-degree-of-freedom system, institute n unique virtual displacements by varying each coordinate separately. This procedure will then given
independent equations of equilibrium from which n unknowns related to the
active forces can he determined. We shall examine several prohlcms to illustrate the method of virtual work and its advantages.
Before considering the examples, we wish to point out that a torque M
undergoing a virtual displacement 6@in radians does an amount of virtual
equal to
work
6wk
-
6<M/,= M 64
(10.15)
The proof of this is asked for in Problem 10.30
L
Four-har
linkage
Pislun and
crank
arrangement
Balance
MULTI DECREE-OF-FREEDOM SYSTEMS
Figure 10.7. Various systems illustrating degrees of freedom.
*The fourth har
ih
the hare.
419
420
CHAPTER 10 METHODS OF VIRTUAL WORK AND STATIONARY POTENTIAL FNEKGY
Example 10.1
A device for compressing metal scrap (a compaclor) is shown in Fig. 10.8.
A horizontal force P is exerted on joint R . The piston at C then comprcsses
the scrap material. For a given force P and a given angle 8, what is the h r c e
F developed on the scrap by the piston C?Neglect the friction hetween the
piston and the cylinder wall, and consider the pin joints to he ideal.
We see by inspection that one coordinate 8 describes the configuration of the system. The device therefore has one degree of freedom. We
shall neglect the weight of the members, and so only two active forces are
present, P and F. By assuming a virtual displaccment 60, we will involve
in the principle of virtual work only those quantities that arc of interest to
us, P. F, and 8.' Let u s then compute the virtual work of the activc forces.
Force P. The virtual displacement 68 is such that force P has a motion in
the horizontal direction of ( 1 68 cos 8) as can readily he deduced from Fig.
10.9 by elementary trigonometric considerations. There is yct another way
of deducing this horizontal motion, which, sometimes, is more desirable.
Using an x)i coordinate system at A as shown in Figs. 10.8 and 10.9, wc
can say Cor the position of joint B:
)in
= 1sin0
(a)
I
,
!
Figure 10.8. Cantpilcting devicc.
"It we had used a free~hodyapproach.wc would h a w had to bring in force component\
at A and at C , and we would have had lo dirmemher thc iysten,. To appreciate the mclhod nf
v i m a l work even for this simple prohlcm. we urge you t o nt least set up the prohlcm hy the
use ot free-body diagrams.
Figure 10.Y. Virtual m(ivement ni
leg AB.
SECTION 111.4 DEGREE OF FREEDOM AND THE SOLUTION OF PROBLEMS
Example 10.1 (Continued)
Now take the differential of both sides of the equation to get
dya = 1 cos 0 d8
(b)
A differential of a quantity A , namely dA, is very similar to a variation of
the quantity, 6A.The former might actually take place in a process; the latter takes place in the mind of the engineer. Nevertheless, the relation
between differential quantities should be the same as the relation between
varied quantities. Accordingly, from Eq. (b), we can say:
Sy, = 1 cos 8 68
(C)
Note that the same horizontal movement of B for Seis thus computed as at
the outset using trigonometry.
For the variation GOchosen, the force P acts in the opposite direction
of Sy,, and so the virtual work done by force P i s negative. Thus, we have
G('7dvin)p
= -PI COS 8 60
(d)
Force F. We can use the differential approach to get the virtual displace.
ment of piston C. That is,
xc = 1 c o s 8 + i c o s e = 2 1 ~ 0 s ~
dx, = -21 sin 0 d 0
Therefore,
SX,
= -21 sin 8 60
(e)
Since the force F is in the same direction as SX,, we should have a positive
result for the work done by force F. Accordingly, we have
6(wvifi)F = F(21 sin 0 )68
(f)
We may now employ the principle of virtual work, which is sufficient here
for ensuring equilibrium. Thus, we can say that
-PI cos 0 60 + F(21 sin 8 )60 = 0
(g)
Canceling 1 60 and solving for F, we get
For any given values of P and 8, we now know the amount of compressive
force that the compactor can develop.
42 1
422
CMhPTER I l l
MLIIHODS OF VIRTUAL W O R K ANI) STATIONARY POTENTIAL ENERGY
r Example 10.2
I
A hydraulic-lift plallurin for loading trucks is shown i n Fig. 10. IO(n). Only
one side ofthc system is shown: Ihc other side is identical. If thc diameter of
the piston in the hydraulic rani is 3 in., what pressurcp is nceded to support
ii load W OS S.000 Ib when 8 = 60"? The lollowing additional data apply:
I = 24 in.
(1 = 60 in.
<' = 10 in.
Pin A is at Lhc center OS the rod.
W c have hcre n system with om d e p e of fi-redum characterized hy
the angle 8. The iictive forces that do work during a virtual displacement
60 are the weight W a n d thc l i m e from the hydraulic rani. Accordingly.
the virtual rnovenicnts of both the platform E and ,joint A of the hydraulic
ram mu?t he found. Using referencr ~x,:
xp, = ? I sin 8
Therefore.
a?,:
=
2 l c o s 0 SB
(ill
For thc ram forcc, we want tlic nioverncnt of pin A i n the direction ofthe
axis 01 the pump. namely SI] where is shown i n Fig. IO. IO(a). Ohserving
Fig. 10. IO(h) we cm say Sor r):
~
A("+ CH?
= ( I sin 8
-
PI' + ((1 - I c o u 8 1 ~
!h)
Hence. we ha\e
2 q 67
= 2iIsin 8
-
eI(lcos6')SO
+ 2id - /cos@)(/sin0 ) S 8
(c)
Solving for 6q. we get
67 =
=
'I
I(/sin 8 - c ) cos 8 + ( d
-
I coc 0) sin 8168
-I( I s i n H c o s B . - t . c t ) s B + d s i n B - I s i n B c o s 8 ) S 0
r)
(d)
I
= -(dsin8-ecos8)68
17
The principle CISvirtual woi-k is now applied to ensure equilihriurn.
Thus, considering one side of the system and using half the Inad, we have
SECTION 10.4 DEGREES OF FREEDOM AND THE SOLUTION OF PROBLEMS
i
Example 10.2 (Continued)
Hence,
1
60= 0
-(2,500)(21~0~~6O)+p(41r)
(e)
Figure 10.10, Pneumatic loading platform
The value of q at the configuration of interest may be determined from Eq.
(b). Thus,
qZ = [(24)(.866) - 1012 + [60 - (24)(.5)]*
Therefore,
q
= 49.2 in.
Now canceling 60 and substituting known data into Eq. (e), we may then
determine p for equilibrium:
-(2,500)(2)(24)(.5)
Therefore.
+ p(4~){$[(60)(.866)
- (lO)(.S)]}
=0
423
In a few of the homework prohlcnih. y o n I i i i w t(i iiw siniplc kinein:rrics
a cylinder rnlling without slipping (see Fig. 10. I I j . Y o u will r c c d li~liii
physics thai ihc cylinder i s iiciually rntitliiig ahout ilic piiitil o1 c ~ i i t i i c r-I. I i
Ihc cylinder rotate\ an angle SH then N' =
68. We sli:ill con>idcr h i m niatics of rigid hoclics i n detail li11cr i n IIK ICXI.
(1i
-,'
/
Figure 10.12. Virtual displacemcnt
WCIOIK
SECTION
We may also introduce the concept of the varied,function
given a function G(.x,y. 7) we may form
G
=
G(x
+&
y
+ 6,
z+6i,)
10.s LOOKING AHEAD: DEFORMABLE SOLIDS
425
such that
(IO. 17)
where SX, s\, and 6z are components of Sr"'. We now define the variation of
G , denoted as SG, as
6G=G-G
(IO. 18)
In a deformable body we can give the movement of each point in the
body when deformation occurs by using the displacement ,field u(x, y, z).
Specifically, when coordinates of a specific point in the underformed geometry are suhstitutcd into the particular vector function u(x,y. 7) depicting a specific deformation, we get the displacement of that point resulting from this
deformation.
We now extend the concept of a virtual displacement of a point to that
of a virtual displacement field, which is a single-valued, continuous vector
field representing a h.!qxitherirml dpfimmble body movement consistent n,ith
the constraints pre.wnt. We shall restrict ourselves here to virtual displacement fields, which result only in infinitesimal deformation." We have shown
the gross exaggeration of a virtual displacement field in Fig. 10.13. wherein
you will notice that tlie constraints have not been u,iolated. It should now be
clear that we: can conveniently set fonh a virtual displacement field by
employing the so-called variational operator 6. Thus Su may be considered as
a virtual displacement field from a given configuration to a vaned configuration; the constrainls present being taken into account by imposing proper conditions on the variation.
The virtual work concept can now he extended to the case of a
deformable body. We compute the work of the external forces during a virtual displacement of the body with the proviso that these external forces he
maintained constant. With a total body force distribution B(x,y,z ) and a total
traction force distribution T(x,?,z), we can then give the virtual work,
denoted as 6M1,,,, as follows:
S Wv,r, =
111B
V
*
6u d~ +
#T
6u dA
(10.19)
c
the virtual ~ ' o r khad to he zero for equilibrium. For
bodies tliis is no longer lrue. Instead, for equilibrium the external
virtual work SW,,, given above must he equal to infernal virtual work, which
For rigid bodies,
d&irmuhlr
"The changes in thc coordinates x, y, and: are nut linked tu lime through thr basic laws of
physics a$ would be the case i f we wcrc considering G 10 represent some physical quantity in
*omc real proccss.
1 8 0 n enerd not s o rrslrict onehell. That is. we ~ i l nwork wiih virtwd displacement fields for
/inire deformation and fonnulatc a principle o f v i i l u a l work. Thia would lilkc us beyond the scopc
oithis hook. however.
Figure 10.13. Virtudl displacement field
cotisistent with constraints.
426
CHAPTER IO METHODS OF VIRTUAI. WORK AND STATIONARY POTENTIAL. ENERGY
must he zero for rigid bodies but which is nut necessrily zero for deformable
solids. In your solid mechanics course you will learn that the internal work for
a deformable body is given by
T,,)d€,,d i : wherc the indices i and.j
I
.I
range over 1,y, and i forming terms i n the integrand such as rtL6ttt,
rAV6t,%.
etc., (nine expressions). The satisfaction of the resulting formulation is a nccessary and sufficient condition for equilibrium and can he used in place of the
familiar equations of equilibrium." Why would one want to do this'! Actually, as we pointed in the Looking Ahead section in the chapter on structural
mechanics, we can readily solvc certain types of problems using virtual work.
and the theorems derived from virtual work, whereas the approach for these
problems using the equilibrium equations i s extremely cumbersome. One
important case is the solution of indeterminate truss problems. You will come
to these problems later in your studies of structures and in your studies ol
machine design.
Virtual work and two other theorems derivable from it are ciilled
enerEy di,vp/acemenf merhod.? because of the use of the virtual displacement." There is an equally useful set of three formulations analogous to the
three energy displacement methods and they are called rrwrgvforcv methods.
wherein we hypothetically vary the forces instead of hypothetically varying
the deformation.
Before moving [in, the author would like to share a philosiiphical thought
with you. In science we often physically disturb cettain surroundings in the laboratory and carefully observe resulting behavior to learn to understand natural
phenomena. We perhaps unwittingly mimic this approach here in the study of
mechanics. That is. we have instituted mathematical "disturbances" and evaluated the results in order to Understand certain vital analytically userul consequences. Thus, we instituted the mathematical "disturbance" of the virtual
displacement field to amve at extremely useful conclusions which form thc
basis of a considerable amount of structural mechanics. Also, we pointed out
that we can institute varied force fields as our mathematicdl disturhanccs.
Again, vital and useful conclusions follow.
"The second energy displacemcm method i s called the rncrhod of iorulporrniial cnryqy. It
wab this principal that was prcsenlcd in the "Looking A h e a d Section 6.2 fur determining the pin
deflections uf simple t ~ u s c s The
.
\pccial ca<e 01 this principle lur c o 1 1 s c ~ v ~ 1iorcc
1 ~ c fields iict~
ing on pwiicle.s and rifiiil hodics IS divrlopcd in Pan R o l this chapter. 'The third energy dis
placement mclhotl deiivahle lrum the secontl i s the h n i Cbsngbono I%eorrm.
For a Ihoruugh dewlopinen1 0 1 rhesc six principles with many applications and which i s
w i t h h reach of atudrnts who have ahsorhcd thc key C O I ~ I Cof
~ ~the stdies portion u l this lex<. see
Shames, I.H., Inrmducrion V J Soiid Mechanic.>.seconded., Prentice-Hall. Inc.. Cnglrwootl Clitti.
N.J., Chapters 1 X and 19. A good grasp of ibex six principles is vital l i i iiiuic advanced work in
solid and structuiill mcchimics. not 111 speak of machine design.
10.1, How many degrees of freedom do the following systems
possess'! What coordinates can be used to locate the system?
(a) A rigid body not constrained in space.
(b) A rigid body constrained to move along a plane sutiace.
(c) The board AB in the diagram (a).
(d) The spherical bodies shown in diagram (b) may slide
along shaft C-C, which in turn rotates about axis E-E. Shaft C-C
may also slide along E-E. The spindle E-E is on a rotating platform. Give the number of degrees of freedom and coordinates for
a sphere, shaft C-C, and spindle E-E.
10.3. What is the longest portion of pipe weighing 4M)Ib/ft that
can be lifted without tipping the 12,000-lb tractor?
T
i-
B
A
Figure P.10.3.
10.4. If W, = 100 N and W, = 150 N, find the angle 0 for
equilibrium.
E
*
I
Figure P.10.4.
I +
C
10.5. The triple pulley sheave and the double pulley sheave
weigh 150 N and 100 N, respectively. What rope force is necessary to lift a 3,500-N engine'?
B
(b)
Figure P.10.1.
10.2. A parking-lot gate arm weighs 150 N. Because of the
taper, the weight can be regarded as concentrated at a point 1.25
m from the pivot point. What is the solenoid force to lift the gate?
What is the solenoid force if a 300-N counterweight is placed .25
m to left of the pivot point?
r__--
I
. I 5 m'
Figure P.10.2.
Figure P.10.5.
42;
10.6. What weight W call hc lifted with Ihc A-frame hoist in the
position shiiwn i f thc cable tension is 7''
10.9. Whal is the tencion in the cahles ofa IO-ft-wide 12-ft-l<mg
h000-1h drawhridge whrn the hndge is first raised? When thc
hridge is at 35"''
3m
,
Figure P.I0.6.
10.7. A small hoist has a lifting capacity of 20 kN. What is thc
innximuin possihle cahlc tension load'!
Figure P.10.9.
10.10. Assuming frictionless cmtacts, determine the magnitude
lor cquilihrium.
if P
.h 111
Figure P.10.7.
Figure P.I0.10.
10.11. A ruck cruhher i'. \how" i n aclivn. If,>, = 50 psiig and p,
on the rock at the configuraiion
\hewn! Thc dinmeter uf the pihlons is 4 in.
= 100 psig. what is the forcr
10.8. I f W = 1.000 N and P = 300 N, find the anglc H For
equilibrium.
W
Figure P.10.8.
Figure P.lO.ll.
10.12. A 20-lb-ft torque is applied to a scissor jack. If friction is
disregarded throughout, what weight can be maintained in equilibrium? Take the pitch of the screw threads to be .3 in. in opposite
senses. All links are of equal length, I ft.
10.15. A hydraulically actuated gate in a 2-m-square water-carrying tunnel under a dam is held in place with a vertical beam AC.
What is the force in the hydraulic ram if the specific weight of
water is 9818 N/m3?
Figure P.10.15.
Figure P.10.12.
10.16. Find the angle p for equilibrium in terms of the parameters
given in the diagram. Neglect friction and the weight of the beam.
10.13. The 5,000-lb van of an airline food catering truck rises
straight up until its floor is level with the airplane floor. What is
the ram force in that position'?
Figure P.10.16.
10.17. Do Problem 5.54 by the method of virtual work
Figure P.10.13.
10.14. What are the cable tensions when the arms of the power
shovel are in the position shown? Arm AC weighs I3 kN, arm DF
weights 1 1 kN, and the shovel plus the payload weigh 9 kN.
A
,/ I '
C.G.
'
10.18. Do Problem 5.55 by the method of virtual work.
10.19. What is the relation among P, Q, and 0 for equilibrium'?
1p
H
2.12m
Figure P.10.14.
Figure P.10.19.
425
--+k i c k hcrc
Figure P.lO.23
.
Figure P.IO.ZI.
Fizure P.10.25
10.26. If A weighs SO0 N, and if B weighs 100 N, determine the
weight of C for equilibrium.
Figure P.10.26.
10.29. Rod ABC is connected through a pin and slot to a sleeve
which slides on a vertical rod. Before the weight W of 100 N is
applied at C, the rod is inclined at an angle of 4 5 ~ If
. K of the
spring i s 8,000 N/m, what is the angle 0 for equilibrium? The
length of AB is 300 mm and the length of BC is 200 mm when
0 = 45'. Neglect friction and all weights other than W. (Note:
The force from the spring is K times its contraction.)
10.27. An embossing device imprint,s an image at D on metal
stock. If a force F of 200 N is exerted by the operator, what is the
force at D on the stuck? The lengths of AB and BC are each 150 mm.
10.30. Show that the virtual work of a couple moment
rotation &$ is given as
M for a
sw = M . 6 6
Figure P.10.27.
[Hint DeL-umpoSK
with
M into components normal to and collinear
&$.I
M
Sd,
10.28. A support system holds a 500-N load. Without the load,
0 = 45* and the spring is not compressed. If K for the spring is
10,000 N/m, how far down d will the 500-N load depress the
upper platform if the load is applied slowly and carefully'? Neglect
all other weights. DB = BE = AB :CB = 400 mm. (Note;
The force from the spring is K times its contraction.)
Figure P.10.30.
43 1
432
CHAPTER I O METHODS OF VIRTUAL WORK A N D SThTIONAIIY P0TI:NIIAI. E N I t K i Y
Part B: Method of Total Potential Energy
10.6
W
Y,
-
Conservative Systems
We shall restrict oiirscI\~csin this v c t i o n to cL'rtiiin types ol active tirrces. Thih
restriction will pcrmit LIS to arrivi: at s o m e additional very uscful relations.
Conzider first i i hody x t c d on only hy gswity lorce W a s an active
Ibrcc iiiid moving iiloiig a fi-icrionlcss p i t h lrom position I to posilioii 2. iih
shown in Fig. 10.14. The work ~ l o n chy grabity.
$F.ilr
=o
'71.',
~,i s
then
(10.23i
How i s the piilcnlial enel-g) l'uiiction V rclnted to F?'10 iimwci- this
query. consider tliat iui w h i t r q iiiljnittlsinial path segment rlr tliirts from
point I . We can then f i v e Eg. 10.12 iis
1.' * ilr = -(I\)
(10.?41
SECTION 10.6 CONSERVATIVE SYSTEMS
av
F =- I
(10.26)
&
Or, in other words,
!
a . a
=-[ja
; i +. z ~ + z k
(10.27)
-grad V = -VV
The operator we have introduced is called the gradient operator and is given
as follows for rectangular coordinates:
=
grad =V=-ta
A
. + a J. + -ak
&
~
&
(10.28)
We can now say, as an alternative definition, that a conservative force
field must be ( I function of position and expressible as the gradient of a scalar
,field function. The inverse to this statement is also valid. That is, ifa force
,field is a function of position and the gradient of a scalarfield, it must then be
a ~f~il,rservatiw~lrcrfield.
Such are the following two force fields.
Constant Force Field. If the force field is constant at all positions, it can
always he expressed as the gradient of a scalar function of the form V = -(M
+ by + C Z ) . where a, b, and c are constants. The constant force field, then, is
F = ai + bj + ck.
In limited changes of position near the Earth's surface (a common situation),
we can consider the gravitational force on a particle of mass, m, as a constant force
field given by .-mgk. Thus, the cunstants for the general force field given above are
a = b = 0 and c = -mg. Clearly, V s P.E. = mgz for this case.
Force Proportional to Linear Displacements. Consider a body limited by
constraints to move along a straight line. Along this line a force is developed
directly proportional to the displacement of the body from some point on the
line. If this line is the x axis, we give this force as
F = -Kxi
(10.29)
where x is the displacement from the point. The constant K is a positive number, so that, with the minus sign in this equation, a positive displacement x
from the origin means that the force is negative and is then directed hack to the
origin. A displacement in the negative direction from the origin (negative x)
means that the force is positive and is directed again toward the origin. Thus,
the force given above is a restoring force about the origin. An example of this
force is that sesulting from the extension of a linear spring (Fig. 10.15). The
force that the spring exerts will be directly proportional to the amount of elongation or compression in the x direction beyond the unextended configuration.
This movement is measured from the origin of the x axis. The constant K i n
this situation is called the spring constant. The change in potential energy due
to the displacements from the origin to some position x, therefore, is
433
434
CHAPTER IO
METHOIX 01: VIRTUAL WORK AND STATIONARY POTENTIAL ENERGY
since -V
~
(K;2)
=
-K.ri
Thc chungr in potential energy has been defined as the nr,yufive of (he
wurk done by a conservative force as we go from one position to another.
Clearly, the potential energy change i s then directly eyuul to the work done by
the t-~actiorrIO the conservalive force during this displaccment. I n thr case of
the spring, the reaction force would he the furce,fr(im the surroundings acting
on thc spring at point A (Fig. 10.15). During extension or compression o f the
spring from the undefotmed position, this force (from the surroundings) clearly
inus1 do a positive amuunt o f work. This work must, as noted above, equal the
potential energy change. We now note that we ciin consider t h i s work (or in
other words the change in potential energy) to be a rneasurc of the energy
~ / ~ / ~ / ~
K / ~ / ~ / ~ / ~ / ~ / ~ srored
/ ~ / ~in/the
~ /spring.
~ / ~That
/ ~ is,
/ ~when allowed to return to its original position, [he
H
spring will do t h i s aniuunt o f positive work on the surruundings at A , provided
that the relum motiun i s slow enough tu prevent oscillations, etc.
1
~~~~
,
Figure 10.15. Linear spring
10.7
Condition of Equilibrium
for a Conservative System
L e t 11s now consider a system of rigid bodies that i s ideally constrained and acted
on by conservative active fixes. For a virtual displacement from a configuration
of equilibrium, the virtual work done by the active forces, which are maintained
constant during the virtual displacement, must bc zero. We shall now show thac
the condition of equilibrium can be stated in ye1 another way for this system.
Specifically, suppose that we have 11 conservative lbrces acting on the
system of bodies. The incremenl of work for a real infinitesimal movement of
the system can be given iis follows:
where V without subscripts refers to tofu1 potential energy. B y treating 6r
I'
likc dr,, in the equations abovc, we can express the virtual work 8711vrr,
as
SECT10N 10.7 CONDITION OF EQLLIBRRIM FOR A CONSERVATIVE SYSTEM
= 0,
But we know that for equilibrium
for equilibrium:
a i
and so we can similarly say
0
a"=
435
(10.31)
,,
Mathematically, this means that the potential energy has a stationary or an
extremum value at a configuration of equilibrium, or, putting it another way,
the variation (fV is zero at a configuration of equilibrium.15 Thus, we have
another criterion which we may use to solve problems ofequilibrium for conservative force systems with ideal constraints.
For solving problems, determine the potential energy using a set of independent coordinates. Then, take the variation, 6, of the potential energy. For
example, suppose that V is a function of independent variables 4,. q2'...,q,,
thereby having n degrees of freedom. The variation of Vthen becomes
av
6 V = -&q,
aY1
av
+ -6q,
ay,
av
+ ... + -6q,
ay,
(10.32)
For equilibrium, we set this variation equal to zero according to Eq. 10.31.
For the right side of the equation above to be zero, the coefficient of each 6qc
must be zero, since the &. are independent of each other. Thus,
We now have n independent equations, which we can now solve for n
unknowns. This method of approach is illustrated in the following examples.
15Tofunher understand this, consider Vas a iuncrion of only one variable,x. A s r a f i ~ w ~ y
value (or, as we may say, an extremum) might be B local minimum (ain Fig. 10.16). a local m a imum ( b in the figure), or an inflection point (c in the figure). Note far these points that for a diiferential movement. hr. there is zero first-order change in V ( i . L 6v = 0).
1
IPJ
(bJ
(CJ
Figure 10.16. Stationary or extremum
points'
436
CHAPTER I O METHODS OF VIRTUAL WORK AND STATIONARY POTENTIAL ENERGY
Example 10.3
A block weighing W Ib is placed slowly on a spring having a spring constant of K IMft (see Fig. 10.17). Calculate how much the spring is com-
pressed at the equilibrium configuration.
Figure 10.17. M a w placed on
B
linear spring.
This is a simple problem and could be solved by using the definition of
the spring constant, but we shall take advantage of the simplicity to illustrate
the preceding comments. Note that only conservative forces act on the block,
namely the weight and the spring force. Using the unextended top position of
the spring as the datum for gravitational potential energy and measuring x
from this position we have, tor the potential energy uf the system:
V = -Wx
+ ;Kx2
Consequently, for equilibrium, we have since there is only one degree of
freedom
L
V = - w + Kx
dx
Solving for x, we have
=0
SECTION 10.7 CONDITION OF EQUILIBRIUM FOR A CONSERVATIVE SYSTEM
431
Example 10.4
A mechanism shown in Fig. 10.18 consists of two weights W, four pinned
linkage rods of length a, and a spring K connecting the linkage rods and
which rides along a stationary vertical rod. The spring is unextended when 0
= 45". If friction and the weights of the linkage rods are negligible, what are
the equilibrium configurations for the system of linkage rods and weights?
Only conservative forces can perform work on the system, and so we
may use the stationary potential-energy criterion for equilibrium. We shall
compute the potential energy as a function of 0 (clearly, there is but one
degree of freedom) using the configuration 8 = 45' as the source of datum
levels for the various energies. Observing Fig. 10.19, we can say that
+ f K(2d)'
V = -2Wd
(4
As for the distance d, we can say (see Fig. 10.19)
d=a~os45~-acos0
Figure 10.18. A mechanism.
(b)
Hence, we have, for Eq. (a),
V = -2Wa(cos45"
- cos0) + f K4a2(cos45" - c 0 s 0 ) ~
For equilibrium, we require that
__
dV
0 = -2Wasin0 + 4KaZ(cos4So- cosO)(sin0)
d0
We can then say
[
sin0 -W-2Ka
(
cos0--
Ail
(C)
=O
We have here two possibilities for satisfying the equation. First, sin 0 = 0
is a solution. so we may say that 0 - 0 (this may not be mechanically pos.
T
sible) is a configuration of equdlbnum. Clearly, another solution can be
reached by setting the bracketed lerms equal to zero:
i
-
t
.'
(
h1
-W - 2Ka cos0 - - = 0
Therefore,
The solutions for 0 then are
We have here two possible equilibrium configurations.
Figure 10.19. Movement of mechanism as
determined by 0.
10.31. A SO-kg block is placed carefully on a spring. Thc spfing
is nonlinear. The force ta deflect the spring a distance 2 rnm is
proportional to the squarz of x. Also, we know that 5 N deflects
the spring I mm. By the method of stationary potential energy.
what will he the comprcssion of the spring? Check the result using
a simple calculation hased on the hehavior of the spring.
10.33. Find the equilibrium configurations for the syslern. The
bars are indentical and each hac n weight W , a length of 3 m, and
a mass o f 2 5 kg. The Fpring is unstrerched when the bars arc horimntal and has a Tpnng const:mt of 1.500 Nlm.
Figure P.10.33.
< Nonlinear spring
<
<
1
4
7
The springs of the mechanism are unstretched when 6 =
1 S . Y W when the weight W is added. Take W =
S O 0 N, o = .3 m, K , = I Nlmm. K2 = 2 Nlmrn, and e,, = 45~.
Ncglect the weight of the memherc.
10.34.
e,,.Show that 0 =
c
Figure P.10.31.
10.32.
der of radius 2 ft has wrappr iround it il light.
"extensible cord which is tied to a iocj-ib block n on a 70'
nclined surface. The cylinder A is connected to a ror.cionrrl rpring.
rhis spring requires a torque of IS100 ft-lblrad of rntntion and it is
inear and, of c o m e , restoring. If B is connected to A when the tor;ional spring is unatrained. and if R is allowed to move h w l y
h w n the incline, what distanced do you allow il to move to reach
m equilibrium configuration? Use the method of stationan.
, .notenial energy and then check the result by more elementary reasmine.
I
Figure P.10.34.
Inextensible cord
\
Figure P.10.32.
10.35. At what elevatiim h must body A he for equilibrium'!
Neglect friction. [ H i n t What is the differential relation hetwren 6
and 1 defining the positions VI
thc blocks alrrng the surface? Integrate to get the relation ihelf.1
,100 Ih
Figure P.10.35.
10.36. Show that the position of equilibrium is 8 = 77.3- for
the 20-kg rod AB. Neglect friction.
10.39. Work Problem 10.28 using the method of total potential
energy.
10.40. Work Problem 10.29 using the method of total potential
energy.
Figure P.10.36.
10.41. Do Problem 10.25 by the method of total potential
energy. [Hint: Consider a length of cord on a circular surface. Use
the top p m of the surface as a datum.]
10.37. A beam BC of length 15 ft and weight 500 Ib is placed
against a spring (which has a spring constant of 10 Ihlin.) and
smooth walls and allowed to come to rest. If the end of the spring
is 5 ft away from the vertical wall when it is not compressed,
show by energy methods that the amount that the spring will be
compressed is ,889 ft.
10.42. If member AB is 10 ft long and member BC is 13 ft long,
show that the angle R corresponding to equilibrium is 3 4 . 5 ~if the
spring constant K is 10 Iblin. Neglect the weight of the members
and friction everywhere. Take R = 3 0 ~for the configuration
where the spring is unstretched.
IS'
K=10 Iblinch
Figure P.10.42.
Figure P.10.37.
10.38. Light rods AB and BC support a 500-N load. End A of rod
AB is pinned, whereas end C is on a roller. A spring having a
spring constant of 1,000 Nlm is connected to A and C . The spring
is unstretched when R = 4 5 ~ Show
.
that the force in the spring is
1,066 N when the 500-N load is being supported.
R
Figure P.10.38.
10.43. A combination of spring and torsion-bar suspension is
shown. The spring has a spring constant of 150 Nlmm. The torsion bar is shown on end at A and has a torsional resistance to
rotation of rod AB of 5,000 N - d i a d . If the vertical load is zero,
the vertical spring is of length 450 mm. and rod AB is horizontal.
What is the angle a when the suspension supports a weight of 5
kN? Rod AB is 400 mm in length.
End view of
torsion bar
Figure P.10.43.
439
10.44. Light rods AB and CB are pinned together at B and pasr
through frictionless hearings IJ and E. These bearings arc connected to the ground by ball-and-socket connections and are free
to rotate about these joints. Sprinzs, each having a s p k f constant
K = 800 N/m, restrain the rods as shown. The springs are
unstretched when 8 = 4 5 ~ Show
.
that the deflection of B i s 440
m when a 500-N load i s attached slowly to pin 8.The rod.. are
each I m in length. and each unstretched spring i s ,250 m in
length. Neglect the weight of the rods. Rods are weldcd to small
plates at A and C
10.47. In Problcm 111.46. the hand i\ l i n t strctched and then tied
while strstched lu supports A and H w that there i s an initial tellsion in the hand 0 1 I
N. What i':then the deflection (, caused hy
the Ill-N load'?
10.48.
A ruhher hand of length .7 m i i stretched t o connect t o
points A and 13. A tcnsim force , > I 30 N i s thereby de~~elopeil
in
the hand. A 211-N \r,eight i s then attached t o the hand at C. Find
thc distance (I that point C iiioves downward if the 20-N weight i\
cnnstraincd t o move vertically downward along a frictionless rod.
[Hbir: I i y o u considcr pan rfthc hand, the "cprinp conwnt" for i t
will he greater than that nf the whole hmd.]
Figure P.10.44.
10.45. Do Problem 10.2h by the method of tntal potential
mergy. [Hint; Use E as a datum and get lengths EJ. K P , and PN
in terms of length H E , including unknown constants.]
10.46. An elastic band is originally 1 m long. Applying a tension force of 30 N, the hand will stretch .8 ni in length. What
jeflection a does a 10-N load induce on the band when the load i s
%ppliedslowly at the center of the baud? Consider the l o r e vs.
dongation of the band to be linear like a spring. [Hint: If you consider half of the band, you double the "spring constant."l
Figure P.10.48.
10.49. 'The spring connecling hodies A and H has a spring cnnt m t K oi. 3 Nlmm. The unstrctched length o f the spring i\ 450
m m . IIhody A weighs 60 N and hody B weighs 90 N, what is the
metched length of the spring for cquilihrium'! [Hinr: I/ w i l l hc il
Fhction of two varinhles.1
1-m1
- i
Figure P.10.46.
Figure P.10.4Y
SECTION 10.8
10.8
Stability
Consider a cylinder resting on various surfaces (Fig. 10.20). If we neglect
friction, the only active force is that of gravity. Thus, we have here conservative systems for which Eq. 10.31 is valid. The only virtual displacement for
which contact with the surfaces is maintained is along the path. In each case,
dyidx is zero. Thus, for an infinitesimal virtual displacement, the first-order
change in elevation is zero. Hence, the change in potential energy is zero for
the first-order considerations. The bodies, therefore, are in equilibrium,
according to the previous section. However, distinct physical differences
exist between the states of equilibrium of the four cases.
The equilibrium here is said to he stable in that an actual displacement from this configuration is such that the forces tend to return the body to
its equilibrium configuration. Notice that the potential energy is at a minimum
for this condition.
Case A.
Case B. The equilibrium here is said to he unstable in that an actual displacement from the configuration is such that the forces aid in increasing the
departure from the equilibrium configuration. The potential energy is at a
maximum for this condition.
Case C. The equilibrium here is said to be neutral. Any displacement
means that another equilibrium configuration is established. The potential
energy is B constant for all possible positions of the body.
Case D.
This equilibrium state is considered unstable since any displacement to the left of the equilibrium configuration will result in an increasing
departure from this position.
How can we tell whether a system is stable or unstable at its equilibrium
configuration other than by physical inspection, as was done above? Consider
again a simple situation where the potential energy is a function of only one
space coordinatex. That is, V = Vix). We can expand the potential energy in
the form of a Maclaurin series about the position of equilibrium." Thus,
( I 0.34)
x
x
Figure 10.20. Different equilibrium configurations.
at x
IhNote that in a Maclaurin series the coefficients of the independent variable x a ~ evalualed
e
position. We denote this position with the subscript eq.
= 0,which for us is the equilibrium
STABILITY
441
442
CHAPTER I O METHODS OF VIRTUAL W O RK AND STATIONARY POTENTIAL ENERGY
We know from Eq. 10.33 applied to one variable that at the equilibrium configuration ( d V / d ~ )=~ 0. Hence, we can restate the equation above:
For small enough x. say xo. the sign of AV will be determined by the
x2," For this reason this
sign of the first term in the series, (l/2!)(d2V/dr2Je4
term is called the dominant term in the series. Hence, the sign of (d2V/dxZ)eq
is vital in determining the sign of AV for small enough x. If (d2V/dx2),q IS
positive, then AV is positive for any value of x smaller than .q,. This means
that Vis a local minimum at the equilibrium configuration as can be deduced
from Fig. 10.2021, and we have stable equilibrium.'8 If (d2V/dx2)eqis negative, then V is a local maximum at the equilibrium configuration and from
Fig. 10.20 we have tinsfable equilibrium. Finally, if (d2V/dx2)eqis zero, we
must investigate the next higher-order derivative in the expansion, and s o
forth.
For cases where the potential energy is known in terms of several variables, the determination of the kind of equilibrium for the system is correspondingly more complex. For example, if the function V is known in terms
of x and v. we have from the calculus of several variables the following.
For minimum potential energy and therefore for stability:
av dv
(10.36a)
(lO.36b)
(10.36~)
For maximum potential energy and therefore for instahility:
(10.37a)
The criteria become increasingly more complex for three or more independent variables.
"As x gets smaller than unity, xz will become increasingly larger than 2' and powcrs of I
higher than 3. Hence. depending on the values uf derivatires of Vat equilibrium, them will h e a
value of *--say x,,-For which the first tern in the series will he larger than the sum of a l l <,lhci
terms
for ValUeE of r < I,,.
"That is. if the body is displaced a distance x < x,? the hody will return to equilibrium on
TdCVSC.
SECTION 10.9 LOOKING AHEAD: MORE ON TOTAL POTENTIAL ENERGY
Example 10.5
A thick plate whose bottom edge is that of a circular arc of radius R is
shown in Fig. 10.21. The center of gravity of the plate is a distance h above
the ground when the plate is in the vertical position as shown in the diagram. What relation must be satisfied by h and R for stable equilibrium?
The plate has one degree of freedom under the action of gravity and
we can use the angle O(Fig. 10.22) as the independent coordinate. We can
express the potential energy V of the system relative to the ground as a
function of 8 in the following manner (see Fig. 10.23):
V = W[R - ( R - h ) COS s]
(a)
where W is the weight of the plate. Clearly, 8 = 0 is a position of equilibrium since
Now consider d2Vld8’ at
= 0. We have
($$) e o
=
W(R-h)
Clearly, when R > h, (dZVld8Z), = is positive, and so this is the desired
requirement for stable equilibrium. Thus for stable equilibrium, R > h.
R-(R-h)
Figure 10.21. Plate with circular
bottom edge.
10.9
Figure 10.22. One degree of
freedom.
Looking Ahead: More on Total
Potential Energy
When we have conservative forces acting on pruticles and rigid bodies, we
found earlier that for establishing necessary and sufficient conditions of equilibrium we could extremize the potential energy Vassociated with the forces.
That is, we could set
6V = 0
to satisfy equilibrium. Recall that this result was derived from the method of
virtual work. We have a similar formulation for the case of an elastic (not necessarily linearly elastic) body wbeEby we can guarantee equilibrium. This more-
COS
0
Figure 10.23. Position of C.G,
443
444
CHAPTER I O
METHODS OF VIRTUAL WORK AND STATIONARY POTEiYTI,\L liNFR(;Y
general principle is derivable from the more-gencral virtual work principle
mentioned earlier i n section lO.5. Hcre. we 11iu\t extrenii7.e an exprcssinn inore
complicated (as you might expect) than V . This expi-ession is denotcd as nwith
no relation to the number 3.1416.. . . The expression n is what we CJII
a.fim.tiond, whercin for the substitution of each functim such as v(~r),into the functional a nunihcr is established. A simple example of ii functional I is as follows:
where F is a function o f r (the so-called independent \iiriahIe), y, and clyldr.
Subtilutinn of. a function y(.C into F fiillnwed hy iiii integration hetween the
fixed limils. yields for this funclion y(.i) :I number h r 1. Fonctionals per\,adr
the field of mechanicc and most other analytic Ciclds o l knowledge. A vital
step is to find the function ~ ( n that
) w i l l cxtremix I . This function then
becomes knnwn aq the urrrrriral/un,.fion.'The ciiIcuIus Ir,r doing this is called
the w l d u s 0 1 variations. Thc particular functional for the method of tntill
potential cncrgy is h'
riven as
n
=
-111B
* u 0'1. -
Ii
8
7
u d.4
+ L'
.\
The function to he adjusted to cxtremize ii is u(.x,J, z) taking the placc of ~ ( ~ 1 )
in the preceding functional. and iinw the independcnt \,;iriablcs are .x. y. and 2
in place of just .r in thc preceding functional. The expression 11 is the energy
of defnrmation that you will study l a t a in your solids ciii~rseand presented
cai-lier in Section 10.5 aTi,&,,
</I,. The principle of total potential
111
1'
t
I
energy for the case of elastic hodies is written i n the following deceivingly
simple looking lormulation:
6n = 0
What is most intriguing about this imoccnt looking equation is that it is oftcn
considercd to he thc most powerful equation in solid mechanics!
We have touched on a h r i d area of sludy. namely vari;itinnal methods.
park of which you will encounter in many o f y 0 u r htudies. For cxamplc. j u t
the method (it' total potential energy has thc following ma,jnr uses:
1. It plays a n important role i n opfiinixiti,Jii iIimii:i,.
2. It can he used profitably to dcrive the p r q w r q u a t i o m and thc houndor?
cnnditiuiis for many areas of vital importance such a h plate theory, elastic
stability theory, dynamics of plates and beams. torsion theory, etc,
3. From it we can develop a numher of vital approximation methods. The
most prominent of these methods is the methiid of.finirrd r m m t s .
Clearly this
iy
an impressive
'vYou can aiudy 111 wme drlail rhr c o n t e i i l i 01 rhc lwi I.imking Ahead scc~ionsof l h i \
chaps, i n I . R . Shamea. lnlrwluclio,~10 .Solid M<dwnU.r. 2nd ed.. 1%9. Prcnlice~Hnll.Inc..
Englewood Cliffs. N.J. Thih i i i a t ~ r i i iihoukl
l
he wiihiii thc wach of hrig,hi *ccimd-semester sopho~
imorcs and cerlninly of juniors. Thc author ha\ raught (hi\ maiciiiil hi iriany year, 1 0 seconilsentester sophomores OUI of the aiwemesrioned hook. See Chiqxcia 18 iind 1'1.
10.50, A rod AB is connected to the ground by a frictionless
ball-and-sockrt connection at A . The rod is free to rest on the
inside edge of a horizontal plate as shown in the diagram. The
square abcd has its center directly over A. The curve eJg is a semicircle. Without resorting to mathematical calculations, identify
positions on this inside edge where equilibrium is possible fur the
rod AB. Describe the nature of the equilibrium and supply supporting arguments. Assume the edge of plate is frictionless.
Y
I
Figure P.10.54.
10.55. A system of springs and rigid bodies AB and BC is acted
on by a weight W through a pin connection at A. If K is 50 N/mm,
what is the range of the value of W so that the system has an
unstable equilibrium configuration when the rods AB and BC are
collinear'? Neglect the weight of the rods.
Figure P.10.50.
P,"
connecuon
10.51. In Problem 10.50, show mathematically that position h is
a position of unstable equilibrium for the rod.
10.52. Rod AB is supported hy a frictionless ball-and-socket
joint at A and leans against the inside edge of a horizontal plate.
What is the nature of the equilibrium position a for the rod?
Assume that the edge of the plate is frictionless.
i
Figure P.10.55.
10.56. A weight W is welded to a light rod AB. At B there is a
torsional spring for which it takes 500 ft-lh to rotate 1 rad. The
torsional spring is linear and restoring and is, for rotation, the analog of the ordinary linear spring for extension or coneaction. If
the torsional spring is unstrained when the rod is vertical, what is
the largest value of W for which we have stable equilibrium in the
vertical direction?
Figure P.10.52.
10.53. Consider that the potential energy of a system is given by
the formulation: V = 8x3 + 6x2 - 7x. What are the equilibrium
positions? Indicate whether these positions are stable or not.
10.54. A section of a cylinder is free to roll on a horizontal surface. If yof the triangular portion of the cylinder is 180 Ib/ft3 and
that of the semicircular portion of the cylinder is 100 IMft', is the
configuration shown in the diagram in stable equilibrium?
Figure P.10.56.
44
I
10.57. A light rod A B is
Also at A are two identical
pinned to a hlock of weight W at A .
springs K. Show that, for W less than
2K1, we have stable equilibrium in the vertical position and, for W
> 2K1, we have unstable equilibrium. The value W = 2KI is Called
il cririrul load for reasons that are explained i n Problem 10.5X.
Figure P.10.58.
10.58. I n Problem 10.57, apply a small transverse force F to
body A as shown. Compute the horizontal deflection 6of paint A
for a position of equilibrium by using ordinary statics as developed
i n earlier chapters. Now show that when W = 2K1 @e.,the critical
weight), the deflection 6mathemdtically blows up to infinity. This
shows that, even if W < 2KI and we have stable equilibrium with
I.’ = 0, we get increasingly very large dcflections as the weight W
approaches its critical value and a side load F , however small, is
introduced. The study of stability of equilibrium configuration
therefore is an important area of study in mechanics. Most of you
will encounter this topic in your strength of materials course.
10.10
10.59. Cylinders A and B have semicircular cross-sections.
Cylinder A supports a rectangular solid shown as C. If p, =
1,600 kg/mz and p< = 800 kg/m’, ascertain whether the arrangement shown is in stable equilibrium. [ H i m Make use of point 0
in computing V.1
I
6m
-
t
Closure
In this chapter, wc have taken an approach that differs radically from the
approach used earlier in the text. In earlier chapters, we isolated a hody for
the purpose of writing equilibrium equations using all the forces acting on the
body. This is the approach we often call vectorial mechanics. In this chapter,
we have mathematically compared the equilibrium configuration with admissible neighboring configurations. We concluded that the equilibrium configuration was one from which there is zero virtual work under a virtual
displacement. Or, equivalently for conservative active forces, the equilibrium
configuration was the configuration having stationary (actually minimum)
potential energy when compared to admissible configurations in the neighborhood. We call such an approach variational mechanics. The variational
mechanics point of view is no doubt strange to you at this stage of study and
far more subtle and mathematical than the vectorial mechanics approach.
Shifts like the one from the more physically acceptahle vectoriul
mechanics to the more abstract variational mechanics take place in other
engineering sciences. Variational methods and techniques are used in the
study of plates and shells, elasticity, quantum mechanics, orbital mechanics,
statistical thermodynamics, and electromagnetic theory. The variational
methods and viewpoints thus are important and evcn v i t d in more advanced
studies in thc engineering sciences, physics, and applied mathematics.
446
10.60. At what position must the operator of the counterweight
crane locate the 50-kN counterweight when he lifts the IO-kN
load of steel?
10.63. The spring is unstretched when 0 = 3LT. At any position
of the pendulum, the spring remains horizontal. If the spring constant is 50 Iblin., at what position will the system be in equilibrium?
20 m
Figure P.10.63.
10.64. If the springs are unstretched when 0 = 0,. find the
angle 0 when the weight W is placed on the system. Use the
method of stationary potential energy.
Figure P.10.60.
10.61. What is the relation between P and Q for equilibrium?
Figure P.10.61.
10.62. A 50 Ih-ft torque is applied to a press. The pitch of the
screw is .5 in. If there is no friction on the screw, and if the base
of the screw can rotate frictionlessly in a base plate A, what is the
force P imposed by the base plate on body B?
-ft
Figure P.10.62.
Figure P.10.64.
10.65. A mass M of 20 kg slides with no friction along a vertica
rod. Two internal springs K, of spring constant 2 Nlmm and ai
external spring K2 of spring constant 3 Nlmm restrain the weigh
W. If all springs are unstrained at 6' = 3LT, show that the equilib
rium configuration corresponds to 0 = 27.8'.
Figure P.10.65.
10.66. When rod AH i h in thc vaticill position, the spring
attached to the wheel by a flexible cord is unstretched. Determinc
all the possible angles L? for equilibrium. Show which are stable
and which are not stable. The spring has a spring constant 01 8
Ib/in.
Figure P.10.67.
Figure P.10.66.
10.67. Two identical rods are pinned together :it H and arc
pinned at A and C . At H there i s a torsinnal spring requiring SO0
N-mlrad of rotation. What is the maximuin weight W thal ciich Ii,d
can have for a c11w uf stable equilibrium when thc ~riids arc
collinear?
448
IO.6X. A rectangular xdid body (11 height h rests on ii cylinder
with a semicircular scctiun Set u p criteria for \lahlc and unstable
riluilihrium iii trrnis n i ii and X 1 1 1hc
~ pasilion shoun
Figure P.10.68.
Dynamics
Kinematics
of a ParticleSimple Relative Motion
11.1 Introduction
Kinemutics is that phase of mechanics concerned with the study of the motion
of particles and rigid bodies without consideration of what has caused the
motion. We can consider kinematics as the geometry of motion. Once kinematics is mastered, we can smoothly proceed to the relations between the
factors causing the motion and the motion itself. The latter area of study is
called dynamics. Dynamics can be conveniently separated into the following
divisions, mnst of which we shall study in this text:
1. Dynamics of a single particle. (You will remember from our chapters
on statics that a particle is an idealization having no volume but having
mass.)
2. Dynamics of a system of particles. This follows division 1 logically and
forms the hasis for the motion of continuous media such as fluid flow and
rigid-body motion.
3. Dynamics of a rigid body. A large portion of this text is concerned with
this important part of mechanics.
4. Dynamics of a system of rigid bodies.
5. Dynamics of a continuous deformable medium.
Clearly, from our opening statements, the particle plays a vital role in
the study of dynamics. What is the connection between the particle, which is a
completely hypothetical concept, and the finite bodies encountered in physical
problems? Briefly the relation is this: In many problems, the size and shape of
a body are not relevant in the discussion of certain aspects of its motion; only
the mass of the object is significant for such computations. For example, in
towing a truck up a hill, as shown in Fig. 11.1, we would only be concerned
45 I
452
('HAPTRK I I
K1NEM:YI'ICS OF A PARTICLE-SIMPLE REl.AT1VE MOTION
W
i+
Figure 11.1. 'liwck uimsidcrrd :is u pxiiclc.
with the mass of the truck and 1101 with i t s shape or size (if we neglect force!,
from the wind, etc.. and the riitatiiinal cffects (if the wheels). The truck can
just a, well he considered ii p;uticle i n computing the necessary towing force.
We car present this relationship more precisely i n the following ntaiiner. As will he learned in tlie next chapter (Section 12.10). the equation iii
inotioii o f the center of nias\ of any hody can he foriiied by:
1. C(mccntriiting thc cntirc mass ill thc miss cciitcr of thc h d y .
2. Applying the
total rc\LiIIant lorce acting [in thc hiidy t o this hypothetical
particle.
When the iniition of the mass ccntcr charactcrizcs a l l w e nced to know ahout
the motiiin (if tlie hody. we employ the particle concept (i.e,, we find the
motion of the m i s s center). Thus. if all points of a hody have the same vclucily at airy tinie I ([hi\ i s callcd rrun.rl~iror~
i n o l i o i i ) . we necd iinly know the
miition o i t l i e mass center to liilly ch;ir;~cterize the motion (This was the case
fix thc truck. where the rotational inertia of the wheels was neglected.) If.
additiiinally, the sizc i i f a hiidy i \ \mall compared to itr trajectory (as i n plarie l x y niiitiiin. for example). the motion of the center o f n i x \ i s all that might
he needed. and so again we can use the particle conccpt ior such hodics.
Part A: General Notions
11.2
Differentiation of a Vector with
Respect to lime
In the study of statics. we dealt with vector quantities. We found i t con\cnient
to incorporale the directional naturc of thcsc quantities i n ii certain n~itatiiin
and set of opevatiiiii\. We called the tolalily iiithese \ery useful foi-intilalions
"\ector algehra." We s h a l l again enparid our thinking iron1 scalars to vectors-hi\
tinic for the operations of differenti;ition and intcgration with
rcspcct t o any scalar variable I (such as timr).
SECTION I I .2 DIFFERENrIATION OF A VECTOR WITH RESPECT TO TIME
For scalars, we are concerned only with the variation in magnitude of
some quantity that is changing with time. The scalar definition of the time
derivative, then, is given as
(11.1)
This operation leads to another function of time, which can once more be differentiated in this manner. The process can he repeated again and again, for
suitable functions, to give higher derivatives.
In the case of a vector, the variation in time may he a change in magnitude, a change in direction, or both. The formal definition of the derivative of
a vector F with respect to time has the same form as Eq. I I . I :
(11.2)
If F has no change in direction during the time interval, this operation differs
little from the scalar case. However, when F changes in direction, we find for
the derivative of F a new vector, having a magnitude as well as a direction,
that is different from F itself. This directional consideration can be somewhat
troublesome.
Let us consider the rate of change of the position vector for a reference
xyz of a particle with respect to time; this rate is defined as the vrlucily vector,
V , of the particle relative to q z . Following the definition given by Eq. 11.2,
we have
The position vectors given in brackets are shown in Fig. 11.2. The subtrxtion
I
..
Path of
Figure 11.2. Particle at times f and
f
+ Al
between the two vectors gives rise to the displacement vector Ar, which is
shown as a chord connecting two points As apart along the trajectory of the
particle. Hence, we can say (using the chain rule) that
453
wherc M'C have multiplied and di\'idrd hy As i n thc liist cxpression As A/
goes to Lerci. the direction ol Ar approaches tangency t o the trajectory at position r ( / )and apprixiches A,\ i n magnitude. Consequently. i n the limit. ArlAs
hecoines a unit w c t w e,. tangent to the traieclor>. Tliat i \
( I 1.3)
We c i ~ then
i
sii~
Therefore. drlih leads to a vector having n magnitudc equal to the speed 01
the pailicle and a dii-ection ~aiigentto the trajectory. Kccp i i i mind that there
can he any ansle hetueen the p(isition Vector and the velocity vector. Students seem to \'ant t o limit this anglc to 90c',which actually restricts you to ii
cit-culiir p a t h The acceleration ~ e c t o (il
r a Ixirtick ciin then be given as
The dillcrcntiation and integration of vector( r, V . and a will concern
thniuphout tlic text.
LI'I
Part 6: Velocity and Acceleration
Calculations
11.3
Introductory Remark
As you know lroni statics. w c ciin expres? ii vector i n many ways. For illstance, we can use rectangular compiinents. or. as we udl hhortly explain, we
can use cylindrical components. In evaluating dcriviitivcs of vectors with
respect 10 time. we mu51 pi-oceed i n accordance with the nianner in which the
vector has heen expre\rcd. I n Part B of this chapter. wc will therefore exaiiiine certain diffcrcntiation processes tliiit iirc used extensively i n mcchanics.
Other diftkrenti;ition proccsses will he exaniined l a t ~ at
r appropriate tiiiies.
We have already carried out a derivative operation in Section 11.2
directly on the vector r . You will scc in Section 11.5 that llic approach u x d
gives tlic dcrivative iii terms 01' potti vorilrblr~c.This approach will he one o S
several that we shall iiow examine with some care.
SECTION 11.4 RECTANGULAR COMPONENTS
11.4
Rectangular Components
Consider first the case where the position vector r of a moving particle is
expressed for a given reference in terms of rectangular components in the following manner:
r ( t ) = x(r)i + y ( r ) j + z(t)k
( I 1.6)
where x ( f ) ,y ( t ) , and z(t) are scalar functions of time. The unit vectors i, j , and
k are fixed in magnitude and direction at all times, and so we can obtain drldt
in the following straightforward manner:
dr = V ( t ) = -dx(t)
' + dy(t)j
t
dt
dt
k = i ( t ) i + y(t)j + i(r)k
+ wdt
( I 1.7)
A second differentiation with respect to time leads to the acceleration vector:
df2
=a =
i ( t ) i + j i ( t ) j + Z(t)k
(11.8)
By such a procedure, we have formulated velocity and acceleration vectors in
terms of components parallel to the coordinate axes.
Up to this point, we have formulated the rectangular velocity components and the rectangular acceleration components, respectively, by differentiating the position vector once and twice with respect to time. Quite often,
we know the acceleration vector of a particle as a function of time in the form
a(t) = X ( t ) i
+ s ( t ) j + Z(r)k
( I 1.9)
and wish to have tor this particle the velocity vector or the position vector or
any of their components at any time. We then integrate the time functionx(t),
y ( t ) , and i ( r ) , remembering to include a constant of integration for each integration. For example, considerx(t). Integrating once, we obtain the velocity
component V,(r) as follows:
Vx(t) =
1
f ( t ) dr
+ C,
(11.10)
where C, is the constant of integration. Knowing \ at some time to, we can
determine C, by substituting to and (V,),, into the equation above and determining C , . Similarly, for x(r) we obtain from the above:
x(t)
= j[jf(r)dt]dt+C,t+C?
(11.11)
where C, is the second constant of integration. Knowing x at some time f , we
can determine C , from Eq. 1 1.1 I . The same procedure involving additional
constants applies-to the other acceleration components.
We now illustrate the procedures described above in the following
series of examples.
455
456
('tIA1Tl:tt
II
K
Example 11.1
Pins A and H must aluayh remain in the vrrtical sIo1 01yrAc C'. which
tno\'e\ ttr tlic right at a c ~ t i s t i i n tspccd o f 6 I t l s e c i n Fig. I I ..?. titrthcrmorc.
the pins cannot leave the elliptic slot. (a) What is the speed at which the
pins apprmach each whet- wlieti the yoke s l o t is at ~r = 5 ft? ( h ) What i s the
rate olch;nrge of \peed k i w m l ciicli nlhcr when the yoke \lot i\ at .i~
= 5 ft'!
Thc cqu:ltioti of Ihc clliptic path iii which the pins iniiict ii~ovci \ seen
by inspection to he
.\z
~~~~
+ >r- _ I
IO'
j
t
!
t
i
j
(:I)
6'
clciirly. iIcoordinates (.1.y ) are to represent tlic cii(irdinatcs of pin R. tiicy
m u ~he
t time functions iiich Ih:it for any tinie i llie v;iliies .r!i) and ? ( I ) sillisfy Eq. (a). Also..i(/) andy(i) tiiust hc such that piti II t i i m e s at all times
i n the elliptic path. We cati s i t i d y thesc rcquircnictits hy first dilrcrcnlialiiif Eq. (a) with respect to tiiiic. Cancelins (lie factor 2. we ohtain
N o w ~ ~ (Y(II..~(I),
0.
andy(i) ~ i i i i ssati\fy
~
Ikl. ( h ) fur iill viilucs o f i 1 0 cnsurc
that H rciniiiiis iii the elliptic path
We can now proceed to calve par1 ( a ) of this pi-ohleln. We !wow that
pili H must have a velocity .i = h fllscc hcciiiisc (IItlic yokc. Furthcrniiirc.
when ~c = 5 It. we know from E[]. (a) that
:.
v = 5.20 ft
N o w going tu Eq. (b). we can solve f o r i at the
iii\tiiiil
01 intcrcit.
!
ii spccd of ?.OX ftlccc. Clcarly. pili A
n i u s t move u p w i r d with the s m i c speed nf 2.08 f t l x c . Thc pins apprmacli
each olhcr at the itistatit o f intcrcsl at ii speed 014.10 I'tlscc.
Thus. pin H tiioves downward with
;
Figure 11.3. Pin \ l i d o in \IDI ; ~ n d)rihc.
SECTION 11.4
Example 11.1 (Continued)
To get the acceleration? of pin H , we first differentiate Eq. (b) with respect
to time.
The accelerations.? and? must satisfy the equation above. Since the yoke
moves at constant speed, we can say immediately that x = 0. And using for
x, y , k, and j known quantities for the configuration of interest, we can
solvc for? from Eq. (d). Thus,
o+65 + 5 . 2 0 +~ 2.0X2 = 0
~~
102
62
Therefore
ji = -3.32 ft/sec*
Pin B must be accelerating downward at a rate of3.32 ft/sec2 while pin A
accelerates upward at the same rate. The pins accelerate toward each other,
then, at a rate of 6.64 ftt/secz at the configuration of interest.
In the motion of particles near the earth’s surface, such as the motion of
shells or ballistic missiles, we can often simplify the problem by neglecting
air resistance and taking the accelcration of gravity g as constant (32.2 ftlsec?
or 9.XI mlsec’). For such a case (see Fig. I1.4), we know immediately that
y(r) = -g andi(r) = i ( t ) = 0 . On integrating these accelerations, we can often
determine for the particle useful information as to velocities or positions at
certain times of interest in the pmhlem. We illustrate this procedure in the
following examples.
L
X
Figure 11.4. Simple ballistic motion of a shell
RECTANGULAR COMPONENTS
457
458
('IIAPTER I I
K I N I M A l I C S OF A PARTICLE- Slh.1I't.ti KEIATIVI'. !KITION
Example 11.2
Ballistics Problem 1.
A shcll i s fired from a hill 500 I t ahove n plain.
Ttic angle (1 of firing (see Fig. I I . S I i s 15' ahovc thc horimntal. and the
muizle velocity
i s 3.000 ftisec. At what horizontal distance. d, w i l l lhc
shcll hit the plain if wc ncglcct frictioti of the air'?What i s the nhaximun~
hcighl of the shell ahovc LIE plain? l:iti;illy, determine the trajectory of thc
shell 1i.e.. find =.f(.xll.
v)
v
We know imnicdiately ttiat
We need not hothcr with ?(f). since the motion i s coplanar with i(11 = := 0
at all times. We iicxt separate the velocity variahles froni the timc viiriahles hy bringing rlf to Ihe right sidc.: 0 1 the previous cqoations. Thus
dVv
=
"V, =
-32.2dr
Odt
Inlegrating the ahove equations. we get
V > [ J )= -32.21
V)(I) =
+
<.,
We shall lake f = 0 211 the iiistaiit the camion is fired. At this instant.
k n o w L; and V, ;rnd can dctcrmine C , atid C,. Thus.
V$Ol = 3,000sin 15" = ( - 3 2 . ? ) ( 0 ) + C',
WL'
SECTION 11.4
Example 11.2 (Continued)
Therefore,
C, = Vv(0) = 776ft/sec
Also
V\(0) = 3,000cos1S" = C,
Therefore,
C2 = b'a(Oj = 2,900 ft/sec
We can give the velocity components of the shell now as follows:
dv
V,(t) = d t = -32.2r + 776 ft/sec
dx
V,(rj =
= 2,900 ft/sec
dt
~
Thus, the horizontal velocity is constant. Separating the position and time
variables and then integrating, we get the J and y coordinates of the shell.
y ( t ) = -32.2
t2
+ 776t + C,
2
*(I) = 2,900t + C,
~
(e)
(f)
When f = 0, y = x = 0. Thus, from Eqs. (e) and (f), we clearly see that C,
C, = 0. The coordinates of the shell are then
y ( f ) = -16.1t2
=
+ 776f
x ( f ) = 2,900t
To determine di.stancp d, first find the time f for the impact of the shell
on the plain. That is, set y = -500 in Eq. (g) and solve for the time t. Thus,
-500 = - l h . l t 2
+ 776f
RECTANGULAR COMPONENTS
459
460
CHAPTER I I KINEMATICS OF A ['AKrlC'l E
C I h I I ' I E K t l U I V F hlOTlON
Example 11.2 (Continued)
Therefore,
Using the quadratic forintila, we get for I :
f = 48.8 vi.
Substituting this value of I i n l c Fq. lhl. we gc'l
d = (2,9OO)t48 8 )
To get the moximirm hpiRhz
from Eq. (c) we get
141,500 ft
~
x,,,,,,. firs1 iind Itre tirile I
1) -
12 2:
;\.hi.ii
776
I
Therefore,
I =
24.1
Now suhstirute t .:21.1 sei' iiilo F.q. ( € 1 .
:.
y-
= 9,350
St'C
'ihih
gtt
<:,
I , \ !, ,
ft
'Thercfnre.
y = -1.917 X 10"x2
Clearly, the tra.jectoi-y is that of a puwholn.
+,268~
,.
\; = 0.Thus
SECTION 11.4 RECTANGULAR COMPONENTS
Example 11.3
Ballistics Problem 2. A gun emplacement is shown on a cliff in Fig. 11.6.
The muzzle velocity of the gun i s 1,000 d s e c . At what angle a must the
gun point in order to hit target A shown in the diagram'? Neglect friction.
y
Figure 11.6. Find a to hit A .
Newton's law for the shell is given as follows for a reference n)'
having its origin at the gun.
j f t ) = -9.81
i(t) =
0
Integrating, we get
s(/) = Vv(tl = -9.81r
X(t) =
+ C,
(a)
(bl
V\(t) = C,
When f = 0, we have j = 1,000 sin a and j; = 1,000 cos a. Applying these
conditions to Eqs. (a) and (b), we solve for C, and C,. Thus,
1,000 sin a = 0
+ C,
Therefore,
C,
=
1,000 sin a
Also,
1,000 cos a =
Therefore.
c, =
c2
1,000 cos a
Hence, we have
+
$ ( t ) = -9.81t
1,000sina
i ( t ) = 1,000cosa
Integrating again, we get
~(f)=-9.811+1.000sinaf+C,
f2
x(t)
= 1,ooocosa/+C,
461
Example 11.3 (Continued)
Using the quadratic lormula. wc linii the liillowing anglcs:
ccI = 8.17"
"2 = 81.44"
Therc tire thus two possible firing angles thal w i l l pennil the rtiell 10
hit the target, as shown i n Fig. I I .7.
I
SECTION 11.4 RECTANGULAR COMPONENTS
Example 11.4
The engine room of a freighter is on fire. A fire-fighting tughoat has drawn
alongside and is directing a stream of water to enter the stack of the
freighter as shown in Fig. II.8. If the initial speed of the jet of water is 70
ft/sec, is there a value of a of the issuing jet of water that will do the job?
If so, what should a be'?
Figure 11.8. Fire-fighting tughuat directing a jet of water into the stack
of a freighter.
Consider a particle within the stream of water. Neglecting friction.
Newton's law for the particle is given as follows:
y = -32.2 ftlsec'
Y = 0 ft/sec?
Integrating twice, and using initial conditions at A , we get
$ = -32.2f
y =
+ 70 sin a ft/sec
-16.1t2 + 7 0 s i n a t ft
(a)
i = 70 cosa ft/sec
(h)
I=
7 0 c o s a t ft
(c)
(d)
Solve for f from Eq. (d) and substitute into Eq. (b) to get
y = -16.1[-L]
2 + 7 0 ~ i n a [ - ~ ]
70 cos a
70 cos a
Replace cos2 a by I/( I + tan? a) and (sin a/cos a) by tan a in the previous
equation and then substitute the coordinates of point B at the stack where the
water is supposed to reach. That is, set x = 40 ft and y = 30 ft. We then get
30 = -(3.29 x 10-')(402)(1 + tan2 cy) + 40 t a n a
:_
tan2 a - 7.61 t a n a + 6.71 = 0
Using the quadratic formula we get
463
464
C'HAPTRII I I
KIK\FMA'I'I('S OF A l ' ~ ~ l ~ ~ l l ~ l , ~ - ~ S
KI11.:\lIVE
l ~ l l ~ l . l V~l .l I l O N
Example 11.4(Continued)
We thus havc IMW
aiiglcs h r a, ciicli i i l w t i i c t i will the,rl-clically ciiusc the
streiilii 10 go ttr piiiiit I1 o i the .;tack. Thew iingles :ire
(1,
i
= 4?.?0'
a, = x 1.37"
Docs 1 1 1 ~ .11011c. or hotti angles ;ihovc yield i i s ~ r c i i i ii)t
i wiiter that will
down iil R YO iis t o enter the stxk'! We i'iiii detzriiiinc t h i s hy liiiding
thc iiiiixiiiiuiii viilue oI J aiid locating [tie p(isilioii t Iiir chi, iiiiixiriiuiii
valuc. To do this, we bel? = 0 and x i l v e for f using each IY. ' I l i i i s u'c have
c(ii1ie
'
:
i
:.
i
i
1
1
I =
I.?? I
{ 2.14')
)
scc
A Acwh (11 the t w o possihlc Ird.jectories i\ shown i n 1:ig. I LO. ClcarIy the
shallow trajcclory will hi?the side of tlie slack nnd i s uiiacceplahlr. while
?hc high trajcct(iry will deposit water iiisidc the stack and i s thus tlic
desired ti-ajcckiry. Thus.
a = 81 37"
SECTION 11.5
VELOCITY AND ACCELERATION IN TERMS OF PATH VARIABLES
465
We do not always know the variation of the position vector with time in
the form of Ey. I I .6. Furthermore, it may he that the components of velocity
and acceleration that we desire are not those parallel to a fixed Cartesian reference. The evaluation of V a n d a for certain other circumstances will be considered in the following sections.
11.5
Velocity and Acceleration
in Terms of Path Variables
We have formulated velocity and acceleration for the case where the rectangular coordinates of a panicle are known as functions of time. We now explore
another approach in which the formulations are carried out in terms of the path
variables of the particle, that is, in terms of geometrical parameters of the path
and the speed and the rate of change of speed of the particle along the path.
These results are particularly useful when a particle moves along a path that
we know apriori (such as the case of a roller coaster).
As a matter of fact, in Section 1 1.2 (F4. I I .4) we expressed the velocity vector in terms of path variables in the following form:
ds
dt
V = - E
(11.12)
where drldt represen& the speed along the path and E, = dr1d.s (see Eq. 11.3)
is the unit vector tangent to the path (and hence collinear with the velocity
vector). The acceleration becomes
(11.13)
Replace dq/dr in this expression by (dc,lds)(drldt), the validity of which is
assured by the chain rule of differentiation We then have
(11.14)
Before proceeding further, let us consider the unit vector E , at two positions that are A,s apart along the path of the particle as shown in Fig. 11.10. If.
A.7 is small enough. the unit i:ector$ e,(.s) and +[.s + A.s) can be considered to
intersect and thus to form a plane. If As + 0, these unit vectors then form a
limitirrg plane. which we shall call the osculating plune.' The plane will have
an orientation that depends on the position s on the path of the particle. The
osculating planc at r(t) is illustrated in Fig. 11.10. Having defined the oscukiting plane, let us confinur discusxion of Eq. 11.14.
IFrum Ihe definition. it should be apparent that the usculating plane at position .Y along a
act~illlyiorixe,il Lo Lhc curve at posirion J. Since usculalc means lo kiss. the plane
C U W ~ is
"kisac.,"thc curie. its it were. at
.$.
&
m
s
,*E,(\)
r ( f+ Ai)
€,(S
+ A,7)
Figure 11.10. Osculaling plane arr(f).
Since we h a w not fiirniiilly carried out thc differentiation or a \'cctor
with respect 10 a spatial coiirdinatc. we shall carry out the derivative dc,lil.v
riecdcd i n Eq. 11.14 from thc basic definition. Thus.
l h e vccmrs ~ p and
) E,(S + A.vj are shown i n Fig. I I. I I(3) along the path and
arc d s i i shown (cnliirgcd) with AE, as a \cctor triangle in Fig. 1 1 . 1 I(h). Ac
piiintcd out earlier, lirr small enough A.s the lincs OS action of thc unit vector5
eJ.sj and E,(.$ + As) w i l l intersect tn form a plane as shown i n Fig. 1 1 . 1 I(a1.
Now i n this plane. draw nonilill line:, to thc aSiircmentioned vcctnrs at the
respcctiivc positions s and ,s + A.s. These lines will intcrscct at some point 0. a s
shriwn in the diagram. Next. concidcr what happens to the plane and to point 0
as A s --f 0.Clearly. the limiting planc is our osculating plane at .s [see Fig.
I I , I I IC)].
Furthermore. the limiting position airived at for point 0 is fir tlrr
o.sdatiiig plonc and i s callcd the c'en/er o { m , ~ ~ i t for
i r ~the
~ path at s . The distaiicc hctwccn 0 and .s i\ denotcd iis R and i s called Ihc riu1iu.s o i ' i ' ~ w w t i m .
Finaly, the vector Ac, (see Fig. I I. I I ih)). in the limit a s As + 0, ends up i n the
osculating plane norind to the path at c and directed toward thc center of cu-viiturc. 'The unit vect~rciillincarujitli the linritirlg vcctor for AE, i\ dcniitcd as e,,
and i s called the p r i i i c i p ~ tnorniul
l
wi'ior.
A
/C/
Figure 11.1 I . Devcli~pmentillthe osculating plane and the center of cunalurc.
With the limilitig diwctioii o f Ae? cstablihed. we next evaluate the
iis an approximate value that hecomes corrcct as Av --f 11.
Ohserving the vector triangle i n Fig. I I.II(h). we can uxiirdingly say:
imi,ytz;fudo o f AE,
jAe,I =
IE,~AO
A*
tIl.lh)
SECTION I 1.5 VEI.OCITY AND ACCELERATION I N TERMS OF PATH VARIABLES
Next, we note in Fig. 1 I.I I (a) that the lines from point 0 to the points s and
s + A,Talong the trajectory form the same angle A@ as is between the vectors
E,(s) and E,(S + A,?) in the vector triangle, and so we can say:
A+ = A.s = A'!
Os
R
~
(11.17)
Hence, we have for Eq. I I . 16:
IAEI =
As
(11.18)
We thus have the magnitude of AE, established in an approximate manner. Using E,,, the principal normal at 5 , to approximate the direction of AE,
we can write
AE, =
A.7
~
R
en
If we use this result in the limiting process of Eq. 1 I . 15 (where it becomes
exact), the evaluation of dr,ld.s becomes
When we substitute Eq. I 1 . I9 into Eq. I I . 14, the acceleration vector becomes
d2s
(ds/dt)*
R
'fl
dt2 e+-
a=-
(II.20)
We thus have two components of acceleration: une cwmimnenr in u dirrctimrr
tangent to the path und one component in the osculating plane uf right anglc,s
to the puth und poinfinji toward the center ifcun,uture. These components
are of great importance in certain problems.
For the special case of a plane curve, we learned in analytic geometry
that the radius of curvalurt. R is given by the relation
Furthermore, in the case of a plane curve, the osculating plane at every point
clearly must correspond to the plane of the curve, and the computation of unit
vectors E,, and is quite simple, as will be illustrated in Example 11.5.
How do we get the principal normal vector en,the radius of curvature R,
and the direction of the osculating plane for a three-dimensional curve'? One
procedure is to evaluate E , as a function of s and then differentiate this vector
with respect to s . Accordingly, from Eq. 1 I . 19 we can then determine E, as well
as R. We establish the direction of the osculating plane by taking the cross
product E, x E,, tn get a unit vector normal to the osculating plane. This vector
is called the binormul vector. This is illustrated in starred problem I 1.7.
461
Example 11.6
..
A particle i s tnovirig iii ilic ~ A pl;ine
Y
almg ii parabolic path giveii a.;
Y = 1.22: .I (see Fig. I 1 . 1 3 J with~riiiicl?in inclcrs. AI position A . the particle ha\ a spccd (11 3 misec
d h a \ ii rate 01 change of speed n l 3 miscc'
iiloiig ilic path What i s Ihc ii
ior o f llic p r l i d c i i thih
~
position'?
SECTION 11.5
VELOCITY AND ACCELERATION IN TERMS OP PATH VARIABLES
Example 11.6 (Continued)
where
At the position of interest (x = I .5 m ) we have
Therefore,
u = 26.5"
Hence,
.st = .X95
+ ,4461
(C)
As for e,,, we see from the diagram that
en = S i n a i - c o s a j
Therefore,
en = ,4461' - ,895j
(d)
Next, employing Eq. I 1.21, we can find R. We shall need the following
results for this step:
dY = . 6 1 0 ~ - ' / ~
(e)
C ~.~* Y- -,305x-i/2
dr? -
(f)
Substituting Eqs. (e) and (f) into Eq. 11.21, we have for R:
[i
R=
+ (.6inx-l/2)zr
305*-1/2
(SI
At the position of interest, x = 1.5, we get
R = 8.40 m
(h)
We can now give the desired acceleration vector. Thus, from Eq.
I 1.20, we have
a = 3(.895i
:.
9
+ ,446j) + ,,(.4461'
+
a = 3.16i ,3791
- ,895.j)
0)
469
470
CHAPTER I I
KINEMATICS OF A I'ARTICl.r.-SIMPI
I:. RFl./\TlVE hlOTIOK
*Example 11.7
A particle i s made t o m o w ; h n g a spirzil palh, iis i s shown in Fig. I I .13.
I
%
/
j
1
'
1
I
Figure 11.14.
Thc equations rcprcscnting thc p ~ arc
h given parametricall? iii terins o i
Ihe uriahle r iii thc Solliiwing iiiannc.l-:
.xi, = A 'in r l r
y,, = A U I \ qr
- =cT
( A . q.
arc h n i ~ \ v ncwistant,)
iill
',,
where the cuhcci-ipt 11 is t i 1 rcniind the rcader that these reliifioti'r rcfcr t o ii
fixed path. When thc p a r l i c k i s iit the .rp pliiiie ( 7 = I)). i l ha\ ii spccd c ) l
V,, I'tJhec and ii riitc: ol'changc o i \peed of N fticec.'. What i s the ~ICCCICRItioii of the partick iii this position?
To answer this. \be must asccrtiiiii E , . E ,,. and K . To pct E , M'C w r i ~ c :
,'. (lp
i
<'
Solving lor t h c ilillcrcntials dr
into Eq. (d). wc gct:
I'
, ;itid
(/:
I r w n Eq. ( a ) and iuhstiluting
SECTION 11.5 VELOCITY AND ACCELERATION IN TERMS OF PATH VARIABLES
Example 11.7 (Continued)
in which we replaced (cosz q7+ sin2 qzJ by unity. Returning to Eq. (cJ. wc
can thus say:
E
To get
E"
=
I
[Aq(cosqzi - Sinqz j J + Ck]
( A 2 q 2 + C 2Jll2
(gJ
and R we employ Eq. I 1.19, but in the following manner:
in which we have replaced dsldz using Eq. (f). We can now employ Eq.
(g) to find de,/dz:
When we substitute this relation for d q l d z i n Eq. (h), the principal normal
vector becomes:
If we take the magnitude of each side, we can solve for R:
We now have e, and E, at any point of the curve in terms of the
parameter z. As the particle goes through the *y plane, this means that the
z coordinate of the position of the particle is zero and zp of the path corresponding to the position of the particle is zero. When we note the last of
Eqs. (a), it is clear, that zmust be zero for this position. Thus en and E, for
the point of interest are:
E, =
I
( A q i + CkJ
(A2q2 + C2)'/*
where we have used Eq. (k) to replace R in Eq. (m). We can now express
the acceleration vector using Eq. 11.20. Thus:
The direction of the osculating plane can be found by taking the
cross product of E , and E,,.
41 I
Figure P.lI.2.
472
What is the acceleration of the particle at f = 3 sec? What distance
has been traveled by the particle during this time? [Hint: Let
dr = <d.r2 + d?' + d:' and divide and multiply by dr in second
half of problem. Look up integratiun f h n
pendix I.]
I
+ f'df
vn'
!~
~
in Ap-
~~
11.7. In Example I I . I , what is the acceleration vector for pin B
if the yoke C is accelerating at the rate of I O ft/sec2 at the instant
of. inlerest?
Figure P.11.K.
11.9. The face of a cathode ray tube is shown. An electron is
made to move in the horizontal (x) direction due to electric fields
in the cathode tube with the following motion:
x = A sin Of mm
Also, the electron is made to move in the vertical direction with
the following motion:
y = Asin (cot + a ) m m
?
IO ftlsec'
X
6 fUsec
Show that for a = 1112, the trajectory on the scrcen is that of a circle of radius A mm. If a = 11, show that the trajectory is that of a
straight line inclined at -45" to the xy axes. Finally, give the formulations for the directions of velocity and acceleration of the
electron in the .xy plane.
Figure P.11.7.
11.8. Particles A and B are confined to always be in a circ u l a r groove of radius 5 ft. At the Fame time, these particles m u s t also be in a slot that has the shape of a parabola.
The slot is shown dashed at time I = 0 . If the slot moves to
the right at a constant speed of 1 ftlsec, what are the speed
and rate of change of speed of particles toward each other
at f = I sec?
x
Figure P.11.9.
413
I
.--
-I
I
Figure P.11.IZ.
Figure P.II.10.
-.
,
Figure P.11.13.
11.14. A churgcd panicle i\ \hot a1 tirric f = 0 at an angle of45"
w i t h ii q ~ " e d10 ftlsec. It an r l r c l i i c iirld i s such L l i i l t [lie body har
an ~ ~ c c u l c l i i t i i-i rX W j f ~ I w c ' , what i \ the cquation for the trajectory'! What IS thi. value < > i d
f w impact'.'
Figure P.ll.11.
ti
I'igure P.11.14.
474
11.15. A projectile is fired at a speed of 1,000 mlsec at an
angle E of 40" measured from an inclined surface, which is at an
angle $ of 20" from the horizontal. If we neglect friction, at
what distance along the incline does the projectile hit the
incline?
Figure P.11.15.
(b)
Grain is being hlown into an open train container at a
speed !
I
of,
20
)
fthec. What should the minimum and maximum
elevations d be to ensure that a11 the grain gets into the train?
Neglect friction and winds.
Figure P.11.17.
11.16.
\
..: ..
..
V
= -St'
+ 27.8 mls
.:.
..,
6-
11.18. In the previous problem, the vane has a velocity given
relative to the ground reference X Y as
.i
IS'
What is the distance 6 between the vane and the position of
impact of the water that left the vane at time I = 0. Use the trajectory of the preceding problem, which relates x and s for a refercnce xy attached to the vane and moving to the left at I = (1 at a
speed of 100 kmlhr = 27.X mls. The trajectory of the water after
leaving the vane at I = 0 is
y = -.00721x2
+ , 3 6 4 m~
with x in meters.
i
11.19.
l ( 1 ' 4
Figure P.11.16.
A fighter-bomber is moving at a constant speed of 500
mi* when it fires its cannon at a target at 8.The cannon
hila a
m u d e velocity of 1,000 mls (relative to the gun barel).
(a) Determine the distanced. Use reference shown.
11.17. A racket-powered test sled slides over rails. This test
sled is used for experimentation on the ability of man to
undergo large persistent accelerations. To brake the sled
from high speeds, small scoops are lowered to deflect water
from a stationary tank of watec placed near the end of the
run. If the sled is moving at a speed of 100 kmlhr at the
instant of interest, compute h and d of the detlected stream
of water as seen from the sled. Assume no loss in speed of
the water relative to the scoop. Consider the sled as an inertial reference at the instant of interest and attach xy reference
to the sled.
(b) What is the horizontal distance hetween the plane and
position B at the time of impact'!
Muzzle velocity
.,
=
1.000mls
Plane velocity = SO0 mls
,
..
,
Figure P.11.19.
475
..
Figure 1'.11.20.
,.* . ..
.
11.28. In the preceding prohlem, a second smaller destroyer is
firing at the target as shown in the diagram. The data of the preceding problem applies with the following additional data. Due
to strong wind and current, the destroyer has a drift velocity
of 6 km/hr in a northeast dircctiun in addition to its full speed o l
75 kmihr. Form two simultaneous transcendcntal equations for
a and p and verify that a = 2 1.39" and that p = 10.727".
y
I
1
N
)
S
12,000 111
.
(in Same
horizontal
plane as
Figure P.11.26.
11.27. A destroyer is making B run at full speed of 75 kndhr.
When abreast of a missile site target, it fires two shells. The target
is 12,000 m fioin the destroyer. If the muzzle velocity is 400
misec. what is the angle of firing a with the horizontal that the
computer must set the guns? Also, what anglc pmust the turret be
rotated relative to the line of sight at the instant of firing? [ H i m
To hit target, what must V, of the shell be? Result: a = 23.7" and
p = 3.260.1
Figure P.11.28.
*11.29. A Jeep with an archer is moving at a speed of 30 mi/hr.
At 100-yd distance and moving at right angles to the Jeep is a deer
running at a speed of 15 mihr. If the initial speed of the arrow shot
hv the archer to bae- the deer is 200 ftisec. what inclinatiiin a must
the shot have with the horizontal and what angle p must thc vertical
projection of the shot ontn the ground have relative to the line A n ?
y
I
AI
nonzontai piane
as guns )
I
I
@ r l S milhi
n
Figure P.11.27.
Figure P.11.29.
477
I'ipurc P.11.36.
,,
I
I
!
Cockpit
Figure P.11.34.
11.43. A passenger plane is moving at a constant speed of 200
kmihr in a holding pattern at a conslant elevation. At the instant of
interest, the angle p between the velocity vector and the x axis is 30".
The vector is known through o n - b o d gyroscopic instrumentation
to be changing at the rate p of -5"Isec. What is the radius of curvature of the path at this instant?
1
'\\,,
..J .',,('
?'
X
A
Figure P.11.40.
11.41. A panicle moves with a constant speed of 3 mlsec along
the path. What is the acceleration a at position x = I .5 m'?Give the
rectangular components of a.
x
Figure P.11.43.
y
kx
Y
= 3x2
ex
11.44. At what position along the ellipse shown does the normal
vector have a set of direction cosines (.707, .707, OY! Recall that the
equation for an ellipse in the position shown is *2iu2 + v2/h2= 1.
Figure P.11.41.
h=5'
at
hasA ,a what
11.42.
speed
A iparticlc
softhe10magnitude
ft/sec
moves
andalong
of
a rate
thea acceleration'?
of
sinusoidal
change of
path.
speed
What
If the
is
of the
5particle
ft/sec2
magnitude and direction of the acceleration of the particle at E , if it
has a speed of 20 ft/sec and a rate of change of speed of 3 ftlsec'
at this paint?
a = IO'
Figure P.11.44.
y
I
11.45. A panicle moves along a path given as
I
I
x=5
4
5'
1
I
x-IO
2
0 '
Figure P.11.42.
i~~
x
4
The projection of the particle along the x axis varies as d.2t2 ft
(where t is in seconds) starting at the origin at t = 0. What are the
acceleration components normal and tangential to the path at f = 2
sec? What is the radius of curvature at this point?
479
4x0
SECTION 1 I.6 CYLINDRICAL COORDINATES
Figure 11.15. Cylindrical coordinates.
Unit vectors are associated with these coordinates and are given as:
e7, which is parallel to the i axis and, for practical purposes, i s the same as k .
This is considered to be the aria1 direction.
e?, which i s normal to the z axis, pointing out from the axis, and is identified
as the radial direction from z .
eP which is normal to the plane formed by e?and eiand has a sense in accordance with the right-hand-screw rule for the permutation z, 7,8. We call
this the transver.se direction.
Nutc thut ei rind eo will change direction as the particle mows relntive
to the xyz reference. Thus, these unit vectors are generally Junctions of lime,
whereas e7 is a constant vector.
Using previously developed concepts, we can express the velocity and
acceleration of the particle relative to the xyz reference in t e r m uf cunrponentr ulwuys in the transverse, radial, and axial directions und can use cylindrical coordinates exclusively in the process. This information is most useful,
for instance, in turbomachine studies (i.e., for centrifugal pumps, compressors, jet engines, etc.), where, if we take the z axis as the axis of rotation, the
axial components of fluid acceleration are used for thrust computation while
the transverse components are important for torque considerations. It is these
components that are meaningful for such computations and not components
parallel to some x y z reference.
The position vector r of the particle determines the direction of the unit
vectors ei and ee at any time f and can be expressed as
r
= ?ei
+ ze.
To get the desired velocity, we differentiate r with respect to time:
( I 1.23)
48 1
4x2
I'IiAl'ItX I I
KINI(MATICS OF h I ~ \ K I I ( ' L t ; ~ S I M ~ l . IKELATIVE
I
MOIION
Our t a h k here i s to evaluate ET On consulting Fig. 11.16. we see clearly that
changes iii direction 01 e, occiir only wtieii the ctrordinatc (if thc particlc
changcs. Hence. rememhering ltial the m:ignirude of e, i s aluays c~inskinl,n e
hiivc Ibr E7 using the chain rule:
6
'
8
der
~
ill
<kr[It)
'10
(If
= dc,
e
ilH
c,(H
I
( I 1.24)
AH)
i~
i
Tu cvaluatc [ I t r IdH. we have shown i n Fig. I l.l6(a) the ve:ctor e7 for a g i w n
?and :.:it position\ correspoiiding to H nnd ( O + AOL In Fig. I l.l6(h), iurthermore. w e have loriiied an enlarged vector triangle from these vectors md. in
this way. we have shown the
tor Ac, (i.c.. Ihc change i n E , durinf a chanfc
i n the coordinate 8). From the vector triangle. we see thal
!A€,
= le, ~
A e= A 0
( 1 1.25)
Furthcrinorc. as A 0 ~i
0 wc see. on consulting Fig. I I. 16. that tlic dircwlion
ill Ae, appn,achc\ Ih:it 01the unit veclor
and \o wc can approximate Ac, as
Ae, = IAe, 'e,# = A 0 e,,
( I I .xi
where we habe u w d tiq. I I .25 i n the last step. Going hack ti) liq I 1.24. we
titilix the preceding result to write
SECTION 1I.h CYLINDRICAL COORDINATES
In the limit, as A 8 + 0, all the previously made approximations become exact
statements and we accordingly have
ds
.
(11.28)
2 = BE,
dr
The velocity of particle P is, then,
(11 29)
To get the acceleration relative to q z in terms of cylindrical coordinates
and radial, transverse, and axial components, we simply take the time derivative of the velocity vector above:
a = dV = ?ei
dr
~
+ ?67 +
+ roes + r f j ~+, ?si
(I 1.30)
Like si, the vector E, can vary only when a
We must next evaluate
change in the coordinate 0 causes a change in direction of this vector, as has
0 have been
been shown in Fig. I l.l7(a). The vectors E&@) and ~ ~+ (A@
shown in an enlarged vector triangle in Fig. I1.17(b) and here we have shown
A€@ the change of the vector E , as a result of the change in coordinate 8. We
can then say, using the chain rule,
(11.31)
(a)
Figure 11.17. Change of unit vector
(b)
483
484
CHAPTER I I
KINEMATICS O F 4 PARTICLE - SIMPLE K E L A I I V
hE
t(~i'im
As A 0 + 0, the dirrctiiin of A E hecomes
~
that of -ti and the nugnitude of
A E on
~ consulting tlie vectoi- triangle, clearly approacheh 1 t 8 , / A 0= A 0 Thus,
the vector A t , become\ appriiximalely -A0 E ~ I. n the limit, wc' then get fiir
Eq. 11.31:
=
-&,
(11.321
lising b:qs. I I .2X ;ind I I .32. we find thal Eq. I I .30 now hecmnes
u =
?E,
+ i&, + i.&, + r&"
-
r@'Er
+ :E
Collecting coinponciits. wc write
a = ( i -rg")et
+ (TO + 2PB)c, + re,
(11.33)
Thus;. we have accompli.;hed the desired tahh. A similar procedure can
he followed to reach corresponding forinulaliiins kir spherical coordinates.
By now you should hc nhlc to produce the preceding equations readily from
the foregoing habit principle\.
I:or mntion i n R c.i,v.lr in the .r? plane. notc 1ha1? = := 0 , and 7 = r. We
yet the folloning simplifications:
Furthermore. the unit vector E~ is rangent 10 llic palh. and t l i e unit vector E , i\
normal ti1 the path and points away from thc ccnler. Thcrcforc. when we c ~ r r i pare Eq. I I .34h with tliiise cteiiiiriing lr(riri considerations irf path variable,
(Section 11,s). c l e x l y for c i r c u l x motioii i n the .xy coordinate plane iif u
right-hand triad:
E , ~- E,
(for counterclockwise motion
ii\ sccii from
:)'
E ~ ,= -E.
(for clochu,i\e miticin
+
( I 1.351
Thus. Eqh. I1.34h and I I . X I are equally risclul for quickly expressing the
acceleration 0 1 ii particle moving in a circular path. YIIUprohahly remember
these formulas from carlicr physics courses and may umii 1 0 use them i n t l ~ c
rnsuing work iif thir chapter.
SECTION 11.6 CYLINDRICAL COORDINATES
Example 11.8
A towing tank is a device used for evaluating the drag and stability of ship
hulls. Scaled models are moved by a rig along the water at carefully controlled
speeds and attitudes while measurements are being made. Usually, the water is
contained in a long narrow tank with the rig moving overhead along the length
of the tank. However, another useful setup consists of a rotating radial a m (see
Fig. I I . IS), which gives the model a transverse motion. A radial motion along
the arm is mother degree of freedom possible for the model in this system.
Figure 11.18. Circular towing tank
Consider the case where a model is being moved out radially so that
in one revolution of the main beam it has gone, at constant speed relative
to the main beam, from position ?= 3.3 m to i =4 m. The angular speed of
the beam is 3 rpm. What is the acceleration of the hull model relative to
the water when i =4 m ?
In order to find the radial speed of the model, note that one revolution of the arm corresponds to a time tevaluated as:
r = 1 = 20 sec
~
60
Hence, we can say for P:
r = -~4 - 3.3 - ,035d s e c
~
T
We can now readily describe the motion of the system at the instant
of interest with cylindrical coordinates as follows:
r=4m,
0
i; = ,035 d s e c ,
i = 0,
L=O
= 3(2”)
60
=
,314 rad/sec
Using Eq. 11.33, we may now evaluate the acceleration vector,
a := [0 ( 4 ) ( . 3 1 4 ) 2 ] ~+~[0 + ( ~ ) ( . 0 3 5 ) ( . 3 1 4 )+] ~[0]5
~
= -.394e, + , 0 2 2 ~
m/sec2
~
~
Finally,
1
0
1=
,395 mlsecz
Note that we could have used notation r instead of i here. (Why?)
485
Example 11.9
~-c*
A firetruck hac ii lclcsc<iping hiwin holding a firefi$itcr :is shown in Fi?.
1 1 . 19. At linic I . lhc boom i s cutciiding at the ratc 010 niis ;~ndincrcnsing
i t s riitl: 01 cxLcnsioii at .3 ni/s2. i\ls<i at time I . I = 10 n i and [j = 30'. II ii
velocity component o f the firefighter in the vertical d i r c c i i m i s 1 0 he
: 3.3 mlscc at this instant. what \hould/j h c ? Also. if iit this instant the wrtical ~~cccIc~..~tion
01 ilic firelighter i s t o he I .7 Inis?. wliiii hhould fl he'!
N o t thal thc nioiiiiii o f thc firefighter i s that u1,joint A .
i
We first inwrt skilionwy rcScrence.~y: at the h a w ( i l t l i c hiuim u i i h tlic
hoom in thc ~y plane iis shown in Fig. I1.20. C'lcarly. tlic hoiini i s rotating
!
SECTION 1 I.6 CYLINDRICAL COORDINATES
Example 11.9 (Continued)
about the I axis and so this axis is the axial direction for the system. Furthermore, the bourn being normal to the z axis must then be in the radial direction. It should be obvious that there is no motion of the firefighter in the axial
direction. We now give the velocity vector as follows, noting that 0 = p here.
-
v = ?e, + F O E . + Z E ;
= (.6)e, + ( l O ) ( b ) ~ ,+ O E m/s
~
We require that V j = 3.3 d s . We thus have for this purpose
3 . 3 = ( . 6 ) ( ~ ; j)+(IO)(B)(EB.j)=.6sinP+IObcosp
where we have used the fact that the dot product between two unit vectors
is simply the cosine of the angle between the unit vectors. Noting that at
the instant of interest p= 30". we can readily solve forb. We get
Now we write the equation for the acceleration.
a = (P
-r@2)Er
+ ( r e + 2rLB)e, + OE*
+
= (.3 - (lO)(.346)*)er (lop
= -.900er + ( l o p
+ (2)(.6)(.346))ee
+ , 4 1 6 ) ~m/s2
~
We require that a j = I .I d s z . Hence
1.7 = -.900(~, j ) + (lop + .416)(c0 * j )
= -.900sin30"+(IO~+.416)cos30"
We then have the following desired information:
Again, as in the preceding example, we could have used r instead of i.
487
Vigore P.Il.52.
Figure P.11.51.
p
=
1.3-
I
I() I X I
v
N
x
1 1
4 C 4 0 m m = radius
I
Figure P.11.56.
N
Fizure P.11.54.
11.55. A paiticle moves with a constant speed of 5 ft/sec along a
straiphl line having direction cosines I = .5, rn = .3. What are the
cylindrical coordinates when lrl = 20 ft? What are the axial and
transverse velocities of the panicle at this position'!
11.57. A plane is shown in a dive-homhing mission. It has at the
instant of interest a speed of 4x5 km/hr and is increasing its speed
downward at a rate of 81 kmhrlsec. The propeller is rotating at 150
rpm and has a diameter o f 4 m. What is the velocity of the tip of the
propeller shown at A and its acceleration at the instant of interest?
Use cylindrical velocity components.
A
Z
X
Figure P.11.57.
Figure P.11.55.
11.56. A wheel i s rotating at time f with an angular speed o o f 5
radlsec. At this instant, the wheel also has a rate uf change of
angular speed of 2 radlsec'. A body H is moving along a spoke at
this instant with a speed of 3 mlsec relative to the spoke and is
increasing in its speed at the rate of 1.6 dsec'. These data me
given when the spoke, on which B is moving, is vertical and when
H is .h m from the ccnter of the wheel, as shown in the diagram.
What ;we the velocity and acceleration of B at this instant relative
to the fixed reference XK?
11.58. The motinn of a panicle relative to a reference xyz is
given as follows:
i = .2 sinh f m
13 = .5 sin lif rad
z = 6tZ m
with f in seconds. What are the magnitudes of the velocity and
acceleration vectoT6 at time = set. Note that
= 3.h269
and cash
= 3,7622,
11.59. Given the following cylindrical coordinates for the
motion of a panicle:
i = 20 m
0 = 2nr rad
z = 51 m
with f in seconds. Sketch the path. What is (his curve? Determine
the velocity and acceleration vectors.
489
Figure P.11.68.
Figure P.11.65.
11.66. A simple garden sprinkler is shown. Water enters at the
base and leaves at the end at a speed of 3 mlsec as seen from the
rotor of the sprinkler. Furthermore, it leilves upward relative to the
rotor at an angle of 60" as shown in the diagram. The rotor has an
angular speed w of 2 radlsec. As seen from the ground. what are
the axial, transverse. and radial velocity and acceleration components of the water just as it leaves the rotor?
11.69. Undcrwater cable is being laid from an ocean-going ship.
The cable is unwound from a large spool A at the rear o l the ship.
The cable must be laid so that is no1 drugged {in the ocean bottom If
the ship is moving at a speed of 3 knots, what is thc necessary angular speed wof the spool A when the cable is coming off at a radius of
3.2 m? What is the average rate of change of w for the spool
required for proper operation? The cable has a diameter of I S 0 mol.
Figure P.ll.69.
11.70. A variable diameter drum is rotated by a motor at a constant speed w of I O 'pm.A rupe of diarnzter d of .5 in. wraps
around this drum and pulls up a weight W . It is desired that the
velocity ofthe weight's upvard movement he given as
12
x = . 4 + - 8,000 ftisec
Figure P.11.66.
11.67. The acceleration of gravity on the surface of Mars is ,385
times the acceleration of giaviry on earth. The radius R of Mars is
ahout ,532 times that ofthe earth. What is the time of flight of one
cycle for a satellite in a circular parking orbit 803 miles from the
surface of Mar?'? [Nore: GM = sK2.1
11.68. A threaded rod rotates with angular position I3 = ,3151'
rad. On the rod is a nut which rotates relative to the rod at the rate
w = .4f radisec. When I = 0, the nut is at a distance 2 ft from A .
What is the velocity and acceleration of the nut at I = 10 sec'! The
thread has a pitch of .2 in (see foatnole 4). Give results in radial
and transverse directiuns.
where for r = 0 the rope is just about to start wrapping around the
drum at Z = 0. What should the radius i of the drum he as a function o f Z to accomplish this'! What are the velocity components Y
and i of the weight W when f = I O 0 sec?
X
Z
Figure P.11.70.
49 1
492
CHAPTER I I
KINFMATI('S ( I P A P.4RlICI I S l h l P I t: REI A l ' l V t ! hI(1TIOK
Part C: Simple Kinematical Relations and
Applications
11.7
Simple Relative Motion
Up to ~ncnv,u c h a w comidercd only a hiiiplc rclerence ill our kinemalical colisideralioiis. Thcrc ai-c rime.; when IUO or morc references may he protitahiy
einploycd in describing llic inmior of a particle. Wc shall considcr in (hi\ scction a ver)' siinple case lliiil w i l l fulfill our needs iii tlic early portiiin of the l e x f .
As a first step consider twn rcierence, . r m~d XYZ (Fig. I I .2 I)moving
in such a way l h a i Ihc direction OS the a x e OS ty: d w a y s rctain the same
orientation relative IO XYZ such iis has heen suggeslrd hq the dashed references giving Eiicccssive posilion\ d v v : . Such ii Iiiotioii o l . r y relative io XYZ
i s called rmndnrio,i.
Y
Figure 11.21. Axcc ~ i y are
r translating rel;iti\c io XY/.
Suppow iiow lliilt we have ii vccliir A l t j wliicli varies with time. Now i n
ihe gciirral case. the time wriatioii o S A will depcnd mi from which rcScrcnce
we iire observing the time nriatinii. Fur this reason. UT otten include \uhscripr\
10 identify Ihc refcrence r e l a l i v r 10 which (he time viiriiilioii i\ Liken. l l u s , w e
have ( d A l d l ) , ~and
z ldA/i/f)xy, iis time derivatives o f R iis seen lroni the .SK and
XYZ axes. respectively. Hou' iire these dcrivatiws related for axes ~rv: and XYZ
that are lranslating rclativc to eiicli ollicr'! For this purpose. consider (dA/dljx,L
We w i l l decompose A into components parallel to the . x y axes and s o us h a c
whei-e A , , . 4 , , and /I. arc thc scalar mmponcnts of A a l m g the .ry; axes.
Because x?:. trandatcs rclative to XYZ ( w e Fig. I I.?II, the unit vectors o l . i - r .
SECTION 11.7
which we have denoted as i , j , and k, are constant vectors as seen from XYZ.
That is, whereas these vectors may change their lines of action, they do nor
change magnitude and direction as seen from XYZ and are thus constant vectors as seen from XYZ. We then have, for the equation above:
But A,. A,y, and A. are scalurs and a time derivative of a scalar, as you may
remember from the calculus, is not dependent on a reference of observation.5
We could readily replace (dALdt),, by (dA;dt)xyz, etc., with no change in
meaning--or we could leave off the subscripts entirely for these terms. Thus,
we can say now:
Now consider (dAldt)xsz.Again, decomposing A into components along the
q z axes and noting that i ,j , and k are constant vectors as seen from xyz, we
can conclude that
( I 1.39)
where as discussed earlier we have dropped the xyz subscripts. Observing
Eqs. 11.38 and 11.39, we conclude that
( I I .40)
That is, the rime derivative o f a vector is the same f o r all reference axes that
are rransluting relative to each uther.
Note in the discussion that the fact that the unit vectors of q z were constunt relative to XYZ resulted in the simple relation 11.41. If xyz were rotuting
relative to XYZ, the unit vectors of xyz would not be constant as seen from
X Y Z and a more complex relationship would exist between (dAldt),,
and
(ddldr)~~,,z.
We shall develop this relationship later in the text.
‘Clearly. the time variation of the temperalure T(x,y,z),
a scalar, at any position in the class^
room does not depend on the motion of an observer in the classroom who might he interested in
the temperature at B particular position at a particular time.
SIMPLE RELATIVE MOTION
493
11.8
Motion of a Particle Relative
t o a Pair of Translating Axes
A pair of references .LK and XY7. are shown now i n Fig. 11.22 moving in
translation rclati\z t o eiicli tither. The wlo<.ii? i ' w i o I 01' any particle P
depends 011 thc referencc fr,irii wliicli the ~ntrliiiinic ohserved. More precisely,
wc say that the velocity ol' particle l'relativc 10 refcrcnce X U is the time rate
oichange OS the piisition veclor r Siir this rcfxeiicc. where this rate of change
i\ Yiewed froti? the XYZ refercncc. Thih can h e slated mithematically as
Similarly. ior ~ l i c\:elocity ofpurliclc Pa' x c n froni rclcrcnce xy. we have
wherc u'e iiow iise pmitiim wctnr p firr refcrence .ryz and view the change
from the .rxz. rclcrence (scc Fig. I I .22). Ry thc same tohcn. (dRidr),, i s the
velocity of h e iirigiii O i tlic .n.; rekrcnce as scen froni XYZ. Since all points
of rhc .y; refcrcnce h a w the same velocity rcliirive 10 X Y Z at any time I for
this c i i i e (translation of .XK], u'e can say that (dRldr),y,,is the velociry or ref-
Y
Figure 11.22. Axe..
. \ K we
transl:ning r c l a t i i r 11) XY'Z
From Fig. I 1.22 we can relate position vectors p and r hy thc equation
r = K + p
(11.44)
N o w tahc the lime
rille 11Sclimgc
01thew Yeclor\
:I\
iccn from XYZ. We get
The tcrin on the Ich side of this equiition is VxyL,iis indicated earlier, and we
shall use rlie IIoLiitiiiiiR Sor (dRldt),lr We ciiii replace the last tern by the
SECTION 11.8
MOTION OF A PARTICLE RELATIVE TO A PAIR OF TRANSLATING AXES
derivative (dp/di)xycin accordance with Eq. I I .41 since the axes are in translation relative to each other. But (dpldt),,? is simply yvz,
the velocity of P relative IO .xyz. Thus, we have
,v
En
V,‘
( I 1.46)
By the same reasoning, we can show that the acceleration of particle P
is related to references X Y Z and xv? as followsh:
( I I .47)
Equations I I .46 and 11.47 convey the physically simple picture that the
motion of a particle relative to XYZ is the sum of the motion of the particle
relative to xyz plus the motion of g z relative to XYZ.
It must be kept clearly in mind that the equations which we have developed
apply only to references which have a translatory motion relative to each other. In
Chapter 15 we shall consider references which have arbitrary motion relative to
each other. (Since a reference is a rigid system, we shall need to examine at that
time the kinematics of rigid bodies in order to develop these general considerations of relative motion.) The equations presented here will then be special cases.
How can we make use of multiple references’? In many problems the
motion of a particle is known relative to a given rigid body, and the motion of
this body is known relative to the ground or other convenient reference. We
can fix a reference n)’z to the body, and if the body is in translation relative to
the ground, we can then employ the given relations presented in this section
to express the motion of the particle relative to the ground.
If, in ensuing chapters, we talk about the “motion of particles relative to
a point,” such as, for example, the center of mass of the system, then it will be
understood that this motion is relative to a hypothetical rqfrrence moving
with the center of mass in a translatory manner or, in other words, relative to
a nonrotating observer moving with the center of mass.’
We illustrate these remarks in the following examples.
“s
you
110
douht will anticipate. the acceleration of a panicle as seen from reference XLZ i s
Similarly. we h i v c fororl,;.
’Using a p i n t to convey information ahout relative motion o r a panicle only allows you to
convcy information as to how far or how near the panicle is Io the paint and also as to the rpccd
and rate of change of speed of the panicle toward or away from the point. The imponant information regarding direction i s entirely left out, requiring a reference frame in order to give this
kind of information.
495
.I___
...
..
..___
...
,
-.
.
. ., , . .
-
.I .
~ . .
SECTION I 1.8 MOTION OF A PARTICLE RELATIVE TO A PAIR O F TRANSLATING AXES
Example 11.10 (Continued)
1
Thus, denoting the total force from the airplane as FpIanc.
and remembering
that the gate valve weighs Ib, we have
4
I
F*l',"e
~
zI k
= I (- 44j
~
89.5k3
(e)
wherc - i k is the fbrce of gravity. Solving for Fpl:~~,c
we get
FDlanc
= -.684j - .890k Ib
(f)
Example 11.11
The freighter in Fig. 11.24 is moving at a steady speed V, of 15 k d h r relative to the water. The freighter is 200 m long at the waterline with point
A at midship. A stalking submerged submarine fires a torpedo when the
submarine and freighter are at the positions shown in the diagram. The torpedo maintains a steady speed V, of 40 kmlhr relative to the water. Will
the torpedo hit thc freighter?
V , = 15 km/hr
V2 = 40 km/hr
Freighter length = 200 m
Figure 11.24. A torpedo is fircd toward n freighter. Does it hit or miss'!
A key feature in solving this problem (and othcrs like it) is that we
can readily tell whether there is a hit or a miss and, if there is il hit. exactly
where this takes place. This is done by simply observing the torpedo from a
vantage point of the freighter. The torpedo vclocity relutivr I O the.freighter
(Le., !.he motion seen by an on-board observer) will point directly to the
position of potential contact with the freighter or will indicate a miss.
491
398
CllAPTER I I
K I N l i M A ~ I K ' SOF A PARTICLE-SIM1'I.I
KI:I.ATIVE MOTION
Example 11.11 (Continued)
We accordingly make thc following referencc lincs:
Fix xyz to the freighter
Fix XYZ to the water
This is shown i n Fig. I 1.25. The velocity o f - ~ yand.
. hence the freighter,
relative to XYZ(i.c.,H) is (-15 cos 3O'i + I ? sin XYj) hmlhr. 'The vclocity
of the torpedo relative tu XYZ is 40j kmlhr. We can t l u s;ly
v,,,
v,, + R
=
Figure 11.25. Veliicity v c c t r m arid relerencc? lor the cnpgcrncnt
Hence.
3Oj = Vt~yy
-- I S coc 30"; + I S sin 3O"j
:. V , ! : = IL.YY; + 1 2 . S j kmlhi-
(a)
To just miss the freightcr, thc vclocity veclor of thc torpedo relative to the
freighter, Vlyz,must havc ii coursc such that this vcctoI forms an angle
with the horizontal axis given a\ (see Fig. 11.261
o(,
pa = n + 60" = tall
I
1 0 0 ~+~6(,"
6,000
= h(J.')5"
Now go hack to Eq. ( a ) to ohtain thc actual angle.
f o r the actual relative velocity w c t o r VI:.
p,,,.,
(h)
(sce Fig. I I .25).
Thus we may all relax: the torpedo.just niisses the freightcr m c e p,.,,~..,
>
o,,,
Figure 11.26. Relative vclocily wxtw
V , . fm.juri I n k + the frcightcl-.
11.71. Two wheels rotate about s t a t i o n q axes each at the same
angular velocity, 8= 5 mdJsec. A particle A moves along the spoke of
the larger wheel at the speed V, of 5 fdsec relative to the spoke and at
the instant shown is decelerating at the rate of 3 fdse? relative to the
spoke. What are the velncity and acceleration of particle A as seen by
an observer on the hub of the smaller wheel? What are the velocity
and acceleration of paticle A as seen hy an observer on the huh of the
smaller wheel if the axis of the larger wheel moves at the instant of
interest to the left with a speed of 10 ftJsec while decelerating at the
rate of 2 ftJsecz'?Both wheels maintain equal angular speeds.
11.73. A sled. used by researchers ti1 test man's ahility to perform during large accelerations over extended periods o f time, is
powered by a small rocket engine in the rear and slides on lubncated tracks. If the sled is accelerating at 68, what force does the
man need to exert on a 3-ounce body to give it an acceleration relative to the sled of
30i
+ 2Oj ft/secZ
Figure P.11.73.
Figure P.11.71.
11.72. Four particles of equal mass undergo coplanar motion in
the ,xy plane with the following velocities:
11.74. On the sled of Problem I I .73 is a device (see the diagram)
on which mass M rotates about a horiLontal axis at an angular speed
w of5.000 rpm. If the inclination Oof the arm BM is maintained at
30" with the vertical plane C-C, what is the total force on the mass
Mat the instant it i s in its uppermost position? The sled is undergoing an acceleration of 58. Take M as having a mass of .I5 kg.
v,= 2 d s e c
v, = 3 d s e c
V, = 2 d s e c
V, = 5 d s e c
We showed in Section 8.3 that the velocity af the center of mass
can be found as follows:
A
c
Figure P.11.74.
where is the velocity of the center of mass. What are the velocities of the particles relative to the center of mass?
11.75. A vehicle, wherein a mass M of I Ihm rotates with an
angular speed w equal to 5 radlsec, moves with a speed V given as
V = 5 sin CLI ftlsec relative to the ground with f i n seconds. When
f = I sec, the rod AM is in the position shown. At this instant, what
is the dynamic force exerted by the mass M along the axis of rod
AM if R = 3 radlsec?
Figure P.11.75.
499
11.79. A rackrl ninve? at a ?peed of 700 mlsec and accalciatus
at :i rate of 5~ relatiw to the g w n d refercncc XYZ. The products
of combustion at A leave thc sockct at B spccd of 1.700 mlsec reliiti\,c t o thc rocket and are acczlcrating at the ratc of30 rnlscc' re+
ativc 10 thc rricket. What arc' thc spccd and acceleration of a n
cleincnt o f thc c ~ m b u ~ t i npruducri,
n
a i heen from the ground? Thc
iriichel mow, along ii straight-line path w,hosc dircctirin cosines
fcir thc XYZ reference iirr I = .h and ,n = .6.
L
.
i
.I
x
Figure P.11.79.
X-----Figure P.Il.78.
Figure P.1 I.80.
11.81. A train is moving at a speed of 10 km/hr. What speed
should car A have to just barely miss the front of the train? How
long does it take to reach this position? Use a multireference
approach only.
connected to a massless rod. At the instant shown, o = 2 radlsec
and 0 = 3 radIsec> both relative to the balloon. What force does
the rod exert on the device at this instant? Give the result in vector
and scalar form. The rod is in a horizontal position (see elevation
view) at the instant shown and has a length of .5 m.
Plan view
Elevation view
Figure P.11.83,
Figure P.11.81.
11.82. A Tomahawk missile is being tested for its effect on a
naval vessel. A destroyer is towing an old expendable naval
frigate at a speed of 15 knots. The missile is shown at time f moving along a straight line at a constant speed of 500 miihr, the guidance system having been shut off to avoid an accident involving
the towing destroyer. Does the missile hit the target and if so
where does the impact occur? The missile moves at a constant eleYation of I O ft above the surface ofthe water.
11.84. A submarine is moving at a constant horizontal speed of
I5 knots below the surface of the ocean. At the same time, the sub
is descending downward by discharging air with an acceleration
of ,023g.s while remaining horizontal. In the submarine, a flyball
governor operates with weights having a mass of 500 g each. The
governor is rotating with speed w of 5 radlsec. If at time f, 0 = 30",
8 = .2 radlsec, and d = I rad/sec2, what is the force developed on
the support of the governor system as a result solely of the motion
of the weights at this instant'! [Hint: What is the acceleration of
the center of r n a ~of
. ~the spheres relative to inertial space'?]
15 hots
I * W l f
5W mi/hr
!lA
z
Figure P.ll.82.
11.83. On a windy day. a hot air balloon is moving in a translatory manner relative to the ground with the following acceleration:
a = 2i - 5 j
+ 3k
m/s2
Simultaneously. a man in the balloon basket is swinging a small
device frw measuring the dew point. The device of mass 5 kg is
Figure P.11.84.
501
~
Y
11.90. Mass M of 3 k g rotates about point 0 in an accelerating
rocket in the xy plane. At the instant shown, what i s the force from
the rod onto the mass? Include the effects of gravity if fi = 7.00
mls' at the elevation of the rocket.
t
The helicopter blade i s rotating relative to the helicopter in the following manner at the instant of interest:
o,= 1OOrpm
0,= 10.3 rpmlsec
The blade i s I O m long. What i s the velocity and the acceleration
nf the tip B relative to the ground reference X T L ? Give your
results in meters and seconds. The blade i s Darallel to the X axis at
the instant of interest.
m1s2
n
V
Om
45"
R
-x
X
o = 5 rad/s2
L, =
- V+
2 radlsz
a
z
Figure P.1 1.92.
Z
11.93. A destroyer in rough seas has thc following translational
acceleration as s e w froin inertial relerencc X Y Z when it is firing
its main battery in the Y Z plane:
Figure P.11.90.
11.91. A light plane
a = Sj
i s approaching a runway in a crnss-wind.
This cross-wind has a uniform <peed y, of 33 milhr. The plane has
a velocity component V, parallel to the ground of 70 milhr relative
to the wind at an angle p of 30'. The rate of descent i s such that the
plane will touch down somewhere along A- A. Will this touchdown
occur on the runway or off the runway for thc data given'?
+ 2k
mls'
What must m, and OJ, of the gun harrcl he relative to the ship at
this instant S I , that tip A of the barrel has zero acceleration relative
to XYZ?
-
A
Figure P.11.93.
Figure P.11.91.
11.94. A small clewtor E i n an nccan-going vessel has thc following mntion relative tn the ship:
11.92. A helicopter i s shown moving relative to the ground with
the following motion:
a ,.,,,, = .2gR m/s2
V
= 13Oi
a = IOi
+ 70j + 20k
+
kmlhr
16j + 7k kmlhrls
'The ship has thc following motion relative to nearby land:
a, ,,,,, =
.2i + 3j + .6k mls'
503
11.96. A particlc at position (3, 4, 6) ft at time I(,= 1 sec i s given
a constant acceleration having the value 6i + 3 j ftisec'. If the
velocity at the time I,) i s I6i 20j 5k ftlsec, what i s the velocity
of the panicle 2.0 sec later'! Alsti give the position of the panicle.
+
+
11.97. A pin i s confined to slide in a circular slot uf radius 6 m.
The pin must also slide in a straight slot that moves to the right at
B cmstant speed, V , of 3 mlsec while maintaining a constant angle
of 30' with the horizontal. What are the velocity and acceleration
of the pin A at the instant shown?
11.99. A light linc attachcd to a streamlined weight A i s "shut"
by a line rifle from a small boat C to a large boat D in heavy seas.
The weight must travel a distance of 20 yd horizontally and reach
the larger boat's deck, which i s 20 ft higher than the deck of hoat
C. If the angle a o f firing i s 40", what minimum velocity V,, i s
needed! At the instant of tiring, buat C i s dipping down into the
waler at a speed of 5 ftlsec. Assume that the larger boat remains
essentially fined at constant level.
&--'I
2 0 yd
y
I
Figure P.11.99.
I1.1W. A projectile is fired at an angle of 60" as shown. At
what elemtion ? does i t strike the hill whose equation has been
estimated as y = IO-'.$
m ? Neglect air friction and take the muzle velocity as 1,llOO mlscc.
Figure P.11.97.
I.
11.98. A freighter i s i n w i n g in a river at a speed of 5 knots relative to the water. A small boat A i s moving relative to the water at a
speed of 1 knots in a direction as shown in the diagram. The river i s
moving at a unifbrm sped of .h knots relative ti) the ground. Will
the hoat hit the freighter and, i f so, where will the impact occur?
-
Figure P.11.100.
5 hots
+
+
3.ooo'
4
.6 knots
+
~ , ,: "
Figure P.11.98.
,
,..?
11.101. A proposed space laboratory, in order to simulate gravity,
iotiltes rclative to an inertial reference XTL at a rate wI. For occupant A in the living quarters ti) be comfortable. what should the
approximate value of w , be'! Clearly, at the center, there i s close to
,em gravity for zero-g experiments. A conveyor connects the living
quarters with thc zero-g laboratory. At the instiant of interest, a
package D hei a ?peed of 5 m l s and a iilte of change of speed of 3
m/s2 rchtivc to the Space station, both toward the laboratory. What
arc the axial. transverse, and rddial velocity and acceleration components at the instant of interest relative to the inertial reference?
What are the rectangular cumponents of the acceleration vector?
505
/-------i------.
k
I tl
Id
i>i,ii
Figure P. I 1.104.
Figure
P.II.IUh.
ll.lU7. Pilots of tighter planes war special suits designed til
prevent blackouts during a severe maneuver. These suits tend to
keep the blood from draining out of the head when the head i s
accelerated in a dil-ectiuo from shoulders to head. With this suit, B
flier can take 5 ~ ' so f acceleration in the aforementioned direction.
If a tlicr is diving at a speed of 1,001) kmlhr, what i s the minimum
radius of curvature that he can manage at pullout without suffcrillg had physiological effects'?
11.108. A particle mwes with constant speed uf 1.5 mlyec along
11.110. A mechanical ''aim" for handling radioactive matcriiils
i s shown. The distance Fcan he varied by telescoping action of the
arm. The ann can he rotated about the vertical axis A- A. Finally,
the arm can he raised or lowered by a worm gear drive (not
shown). What i s the velocity and acceleration of the object C if
the end of the arm mvves U U radially
~
at a rate of 1 ftJsec while the
arm turns at a speed w of 2 r a d s e c ? Finally, the arm i s raised at a
rate of 2 ftJsec. The distance T at the instant of interest i s 5 ft.
What i s the acceleration in the direction E = .Xi ff!
+
a path given as x = j2 In y m. Give the acceleration vector of the
particle in terms o f rectangular components when the partick i s at
position y = 3 m. Do the problem by using path coordinate techniques and then by Cartesian-component techniques. How many
x ' s of acceleration i s the particle subject to'!
L
~
I
A
ll.109. A submarine i s moving in a translatory manner with
the following velocity and acceleration relative to an inertial refercnce:
V
=
hi
+ 7.5j + 2k
knots
a
=
.2i
~
.24j
+ .X?k knotils
A device inside the submarine consists of an arm and a mdSb at die
end of the arm, At the instant of interest, the arm i s rotating in a
vertical plane with the following angular speed and angular acceleration:
w = It) r a d s
dJ = 3 rad/s*
The ann i s vertical at this instant. Thc mass at the end of the m d
may he ctrnsidered t o he a particle having a mass of 5 kg. What
are the velocity and acceleration vectors for the motion of the particle at this instant relative to the inertial reference? Use units of
meters and seconds. What must he the force vector from the arm
onto the particle at this instant?
x
/ "
Figure P.ll.llO.
11.111. A top-section view of a water sprinkler i s yhown. Water
four pas-
cnteis at thc center Srom helaw and then goes thrvugh
~agewayiin an impeller. The impeller is rotating at constant speed
w of 8 rpm. As xzii from the impeller. the water ICBVCS
at a speed
o f I O ft/sec at an angle uf 3U" relative to r. What is the velucity
and acceleration as seen from the ground of the water as i t leaves
the impeller and becomes free of the impeller? Give results in the
radial, axial, and transverse directions. Use one reference only.
Y
I
Figure P.11.109.
Figure P . l l . l l l .
507
11.112. A luggdgc dispcnscr at ;maiiyort rcxmblch a pyramid with
iix flat scgrncnts as sidcs as \liowri in the diagram The S Y ~ ~ ~ IOLLLI~S
ITI
with ail angular spced w u l 2 rpm. Luggagc is dropped from above
ind slidzb duwii lhc lilccs t u bz pickcd up by I w c I e r s at the ~ J W
A piecc (if luggage IS shmw on a f x r . It ha\ j u c t heeri
Jropped at the position indicaled. If has at thir instant ieru vclocity as \een from the rotating face but has at this instmt and thereiftci an acceleration crf .2g along the f i c c . What i s the tovill
iccrlcriltiun, as seen from the ground, of the l u ~ ~ n ea>
t .i t r c a c l i c ~
he hasr
ill H'I
Use one reference only.
Figure P.11.113.
11.1 14. A jet uf wiltet Ihiu 3 s p e d ai the n o u l e of 20 mi\. At
what position dues i t lhit thc parabolic hill? What i s i t & speed at
lliilt puiril'! Dc IOI include I l i c t i w i .
Sidc view
Figure P.11.112.
A landing craft i s in the pn,ce\r 01 landing o n M:
i.herr the accclcratmn of gravity i s . 3 X S times that 01 the c a
I'he craft has thc followmg :iccelerntiiin relative 10 the landing
11.11
i u r f k x at the instant 01interest:
a
=
.2gi
i .4gj
~
2xk rni\cc'
1s the :icceleretirm 01gravity o n thc with. At this ~iisliliit.
astnrnaut is raising a harid c i m c w weighing 3 N on the earth. I1
he i \ giving the ciliiirril an upward :I
Icrution 0 1 3 mdscc' relalive tu the landinp craft, whiit forcc i i i u ~the
t igtimnilut cxcrt (in [he
:aniera at the instant of intercct'.'
where fi
XI
iO8
11.119. A tube, must of whose centerline is that of an ellipse
given as
?2
?I
-+--I
1.8’
.122
Figure P.11.116.
*11.117. A particle has a variable velocity V(t) along a helix
wrapped around a cylinder of radius e. The helix makes a constant
angle a with plane A perpendicular to the z axis. Express the
acceleration a of the particle using cylindrical coordinates. Next,
express 6, using cylindrical unit vectors and note that the sum of
the transverse and axial components of a (lust computed) can be
given simply as VE,. Next, express the acceleration of the particle
using path coordinates. Finally, noting that E , , = --E-, show that
the radius of curvature is given as R = ?-cos2 a
has a cross-sectional diameter D = 100 mm. The tube has the following rotational motion at the instant of interest:
w = . I 5 radis
W = ,036 radls’
Water is flowing through thc tube at the following rate at the
instant of interest:
Q = . I 8 Lis
Q = ,025 L/s’
The tube is in the vertical plane at the instant of interest. What is
the acceleriltion of the water particles at the centerline of the tube
at point C using cylindrical coordinates and cylindrical components’! Assume over the cross-section of the tube that the water
d o c i t y and acceleration are uniform.
’I’
Figure P.11.119.
Figure P.11.117.
11.118. An eagle is diving at a constant $peed of 40 ftlsec to
catch a IO-ft snake that is moving at a constant speed of 15 fl/sec.
What should ff be so that the eagle hits the small head of the
snake‘? The eagle and the snake are moving in a vertical plane.
11.120. A World War I fighter plane is in level tlight moving at
a speed of60 k d h r . At time lo it has an acceleration given as:
a = .2gi
~
3gj
+2
k m/s2
Also at this time, the co-pilot is raising a camera upward with an
acceleration of 0. IRrelative to the plane. If the camera has il m a s
of .01 kg, what forcc must thc co-piloi exert on the camera to give
it the desired motion at time t,,? Note that the plane never rotates
during this action. Take g = Y.81 d s ’ .
X’
Figure P.ll.118.
Figure P.II.120.
509
Particle
Dynamics
12.1
Introduction
In Chapter 1 I , we examined the geometry of motion-the kinematics of
motion. In particular, we considered various kinds of coordinate systems: rectangular coordinates, cylindrical coordinates, and path coordinates. In this
chapter, we shall consider Newton’s law for the three coordinate systems
mentioned above, as applied to the motion of a particle.
Before embarking on this study, we shall review notions concerning
units of mass presented earlier in Chapter 1. Recall that a pound mass (Ibm)
is the amount of matter attracted by gravity at a specified location on the
earth’s surface by a force of I pound (Ibf). A slug, on the other hand, is the
amount of matter that will accelerate relative to an inertial reference at the
rate of I ft/secz when acted on by a force of 1 Ibf. Note that the slug is
defined via Newton’s law, and therefore the slug is the proper unit to be used
in Newton’s law. The relation between the pound mass (Ibm) and the slug is
M (Ibm)
(12.1)
M (slugs) =
32.2
Note also that the weight of a body in pounds force near the earth’s surface
will numerically equal the mass of the body in pounds mass. It is vital in
using Newton’s law that the mass of the body in pounds mass be properly
converted into slugs via Eq. 12. I
In SI units, recall that a kilogram is the mass that accelerates relative to
an inertial reference at the rate of I meter/sec2 when acted on by a force of I
newton (which is about one-fifth of a pound). If the weight W of a body is
given in terms of newtons, we must divide by 9.81 to get the mass in kilograms needed for Newton’s law. That is,
(12.2)
M ( k g ) = W (N)
9.81
We are now ready to consider Newton’s law in rectangular coordinates.
~
511
Part A:
12.2
Rectangular Coordinates:
Rectilinear Translation
Newton's l a w for Rectangular
Coordinates
In iect,ingul,ii c i i ~ i i c I i n ~ i l cwc
\ ciin expic\\ Ncuton
\
I,\\\ d\ tollow\
If the notion i s hnown rclalivc IC) mi incrtiill rclercncc. we can easily
~ I v for
c the rcckingulai- ciiiiiprincnts oc (lie r c s t t l ~ i i nforce
~
on the palticlc.
The eqoatii)ns IO he s o l v e d arc j u s 1 algebraic cqiiations. The iiii.er.re o f this
problem. wherein the forces iirc known ovcr ii l i m e inlcrviil and thc motion i s
desired during this iiitcr\~~iI,
i s no1 s o himplc. For the in\,crsc case. wc tnu\t
get inwilved generally with intcgrxtiiin procedures.
I n the next section. w e s l i i i l l considcr siiuatioiis in which the resultiiiit
lorce on a piisiiclc hiis thc \amc dirccliiin and line 0 1 d o n tit all times. Thc
resulting niotiiiii i s then confined to ii sll-ai:hc line and i s usually called r w l i ~
llmYrr fr',~l.~l~~fio~~.
12.3
Rectilinear Translation
For reclilinciir tran&lalion. wc ma) considcr llic line of action 01the tnotion to
be collinew with one a x i s 01a rectilinear coordiiiale \yhlein. Newton'\ law i \
then one of the cquatiiins (11 the set 12.3. Wc shall use the .r axi.; to coincide
with the line 01 iiclion (11the nici1ioii. The reculinnt lorce I- (we shall not
bother with the .t siihxript hcrc) can hc a conslaiit. a functioii of time, ii function 01 speed, ii luticlion o f piisitioii. or any comhinalion of these. At thic
time, u'c shall enamine some 0 1 these c a e s . leiiving others to Chapter 19.
where. with the aid of the \tudcnts' knowlcdgc (if differential equations,' wc
s h a l l he iniorc prcparcd l o crinsidcr Ihcni.
Case 1. Force Is a I'unction ofl'imc or a Constant. A piirticlc of mass t,z
acted oii by ii limc-\arying force F ( r ) i s shown i n Fig. 12.1. The plane on
which the hod? iiio\cc\ i\ lrictiiinlcss. Tlic Ihrcc 01 gravity i s equal and opposite
SECTION 12.1 RECTILINEAR TRANSLATION
Figure 12.1. Rectilinear translation.
to the normal force from the plane so that F(t) is the resultant force acting on
the mass. Newton’s law can then be given as follows:
Therefore,
d2.x
-
dr2
F(t)
m
~~
(12.4)
Knowing the acceleration in the x direction, we can readily solve for F(f).
The inverse problem, where we know F ( t ) and wish to determine the
motion, requires integration. For this operation, the function F(t) must be
piecewise continuous? To integrate, we rewrite Eq. 12.4 as follows:
Now integrating both sides we get
where C, is a constant of integration. Integrating once again after bringing dt
from the left side of the equation to the right side, we get
(12.6)
We have thus found the velocity of the particle and its position as functions of time to within two arbitrary constants. These constants can be readily
determined by having the solutions yield a certain velocity and position at
given times. Usually, these conditions are specified at time t = 0 and are
then termed initial conditions. That is, when t = 0,
V
=
V, and
.x = xo
(12.7)
These equations can be satisfied by substituting the initial conditions into
Eqs. 12.5 and 12.6 and solving for the constants C, and C,.
Although the preceding discussion centered about a force that is a function of time, the procedures apply directly to a force that i s a constant. The
following examples illustrate the procedures set forth.
>That is, the function haa only a finite number d finice discontinuities.
5 13
5 14
CHAPTER 12 PARTICLE DYNAMICS
Example 12.1
A 100-lb body is initially stationary on a 45" incline as shown in Fig.
12.2(a). The coefficient of dynamic friction pd between the block and
incline is .5. What distance along the incline must the weight slide before
it reaches a speed of 40 fthec?
A free-body diagram is shown in Fig. 12.2(b). Since the acceleration
is zero in the direction normal to the incline, we have from equilibrium
that
100~0~4
= 5N~= 70.7 Ib
(a)
Now applying Newton's law in a direction along the incline, we have
Therefore,
2
= 11.38
Rewriting Eq. (b) we have
d ( 2 ) = I1.38dt
Integrating, we get
2
+ C,
- = I1.38f
t2
s = 11.38-
2
+ C,t + C,
(C)
(4
When f = 0, s = d d d t = 0, and thus C, = C, = 0. When d d d i = 40
ft/sec, we have for t from Eq. (c) the result
40 = I1.38t
Therefore,
f
= 3.51 sec
Substituting this value off in Eq. (d), we can get the distance traveled to
reach the speed of 40 ftlsec as follows:
Figure 12.2. Body slides on an incline.
SECTION 12.3 RECTILINEAR TRANSLATION
rn
Example 12.2
"
A charged particle is shown in Fig. 12.3 at time f = 0 between large parallel condenser plates separated by a distance d in a vacuum. A time-varying voltage V (notation not to he confused with velocity) given as
V = 6 sin ut
(a)
is applied to the plates. What is the motion of the particle if it has a charge
q coulombs and if we do not consider gravity'?
As we learned in physics, the electric field E becomes for this case
The force on the particle is qE and the resulting motion is that of rectilinear translation. Using Newton's law we accordingly have
dZx
z =
q
6sinot
m
d
dx
6 sin wt
df
x)
= 4 7-
Integrating, we get
d x - -- 6q c o s w t + c ,
dt
wmd
Applying the initial conditions x = b and dx/dt = 0 when f = 0, we see
that C, = hq/mwd and C, = b. Thus, we get
The motion of the charged particle will he that of sinusoidal oscillation in
which the center of the oscillation drifts from left to right.
Case 2. Force Is a Function of Speed. We next consider the case where
the resultant force on the particle depends only on the value of the speed of
the particle. An example of such a force is the aerodynamic drag force on an
airplane or missile.
We can express Newton's law in the following form:
~dV
~
dr
F(V)
m
-
Figure 12.3. Charged particle
(C)
Rewriting Eq. (c). we have
d(
y
(12.8)
between condenser plates.
5 15
516
CHAPTEK I? PAKTICLL DYNAMICS
where F ( V ) is a piecewise continuous function reprcsenting the force in the
positive x direction. If we rearrange the equation in the following manner
(this is called separnrion of vuriuhlrs):
we can integrate to obtain
(12.9)
The result will give I as a function of V. However, we will generally prefer to
solve for V in terms o f f . The result will then have the form
v
= H(t,
c,)
where H is a function of t and the constant of integration C,. A second integration may now he performed by first replacing V by dxldt and bringing dt
over to the right side of the equation. We then get on integration
x =
j H ( r , C,
rir
+ c2
( 12. I O )
The constants of integration are determined from the initial conditions of the
problcm.
rn
Example 12.3
A high-speed land racer (Fig. 12.4) is moving at a speed of 100 mlsec. The
resistance to motion of the vehicle is primarily due to aerodynamic drag.
which for this specd can he approximated as .2V2 N with V in mlsec. If the
vehicle has a mass of 4,000 kg, what distance will it coast before its speed
is reduced to 70 mlsec?
We have. using Newton's law for this case.
1
SECTION 12.3 RECTILINEAR TRANSLATION
Example 12.3 (Continued)
Integrating, we have
- _I - -5 x lo-% + c,
V
(c)
Taking f = 0 when V = 100, we get C, = -1/100. Replacing V b y dx/dt,
we have next
df
5 x IO-%
V - & =
I
+100
(4
Separating variables once again, we get
dt
- dx
5 x 10-5t + (1/100)
To integrate, we perform a change of variable. Thus
lJ = 5 x IO-% + (1/100)
:.
dlJ = 5 x Io-sdf
We then have as a replacement for our equation
3 = 5 x 10-3 dx
II
Now integrating and replacing lJ, we get
(
)'
= 5 x lo-%+
In 5 x 1 0-5 f + l m
c,
When t = 0, we take x = 0 and so C, = In (l/lOO), We then have on
combining the logarithmic terms:
ln(5 x
+ I ) = 5 x 10-5.r
(e)
Substitute V = 70 in Eq. (d); solve for f. We get t = 85.7 sec.
Finally, find x for this time from Eq. (e). Thus,
In[(5 x 10-3)(85.7) + I] = 5 x l 0 - h
Therefore,
The distance traveled is then 7.13 km.
5 17
518
CHAPTER I 2
I'AKI'ICLb; IUYNAMICS
Example 12.4
A conveyor i s inclined 20" from the horizontal ah shown in Fig. 12.5. As a
i-esult o l spillagr ciluil (in l l i z belt. Ihcrc i s a viscous friction Scirce hctwzen
body 11 and the helt. This foi-ce equals I).
I Ihf per unit relative velocity
between body L, and thc hell. Thc helt moves at il conhtant speed VNup the
conveyor while inilially hrdy I ) hac a speed (V,,),, = 2 Il/scc relative 10
the ground i n :I direction doum the conveyor. What \peed V;, should the
belt have in order for hody 1) l o be ahlc to cvciitually approach a Lero
velocity rclatiw t o tlic ground? For hell speed VI,. and Sor the giuen initial
speed o l body 0, namely IV,,),, = Z ftlsec, detcmiinc the time when body
1) attains a !,peed of I ftlsec relative to the ground. l h c mass of ll i s 5 I b m
I.'iFure 12.5. A I h d y \lides ilvwn a convcyoi~hell
WCI
with oil.
We hegin hy assigning axes for the prohlem a s I i l l i i u , s (see Fig. 12.6):
Y
Figure 12.6. Friction force f
hrtwzcn hody D and the hrlr.
From kinematics we can say
15
0.1 limes the relati\c vclority
SECTION 12.3 RECTILINEAR TRANSLATION
Example 12.4 (Continued)
For the friction forcefwe have
f
=
-(.MVD)qz
=
-(.lW,
+ VB)xni
We may now use Newton's law for body D in the x direction. Since
all velocities from here on will be relative to the ground, we can dispense
with the reference subscripts. Thus
When body D attains a theoretical permanent zero velocity relative to the
ground, V, and dV, are equal to zero. This gives us (noting that V, now
dr
becomes V i )
Now determine the time for body D to attain a velocity of 1 ft/sec
relative to the ground for a belt speed of 17.10 ft/sec. For this we go back
to Eq. (a).
dVD
dt
:.
- (-.l)(V, + 17.10) + 5sin20'
In V, = - -r3.22
5
=
-.lVD
+ C,
When
f = 0,
V, = 2 ftlsec,
.:
Ct = l n 2
Hence, on combining log terms3
In
[$1
=
-.644r
Set V, = 1 ftlsec. Solve for f .
f
'
= --]
n(S00) =
,644
'Note from this equation that V , = 2e4.M4' and that VD = -1.288e4.w' and so we
see that as t approaches infinity both of these quantities approach zero. Thus, theoretically
body D could appmach a permanent zero velacity relative to the gmnnd.
5 19
Case 3. Force Is a Function of Position.
As the final case of this series.
we now consider the rectilinear motion of a body under the action of a force
that i s exprcssihlc cis ii function of pocilion. Pcrhaps the simplest example of
such a case i s Ihc frictionless miss-spi-ing system shown in Fig. 12.7. The
body i s shown at a position where the spring i s unstrained. The horizontal
force froni the spring at d l positions of the body clcarly w i l l be a function of
position .c,
Figure 12.7. Mn\s-spring systcm.
Nrwfi,,!',T In),. for position-dependent forces can hc giber] as
We caiiiiol scpur;ilc the variables for this form o l thc cquiltion as in previous
c:iscs since there are three variables (V. f. and.r). However, by using the chain
rulc of differentiation. w e can change thc left side of the equation 10 a inore
desirable tnrrn in the Ibllowinp manncr:
We ciin now scparate the variable? in
Eq. 12.1 I as follows:
r17VdV = F ( s ) r l . r
Intcgrating. we pet
(12. I ? )
Solving 1,r Vand using rl.rldt i n i t s pliicc, we get
SECTION 12.3 RECTILINEAK TRANSLATION
Separating variables and integrating again, we get
For a given F(x), V and x can accordingly be evaluated as functions of time
from Eqs. 12.12 and 12.13. The constants of integration C, and C, are determined from the initial conditions.
A very common force that occurs in many problems is the lineur restoring,forct!. Such a force occurs when a body W is constrained by a linear spring (see Fig. 12.7). The force from such a spring will be
proportional to x measured from a position of W corresponding to the
undefiirmed configuration of the system. Consequently, the force will have
a magnitude of IKxI, where K, called the .spring consfunf, is the force
needed on the spring per unit elongation or compression of the spring. Furthermore, when x bas a positive value, the spring force points in the negative direction, and when x is negative, the spring force points in the
positive direction. That is, it always points toward the position x = 0 for
which the spring is undeformed. The spring force is for this reason called a
restoring force and must be expressed as -Kx to give the proper direction
for all values of x .
For a nonlinear spring, K will not be constant but will be a function of
the elongation or shortening of the spring. The spring force is then given as
SpmE
(12.14)
In the following example and in the homework problems, we examine
certain limited aspects of spring-mass systems to illustrate the formulations
of case 3 and to familiarize us with springs in dynamic systems. A more
complete study of spring-mass systems will be made in Chapter 19. The
motion of such systems, we shall later learn, centers about some stationary
point. That is, the motion is vibratury in nature. We shall study vibrations in
Chapter 19, wherein time-dependent and velocity-dependent forces are present simultaneously with the linear restoring force. We are deferring this
topic so as to make maximal use of your course in differential equations that
you are most likely studying concurrently with dynamics. It i s important to
understand, however, that even though we defer vibration studies until later,
such studies are not something apart from the general particle dynamics
undertaken in this chapter.
521
522
CHAPTER 1 2 PARTICLE DYNAMICS
Example 12.5
A cart A (see Fig. 12.8) having a mass of 200 kg is held on an incline so as
to just touch an undeformed spring whose spring constant K is SO N/mm.
If body A is released very slowly, what distance down the incline must A
move to reach an equilibrium configuration'? If body A is released suddenly, what is its speed when it reaches the aforementioned equilibrium
configuration for a slow release'!
1
u
-1
Figure 12.8. Cart-\pring \y\tem
As a first step, we have shown a free body of the vehicle in
Fig. 12.9. To do the first part of the problem, all we need do is utilize the
(200)'(Y
xI
Figure 12.9. Free-hody diagram of cart
definition of the spring constant. Thus, if. 6represents the compression of
the spring, we can say:
SECTlON 12.3 RECnLlNEAK TRANSLATION
Example 12.5 (Continued)
Therefore,
Thus, the spring will be compressed ,01962 m by the cart if it is allowed to
move down the incline very slowly.
For the case of the quick release, we use Newton’s law. Thus, using
x in meters so that K is (SO)( 1,000) N/m:
200x = (200)(9.81)sin 30” - (5O)(l,OOOJ(x)
Therefore,
f
= 4.905 - 250x
Rewritingi, we have
dV
V;
i
;= 4.905 - 250x
Separating variables and integrating,
-v_2 - 4.9051
2
~
1 2 5 ~ ’+ C,
To determine the constant of integration C , , we set x = 0 when V = 0.
Clearly, C; = 0. As a final step, we set x = ,01962 m and solve for V.
V = {2[(4.905)( ,01962) - (125)( .01962)’]\”
’
The following example illustrates an interesting device used by the U S .
Navy to test small devices for high, prolonged acceleration. Hopefully, the
length of the problem will not intimidate you. Use is made of the gas laws
presented in your elementary chemistry courses.
523
524
CHAPTER 1 2 PARTICLE DYNAMICS
Example 12.6
An air gun is used to test the ability of small devices to withstand high prrr
longed accelerations. A “floating piston” A (Fig. 12.10).on which the device
to he tested is mounted, is held at position C while region I> is tilled with
highly compressed air. Region E is initially at atmospheric pressure hut is
entirely sealed from the outside. When “fired,” a quick-release mechanism
releases the piston and it accelerates rapidly toward the other end of the gun,
where the trapped air in E “cushions” the motion so that the piston will begin
eventually to return. However, as it starts back. the high pressure developed
in E is released through valve F and the piston only returns a short distance.
Suppose that the piston and its test specimen have a combined mass
of 2 Ihm and the pressure initially in the chamber D is 1,000 psig (above
atmosphere). Compute the speed of the piston at the halfway point of the
air gun if we make the simple assumption that the air in D expands according t o p = constant and the air in E is compressed also according to p t > =
~ o n s t a n t .Note
~ that I ’ is the specific volume (Le., the volume per unit
mass). Take / ‘ o f this fluid at D to he initially ,207 ftg/lbm and oin E to he
initially 13.10 ft’llbm. Neglect the inertia of the air.
The force on the piston results from the pressures on each face, and
we can show that this force is a function o f x (see Fig. 12.10 for refercnce
axes). Thus, examining the pressure plj first for region D , we have, from
initial conditions,
(pf,~’fj=
) c [ci,cno
,
+ 14.7)(144)](.~07)
=m
oo
(a)
Furrherrnore, the mass of air D given as MI, is determined from initial data as
where (V,,), is the volume of the air in D initially. Noting that p7) = const.
and then using the right side of Eq. (a) for pf,tiL, as well as the first part of
Eq. (h) for cD, we can determine pf, al any position x of the piston:
‘YOU should r ~ dliom
l your earlier work 10 physics and chemistry thilr “U we using 11em
the isothermal form of the quiltion of state for a perfect gas. Two factors of caution should be
pointed out ~ ~ I a t i vtoethe use of chis expression. First. at the high pressures involved in p‘uf ofthe
expansion, the pedcct gas m o d i is only an approximation for the gas. and so the yuarion ol state
of a perfect gas that gives uspri = Consrant is only apprurimale. Fnnhemorr. the assumption 01
isotherm4 expansion gives only an approximatim of the actual process. Perhaps il better approximation is to ilssumr an adiaharic expansion (i.e. no heat wmler). This is dune in Prohlem 12.130,
Val
Figure 12.10. Air gun.
SECTION 12.3 RECTILINEAR TRANSLATTON
Example 12.6 (Continued)
We can similarly get p, as a function of x for region E . Thus,
( p E u E ) o= (14.7)(144)(13.10) = 27,700
and
Hence, at position x of the piston
27,700 - 27,700 27,700
vE
V E / M , - (rr/4)(Iz)(50 - x)/2.88
PE=---
Therefore,
101,600
PE
=
5O-x
Now we can write Newton’s law for this case. Noting that V without
subscripts is velocity and not volume,
M V -dV
zl2
--
50 - x
(4
where M is the mass of piston and load. Separating variables and integrating, we get
M V Z = $[293,00OInx+ 101,6001n(50-x)]+C,
2
To get the constant C,, set V = 0 when x = 2 ft. Hence,
C, = -
$ (293,000 In 2 + 101,600In 48)
Therefore,
C, = 468,000
Substituting C, in Eq. (e), we get
(6) { ~ [ 2 9 3 , 0 0 0 l n x + 1 0 1 , 6 0 0 l n ( 5 0 ~ x ) ] - 4 6 8 , 0 0 0
In
V =
We may rewrite this as follows noting that M = 2 Ibdg:
V = 566[23 In x
+ 7.98 ln(50 - x) - 46.81”’
At x = 25 ft, we then have for V the desired result:
V = 566(231n 25 + 7.98 In 25 - 46.8)”’
(e)
525
526
CHAPTER I ?
PARTICLE DYNAMICS
*Example 12.7
A light stiff riid i s pinned at it and i s constrailled by two linear springs.
K, = 1.000 Nitn and K?
- = 1.2(X) Nim. The spring are urirtretched when the
rod i s horimntal. At the right end o f the rod. ii mass M = 5 kg i s attached. If
Ihc rod i s rotated 12" doc.kwi.sc front a horizontal ciintiguration and thcn
released, what i s the spccd of the mass when the rod returns to a position corresponding to the .s/& ~ ~ q ~ r i l i h r iposilion
i r n ~ with miss M attached'!
Fieure
weightless
A free-hody diagram 01 the systcin for pi.si/iw H i s shown
i n Fig. 12.12(;1) end a free-hody diagram o f the pailicle M i s shown i n
F . mI
F.B.U. I I
(a,
(hl
Figure 12.12. Frcc-body iliagrams of thc system and the pnniclr il,r positive c)
Fig. I2.12(b). The spring forces
for small positive ro1iitions H :
4
=
F;
aiid F,
011
tlic rod arc givrn as follows
( . 3 ) ( H ) ( K ,=
l 3000N
F2 = - ( . I ) ( B ) ( K ? )= - 1 2 0 O N
(ai
where 8 is in radians. I n the first free hody. we w i l l think 01 the rod as a
illassless perfectly irigid Icvcr iih sludied i n high school or perhaps even
SECTION 12.3 RECTILINEAR TRANSLATION
Example 12.7 (Continued)
earlier. Then we can say for the forces on the secund of our free bodies
stemming from the springs5
(fromF;) =
-4
= -300ON
1
(fromFz) = ? ( F , ) = 4 O N
F;
F;
We can now give Newton's Law for M as follows using y for the
vertical coordinate of the particle:
5 y = -5g-300e-40e
... j i =
- g - 3 4 50
e
(b)
Next, from kinematics, we can say for small rotation
Now going back to Eq.(b) we replace y by
in order to be able to separate variables. Also replace O by yl.3. We then
may say
ydy=(-227y-g)dy
Integrating
2
When O = - 12" = -(&)(2n)
rad = - ,2094 rad,8 = y = 0. We can
360
then solve for the constant of integration using y = .3 8.
L
J
Hence,
SNote that a positive B gives negative values for F ; and F; on M and vice versa. It is for
this reason that we require the minus signs.
521
528
C H A PTER 1 2 PAKTICL~.U Y N A M I ~ S
Example 12.7 (Continued)
For the stiitic equilihrium configuratior of the rod. we require firoin
Fig. 12.12(al
Suhstituting values irim Eqs. (a) and noting that we are only using the
magnitudes of the forces ahiive ii)r the required ncSativc ino~iiciitswc get
-(SI(9,Xl)(.3)
-
(300B,/ j(.3) - ~ 1 2 0 B , ~ , l l ~=
. l l0
Solving for (I,,(,
BE</ = -. 1-1-13 rad
Hcncc
y
,,,, = (.3)(-.
1443) = -.0432x
Now gii to Eq. ( d j and suhstitutc :,/.
y,2<#=
,r[(-l13.5)(-.0432X)'
111
Wc get
112
-
('~.X1~(-.04328)- ,16831
The desired result i s then
YE. = 0.968 m / s
._
12.4
..
.. .
A Comment
111 Part A . we haw considcrcd only rccliliiiwr inotiiins iif particles. Actu;rlly
i n Chapter I I, we coiisidercd the coplanar tiiotiiiii of particle having a ciinstili11 acceleriition ( 1 1 grabity i n tlic minus :
dircctiiin nnd x r o ;rcceleration iii
the I. direction. These were the hulli.sfic. prohlcms. We treated them earlicr i n
Chapter I I because rhc considcl-;itioiis were priniarily kinematic i n iiature. In
[lie p r e x i i t chaplcr. thcy ciirrcspond to the coplanar niotiiiti 0 1 a particle having ii constilnl iorce in the ininus :direction ;rlong with an initial velocity
coniponent iii this direction, plus a zero tioire i n thc I. dircctiiin, with a possible initial veliicity compiinent i n (his dircctioii. Therelure. i n Ihc cimtcxt ( i t
Chapter 12 we would hn\c intcgratcd two scalar equations of Newton's law
i n rectangular ciimpiinents (Hq\, 12.3) lor ii single particlc. The resulling inw
tion i s soinctiiiics called <.urvilim,,ir traiislation.
12.1. A particle of mass 1 slug is moving in a constant Sorce
field given as
F = 3i+ l 0 j Sklb
The particle starts from rest at position (3, 5 , -4). What is the
pasitian and velocity of the particle at time I = 8 sec? What is the
position when the particle is moving at a speed of 20 ftlsec?
12.6. Do Problem 12.5 with the belt system inclined 15" with
the horizontal so that end B is above end A.
~
12.2. A particle of mass m is moving i n a constant force field
given as
F = 2mi
~
I2mjN
12.7. A drag racer can develop a t u q u e of 200 Ft-lb on each of
the rear wheels. If we assume that this maximum torque is maintained and that there is no wind friction, what is the time to travel
a quarter mile from a standing start'? What is the speed of the vehicle at the quarter-mile mark? The weight of the racer and the driver
combined is 1,600Ih. For simplicity, neglect the rotational effects
of the wheels.
Give the vector equation for r(r) of the panicle if, at time r = 0, it
has a vzlocity y) given as
y , = hi +
Also, at time
I
I2j
+
3k misec
= 0. it has a position given as
ri, = 3i
+ 2 j + 4k m
Figure P.12.7.
What are coordinates of the body at the instant that the body
reaches its maximum height, yn,,,'!
12.3. A block is permitted to slide down an inclined surface.
The coefficient of friction is .05. If the velocity of the block is 30
ftJsec on reaching the bottom of the incline, how far up was it
released and how many seconds has it traveled?
12.8. A truck is moving down a 10" incline. The driver strongly
applies his brakes to avoid a collision and the truck decelerates at
the steady rate of I d s e c 2 . If the static coefficient of friction p,
between the load Wand the truck trailer is .3, will the load slide or
remain stationary relative to the truck trailer? The weight of W is
4,500 N and it is not held to the truck by cables.
n
Figure P.12.3.
12.4. An arrow is shot upward with an initial speed of 80 ftisec.
How high UP does it go and how long does it take to reach the
maximum elevation if we neglect friction'?
12.5. A mass D at I = 0 is moving to the left at a speed of .6
mlsec relative to the ground on a belt that is moving at constant
speed to the right at 1.6 misec. If there is coulombic friction p~
sent with N,, = 3,how long does it take before the speed of D relative In the belt is .3 mlsec to the left?
A
Figure P.12.8.
12.9. A simple device for measuring reasonably uniform accelerations is the pendulum. Calibrate Oof the pendulum for vehicle
accelerations of 5 St/sec2, 10 ftlsec2, and 20 f!Jsec2. The bob
weighs I Ih. The bob is connected to a post with a flexible string.
B
Figure P.12.5.
Figure P.12.9.
529
12.10. A piston is bcing moved through a cylinder. The piston is
moved at a canstant speed ol .6 mlsec relative to the ground hy
a force F,The cylinder is free to move along the ground on small
wheels. There is a coulombic friction fbrce between the piston and
the cylinder such that p,j = .3. What distance d must the piston
move relative to the ground to advancc .01 m along the cylinder if
the cylinder is stationary at thc outset? The piston has a mass of
2.5 kg and the cylinder has a mass 5 kg.
y,
iILd = .4
30
', F -
IOON
Figure P.12.14.
12.15 A block A of niass M is heing pullcd up an incline by a
force F. If p,j is 3,at what angle a will the force F cause the maximum steady accrlrrafirm'?
Figure P.12.10.
12.11. A force F of 5,000 N is suddcnly applied to mass A .
What is the speed after A has moved . I m? Mass B is a triangular
block of uniform thickness.
&f4= 20 kg
+d
F
=
=
.3
5,000 N
Figure P.12.15.
Figure P.12.11.
12.12. A fighter plane is moving on the ground at a speed of 350
k d h r when the oilot deolovs the brakine oarachutc. How far does
the plane move to get down to a speed of 200 km/hr'! The plane has
a mass of 8 Mg.The drag is 27.SV2 with Vin m/s ( 1 A4g = IO'kg).
L
I
- I
to hody B whose mass is 15 kg.
12.16. A IO-kN force is applied
..
Body A has a mass of 20 kg. What is the speed of B after it moves
3 m? Take p(, = .2X. The center uf gravity of body A is at its geometric center
12.13. Blocks A and A are initially stationary. How far does A
move d u n g B if A moves .2 m relative 10 the ground'
Figure P.12.16.
12.17. A constant fnrcc F is applied to the hody A when 11 is in
thc position shown What should F be if A is to attain a velocity of
2 mli after moving I m'!The spring is unstretched at the position
shown.
v
K
I
,.'
,A
Figure P.12.13.
12.14. A 30-N block at lhe vmition \how" ha* a furce t = Io0 N
applied suddenly. what is i t q velocity after moving I m? Also, how
far does the block move before st~pplllg?Member AB weighs 200M.
530
x
Figure P.12.17.
K = 5,000 Nlm
W, = 480 N
IiL = 775 N
u , = 36
12.18. Two slow moving steam roller vehicles are moving in
opposite directions on a straight path. They start at A and B at the
time f = ,O. How far from point A do they pass each other? What
are their speeds when this happens? [Hint: Show that the time for
this is 1.5 hours.] Note f is in hours.
22,695 km
L
r
A
V, = 6f i.q'?;
~
-
5 sec
VB
'A
+ 3 kmlhr
V, = 5
Force
+ t2" + 0.Sr"'
10sec
Figure P.12.21.
8
kmihr
Figure P.12.18.
12.19. As you learned in chemistry, the cueficirnt ofviscosity g
is a measure, roughly speaking, of the "stickiness" of a fluid. To
measure this property for a highly viscous liquid-like oil, we let a
small sphere of metal of radius R descend in a container of the liquid. From fluid mechanics, we know that a drag force will be
developed from the oil given by the formula
F = hnpVR
This relation is called Stoke's luw. The other forces acting on the
sphere are its weight (take the density of the sphere as psp,,,j and
the buoyant force, which is the weight of the oil displaced (take
the density of the oil as poJ. The sphere will reach a constant
velocity called the tenninul velocity denoted as VTerm,.Show that
p = -9"
g Rk
Z (Pspherc
"
-Poi,)
12.20. A force F is applied to a system of light pulleys to pull
body A . If F is 10 kN and A has a mass of 5,000 kg, what is the
speed of A after 1 sec starting from rest'?
Time
12.22. A body of mass I kg is acted on by a force as shown in
the diagram. If the velocity of the body is zero at r = 0, what is
the velbcity and distance traversed when r = 1 min? The force
acts lor only 45 sec.
Force
10 sec
30sec
Figure P.12.22.
45 sec
Time
12.23. Three coupled streetcars are moving down an incline at a
speed of 20 k d h r when the brakes are applied for a panic stop.
All the wheels lock except for car B, where due to a malfunction
all the brakes on the front end of the car do not operate. How far
does the system move and what are the forces in the couplings
between the cars? Each streetcar weighs 220 kN and the coefficient of dynamic friction gd between wheel and rail is 30. Weight
is equally distributed on the wheels.
Figure P.12.23.
12.24. A body having a mass of 30 Ibm is acted on by a force
given by
F = 30tz
+
e-' Ib
If the velocity is 10 IVsec at r = 0, what is the body's velocity
and the distance traveled when r = 2 sec?
Figure P.12.20.
12.21. A force represented as shown acts on a body having a
mass of I slug. What is the position and velocity at f = 30 sec if
the body starts from rest at f = O?
12.25. A body of mass 10 kg is acted on by a force in the x
direction, given by the relation F = 10 sin 61 N. If the body has a
velocity of 3 d s e c when f = 0 and is at position x = 0 at that
instant, what is the position reached by the body from the origin at
f = 4 sec? Sketch the displacement-versus-time curve.
531
12.26. A water skier is shown d a n g l i q from a kite that is towed
via a light nylon cord by a powerboat at a constdnt speed of 30
mph. The powerhoat with passenger weighs 700 Ih and the man
and kite together weigh 270 Ih. If we neglect the mass of the
cable, we can take it as a straight line as shown in thc diaeram.
The horizontal drag from the air on the kite plus man is estimatcd
from fluid mechanics to he 80 Ih. What is the tension in the cahle'?
If the cable suddenly snaps, what is the instantznerus hurimntal
relative acceleration hetween the kite system and the powerhoar?
12.29. A hlock A of m d S S 500 kg is pulled by a force of 10,000
N as shown. A second block R of mass 200 kg rests on small frictionless rollers on top of block A . A wall prevents block B from
moving 10 the left. What is the speed of hlock A after I sec starting from a stationary position? The coefficient of friction pd is .4
hetween A and the horizontal surface.
Figure P.12.29,
Figure P.12.26.
12.27. A mdSs M is held hy stiff light telescoping rods that can
elongate or shorten Sreely hut cannot hcnd. Each rod is pin connected at the ends A , R . C. and 11. The system is o n a horimntal.
frictionless surface. Two linear springs having spring constants K ,
= 880 N/m and K2 = 1,400 N/m are connected to the rods as
shown in the diagram. If mass M = 3 kg is moved ,003 m to thr
right and i s released from rest, what is the equation for the vclocity in the x direction as a function of*'? What is the speed of the
mass when it returns to the vertical position of the rods?
12.30. Block B weighing SOII N rests on block A, which weighs
300 N. The dynamic coefficient of friction hetween contact surfaces i s .4. At wall C there are rollers whose friction we can
neglect. What is the acceleration of body A when a force F of
5,000 N is applied?
y
Figure P.12.30.
K , = 880 Niin
Kz = 1,400 Nlm
M=lkg
12.31. A hndy A of mass I Ibm is forced to move by the device
shown. What total force is exerted on the body at time f = 6 sec?
What is the maximum total force on the body, and when is the first
timc this force is developed aftzrr = O?
1
cos 2t ft/sec
Figure P.12.27,
12.28. A force given as 5 sin 3f Ib acts on a mas!, 01 I slug. What
is the position o f t h e mass at f = 10 sec'? Determine the total distance traveled. Assume the motion started from rest.
532
21' fU%C
Figure P.12.31.
12.32. Do Problem 12.10 for the case where there is viscous
friction between piston and cylinder given as 150 NImJsec of relative speed. Also, what is the maximum distance 1 the piston can
advance relative to the cylinder?
12.37. Mass R is on small rollers and moves down the incline. It
is connected to a linear spring, which at the position shown is
stretched from its undeformed length of 2 m to a length of 5 m.
What is the speed of R after it moves 1 m? Use Newton's law as
well as the x coordinate shown in the diagram.
12.33. The high-speed aerodynamic drag on a car is .02V2 Ib
with V in ftlsec. If the initial speed is 100 mihr, how far will the
car move before its speed is reduced to 60 miihr? The mass of the
car is 2.000 Ibm.
12.34. A block slides on a film of oil. The resistance to motion
of the block is proportional to the speed of the block relative to the
incline at the rate of 7.5 N/m/sec. If the block is released from
rest, what is the rerminul sneed? What is the distance moved after
I O sec?
0"
MA= 40 kg
M B = 20 kg
p d = .2
lo = 2 m (""stretched length of spcing)
K = 20 N/m
Figure P.12.37.
Figure P.12.34.
am th the
1 35. When you study fluid mechanics, you w
drag Don a hody when moving through a fluid with mass density
I
p i s given as ?C#Vz A where Vis the velocity of the body relative to the fluid A is the frontal area of the object: and C,, is the
so-called coeficienr of drag usually determined by experiment.
A racing plane on landing is moving at a speed of 350 k d h r
when a braking parachute is deployed. This parachute has a
frontal area of 70 mz and a C, = I .2. The plane has a frontal area
of 20 m2 and a C, = 0.4. If the plane and parachute have a combined mass of 8 Mg, how long does it take to go from 350 k m h r
to 200 k m h r by just coasting? Take p = 1.2475 kglm' and
neglect rolling resistance from the tires. There is no wind.
12.38. A wedge of wood having a specific gravity of 0.6 i s
forced into the water by a 150-lb force. The wedge is 2 ft in width.
(a) What is the depth d?
(h) What is the speed of the wedge when it has moved
upward 0.48 ft after releasing the 150-lb force assuming the
wedge does not turn as it rises? Recall, a buoyant force equals the
weight of the volume displaced (Archimedes).
X
Figure P.12.35.
12.36. In the previous problem, what is the largest frontal area
of the braking parachute if the maximum deceleration of the plane
is to be 58's when at a speed of 350 k d s the parachute is first
deployed?
w
I
w
Y
Figure P.12.38.
533
12.39. A poistm dart pun i c chrwn. The cross-sectional area
inside the tuhe i s I in?. Thc dart heing hlown wcighs 3 I M The
dart gun how haq a viscous icsistilnce given LEI .1 WL pcr (init
velocity in ftlszc. The hunter applies a constant pressure p at the
mouth i f the gun. Exprc\q the d a t i o n between p, V (velocity).
;md f. What cirnstmt pressure p i\ nccdcd t o cause the dart t u
rcach a speed of hO ftlscc in 2 x c ' ! Aswine Ihe dart pun i\ long
12.44. The spring shnwn i s nrmlmcar. That is, K i s not a COIIstant. hut i s a function of the extension of the qpring. If K = 2r
3 Ihlin. with ~I ineastired in inches, wh;e i s the speed of the m a ~ s
when x = 0 alter i t i s released from B state of rest at a poaition 3
in. firom the equilihrium position?Thc mass olthe body i h I slug.
+
Nonlinear
enough.
I'
-
Smooth
Figure F.12.44.
Dart gun
Figure P.12.3Y.
12.45. A particle nf mass m i c suhiect
tu the following furce
firld:
12.40. Using the diagram for Pnrhlem 12.5. assume that there i s
a luhricant hctwcen the body I ) of maw 5 Ihm and the hclt such
thnl there is a viscous friction Iurce given a~ .I Ih pcr unit izlativr
velocity between the hodv and the hclt. The helt moves at a uniform speed of 5 ftlscc to thc right m d initially the hndy has a
speed to the lcft of 2 il/ec relatiw to ground. At what time later
docs thc hody have a iero in\lanlaneous vclrrcity relative to the
ground'!
cpeed. how long docs i t take lor the hody to \low down to half of
i t s initial speed o f 2 ftlsec r e l a t i w t o the ground'!
12.42. One of thc largest (rt the supertankers in the world today
i s the S.S.Clohfik Londo,r. having a weight when fully laaded of
476,292 tons. The thrust needed to keep this ship moving iit I O
knots i s 50 kN. If the drag on the ?hip from thc watcr i s proportional t o the speed, how long w'ill it take for thic ship to slow down
from I O knots IO 5 knots :~llerthe engines arc chut down'? (Thr
a n s ~ e may
r
make you wonder nhmt thc safety of such ships.)
12.43. A cantilwcr heam i s shrwn. It i s ohmved that thc vertical detlection of the rnd A i s dircctly p n p r t i o n i t l 10 I I vertical tip
load F provided that thih load i s no1 too exce\\ive. A hody H of
inass 200 kg, when attached t i l the end of the heam with F
removed, C ~ U S C Sa dcilcction of 5 inni there after all motion has
ceased. What i s the spccd o f this hody if i t i s attachcd suddcnly t o
the heam and has dcsccnded 3 mm'!
F
A
Figure P.12.43.
534
F = mi
I n addition.
+
4mj
+
lhmk Ih
i t i s suhlectcd to a frictional frxcc f givcn
J = -mii
-
myj
+
as
2 r n 3 Ih
Thc particle i s stationary at the origin at time I = 0. What i s the
podion of the particlc at time I = I SCC?
12.47. I f in the previous problem, the heehcc has reached a maximum height of 92.75 ft, what i s the speed when i t returns to the
ground, assuming it docs !not reach its teimiiial velocity'? I f i t has
ieachcd thc terminal velocity, what i s your answer'?
12.48. A rocket weighing 5.000 Ih i s fired venically from a te\t
Eland on thc ground. A constant thrust of 20,000 Ib is developed
for 20 scconds. l f j u s t as an exercise, we do not tdke intu account
the amount of fucl hurned, and if wc ncglect air resistance, h i w
high up does this hypothctical rocket g a ? Note that neglecting luel
conrumption i \ a ceiious error! In the next prohlem WE will investigaie the case of the variahle mas? pmhlem.
*12.49. Calculate the velocity after 20 seconds for the case
whcrc there i s a rlrrrm.sr of mass of a rocket of 100 Ihmlsc
result of cxhaust combustion products leaving the rocket at a spzed
of 6,IH)O ft/sec iclativr til thr rockzt. At the outset the mckct
weighs 5,000 Ih. [Hint; Stan with Newton's law in the form F =
(d1dI)fmV)where F i s the weight, a vziriahla that decreases as fiiel
i s hurned. The first term o n the right side of this equation i s
m(dV1dl) whcrc m i s the instantaneous mass of thc rockct and u n ~
hurned fuel. Now there i s a f m c c on the IO(1 Ihmlsec ofcomhustion
products heing expelled from the rocket at a speed relativc to the
rocket of 6,000 ft/sec. The rate of change of linear momentum 12.52. An electron having a charge of -e coulombs is moving
associated with this force clearly must be- (dm/dt)(6,M)o). The between two parallel plates in a vacuum with an impressed voltage
reaction to this force for this momenNm change is on the rocket in E. If at t = 0, the electron has a velocity V, at an angle a. with the
the direccion of flight of the rocket and must he added to m(dV/dt). horizontal in the q plane, what will be the trajectory equation takThe force exerted by the exhaust gases on the rocket is a propul- ing the initial conditions to be at the origin ofxy? Show that
sive force and is called the thrust of the rocket. Again, neglect drag
of the atmosphere since it will be small at the outset because of low
y = - eE
+
x2
velocity and small later because of the thinness of the atmosphere,]
+xtanao
2m
12.50. We start with a cylindrical tank with diameter 50 ft containing water up to a depth of 10 ft. Initially the solid movable
cylindrical piston A having a diameter of 20 ft and a centerline
colinear with the centerline of the tank is positioned so that its top
is flush with the bottom of the tank. Now the cylinder is moved
upward so that the following data apply at the instant of interest
assuming the free surface of the water remains flat:
4 =2ft
& = 5 ft/sec
(voc o ~ a , ) ~
where m is the mass of the electron. Note we have neglected gravity here since it is very small compared with the electrostatic force.
i;z = 3 ft/sec2
What is the external force from the ground support on the water
needed for this condition not including the force required to support the dead weight of the water?
Y
_r
-
acuum
Figure P.12.52.
Figure P.12.50.
12.51. A sleeve slides downward along a pipe on which there is
dry friction with pd = .35.A wire having a constant tension of 80
N is attached to the sleeve and moves with it always retaining the
same angle a with the horizontal. If the sleeve weighs 60 N, what
should a be for the sleeve to move for I O seconds before stopping
after starting downward with an initial speed of 5 d s ?
12.53. A system of light pulleys and inextensible wire connects
bodies A, B, and C a s shown. If the coefficient of friction between
C and the support is .4, what is the acceleration of each body?
Take 4 as 100 kg, MB as 300 kg, and h& as 80 kg.
I
Y
Figure P.12.51.
Figure P.12.53.
535
536
CIIAPTER 12 PARTICI.K IIYNAMICS
Part B: Cylindrical Coordinates: Central
Force Motion
12.5
Newton's Law
for Cylindrical Coordinates
In cylindrical coordinates we caii cxpress Newton's law as follows:
I;_ =
F"
,n(l+
702)
~
= m(T4
~
2i6~
I: =
If the motion is known. it is a simple iniitter lo ascertain tlic lorce coniponents using Eqs. 12.15. Thc inverse prohlcni of determining the inotion
given the forces is particulai-ly difficult i n this casc. The reason lor this difficulty, a s you may havc alrcady learned in your differential equations course.
"
all lorce fnnctions. For this
is that Eqs. 12.15a and 12.15b iirc I I , I I I / ~ I W O ~ fcir
reason, we cannot present integration proccdures as in Part A of this chapter.
The following example will sene t o illustrate the kind of prihlem w e are able
to solve with the methods thus far presented in t h i h chapter.
Example 12.8
A platform shown in Fig. 12.13 has a constant angular velocity coequal to
5 radlsec. A mass L( of 2 kg slides in a frictionless chute attached to the
plaU'orm. .The mass is connected via a light inextensible cahle to a linear
spring having a spring constant K of 20 Nlm. A swivel connector at A
allows the cable to turn freely relative to the spring. The spring is unstretched when the mass B is at the center ('of the plalltirm. If the mass B
is releascd at I = 200 m m from il stationary position relative to the piaform, what i s its speed relative to the platform when it has m o v c d Lo position r = 400 min'? What is the transverse force on the hody B at this
position'!
We have hcrc a coplanar motion for which cylindrical coordinates iirc
iniist useful. Because the motion is coplanar. we can use r instead of T with
no ambiguity. Applying Eq. 12.151 first, we have
-20r = 2 ( r - 2 5 r )
Figure 12.13. Slider on mvaiing platform.
j
S E C TIO N 12 5
I
NEWTON'S L AW FOR CYLINDRICAL COORDINATES
Example 12.8 (continued)
Therefore,
r = 15r
(a)
As in Example 12.5, we can replace i. so as to allow for a separation of
variables.
Therefore,
Vr dV, = 15rdr
Integrating, we get
- 15r2
I
-
2
2
+
c,
To determine C,. note that, when r = .20m,
= 0. Hence,
600
cl
=-2
Equation (b)then becomes
Vj
= 15rZ
~
,600
(C)
When r = .40 m, we get for VFfrom Eq. (c):
v, =
1.
This is the desired velocity relative to the platform.
To get the transverse force G, go to Eq. 12.15b. Substituting the
known data into the equation, we have
Fs = 2[(.40)(0) + (2)(1.342)(5)]
F, = 26.W.N'
This is the transverse force on the mass B.
531
538
CHAPTER 12 PARTICLE DYNAMICS
Although you will he asked to solve problems similar to the preceding
example, the main use of cylindrical coordinates in Part B of this chapter will
be for gravitational central force motion. We shall first present the basic
physics underlying this motion expressing certain salient characteristics of
the motion, and then we shall amve at a point where we can effectively
employ cylindrical coordinates to describe the motion.
12.6
Central Force MotionAn Introduction
At this time, we shall consider the motion of a particle on which the resultant
force is always directed toward some point.fired in inertial space. Such forces
are termed centrul forces and the resulting motion of this particle is called
central force motion. A simple example of this is the case of a space vehicle
moving with its engine off in the vicinity of a large planet (see Fig. 12.14).
i
Figure 12.14. Body m rnovins about B planet.
The space vehicle is very small compared to the planet and may be considered to be a particle. Away from the planet's atmosphere, this vehicle will
experience no frictional forces, and, if no other astronautical bodies are reasonably close, the only force acting on the vehicle will be the gravitational
attraction of the fixed planet.' This force is directed toward the center of the
planet and, from the gravitational law, is given as
In the ensuing problems for this chapter and also for Chapter 14, we
shall need to compute the quantity GM in the equation above. For this purpose, note that, for any particle of mass m at the surface of any planet of mass
M and radius R, by the law of gravitation:
'We are neglecting drag developed from collisions of the space vehicle with solar dwt
partid.%.
SECTION 12.7 GRAVITATIONAL CENTRAL FORCE MOTION
539
where g is the acceleration of gravity at the surface of the planet. Solving for
GM, we get
GM = g R 2
(12.17)
Thus, knowing g and R for a planet, it is a simple matter tu find GM needed
for orbit 'calculations around this planet.
As pointed out earlier, the motion of a space vehicle with power off is
an important example of a central force motion-more precisely a gravitutionul cehtral force motion. The vehicle is usually launched from a planet and
accelereFd to a high speed outside the planet's atmosphere by multistage
rockets (8ee Fig. 12.15). The velocity at the final instant of powered flight is
called the burnout velocity. After burnout, the vehicle undergoes gravitational
central force motion. Depending on the position and velocity at burnout, the
vehicle can go into an orbit around the earth (elliptic and circular orbits are
possible), or it can depart from the earth's influence on a parabolic or a
hyperbolic trajectory In all cases, the motion must be coplanar.
J
Space
vehicle
Figure 12.15. Launching a space vehicle
In the following sections, we shall make a careful detailed study of
gravitational central force trajectories. Those who do not have the time for
such a detailed study of the trajectories can still make many useful and interesting calculations in Chapter 14 using energy and momentum methods that
we shall soon undertake.
"12.7
d--
Gravitational Central Force Motion
For gravitational central force motion, we shall employ an inertial reference
q in the plane of the trajectory with the origin of the reference taken at the
point P toward which the central force is directed (see Fig. 12.16). We shall
use cylindrical coordinates r a n d 9for describing the motion. Because z = 0
at all times, these coordinates are also called polar coordinates. Since the motion is coplanar in plane xy, we can delete the overbar used previously for r
with no danger of ambiguity.
Figure 12.16. xy is ineltial reference in plane
of the trajectory.
540
('HAPIER 12 I'AKTICI F DYNAMICS
Imt us consider Nrwron's Iuw for ii body of m i s s
near a star OS m a s s M:
in
dV
-
dl
-(;
in,
which i s moving
Min
r2
(12.18)
~~~~~
Canceling m and using cylindrical coordinates and componenls, we can express the cquatiuti above i n the following manner:
+ i r 8 + 2r8)t,
(i' - r d 2 ) E ,
= -
GM
r2
( 12. I9 1
Since trand i arc identical vecturs, the scalar equations o f the preceding cquation become
r
=
&2
(12.20a)
-GMIr?
,-B + 2i-8 = 0
(12.20hi
Equation 12.20b can he expressed i n the lumi
1
Y
ah
dl
(I2.2 I )
(r28)= 0
you can readily verify. We ciin conclude from Ey. 12.21 that
. .
r 2 0 = constant = C
(12.22)
Equation 12.22 leads to an iniportant conclusion. To estahlish this, colisider the arca swept nut by r during a time dr, which in Fig. 12.17 i s the shaded
&--
I
\
Figurc 12.17. Purriclc r w r r p
area. B y cnnsidcring Ihis area to hc
tliiit
UUI
awn
OS a triaiigle, we can cxprcss it as
Dividiiig through hy dr. we have
Now dAId1 i s the rate
iit which area i s heing swept nut by r : i t i s called uiwril
And, sincc $8 i s a cniistiint Sur each gravitational central force
rnolion (see t k . 12.22). wc can conclude that the areal velocity i s a c o i l m i i t
i.?lwit?.
SECTION 12.7 GRAVITATIONAL CENTRAL FORCE MOTION
for each gravitational central force motion. (This is Kepler’s second law.)
This means that when r is decreased, 0 must increase, etc. The constant, understand, will be different for each different trajectory.
In order to determine the general trajectory, we replace the independent
variable f of Eq. 12.20a. Consider first the time derivatives of r :
(12.23)
where we have used Eq. 12.22 to replace d0ldt. Next, consider r in a similar
manner:
(12.24)
Again, wing Eq. 12.22 to replace dO/dt, we get
(12.25)
For convenience, we now introduce a new dependent variable, u = l / r , into
the right side of this equation
By replacing i: in this form in Eq. 12.20a and
12.22, and finally, r by l/u, we get
e2 in the form C2u4from Eq.
Canceling terms and dividing through by C 2 , we have
GM
C2
(12.26)
This is a simple differential equation that you may have already studied in
your diffqrential equations course. Specifically, it is a second-order differential equation with constant coefficients and a constant driving function
GMIC2. We want to find the most general function .(e), which when substituted into the differential equation satisfies the differential equation-i.e.,
renders it an identity. The theory of differential equations indicates that this
general solution is composed of two parts. They are:
541
542
CHAPTER 12 PARrlCLE DYNAMICS
I . The general solution of the diffcrrntial equation with the right side of the
differcntial equation set equal tu 7e1-(Iand hence given a s
This solution is called thc ~ n i n ~ i l e i n ~ ~ r(or
i l ul ito~~~r r ~ i , g ~ ~ r rsolution,
ous)
u,
2. An? solution u,, that satisfies the fiill differential cquation. This part is called
thc purtir.alur .solutio~i.
The desired general solution is then the sum of thc cornplcmentary and particular solutions. It is a simple matter to show by substitution that the functiun
A sin e wtisfies Eq. 12.27 for any value of A . This is similarly truc for H cos
0 fcir any value of 8.'The theory 01differential equations tells u s that therc arc
two independcnt functions lor thc solutiun ul' Eq. 12.27. The general complementary solution is then
11,
= A sin
e + Hcos e
(12.28)
where A and B arc arbitrary constants (I!' integration. Considering the full
differential equation (Eq. 12.26). we sec by inspection that a particular
solution is
The general solutinn to the differential equation (Eq. 12.26) is then
By simple trigonoinetiic considerations. we can put the complementary
solution in the equivalent form, D cos ( 0 fi). where I1 and /3 are then
the constilnts of integralion.' Wc then have as an alternative fiir~nulation
~
f o r u (=
1/19:
You may possibly recognize this equation as the general conic equutiun
i n polar coordinates with the focus at Ihe origin. In your iinalytic geo-
543
SECTION 12.7 GRAVITATIONAL CENTRAL FORCE MOTION
metry class, you probably saw the following form for the general conic
eq~ation.~
1
1
'
q
- = -
1
+ -cos(@
- p,
P
(12.37)
where E is the eccentricity, p is the distance from the focus to the directrix, and
pis the angle between the x axis and the axis of symmetry of the conic section.
Comparing Eqs. 12.31 and 12.37, we see that
I
(1238a)
P=o
E = - DC2
(1 2.3%)
GM
'A conic section is the locus of all points whose distance from a f i e d p o i n l has a consmnl
rnlio to the distance from a fued line. The fixed point is called the/ocus (or focal point) and the
line is termed the direcfrir. In Fig. 12.18 we have shown paint P, a directrix DD,and a focus 0.
For a conic Section to be traced by P, it must move in a manner that keeps the ratio rim, called
the rccenfricify,a fixed number. Clearly, for every acceptable position P , there will be a mirror
image position P (see the diagram) about a line normal to the directrix and going through the
focal point 0. Thus, the conic section will be symmetrical about axis OC.
Using the letter c to represent the eccentricity, we can say:
r
I _
(12.32)
D
C
m-€=
p + ,cos
Pp
P
\
where p is the distance from the focus to the directrix. Replacing cos q by --cos (e - p), where
0 (see Fig. 12.18) is the angle between the x axis and the axis of symmetry, we then get
r
=c
(12.33)
- rcas(e - p )
\
\
\
\
D ------------------&p'
X
Figure 12.18. rlw =
(12.34)
+).z
SP
P
p
section.
+?2
Simple algebraic manipulation permits us to put the preceding equation into the following form:
(I - .')x2
+ yz + 2 p r ' x
- czp2 = 0
Dtb
(12.36)
If c > I , the coefficients of xz and y2 are different in sign and unequal in value. The equation then represents a hyperbola
If c = I, only one of the squared terms remains and we have aparnhola.
If c < I,the coefficients of the squared terms are unequal but have the same sign. The
curve is that of an ellipse
\
Directrix
Now. rearrangins the terms in the eauation. we anive at a standard formulation far conic sections:
Jx2
\
Focus
x
C.
E E
constant for conic
544
WAPTliK 12
P,\KTICI.I
DYNAMIC'S
From our knowledge if conic sections. we can then iay that i f
1":
> 1.
'Ic'
= I, the trajectory i s a piirahiilii
(l2.39h)
I"'
< I. the trajectiiry
(I2.39~)
GM
~~
GM
~~
~
GM
Fb
the trajectiiry
i h ii
ih
hypcrhola
an ellip\e
= 0. the trajectory i s a circle
(12.3%)
(12.3YlI)
Clearly, L)CWM, ~ I i cecccnwicitj. i\ :in exlrcniely iinportant quanlity. Wc
shall next look into the practical applications of the preceding fcrieral theory
to prohlems i n space niechiiiiics.
x
"12.8
Applications to Space Mechanics
We sliull iiow employ the theory scl rorth i n the previous section to study Ihe
motion o f space vchicles--a prohlciii of great present-day interest. We shall
\
.'..
Figui:e 12.20. Launching at axis of symmetry.
assume that at the end of powcred flight the puhition r,) ;mil velocity I;;of the
vehicle are known from rocket calculaliotis. The reference cmploycd w i l l he
an inertial relcrence at the ceiitcr of the planet and s o the reference w i l l Iranslate with the planet relative tu the "fixed stars." Accordingly. the earth hill
rotate one cycle per day for such a reference. We know that the trajectory of
the hody w i l l l h i ii plme fixed i n iriertiitl space and so, forci~nreniencc.we
take the .r? plane of lhc relerence to he the plane of thc trajectory. I t i s the
usual practice to chixist. the .x axis to hc the axis of cymmctry for the tra,jectory. Ifthere i s a zt'ro rodid vrlociry component at "hurnout," then the
launching clearly occurs at a position along the a x i s of symmetry o f the trajectory (i.e., along the .r axis). This ciise lias hcen shown in Fig. 11.20.
wherein the suhscript 0 denotes launch d;ita. If. on thc other hand, a ]radial
component
i s present at hurnouc. then the launch condition occurs ill
somc piisition O,,from the .x axis. as shown i n Fig. 12.21. We generally do not
know
a priori, since its value depends un the cqu;itioii of thc trajectory.
Finally, the angle n s h o k n i n the diagram w i l l he called the l i i i r n d ~ i n j iL i i i , y I e
in the ensuing discussion.
Since the .r axis has heen chosen to he the a x i s of symmetry. the q u a tion of motion of the vehicle aflcr powercd Ilight i s given in lcrnis of x h i Irary conslants C and D h y Eq. 12.31 with the anglc
set equal to zero.
Thus. we have
(v),,
e,,
\
\
'\
.'.
Figure 12.21. Burnout with radial velocity
present.
1 =
r
(-2
+ Dco,H
(12.401
StCTION I2 8
The problem is to find the constants C and D from launching data. We shall
illustrate this step in the examples following this section. Note that when
these constants are evaluated, the value of the eccentricity E = DC2/CM is
then available so that we can state immediately the general characteristics of
the trajectory.
Furthermore, if the vehicle goes into orbit, we can readily compute the
orbital time z for one cycle around a planet. We know from the theory that
the aerial velocity is constant and given as
But r%equals the constant C in accordance with Eq. 12.22. Hence,
c
2
dA = - dt
(1 2.42)
The area swept out for one cycle is the area of an ellipse given as nub, where
u and b are the semimajor and semiminor diameters of the ellipse, respec-
tively. Hence, we have on integrating Eq. 12.42:
Therefore,
2nab
z = -~
C
(12.43)
We have shown in Appendix I11 thar
a=EP
I -$
b = u(l - € * ) ‘ I 2
(12.44~1)
(12.44h)
Replacing p by 1/11in accordance with Eq. 12.38a, we then get
(12.45a)
( I 2.45b)
Thus, we can get the orbital time T quite easily once the constants of the
trajectory, D and C , are evaluated.
AI’PLICATIONS TO SPACE MECHANIC.\
545
54h
C'HAP'JER 12 PAKrlCI,F. UYNAMICS
To illustrate many of the previous general remarks in a most simple
manner, we now examine the special case where, a s shown in F i g 12.22, various launching.; ii.c., burnout conditions) are made from a given point a such
that the launching angle a = 0. Clearly ( V ) , , = 0 f(ir these cases and the
launching axis corresponds to the axis of symmetry (if the various trajectories.
Only will be varied in this discussioii.
v,
.x
3
I
\
I
Figure 12.22. Various laurrchinys liom the earth or some other planer.
The c ~ ~ i s t a i iC
t sand U arc readily available for thesc trajectories. Thus.
we have from Eq. 12.22:
v
C = r ? @ = r HV -- r 11 n
And Sroni ELI. 12.40. setting r = q) when I3 =
(12.46)
e,, = 0. we gct, on solving for 0:
(I2.47)
Since C and I1 ahove, for a given ro. depend only on V,,, we conclude that the
eccentricity here is dependent only on C;, lor a given ro.
If V,, is so large that DCZ/GM exceeds unity, the vehicle will have the
trajcctory 111 a hyperbola (curve I ) and will eventually leavc the intluence of
the earth. If V,, is decreased to a valuc such that the eccentricity is unity. the
trajcctory becorncs a parabola (curve 2 ) . Sincc a further decrease in the value
of yl will cause the vehicle tu orbit. curve 2 is the limiting trajectory with our
launching conditions for outer-space night. The launching velocity for this
case is accordingly called the rscapr ~elocir\ and is denoted as
We can
SECTION 12.8 APPLICATIONS 1'0 SPACE MECHANICS
solve for ( y J Efor this launching by substituting for C and D from Eqs. 12.46
and 12.47 into the equation DC21GM = I . We get
(12.48)
a result that is correct for more general launching conditions (i,e., for cases
where launching angle a # 0). Thus, launching a vehicle with a speed equaling or exceeding the value above for a given r,, will cause the vehicle to leave
the earth until such time as the vehicle is influenced by other astronomical
bodies or by its own propulsion system. If V, is less than the escape velocity,
the vehicle will move in the trajectory of an ellipse (curve 3). The closest
point t o the earth is called perigee; the farthest point is called apogee.
Clearly, these points lie along the axis of symmetry. Such an orbiting vehicle
is often called a space satellite. (Kepler, in his famous first law of planetary
motion, explained the motion of planets about the sun in this same manner.)
One focus for the aforementioned conic curves is at the center of the planet.
Another f o c u s , r now moves in from infinity for the satellite trajectories. As
the launching speed is decreased,f' moves t o w a r d j When the foci coincide,
the trajectory is clearly a circle and, as pointed out earlier (see Eq. 12.39d).
the eccentricity E is zero. Accordingly, the constant D must be zero (the constant C clearly will not be zero) and, from 9.12.47, the speed for a circular
orbit (l(JC
is
For launching velocities less than the preceding value for a given ro, the
eccentricity becomes negative and the focus f' moves to the left of the earth's
center. Again, the trajectory is that of an ellipse (curve 5). However, the satellite will now come closer to the earth at position b, which now becomes the
perigee, than at the launching position, which up to now had been the minimum distance from the earth." If friction is encountered, the satellite will
slow up, spiral in toward the atmosphere, and either burn up or crash. If Vu is
small enough, the satellite will not go into even a temporary orbit but will
plummet to the earth (curve 6). However, for a reasonably accurate description of this trajectory, we must consider friction from the earth's atmosphere.
Since this type of force is a function of the velocity of the satellite and is not
a central force, we cannot use the results here in such situations for other than
approximate calculations.
"'Note that with Ihr positive I axis going through perigee, I i s minimum when 0 = 0.
From Eq. 12.40. we can conclude for this case (0 is measured here from perigee) thil, to m i 6
miLe r. the conatam D mUSl be posilive. me eccentricity must then he positive for 0 measured
from perigee. If the positive x axis goes through apogee, then r is marimurnwhen 0 = 0. From
Eq. 12.40 we can conclude that D must be negative for this case (0 is here measured from
apogee). Thus, the eccentricity is negative fof 0 measured from apogee. This is clearly the case
for curve 5.
547
Example 12.9
The first American satcllile. the V a n p a r d . win launched at
ii velocity 01
18.000 milhr at an altitude 01 400 mi (see Fig. 12.23). I I the "humout"
velocity of the last stage is parallel to the earth's \iirface. compute tlic
inaximum iiltitudc from the a i - t h ' s surface that the Vanguard iatcllite will
reach. Consider the earth to hc perfectly spherical with a radius O S 3.960
mi (ti) is therefore 4.360 mi).
Figure 12.23. 1.aunching 01ilie V : i n p a r ~ lsiitrllit~.
We m u h t now compute thc quantitich GM, C. and D lrom thc initial
data and other known data. To detcrmint. GM. we employ Eq. 12. I 7 and i n
units of iiiilcs and hours wc get
terms
= 1.1.3Lj
x [(I"
inii/hr'
The ccinstmt C i \ readily dctcriiiincd directly froin initial data a:,
<' = q,y, = (4.3h0)~18.001))
= 7.85 x 10'mi2/hr
Finally. the cimstaiit I1 i:, available Srom Eq. 12.47:
=
. Z Xx~ ((1 n i i ~I
The eccentricity 1)C'IGM can nriu he computed
iis
The Vanguard will thus definitely riot e x a p e into outel- s p x z
The tra,jectory 01. this n i v l j ~ u iis fiirnied from tk.12.41:
__..x__I_
..",,
.....
-,
.
.
.... .
.
.
"___
.
SECTION 12.8 APPLICATIONS TO SPACE MECHANICS
I
Example 12.9 (Continued)
Therefore,
~
I = 2.01 x
i o 4 + ,283 x i o 4 case
r
(e)
We can compute the maximum distance from the eaith’s surface by setting
0 = Kin the equation above:
-~
I - (2.01 - ,283) X IO-‘
= 1.727 x 10” mi-’
rn,,
Therefore,
r-
= 5,790 mi
By subtracting 3,960 mi from this result, we find that the highest point in
the trajectory is 1,830 mi from the earth’s surface.
I
Example 12.10
In Example 12.9, first compute the escape velocity and then the velocity
for a dircular orbit at burnout.
Using Eq. 12.48, we have for the escape velocity:
\
‘il
2(1 239 X IO”)]”’
4,360
(V& = 23,840 mi/hr
For a circular orbit, we have from Eq. 12.49:
(V& =
E
=1
milhr
Thus, the Vanguard is almost in a circular orbit.
549
550
CllhPTER I ? PAKTICL.1: IDYNAMICS
Example 12.11
I
i
Determine the orhital time i n Examplc 12.1) Sor the Viinguard siitellitc.
Wc employ Eys. 12.44 for the semimajor and semiminor axe\ OS the
elliptic orbit. Thus, recallinp t h a t l ~= lill we have
x
/I
5,080 ini
= dl - e l ) ’ / ? = 5,080(1 -,1408?)”’
Theretiire, from E q 12.43 wc have f o r the oi-hitiil time:
z = 2.05 hr = 122.7 min
Example 12.12
A space vehicle i \ i n ii circular “parking” orhit around the pliinet Vcnus.
320 km ahovc the surtace o f thiq planet. The radius o f Venus i a 6.160 kin,
and the escape velocity at the surticc i s 1.026 x IOi mlsec. A retro-rockct
i s fired to a l i w the vehicle \n that i t w i l l cnine within 12 kin of the planet.
If we consider that the rocket changes the speed d t h e vehicle w e r a comparatively short distance ot its tmvcl, what i s this change of apeed? What
i s the speed ofthe vehicle at i t s closest position tu the s u r f x c of Venua’!
We show the vehicle in a circular parking nrhit i n Fig. 12.24. We
shall considcr that the rctro-rocketa are fircd at position A SO as Lo estahlibh
a new elliptic orhit with apogee at A and perigee at l j .
As e first step, we shell compute GM using the escape-velocity equation 12.48. Thus, we havc
Thereforc,
=
4.20 x IO’? krn’ihr’
,
Figure 12.24. Ch,ingc “1 orbit
SECTION 12.8 APPLICATIONS TO SPACE MECHANICS
Example 12.12 (Continued)
The equation for the new elliptic orbit is given as
~
I - -GM + D c o s e
r
cZ
(a)
Note that when
e = 0,
e = ff,
r = r,, = 6,480 km
(h)
r = r,,, = 6,192 km
(C)
To determine the constant C, we subject Eq. (a) to the conditions (b) and
(c). Thus,
1 4.20 x IO1*
c*
6,480 =
1 4.20 x IO'*
6,192 =
+
-
c2
Adding these equations, we eliminate D and can solve for C. Thus;
8.40 x loLz- I
1
C*
- 6,480'6,192
~
Therefore,
C = 1.631 x 108kmz/h~
Accordingly, for the new orbit,
r,V, = 1.631 x lo8
Therefore,
V, =
25,168 km/hr
For the circular parking orbit the velocity
v c =j#EM
r,
I-=
y. is
~~~
14.20 x loLz
6,480
1
= 25,458 km/h~
The change in velocity that the retro-rocket must induce is then
AV = 25,168 - 25,458 =
The velocity at the perigee at B is easily computed since
rBVB= C = 1.631 x IO8
Therefore,
V, = (1 631
X
108)/6,192 =
55 1
Now l e t us consider more general launching conditions where the
launching angle a i\ not 7,ero (see Fig. 12.25. The canslant C is still eahily
cvaluated (sce Eq. (12.46)) in tcnns of launching data as ro(K,)l,. To get D.
we write Eq. 12.40 lor lauiichinf condition\. Tliur.
1
Figure 12.25. I.aunch witti d i a l \elricify.
The value of 4, i\ not yet known. Thus. we have two unknown quantities in
this equation, namcly I1 and
Differentiating Eq. 12.40 with respect to time
and siilving for r, we get
qr
i =
W8sinH
=
DCsinH
(12.51)
Noting Lhatk i s equal t o ", and submitting the preceding equation t o launching
conditions. we then form a accond equation for the e\'aluation n l the unknown
coiistiint\ D and 4,. Thus.
(y
= DCsiii B,,
( I2.52 1
w
= ,g c<,\ e,,
C?
(12.51)
Rearranging Eq. 12.50. we havc
I
5)
~
Divide both sides 01 Eq. 12.52 by C. Now. squaring Eqs. 12.52 and 12.53,
adding terms, and using the lact tliat sin2 e,, + cos' Olj = I. we get for the
constant Ll the result : I '
SECTION 12.8 APPLICATIONS TO SPACE MECHANICS
Having taken the positive root for D,we note (see footnote #IO on page 545)
that 0 is to be medwred from perigee. The eccentricity is
First, bringing C’ into the bracket and then replacing C by r0(V,), in the entire
equation, we get the eccentricity conveniently in terms of launching data:
One can show, using the preceding formulations, that the equation for
the escape velocity developed earlier, namely
is valid for any launching angle a.Remember that % in this equation is measured from a reference xyz at the center of the planet translating in inertial
space. The velocity attainable by a rocket system relative to the planet’s surface does not depend on the position of firing on the earth. but depends primarily on the rocket system and trajectory of flight. However, the velocity
attainable by a rocket system relative to the aforementioned reference xyz
di~esdepend on the position of firing on the planet’s surface. This position,
accordingly, is important in determining whether an escape velocity can be
reached. The extreme situations of a launching at the equator and at the North
Pole are shown in Fig. 12.26 and should clarify this point. Note that the
motion of the planet’s surface adds to the final vehicle velocity at the equator.
but that no such gain is achieved at the North Pole.
Figure 12.26. Launching at equator and North Pole.
553
554
CHAFTER 12 PARTICLE DYNAMICS
Example 12.13
Suppose that the Vanguard satellite in Example 12.9 is off course hy an
angle a = 5" at the time of launching but otherwise has the same initial
data. Determine whether the satellite giies into orbit. If so, determine the
maximum and minimum distances from the earth's surface.
The initial data for thc launching are
y, = 18,000 miihr
7, = 4,360 mi,
Hence,
(V,),,
=
(18.000)sinrx = (1X,OOO~(O.O872j
= 1.569
[V,j,,
mi/hr
= (lX.Il00)cosa = (IX,OOO)(O.Y96i
=
17,930 milhr
To determine whether we have an orbit, we would have to show first
that the eccentricity E is less than unity. This condition would preclude the
possibility of an escape from the earth. Furthermore. we must he sure that
the perigee of the orbit is far enough from the carth's surface to ensure a
reasonahly permanent orbit. Actually, for both questions we need only cdl= ?I. An infinite value of one of the r ' s will
culate r for B = 0 and
mean that we have an escape condition, and a value not sufficiently large
will mean il crash or a decaying orbit due to atmospheric friction.
Using the value of GM as 1.239 x 10" mi3/hrz from Example 12.9
and using Eq. 12.54 for the constant D ,we can express the trajectory of the
satellite (Eq. 12.40) as
e
SECTION 12.8 APPLICATIONS TO SPACE MECHANICS
Example 12.13 (Continued)
Therefore,
~
I = 2.03 x I O F 4
+ 3.33 x IO-'
cosomi-l
(a)
Set 0 = 0:
= 20.3 x IO-'
4)
+ 3.33 x lO-'mi-'
Hence,
r', , = 4
ni
Thus, after being launched at a position 400 mi above the earth's surface, the satellite comes within 270 & of the earth as a result of a 5"
change in the launching angle. This satellite, therefore, must he
launched almost parallel to the earth if it is to attain a reasonably permanent orbit.
Now, setting 0 = n,we get
Hence,
'ma,
= 5,893 mi
Obviously, the maximum dntance from the earth's burface is
555
54. A d w i c e uscd at amucemcnt parks cnnsists of a circular
m that i s made t o r e ~ o l v eahout its :,xi\ 01 \ymmctry. People
~d up against the wall, as shown i n the diagram. Altcr rhc whole
m has been hmught up t o spccd, thc flooi~i\ lowcrcd. What
timum angular s p e d i s required to c n s ~ i i ethat a p c ~ n w
n ~ l not
l
, dciwn the w a l l when the tloor i s Iuwrred? 'lake pb = .3.
ueight W. What i s the distancc of thr plane otthc tra.jectrry of the
hoh lrnrn thc support at o?
,
I'
I
-_- ,,
-.
Figure P.12.56.
Figure P.12.54
12.57. A \haft AH mt:itcs 11 an angular velocity nf 100 rpm. A
hody E of m a s IO kg can move without friction along rod CI)
lixcd to AI(. I l t h c hody E i\ to remain mtionary d a t i v e 10 ('1) at
any position along CD. how m i i s 1 the spring cnnstant K vary'? The
tlistancc 5, frum the a x i i i\ thc onstl-ctchcd length n f t h c spring.
-
55. A tlywhccl is rotating at a s p e d 1110, = IO radlceu and
at this instant a rate n l changu o l spceil 0 of 5 rsdlsec'. A
m o i d at thi5 instnnt m w c s a valve toward the centerline i f t h c
,wheel at a speed 0 1 I .S imlrcc and i s decelerating at rhc rate 01
nlsec'. The \-alve haa il mils\ OS I kg and i c .3 tm f r o m the axis
rotatian at the timc of interest. What i\ the total forur on thc
A
c
ve'!
C
> lO(1 rpm
A
-
__
Figure P.12.57.
12.58. A drvicc u m \ i s t \ 01three m a l l miisics. thrzc wrightle\r
mk. and ii linrilr y i n p with K = 200 N l m The system i s rot at^
trig in thc g v r n fixed configuration at :i conctant rpecd w = 10
rad/\ i n a lhorimntiil plme. 'The li>llowingdata apply:
I
Figure P.12.55.
:mica1 pendulum of Icngth 1 i s
> w nThe
nadc tn rotate ill a constant angular spccd of wahoo
axis. Compute the tension in the cord if thc pcndulum huh ha!
56.
5
M , = 2 kg
M,j = 3 kg
M,. = 2 kg
If thr spi~inpi\ \trctchcd h) an a m r u n t .(I25 m. dcterminc the total
lurce components :sting o n thc miisr at ('and the tenqile force in
mcmher /),4 [ I l i n i ('onwicr a single panicle. then a \y\tem 0 1
pmticIc\; I;\ and OH arc pin connec1ed.l
12.61. A platform rotates at 2 radhec. A body C weighing 450 N
rests on the platform and is connected by a flexible weightless
cord In a mass weighing 225 N, which is prevented from swinging
out by pan of the platform. For what range of values o f x will bodies C and B remain stationary relative to the platform? The static
coefficient of friction fix a11 surfaces is .4.
x-
_-- .',
Figure P.12.58.
12.59. In the preceding problem consider that member DA is
welded to the sphere at A . Now, at the instant of interest, there is
also an angular acceleration of the system having the value of .28
rad/sec* counterclockwise. What are the force components acting
on panicle C. and what are the force components from rod AD
acting on panicle A'? See hint given in the preceding problem.
12.60. A device called a flyhall governor is used to regulate the
speed o f such devices as steam engines and turbines. As the governor is made to rotate through a system of gears by the device to
he controlled, the halls will attain a configuration given by the
angle 6'. which is dependent on both the angular speed w of the
governor and the force P acting on the collar hearing at A. The upand-down mntion ofthe bearing at A in response to a change in w
is then used to open or close a valve to regulate the speed of the
device. Find the angular velocity required to maintain the configuration of the flyhall governor for 8 = 30". Neglect friction.
wL,=.4
for 811 surfac
Figure P.12.61.
12.62. A particle moves under gravitational influence about a
body M, the center of which can he taken as the origin of an inertial reference. The mass of the particle is 50 slugs. At time t, the
particle is at il position 4,500 mi from the center of M with direction cosines I = .5, m = -3. n = ,707. The panicle is moving
at a speed of 17,000 milhr along the direction E, = .Xi + .2j +
.Shhk. What is the direction of the normal t o the plane of the
trajectory?
Figure P.12.62.
Figure P.12.60.
12.63. If the position of the particle in Problem 12.62 were to
reach a distance of 4,300 mi from the center of hody M, what
would the transverse velocity V, of the particle be?
551
12.64. Use Eqs. 12.38h and 12.40 t o show' that i f t h c eccentricity
is x r u , the trujectoq nmsl he that of a circle.
12.65. A satellite ha\ at m c t i m e during its flight around thc
caith a radial crrmpment 01velocity 3.200 krnlhl- and a tlans\.cr\e
cm~poneril0 1 2h.XIO km/hr. I1 the s;ilrllitc i \ at a distanvc 01
7.040 krn frtmm tlic cciitzr 01 the eimrth. u'hal is i l \ arral velocity'!
12.66. Compute the c s u p e velocit) ;it :I pnsition X.000 In, f i r m
the center of the sarlh. What q x c d i s needcd to inninlain il circw
tar orhit at that di.;tance from the earth's center? Derive Ihc c q u a ~
tiaii for the speed riccdcd fur il circular orhit direclly from
Neuston's law without ucing infill-malioo iibrrut ecccnlricilieh. ctc.
z
Figure P.12.69.
sonnection
I
Figure P.12.67.
i5X
12.72. Consider a satellite of mass m in a circular orbit around the
earth at a radius R, from the center of the earth. Using the universal
law of gravitation (Eq. 1.1 I ) with M as the mass of the eanh and
using Newron's law in a direction normal to the path, show that
-
IGM
"ctrc.
12.78. The satellite Hyperion about the planet Saturn has a
motion with an eccentricity known to he ,1043. At its closest distance from Saturn, Hyperion is 1.485 x IO6 km away (measured
from center to center). What is the period of Hyperion about Saturn? The acceleration of zravitv of Saturn is 13.93 m/recz at its
surface. The radius of Saturn is 57,600 km
"
I
= '-
4 Ro
for a circular orbit. Now at the earth's surface use the
gravitational law again and the weight ,f the body, to show that
CM = gRE.ar,,, where y is the acceleration of gravity.
12.73. The acceleration of gravity on the planet Mars is about
,385 times the acceleration of gravity on earth, and the radius of
Mars is about ,532 times that of the earth. What is the escape velocity from Mars at a position 100 mi from the surface of the planet?
12.79. Two satellite stations, each in a circular orbit around the
earth, are shown. A small vehicle is shot out of the station at A
tangential to the trajectory in order to "hit" station B when it is at
a position E 120" from the x axis as shown in the diagram. What
is the velocity of the vehicle relative to station A when it leaves?
The circular orbits are 200 miles and 400 miles, respectively, from
the earth's
12.74. In 1971 Mariner 9 was placed in orbit around Mars with
an eccentricity of .5. At the lowest point in the orbit. Mariner 9 is
320 km from the surface of Mars.
(a) Compute the maximum velocity of the space vehicle
relative to the center of Mars.
(b) Compute the time of one cycle.
x
Use the data in Problem 12.73 for Mars.
12.75. A man is in orbit around the earth in a space-shuttle vehicle. At his lowest possible position, he is moving with a speed of
18,500 mihr at an altitude of 200 mi. When he wants to come
back to earth, he fires a retro-rocket straight ahead when he is at
the aforementioned lowest position and slows himself down. If he
wishes subsequently to get within SO mi from the earth's surface
during the first cycle after firing his retro-rocket, what must his
decrease in velocity be? (Neglect air resistance.)
12.76. The Pioneer 10 space vehicle approaches the planet
Jupiter with a trajectory having an eccentricity of 3. The vehicle
comes to within 1,000 mi of the surface of Jupiter. What is the
speed of the vehicle at this instant? The acceleration of gravity of
Jupiter is 90.79 ftlsec' at the surface and the radius is 43,400 mi.
12.77. If the moon has a motion about the earth that has an
eccentricity of ,0549 and a period of 27.3 days, what is the closest
distance of the moon to the earth in its trajectory?
Figure P.12.79.
12.80. In Problem 12.79, determine the total velocity of the
vehicle as it arrives at E as seen by an observer in the satellite B .
The values of C and D for the vehicle from Problem 12.79 are
mi-', respectively.
7.292 x IO' m i 2 h and 7.373 x
12.81. The Viking I space probe is approaching Mars. When it is
80,650 km from the center of Mars, it has a speed of 16,130 kmihr
with a component (V,) toward the center of Mars of 15,800 kmJhr.
Does Viking I crash into Mars, go into orbit, or have one pass in
the vicinity of Mars? If there is no crash, how close to Mars does it
come? The acceleration of gravity on the surface of Mars is 4.13
m/sec2, and its radius is 3,400 km. Do not use formula for D as
given by Eq. 12.54, but work from the trajectory equations.
559
12.83.
110 I'mhlcm 12.U2 u'ilh the
ilitl
of Eq. 12.14.
Figure l'.l2.87.
12.85.
Do Prohleni 12.84 with the aid of Eq. 12.54.
Figure P.12.88.
SECTION 12.9 NEWTON’S LAW FOR PATH VARIABLES
Part C:
12.9
Path Variables
Newton’s Law for Path Variables
We cm express Newton’s law lor path variables as follows:
< = m , d2s
dt
(12.57a)
Notice that the second of these equations is always nonlinear, as discussed in
Section 12.5.’’ This condition results from both the squared term and the
radius of curvature R. It is therefore difficult to integrate this differential equation. Accordingly, we shall be restricted to reasonably simple cases. We now
illustrate the use of the preceding equations.
‘%quation 12.57a could also he
nonlinear, depending on Lhe nillure of the function $.
Example 12.14
A portion of a roller coaster that one finds in an amusement park is shown
in Fig. 12.27(a). The portion of the track shown is coplanar. The curve from
A to the right on which the vehicle moves is that of a parabola, given as
(y -
=
loox
(a)
,’
,
,
v
I:1
loop
,
,
,
,
/ / I ’
n
n
x
(a)
Figure 12.27. Roller coaster trajectory.
P,
(b)
561
562
CHAPTER 12 PARTI(’1.E DYNAMICS
Example 12.14 (Continued)
with x and y in feet. If the train of cars is moving at a speed of 40 ft/sec
when thc front car is 60 fl above the ground, what is the total normal force
exerted hy a 200-1h occupant nf thc front car on the seat and floor of
lhe car’!
Since we requirc only the force F normal to the path, we need only
he concerned with u,~.Thus, we have
Wc can compute R from analytic geometry as follows:
wherein from Eq. (a) we have
d\
SO
&=m
Substituting into Eq. ( c ) ,we have
AI the p o w o n of intere\t. we get
SECTION 12.9 NEWTONS LAW FOR PATH VARlABLES
Example 12.14 (Continued)
Accordingly, we now have for y,, as required by Newton's law:
Nnte that I;; is the total force component normal to the trajectory
needed on the occupant for maintaining his motion on the given trajectory.
This force component comes from the action of gravity and the forces from
the seat arid floor of the car. These forces have been shown in Fig. 12.27(h),
where P, and p a r e the normal and tangential force components from the
car acting on the occupant. The resultant of this force system must, accordingly, have a component along n equal to 94.6 Ib. Thus,
-200 cos
To get
+ P,
= 94.6
(h)
p, nnle with the help of Eq. (d) that
Therefore.
/3 = 51.3"
Suhstituting into Eq. (h) and solving for P,. we get
9, =
200 COS S I.30
+
94.6
This is the force component from the vehicle onto the passenger. Tne reaction to this force is the force component from the passenger onto the vehicle.
563
564
CHAPTER I ?
PARTICLE DYNAMICS
Part D: A System of Particles
12.10
The General Motion of a System
of Particles
I.et us examine a system of n particles (Fig. 12.28) that has inlcractions
between thc particles fiir which Newton '.s thit-d lobe of motion (action equals
Figure 12.28. Force\ on ith pairiclr of the system
reaction) applies. Newror~'.ssrt.ond /ai?.for any particle (lcl us say thc ith particle) is then
(12.58)
,*I
wherc4.;. is the force on particle i from particle j and is thus considered an
internal force for the system of particles. Clearly. the,, = i term of thc sunimation musf be deleted since tlic ith particle cilnnoi exert force iin itself. The
force F, reprcscnts the resultant force on the ilh particle from the forccs < w c , i nrrl to thc syslem of paniclcs.
If thehe equations are added for all n particlcs, we have
(12.59)
Carrying out the double summation and excluding terms with rcpeakd indexes,
such as&,,&,. etc., we find that lor each term with any one set o f indexes
there will he a term with Ihe reverse of thcse indexes present. For example.
lor the force.(,, it lorcef2, will ex is^. Considering the ineming iifthe indexes.
we see thatf..,.I andJ, represent action and reaction lorccs between a pair of
particles. Thus, as a result of N e w o n ' s third Irrw. thc double summation i n
Eq. 12.59 should add LIPt u zero. N < w t r n i ' .srmrtd
~
law lor a system iif paiticlcs
then becomes:
I2.60 1
where F now rcprehenls the vector sum of all the r,rtr,ruol forccs acting on all
the panicles of the system.
SECTION 12.i o THE GENERAL MOTION OF A SYSTEMOF PARTICLES
To make further useful simplifications, we use the first moment of mass
of a system of n particles about a fixed point A in inertial space given as
first moment vector =
C miri
i=l
where 7 represents the position vector from the point A to the ith particle
(Fig. 12.29). As explained in Chapter 8, we can find a position, called the
Figure 12.29. Center of mass of system
center of mass of the system, with position vector r,, where the entire mass of
the system of particles can be concentrated to give the correct first moment.
Thus,
Therefore,
(1 2.61)
Let us reconsider Newton's law using the center-of-mass concept. To
m,r( by Mr, in Eq. 12.60. Thus,
do this, replace
(12.62)
We see that the center ofmass of any aggregate ofparticles has a motion that
can he computed by methodr already ser,forth, since fhis i.7 a problem involving a single hypothetical particle of mass M . You will recall that we have
alluded to this important relationship several times earlier to justify the use of
the particle concept in the analysis of many dynamics problems. We must
realize for such an undertaking that F is the total external force acting on all
the particles.
565
566
CHAPTER 12 PARTlCLt DYNAMICS
Example 12.15
Three charged particles i n a vacuum are shiiwn in Fig. 12.30. Particle I
has a mass of 1W5 kg and a charge o f 4 x IO-' C (coulomhs) and i s at the
origin at the instanl (ifinterest. Particlcs 2 and 3 each have a mass of 2 x
1 0 kg and a charge o f 5 x 10 C and are located. respectively, at the
rP?
Figure 12.30. Char&
particles in field E .
inslant of interest I ni d i m g t h e y axis and 3 m along the :axis. An electric field E given a s
E = 2.ti
+ 3 r j + 3(r + .?)k
NIC
(a)
i s imposed from thc nutsidc. Compute: (a) the position [ifthe center 01
mass for the system. (h) the acceleration o f the center o f mass, and (c) the
acceleratinn o f particle I
To get the position [if the ccnter of mass, we merely equate moments
of the masses ahnut the origin with that of R particle having a mass equal
to the sum o f masses of the system. Thus.
(I+ 2
+ 2)x
IO ' r , = ( 2 x 10 i ) j+ ( 2 x 1Wi)3k
Therefore,
rc = .4j+ 1.2km
(h)
To get the acceleration of the mass center. we must find the sum of
the extrmal forccs acting on thc particles. T w o cxternal forccs act on each
particle: the force o f gravity and the electrostatic force from the external
field. Recall from physics the1 this electrostatic force i s given as 4E, where
4 i s the charge on the particle. Hcnce, the total exlemal force for each particle i s given a s Sollows:
Fl
=
-(9.X1)(11i~7)k+ 0 N
Fz =-(9,Xlj(2x 1 0 ~ ~ j k + ( 5 ~ 1 0 ~ ~ j ( 3 k ) N
F3 = -(9.81)(2 x
jk
+ (5 x I O s ) ( 9 j + 27k) N
(C)
(d)
(e)
SECTION 12.10 THE GENERAL MOTION OF A SYSTEM OF PARTICLES
Example 12.15 (Continued)
The sum of these forces FT is
F~ = 45 x
io-5j
+ 100.9 x 10-5k N
(f)
Accordingly, we have for<:
..
rc =
45 x 10-5j + 100.9 x 10"k
s x 10-5
Finally, to get the acceleration of particle I , we must include the
coulombic forces from particles 2 and 3. As you leamed in physics, this
force is given between two particles a and b with charges q, and qb as follows:
where P is the unit vector between the particles, and eo is the dielectric
constant equal to 8.854 x lO~'*F/m (farads per meter) for a vacuum. Note
that the coulombic force is repulsive between like charges. The total coulombic force 4 from particles 2 and 3 is
= - I S j - 2kN
--
The total force acting on particle 1 is then
.
( FI )T -- -(9.81)(10-5)k
,
,+
from
weight
0
+ (-18j
- 2k)N
from
from
external
l"f0rnal
field
field
(i)
Clearly, the internal field dominates here. Newton's law then gives us
fl
-18j - 2k
10-5
We see here from Eqs. (g) and (j) that although the particles tend to
"scramble" away from each other due to very strong internal coulombic
forces, the center of mass accelerates slowly by comparison.
561
568
CHAPICK I 2
PAKllCLE DYNAMICS
Example 12.16
A young man i s standing i n a canoe awaiting a young lady (Fig. 12.3 I ) .
The !man weigh:, 150 Ih. and, a s shown. i s positioncd ticiir the end of the
canoe, which weighs 200 Ih. When the ynung lady appear:,, he quickly
scramble:, forward to greet her, hut when hc has moved 20 fl 10 the f i r ward end nf thc canoe. l i c finds (nnt having htudied mcclianich) that he
cannot reach her. How far i s the tip of the ciinoc from the dock after our
hem has made the 20-ft dash? The ciinoc is i n tin way ticd to the dock and
there are no water currents. Neglect friction frnm (lie water on the canoe.
Figure 12.31. Man in i.anor awaits hi, date.
The center o l iiias'i ofthc inan plus the canoe cannot change posilioii
during this action sincc there i s 110 iict external lorcc acting nn (hi:, hystem
during this action. Hencc thc firht mnincnt of mass about a n y fixed position r n u b t rcmain constant during this action. I n Fig. 12.32 we have shnwn
the man i n the forward position and we choose the position at the tip of the
dock to equate ~ i ~ o ~ i i coft i tilass
t
iil the beginning <if the action and ,iu\t
when the inan has moved the 20 fc. We then can say, noting that we arc
denoting the unspecified distance hetween the tip of the canoc and the forward position o1'Lhe iiiaii as d as shown i n Figs. 12.31 and 12.32.
Conccling terms where possible
.
7511, = 7,0._
0., .
0.
we then
...
liiive
,.
I
-.- ....
R571
,
ft
12.90. A wamor of old is turning a sling in a vertical plane. A
rock of mass ..3 kg is held ill the sling prior to releasing it against an
enemy. What is the minimum speed u t o hold the rock in the sling?
F _
with x and y in feet. A small one-passenger vehicle is designed to
move along the catenary to facilitate repair and painting of the
bridge. Consider that the vehicle moves at uniform speed of 10
ft/sec along the curve. If the vehicle and passenger have a combined mass of 250 Ibm, what is the force normal to the curve as a
function of position x ?
Figure P.12.90.
12.91. A car is traveling at a speed of 55 mi/hr along a banked
highway having a radius of curvature of 500 ft. At what angle
should the road he bdnked in order that a zero friction force is
needed for the car to go around this curve?
12.92. A car weighing 20 kN is moving at a speed V v f 60 k d r
on a road having a vertical radius vf curvature of 200 m as shown.
At the instant shown, what is the maximum deceleration possible
from the brakes along the road for the vehicle if the coefficient of
dynamic friction between tires and the road is .55?
Figure P.12.94.
12.95. A rod CD rotates with shaft G-G at an angular speed u
of 300 rpm. A sleeve A of mass 500 g slides on CD. If no friction
is present between A and CI), what is the distances for no relative
motion between A and CD?
200 ni
Figure P.12.92.
12.93. A particle moves at uniform speed o f 1 d s e c along a
plane sinusoidal path given as
y = 5 sin lix m
What is the pusition between x = 0 and x = 1 m for the maximum fbrce normal to the curve? What is this force if the mass of
the particle is I kg?
12.94. A catenary curve is formed by the cable of a suspension
bridge. The equation of this curve relative to the axes shown can
he given it\
y =
5 (e" + e-")
= a cosh M
G
Figure P.12.95.
12.96. In Problem 12.95. what is the range of values for S for
which A will remain stationary relative to CD if there is coulomhic friction hetween A and CD such that p,5 = .4?
12.97. A circular rod EB rotates at constant angular speed u of
50 rpm. A Fleeve A of mass 2 Ibm slides on the circular rod. At
what position B will sleeve A remain stationary relative to the rod
ELI if there is no friction?
569
what is the velocity of each particle relative to the center of mass
nf the system after 2 sec have elapsed'! Each particle has a weight
of . I O L .
12.101, A stationary uniform block of ice is acted on by forces
that maintain constant magnitude and direction at all times. If
F;
= (25,y)N
5
= (log) N
F? = (15g) N
what is the velocity ofthe center of mass ofthe block after 10 sec'!
Neglect friction. The density OS ice is 56 Ihm/fti.
C
Figure P.12.97.
12.98. In Problem 12.91 assume that thcre is coulombic friction
between A and EB with p,$= .3. Show that the minimum value
of Ofor which the sleeve will remain stationary relative to the rod
is 75.45".
12.99. The fbllowing data for a system of particles are given a1
time t = 0
M , = SO kg at position ( I , 1.3, -3) m
M2 = 25 kg at position (-.6, 1.3, -2.6) m
M3 = 5 kg at position (-2.6, 5.3, 1 ) m
The particles are acted an by the following respective external
forces:
+
l0rk N (particle 1 )
F2 = 50kN
(particle 2)
F, = 5t2i N
(particlc 3)
6=
5Oj
Figure P.12.101.
12.102, A space vehicle decelerates downward ( Z direction) at
1 ,h I3 kmlhrisec while moving in a translalory manner relative to
inertial space. Inside the vehicle is a rod BC rotating in the plane
of the paper at a rate of SO radisec relative to the vehicle. Two
masses wtate at the rille of 20 radisec around BC on rod EF. The
masses are cach 300 mm from C. Determine the force transmitted
at C hetween BC and EF if the mass of each of the rotating bodies
is 5 kg and the mass of rod E F is I kg. BC is in the vertical position at the time of interest. Neglect gravity.
What is the velocity of M , relative to the mass center after 5 sec,
assuming that at f = 0. thc particles are at rest?
*12.100. Given the following force field:
F = -2xi
+
3j - zk lbislug
what is the force on any particle in the field per unit mass of the
particle. If we havc two particles initially stationary in the field
with position vcctors
570
r , = 3i
+
2j ft
r2 = 4i
-
2j
+ 4k ft
Py
I
X
Figure P.12.102.
12.103. Two men climb aboard a barge at A to shift a load with
the aid of a fork lift. The barge has a mass of 20,000 kg and is 10
m long. The load consists of four containers each with a mass of
1,300 kg and each having a length of 1 m. The men shift the containers tv the opposite end of the barge, put the fork lift where
they found it, and prepare to step off the barge at A , where they
came on. If the barge has not been constrained and if we neglect
water friction, currents, wind, and so on, how far has the barge
shifted its position? The fork lift has a mass of 1,000 kg.
Figure P.IZ.104.
Figure P.12.103.
12.105. Two identical adjacent tanks are each 10 ft long, 5 ft
high, and 5 ft wide. Originally, the left tank is completely full of
water while the right tank is empty. Water is pumped by an internal pump from the left tank to the right tank. At the instant of
interest, the rate of flow Q is 20 ft3/sec,whileQ is 5 ft3/sec2.What
horizontal force on the tanks is needed at this instant from the
foundation? Assume that the water surface in the tanks remains
horizontal. The specific weight of water is 62.4 Iblft’.
I--IO’- 1 -
1
0
’
4
12.104. An astronaut on a space walk pulls a mass A of 100 kg
toward him and shonens the distance d by 5 m. If the astronaut
weighs 660 N on earth, how far does the mass A move from its
original position? Neglect the mass of the cord.
Figure P.12.105.
12.11
Closure
In this chapter, we integrated Newton’s law for various coordinate systems.
Also, with the aid of the mass center concept, we formulated Newton’s law
for any aggregate of panicles. In the next two chapters, we shall present alternative procedures for more efficient treatment of certain classes of dynamics
problems for particles. You will note that, since the new concepts are all
derived from Newton’s law, whatever problems can be solved by these new
methods could also be solved by the methods we have already presented. A
separate and thorough study of these topics is warranted by the gain in insight
into dynamics and the greater facility in solving problems that can be achieved
by examining these alternative methods and their accompanying concepts. As
in this chapter, we will make certain generalizations applicable to any aggregate of particles.
57 1
-
12.106. A hlock A of mass 10 kg rests on a second block R of
mass 8 kg. A force F equal to 100 N pulls block A . The coefficient
of friction between A and A is .S; between B and the ground. . I
What is the speed of block A relative to block B in 0.1 sec if the
-"
system starts from rest?
Figure P.12.108.
12.109. A ipring requires a fiircr x2 N for a deilection UIX
~mm,
where 1 is the deflection of the spring from the undeiurmcd georriet?. Because the deflection i'i n a l propottimiil to .li t o t h r Iirsi
power, the spring is C d k d a rLo,ilitieor spring. If a I(lO-kg hlock is
suddenlv, released on the undetnrmrd sormg.
-. what i s the meed of
the block after il has descended I O mn'!
~~
~~
~
~~~~
~
~~
Figure P.12.106.
*12.107. A block B slides from A to F along a rectangular chute
where there is coulombic friction on the faces of the chute. The
coefficient of dynamic friction is .4. The bottom face of the chute
is parallel to face EACF (a plane surface) and the other two faces
are perpendicular to EACF. The body weighs 5 Ib. How long does
it take R to go from A to F starting from rest'?
Figure P.12.109.
1Z.IJU. A horifontal platform is rotating at il coilstant angular
speed 0 ot 5 radls. Fixed to the platiorni is a triclionlcw chute i n
which two identical masses each of 2 kg are constrained hy a pair
01lincar spring, cach 01 q x i n p conitant K = 250 Nlm. If the
unstrctched length lo of rach o i thc qxings i b . I K m, \how that at
?teddy stiltc the anglc 0 must hilvc the value 36x7". Springs are
fixed to the platform at A.
/'Z
K = 250 Nlm
Figure P.12.107.
12.108. A tugboat is pushing a barge at a steady speed of X
knots. The thrust from the tugboat needed for this motion is 800
Ih. The barge with load weighs I00 tons. If the wilier resisrancr 10
the barge is proportional to the speed of the barge. how l w p will
it take the barge to slow to 5 knots after the tugboat ceascs to
push? (Note: 1 knot equals 1.152 milhr.)
572
Figure P.12.110.
12.111. What is the velocity and altitude of a communications
satellite that remains in the same position above the equator relative to the earth's surface?
12.116. The following data are giveii for the flyhall governor
(read Problem 12.60 for details on how the governor works):
I = ,215 m
D = 50mm
12.112. A satellite is launched and attains a velocity of 19,000
o = 300rpm
mihr relative to the center of the earth at a distance of 240 mi
e = 45"
from the earth's surface. The satellite has been guided into a path
that is parallel to the earth's surface at burnout.
(a) What kind of trajectory will it have'?
(h) What is its farthest position from the earth's surface?
(c) 11it is in orhit, compute the time it takes to go from
the minimum point (perigee) to the maximum point (apogee) from the earth's sutiace.
(d) What is the minimum escape velocity for this position of launching?
12.113. A rocket system is capable of giving a satellite a velocity of 35,200 km/hr relative to the earth's surface at an elevation
of 320 km above the earth's surface. What would be its maximum
distance h from the surface of the earth if it were launched ( I )
from the North Pole region or (2) from the equator, utilizing the
spin of the earth as an aid?
12.114. A space vehicle is to change frnm a circular parlang
orbit 320 km above the surface of Venus to one that is 1,620 h
above this surface. This motion will be accomplished by two fKings of the rocket system of the vehicle. The first firing causes the
km above the surface of
vehicle attain an apogee that is
At this apogee, a second firing is accomplished so as
achieve the desired circular orbit. What is the change
- in swed
demanded for each firing if the thmst is maintained in each
instance over a small portion of the trajectory of the vehicle?
Neglect friction. The radius Of Venus IS 6,160 km, and the escape
velocity at the surface is 1.026 x I O 4 mlsec2.
P
Figure P.12.116.
~
What is the force P acting on frictionless collar A if each ball has
a mass of I kg and we neglect the weight of all other moving
members Of the system?
~2.117. A spy
to observe the united states
is put into a
circular orbit about the North and South Poles. The satellite is to
(24 hr), What must he the distance from the
make 10
surface of the earth for this satellite?
12.115. Weights A and B are held by light pulleys. If released
from rest, what is the speed of each weight after 1 sec? Weight A
is 10 Ib and weight B is 40 Ih.
N
Figure P.12.115.
Figure P.12.117.
S
513
12.118. A skylah i s in a circular w h i t ahout the earth at a dirance of 500 km a h w e the earth's surface. A space shuttle has
.endezvoused with the skylab and now, wishing tu dcpart, dcoou,les and fires its rockets to move more slowly than thc skylah. If
he rockets are fircd vver il shon time interval, what should the
.elalive speed between thc spdcc shuttle and skylah he at thc end
~f rocket fire ifthe space shuttle ir tu come as cluse as IO0 k m tu
he earth's surface in auhxqurnt ballistic (rocket motoii off1
light'!
12.119. A space vehiclc is launched at a speed of 19,001)milhi
d a t i v e to the earth's center at a position 250 mi a h w e the earth'\
iutiace. If the vehicle has a radial velocity component of 3,000
nilhr toward the earth's center, what is the eccentricity o i the t r a ~
ectory'! What i s the maximum elevation a h w e the earth's a u r f c c
a c h e d by the vehicle'? Do not usc Eq. 12.54.
12.120. A skier is moving down a hill at a speed of 30 milhr
when he is at the position shown. I i the skicr weighs I X O Ih, what
uta1 force do his skis exert 011 thc m u w surikcr? Assumc that thc
:oefficient of friction is . I . The hill Can hc taken a h a paldhdiu
(h)
Figure 1'. 12.121
surface.
50' ___/
Figure P.12.120.
12.121. A submarine is moving at cunstant speed of 15 knuli
ielow the surface of the ocean. The sub i s at the same time de,tending downward while remaining horimntal wnh an acceleraion of , 0 2 3 ~ In
. the submarine a tlyball gvvernor opcrates with
weights having a mass each of 500g. Thc governor i s rotating with
ipeed w of 5 r d s e c . If at time I. 0 = 30", 0 = .2 radlaec, and 0
= I rad/sec2, what i s the fwce developed on the upport of guverlor system as a result solely of the mution iif the wcights at thia
nstant?
i74
Figure 1'. 12.122.
12.123. Three bodies have the following weights and positions
at timet:
W, =
IOlb,
x, = 6ft,
y , = 10 ft,
12.125. A small body M of mass 1 kg slides along a wire from A
to B. There is coulombic friction between the mass M and the
wire. The dynamic coefficient of friction is .4. How long does it
take to go from A to B?
10 ft
ZI =
W, = 5 Ib,
x2 = 5 ft,
y1 = 6 ft,
z2 = 0
W, = 8 Ib,
x3 = 0,
y., = -4 ft,
zj = 0
Determine the position vector of the center of mass at time t.
Determine the velwity of the center of mass if the bodies have the
following velocities:
+
V,
= 6i
V,
= 1Oi - 3kftisec
3jfUsec
V, = 6k ft/sec
Figure P.12.125.
z
12.126. F o r M = I slug and K = 10 Ibhn., what is the speed at
= 1 in. if a force ot 5 Ib in the x direction is applied suddenly to
the massspring system and then maintained constant? Neglect
the mass of the spring and friction.
x
/L;;.
e
w3
Y
w2
X
Figure P.12.123.
12.124. In Problem 12,123, the following external forces act on
the respective particles:
6
= 6fi + 3 j - 1Ok Ih
(particle 1)
F2 = l5i - 3 j l b
(particle 2)
F,, = OIb
(particle 3)
X
Figure P.12.126.
12.127. A rod B of mass 500 kg rests on a block A of mass 50 kg.
A force F of 10,ooO N is applied suddenly to block A at the position
shown. If the coefficient of friction pd is .4 for all contact sutiaces,
what is the speed ofA when it has moved 3 m to the end of the rod?
What is the acceleration of the center of mass, and what is its position after I O sec from that given initially? From Problem 12.123 at
t = o
rc =
V, =
+ 4.26j + 4.35k ft
4.7% + 1.304j + 1.435k ft/sec
3.1%
Figure P.12.127.
I
12.128. A simply supported beam i s shown You will learn in
your course on strength of materials that a vertical furcc I applicd
at the center cause.: a dellection 6at the center giveii a\
If a mdSS of 200 Ihm, fd\tened to the hsam at i t s midpoint. is suddenly relcased. what will its speed hc when the dellrction i\ in.'!
Neglect the mass of the heam. The lcngth of the ham, L. i s 20 it.
Young's modulus E is 30 x IO" psi. and the moment of incrtia 0 1
thc cross bection I i\ 20 in?.
W(j
Figure P. 12.13I.
L
M,# = KOks
MA = 101!kg
Figure P.12.128.
M , = 5Ohg
12.129. A piston is shown mainlaining air ill a pressure of X psi
ahore thal ofthe almo.sphrre. I I the piston is all~iwrd1,) ~ccsleraleto
the left, Whdl i s the speed of the piston a h it I ~ O Y C S3 in.? Thc piston assembly has a mass 013 lhin. Assume that the air rxpends ndi~
ahuticdly (i.e.. with no heat transfer). This meenh that :it all times
(,Vi = constant, where V i ? thc volume of the gas and k ib a coilstant which f a r air cquals I .4. Neglect the incrtiill elfects of the air
-
I'
I
Figure P.lZ.I.32.
f
Figure P.12.129.
*12.130. In Example 12.6 assume that there are adiabatic expans i o n s and compressions of the gave, (i.c.. that
= constant
with k = 1.4). Compare thc 1esu1ts for thc speed of the piston.
Explain why yuur result should bc higher or l m v w than h r the
isothermal case.
12.131. Body A and hvdy B are connected hy an incxtenrihle
curd as shown. If hoth bodies are released cimultanetiu\ly. what
dislance do they move in
i e c ? Take MA = 25 kg and Ma = 35
kg. The coefficient of friction p,, is .3.
t
i
.
.
f, 30'
I
~
L.x
.I.,
100 kg
M,, .~60 kg
f
Figure P.12.133.
~
3.000 N
12.134. The system shown is released from rest. What distance
does the body C drop in 2 sec? The cable is inextensible. The coefficient of dynamic friction pd is .4 for contact surfaces of bodies
A and A.
Figure P.12.137.
Figure P.12.134.
12.138. A car is moving at a constant speed of 65 kmlhr on i
road pan of which (A-R) is paraholic and part (if which (C-D
is circular with a radius vf 3 kin. If the car has an anti-lock brak~
ing system and the static coefficient of friction LI, between the
road and the tires is 0.6, what is the maximum deceleration p o s i ble at the x = 2 km position and at the .x = I O km position'? The
total vehicle weight is 12,000 N.
12.135. Do Problem 12.134 for the case where there ic viscou\
damping for the cmtact surfaces of bodies A and B given as .5V
Ih, with V i n ftlsec.
12.136. Two biidies A and A are shown having masses of 40 kg
and 30 kg, respectively. The cables are inextensible. Neglecting
the inertia of the cable and pulleys at C and D, what is the speed
of the block A I sec after the system has been released from rest'?
The dynamic coefficient of friction p,, for the contact surface of
body A is 3. [Hint: From your earlier work in physics, recall that
pulley I ) is instantaneously rotating about point ( I and hence point c
lnnves at a speed that is twice that of point h.]
,
I
I O km
2km
Figure P.12.138.
12.139. A mass rrf 3 kg is moving along a vertically oriented
parabolic rod whose equalion i s ? = 3 . 4 ~ ' A
. linear spring with K
= 550 Nim connects to the mass and is unstretched when the
mass is at the bottom of the rod having an unstretched length & =
I m. When the spring ccnterline is 30" from the vertical, a< shown
i n the diagram, the mass is moving at 2.8 mls. At thih instant, what
is the force component on the rnd directed normal tn the rod'!
I
L
2 4r'
..
~
M-3kg
Figure P.12.136.
K
12.137. Bodies A , A, and C have weights,
of 100 Ib, 200 Ih, and
150 Ib, respectively. If released from rest, what are the respective
speeds of the bodies after I sec? Neglect the weight of pulleys.
=
sso N/,,,
- -x
Figure P.12.139.
577
12.140.
A heated cathode gives off electnins which are attracted
to thc pusitivc anode. Some go through a s m d 1 hole and enter the
parallel plales at an anglc with the hori/ontal o f a,, = 0 and a
velcrcily of
Dctrrrninc the horizontal and ve!iical motiorls of
the declron inside the plates as a function o f time. Letting r = /.
find thc time that the clectron i s in the parallel plalc rcgion and
then "blain thc exit vertical velucily. Assuming straight-line motion
until the electrm~hits thc screen. show that the vcnicill position of
impact, assuming thc screen i s flat. i s
vr
*12.141.
A weightless cord supports two identical milh\es cauh
W . The cord i s heing pulled at a constdnt speed y, by a
force I.'. Formulate im equation for F ill terms of l{j,
L, i. W , and h .
Detmmine t for the filllowing condition\:
o l weight
\{, = 2.2 nil\
L = 3.3
111
/ = .2h
111
W = $0 N
I, = .I6 m
Location of
sinall hole
Figure P.12.140.
Figure P.12.141.
Energy Methods
for Particles
Part A
13.1
Analysis for a Single Particle
Introduction
In Chapter 12, we integrated the differential equation derived from Newton's
law to yield velocity and position as functions of time. At this time, we shall
present an alternative procedure, that of the method of energy, and we shall
see that certain classes of problems can be more easily handled by this
method in that we shall not need to integrate a differential equation.
To set forth the basic equation underlying this approach, we start with
Newton's law for a particle moving relative to an inertial reference, as shown
in Fig. 13.1. Thus,
z
1
F = m -d z r dt?
mdV
dt
(13.1)
Multiply each side of this equation by dr as a dot product and integrate from
r, to rr along the path of motion:
X
Figure 13.1. Particle moving relative to an
inertial reference.
In the last integral, we multiplied and divided by dt, thus changing the varable of integration to t . Since drldt = V . we then have
519
580
CHAPTER I3 ENERGY METHODS FOR PART1CLb.S
On carrying out the integration, we arrive at the familiar equation
(I3.2)
where the left side i s the well-known expression for work (to be denoted :it
1
times as ?li-2)l
and the right side i s clearly the change in kinetic. enrr,q? as
the mass moves from position rI to position r2.
We shall see in Section 13.7 that for any system ofparticles, including. of
course, rigid bodies. we get a work-energy equation o f the form 13.2, where
the velocity i s that of the mass center, the force i s the resultant external force on
the system, and the path of intcgration i s that of the mass center. Clearly, then.
we can use a single particle model (and consequently Eq. 13.2) for:
I. A rigid hody movirig without nmtion. Such a motion was discussed in
Chapter I I and i s called traiislatioii. Note that line:, in a translating body
Figure 13.2. lranslating hody.
remain parallel lo their original directions, and points i n the body move
over a path which has identically the same form for all points. This condition i s illustrated in Fig. 13.2 for two points A and B. Furthermore, each
point in the body has at any ins1:int oftiinc Ithe same vel(ici1y as any other
point. Clearly the motion o f the center o f mass fully characterizes the
motion o f the body and Eq. 13.2 w i l l hc used often for this hituation.
2. Sometimes for a bod)>whose s i i e is sniull c.o,izporrd to i t s t r ( ~ j r l r c t i ~Here
r~.
the paths of points in the body d i l t r very little from that of the mass center and knowing where the center o i i n a s s i s tells us with sufficient accuracy all we need tu know about the position o f the body. However, keep in
mind that the v e k ~ i t yand ucwl6'mliivJ 1-el;itive to the center of mass of a
pan of the body may be Very large. irrespective o f how small the body
may be when compared to the trajectory OS i t b center of mass. Then, information about the velocity and acccleration of this part o f the body relative
to the center of mass would require a more detailed consideration beyond
a simple one-particle model centered around the ccnter o f mass.
Thus, as i n our considcrations or Newton':, law in Chapter 12. when the
motion of the in as^ center characterizes with sufficient accuracy what we
want to know about thc motion of a body. we USE a particle at the mass center for energy considerations.
Next, suppose that we have a component OS Newton's law in one direction. say the x direction:
SECTION 13.1 INTRODUCTION
Taking the dot product of each side of this equation with h
i + dyj
(= dr), we get, after integrating in the manner set forth at the outset:
+
dzk
Similarly,
(13.3b)
Thus, the foregoing equations demonstrate that the work done on a particle in
any direction equals the change in kinetic energy associated with the component of velocity in that direction.
Instead of employing Newton’s law, we can now use the energy equations developed in this section for solving certain classes of problems. This
energy approach is particularly handy when velocities are desired and forces
are functions of position. However, please understand that any problem solvable with the energy equation can he solved from Newton’s law; the choice
between the two is mainly a question of convenience and the manner in
which the information is given.
ExamDle 13.1
An automobile is moving at 60 mihr (see Fig. 13.3) when the driver jams
on his brakes and goes into a skid in the direction of motion. The car weighs
4,000 lb, and the dynamic coefficient of friction between the rubber tires
and the concrete road is .60. How far, 1, will the car move before stopping?
A constant friction force acts, which from Coulombs law is pdN =
(.60)(4,000) = 2,400 Ih. This force is the only force performing work, and
c l e d y it is changing the kinetic energy of the vehicle from that corresponding to the speed of 60 mihr (or 88 ft/sec) to zero. (You will learn in
thermodynamics that this work facilitates a transfer of kinetic energy of
the vehicle to an increase of internal energy of the vehicle, the road, and
the air, as well as the wear of brake parts) From the work-energy equation 13.2, we get2
1 4,000
-2,4001 = - -( 0 - 88*)
2 g
Hence,
I =
(Perhaps every driver should solve this problem periodically.)
‘Note that the sign of the work done is negative since the friction force is opposite in
s m s e to the motion.
4.oM) Ib
Figure 13.3. Car moving with brakes
locked.
581
582
CHAPTER I 3 ENERGY METHODS FOR PARTK'LES
Example 13.2
Shown in Fig. 11.4 i s a light platSorm R guided hy vertical rods, Thc platform i s positioned so that the spring has been compressed 1 0 mm. In thir
configuration a body A weighing 100 N i s placed on the platiorm and released suddenly. If the guide rods give a total ciinstant resistance forccf to
downward movement CIS the platform 1ir 5 N, what i r the largest distancc
that the wcight f i l l s ' ? The spring used here i s il rionliricor spring requiring
.Sx2 N of f i x x for il deflection of x mm.
We take as the position OS interest for thc body the location 6 helo\u
the initial configuration at which Iiication thc body A reaches x r o velocity
for the f i r s t lime alter having been released. The change i n kinctic encrgy
ovci- the interval i s accordingly zerii. Thus, zero net work IYd i m hy the
fiirces acting on the body A during diydacemcnl 6. Thcse forces comprihc
the force ofgravity. the friction force Sroin the guides. and finally the force
from the spring. Using as thc origin for our tne:tsuretnenls the rindrfiwmi4
top end position 01 the spring,' we can say:
Therefiirc,
6'+ 306'
- 2706 = 0
(L )
One solution to Eq. (c) i s 6 = 0. Clearly, no work i s done if there i s no
detlection. But this solution has no meaning Tor this problem since the force
in the spring i s only .5x2 = .5(10)2 = SO N, when the weight of 100 N i s
released. Therefore, there must be a non7,erii positive valuc of 6that satislics
the equation and has physical meaning. Factoring out onc 6 from the equation, we then set the resulting quadratic expression equal 10 7.erci. Two roots
result and the positive root 6 = 7.25 mm i s the one with physical meaning.
6 = 7.25 mm
'Since the force in the apring i \ il lunction 01 lhc rlongalion ut the ipiiiig from i t \ i,n,k,~
formed gmnetry. we must pul Ihe origin ot mu referenre at a p a s i l i i m c ~ r m p m ~ l i n10p ilic
undelormcd geometry. At this position. both I and the bpriog h l ~ arc
c KIC ~~,,,~,l,~,,,~,,~,~l~
In the f ~ i l l o w i n gexample we deal with two bodies which can he ciiiisidrred as pnrticles, lathcr than with one body as lids been the case in the previous examples. We shall deal with these biidies separately in t h i s example.
Later i n the chapter. we shall consider s?.srenn of particles, and in that context we w i l l he ahlc t i i consider this problem a\ a system of particles with less
work needed ti1 reach a siilutiiin.
SECTION 13.1 INTRODUCTION
Example 13.3
In Fig. 13.5, we have shown bodies A and B interconnected through a block
and pulley system. Body B has a mass of 100 kg, whereas body A has a
mass of 900 kg. Initially the system is stationary with B held at rest. What
speed will B have when it reaches the ground at a distance h = 3 m below
after being released? What will he the corresponding speed of A? Neglect
the masses of the pulleys and the rope. Consider the rope to he inextensihle.
'r I
I
r
c
h
'XB
Figure 13.5. System of blocks and pulleys.
You will note from Fig. 13.5 that, as the bodies move, only the distances
l, and lA change; the other distances involving the ropes do not change. And
because the rope is taken as inextensible, we conclude that at all times
l,
+ 41A = constant
(a)
Differentiating with respect to time, we can find that
l,
+ 41,
=0
Therefore,
in
= -4i,
(b)
On inspecting Fig. 13.5, you should have no trouble in concluding that
= -V, and that in = V,. Hence, from Eq. (h), we can conclude that
v,
= 45
(C)
Next take the differential of Eq. (a):
dl,
+
4dlA = 0
Therefore,
dlB = -4dl,
(d)
583
584
CHAPIBR I3
ENEKGY M t T I l O D S I'OK PARTICLES
Example 13.3 (Continued)
?
i8
I
:
4
N o k that d ~ =, (/IH
~ and lhat 'It,, = -dl,l. Hencc we szc Irom Eq. (dl that
:I niovcnicnt magnitudc A, 0 1 hody A rcsults i n a no\clnent ningnilude.
4A,. of hwly H :
A,j = 4A,,
IC)
With these kinemaLic;ll conclusions as a hackground. we arc now rcady to
pnicced with llie workbenergy considerations.
For lhis purpose, wc liave shown a lrcc-hody diagram of body /I
i n Fig. 13.6. Thc work-energy equation for hody /I can then he givcn 21s
Therefore.
(981 ~-T)i3) =
p
?
!
IOOV;
if)
Now consider thc Iree-body diagram of hody A in Fig. 13.7. The
work-energy equation for hotly A i s then
(47)iA,,)
I
r
'3
I~lll~>w5:
=
i9ilOV:
cg)
And according to Eq. IC).
\/
.I
L \/
~I
R
Suhslituting thc results Ironr Eqs. ( h ) and (i) into (gl. MC gel
(i)
SECTION 13.2
i
Example 13.3 (Continued)
Adding Eqs.(f) and (j), we can eliminate T to form the following equation
with V, as the only unknown:
(981)(3) = f ( V i ) ( l O O
+ F)
Therefore,
V,, = 6.14 d s e c downward
Hence,
V, = 1.534 d s e c to the left
13.2
Power Considerations
The rate at which work is performed is calledpowrr and is a vely useful concept
to represent work, we have
for engineering purposes. Employing the notation
<wk
dwk
- 4.,
Since
for any given force F. is <.
we can say that the power
being developed by a system o f n forces at time f is, for a reference xyz,
(13.5)
where is the velocity of the point of application of the ith force at time I as
seen from reference . ~ y z . ~
In the following example we shall illustrate the use of the power conccpt. Note, however, that we shall find use of Newton’s law advantageous in
certain phases of the computation,
“We could haw defined work
%
in terms
iof power as follows:
When the force acts on a particular particle. the result ahove hecomcs the familiar
5,:
F * dr.
where I is the pohition vestor uf the panicle since V dr = dr. Thcrc are times when the force acls
on ionrinuourly <hanging particles as time pasaer (see Section 13.8).The more general fur mu la^
lion above can then be used effectively.
POWER CONSIDERATIONS
58s
586
CHAPTER 13 ENERGY METHODS FOR PARTICLES
Example 13.4
In hilly terrain, motors of an electric train are sometimes advantageously
employed as hrakes, particularly on downhill runs. This is accomplished
hy switching devices that change the electrical connections of the motors
so as 10 comespond to connections for generators. This allow? power
developed during hraking to he returned to the power source. In this way,
we save much of the energy 1051 when employing conventional hrakcs-a
considerable saving in every round trip. Such a train consisting of B single
car is shown in Fig. 13.8 moving down a 15" incline at an initial speed of
3 mlsec. This car has a mass of 20,000 kg and has a cogwheel drive. If the
conductor maintains an adjustment of the fields in his generators s o as to
develop a constant powcr O M ~ ~ of
U I 50 kW, how long does it take before
the car moves at the rate of S nilsec'? Neglect the wind resistance and rotational effects of the wheels. The efficiency of the generators is 90%
Figure 133. Train moving downhill with generators acting as brakes
We have shown all the forces acting on the car in the diagram.
Newton's law along the direction of the incline can he given as
W s i n l Y - / = M dV
dt
(a)
where ,f is the traction force from the rails developed by the generator
action. Multiplying hy V to gel power, we get
W s i n 1 5 ° V - , f V = M Vd ~V
dt
(h)
If the efficiency of the generators (i.e., the power output divided by the
power input) is .YO. we can computc.fV. which is the power input to
the generators from the wheels. in the following manner:
~
generator output
.90
~~~~~
-
.fV
(c)
SECTION 13.2 POWER CONSIDERATIONS
Example 13.4 (Continued)
Hence
Equation Ibj can now be given as5
(2O,OoO)(Y.81)(.25Y)V - 55,560 = 20,000 V$
Therefore.
2.54V - 2.78 = V-dV
(e)
dt
We can separate the variable5 as follows:
dt =
V dV
2.5411 - 2.78
Integrating, using formula I in Appendix I, we get
f =
1
-[2.541/
2.54=
- 2.78
+ 2.78ln (2.54V - 2.78)] + C
To get the constant of integration C, note that when
Hence,
1
0 = -{(2.54)(3)
2.54=
- 2.78
t
= 0, V = 3 d s e c .
+ 2.78 In [(2.54)(3) - 2.78]} + C
Therefore.
C = -1.430
We thus have for Eq. (g):
f = -
2.54=
[2.54V - 2.78 + 2.78 In(2.54V - 2.78)] - 1.430
When V = 5 mlsec, we get for the desired value off:
f=-- I
2.542
{(2 .54)(5) - 2.78 + 2.78In[(2.54)(5) - 2.78]} - 1.430
t = l
watt (W)
(g)
is I Jlsec, where J =joule, which in turn is I N-m.
587
13.1. What value of constant force P is required to bring the
100-lb body, which starts fmm rest, to a velocity of 30 ft/sec in
20 f t ? Neglect friction.
Figure P.13.1.
Figure P.13.4.
13.2. A light cable passes over a frictionless pulley. Determinz
the velocity of the 100-lh block after it has moved 30 ft from rest.
Neglect the inertia of the pulley. The dynamic cocfficienl of friction between block and incline i s 0.2
13.5. A 50-kg mas5 on a spring is maved so that it extends the
spring 50 mm frnm its unextended position. If the dynamic cvefficient of friction hetwecn thc mass and the supporting surface is 3.
(a) What is the velocity of the mass as it returns to the
undeformed configuration of the spring?
(b) How far will the spring be campressed when the mass
stops inskmtaneously beforc starting to the left'?
F,, = .3
Higure P.13.5.
F,,
=
2
Figure P.13.2.
13.3. In Problem 13.2, the pulley has a radius o f I ft and has a
resisting torque at the bearing af I O Ib-ft. Neglect the inertia of the
pulley and the mass of the cable. Compute the kinetic energy of
the 100-lb block after it has moved 30 ft frvm rest.
13.4. A light cahle is wrapped around two dmms fixed between
a pair of blocks. The system has a mass of 50 kg. If a 250-N tension is exerted on the free end of the cable, what is the velocity
change of the system after 3 m of travel down the incline? The
for all surfaces as .OS.
body starts from rest. Take u~,
588
13.6. A truck-trailer is shwm carrying three crushed junk automobile cubes each weighing 2,500 Ih. An electromagnet i s used to
pick up the cubes as the truck moves by. Suppose the truck starts
at pasition I by applying a constant 600 in.-lb total torque on the
drive wheels. The magnet picks up only one cuhe C during the
process. What will the velocity of the truck he when it has maved
a total of 100 It? l h e truck unloaded weighs 5,0130 Ib and has a
tire diameter of I 8 in. Neglect the rotational effects of the tires
and wind friction.
25'
A
-I
Position I
' l " ~ l 0 '
Figure P.13.6.
13.7. Do Problem 13.6 if the first cube B and the last cube D are
removed as they go by the magnet.
13.8. A passenger ferry is shown moving into its dock to unload
passengers. As it approaches the dock, it has a speed of 3 knots
(1 knot = ,563 d s e c ) . If the pilot reverses his engines just as the
front of the ferry comes abreast of the first pilings at A, what constant reverse thrust will stop the feny just as it reaches the ramp
B? The ferry weighs 4,450kN. Assume that the ferry does not hit
the aide pilings and undergoes no resistance from them. Neglect
the drag of the water.
13.11. A 1,000 N force is applied to a 3,000-N block at the position shown. What is the speed of the block after it moves 2 m?
There is Coulomb friction present. Assume at all times that the
pressure at the bottom of the block is uniform. Neglect the height
of the block in your calculations. Roller at right end moves with
the block.
p n = .2
Figure P.13.11.
13.12. Two blocks A and B are connected by an inextensible
chord running over a frictionless and massless pulley at E. The
system starts from rest. What is the velocity of the system after it
has moved 3 ft? The coefficient of dynamic friction fid equals .22
for bodies A and U .
W, = 50lb
Figure P.13.12.
Figure P.13.8.
13.9. Do Problem 13.8 assuming that the ferry rubs against the
pilings as a result of' a poor entrance and undergoes a resistance
against its forward motion given as
f = 9(x
+
13.13. What are the velocities of blocks A and B when, after
stalting from rest, block U moves a distance of .3 ft? The dynamic
coefficient of friction is .2 at all sudaces.
Y
50) N
where I is measured in meters from the first pilings at A to the
front of the ferry.
13.10. A freight cdr weighing 90 kN is rolling at a speed of
1.7 d s e c toward a spring-stop system. If the spring is nonlinear
such that it develops a .0450x2-kN force for a deflection of x mm,
what is the maximum deceleration that the car A undergoes?
X
Figure P.13.10.
Figure P.13.13.
589
13.14. A particle of mass 10 Ihm is acted on hy the following
Forcc field:
F = 5xi +
(Ih
+
2y)j
+
2Ok Ih
When it is at the origin. the particle has a vclricity
v,
=
Si
+
l0j
+
y , given iis
Xkftisrc
What is ils kinetic energy when it reaches positiun (20. 5 . I O )
while moving along a frictionless path'! Does the shape of llie path
xtween the origin and (20, 5, I O ) affect the resull'?
Figure P.13.17.
13.15. A plate AA is held down by screws C and 11 xi that a
'nrce of 245 N is developed in cach spring. Mass M of 100 kg is
ilaced on plate AA and released suddenly. What is the inaxiiniim
iistance that plate AA descends if the plate can slide freely down
:he vertical guide rods'! Take K = 3,600 Nlm.
13.1% A classroom demonstration unit is used i o illustrate
vihration': and interactions of bodies. Body A has a mass 01.5 kg
and is moving to the left at a s p e d of 1.6 iiilsec at the position
indicated. 'The body rides o n il cushion of air supplied from thc
t u k H through \mall openings in the tube. If there is a ~ ~ n s i a n i
frictioii force of . I N, what speed will A have when it returns ID
thc position shown in the diagram'! Thcrc are two spring? at (',
each having a ~ p n t i pconstant nf 15 Nlm.
A
A
Figure P.13.15.
13.16. A 200-1h block is dropped o n the system of springs. I f
and K2 = 200 Iblft, what is the niaxinium force
leveloped (in the body'!
Y, = 600 Iblft
5'
I
Figure P.13.18.
13.19. A n clcctmn mwes i n a circulx orbit in a plane at right
anglcs to Ihe direction of a uniform magnetic field N . I S the
strength of I3 is slowly changed si) that thc radius of the orbit is
halved, what is the I d h OS the final t o the initial angular speed of
the electron'! Explain the steps you take. The force F on a chu-ged
particle is yV x H , where q is the chargc and Vis the veloclty d
the p;sticle.
H
X
X
X
x
x
X
Magnaio
x
Figure P.13.16.
13.17. A block weighing SO Ih i'i shown on an inclined surfacz.
The block is released at the position shown at a rest condition.
Nhat is the maximum compression of the spring'? The spring har
I spring constant K of 10 Iblin., and the dynamic coefficient 0 1
,.
riction between the hlock and the incline is .3.
i90
X
X
x
x
x
x
J Field
X
x
x
x
x
Figure P.13.19.
13.20. A light rod CD rotates about pin C under the action of
constant torque T of 1.000 N-m. Body A having a mass of 100 kg
slides an the lhorizontal surface for which the dynamic coefficient
of friction is .4. If rod CD starts from rest. what angular speed is
attained in one complete revolution? The entire weight of A is
borne by the horizontal surface.
'1
\
Figure P.13.22.
13.23. A conveyor has drum D driven by a torque of 50 ft-lb.
Bodies A and B on the conveyor each weigh 30 Ib. The dynamic
coefficient of friction between the conveyor belt and the conveyor
bed is .2. If the conveyor starts from rest, how fast along the convevor do A and B move after traveline 2 ft? Drum C rotates freely,
and the tension in the belt on the underside of the conveyor is
20 Ib. The diameter of both drums I S 1 ft. Neglect the mass of
drums and belt. A and B do not slip on belt.
~
1 Top view
Figure P.13.20.
13.21. An astronaut is attached to his orbiting space laboratory
by a light wire. The astronaut is propelled by a small attached
compressed air device. The propulsive force is in the direction of
the man's height from foot to head. When the wire is extended its
full length of 20 ft, the propulsion system is started, giving the
astronaut a steady push of 5 Ib. If this push is at right angles to the
wire at all times, what speed will the astronaut have in one revolution about A ? The weight on earth of the astronaut plus equipment
is 250 Ib. The mass of the laboratory is large compared to that of
the man and his equipment.
Figure P.13.23.
13.24. Bodies A and B are connected to each other through two
light pulleys. Body A has a mass of 500 kg, whereas body B has
a mass of 200 kg. A constant force F of value 10,000N is applied
to body A whose surface of contact has a dynamic coefficient of
friction equal to .4. If the system starts from rest, what distanced
does B ascend before it has a speed of 2 mlsec'! [Hinr: Considering pulley E, we have instantaneous rotation about point e. Hence,
v, = fy.1
Figure P.13.21.
13.22. Body A , having a mass of 100 kg, is connected to body B
by an inextensible light cable. Body B has a mass of 80 kg and is
on small whcels. The dynamic coefficient of friction between A
and the horizontal surface is .2. If the system is released from rest,
how far d must B move along the incline before reaching a speed
of 2 d s e c ?
e
F
Figure P.13.24.
591
Vigure P.13.28.
Figure P.13.29.
Figure P.1.3.27.
the path. The spring is unstretched when P is released. Neglect
friction and find how far P drops. Take 7 = 7~12,A = C = 1
13.34. A 15-ton streetcar accelerates from rest at a constant rate
a,, until it reaches a speed V , , at which time there is zero acceleration. The wind resistance is given as KV2. Formulate expressions
for power developed for the stated ranges of operation.
.
Figure P.13.34.
Figure P.13.30.
13.31. A body A of mash 1 Ibm is moving at time f = 0 with a
speed V gf I ft/sec on a smooth cylinder as shown. What is the
speed of (he body when it arrives at E! Take I = 2 ft.
Figure P.13.31.
13.32. 4 n automobile engine under test is rotating at 4,400 rpm
and develops a torque of 40 N-m. What is the horsepower developed by *e engine'? If the system has a mechanical efficiency of
.90, wha( is the kilowatt output of the generator'? [Hinr: The
work of torque equals the torque times the angle of rotation in
radianq.1
4
~~
13.35. What is the maximum horsepower that can be developed
on a streetcar weighing 133.5 kN? The car has a coefficient of sta-
tic friction of .20 between wheels and rail and a drag given as
32V2 N, where Vis in d s e c . All wheels are drive wheels.
13.36. A 7,500-kg streetcar starts from rest when the conductor
draws 5 kW of power from the line. If this input is maintained
constant and if the mechanical efficiency of the motors is 90%
how long does the streetcar take to reach a speed of 10 k d h r ?
Neglect wind resistance. ( I kW = 1.341 hp.)
13.37. A children's boat ride can be found in many amusement
parks. Small boats each weighing 100 Ib areTotated in a tank of
water. If the system is rotating with a speed 0 of 10 rpm, what is
the kinetic energy of the system? Assume that each boat has two
60-lb children on board and that the kinetic energy of the supporting structure can be accounted for by "lumping" an additional
30 Ibm into each boat. II a wattmeter indicates that 4 kW of power
is being absorbed by the motor turning the system, what is the
drag for each boat? Take the mechanical efficiency of the motor to
be 80% (1 kW = 1.341 hp.)
~~
Figure P.13.32.
13.33. A rocket is undergoing static thrust tests in a test stand. A
thrust of h ( H ) , 0 0 0 Ib is developed while 300 gal of fuel (specific
gravity 2)is burned per second. The exhaust products of combustion ha"$ a speed of 5,000 Wsec relative to the rocket. What
power is emg developed on the rocket'! What is the power d e w oped on t , e exhaust gases'! ( I gal = ,1337 ft'.)
b
Figure P.13.37.
593
594
,'
CHAPTER I? t:NER(iY METHOIS FOR PARTICLES
13.3
W
In Section 10.6 we discussed an important class [if fnrccs called conservatiw
fiirces. For convenience. wc shall now repeat this discussion.
Consider first a hody acted on only by gravity W a s an active force (i.c.. ii
fiirce that citn do wnrk) and moving along a frictionless path from position I
to position 2, a s shown in Fig. 13.9. The vmrk done by gravity Tt:
i s tllcn
i',
2'
Conservative Force Fields
0
2
x
= f l Z ~ . ( ~ r = ~ 1 2 ( - ~ ) . ( ~ r ~ = ~
i*~)~
Figure 13.9. Panicle moving along
frictimless path.
= =W(y, - V I ) =
W(.,
=
,.?I
(13.6)
Note that the work done dops iiof d(,pmil on the path, but depends only oil the
S P like , y ~ - ~ i b , i f ?i.\
positions of the end points of the path. f:orw.fieldy ~ ~ O work
iridependmr of fhf, p a f h are c.a/led wnscrvative ,/orce firld,s. I n general. we
can say f o r conservative force field F(n. y, :) that, along any path hetween
positions I and 2, the work i s
~
~
=1
~~ ~ . i 1l r = =~I , ~. . (.2~. )r- \ ? ( x1 , y , ; )
(13.7)
where \?is a function [if position of the end points and i s called the poferifiol
f~mction."We may rewrite Eq. 13.7 as follows:
(13.R)
-Jl'F.dr=\-:(n,
Note that the potential energy, X'(.x, y, z ) , depends on the reference .qused or.
a s we shall often say. the dmum used. However, the <.hungein potential cncrgy,
AX), i s in&pmdmr o f the datum used? Since we shall hc using the change in
potential energy. the datum i s arbitrary and i s chosen f(ir convenience. From Eq.
I3.X, we can say that the chongP in potenrial energy, A I ' ( = I
.! of it
conservative force field is rlir nrgutive of r l i ~work (lone hv this coir.wn'arii,e
lorwfield on i i purfide in p i n g from pmifiim I fo po.yitio,, 2 nlon,? any patli.
For any do.sed path. clearly the work done hy a cnnservative lbrce field F i s then
$ ~ . d r= O
( I3.9)
Hence. this i s a second way to define a conservative force field. How i s
the potential energy function %)related to F ? To answer this query, consider
that an infinitesimal path dr stiifls from point I . We can then give Eq. 13.8 a s
F
-
dr = -dl i
(13.10)
SECTION 13.3 CONSERVATTVE FORCE FIELD
E x p r e s s ~ gthe dot product on the left side in terms of components, and
expressing d v a s a total differential, we get
,
(13.11)
FA&+Fydv+Fd7=-
We can tonclude from this equation that
(13.12)
In other Lords.
av. av
-JrV+ .- - ~ + - k
ax
ay
az
(I 3.13)
The opetator grad or V that we have introduced 13 Cdlled the gradient operator8 and IS given as follows for rectangular coordinates:
(I 3.14)
We can now say as a third definition that a conservariveforcejield must
be a fun+tiun of position and expressible as the gradient of a scalarfunction.
The invebse to this statement is also valid. That is, $a forcefield is a function
of positi n and the gradient of a scalarfield, it must then be a conservative
,force&
T d o examples of conservative force fields will now be presented and
discusse#.
C n n s t a j t Force Field. If the force field is constant at all positions, it
can alw ys be expressed as the gradient of a scalar function of the form
7 / = -( + by + cz), where a, b, and c are constants. The constant force
field, then, is F = ai + bj + ck.
In limited changes of position ne% the earth’s surface (a common situation), d e can consider the gravitational force on a panicle of mass, m, as a
4
The~gradientoperator comes up in many situations in engineering and physics. In short.
the gradie t repre\enta a driving action. Thus. in the present case. the gradient is a driving action
to cause A s s to k v e . And, the gradient of temperature causes heat to tlow. Finally, the gradient ofelec(ric potential causes electric charge to tlow.
595
596
CHAPTER I 7
ENRROY METHODS FOR PARTICLES
constant force field given by -ingk (or -Wk). Thus, the constants for the
general force field given abovc are u = h = 0 and c = -mg. Clearly.
PE = m g for this case.
Force Proportional to Linear Displacements. Consider a hody limited by
constraints to move along a straight line. Along this line is developed a force
dircctly proportional tu the displacemcnt of the body from some position 0 at
x = 0 along the line. Furthermore, this force is always directed toward point
0; it is then termed a wstoriiig force. We can give this force as
F
=
-Kxi
(13. IS)
where .x is the displacement from point 0. An example of this force is that of
the linear spring (Fig. 13.I O ) discussed in Section 12.3. The potential energy
of this force field is given as follows wherein x is measured from the undeformed geometry (don't forget this important factor) of the spring:
configuration
Figure 13.10. Linear spring.
What is the physical meaning of the term PE'? Note that the change in
potential energy has been defined (>reEq. 13.81 as the negative of the work
done by a conservative furce as the particle o n which it acts goes from one
position to another. Clearly, the change in the potential energy is then direct/!
equal 10 the work donc by the rmction to thc conscrvative furce during this
diqplacenient. In the case of the .spring. the rcaction force would be the furce
,fron7 the surroundings acting 0 1 7 the spring at point B (Fig. 13.10). During
extension or ciimprcssion of the spring from thc undeformed position, this
force (from the surroundings) does a positive amount of work. This work can
bc considered as a nieasure uf the energy storm1 in the spring. Why'! Because
when allowed to return to its original position, the spring will do this amwnt
01 positive work on the surroundings at B , provided that the return motion is
slow enough to prevent oscillations; and so on. Clearly then. since PE equals
work of the surroundings on the spring, then PE is in effect thc stored energy
in the spring. I n a general case, PE is the energy stored in the force field as
mcasurcd from a given datum.
SECTION 13.3 CONSERVATIVE FORCE RELD
In' previous chapters, several additional force fields were introduced
the gra$tational central force field, the electrostatic field, and the magnetic
field. Let us see which we can add to our list of conservative force fields.
Cinsider first the central gravitational force field where particle m,
shown ib Fig. 13.11, experiences a force given by the equation
Mm
F =4 - f
r2
(13.17)
Figure 13.11. Central force on m.
Clearly, this force field is a function of spatial coordinates and can easily be
expressed as the gradient of a scalar function in the following manner:
F = -grad
(-F)
(13.1 8)
Hence, this is a conservative force field. The potential energy is then
GMm
PE = -~
r
(13.19)
Next, the force on a particle of unit positive charge from a particle of
charge q, is given by Coulomb's law as
(13.20)
Since this equation has the same form as Eq. 13.17 (Le., is also a function of
l/r2), we see immediately that the force field from q , is conservative. The
potential energy per unit charge is then
(13.21)
The remaining field introduced was the magnetic field where
F = qV x B. For this field, the force on a charged particle depends on the
velocity of the particle. The condition that the force be a function of position
is not satisfied, therefore, and the magnetic field does not form a conservative
force field.
597
598
CllhPTEK 13 ENERGY METHODS FOK PARTICLES
13.4
Conservation of Mechanical Energy
1x1 tis iiow consider the niotion of a particle upon which only a co~iscr~~ittivc
fiirce field does work. W e btart with Eq. 11.2:
F. dr = InjV?
2
2
~
(13.221
tnlV’
I
-
Using the definitiiin of potential energy, we replace the left side of the equatioii in the following manner:
(PE),
~
(PE),-= !,mVt
.
.
~~
(11.231
imV;
Rearranging terms. wc rcach the fdlowing tr~cltrIrehti(1n:
-$nq = (PE), + $rnq
( 13.24)
Since positions I and 2 are arhitrdry, obviously rlrr .sitni ~ f r l w
I i m i i t i r i l enrr,y?.
und the kinetic. ow’rgy,fi,r a purricle rmrnim < oiisfaiit ut r i l l rimes d u r i q the
morion of I h P p r r i d e Thir statement is sometimes called the Iukc o / ( ’ o r i . s ~ r ~
varion of n z ~ h u , i i c u lm e r g y , f i i r ~nn.s(,ri,Nrivt,,swe,n.s. The usefulness 01 this
relation ciiii he denionstrated by the lollowing examples.
Example 13.5
A particle is dropped with Levo initial rclocity down a frictionless chute
(Fig. 13.12). What is the magnitude of its velocity if the vertical drop during the niotion is h ft’?
For small trajectorics. we cat1 assume a unifiirm force field -m.yi.
Sincc this is the only force that can pertomi work on the particle (the normal force lrom the chutc docs 110 work). we can employ the conservationof-mechanical-energy equation. I f we take positii~n2 as a datum. wc then
havc from Eq. 13.24:
mg/i
+ 0 = 0 + -I inVi
Solving for V’, we get
v, = SZaR
The advantages o f the energy approach fiir conservative ficlds
become appai-ent li-om this prohleni. That is. not all thc lirrccc need he
considered in computing velocities, and the path, howevcr complicated. is
of n o conccrn If friction were present. a n ~ n c ~ n ~ e r v i i tfwce
i v c would perform work, and we would have ti1 g11 back t~ the pencral relation given hy
Eq. 11.2for thc analysis.
r)atum
-0
Figure 13.12. Paiticlc
I
on
irictionlrrs chute
SECTION 13.4 CONSERVATTON OF MECHANICAL ENERGY
I
EXafnple 13.6
A mass is dropped onto a spring that has a spring constant K and a negligible
mass (see Fig. 13.13). What is the maximum deflection S? Neglect the effects
of permanent deformation of the mass and any vibration that may occur.
In this problem, only conservative forces act on the body as it falls.
Using the lowest position of the body as a datum, we see that the body
falls a distance h + 8. We shall equate the mechanical energies at the
uppermost and lowest positions of the body. Thus,
mg(h+S)+
PE gravity
+
+
+ 0 = 0
fKSz
0
(a)
+
PE spfinng
KF.
PEgravify
PE spnDg
KE
0
Rearranging the terms,
We may solve for a physically meaningful Sfrom this equation by using the
quadratic formula.
EXafnple 13.7
A ski jumper moves down the ramp aided only by gravity (Fig. 13.14). If
the skier moves 33 m in the horizontal direction and is to land very
smoothly at B, what must be the angle 0 for the landing incline? Neglect
friction. Also determine h .
Y
Figure 13.14. A ski jump with a landing ramp at an angle 0 to be determined.
We first use conservation of mechanical energy along the ramp. Thus
( m g ) ( 1 7 )= i m V 2
:.
_-
V = 4(2g)(17) = 18.26 d s
Figure 13.13. Mass dropped on spring
599
600
C'HAPI'EK I3
ENFR(;Y MKIHOOS TOR PARTI('I.FX
Example 13.7 (Continued)
Using a reference
t A a s hhown in Fig. 13.14 and nieasuring time fnin1
the instant thal thc nkiei- is at the origin. we tiow use Newton's law for the
frcc flight. Thus
i; = -9.Xl
c,
i= -9.x I1 +
1' = -<).XI
When
I =
0. \.
=
0. and we take 1'
<, I
=
I2
~
2
+ ( ' ] I + <;
0.Hcncc.
= (,., = 0
Also.
j; =
=
~X
Whcn
f
0
c
= C,l
+ C,+
= 0,.i= 18.26, ;ind .Y = 0,
r, = o
:_
C3 = 18.26
Thus wc have
i. = - 0 . X l l
1' = -9.81
To get 11, h e t
.X
12
-
(21)
i = IX.26
(C)
(b)
x = IX.26t
(d)
= 33 iii Eq. (d) and solve Ihr lhc h i e t.
:. 33 = IX.?6/
Hence, going t o Eq. ( h ) we get
/I
= -9 81('
1 =
1.807 scc
'r2)~
=
i6.01m
Now gct $ at landing. Using Eq. (a) we havc
i= -(9,Xl)(I,Xl~7)= -17.73 rnls
Also, we have at all tiinch
i = 18.26 1111s
Fur best landing, Vis parallel to incliiic
t3 = 44.15"
SECTION 13.4 CONSERVATION OF MECHANICAL ENERGY
601
Example 13.8
A block A of mass 200 kg slides on a frictionless surface as shown in Fig.
13. IS. The spring constant Kl is 25 Nlm and initially, at the position shown,
it is stretched .40 m. An elastic cord connects the top support to point C on
A . It has a spring constant K2 of 10.26 Nlm. Furthermore, the cord disconnects from C a t the instant that C reaches point G at the end of the straight
portion of the incline. If A is released from rest at the indicated position,
what value of 0 corresponds to the end position B where A just loses contact with the surface? The elastic cord (at the top) is initially unstretched.
,~,~-~
I
(.Y2)(707) mrn
-
Datum
Figure 13.15. Mass A slides along frictionless surface.
We have conservative forces performing work on A so we have
conservation of mechanical energy. Using the datum at 6 and using 1, as
the unstretched length of the spring with 6 as the elongation of the spring,
we then say that
where the last term is the energy in the elastic cord when it disconnects at
G . Therefore, noting that
= .94 m and that OB = .92 m, we have on
ohyerving vertical distances in Fig. 13.15:
(.200)(9.SIJ[(.Y2)(.707)
=
+ (.92)(.707) + (.92Jsin@]+ 0 + $(25)(.40)Z
0 + 1(.20)V:
+ &(25)(.92- /o)2 + 4(10.26)(.94)2
(a)
602
CHAPTER 13 ENERGY METHODS FOR PARTICLES
Example 13.8 (Continued)
To get I , , examine the initial configuration of the system. With an initial
stretch of .40 m for the spring, we can say observing again vertical distances in Fig. 13.15:
1, = [(.92)(.707) + (.92)(.707)] - .40
= ,901 m
Equation (a) can then he written as
V: = .I490 + 18.05sin0
(b)
We now use Newton’s law at the point of interest B where A just
loses contact. This condition is shown in detail in Fig. 13.16. where you
will notice that the contact force N has been taken as zeru and thus deleted
from the free-body diagram. In the radial direction. we have
Figure 13.16. Contdcl I \ fir71 lost at 0
Therefore,
This equation can be written as
V: = 2.20
~
9.03 sin 8
Solving Eqs. (b) and (c) simultaneously for 0, we get
e = 4.349
SECTION 13.5 ALTERNATIVE FORM OF WORK-ENERGY EQUATION
13.5
603
Alternative Form
of Work-Energy Equation
With the aid of the material in Section 13.4, we shall now set forth an alternative energy equation which has much physical appeal and which resembles
thefirst law of thermodynamics as used in other courses. Let us take the case
where certain of the forces acting on a particle are conservative while others
are not. Remember that for conservative forces the negative of the change in
potential energy between positions 1 and 2 equals the work done by these
forces as the particle goes from position 1 to position 2 along any path. Thus,
we can restate Eq. 13.2 in the following way:
f F dr
~
AW‘E)l,2 = A ( W l , 2
(13.25)
where the integral represents the work of nunconservative forces and the A
represents the final state minus the initial state. Calling the integral 5%-2, we
than have, on rearranging the equation:
(13.26)
In this form, we say that the work of nunconservative forces goes into changing the kinetic energy plus the potential energy for the particle. Since potential
energies of such common forces as linear restoring forces, coulombic forces,
and gravitational forces are so well known, the formulation above is useful in
solving problems if it is understood thoroughly and applied properly.’
YEquatim 13.26. you may notice. is actually a form of the first law of thermodynamics for
the case of no heat transfer.
Example 13.9
Three coupled streetcars (Fig. 13.17) are moving at a speed of 32 km/hr
down a 7” incline. Each car has a weight of 198 kN. Specifications from
Figure 13.17 Coupled streetcars.
I
604
CHAPTER 13 ENERGY METHODS FOR PARTICLES
Example 13.9 (Continued)
the buyer requires that the cars must stop within SO m beyond the position
where the brakes are fully applied so as to cause the wheels to lock. What
is the maximum number of brake failures that can he tolerated and still
satisfy this specification? We will assume for simplicity that the weight is
loaded equally among all the wheels of the system. There are 24 brake
systems, one for each wheel. Take p,, = .45.
The friction force f on any one wheel where the brake has operated
is ascertained from Coulomb’s law as
We now consider the work-energy relation 13.26 for the case where a
minimum number of good brakes, n, just causes the trains to stop in SO m.
We shall neglect the kinetic energy due to rotation of the rather small
wheels. This assumption permits us to use a single particle to represent Lhe
three cars, wherein this particle moves a distance of 50 m. Using the end
configuration of the train as the datum for potential energy of gravity, we
have for Eq. 13.26:
+ APE = wl-2
AKE
= -(w)(l 1,050)(50)
>I
= 10.89
The number of brake failures that cdn accordingly be tolerated is
24 - I I = 13.
N
e fa
13
Another example of conservation of mechanical energy will he in the
next section (Example 13.1 I ) for the case of a system of particles.
13.38. A railroad car traveling 5 kmmr runs into a stop at a rairoad terminal. A vehicle having a mass of 1,800 kg is held by a
linear restoring force system that has an equivalent spring constant of 20.000 Nlm. If the railroad car is assumed to stop suddenly and if the wheels in the vehicle are free to turn, what is the
maximum force developed by the spring system'? Neglect rotdtional inertia of the wheels of the vehicle.
5 km/hr
13.42. In Problem 13.41, a weight W of 225 N is released suddenly from rest on the nonlinear spring. What is the maximum
deflection of the spring?
13.43. A vector operator that you will learn more about in fluid
mechanics and electromagnetic theory is the curl vector operator.
This operator is defined for rectangular coordinates in terms of its
action on V a s follows:
(2 2).
curl V(x, y. z ) = -- -
Figure P.13.38.
13.39. A mass of one slug is moving at a speed of 50 ft/sec
along a horizontal frictionless surface, which later inclines upward
at an angle 45". A spring of constant K = 5 Iblin. is present along
the incline. How high does the mass move,?
Figure P.13.39.
13.40. A block weighing 10 Ib is released from rest where the
springs acting on the body are horizontal and have a tension of
10 Ib each. What is the velocity of the block after it has descended
4 in. if each spring bas a spring constant K = 5 Iblin.?
A
(When the curl is applied to a fluid velocity field V a s above, the
resulting vector field is twice the angular velocity field of infinitesimal elements in the flow.) Show that if F is expressible as
V 4 ( x , y , z ) , then it must follow that curl F = 0. The converse is
also true, namely that ifcurl F = 0, then F = V$ (x, y. z ) and is
thus a conservative force field.
13.44. Determine whether the following force fields are conservative or not.
(a) F = (1Oz
(b) F
ltl(Y'+lo+
B
I
=
+ y j i + (15yz + x ) j +
( z sin x
[
lox
+ I:")
k
~
+ y ) i + ( 4 y z + x ) j + ( 2 y 2 - 5 cos x ) k
See Problem 13.43 before doing this problem.
13.45. Given the following conservative force field:
F = (102
Figure P.13.40.
13.41. A nonlinear spring develops a force given as ,063 N,
where x is the amount of compression of the spring in millimeters.
Does such a spring develop a conservative force? If so, what is the
potential energy stored in the spring for a deflection of 60 mm?
+ y ) i + (15yz + x ) j +
lox
+ 9k N
2 1
find the force potential to within an arbitrary constant. What work is
done by the force field on a particle going from r, = 1Oi + 2 j +
3k m to r2 = -2 + 4 j - 3k m? [Hint: Note that if w d x equals
some function (xy2 + I), then we can say on integrating that
4
X?jJ
=
2 + ui + P(Y, 2 )
where ~ ( y z ,) is an arbitrary function of y and z. Note we have
held y and z constant during the integration.]
13.46. If the following force field is conservative,
Figure P.13.41.
F =
(52
sin x
+ y)i +
(4yz
+ x)j +
(2y2
-
5 cos x)k Ib
605
(where x. y . and I arc i n ft), find the lorce potential up to an arhitrary ciinstanl. What i s thc wtirh doric on B partick \tarring at the
origin and moving in a circular path of radius 2 ft tu 1mn a s e m circle alung the positive x axis'! (See the hint in Problem 13.15.)
13.47. A body A can slide i n a frictionlesi manner along it s t i l t
rod CD. At the position shown, the Ypring along CD has been
compressed 6 in. and A is at ii d i m n c e o f 4 ft from 11. 'The spring
connecting A to E has been elongated I in. What is the spccd 01
A after it moves I ft? .The Fpring cmstanla are K , = 1.0 lhiin.
and K , = .5 Iblin. The mass oSA is 30 I b m
Figure P.13.47.
13.48. A collar A uf i n a h s 10 Ihm slides un il frictionless tubs.
The collar is cimnectetl to a IiiiCar spring whose spring cnnstant
K is 5.0 Ihiin. If the c(illiir i s released frum rest at the position
shown, what i?i t s speed when the \ p i n g is at c l w i i l i m h Y ? The
spring is stretched 3 in. at thc initial posilinn of the collar.
tiigiire P.13.50
13.51. A slotled rod A i \ ~ m w ~ i IO
i glhe lclt iit L: 5pccd 012 i n k c .
Pins arc moved 10 the left by this rud. 'I h e w pin\ must slidr in a
slnt under the cod as shown i n the diagram. The pins are coilnccted hy a spring having ii spring constant K 011.500 Nim. Thc
spring is uiihtrctched in thr conliguration shown What distance d
dc the pills rcach heflirt. \lnppinp iiistantanci,usly'? 'Ibe niaw of
the slutted rod i\ IO kg. 'The spring is held i n thc .;lotled rod XI a
inot to buckle outw:ird. Neglect the mass 01lhc pin\.
Figure P.13.48.
13.49. A mass M of20 kg slides with no friction along a vertical
rod. Two springs each of spring constant K , = 2 Nimm and a
third spring having a spring constant K , = 1 Nlmm are attached
to the mass M .At the starting position when 6 = 30". the springs
are unstretched. What is thc velocity of M after it descends a distance d of ,112 in?
606
Fipure P.13.51.
13.52. The top view of a slotted bar of mass 30 Ibm is shown.
Two pins guided by the slotted bar ride in slots which have the
equation of a hyperbola xy = 5, where x and y are in feet. The
pins are connected by a linear spring having a spring constant K
of 5 Iblin. When the pins are 2 ft from the y axis, the spring is
stretched 8 in. and. the slotted bar is moving to the right at a speed of
2 ftlsec. What is V ofthe bar? [Him: Differentiate energy equation.]
v
13.55. A meteor has a speed of 56,000 k d h r when it is 320,000
km from the center of the earth. What will be its speed when it is
160 km from the earth's surface?
13.56 Do Problem 13.2 using the energy equation in the usual
form of the first law of thermodynamics.
13.57. Do Problem 13.5 using the energy equation in the usual
form of the first law of thermodynamics.
13.58. Do Problem 13.17 usine the enerev eauation in the usual
form of the tirat law of thermodynamics.
I
I
,
1
13.59. Do Problem 13.18 using the energy equation in the usual
form of the first law of thermodynamics.
13.60. A constant-torque electric motor A is hoisting a weight W of
30 Ib. An inextensible cable connects the weight W to the motm over
a stationary drum of diameter D = 1 ft The diameter d of the motor
drive is 6 in., and the delivered torque is I50 Ib-ft. The dynamic coefficient of friction between the drum and cable is .2. If the system is
started from rest, what is the speed of the weight W after it has been
x
I
r a i d 5 ft?
Figure P.13.52.
13.53. In Problem 13.52, what is the speed of the slatted bar
when x = 2.25 ft?
13.54. Perhaps many of you as children constructed toy guns
from half a clothespin, B wooden block, and bands of rubber cut
from the inner tube of an automobile tire [see diagram (a)]. Rubber band A holds the half-clothespin to the wooden "gun stock."
The "ammunition" is a rubber band B held by the clothespin at C
by friction and suetched to go around the block at the other end.
The rubber band B when laid flat as in (b) has a length of 7 in. To
"load the ammunition'' takes a force of 20 Ib at C. If the gun is
pointed upward, estimate how high the fired rubber band will go
when " tired if it weighs 4 oz. To "fire" the gun you push lowest
part of clothespin toward the nail (see diagram) to release at C.
1Nail
Half
clothespin
Figure P.13.60.
13.61.
A weighing 10 Ib, can slide i ig a fixed rod
B-B. A spring is connected between fixed point C and the mass.
AC is 2 ft in length when the spring is unextended. If the body is
released from rest at the configuration shown. what is its speed
when it reaches the y axis? Assume that a constant friction force
of 6 OL acts on the body A. The spring constant K is 1 Ih/in.
I
14"
7
(a)
x
(b)
Figure P.13.54.
B
C
Figure P.13.61.
607
13.62. A hody A is releascd from rcst on il vcitical circular path
as shown. IIa constant resisrance force 01 I N acts along thr path.
what i c the speed of the hndy when i t reaches li? The rnasr of thr
bod!' is .S kg and the radiw r ofthc path i s I .h In.
E
(,ioule)
A
Figure P.13.62.
13.63. A cylinder slides down a rod. What is the distance fithat
the qpring is deflected at the instant that the disk stops inFtnntanewsly'?T a k r p,, = .i.
W
8
'
K
=
~
500 N
IO.II(X1 N l
Figure P.13.63.
13.64. In ordnancc work a very v i k d tebt for equipmcnl i s the
attack teit, in which a piccc of equipment i i whjectcd t o a certain
Figure P.13.65.
l e v d of accrleiiltiun of shoii duration. A comrnor technique for
thi? test is Ihe drop fmf.The specimen is mounled o n a ripid carriage, which upon release is dropprd along guide rods onto il XI 13.66. Suppose in Example l j 4 thal wI! tlic hr&?s t i t i train 11
opcratc and lock. What 1s thc distance il helore stopping? A l w .
of lead pads resting irn a heairy rigid anvil. 'The pads dchrrn and
ahiorh the energy 01 the carriage and specimen. We estiniatc dctermine the lorcc in each coupling o i the system.
through other tests that the energy F ahsorhed hy a pad \ersii%
compression distvncc S is given as shown, where the curve can bc
takcn as a pwabala. For four such pads, each placed dircclly on
the anvil, and a height h of 3 111. what is the compressinn 0 1 the
pads'! The carriage mil cpecimen together wrigh SO,? N. Ncglrct
the friction of thc guides. (NoIP: I I = I N-m.)
Figure P.13.67.
608
SECTION 13.6 WORK~ENERGYEQCJATIONS
Part 6: Svstems of Particles
13.6
Work-Energy Equations
We shall now examine a system of particles from an energy viewpoint. A
general aggregate of n particles is shown in Fig. 1.3. I X. Considering the ith
particle, we can say, by employing Eq. 13.2:
where, as in Chapter 12, Aj is the force from the j t h particle onto the ith partcle, as illustrated in the diagram, and is thus an internal force. In contrast, F
represents the total external force on the ith particle. In words, Eq. 13.27 says
that for a displacement between r1 and rz along some path, the energy relations for the ith particle are:
external work
+
internal work
= (change in kinetic energy relative to X Y Z )
( I 3.28)
Furthermore, we can adopt the point of view set forth in Section 13.5 and
identify conservative forces, both external and internal, so 3s to utilize potential
energies for these lorces in the energy equation. To qualify
force, an internal force would have to be a function of only the spatial configuration of the system and expressible as the gradient of a scalar function. Clearly,
forces arising from the gravitational attraction between the particles, electrostatic forces from electric charges on the panicles, and forces from elastic connectors between the particles (such as springs) are all conservative internal forces.
We now sum Eqs. 13.27 for all the particles in the system to get the
energy equation for a .v.vvstem ofparricles. We do nor necessarily get a cancellation of contributions of the internal forces as we did for Newton’s law in
Chapter 12 because we are now adding the work done by each internal force
on each particle. And even though we have pairs of internal forces that are
equal and opposite, the movemem of the corresponding particles in general
are nut equal. The result is that the work done by a pair of equal and opposite
internal forces is not always zero. However, in the case of a rigid body, the
contact forces between pairs of particles making up the body have the same
motion, and so in this case the internal work is zero from such~forces.1°Also, if
there is a system of rigid bodies interconnected by p i n y 6 a l l joint connections,
and if there is no friction at these movable connecf’ons, then again there will
be no internal work. (Why?) We can then say for the system of particles that
A(KE
+
PE) =
‘M.j-,
“’We shall show this more directly in Chapter 17.
i 13.29)
Z
e
x/
Inertial reference
Figure 13,18. System of
609
6I O
~ZNEKGY
ivtimtoiis FOK P A RTIC L E S
CHAITEK I 3
I.
where 'kL:~represents the net work done by internul und exremu/ nonconservalive forces. and PE represents the total potcntial energy of the conscrvative
inlrrnul u r d exlrmol forces. Clearly. i f thzre are no nonconselvative forces
present then Eq. ( 13.291 degenerates to the conservation-of-mechanic;~l-eiiergy
principle. As pointed out earlier. since wc are employing the i~kangein potentiiil energy. the dalums choscn for measuring PE are of littlc significance
herc." For instance. any convenient datum for meahusing the potential energy
due to gravity of the carth yields the same rcsult for thc term APE.
Looking hack on Eq. ( 13.27). which o n buiiitnatiun over a11 thc particles
pave rise to Eq. ( 13.291, namely the equation 10 he used fora system O S particles, we wish to make the following point. I t is the fact that the work contribution of each force Stems from the mnwineirr of rudi f i ) r c r wilh ill .spcific
puiri~o/uppliwlioii. This should he clear from the 11% oSdr,. with i identilying tach particlc. This will be an important considcration later.
Let us now consider the action of gravity on a system of particles. The
potcntial energy relative to a datum pI"ne, q,
for such a system (see Fig.
13.19) is simply
PE = C n l , ~ ; ,
Figure 13.19. Pil-ticlei a b o w refcnencr
planc.
Notc that the right side o f this equation represents the first moment of the
weight of the system ahoul the .rj plane. This quantity can he given i n teriiis
of the center of gravity and the entire weight of W a s follows:
PE = Wr,
where :, is the vertical distance from thc datum plane to the center of gravity.
Note that if g is comtant, the center of gravity corresponds to lhe center of
mass. And so for any Yystem of particles. the change in potential energy i\
readily found hy concentrating the entire weight at the center of gravity or. as
is almost always the case, at the center OS mass.
Before proceeding with the problems we wish to emphasize certain
salient features governing the w o r k m e r g y principle for a system of
particles.
5 the mass center.
SECTION 13.6 WORK-kNEKGY EQIJATIONS
61 I
Example 13.10
In Fig. 13.20, two blocks have weights W, and W,. respectively. They are
connected by a flexible, elasrii. cahle of negligible mass which has an
equivalent spring constant of Kl. Body I is connected to the wall by a spring
having a spring constant K2 and slides along a horizontal s u r f x e for which
the dynamic coefficient of friction with the body is p8 Body 2 is supported
initially by some external agent so that, at the outset of the problem, the
spring and cable are unstretched. What is the total kinetic energy of the
system when, after release, body 2 has moved a distance d2 and body 1 has
moved a smaller distanced,?
Use Eq. 13.29. Only one nonconservative force exists in the system.
the external friction force on body I . Therefore, the work term of the equation becomes
w2
= -W,P<,dl
(a)
Three conservative forces are present; the spring force and the gravitational force are extrrnal and the force from the elastic cable is inrrrnul.
(We neglect mutual gravitational forces between the bodies.) Using the
initial position of W, as the datum for gravitational potential energy, we
have, for the total change in potential energy:
APE
=
[$K,d: - 01 + [ i K , ( d ,
~
d , )2
~
01 + [0 - W,d,]
We can compute the desired change in kinetic energy from Eq. 13.29 as
As an additional exercise, you should arrive at this result by using
the basic Eq. 13.28, where you cannot rely on familiar formulas for polential energies.
In the following example, we will see how using a system of particles
approach can make for great simplification in a problem over the procedure
of dealing with particles individually. Also we will have a case where there i n
no nonconservative work, which then results in a conservation of mechanical
energy.
'I
Figure 13.20. Elastically connected badizs.
Example 13.11
Masses A and H . each having a mass of 7 5 kg, are constrained Lo move i n
frictionless slots (see Fig. 13.21).They are connected hy ii light rod of length
I = 300 mm. Mass B is connected LO two niassless linear springs, each
having a spring conslant K = 900 Nlm. The springs are unstretched when
the connecting rod LO massesA m d l j i\ vertical. What are thc velocities of
H and A when A descends 3 distance of 75 mni? There Is no friction in the
end connections of thc rod."
Figure 13.21. 'IV,"
hy Iinciir
intrrconrirctrd
IIIBSSCS
cwnlrained
\pring\.
W e shall use a system of particles approach f o r this case. This will
eliniinatc the need to calculate thc work of the rod on cach ma\\ that WL'
would need had we elected tu dcal with cach mass separately. For a syhtrm of
particles approach, this work is inlemal hetveen rigid hodies, tiwing 7ero
value as a r e d ( of Newton's third law and having ideal pin-conoccted ,joints.
Wc n e x t note that only coiisciwa~iveforces arc prcsenl (giravilalional
force and spring forces;) s o the first law form of our energy equation degenerates to conhervation of mcchanicid energy. I n Fig. 13.22. we show
SECTION 13.6 WORK-ENERGY EQUATIONS
Example 13.11 (Continued)
the system in a configuration where mass A has dropped a distance of
,025 m. We can then say for the beginning and end configurations using
the datum shown in Fig. 13.22,
A[KE
($MAV2- 0 ) + (;M,V;
+
- 0)
PEI =
w_2=
0
+
[Mag(.3 - ,025) - M,g(.3)]
+ [2f(900)(SZ)
~
0,
:
0 (a)
We have three unknowns here. They are 6, V,, and V,. Observing the
shaded triangle in Fig. 13.22 and using the Pythagorean theorem, we have
l;
+ 62
= .32
(h)
Taking the time derivative we get
ziAi,
We note that
equation that
i,
= V, and that
+ 268 = o
S = 4. We
then see from the preceding
Now, returning to Fig. 13.22, we can compute 6 for the case at hand.
Noting that A has descended a distance of ,025 m, thus making
1, = .3
,025 = ,275 m, we next go to Eq. (h) tu get 6. Thus
~
(.275)'
:.
+ S2 = .3'
S = . I I99 m
Substituting data from Eqs. (c) and (d) into Eq. (a) (conserving mechanical
energy) we get
V, =
-. 1524 m l s
:_
from (c), V, =
- .I199 VA
= ,3496 m l s
Thus,
V,
f ,1524 m l s down
V, = ,3496 m l s to the right or the left
61 3
SECTION 13.7 KINETIC ENERGY EXPRESSION BASED ON CENTER OF MASS
the relation above into the expression for kinetic energy, Eq. 13.30,
we get
K E = $ ~ ~ , ( v+P,,)*
~
= ~ ~ r n ~ ( V ~ + P ~ ~ ) . ( v ~ + P ~ ~ )
,=I
,=I
Carrying out the dot product, we have
Since y. is common for all values of the summation index, we can extract it
from the summation operation, and this leaves
Perform the following replacements:
We then have
KE
=
$MV:
+ V;
c
C mip2i
,=I
i=l
d n m,pci + f
n
(13.34)
But the expression
represents the first moment of mass of the system about the center of mass for
the system. Clearly by definition, this quantity must always be zero. The
expression for kinetic energy becomes
(13.35)
Thus, we see that the kinetic energy for some reference can be considered to
be composed of two parts: ( I ) rhe kinetic energy of the total mass moving relative to that reference with the velocity of the mass center, plus ( 2 ) the kinetic
energy of the morion of the particles relative io the mass center.
615
6 16
('HAI'I'CR 13
1:NliKGY M1:'IHODS FOK I'ARTICl.tS
p Example 13.12
I
j
:
i
i
I
A hyprithctical vchiclc
i s niiiving at specd ",,in Fig. 13.24. (111 this \chicle
are l w o hodics ciich of i i i a s s 111 diding :dong a liorizonlal rod at a speed 1 '
relative to the rod. This rod i s rotating at an angular specd w l a d k c relativc to the vehicle. What i s the kinetic cnergy of the two hodies relative 1 0
the g r w n d (XYZ) when they arc iit II clistancc I- froin point A?
Clearly. the cciitcr 01 niass c(irresponds to point A and i s thus miwing
at a spccd V, relatiYc to tlie ground. Hence. we liiivc tis part of thc kinetic
cnergy the tei-m
2!MV',
7
,nVi
(:I)
The velocity o f cach hill rcliitive to thc center of iiiass i s e x i l y
f1,rmcd using cylindrical components. Thus. imagining a reference .x?: at A
translating with thc vehicle relative to XYZ. we h a w lor the vclocity of
each hall rcl;ilive to s ; r
P'
The
t01iiJ
= p2
kinetic cncrg!, of thc
+
( o r ) ' = 1.2
~ M : Oni;isse>
KE = rnVi + rnlt;'
+ (OJI.)?
(hJ
r c l a l i x Lo tlic griw~idi s thcn
+ (Wr)']
(e)
I
111 k;xiiniple 13.12, wc co~rsidcrcdii casc where thc hodics involved
constituted a finitc number of d i w r c , l ~ ,particle\. In the nexl cxamplc. we coilsidcr ii case wliei-e w c h a w a m,zlitzuuni o l particles forming ii rigid hody.
The formulatiiin given by Eq. 13.35 can still he uscd hut iiow. iiistcad of summing 1ir a finite nuinher 01discrete particles. we imust integratc to accuunt
f o r the infinite numher of infinitesimal particles comprising thc syslem. We
arc thus tiiking a glimpse. 101- simple ciises, 11l' rigid-hody dynamick 10 be
sludicd latei- i n the tcxt. Those that iki nut have time fiir studying such energy
prohlenx i n detail w i l l hc able to solvc simple hut useful rigid-body dynamics prohlcins on the hiisis of thcsc examples as ucll as liitet- exaiiiples in this
chapter.
617
SECTION 13.7 KINETIC ENERGY EXPRESSION BASED ON CBNTEK OF MASS
Example 13.13
A thin uniform hoop of radius R is rolling without slipping such that 0,the
mass center, moves at a speed V (Fig. 13.25). If the hoop weighs W Ib,
what is the kinetic energy of the hoop relative to the ground?
Clearly, the hoop cannot be considered as a finite number of discrete
finite particles as in the previous example, and so we must consider an
infinity of infinitesimal contiguous particles. It is simplest to employ here
the center-of-mass approach. The main problem then is to find the kinetic
energy of the hoop relative to the mass Center 0, that is, relative to a reference q translating with the mass center as seen from the ground reference
X Y (see Fig. 13.26). The motion relative to xy is clearly simple rotation;
accordingly, we must find the angular velocity of the hoop for this reference. The no slipping condition means that the point of contact of the hoop
with the ground has instantaneously a zero velocity. Observe the motion
from a stationary reference X Y . As you may have learned in physics, and
as will later he shown (Chapter 15), the body has a pure instantaneous
rututioizal motion about the point of contact. The angular vclocity w for
this motion is then easily evaluated by considering point 0 rotating about
the instantaneous center of rotation A . Thus.
A
Figure 13.25. Rolling hoop
Since reference .ry translures relative to reference XU, an observer on
xy sees the ,same angular velocify w for the hoop as the observer on
X Y . Accordingly, we can now readily evaluate the second term on the
right side of Eq. 13.35. As particles, use elements of the hoop which are
R d0 in length, as shown in Fig. 13.26, and which have a mass per unit
length of Wl(27rh'g). We then have, on replacing summation by integration,
the result
A
Figure 13.26. . ~ yLranblaleb with 0
relative to X Y .
The kinetic energy of the hoop is then in accordance with Eq. 13.35:
X
6 18
CHAPTER I3 ENERGY METHODS FOR PARTICLES
Example 13.13 (Continued)
Suppose that the body were a generalized cylinder of mass M (see Fig.
13.27) such as a tire of radius R having 0 as the center of mass with
iixisymmetrical distribution of mass ahout the axis at 0 . Then, we wnuld
express Eq. (h) 21s follows:
$ T m , P ? , = +JJ&m,iroP
I
id)
You will recall from Chapter 9 that
JJI,
dm
IZ
is the sccond mornrnl of inrrtiu of the body taken ahout the
That i s .
Iaxis
at 0.
Figure 13.27. Rolling generalized cylinder
of mass M.
Thus, we have lor the kinetic energy of such a body:
K E = - IM V 2 + ~11 ; z o 2
2
(e)
You may also recall from Chapter 9 that we could employ the mdius /$
,~vrufionk to express IC; as follows:
i-. = k2M
cn
Hence, Eq. ( e )can he given as
We shall examine the kinetic energy formulations of rigid hodies carefully in Chapter 17. Here, we have used certain familiar results Srom physics
pertaining to kinematics of plane motion of B nonslipping rolling rigid body.
For a more general undertaking. we shall have to carefully consider more
general aspects of kinematics of rigid-body motion. T h k will be done in
Chapter I S . Also, in the last example we see one term of the inertia tensor I,,
showing up. The vital role of the inertia tensor i n the dynamics o l rigid hodies will soon be seen.
SIXTION 13.8 WORK~KINETICENERGY EXPRESSIONS BASED ON CE<NTEROF MASS
13.8
Work-Kinetic Energy Expressions
Based on Center of Mass
The work-kinetic energy expressions of Section 13.6 were developed for a
system of piitticles without regard to the mass center. We shall now introduce
this point inio the work-kinetic energy formulations. You will recall from
I Z ' .for
~ the mass center of an) system of particles is
Chapter 12 that N ~ W I O law
F = Mi,
(13.36)
where F is the total external forcc on the system of particlcs. By the same
development as presented in Section 13. I , we can readily amive at the following equation:
p
dr, = (&%!fv?)2
- (:MVZ)I
(13.37)
It is vital to understand from the left side of Eq. 13.37. where we note the
term dr , that the e.rternal.fiirces mmr ail movr with the center uJnioss for the
computation of the proper work term in this equation.'d We wish next to
point out that the single particle model represents a special case of the use of
Eq. 13.37. Specifically. the single particle model represents the case where
the motion of the center of mass of a body sufficiently describes the motion
of the body and where the external forces on the body essentially mnve with
the center of mass of the body. Such cases were set forth in Section 13.1.
Before proceeding to the examples, let us consider for a moment the
case of the cylinder rolling without slipping down an incline (see Fig. 13.28).
We shall consider the cylinder as an ufifirefiatrofparticles which form a rigid
hody-namely a cylinder. When using such an appruiach, we require that ull
thej%rce.r horh extrrnal and intenial inlist ~ O V Pwith their reipective points of
applkurion. Let us then consider the external work done on the particles making up the cylinder other than the work done by gravity. Clearly, only particles
on the rim of the cylinder are acted on by external forces other than gravity.
Consider one such particle during one rotation of the cylinder. This pasticle will
have acting on it a friction Corce.fand a normal force N at the insrant when the
particle is in contacr with the inclined surface. The particle will have zero external force (except for gravity) at a11 other positions during the cycle. Now, at the
instdnt of this contact, the normal force N has zero velricity in its direction
because of the rigidity of the bodies. Therefore, N transmits no power and does
no work on the particle during the cycle under consideration. Also, the friction
Figure 13.28. Cylinder on incline
619
620
CHAPTER 13 ENERGY MIiTIIODS FOR PAKTlCl.liS
1orcc.facts on a particle having zero velocity at the instant of contact because of
the no .slipping rondition. Accordingly.,IIransniii\ no power and does 110 wnrk
on the panicle during thi? cycle." This result must he true for each and evcry
pai'icle on the rim of the cylinder. Thus, clcarly,,fand N do no work when the
cylinder rolls down the incline. Also. because (if the rigidity of the body the
internal forces do no wo
pointed out carlisr. Thus. only gravity docs work.
Howevcr. in consi
ng the motion of the w i r e r o f n i ~ i s s C of the cylinder in Fig. I3.2X. wc noie that forcc fnow mow^ with C and hence do^ work.
At the risk of heing repetitive, w c now summarize the key lealures 1 1 r
properly using the center-or-mass approtach.
I. Only external forces are involved.
r of m&ps when computing work (see
2.
8 all move with the
ic energy of the center of maas is used.
Example 13.14
A cylinder with a mass o f 25 kg is released frorn rest on an incline, a h
shown in Fig. 13.29. The diameter of the cylinder is .hO m . If the cylinder
rolls without slipping, compute the speed of thc centerline C after i t has
moved 1.6 m along the incline . Also. ascertain the friction force acting 011
the cylinder. Use the result from Problem 13.76 that the kinetic energy of
a cylinder rotating aboul its own stationary axis is a M R 2 ~ 'where
,
rn i E
the angular speed in radisec.
In Fig. 13.29 we have shown the free budy of the cylinder. We proceed to use thc work-energy equation for a system of particles. Recall
that w e can concentrate the weigh1 at the center of gravity (Scctiiin 13.6).
Accordingly, using the luwest position as a datum and noting from our
earlier discussion that the friction forcc,/diies no work \tc have
AiPE
+ KE) = '71:
~
SECTION 13.x WORK-KINETIC ENERGY EXPRESSIONS BASED ON CENTER O F M A SS
621
Example 13.14 (Continued)
where the kinetic energy of the cylinder is given as the kinetic energy of
the mass taken at the mass center (straight line motion of C ) plus kinetic
energy of the cylinder relative to the center of mass (pure rotation about C ) .
Noting from Example 13.13 that
o=
V -=li
R
we substitute into Eq. (a) and solve for
.30
y.. We get
Now to findf, we consider the motion of the mass center of the cylinEq. 13.37 for the center of mass. Now all external.forces maSt move with the center of mass; thus, .f does work. Since
the center of mass moves along a path always at right angles to N , this force
still does no work. Accordingly, we can say:
der. This means that we use
-.f(l.6)+ W(1.6sin30") = fMVZ
-f'(I.6)
+ (25)(9.81)(1.6)sin3Oo = $(25)(3.23*)
fa
Figure 13.30. Three rigid bodies moving
without slipping at any of the contact points.
Before going further, let us consider the two cylinders and the block in
Fig. 13.30 as simply a system of particles. If there is no slipping between the
block and the cylinders, the velocities of the particles on the block and the
cylinders at the points of contact between these bodies have the same velocity
at any time t. Furthermore, the friction force on the cylinder from the block is
equal and opposite to the friction force on the block from the cylinder at the
point of contact. We can then conclude that there is zero net work done by the
friction forces between block and cylinders when considering them as an
aggregate of particles.
Also, in the next problem, we will consider as a system of particles,
rigid bodies which are joined by rigid connectors with frictionless interconnections.
622
C'HAITLK I 1
I:WR(;Y
MHTHOI)S to^ I'AKI'IC'I.I:S
Example 13.15
An extcriiiil lorquc 7 <,I50 N-in i s applied to ii solid cylindcr I1 ( x c Fig.
13.31). which I i i ~ hi~ tii its$ of 30 hg iiiid ii radius o i .2 111. Thc cylinder roll'
willioiit dippiiig. Block A, hitving ii iiiii%o I 20 kg, i s dr;iggcd LIiJ the 15
incline. Thc dynamic c d f i c i c n t o f iriction ki,, between hloch A and thc
incline i \ .25. The coiiiiectioii\ iit (' iiiid I ) iirr Iriclionlehs.
( a ) What i\ the \cIocity of the \ystcm :iller miiving
ii d i i t i i i i w
il
()i
2I ~ ~ ?
(I))Whal i s tlic lrictiirii llricc on the cyliiider?
Neglect thc
!niihh
o i thc coniiccting rod
Wc show ii Srcc-hody diagrairi o i l l i e !,ywiii in Pig. 13.32. Wc hrgin
by employing ii syslem of particles point o i vicu. Note tlierc :m pair\ o S
intcrnal forccs prcscnt hclwccn the r(id ('11 i i i i d hody 4 ill (tie c o i i t x t point
SECTION 13.x WORK-KINETIC ENERGY EXPRESSIONS BASED ON CENTER OF M A SS
Example 13.15 (Continued)
and similarly between CD and cylinder B. These force pairs are equal and
opposite because of Newton's third law. And because the forces in each
pair move exactly the same distance at the respective points of contact,
there will be zero internal work from these force pairs. Hence, using the
uppermost configuration as the datum,
%:+: = APE + AKE
TO - ( p d N A ) ( d )= [(WA + W,)(d)(sin 15") - 0]
Note that as the cylinder moves without slipping a distance d along the
incline, the circumference of the cylinder must come into contact with the
incline along the very same distance d. Hence, by dividing d by the radius
r, we get the rotation of the cylinder in radians associated with the movement of its center.
0=d
-r
We then get for Eq. (a), on substituting data for the problem
(SO)[
- (.25)[(2Og)(cos 150)](2) = [(5Og)(2)(.2588)]
c
+
+ z(30)(.22)-
(50)1
v2
"'
(.2Y
1
Next, use the center-of-mass approach. We have on noting that a
couple which is translating does no work. (Why?)
f F * drc = +MTo,a,(V; V:)
~
+ W,)(sinIS")(d) + fd
= f(50)(2.158')
-(.25)[(2Og)(cos 1So)](2) (SOg)(.2588)(2)
+ ,f(2) = 116.4
- ( . 2 5 ) ( N A ) ( d )-(W,
~
f = 232.5
623
624
C I I A P T I R I3
E U H t ( ; Y \IETHOlYi FOR l ' A 1 ~ ~ l ~ ' L E S
Example 13.16
:
111 Example 13.15. suppcxe that cylinder II i s .slippitig. What i s Ihc
dyn;iinic ciicSSicierit or Sriclioii ( p,,In hetween the cylinder wid the iiicline cn tliiit the system reaches a speed iif I .5 mlc after miiving il distaiicc
d = 2 111 starting from rest?
Usiiig ~ h ccnter-of-mass
c
apprmrch. w c thaw
(pd)H= ,7122
111
the next example. we shall coiisider a case where i , , / ~ , , ~ , , ' i i , ~ , ~ ~ , ~ , ~
do work.
Example 13.17
A die.;el-powered electric triiiii movex up ii 7'' grirde in Fig. 13.31. If ii
torque of 750 N-m i s developed i i t eiicli of i t s s i x pair, of drive u'heels.
what i s the iricrcasc of spccd of thc tiniii aftcr il ~ i i o v c \100 iii'?l i i i t i i i l l y .
thc lriiin ha\ a spccd 01 16 knilhr. The lrain weighs 00 k N . 'I'hc drive
whccls h a w a diameter of hilO min. Neglect tlie rotational energy of the
drivc whccls.
Figure 1.4.33. Ilic\clLcIccIric Iraiii.
Wc shall consider thc train as ii
tern of particles including thc
h pairs of whccls arid the body. We have shown tlic I n i n iii Fig. 13.34 \\ilh
tlic cxtcriiiil 1orces. W. N. and/: In addition. %'e ha\c shiiwii cc~rliiiiiiiilcriial
!
13,34, L:xlcln;,l
:mi r ~ r q u c ~ .
i,,rL.r,,~,l fc,rccs
SECTION 13.8
WORK-KINETIC ENERGY EXPRESSIONS BASED ON CENTER OF MASS
Example 13.17 (Continued)
torques 7:lhThe torques shown act on the rutor,y of the motors, and, as the
train moves, these torques rotate and accordingly do work. The reucfiuns to
these torques are equal and opposite to T according to Newton's third law
and act on the sfuturs of the motors (Le.. the field coils). The stators are stationaly, and so the reactions to T do nu work as the train moves. Thus we
have an example wherein, using a system of particles approach, internal
forces pelform a nonzero amount of work. We now employ Eq. 13.29. Thus
APE + KE =
w,,,,
(a)
For the rolling without slipping condition, the friction forces f do no work.
We then have
[(90,000J(100sin7"- O)]
I 90,000 (16)(1,000)
2 s
3,600
[-
+
]?)
(b)
= (6)(750)(8)
where 0 is the clockwise rotation of the rotor in radians. Assuming direct
drive from rotor to wheel, we can compute Bas follows for the 100-m distance d over which the train moves:
d = 100 = 333.3 rad
B=-
r
~~~
(C)
.3
Substituting into Eq. (b) and solving for V, we get
V = 10.38 m/sec
Hence, the increase of speed of the train is
To determine the friction f o r c e s j we now adopt a center-of-mass
approach. Thus all forces now move with the center of mass. And they
must be e.rtemul,fi,rces. Accordingly we have
[6f - (90,000)(sin 7")](1OOJ
I 90,ow
I 90 000 (16)(1,000)
- -~
(10.38), --A
2 9.81
'"Figure I?34 accordingly.
19
not a free-body dngram
2 9.81
(
3,600
625
13.68. A chain of total length L is relcased from res1 on a
smooth support as shown. Delerminr the veliicily of tlic chain
when the last link m w e s off the horizonVal wrtacr. In this
prohlem, neglect the friction. Also, d o n o t attempt to accnunt
for centrirugal cffecth ctemining f r < m the chain links rounding
the corner.
Figure P.13.71)
Figure P.13.68.
13.69. A chain is SO ft long and wcighs IN)Ih. A force PUT 80 Ih
has been applied at the configuratian shown. What is the s p d of
the chain after force P has moved IO fl! 'The dynamic coefticiznl
of friction hetween the chain and the cupporting surface i q .3.
Utilize an apprvximatc snlilysis.
13.71. A device is mountcd mi 3 plntfnrm that I S rotating with an
anguliir s p d 01 10 nldlarc. The devicc consists of t w i masse\
(each i \ ,I 'lug) rotating o n a spindle with an angular specd of
5 rad-crc relativr to the platfbmi. The m a w s arc iiwving radially
outward with a ?peed of I O ftlsrc, and the cntire platf<>lmis being
raised at a speed rrf S fllsec. Compule thc kinetic encrgy n1 the
system of two particles when they are I ft from the qpindlc.
radlsec
+,IC
A
~
Each m a u
~~
S radlsec
0. I d u g
Figure P.13.71.
13.72. A hoop. with four spoke&.rolls without slipping such that
Ihc ccntcr C moves at a bpced V of I .7 mlsec. The diameter of the
hixrp i h 3.3 m and Ihe weight pcr unit length o l t h e rim is 14 Nlni.
The y o k c 5 arc uniform rods also having a wright p ~ l unit
- length
of 14 Nlm. A w m c that rini and spokes arc t h i n What is the
kinetic energy ot thu body'!
Figure P.13.69.
13.70. A hullet of weight
is fired into a hlock of wood weigh
ing W, Ih. The bullet lodges in the wood, and both hodies then
move to the dashed Dosition indicated in the diamam before
falling back. Compute the amount 01 internal work done during
the action. Discuss the effects of this work. .The hullet has il apecd
V,, hefore hitting the black. Neglect the mass o f t h e supporting rrid
and friction at A .
626
Figure P.13.72.
13.73. Three weights A , L1, and C slide frictionlessly along the
system of connected rods. The bodies are connected by a light,
llexihlc, inextensible wire that is directed by frictionless small
pulleys at E and F . If the system is released from rest, what is its
speed after it has moved 300 mm? Employ the following data firr
the body mas5e.s:
Body A :
5 kg
Body H :
4 kg
Body C
7.5 kg
13.75. A tank is moving at the speed Vof 16 k d h r . What is the
kinetic energy of each of the treads for this tank if they each have
a mass per unit length of 300 kgim'?
-l
Figure P.13.75.
13.76. A cylinder of radius R rotates about its own axis with an
angular speed of w. If the total mass is M , show that the kinetic
energy is iMRW.
13.77. Cylinders A and C each weigh 100 Ih and have a diameter of 2 ft. Body A, weighing 300 Ih, rides on these cylinders. If
there is nu slipping anywhere, what is the kinetic energy of the
system when the body A is moving at a speed Vof 10 fvsec'? Use
result of Problem 13.76.
Figure P.13.73.
13.74. B d i c s E and F slide in frictionless grooves. They are
interconnected by a light, flexible, inextensible cable (not shown).
What i s the speed of the system after it has moved 2 ft? The
weights of bodies E and F a r e 10 Ih and 20 Ih, respectively. A is
equidistant from A and C. E remains in top groove.
Figure P.13.77.
13.78. A pendulum has a hob with a comparatively large uniform disc of diameter 2 ft and mass M of 3 Ibm. At the instant
shown, the system has an angular speed e of .3 radlsec. If we
neglect the mass of the rod, what is the kinetic energy of the pendulum at this instant? What error is incurred if one considers the
hob lo be a particle as we have done earlier for smaller hobs'? Use
the result of Problem 13.76.
x
Figure P.13.74.
Figure P.13.78.
627
13.79. In Problem 13.7X compute the maximum angle that the
pendulum rises.
13.80.
Do Example 13.3 hy treating
13.81.
Do Problem 13.27 hy treatmg iis an aggregate ofpwticles.
13.R2.
Do Prohlem 13.22 hy treating a c a n agprrgatc of paniclrc.
a s an aggregate ofpitrliclc\.
13.83. Do Problcm 13.24 hy treating a\ an aggrcptc of p a l t i c k \ .
13.84. A constant force P' is applicd t o the nxic of a cylinder, as
shown, causing thc axis to increasc its speed from I ftisec trr 3 Slliec
in 10 ft without slipping. What i s the friction force acting on the
cylinder'? The cylinder weighs 100 Ih.
Figure P.13.86.
13.87. Cylindcrs A and B w u h havc a mass <if 25 kg and a d i m eter of 100 mm. Block C, riding on A and H. has a mass of 100 kg.
I f the ~ y c t e mis released fmni rc\t at thc configuratinn chown.
what i s the speed of 1.attcr the cylindera h w c made hall a revdiiLion'! Use thc r e w l t 01Pmhlrm I3.70.
Figure P.13.84.
A cylinder with a maw nf 25 kg i\ releaced from rest m :in
incline, a i shown. 'The inner diameter 11 01the cylindcr i s 30(1 mm.
If the cylinder rolls without 'lipping, compute the peed nf the centerline 0 after the cylinder has muwd I .h m almg the incline.
Ascertain the friction Ibrce acting on the cylinder. The radius 01
gyration h at 0 i s .30/\'2 in
13.85.
Figure P.13.X7.
13.88.
Shown arc two iilsnliual hlocks A and H , a c h weighing
SO Ih. A force /.'of 100 Ih i s applied Lo thc lower hlrick, causing it
i n n u x 10 the right. Blrrck A . Ihowever. is rcitrained hy thc wall
('. I f hlock H rcaches a speed of I O ttlscc 1n 2 11 starling from rcit
at thc pri\ition shown in thc diagram. \*hat i~ thc I-cctraining lorce
from thc mall'! The dynamic coefficient 01friction hctwecn H and
the grciund w1.111cci\ .3. L>o this p h l e n , firs1 by using Eq. 1 3 . X
Then check thc r c w l t h) using scparatc lrce-hody diagrams.
and 511 o n
Figure P.13.85.
13.86. A uniform cylinder having a diameter o f 2 ft and a wright
of 100 Ih TOIIS
down a 30" incline without slipping. as shown.
What is the speed o f t h e center after it has mrwcd 20 ft'! Compare
this result with that fnr the case whcn thcre i\ no friction present.
[Hint: Use the result of Problem 13.76.1
I 628
Figure P.13.XX.
13.89. What is the tension T to accelerate the end of the cable
downward at the rate of 1.5 mlsec'? From body C, weighing 508
N, is lowered a body D weighing 1 2 . 5 ~N at the rate of 1.5 m/sec2
relative to body C. Neglect the inertia of pulleys A and R and the
cable. [Hint; From e i l i e r courses in physics, recall that pulley B
i s rotating instantaneously ahout p i n t e, and hence point b has an
acceleration half that of pointf. We will consider such relations
carefully at a later time.]
,Pad
w,= w,= i nwr
K
=
m n Nimm
6 = 20 0 mrn
Figure P.13.91.
(a) What is the velocity of the vehicle after it moves a
distance of I .7 m staning from rest?
(b) What is the total friction f0rce.f on the cylinders from
the ground?
T
"f
e
13.92. A cylinder weighing 500 N rolls without slipping, first on
a horizontal surface and then along a 30" incline.
(a) How far up the incline does it move?
(b) What are the friction forces on the cylinder along the
horizontal surface and along the incline'?
Figure P.13.89.
13.90. Cylinder C is connected by a light rod AR and can roll
without slipping along the stationary cylinder D. Cylinder C weighs
10 N. A constant torque T = 20 N-m is applied to AB when it is
vertical and stationary. What is the angular speed of AB when it
has rotated YI)"? The system of bodies is in the vertical plane.
Recall from physics that a body which is rolling without slipping
has instantaneous rotation about the point of contact.
Figure P.13.92.
13.93. A hoop with four spokes is released from rest from a vertical position.
(a) What is the velocity of point C after it moves 1.3 m?
(h) What is the tension in the wire?
The rim and the spokes each have a weight per unit length of 15 Nlm
and are to be considered as thin. The wire is wrapped around the
hoop and is the sole support
Figure P.13.90.
13.91. An 800-N force F is pulling the vehicle. The cylinders A
and Reach weigh I,OoO N and roll without slipping. The indicated
spring has a spring constant K equal to 50.0 Nlmm and is compressed a distance S of 20.0 mm. The pad slides on the upper
guide with a dynamic coefficient of friction p,, equal to 3. Neglect
all musses except the cylindcrs, whose diameter D is .2 m.
t
5m
Figure P.13.93.
629
I
Figure P.13.W.
l4pure l'.l.3.Y4.
(hi W1i;it arc lhe iricrion ii~l-ccs1cmr11 tlir grliiiiiitl
cylinder A'!
M,,= 2(10 hp
2 M,,= SO kg
M , = 30 kg
Figure P.13.95.
I=
500 N
13.96.
A cylinder is i l h 0 ~ 1t u roll down an incline without slip^
OII
each
SECTION 13.9 CLOSURE
13.9
Closure
In this chapter, we presented the energy method as applied to particles. In
Part A, we presented three forms of the energy equation applied to a single
particle. The basic equation was
J 1 2 F . d r =*' (M V Z) , - f ( M V 2 ) ,
(13.38)
For the case of only conservative forces acting, we presented the equation for
the consemuion of mechanical energy:
(PE), + WE), = (PE), + (KE),
(13.39)
Finally, for both conservative and nonconservative forces, we presented an
equation resembling thefirst law of thermodynamics as it is usually employed:
A(PE
+ KE) = 'Wy.',,
(13.40)
In Pan 8 , we considered a system of particles and presented the above
equation again, but this time the work and potential-energy terms are from
both internal and external force systems.'? Furthermore, all work and potential-energy terms are evaluated by using the actual movement of the points of
application of internal and external forces.
Next, we presented the work-energy equation for the center of mass of
any system of particles:
jlzF-dr,:= $ ( M V Z ) , - ~ ( I W V : ) ~
(13.41)
where F , the resultant external force, moves with the center of mass in the
computation of the work expression. We pointed out that the single particle
model is a special case of the use of Eq. 13.41 applicable when the motion
of the center of mass of a body sufficiently describes the motion of a body
and where the external forces on the body move with the center of mass of
the body.
To illustrate the use of the work+nergy equation for a system of particles, we considered various elementary plane motions of simple rigid bodies.
A more extensive treatment of the energy method applied to rigid bodies is
found in Chapter 17.
We now turn to yet another useful set of relations derived from Newton's
law, namely the methods of linear impulse-momentum and angular impulsemomentum for a particle and systems of panicles.
631
Figurr P.13.98.
..-
Figure P.13.101.
,I
Figure P.13.102.
Figure 1'.13.103.
532
13.104. A self-propelled vehicle A ha? a weight of & ton. A gasoline engine develops toque on the drive wheels to help muveA up the
incline. A countenveight B of 300 Ib is also shown in the diagram.
What horsepower is needed when A is moving UP at a speed of 2 ftlsec
and has an acceleration of 3 ftlsec2? Neglect the weight of the pulley.
[Hint: The pulley rolls along c o d dg without slipping. It therefore has
an instantaneous center of rotation at d. What does this mean about the
relative value of velwity of point h on the pulley and p i n t a?]
13.109. A 100-lh boy climbs up a rope in gym in I O sec and
slides down in 4 sec after he reaches uniform speed downward.
What is the horsepower developed by the boy going up'! What is
the average horsepower dissipated on the rope by the boy going
down after reaching uniform speed'? The diqtance moved before
reaching uniform speed downward is 2 ft.
Figure P.13.109.
Figure P.13.104.
*13.105. Set up an integro-differential equation (involving derivatives and integrals) for B i n Problem 13.31 if there is Coulombic
friction with pz, = .2.
13.106. At what angle QdoesbodyA of Problem 13.31 leave the
circular surface'?
*13.107. Show that the workxnergy equation for a particle can
be expressed in the following way:
13.110. An aircraft carrier is shown in the process of launching
an airplane via a catapult mechanism. Before leaving the catapult,
the plane has a speed of 192 k d h r relative to the ship. If the plane
is accelerating at the rate of I g and if it has a mass of IR,000 kg,
what horsepower is bring developed hy the catapult system at the
end of launch on the plane if we neglect drag? The thrust from the
jet engines of the plane is 100,000 N.
Integrating the right side by parts," and using relativistic mass
m,l<l
V z / c ' , where m,, is the rest mms and c is the speed of
light, show that a relativistic form of this equation can be given as
Figure P.13.110.
~
so
that the rdntivistic kinetic energy i s
KE = n x 2
~
m,c2
*13.108. By combining the kinetic energy as given in Problem
13.107 and m,,? to form E , the total energy, we get the famous
furmula of Einstein:
E = mc2
Therefore
in which energy is equetrd with mass. How much energy is eqUiVdlent to 6 x
Ihm of matter'? How high could a weight of 100 Ib
be lifted with such energy?
The l a i t formulation i?called integration by parls.
633
13.111. Vehiclz B , weighing 25 kN, is to g o down a 30" incline.
The "chicle is connecled to hody A through light pulleys and il
capstan. Whilt should body A weigh if starting from rest it restricts
hody B 111 a speed of 5 mlsec when R m o w c 3 in? There are two
wraps of rope around the capstan.
13.115. Cylinders A and B havc mtas\c~of 50 kg cach. Cylindcr
A can ouly roLil1.e about a stalionary axis while cylinder B rolls
without slipping. Block C has a mass of 100 kg. Slarting from
rest, what is the speed of C after nioviiig . I
and the diainetcr of the cylinders is .2 n,.
it?
Force P is 500 N
Figure P.13.115
F i g u r e P.13.111.
13.112. A jet pussenger phne is moving along the runway for a
takeoff. If cach of ils four engine\ is developing 44.5 kN olthrust.
what is the horsrpowcr dcvcloprd w h m the plimr is moving a1 a
speed of 240 kmlhr?
13.116. A ryslrm o f 4 d i d cylinder\ and a heavy hluck move
vcrtically downward aidcd by a 1,00(1-N forcr F. What is the
angular ?peed of the whecli after thc systcin descends .5 m after
mrting fnrm rest'! What is thc friction friicc Irum Itit. walls on
each wheel'! The wheels roll without slipping.
13.113. Block R, with a n a s i of 200 kg. i s being pulled up an
inclinc. A inotor C' pulls on one cable, developing 3 hp. 'The other
cable is connected t o a counterweight A having a inass of 150 kg.
If B is moving at li s p e d of 2 mlscc, what is its acceleration?
[Hinr: Stalt with Newton's law lor A and B . ]
I
Figure P.13.116.
F i g u r e P.13.113.
13.114. A block G slidcs along a frictionlcss path as shown.
What is the minimum initial speed that G should have along the path
if it is to izniain in contact when it gcts ton, the uppermost positirrn
ofthe path? 'The block weigh\ 9 N. Whilt is the nulmal Surcr ou the
path when forthe condition descrihed the hlock is at position H?
13.117. A wllilr H having a inass uf IO0 g moves along a frktionless curved rod in a vertical plane. A light rubbri~band cunnccts B to
a lixed p i n t A . T h r rubber band is 230 intn in length when
unmetched. A force of 30 N is required io exlend Ihe band 50 mm.It
thc collar is released from rest. what distance muht d he so that the
downward nrrrinal h r c c on the rod at C is 20 N?
I-!'+-!
Figure P.13.Il4.
h34
Figure P.13.117.
13.118. When your author was B graduate student he built a sys- from rest from a position where the elastic cord is unstretchcd,
tem for examining the effects of high-speed moving loads over what is body B's speed after it tnoves 3 m?
elastically supported beams (see the diagram). A "vehicle" slides
along a slightly lubricated square tube guide. At the base of the
vehicle is a spring-loaded light wheel which will run over the
beam (no1 shown). The vehicle is catapulted to a high speed by a
strelched elastic cord (shock cord) which i s pulled back from
position A-A to the position B shown prior to "firing." At A-A thr
shock cord is elongated 10 in., while at the firing position it is
elongated 30 in. A fbrce of I O Ib is required for each inch ofelonEation of the cord. If the cord weighs a total of 1.5 Ib and the v e h i ~
cle weighs 10 O L , what is the speed of the vehicle when the cord
reaches A-A after firing? Take into account in some reasonable
way the kinetic energy of the cord, but neglecl friction.
/H
Figure P.13.119.
S
Side view
(a)
13.120. A collar slides o n a frictianless tube as shown. The
spring is unstretched when i n thr hurimntal position and has a
spring constant of I .O Iblin. What is the minimum weight u f A to
just reach A' when released from rest from the pusitiun shown in
the diagram? What is the force o n the tube when A has traveled
half the distance to A'?
Figure P.13.120.
13.121. A 15-kg vehicle has t w bodies
~
(each with mass I kgj
mounted on it, and there bodies nitate at an angular speed of
50 radsec relative to the vehicle. If a 500-N force acts on the
vehicle for a distance of 17 m, what is the kinetic energy of the
system, assuming that the vehicle starts from rest and the bodies
in the vehicle have constant rotational speed? Neglect frictivn and
the inertia of the wheels.
100 mm 3oll mm
1
6
4
Figure P.13.118.
13.119. A body B of mass 60 kg slides in a frictionless slot on
an inclined surface as shown. An elastic cord connects B to A . The
cord has a "spring constant" of 360 N/m. If the body B is released
Figure P.13.121.
635
13.122. T w o identical solid cylindcrs each weiehing 100 N &upport a load A wcighing SO N. If ii force F nf 300 N acts as shown.
what i q the speed 01 the vehicle after rmwing S m? Alsu. what i s
the total friction iorce on each wheel'! Nrglect the inass nl the
wpporting system connecting the cylindesr. Note that thc kinetic
e n e r g uf the angular m r t i o n of a cylinder about it\ own axis i h
i M K L u ' . l h e sy\tcrn stiirts from st.
13.125. Two discs move ou il horimntnl Srictiunlcss aurface
h ~ w luokiiig
n
down from ahuvc. Each djsc wighs 20 N . A rectangular rncmber B weighing SO N i'i pullrd hy i~ fiwcr F having ii
viiliie ut 2I)O N. If there is no slipping anywhcre cxccpt , i n thc
horimatal \upport surfacc, what 15 thc speed of B aftrr it 1110va
I 8 c m ? Determine the i r i c i i w lol-ce\ t r i m the wall\ onto the
cylinders.
Figure P.13.125.
Figure P.13.123
13.124. 'l'hrcc block, arc conncctcd hy an Inextensible tlexihlr
cahle. The hlocki arc rcleaied from L: r m t configuration with thc
cable taut. I t A can only fall a disrance h equal 10 2 11, what IS the
velocity of bodics c' and B after each har moved a distance u t
3 ft? Each hody beighs 100 Ih. The cozSfiuent of dynamic friction fur body C i s .3 and lor body B i s .2.
Figure P.13.124.
Figure P.13.126.
Methods of
Momentum for
Particles
Part A: Linear Momentum
14.1
Impulse and Momentum Relations
for a Particle
In Section 12.3, we integrated differential equations of motion for particles
that are acted upon by forces that are functions of time. In this chapter, we
shall again consider such problems and shall present alternative formulations,
638 CHAPTER 18 METHODS OF MOMENTlli FOR PARTICLES
Example 14.1
A particlc initially
i s shown graphicall
constrained tv n i w c rectilinearly in the dii-ectinn 01' the l i m e . what i\ the
speed after I S sec?
From the definiliiin of the irnpuhc. the area under the force-time
curve will, i n the one-dirncnhional example. equal the impulse magnitudc.
Thus, we ?imply compulc this area hclween the times I = 0 and I = I C sec:
impulse
=
4
+
-(10)(10)
,llCil
T L . t:-..l
~
.,-l,~,.;+>,
t h , ~ , - ic
I
,/c
(5)(15) = 125 Ih-sec
i
r
i
./,<a 2
,L
I
J
IO\CC
I
15 \ec
:
I
Figere 14.1. l , ~ , ~ ~ ~ ~ ~ ~plot.
, , , , ~ j~ t , , , ~ ~
SECTION 14.1 IMPULSE AND MOMENTUM RELATIONS FOR A PARTICLE
Example 141
II
rZ prticle A with a mass of I ke- has an initiai velocitv I!:1Oi +
,
"
6j d s e c . After particle A strikes particle B , the velocity becomes V =
16i - 3j + 4k d s e c . If the time of encounter is IO msec, what average
force was exerted on the particle A ? What is the change of linear momentum of particle B?
The impulse I acting on A is immediately determined by computing
the change in linear momentum during the encounter:
ZA = (1)(16i - 3j
= 6i - 9j
+
+ 4k) -
( I ) ( l O i + 6j)
4k N-sec
Since
/,:
FAdt =
IAAt
the average force (Fa&becomes
(<JA(O.OIO]
Therefore,
= 6i -
9j
+
4k
639
Example 14.3
Two bodies, I and 2. are connected hy an inextensihle and weightless cord
(Fig. 14.2). Initially, the bodies are at rest. If the dynamic cocnicient of friction is pd for body I on the surface inclined at angle a, compute the velocity
of the bodies at any time f before body 1 has reached the end of the incline.
{,
Figure 14.2. Two bodics corinectsd hy il cord.
Since only constilnt forces exist and since a time interviil has hcen
specified, we can use momentum considerations advantageously. The
free-body diagrams of bodies I and 2 are shown i n Fig. 14.3. Equilihrium
considerations lead to the conclusion that Nl = W, cos a,s o the frictiiin
forcef; is
,fi
Fnr hndv I take
~~~
W; \ i n (I
= p d N , = pdWi cos a
the cnmuonent of the linear impulse-momentum equii-
)*P
N,
W.
SECTION 14.1 IMPULSE 4 N D MOMENTUM RELATIONS FOR A
By adding Eqs. (a) and (h), we can el;m;nate
T and solve for the desired
unknown V . Thus,
V
( - ~ c , W , c o s a + W , s i n n + W , ) r = ~ ( W+, W 2 )
g
Therefore.
Note that we have used considerations of linear momentum for a single particle each time in solving this problem.
Example 14.4
A conveyor belt is moving from left to right at a constant speed V of
1 ftlsec in Fig. 14.4. Two hoppers drop objects onto the belt at the total
rate II of 4 per second. The objects each have a weight W of 2 Ib and fall a
height h of I ft before landing on the conveyor belt. Farther along the belt
(not shown) the objects are removed by personnel so that, for steady-state
operation, the number N of objects on the belt at any time is IO. If the
dynamic coefficient of friction between belt and conveyor bed is .2, estimate the average difference in tension T2 - T , of the belt to maintain this
operation. The weight of the belt on the conveyor bed is I O Ib.
PARTICLE
641
642
CHAP'rER I4
METHODS Or MOMENTUM FOR
PARTICLES
Example 14.4 (Continued)
We shall superimpose the fdlowing effects to get the dcsired result.
1. A friction force from the bed onto the belt results from the static weight
of the ten ohjccts riding on the belt and the weight of the portion of belt
(in the hed.
2. A friction I i r c e froin the bed ontu tlic hclt results from the force in
tlie directinn needed tu change the r r r l i c u l linear tmnmentuni of the
fillling ohjccls from a value corresp<indingto the free-fall velocity just
before impact ( , 2 ~ : / c to
) a value nf z c r ~after impact.
3. Finally, the h e l l must supply a foi-ce in thc .idirection to change the
lm,-i:o,ild liiiear iiinmcmum of the lhlling objects from it value of zero
to it value c w r e ~ p o n d i n pLO tlie speed of the helt.
Thus. wc have for the first contriholicin. which we dunale as A T . the following result:
AT, = ( N W
+
lO)p<(= 1(10)(2) + 10](.2) = 6 Ih
(a1
As for the Yecond con~ributii~ri.
w e can only compute an average value
hy iiiiting that each impacting oh.ject i s given a vertical change in
litiear iiioiiientum equal to
vertical change i n linear rnoiiientiini per object =
=
W
~
fi
7
7
fi
,~~
(,,28/1)
,(2g)(I)
= ,498 Ib-sec
where we have assumed a free kill sparling with zero velocity at the hopper. For fiiur i n i p x t h per second. we have as lhe total vertical change
in h i e a r iiio~iicntuinper second Lhc v ~ l u e(4)(.498) = 1.994 Ih-scc. The
average vertical forcc during thc I-scc interval tu give the impulse needed
for this change i i i linear m~iinenturnis clearly 1.994 Ib. Sincc this result is
enrrect f o r rvzry \econd, I .994 Ih is tlic avcrage normal force that the bed
of the conveyor must Lransmit to
I t for arresting the vertical motion
.. . ,. ...
, .
-.
.
.. ., , ,. . . F
c .
~ . ~
I
~
~
SECTION 14.2 LINEAR-MOMENTUM CONSIDERATIONS FOR A SYSTEM OF PARTICLES
Example 14.4 (Continued)
For four impacts per second we have as the total horizontal change in linear momentum developed by the belt during 1 sec the value (4)(.0621) =
,248 Ib-sec. The average horizontal force during I sec needed for this
change in linear momentum is clearly ,248 Ib. Thus, we have
[(AT),lav = ,248 Ib
(C)
The total average difference in tension is then
(AT)="= 6
14.2
+
399
+
248 =
lb
LineaFMomentum Considerations
for a System of Particles
In Section 14. I , we considered impulse-momentum relations for a single
particle. Although Examples 14.3 and 14.4 involved more than one particle,
nevertheless the impulse-momentum considerations were made on one particle
at a time. We now wish to set forth impulse-momentum relations for a system
of particles.
Let us accordingly consider a system of n particles. We may start with
Newton's law as developed previously for a system of particles:
n
dV.
~ = z mI dt
. -
(14.3)
j=l
Since we know that the internal forces cancel, F must be the total external
force on the system of n particles. Multiplying by dt, as before, and integrating between ti and 5, we write:
Thus, we see that the impulse of the total external force on the system of
particles during a time interval equals the sum of the changes of the linearmomentum vecturs of the particles during the time interval.
We now consider an example.
643
644
CHAPTER 1.1 MET H O D S OF M O ME N T U M FOR PARTICI.FS
Example 14.5
A 3-ton truck is moving at a speed 0160 inilhr. [See Fig. 14.S(a).l The
driver suddenly applies his hrakes at time f = 0 so as to lock his wheels
in a panic stop. Load A weighing I ton breaks loose from its ropes and at
time f = 4 sec is sliding rrlufivi, 10 rhr rrirck at a speed of 3 ftlsec. What
is the speed of the truck at that time? Tdke ,u,/ between thc lires and pavement to hc .4.
Since we do tzot kriow the nature of the forces hctween thc truck and
IiiadA whilc the latter is breaking loose, it is easiest to consider the
01 two particles comprising the truck and the load simultaneously whereby
the af<)rementiorredforccs become iiiterr~dand arc imf considered. Accordingly. we have shown the system with all the external loads i n Fig. 14.Xh).
Clearly. N = (4)(2,000) = 8,000 Ih and tlic friction force is (.4)(8.0001 =
3,200 Ih. We now employ Eq. 14.4 in the ~Idirection as follows:
1% =[?.;.’
2
(3)i 2,000 I
~~~
v, +
1
(I)(?.ilOO)
~~~
,s
~~~~
Ii
(V. + 31
(b)
Figure 14.5. l i u c h u n d c r p i n g panic
-[Zn!vJ]
d
~-
\10p.
(21)
Note that the first quantity inside the first brackets on the right side 01 Eq.
(a) is the inomenturn of h e truck at f = 4 sec. and the second quantity
inside Ihe same hrackels is the miiineiituin o f t h c Imid a1 this instant. We
may readily solve f o r V,:
V, = 35.7 ftlsec
Introducing ni(i.s.s-wtirt’r quantities inti, Eq. 14.4 is easy and sometimes
ad\,antageous. Y o u u’ill reinemher that:
Mr, = C r n , r ,
(14.5)
i=l
Diffcrcnliating with respect lo lime. we get
MV = -pl,V)
(14.6)
!-I
Thus, we see from this cqualion that rlir rorul lirrrar nio~nrrir~trn
o f 0 vysrem of
I ? I U I I I P ? I ~of
I ~ n partidc, ihur has rhr fotul muss 01
/xirlic/e.s c,yriol.s fhr lirieor
SECTION 14.2 LINEAR-MOMENTUM CONSIDERATIONS FOR A SYSrEM OF PARTICLES
the system and that moves with the velocily of the mass center. Using Eq. 14.6
to replace the right side of Eq. 14.4, we can say:
Thus, the total external impuke on a s.ystem ofparticles equa1.T the change in
linear momentum af a hypothetical particle having the mass oj the entire
aggregate and moving with [he mass center.
When the separate motions of the individual particles are reasonably
simple, as a result of constraints, and the motion of the mass center is not easily available, then Eq. 14.4 can he employed for linear-momentum considerations as was the case for Example 14.5. On the other hand, when the motions
of the particles individually are very complex and the motion of the mass center of the system is reasonably simple, then clearly Eq. 14.7 can he of great
value for linear-momentum considerations. Also, as in the case of energy
considerations, we note that the single-particle model is really a special cast
of the center-of-mass formulation above, wherein the motion of the center of
mass of a body describes sufficiently the motion of the body in question.
I
Example 14.6
A truck in Fig. 14.6 has two rectangular compartments of identical size for
the purpose of transporting water. Each compartment has the dimensions
20 ft x 10 ft x 8 ft. Initially, tank A is full and tank B is empty. A pump
in tank A begins to pump water from A to B at the rate Q , of I O cfs (cubic
feet per second) and IO sec later is delivering water at the rate Q2 of 30
cfs. If the level of the water in the tanks remains horizontal, what is the
average horizontal force needed to restrain the truck from moving during
this interval?
Figure 14.6. Truck with tank compartments.
645
646
CHAPTER I ?
METHODS OF MOMFNTIJM FOR PAUTICLES
Example 14.6 (Continued)
In this setup, the mess center 11f the water in the tanks is moving
from left tn right and niiiving noniin(f0rrnly during the time interval 0 1
interest. We show thc water in Fig. 14.7 at home tiinc f where the level i n
v
Water surface
t
area
=
20’ x 8’
tank A has dropped an amnunt ’1 while, hy conservalion of niass, the levcl
in tank B has risen exactly the same amount q. The position x, of the cellter of mass at [his instant can he readily calculated in terms of 7. Thus.
using the basic dcfinition of the center of mass, we can say:
M.t-, = iM,,)(.r,,) + W&,J
1(20)(x)(in)i(pj(~~)
= ~ i z o ) ( x ~ ii o rl)i(p)(io)
+ 1120i(8)(rl~l(P)(~Oi
(>I)
Since we are interested in the time n~tcof change of I< so that wc cat1
profitably employ Eq. 14.7, we next diflerentiate wilh respect t u lime as
follows:
[(20)(8)(IO)](pj(i.)= -[(2O)(Xj
h](p)(IO) + [(ZOjCX)(i~~l(p)(30)
(h)
But (ZOj(8)O is the volume of flow2 from tank A tc> tank B at time
Q lo represent lhis volume flow, we get for the equation above:
~(20)(8)(lolI(~”r<
= -(P)ilO)Q
1.
Using
+ (p)i3o)Q = (2o)(pic)
Solving foric,we have
i
.‘
=I ~
XI)
Q
(C)
‘Remember r h a ~20 fl x X fl 1s the irra of the lop water surf8tcc in each tank. a\ r h u w n
in Fig. 14.7.
SECTION 14.2 LINEAR-MOMENTUM CONSIDE RATIONS FOR A’SYSTEM OF PARTICLES
Example 14.6 (Continued)
Now consider the momentum equation in the x direction for the
water using the center of mass. We can say from Eq. 14.7:
where we have used Eq. (c) in the last step. Putting in Q, = 30 cfs and
Q , = IO cfs, we then get for the average force during the IO-sec interval
of interest on using p = 62.4/g slugdft’:
(e)
This is the average horizontal force that the truck exerts on the water.
Clearly, this force is also what the ground must exert on the truck in
the horizontal direction to prevent motion of the truck during the water
transfer operation.
From another viewpoint, this system is not unlike a propulsion
system like a jet engine to be studied with the aid of a control volume (see
Section 5.4)in your fluids course.
If the total external force on a system of panicles is zero, it is clear from
the previous discussion that there can he no change in the linear momentum
of the system. This is the principle of conservation of linear momentum,
which means, furthermore, that with a zero total impulse on an aggregate of
particles, there can be no change in the velocity of the mass center. If at some
time to the velocity of the mass center of such a system of particles is zero,
then this velocity must remain zero if the impulse on the system of particles
is zero. That is, no matter what movements and gyrations the elements of the
system may have, they must he such that the center of mass must remain
stationary. We reached the same conclusion in Chapter 12, where we found
from Newton’s law that if the total external force on a system of particles is
zero. then the acceleration of the center of mass is zero.’
’Problems 12.103 and 12.104 are examples of this condition.
647
14.3 Impulsive Forces
1x1 its iiow cxaininc the action involved i n the explosiiin o f a h i m h that is i n tially snspmdcil fi-nin a wire. a s shown in Fig. 14.8. Firht, consider thc hituation directly i4fie.r tlic explosion has been set off. Since very large forces are
present from expanding gases. a / r r r p n r f t f of tlic honih receivcs ;in xppreciahlc impul\e during this short time irileiwal. A l w . directly alter the explosion.
the gravitatiiinal forccs itre no longer countcractcd hy the sitppiirting wire. s o
thcrc is an additional impulse acting on the fragmcnt. But sincc the gravit;itional force i h snia11 coniparcd to fcirces from the esplobion. the gravilalional
impul\e on ii friignrent can he considcrcd negligibly small for the short period
of time under discuhsion coniparcd to that [ i f the expanding gases acting (in
the lragiiient. A plot of thc impulsiuc force (fmni the explosiiin) arid the f w c c
o f gravity on a fragmcnt i \ shown i n Fig. 14.1). It is clear from this diagram
SECTION 14.3 IMPULSIVE FORCES
that the impulse from the explosion lasts for a very short time At and can he
significant, whereas the impulse from gravity during the same short time is by
comparison negligible. Forces that act Over a very short time hut have nevertheless appreciable impulse are called impulsiwforces. In actions involving
very small time intervals, we need only consider impulsive forces. Furthermore, during a very short time At an impulsive force acting on a particle can
change the velocity of the particle i n accordance with the impulse-momentum
equation an appreciable amount while the particle undergoes very little
change in position during the time At.'' It is simplest in many cases to consider the change in velocity of a purticle from an impulsive .force to occur
over zero distunce.
Up to now, we have only considered a fragment of the bomb. Now let
us consider a11 the fragments nf the bomb taken as a system of particles.
Since the explosive action is inferno1 to the bomb, the action causes
impulses that for any direction have equal and opposite counterparts, and
thus the total impulse on the bomb due tu the explosion is zero. We can thus
conclude that directly ufter the explosion the center of mass of the bomb ha,y
not moved uppreciubly despite the high velocity of the fragments in all
directions, as illustrated in Fig. 14.8. As time progresses beyond the short
time interval described above, the gravitational impulse increases and has
significant effect. If there were no friction, the center of mass would descend
from the position of support as a freely falling particle under this action
of gravity.
The following problems will illustrate these ideas.
iThis idealization can he explained more preciacly ab fullows. For an impulsive force F
acting on a hody of milss M. we ciln say from the linear momentum equation
The tnvrimuiu iiiovement of the hody M during this time interval according LO N c w a n ' s law on
using the above result for V i s then
;
,.'
Nulc that V,,,,," i h proportimill 10 At while x is proponional to (AI)>, Clearly for a v e q inwli inter^
"ill Ar the value of the movement x of the mass M can he considered ~ e c v n dorder cumpared tu
the value of the velocity V . For simpliciry. with minimal error, we can say that the mash M d w s
mi move while underpin8 ' 8 dzun8r qfveiuciiy in rrrponse to on impuisivr force.
649
650
C H A PTER 1 4 METHODS OF MOME,WTUM OR P A R T I C ~ S
Example 14.7
Some top-flight tennis players hit the ball on a service at the instant that
the hall is at thc top of its trajectory after being released by the free hand.
The ball i s often given a speed V o f 120 milhr by the racquet directly after
the impact i s complete. I f the time 01duration of the impact process is .On5
scc, what is thc magnitude o f the average force from thc racquet on the
hall during this time interval'? Take the weight of the ball as 1.5 OL.
Figure 14.10. Impact of a rciinis hall at service.
We have here acting o n the ball during a very small time interval an
impulsive force and [he force of gravity. We will ignore the gravity force
during the timc of impact and wc will consider that the b a l l achieves a post
impact velocity while not moving. as explained earlier in the model for
impulsive force behavior. As shown in Fig. 14.10, the impulse I generated
o n the hall by the racquet accordingly is
=
..5124[.99hi
~
. 0 8 7 2 j ) Ib-sec
Next. we go to the impulse momentum equation. Thus
(F,\)(.nns) =
.5124(.996i
~
.0872j)
The magnitude of the average furce is l"inally given as follows:
Aster the impact, the hall will havc a trajectory determined hy gravity.
wind furces. and the initial post-impact conditions.
SECTION 14.3 IMPULSIVE FORCES
Example 14.8
A 9,000-N idealized cannon with a recoil spring (K = 4,000 Nlm) fires a
45-N projectile with a muzzle velocity of 625 mlsec at an angle of SV
(Fig. 14.1 I). Determine the maximum compression of the spring.
I
Figure 14.11. Idealized cannon..
The firing of the cannon takes place in a very short time interval. The
force on the projectile and the force on the cannon from the explosion are
impulsive forces. As a result, the cannon can be considered to achieve a
recoil velocity instantaneously without having moved appreciably. Like the
exploding bomb, the impulse on the cannon plus projectile is zero, as a result
of the tiring process. Since the linear momentum of the cannon plus projectile
is zero just before firing, this linear momentum must be zero directly after
firing. Thus, just after firing, we can say for the x direction:
(MY)<a""o"
+
(MV.)proje& =
Using y. for the cannon velocity along the x axis and
for the projectile velocity along the x axis we get
"
(a)
7 = y. + 625 cos S O
%?! y. + 9[(625)(cos 50') + y , ]= 0
R
Solving for
8
y., we get
= -2.00 mlsec
(b)
After this initial impulsive action, which results in an instantaneous
velocity being imparted to the cannon, the motion of the cannon is then
impeded by the spring. We may now use conservation of mechanical
energy for a particle in this phase of motion of the cannon. Denoting 6 as
the maximum deflection of the spring, we can say:
'
-~
9~000(2.00') = i(4,000)(Sz)
2
Therefore.
R
65 1
652
CHAPTER IJ
METHODS OF MOMENTUM FOR PARTICI.ES
Example 14.9
For target practice, a 0-N rock i s thrown inti, the air and fired im by a pistol. The pistiit bullet, of m a s s 57 g and moving with a speed of 312 mlsec,
strikes the rock as i t i s descending vcrrically at a specd of 6.25 nilsec. [See
Fig. 14.12(a).] Bolh the velocity of the hullet and the rock are parallcl to
the ~ryplane. Directly after (he hullet hits the rock, the lack hreakh up inti)
two pieces, A weighing 5.78 N and R weighing 3.22 N. Whal i s the velocity of R after collision for the given coplanar postcollision wliicitics of the
hullet and the piece A shown i n Fig. 14.12(h)'? Thc bodies, thr clarity, are
shown separated in the diagram. Keep i n mind, nevertheless, that they are
very close ti1 cach other at post-impact. I n our model of the impact
process. they would not even have moved relative to each other during lhis
process. The indicated 219-mlsec and 25-mlsec vclocities are in the x?
plane. If we neglect wind resistance, how high up doe? the ccnter of mass
o f the rock and hullet system rise after collision'!
irll
21') rni,rc
A
Figure 14.12. Bullet striking
ii
rock.
Linear momentum is conserved during the collision, 50 we can
equate linear momenta directly belore and directly after collision. Thus,
+ ,866;) + 9 ( - 6 . 2 5 j i
.s
= (.057)(219)(-sin2O"i + c r i c 2 0 " J i
+ -5.78 25(.8(i(ii + .5j ) + 3.22 [(I/,,
(.057)(312)(.51'
~
~~~~
R
fi
i
+ ( v ~ ) , j, ]
SECTION 14.3 IMPULSIVE FORCES
We may solve for the desired quantities (V,), and (V’)y to get
We now compute the velocity of the center of mass just before collision. Thus,
My. =
t 1
-~+ .OS7
9
y. = -(-6.2S)j
+ (.057)(312)(.5i + .866j)
Therefore,
V . = 9.12%
+
9.92j d s e c
Hence, for the center of mass there is an initial velocity upward of 9.92
d s e c just before collision. Directly after collision, since there has been no
appreciable external impulse on the system during collision, the center of
mass srill has this upward speed. But now considering larger time intervals, we must take into account the action of gravity, which gives the center of mass a downward acceleration of 9.81 d s e c . 2 Thus,
jic = -9.81
yc = -9.81t
+ C,
y, = -9.81 - + C,t + C,
2
f2
When I = 0, yc = 9.92 and we take y‘ = 0 for convenience. Hence
we have
-v< ,
= -9.81t
v,=
.<
+ 9.92
t , + 9.92t
-9.81-
2
(a)
(b)
Set 4;, in (a) equal to zero and solve for 1. We get
I =
1.011 sec
Substitute this value of t in Eq. (b) and solve for v(.,which now gives the
desired maximum elevation of the center of mass after collision. Thus,
(C)
653
14.5. It the coefficient of static friction is .S iii Plohlem 14.4 and
an incline o f 3 0 ~while it
moving at SO ft/iec. If the dynamic coefficient of friction is 3, ihe coefficient of dynamic friction i c .i.what is the speed of thc
hlock after 2X scc?
how long before [he hody stops'!
14.1. A body wcighing 100 Ib reachcy
IS
14.6. A hody is dropped lrom reil. (a) Dctermine the liillr
rcquired for i t IO acquire a vclocity of I h mi\ec. ( h ) Determinc thr
timc nerdcd tu increace its velocity from I 6 misec 10 23 m l w
Figure P.14.1.
14.7. A hody having a m a s s u l 5 lhirr i ? actcd mi by thc fkllow
14.2. A particle of mass I kg is initially stationary at thc origin
of a reference. A force having a known variation with time acls on
the panicle. That is,
F(r) = ?i
+
(hi + l0)j
+
I.hi'k N
where r is in seconds. After 10 sec, what is the whcity of thz body?
ing forcc:
F = 8ri + ( 6
Ih
where r is in seconds. What is the velocity 01the hody aftei 5 ccc
i f the initial velocity i?
V,
14.3.
A unidirectional force acting o n B particle of mass I6 kg is
plotted. What is the velocity of the particle at 40 sec'! Initially, the
particle is at rest.
+ 3 ; l ) j + (16 + 3r')k
= hi
+
ij
~
Illk ftlsec'!
14.8. A body with ii masr of If, kg is rl-quirrd t o change
its velocity from V , = 2i + 4 j -. IOk mlsec t o a velocity
V , = IOi - S j + 20k mlsec in I O sec. What average force
over this lime intervill will dc the jrib'!
F,,
14.9. In Prohlem 14.X. determine the forcr as ii fiincliun ollitiir
tor the case where f i ~ c evnrics lincarly with lime slarting with ii
10
20
Time (sec)
Figure P.14.3.
14.4.
30
A 100-lb block is acted 011 by a force P . which varies with
time as shown. What i s the sped of the block atter X O hec'!
Assume that the block stan\ from rest and neglect friction. The
time axis gives timc intervals.
ZCIO
valuc.
14.10. A hockey puck miivcb ai 30 fllwc troin left In right.
The puck is intercepted by a player who whisks it at 80 Illsec
toward g o d A . a\ shoun. The puck i s also risks lrom the ice ill
a rille of I O ftlsec. What is thc impolrc 011 Ihe puck. uhrrhe
weight is 5 o x ' !
30 ftlsec
y
~
Figure P.14.4.
654
1.1
100 Ib
Figure P.14.10.
14.11. Gravel is released from a hopper at the rate of I kglsec.
At the exit of the hopper it bas a speed of . I 5 m i s . The belt is
moving at a constant speed of 3 mis. If there is 20 kg of gravel on
the conveyor helt at all times and if the belt on the conveyor bed
has a weight of 50 N, what is the difference in tension T, - T , for
the belt to maintain operation? The dynamic coefficient of friction
between bed and belt is 0.4. Assume that the gravel drops 0.2 m
from the hopper outlet.
Y
-X
Figure P.l
Figure P.14.11.
14.16. Two boxes per second each weighing 100 Ih land on a
circular conveyor at a speed of 5 ftlsec in the direction of the
chute. If there are 6 boxes on the circular belt at any one time,
determine the average torque needed to rotate the helt at an
angular speed of .2 radlsec. The dynamic coefficient of friction between the belt and the conveyer bed is .3. What horsepower is needed for operating this belt? Neglect the rotational
effect on the boxes themselves as they drop from the chute onto
the conveyor. Also, does the radial change in velocity of the
boxes affect the torque needed by the conveyor? Neglect any
radial slipping of the boxes as they land.
14.12. Do Problem 12.5 by methods of momenhm
vBelt
.
rides on a bed: u = . 3
14.13. Do Problem 12.6 by methods of momentum
14.14. A commuter train made up of two cars is moving at a
speed of 80 k h r . The first car has a mass of 20,000 kg and the
second 15,000 kg.
(a)
If the brakes are applied simultaneously to both
cars, determine the minimum time the cars travel
before stopping. The coefficient of static friction
between the wheels and rail is 3.
(b)
If the brakes on the first car only are applied, determine the time the cars travel before stopping and
the force F transmitted between the cars.
14.15. Compute the velocity of the bodies after 10 sec if they
start from rest. The cable is inextensihle, and the pulleys are frictionless. For the contact surfaces, pd = .2.
2 bores per sec.
6 boxes on belt
Chute
w = 2 mdlw
Figure P.14.16.
655
Figure P.14.M
556
14.21. In Problem 14.20, compute the impulse on the horimntal
surface. A moves 4 ft in 1 sec and WB = 20 Ih.
14.22. An antitank airplane fires two 90-N projectiles at a tank
at the same time. The mum12 velocity a f the guns is 1,000 d s e c
relative to the plane. Ifthe plane before firing weighs 65 kN and is
moving with a velocity of 320 k d h r , compute the change in its
speed when it fires the two projectiles.
14.25. Two vehicles connected with an inextensible cable are
rolling along a road. Vehicle A, using a winch, draws A toward it
so that the relative speed is 5 ftfsec at f = 0 and I O ft/sec at
1 = 20 sec. Vehicle A weighs 2,000 Ih and vehicle A weighs
3,000 Ib. Each vehicle has a rolling resistance that is .01 times the
vehicle's weight. What is the speed of A relative tn the ground
at 1 = 20 sec if A is initially moving tu the right at a speed
of 30 f t h c ' !
14.23. A toboggan has just entered the horizontal pan of its run.
It carries three people weighing 120 Ib, 180 Ih. and 150 Ib, respectively. Suddenly, a pedestrian weighing 200 Ib strays ontn the
course and is turned end for end by the toboggan, landing safely
.
amvnr!
- thz riders. Since the tohorean Dath is icv.
,, we can nedect
friction with the toboggan path for all actions descrihed here. If
the toboggan 1s traveling at a speed of 35 mph just before collision
occurs, what is the speed after the collision when the pedestrian
has become a rider'? The tobuggan weighs 30 Ih.
I
-
Figure P.14.23.
Figure P.14.25.
14.26. Treat Example 14.3 as a two-particle system in the
impulse-momentum considerations. Verify the results of Example
14.3 for V. (Be sure to include all external forces far the system.)
14.27. Determine the velocity of body A and body A after 3 sec
if the system is released from rest. Neglect friction and the inertia
of the pulleys.
14.24. An 890-N rowboat containing a hhX-N man is pushed off
the dock by an 800-N man. The speed that is imparted to the boat
is .30 mlsec by this push. The man then leaps into the boat from
the dock with a speed of .60 mlsec relative to the dock in the
direction of motion of the boat. When the two men have settled
down i i i the boat and before rowing crimmences, what is the speed
of the boat'! Keglect water resistance.
Figure P.14.27.
Figure P.14.24.
A and E . )
657
14.29. A 40-kN (ruck i s cnoving at the speed of 40 km/hr carrying a 15-kN load A . ‘The load is restrained only by friction with the
floor of the truck whcie there i i a dynamic coefficient o f friction of
.2 and thc static cuelficicnt of friction is 3. The driver suddenly
jams his hrahes on si) as t o loch all wheels for I .5 sec. At the cnd
ofthis interval, the hraker arc rslcased. What is the final speed Vuf
the truck neglecting wind reristance and rotational inertia of the
wheels after load A stops slipping’! The dynamic ctrefficient of friction hetwcen the tire? and the road is .4.
Figure P.14.32
Figure P.14.29.
14.30. A I ,300-kg Jeep is carrying thrcc 100-kg passengers. The
Jeep is in Sour-wheel drive and i s under test to see what maximum
speed i s posrihlc in 5 sec from a start on an icy mad surface for
which
= . I .Compute ,,,, at I = 5 sec.
14.33. A device to be detonated i s shown in (a) suspended above
the ground. Ten seconds after detonation, there are four fragments
having the following masses and position vectors relative to reference X Y Z :
5 kg
r ; = l,(A%Ji
$ I . !
m, =
+ XoOj +
90M m
3 kg
ri = 80Oi
+
l,XiKlj
= 4hg
rl = 400i
+
I ,000j + 2,OOiJk m
tn, =
+
2,500k m
ini
Figure P.14.30.
14.31. Two ad.jacrnt tanks A and B are qhown. Both tanks are
rectangular with a width of 4 ni. Cawline from tank A is heing
pumped into tank B . When the level of lank A i s .7 m frum the top,
the rate of flow Q frum A to R is 1 O i l livrslsec, and 1 0 sec Iatcr
i t i s 500 literslsec. What is the average horizontal force from the
fluids onto the tank during this IO-sec timc interval? The density
of thc gasoline i s .R x IO’ hg/m’. Tank A is originally full and
tank H is originally empty.
kg
r, = X 4 i + Y,j
m, = 6
+ Z4k
Find the position ri if thc center of mass of the device is initially
at position r(,,where
T,,
= 600i
+
12iJij
+
2,300k m
Neglect wind reqistance.
4
Figure P.14.31.
14.32. Two tanks A and H are shown. Tank A is originally full of
water ( p = 62.4 Ibmlft’), while tank B ic empty. Water is pumped
from A to R. If initinlly 100 cfs of water is heing pumped and if this
flow incrcasu at the rate of 10 cfslsec’ For 30 sec thereafter. what
is the average verticil1 force onto the tiinks from the watcr during
this timc period, itside from thc static dead weight of thc water!
X
i
Figure P.14.33.
SECTION 14.4 IMPACT
14.4 Impact
In Section 14.3, we discussed impulsive forces. We shall in this section discuss
in detail an action in which impulsive forces are present. This situation occurs
when two bodies collide but do not break. The time interval during collision is
vely small, and comparatively large forces are developed on the bodies during
the small time interval. This action is called impact. For such actions with such
short time intervals, the force of gravity generally causes a negligible impulse.
The impact forces on the colliding bodies are always equal and opposite to each
other, so the net impulse on the pair of bodies during collision is zero. This
means that the total linear momentum directly after impact (postimpact) equals
the total linear momentum directly before impact (preimpact).
We shall consider at this time two types of impact for which certain
definitions are needed. We shall call the normal to the plane of contact during
the collision of two bodies the line cfimpuct. If the centers of mass of the two
colliding bodies lie along the line of impact, the action is called central
impact and is shown for the case of two spheres in Fig. 14.13.5 If, in addition, the velocity vectors of the mass centers approaching the collision are
collinear with the line of impact, the action is called direct central impact.
This action is illustrated by V, and V, in Fig 14.13. Should one (or both) of
the velocities have a line of action not collinear with the line impact-for
action is termed ublique central impact.
example, VI and/or V'-the
In either case, linear momentum is conserved during the short time
interval from directly before the collision (indicated with the subscript i ) to
directly after the collision (indicated with suhscriptf). That is,
(mIVl); + (m,VJ, = ( m l y ) , + (m,V,),
(14.8)
In the direct-.central-impact case for .smooth bodies (i.e., bodies with no friction), this equation becomes a single scalar equation since (V,)fand (V,),are
collinear with the line of impact. Usually, the initial velocities are known and
the final values are desired, which means that we have for this case one scalar
equation involving two unknowns. Clearly, we must know more about the
manner of interaction of the bodies, since Eq. 14.8 as it stands is valid for
materials of any deformahility (e& putty or hardened steel) and takes no
account of such important considerations. Thus, we cannot consider the bodies undergoing impact only as particles as has been the case thus far, hut
must, in addition, consider them as deformable bodies of finite size in order
to generate enough information to solve the problem at hand.
For the oblique-impact case, we can write components of the linearmomentum equation along the line of impact and for smooth (frictionless) bodies, along two other directions at right angles to the line of impact. If we know
the initial velocities, then we have six unknown final velocity components and
only three equations. Thus, we need even more information to establish fully
the final velocities after this more general type of impact. We now consider
each of these cases in more detail in order to establish these additional relations.
'Noncentral or e c w n f r k impact i s examined in Chapter 17 for the case of plane motion.
Plane of
Line of
y'OO
Central impact
Figure 14.13. Central impact of two
'pheres.
659
660
CHAPTER 14 MbTHODS 01.MOMENTUM FOR PARTICLES
Case 1. Direct Central Impact. Let us first examine the direct-centralimpact casc. We shall consider the period of collision to be made up of two
subintervals of timc. The period (?fdeformation refers to the duration of the
collision. starting from initial contact of the bodies and ending at the instant
of maximum deformation. During this period, we shall consider that impulse
ID df acts oppositely on each of the bodies. The second period, covering the
time from the maximum deformation condition to the instant at which the
bodies just separate? we shall term the period ofrestitution. The impulse acting oppositely on each body during lhis period we shall indicate as R dt. If
the bodies are perfec.t/y elastic, they will reestablish their initial shapes during
the period of restitution (if we neglect the internal vibrations of the bodies),
as shown in Fig. 14.14(a). When the bodies do not reestablish their initial
shapes [Fig. 14.14(b)], we say that plastic deformation has taken place.
I
1 deformation
'
'
restitution
Pcrfrctly elastic collisinn
(ai
f
,
Periodof-4
deformation
restitution
Inelastic collision
Penodof
(h)
Figure 14.14. Collision process
Thc ratio of the impulse during the rcstitution period R dt to the
impulse during the deformation period D dt is a number E, which depends
mainly on the physical properties of the bodies in collision. We call this number thc coeficirnr 4 rmtitntion. Thus,
I
impulse during restitution e = ~~
impulw dunng deformation
j Rdr
] ";
(14 9)
We must strongly point out that the coefficient of restitution depends also on
the size, shape, and approach velocities of the bodies before impact. These
dependencies result from the fact that plastic deformation is related to the
magnitude and nature o f thc force distributions in the bodies and also to the
rate of loading. However, values of E have been established for different
materials and can he used for approximate results in the kind of computations
VI(
thcy don'l separate. the end of the second pcriod occun \,hen the hodiet cease to
B process B plaStir impact.
deform. Wc call such
SECnON 14.4 IMPACT
to follow, We shall now formulate the relation between the coefficient of
restitution and the initial and final velocities of the bodies undergoing impact.
Let us consider one of the bodies during the two phases of the collision.
If we call the velocity at the maximum deformation condition (VID, we can
say for mass I :
Ddr = [(m,V,), -(m,V,),] = -ml[(Vl)i -(VI),]
(14.10)
During the period of restitution, we find that
Dividing Eq. 14. I1 by Eq. 14. IO, canceling out m,, and noting the definition
in Eq. 14.9, we can say:
(14.12)
A similar analysis for the other mass (2) gives
In this last expression, we have changed the sign of numerator and denominator. At the intermediate position at the end of deformation and the beginning
of restitution the masses have essentially the same velocity. Thus, (VI), =
(VJD.Since the quotients in Eqs. 14.12 and 14.13 are equal to each other, we
can add numerators and denominators to form another equal quotient, as you
can demonstrate yourself. Noting the abovementioned equality of the
yj terms, we have the desired result:
This equation involves the coefficient E , which is presumably known or estimated, and the initial and final velocities of the bodies undergoing impact.
Thus, with this equation we can solve for the final velocities of the bodies
after collision when we use the linear-momentum equation 14.8 for the case
of direct central impact.
During a perfectly elasric collision, the impulse for the period of restitution equals the impulse for the period of deformation? so the Coefficient of
restitution is unit?, for this case. For inelastic collisions, the coefficient of
restitution is less than unity since the impulse is diminished on restitution as a
’The impulses are equal kcause during the period of restitution the body can k considered to undergo identically the reverse of the process corresponding to the deformation period.
Thus. from a thermodynamics poinl of view, wc are considering the elastic impact to k a
rcvcrsiblr process.
661
662
CHAPTER 14 METHODS OF MOMENTUM FOR PARTICLES
result of the failure of the bodies to resume their original geometries. For a
i ) ~ , ~ ~ ~ r l y / ) l oimpact,
s t i i . e = 0 1i.e.. (V2), = ( V , ) , ] and the bodies remain in
contact. Thus t ranges from 0 tu I
Case 2. Oblique Central Impact. Lct us now consider the case of oblique
central impact. The velocity components along the line of impact can be
related by the scalar component of [he linear-momentum equation 14.8 in this
direction and also by Eq. 14.14, where velocity componcnls along the line of
impact are used and where the coefficient of reslilution may he considered
(for smooth bodies) lo be thc same as for the direct-central-impact case. If we
know lhe initial conditions, we can accordingly solve for those velocity components after impact in the direction of the line of impact. As for the other
rectangular components of velocity, we can say that for smooth bodies, these
velocity components are unaffected by the collision. since no impulses act in
these directions on either hody. That is, the velocity components normal to
the line of impact for cacli hody are the samc immediately after impact as
hefore. Thus, the final velocity components of both hiidies can be established,
and the motions of thc hodies can he determined within the liniils of the
discussion. The following examples are used to illustrate the use of thc preceding fiirmulations.
Note that the mass and iiiaterials of the colliding bodies for both direct
or ohlique central impact can he different from each other.
Example 14.10
Two hilliard hall? (of the same sizc and mass) collide with the velocities of
approach shown in Fig. 14.15. For a coefficient of restilutioii of .90. what
are the final velocities of the halls directly after they part? What is the loss
in kinetic energy?
i
5
t
--x
7,,,7I ftisec
\ I O Rlscc
Figure 14.15. Ohlique CZIIII~Iimpact.
A reference is established si1 that the .x axis is iilong line (if impact
and they axis is in the plane of contact such that the reference plane I S par-
SECTION 14.4 IMPACT
Example 14.10 (Continued)
allel to the billiard table. The approach velocities have been decomposed
into components along these axes. The velocity components Cy), and (V,),,
are unchanged during the action. Along the line of impact, linear-momenturn considerations lead to
5m - 7.07~1= mt(V,)J, + m~W2)J,
(a)
Using the coefficient-of-restitution relation (Eq. 14.14). we have
We thus have two equations, (a) and (b), for the unknown components in
then direction. Simplifying these equations, we have
[(v1)~i,+ ~(VJJ,
= -2.07
(C)
I(V,)J, + [(VJJ,
= -10.86
(d)
Adding, we get
[(V,)J,
Solving for
= - 6.47 ft/sec
[(V,)J, in Eq. (c), we write
[(VJJ-
6.47 = -2.07
Therefore,
[(V,)J,
= 4.40 ft/sec
The final velocities after collision are then
The loss in kinetic energy is given as
(KE), - ( K E ) f = ( $ 5 2
+ 4m1O2) - [im6.472 + 1m(7.072 + 4.402)]
AKE=$m[25 + 100 - (41.9 + 50.0 + 19.33)]
Please note that mechanical energy is conserved only if E is unity (i.e., a
perfectly elastic impact). For all other cases, there is always dissipation of
mechanical energy into heat and permanent deformation. However, all
impacts involve conservation of linear momentum for the system.
663
664
CHAPTER 14
Mtl'HODS O F MOMENIUM FOR PAKTICI.ES
Example 14.11
A pile driver is used to forcc a pile A into the ground (Fig. 14.16) as part
of a program to properly prepare the foundation for a Lall building. The
device consists of a piston C on which a pressurc p is developed from
steam or air.
The piston i s connected t n a I ,000-lh hanrriier R. The assembly is
suddcnly released and accelerates downward a distance h n l 2 ft 10
impact on pile A weighing 400 Ih. If the earth develops a constant
resisting forcc to inoverncnt of 25,000 Ih, what distance d will the pile
niovc Cor a drop involving 110 contribution Crmr 11 (which is then 0 psig).
Take thc impact as plnstic.. The weight of the piston and the connecting
rod is 100 Ib.
We begin hy using conservation of mechanical energy lor the
freely falling system to a position just before impact (preinipact). Using
the initial configuration as the datum we have
0
+ (1.000 + 100)i2) = 7I ('J,o"!vz + 0
I
:,
v=
,'2R/L
=
,'(2)[32.2)(2)= 11.35 ftisec
Now we get to the impact process. We have conservation of linear
at
momentum while the bodies remain hypnthc1ic;llly a1 the p~~silion
which contact is first made. Thus, for plastic impact we can say
Finally. we come to the post-impact process where we shall use thc work
energy equation for the pile driver and the pilc.
where the terin nn the left side must he negative hecause the ~nctforce
on the system (23,500 Ih) is in the nppositc direction to the rno~ion
(see Eq. 13.2). Solving f o r d wc get
d = ,0686 ft = 3'2.3 in
Figure 14.16. S ~ e n n - d l - i v mpilc driver.
SECTION 14.5 COLLISION OF A PARTKLE WITH A MASSIVE RIGID BODY
"14.5
Collision of a Particle
with a Massive Rigid Body
In Section 14.4, we employed conservation-of-momentum considerations and
the concept of the coefficient of restitution to examine the impact of two
smooth bodies of comparable size. Now we shall extend this approach to
include the impact of a spherical body with a much larger and more massive
rigid body, as shown in Fig. 14.17.
\
Figure 14.17. Small body collides with large body
The procedure we shall follow is to consider the massive body to be a
spherical body of i&ite mass with a radius equal to the local radius of curvature of the surface of the massive body at the point of contact A . This condition is shown in Fig. 14.18, The line of impact then becomes identical with
the normal n to the surface of the massive body at the point of impact. Note
that the case we show in the diagram corresponds to oblique central impact.
With no friction, clearly only the components along the line of impact n can
change as a result of impact. But in this case, the velocity of the sphere representing the massive body must undergo no change in value after impact
because of its infinite m a w RWe cannot make good use here of the conservation of the linear-momentum equation in the n direction because the infinite
mass of the hypothetical body (2) will render the equation indeterminate.
However, we can use Eq. 14.14, assuming we have a coefficient of restitution
E for the action. Noting that the velocity of the massive body does not change,
we accordingly get
(14.15)
Thus, knowing the velocities of the bodies before impact, as well as the quantity E , we are able to compute the velocity of the particle after impact. If the
"Otherwise. there would he an infinite change in momentum for chis bphere,
665
666
CHAPTER 1 4 METHODS 01-MOMENTUM
m u PARTICLES
collision is perfectly elastic.
tionary massive body
= I, and we see from Eq. 14. IS that for a sta-
E
I
I
I
I
I
I
I
I
,
I
\
\
\
,,
.
,
,
I
,
,
,
Figure 14.18. Anglc of i m j d m c e and angle of retlection.
This means that the angle of incidence 0 equals the angle of reflection p. For
E c I (i.e., for an inelastic collision), the angle of reflection p will clearly
exceed 0 as shown in Fig. 14.18.
We now illustrate the use of these formulations
Example 14.12
A ball is dropped unto a concrete floor from height h (Fig. 14.19). If the
coefficient of restitution is .90 for the action, to what height h' will the ball
rise on the rebound:)
Here the massive body has an infinite radius at the surface. Furthermore, we have a direct central impacl. Accordingly. from Eq. 14.15 we have
( v ) - 0 = ,/2gh'
<=---l
_
~~~
(VI,- t i
3h = 81h
1
~
$3
Solving for h', we get
h
~
Figure 14.19. Ball dropped on concrete
floor.
SECTION 14.5 COLLISION OF A PARTICLE WITH A MASSIVE RIGID BODY
In the following interesting example as well as in some homework problems, we will have to determine, for a given uniform distribution of stationary
particles in space, how many of these particles collide per unit time with a rigid
body translating through this cloud of particles at constant speed V,. To illustrate how this may be accomplished easily, we have shown a cone-cylinder
moving through such a cloud of particles at constant speed y, in Fig. 14.20.
Figure 14.20. A conexylinder moving through a cloud of particles
During a time interval Ai, the cone A moves a distance V, At, colliding with
all the particles in the volume swept out by the conical surface during this
time interval as shown in Fig. 14.21, where this region is outlined with
dashed lines. This volume can easily be calculated. It is that of a right circular cylinder shown in Fig. 14.22 having a cross section corresponding to the
projected area of the cone taken along the axis parallel to the direction of
motion of the moving body. Clearly, by adding the volume of cone A to the
right circular cylinder along its axis at the forward end and then deleting the
same volume at the rear end, we reproduce the dashed volume in Fig 14.21
during the time interval Ai. In general, the volume swept out by a body during
a time interval can readily he found by using the projected area of the body in
the direction of motion. We then use this area to sweep out a volume during
this time interval. This negates having to deal with the actual more complicated three-dimensional end surface itself. We shall make use of this procedure in the following example.
I-
VOAt
I-
Figure 14.22. Volume swept by cone A
f-
_fi
Figure 14.21. Dashed region is volume
swept oUt by the coneA during At,
661
668
CHAPTER 14 METHODS OF MOMENKJM FOR PARTICLES
Example 14.13
A satellite in the fnrm of a sphere with radius K [Fig. 14.23(a)l is n ~ i v i n g
above the earth’s surface i n a region of highly rarefied atmosphere. We
wish to estimate the drag on the satellite. Neglect the cmtribution from
the antennas.
ihl
Figure 14.23. Satellite imoving at high spccd
iC)
i n space.
In this highly rarefied atmosphere, we shall assume that the average
spacing of the molecules is large enough relative to the satellite that ue
cannot use the continuum approach of fluid dynamics, wherein mntter is
assumed to be continuously distributed. Instead, we must consider collisions of the individual molecules with the satellite, which is a n ~ n c o i i tinuum approach, 21s discussed in Section I .7. The mass per molecule is in
slugs and the number density of the molecules is uniformly 11 moleculesift.’
Since the siltellite is moving with a speed much g r a t e r than the speed of
the molecules (the inolecules move at about the speed of sound), we can
assume that the molecules are staiionary relative to inertial space reference
XYZ and that only the satcllite is moving. Furthcrmore, we assume that
when the satellite hits a inolecule there is an elastic. Irictionlcss collision
To study this prohlem, we have shown ;I section of the wtrllite in
Fig. 14.23(b). A referencexx is fixed tu the satellite at its center. We shall
consider this refercnce a l s o to be an inertial relcrence--a step that for
small drag will introduce littlc error for the ensuing calculations. Kelalive
ti) this refcrmce. the molecules approach the satellite with :I horizontal
as shown lor one molecule. ‘They then collidc with the surlice
velocity
v,
SECTION 14.5 COLLISION OF A PARTICLE WITH A MASSIVE RIGID BODY
Example 14.13 (Continued)
with an angle of incidence measured by the polar coordinate 8. Finally,
they deflect with an equal angle of reflection of 8. The component of the
impulse given to the molecule in the x direction (I,,,), is
(Im0JA= ( m y cos 28)
- (-my)
= (my (I
+
cos 28)
(a)
This is the impulse component that would he given to any molecule hitting
a strip that is R d8 in width and which is revolved around the x axis as
shown in Fig. 14.23(c). The number of such collisions per second for this
strip can readily be calculated as follow^:^
collisions for strip per second
projected area distance the
= of the strip
strip moves
[in x direction ][in 1 sec
[I
number of
molecules per
unit volume
]
(b)
= [(R d8 cos B)(2nRsin B ) ] [ y ] [ n ]
= 2nR2nV7sin 8 cos B d8
The impulse component dl, provided by the stnp in 1 sec is the product of
the right sides of Eqs. (a) and (b). Thus,
dIx = 2nmnR2V3 (sin @cos8)(1
+ cos 28) d8
(c)
Noting that 2 sin OcosB = sin 20, we have
d', = nmnRZV?(sin20+sin2Bcos28)dB
2
Integrating from 8 = 0 to 8 = n/2,1° we get the total impulse for 1 sec by
the where:
The average force needed to give this impulse by the satellite is clearly
nmnRZV:, and so the reaction to this force is the desired drag.
YAs is shown here the volume swept out by the strip in one second will be a right
circular tube of length V, Ac = (<)(I) and thickneas R d 8 cos 8 and having a radius equal to
R sin 8.
# W eintegrate only up to 7112 because collisions take place only on thefront part of the
sphere. (Note also, we are already rotating for any 8 completely around the axis of the
sphere.) This is so since. in our model, the molecules are moving only trom left to right
toward !he sphere with no collisions possible beyond 8 = n12.
RdO cos H
~
669
14.43. Cylinder A , weighing 20 Ib, is moving at a speed of
20 ftlsec when it is at a distance 10 ft from cylinder B, which is
stationary. Cylinder B weighs 15 Ih and has a dynamic coefficient
of friction with the rod on which it rides of .3. Cylinder A has a
dynamic coefficient of friction of .I with the rod. What is the
coefficient of restitution if cylinder B comes to rest after collision
at a distance 12 ft to the right of the initial position?
k
I
O
'
14.46. Two identical cylinders, each of mass 5 kg, slide on a frictionless rod. Each is fastened to a linear spring ( K = 5,000 N/m)
whose unstretched length is .65 m. The spring mass is negligible. If
the cylinders are released from rest by raising the restraints,
(a) What is their speed just after colliding with a coefficient of restitution of .6?
(b) How close do they come to the walls'?
Restraints
4
Q&&@
Figure P.14.43.
M=5kg
e= 6
14.44. A load is being lowered at a speed of 2 d s e c into a
barge. The barge weighs 1,000kN, and the load weighs 100 ldu. If
the load hits the barge at 2 d s e c and the collision is plastic, what
is the maximum depth that the barge is lowered into the water,
assuming that the position of loading is such as to maintain the
barge in a horizontal position? The width of the barge is 10 m.
What are the weaknesses (if any) of your analysis? The density of
water is 1,000 kglm.' [Hint: Recall the Archimedes Principle]
Chains
ej
Figure P.14.46.
14.47. A light ann,connected to a mass A, is released from re51
at a horizontal orientation. Determine the maximum deflection of
the linear spring ( K = 3,000 N/m) after A impacts with body B
with a coefficient of restitution equal to .8.If body B does not
reach the spring, indicate this fact. Note that there is Coulomb
friction between the body B and the floor with pd = .6. Consider
bodies A and B to be small.
Im
B
I-- 3 0 m
.02
f
+
T
Figure P.14.44.
1
.3 m
14.45. A tractor-trailer weighing 50 kN without a load canies a
LO-kN load A as shown. The driver jams on his brakes until they
lock for a panic stop. The load A breaks loose from its ropes.
When the truck has stopped the load is 3 m from the left end of the
trailer wall (see diagram) and is moving at a speed of 4 d s e c relative to the buck. The coefficient of dynamic friction between the
load A and the trailer is .2 and between the tires and road is .5. If
there is a plastic impact between A and the trailer and the driver
keeps his brakes locked, how far d does the truck then move?
MA= l 0 k g
MB=7kg
c = .8
K = 3,000 N/m
Figure P.14.47.
14.48. Mass M,,, slides down the frictionless rod and hits mass
M,, which rests on a linear spring. The coefficient of restitution e
for the impact is .8.What is the total maximum deflection 6 of the
Figure P.14.45.
spring?
67 I
Figure 1'.14.54.
*14.55. A neutrvn N is moving toward a stationary helium
nucleus He (atomic number 2) with kinetic energy I O MeV. If the
collision is inelastic, causing a loss of 20% of the kinetic energy,
what is the angle 0 after collision? See the first paragraph (only)
of Prohlem 14.54. [Hinr: There is no need (if one is clever) to
have to convert the atomic number to kilograms.]
14.57. Masses A and LI slide on a rod which is frictionless. The
spring is initially compressed from .8 m to the position shown.
The system is released from rest. A and LI undergo a plastic
impact. The spring is mdSskss.
(a) What is the speed of the masses after B moves .2 m'!
(h) What is the loss in mechanical energy for the system?
n
I
i t 4 M
.I m
M x = 1 kg
K = 1,000 N/m
Figure P.14.57.
Hefrm collision
After collision
Figure P.14.55.
14.56. Cylinders A and 6: are free to slide without friction along
a rod. Cylinder A is released from rest with spring K , to which it
is connected initially unstretched. The impact with cylinder B has
a coef'ficient of restitution E equal tu .X. Cylinder B is at rest
hefore the impact supported in the position shown by spring K2.
Assume springs are massless.
(a) How much is the lower spring cornpressed initially?
(h) How much does cylinder B descend after impact
before reaching its lowest position?
14.58. A ball is thrown against a floor at an angle of 6 0 ~with
a speed at impact of 16 m/sec. What is the angle of rebound a if
t = .7? Neglect friction.
Figure P.14.58.
14.59. A ball strikes the xy plane of a handhall court at r =
3i + l j ft. The ball has initially a velocity V, = -IOi - l O j 15k fricec. The ctiefficient of restitution is .8. Determine the final
velocity V, after it hounces off the xy, yz, and xz planes once.
Neglect gravity and friction.
K,
=
I.000 Nlm
K z = 11.000Nlm
t =
.x
x
Figure P.14.56.
Figure P.14.59.
673
14.60. A cpace vehiclc i n the shape of a cone ~ c y l i n d r ri s liloving at B \peed V mlcec, many times the spccd of sound thrvugh
highly rarefied atmosphere. I f r a c h molecule of the pas has a mass
m kg and if therc are. on thc average, n moleculcs per cubic meter.
unmputz thc drag i m the cone-cylinder. Thc cone half-angle i b
3V. Take the collision to he perrcctly ~ l i i i t i c .
14.63. C u n d c r a parallel heam of light having a n cnergy
flux of S wilt ti in^^. shining nornial t o a Oat surface that c o n plctely ahsorhr the cnergy. You leiirned in physics that an
impulre d l is devclopcd on the suidacc during titne dl given hy
the formula
where < IS the rpacd of light in V ~ C U Oi n mdsec. If the surfacc
retlecti the light, then we have an impulse ill developed on the
~ r f i u t given
.
as
Figure P.14.60.
14.61. 110 Problem 14.00 lor a cake wheic the cnllisiiins arc
assumed tu be inelastic. Assumr the cocfficieiii of restitution 10 be .E.
14.62. A double-wedge ikiifoil rection for a space g k h i s
s h o w n If the glidcr riiove~in highly rarcfield atmosphere at il
speed V many t i m s grrater than the hpccd d r o u n d . what is thr
drag per unit length o f t h i s airfoil? Asiunie thc ciillision to he pcrl c t l y elastic. There are ,I niolecuIes pcr It'. each having IIn ~ a ' i hti,
in slugs.
t
2
i
4-
V
Figure P.14.62.
614
Compute the fcirce uteinming tiom the rellection of light
\hinine n ~ r r n a l10 il perfectly retlecting mirror having an area of
I tm2, The light has an cnergy flux S 01 20 W i d Take the apeed
(' = 3 x
IOx mlsec. Whet is the iililiation prrsbnre ,,ad on the
mirror'!
*14.64. Thc Echo satellite when put inlo orhit is inllated trr a
4S-iii-diamctcr hphcre having a skiti nude up of a Imindte u l
aluiniiium ovcr mylar ovci alutnioum. .This \kin is highly reflectaiit of light. Bccause of the m a l l mass 111 this satellite. i t may
he nt~lectcdhy s m a l l forces such as that vternming firm tht: relleclion of light. If a parallel h a i n of light having an energy drnrity
S of .SO Wlmm' impinges on the Echo satdlite. what total firer
i \ dcvclq,cd on the hatrllite from this soul-CC?Fmm physics
(hcc Pnrhlcm 14.63). thc radiation pressure, /,r,,cr on a reflecting
rurtice frolr a hcain ollight inrlincd hy 8' from the iwrnial to the
surlxe i\
SECTION 14.6 MOMENT~OF-MOMENTUMEQUATION FOR A SINGLE PARTICLE
Path of particle
3
z
Inertial reference
X
Figure 14.24. Point a fixed in inertial space
If this point ti is positioned at a fixed location in XYZ, we can simplify the
right side of Eq. 14.16. Accordingly, examine the expression (d/dt)(p= X P):
%p,
dr
x P ) = p" x
P+&' x P
(14.17)
But the expression p, x P can he written as p , x m i . The vectors p,, and I
are measured in the same reference from a fixed point a to the panicle and
615
676
CHAPTER 14 METHODS OF M O M I N I U M FOR PARTICLFS
lrom the origin to the particle. respectively (sce Fig. 14.25). They are thus
different at all tinies to the extent o f a constilnt vector&. Note that
r =
GI +
p,,
Figure 14.25. Pocitim vector5
111 ni
and
ii
'Therefore.
r =
b,,
Accordingly. the expression fi', X m i is zcro. Thus.
Ey. 14.11 becomes
and Eq. 14.16 ciiii he Nritteii in the form
Therelcxc
where M. is the torque 01 the total external fiirce about the z axis ;ind H: is the
moment of the rnoiiieiitiiin (or ;rngular moincntuni) ahout the .: axis.
SECTION 14 6
MOMENT-OF~MOMENTUMEQUATION FOR A S1NC;I.E PARTICLE
Example 14.14
A boat containing a man is moving near a dock (see Fig. 14.26). He throws out
a light line and lassos a piling on the dock at A. He starts drawing in on the line
so that when he is in the position shown in the diagram, the line is taut and has
a length of 25 ft. His speed V, is 5 ft/sec in a direction normal to the line. If the
net horizontal force F on the boat from tension in the line and from water
resistance is maintained at 50 Ib essentially in the direction of the line, what is
the component of his velocity toward piling A (i.e., 4) after the man has
pulled in 3 ft of line? The boat and the man have a combined weight of 350 Ih.
We may consider the boat and man as a particle for which we can
apply the moment of momentum equation. Thus,
MA = H;,
(a)
Clearly, here MA = 0 since F goes through A at all times. Thus. HA is a
constant-that is, the angular monientum about A must be constant.
Observing Fig. 14.27, we can say accordingly
r , X mV, = r2 X mV2
Since rl is perpendicular to V, and r2 is perpendicular to
simple scalar product from above. Thus
(V,),,we get
Figure 14.26. Man pulls toward piling
a
(25l(m)(5) = (22l(ml(V,),
Therefore,
(V,), = 5.68 ft/sec
(bl
We need more information to get the desired result V, toward the
piling. We have not yet used the fact that F = 50 Ib. Accordingly, we now
employ the work-kinetic energy equation from Chapter 13. Thus,
j: F * dr = ( ; M y 2 ) ’
1350
( 5 0 ) ( 3 )= - - ( V , ) ’
2 8
-
(k
MV2)
y
I
Figure 14.27. Boa1 a l po\itions I and 2.
I350
---(25)
2 8
Therefore,
V,
= 7.25 Wsec
(cl
Now V, is the total velocity of the boat at position 2. To get the desired
component % toward the piling, we can say, using Eqs. (b) and (c):
v:
(7.25)’ =
Therefore,
+ v;
(5.68)’ + V i
= (VJ?
677
678
CHAPTER I 4
MCTHOlX OF LIOMENTIIM I'OR PhKIICl.ES
14.7
More on Space Mechanics
M a n y problems o i cpacc mechanics can be ~(rlvedby using energy and angiiIar-momentuni method\ of thi5 and thc preceding chapter without considering
the detailed trajectory equations ol' Chapter 12. I.et Lis thereliire set forth
s o m e d i e n t Factors concerning the motion of a space vehicle moving in the
vicinity o i ii pl;inet or star with the cngine s h u t ~ i l i a n d
with negligible I"-iction
from the outside."
After the .;pace veliicle hiis heen propelled ;it great spced by i t s r o c k t
engines ti1 a positiiiii oiitside the planel's atmiiqiherc (the final piiwered
vclocity i\ called the tmmoiif vcliicity), thc vehiclc then imdergiies plane.
gravitational. w i i t r u l : / i ) r w motioii (Section 12.6).If i t continue5 to go around
the planet. the vehicle i s said to go in10 orhi! and the trajcctiiry i s that of a circle or that of iiii ellipse. IF. on the other hand. the vehicle scapes frmn the
influence of thc plaiiet. then the tra,jectol-y will either bc :I parabola or a hyperbola. In the case 0 1 xi elliptic orhit. the pmilion closest 111 the curlicr o l the
planet i s called I w r i p c (see Fig. 14.2X) and the position Fartheqt friini the curI c e of tlic plaiict i s called u/xi,qf'f'. Notice that iit apogcu and perigcc the
vclocity vectors V , and
rilthc veliicle are parallel to the surface oi the planet
and s o at the% points (and m l y at thcsc points)
y,
v
1
L' H ,
", = 0
,-
I n thc case OF ii <,ircuIor orbit OF radius and relocity V..we ciin use Newt m ' s law and the grwit;itional law to state
where M i c the mass of tlie planet and io i c the angular speed of the radius
vector ti1 the vehiclc. K e p l a c i q the acceleration lcrin NO' by V'irand solving
for
we get
y,
vc = l
r
(14.20)
Knowing GM and r. we can readily ciimpute the spced L: for a particular
circular orhit. In Section 12.6 we showed that GM can he ea5ily computed
iisiiip (he r e l i i t i m
CM = gR2
( 14.211
SECTION 14.7 MORE ON SPACE MECHANICS
where g is the acceleration of gravity at the surface of the planet and R is the
radius of the planet.
In gravitational central-force motion, only the conservative force of
gravity is involved, and so we must have conservation of mechanical energy.
Furthermore, since this force is directed to 0, the center of the planet, at all
times (see Fig. 14.28), then the moment about 0 of the gravitational force
must he zero. As a consequence, we must have conservation of angular
momentum about 0."
Y
ee
x
Figure 14.28. Elliptic orbit with perigee and apogee
We shall illustrate in the next example the dual use of the conservation-of-angular-momentum principle and the conservation-of-mechanicalenergy principle for space mechanics problems. In the homework problems
you will he asked to solve again some of the space problems of Chapter 12
using the principles above without getting involved with the trajectory
equations. Such problems, you will then realize, are sometimes more easily
solved by using the two principles discussed above rather than by using the
trajectory equations.
"Those who have studied the Vajeclory equations of Chapter 12 mighl realize that
c=
rva = CDnStant
is actually a statement of the conservation of angular momentum since mrV, is the moment ahout
0 of the linear momentum relative lo 0.
619
680
('HAPTER 1-1
METIIOIX
ni: MOM+.KTLIM rot?P.%KTI('I.I~S
Example 14.15
A space-shutlle vehicle 011 a iresciic ~iiissiiiii(hcc Fig, 14.29) i \ sell1 into a
circular orhil ;it ii distiincc of 1.200 kin above the earth'\ surface. Thih
orbit is inserted so as 10 be in the same plane as that of a spii
rocket engines will not start, thus preventing it from initiating a priicedure
i
T f(1r returning to earth. Thc goal of thc shuttle is to enter a ti-ajcctory that
will permit docking with the disahled spacecraft and then to rescue the
iiccupant\. The timing of insertion of the circtilar orhit of the space shuttle
h;is ho been chosen thal Ihc space shuttle by firing i l h rockets at rhe piisij tion shown can. by thc proper change ot. speed. rcacli apiigec at the siiinl:
I
time and sanie location a\ docs thc cripplcd \pace vehicle. At thih lime.
docking proccdures caii he carried O U I . Considcring tlliit the rocket
engines o l the spau-\huttle vchicle iiperalc during ii w i : v shorl < l i s r m ~ ~ ,
of travclli to achieve the proper velocity V,, fiir the mission. determinc the
i change in speed that thc \pace chuttle intist achieve. The radiuh of the earth
i \ 6,373 kni.
!
j
i
SECTION 14.7 M O RE ON SPACE MECHANICS
Example 14.15 (Continued)
We shall first compute GM. Thus, working with kilometers
and hours,
(6,373)2 = 5.16 x
km’/hr2
The velocity for the circular orbit for the space shuttle is (hen
From conservation of angular momentum for the space-shuttle
rescue orbit we can say
mr,,c;, = m(rV).pop..
(7.573)(c;$ =
(Io’ooo)(v)ap”gc~
Therefore,
c;, =
1.320v>p<>gee
(b)
where V, is the speed of the space shuttle just afer firing rockets. Next, we
use the principle of conservation of mechanical energy for the rescue
orbit. Thus,
Substitute for
yp,>gce
using Eq. (b) and solve for V,.
V,,
We get
= 27,861 k m h r
Hence, using Eq. (a), we can say:
AV = 27,861 - 26,105 =
1,756km/hr
68 I
14.65. A particle mfates at 30 radlscc along a frictionlcss sui'ace at a distance 2 ft from the center. A tlcxihlc card restrains the
iarticlc. If this cord is pulled so that the particle moves inward at
I velocity of 5 ftlsec, what i s the magnitudt: of the total vzlncity
when the panicle is I ft from the center'!
14.69. A hody A weighing I O Ib is moving initially at a speed of
V , n f 2 0 ftisec on ii frictionless surface. An elastic cordA0, which
ha\ a length / nf 20 ft. hccnmes taut hut not stretched at the position shown in the diagram. What is the radial speed toward 0 of
the hody whcn the c r m l is strctched 2 ft? The cord has an cqui\,aleiit spring convtant of 3 Ihlin
"3
0
ts
ftiscc
Figure P.14.65.
14.66.
satellite has an apogee of 7.12X k m I t i ? mov
iatelliteofwhen
;peed
36,4X(1
I =kmlhr.
6,970 km'?
What is the transverse velocity of atthea
14.67. A system is shown rotating lieely with an angular speed
,if 2 rad/sec. A inass A of 1.5 kg is held against II spring such
hat the spring is cmnpressed 100 mm. If thc devicc <iholding the
nass in position is suddenly rrmirved. determine how far taward
he vertical axis of the system the mass will move. 'The spring
:onstant K is ,531 Nlmm. Neglect all friction and inertia of the
u r . The spring is not connected to the m a s .
JI
__
- _ _ _ _ _ _ _ .~
"I
Figure P.14.6Y.
14,70, A
hall
Ih is n,tating
a YertiCal
at a speed (u, nf
The ball is
hearings
shaf, by light inextensible strings
a length , o f ft.
' l h c m g l c 8, is 30'. Wh:a is the angular speed o,of the hall if
hearing A is moved up 6 in:!
. .
so "In?
Figure Y.14.67.
14.68. Do Pmhlem 14.67 fix the case where therr is Coulombic
.riction
.
hctwcen the mass A and the horimnral rod with a constant
I,, cqual to .4.
i82
Figure P.14.70.
14.71. A mass m uf I kg is swinging freely about the I axis at a
speed w , of O
I radlsec. The length I , of the string is 250 mm. If
the tube A through which the connecting string passcs is moved
down a distance rl of 90 mm,what is w2 of the mass'? You should
get a fourth-order equation for w2 which has as the desired root
w2 = 21.05 radlsec.
14.73. A space vehicle is moving at a speed of 37,000 km/hr at
position A, which is perigee at a distance of 250 km from the
earth's surface. What are the radial and transverse velocity components as well as the distance from the eerth's surface at R? The
trajectory is in the xy plane.
~V
Perigee
Figure P.14.73.
i
Figure P.14.71.
14.74. A space vehicle is in orbit A around the eanh. At position
( I ) it is 5,000 miles from the center of the earth and has a velocity
of20,000 milhr. The transvene velocity at ( I ) is 15,000 milhr. At
apogee, it is desired to continue in the circular arbit shown
dashed. What change in speed is needed to change orbits when
firing at apogee?
14.72. A small 2-lb ball B is rotating at angular speed w , of
10 radlsec about a horimntal shaft. The ball is connected 10 the
bearings with light elastic cords which when unstretched are each
12 in. in length. A force of 15 Ih is required to stretch the cord
I in. The distance d , between the bearings is originally 20 in. If
bearing A is moved to shorten d by 6 in., what is the angular
velocity w2 of the ball? Neglect the effects of gravity and the mass
of the elastic <:or& [Hint;You should arrive at a transcendental
equation for t17 whose solution is 54.49~.]
/
I
\
I
\
x
++
4
Figure P.14.72.
Figure P.14.74.
683
14.75. Dc Prohlem 12.75 using the principles ofconserwtion of
momentum and con\eiwiltim o f mcchanical cnerpy.
R = 6,373 km
<I
I,200 km
L.;
5,000 kmlhr
~
1
14.76. 111 Prrrhlcm 12.Xh iind thc radial velocity hy using the
method of wnservatim of angol;il- nimnenlum and mechanical
energy.
Figure F.14.82.
14.77. Do Prohlem 12.82 hy the method 01 conrervatiiin id
angular mnmentuni and mechanical energy.
14.83. A space vzhiclr i s i n a circular parking orhit 300 mile\
ahovc thc w r t a c r ,if the earth IIthe vehicle i c tu reach an aniierr
at locatinti 2 uhich i q 5 0 0 nilc.; a h w r the earth's suriace. what
increaw in velocity mu\1 the vehicle attain hy firing Its rockets f i x
ii \hart timc at Incation I'! The radiw n l the earth i s 3.'9hO miles.
.
In Prohlrm 12.87, find the height of the hulk1 above the
14.78.
surfnce of the
hy the
momentum and mechanical energy.
c,fcilnscrvati,m
14.79. In Problem 12.1 19, find thc maximum e l w a t i m ahove
the canh's surface hy the methods n l cunscwation rrf angular
mmnentuin and mcchanical rncrgy.
14.80. 110 Prohlem 12.1 14 hy m e t h d s ( 1 1 unnserwlinii of anp l a r mnrnentuin a n d mechanical energy. [Hint: The ercape
= ,?_(:MI, = ,, 2 1:
L
0
3 1 0 miles
.I
I
14.81. Do Prohlrm 12.1 13 using the principles of conservation
encrgy.
of angular inmientuni and mechanical
14.82. A \pace vchiclc i c i n a circular orhit 1.200 km a h w e
the suifacc 0 1 rhr earth. A prnjcctiic is \hot liar thi'i space v c h cle at a specd relative t o the vchiclc of 5.000 kmlhr i n 21 radial
direction a \ secn from the vehicle. What are the q q q
nd the
distances from the cenier or the earth lnr the trajector? of
rhc prnjectilc'?
14.84. A cp;ioc stiltion i c i n it circular parking mhit iiruund
the earth ;at a tli'itancc 0 1 5.000 mi from thc crntri. A pl-nicctilr
ih fired ahead in t i dircctiim taugunrial to the tra,jactnry of thc
upace tii it inti with ii \peed of 5.000 milhr relative tn the space
statioii. W h a t i \ the miixiniuin distance frnm earth reached hy
the pn>,jcctilc'!
14.85. A skylab is in a circular orhit about the earth 500 km
above the earth’s surface. A space-shuttle vehicle has rendezvoused with the skylah and now, after disengaging from
the skylab, its rocket engines are fired so as to move the vehicle with a speed of 800 mlsec relative l o the skylab in the
opposite direction to that of the skylab. Assume that the firing
of the rocket takes place over a short distance and does not
affect the skylab. What speed would the space-shuttle vehicle
have when it encounters appreciable atmosphere at ahout 5 0
km above the earth’s surface? What is the radial velocity at
this position?
A
I
,/
,,
/
,
,
-_/’
Figure P.14.86.
14.87. In Problem 14.86, a midcourse correction is to be made
to get the probe within 1,000 mi from the surface of Mars. I f V, at
ro = 50,Wll mi is still to be 10,000 milhr, what should be the
radial velocity component
(v);!
Figure P.14.85.
14.86. A space probe is approaching Mars. When the probe
is 50,000 mi from the center of Mars it has a speed V, nf
10,000 milhr with a component (V,), toward the center of Mars
of 9,800 milhr. How close does the probe come to the surface
of Mars’? If retro-rockets are fired at this lowest position A ,
what change i n speed is needed to alter the trajectory into a
circular orbit as shown? The acceleration of gravity at the surface of Mars is 12.40 ft/sec2, and the radius R of the planet is
2,107 mi.
14.88. The Apollo command module is in a circular parking
orbit about the moon at a distance of 161.0 km above the surface of the moon. The lunar exploratory module is to detach
from the command module. The lunar-module rockets are fired
briefly to give a velocity V , relative to the command module in
the opposite direction. If the lunar module is l o have a transverse velocity of 1,500 miiec when it is 80 km from the surface
of the moon before rockets are fired again, what must V, be?
What is the radial velocity at this position? The radius of the
moon is 1,733 km. and the acceleration of gravity is 1.700
misec’ at the surface.
Figure P.14.88.
685
685
< IIAf’I tR
I4
MTTHODS O r MO”VlrNT11M IO K P A R I I C Lt 5
14.8
I
m
0
MOment-Of-MOmentUm Equations
for a System of Particles
We shall now develop the n i i ~ ~ i i c n t ~ ~ i l ~ t n ~ icquetion\
~ i i c n l i ~ for
~ n an aggrcgatc
(if particlcs. The rcwltiiig cquatiiins w i l l hc of vital imporkincc when we
apply then1 to rigid hodies in later chapters. We shall consider ii number
0
of cases.
’
~
0
Case 1. Fixed Reference Point in Inertial Space. An aggregate of n particles a n d a n incrtiiil rclercnce are shown in Fig. 14.30. The moiiictit
o l tnonicnluni cquetiiin for the ith particle i\ now written about the origin of
thih rcfcrcncc:
X
i
Figure 14.30. Systcrn of ii particles.
i
1
where. a\ uwal.J, i s the internal force froni ttic.jth particle on the ith particle.
We now sum (hi\ equation for all n particles:
,,
cc(r,
/I
+
X
.f;!I
= $ [ g ( r , XI‘,)]
=
k,,,k,l(14.23)
,=I ,=I
Y
X
Figure 14.31. Inlernal equal and
nppmitc forces.
where the suninlation operation hxs heen put after the dillcrcntiation on
the right side (perniishible because of the distrihutive property of difrerentiation with respect to addition). For any pair of particles. the intcrn;tl
forces w i l l hc equal and oppiisitc mil collinear (see Fig. 14.31). Hence,
the forces w i l l have a ~ c r oiiioincnt ahiiut the origin. (This result i s most
easily understood by reinemhering that, for purposes of taking moments
ahout a point. forces w e tmn\mi\\iblc.) W e can Ihcn conclude that the
expi-ession
in Ciis equation i s x I o . Rcaliritif l l i i i t c r , x I ; i s the tiital tiiorneiil of the
exlemal forces ahout the origin, we have iis
Ma = H a
ii
1-esult fiir Eq. 14.23:
(14.24)
SECTION 14.8 MOMENT-OF-MOMENTUM EQUATIONS FOR A SYSTEM OF PARTICLES
where this moment is taken about the aforementioned point a.I4
point can he given as the angular momentum of the center of
(14.25)
ri = rc + pcj
X
0
Figure 14.32. c is center of mass of aggregate
The ar
ir momentum for the aggregate of particles about 0 is then
Carry
.he cross product and extract r‘ from the summations:
4
1 =
But si]
Id
XYZ. H
m
with res
I
is the center of mass, it follows that
u could also be moving with a constant velocity y , relative to inertial reference
er, a would then be fined in another inertial reference X’Y’Z’, which is translating
to XYZ at a speed Yr
687
Going hack to Eq. 14.27, we
that the second and third expressions o n the
righr %ideare t~ he delered and we get then the desired result lor H , ) :
H,,
= r,
x
~ r +El?,
;
x
I I I , ~ ,=
r, x M?,
+ H,
where H i\ the inomeiit ;ihour the center of i n i i c s n i the linear tnoineiituni as
seen from the center iif ]miss for the aggregate." 'l'his n a y he rewritten and
cxprccced for (itty lixcd point (i whcrc, u\ing (,, a\ tlic positiiin vectni from
f i x e d point (1 t o the center of tiiiiss c', we have
(14.2)
I I , , = I I , + ',, x Mi,,,
analogcius to thc c a w 01kinetic enerfy (see Section 13.7).
Thus, i n ii rniiiiiier
the rnoinciit iif riioiiiciiluin ahiiut point ti is llic \uin (if the imiiiiciit (if iiionieiituin rcliiti\'c to the cciitcr o l r i i i i s i plus ihc iiioniciit o i inoiiicntum o i the cc11ter of inash ahiiut point ( I . Note tha1
i s the \'elocity (if ( ' reliitive rn fixed
point <I and i s thus c q u i t o the d o c i V o i the n m s center ~-eliiti\'cti1 XYZ.
Thus. we ciin express k l l . (14.2Xl a s
Furthermore. we have fiir l t , :
H,, = H,
+
r,< x MI;:
whcre wc hiivc u\cd thc tact that r,,( = V to delete one expression Note in
efiect wc h a c put d w ovci-H,, H , and V in Eq ( 14.3) to reach to iihove equatiim. We may now reet:ite E q 14.24 fni- ( i . f i . i d p , ~ i t i ~i
r ils iulliiws o n replacing
I f , uhiiig the iihove equation. Then. uhing a, lir V wc have the desired rcsult:
M, = Hc
+
<,c
X
Ma,
( 14.30)
Case 2. Reference Point at the Center of Mass. We ciin use Eq. 14.30 for
thih purpose. I'int. we will replace M,, usins the left side of Eq. 14.23. BUI i n
s o doing. we w i l l rephce 5 iii the fii-st expression hy (< + tic,).
Note next thiit
Eq. 14.30 calls fix stationary point (1. Wc w i l l want to he the origin 0 o i
XYZ arid si) {,, hecoiiici 'itiiply 5 . l:iiiiilly. \+e replace a, hy i;' i n liq. 14.30
;ind w e have altcr thc\c steps
The internal force\ 'f'.I f i w
earlier and we
Ihii\,c
LCIO contrihutiiin in this cqiiatiiin a s explained
on i-c;irranging the remaining ternis i n thc cqu;itim
SECTION 14.8 MOMENT-OF-MOMENTUM EQUATIONS FOR A SYSTEM OF PARTICLES
~
wton's law for the center of mass, we know that
F, = Mi, and so
on the left and right sides of the equation above cancel. The
on the left side of the equation is the moment about the
external forces. We then get
M, = Hc
(14.31)
the same formulation for the center of mass as for a fixed point
Please note that H,. is the moment about the center of mass
of niass but that the time
Toward or Away from the Mass Center. There
to he considered and that is a point a accelermass center of the aggregate (Fig. 14.33).
0
0
i
Figure 14.33. Point ii accelerates toward or away from c
For such a point, we can again give the same simple equation presented for
cases I a d 2. Thus,
1
where
with poi
lem 14.9
Th
in say th
I
M a = H"
(14.32)
is taken relative to point u (i.e., relative to axes xyz translating
t u ) . Wc have asked for the derivation of this equation in Prob-
component of the equation M,, = Hu for any one of the three cases
x direction,
hf, = H c
ry useful. Here, M,, is the torque about the x axis, and Hr is the
f momentum (angular momenturn) about the x axis. We now examproblem in the following example.
689
690
CHAF'TEK 14
METHODS OF MOMENTIJM FOR PARlICLES
Example 14.16
A heavy chain of length 20 ft lies on a light plate A which is freely rotating
at ;in angular specd of I r a d k c (see Fig. 14.34). A channel C acts as a
guide lor the chain on the plate, and a stationary pipc acts as a guide for thc
chain helow the plate. What is the speed of the chain after it moves 5 fr .staning froin rest relative to the platform'? Neglect friction. thc angular mornentum of the platc, and the angular momentum of the vertical section of the
chain ahout its own axis. The chain weight per unit length, n', is I O IhMi.
We \hall first apply the moment-of-momentum equation about
paint 11 f(ir the chain and plate. Taking the component of this equation
a l m g the I axis. we can say:
M; =
Clearly M: = 0. and so we havc conservation of angular momentum.
That is,
(h)
whcre I and 2 refer to the initial condition and the condition after the chain
moves 5 ft. We can then say:
Therefore,
wr
= 8 radsec
(cl
To find the speed iir iiiovemerit of the chain, we musr next gii 10
energy considcrations. Because only conservative forces are acting here,
we may employ the conservation-of-mechanical-energy principle. In so
doing. we shall use as a datum the end of the chain R at the initial condi-
tion (see Fig. 14.34). We can then say:
(PE), = (IO)(W,)(IOJ
+ (in)(w)(s)
= i.snnSt-lh
Observing Fig. 14.35. wc can say for condition 2:
(PE), = (S)(M.)(IO)
= 875 ft-lb
As for kinetic energy. we have
+ (IO)(w,)(5)
~
(S)(w)(2.5)
Fieure 14.34. Sliding chain
I
SECTION 14.8 MOMENT-OF-MOMENTUM EQUATIONS FOR A SYSTEM OF PARTICLES
691
Exa ple 14.16 (Continued)
wher the first two expressions on the right side give the kinetic energy
from the motion relative to the channel and pipe, respectively. The lasl
expr ssion is the kinetic energy due to rotation of that part of the chain that
is i n he channel. Clearly, V&rl
= V . = 0 initially, and so we have
w e
I
(KE), = +(lJ2[Y)j i " r Z d r = 51.Xft-lb
Furthbrmore, at condition 2, we have (see Fig. 14.35)
Notelhat
(yhannr,)2
= ( V . ) Simply calling this quantity V2, we have
P W 2'
i
1 :
=3.11V:+414
Datum
We c n now state
i
WE), + WE), = (PW, + WE),
1,500 + 51.8 = 875 + (3.11 V;
I
+ 414)
Ther fore,
v*
i
:
9.19
a speed of 9.19 ft/sec along
that the plate A is rotating al
i
M
body. T
of conti
The fin
in mech
In
several
M, =
now illu
ch time will he spent later in the text in applying Ma = fie to a rigid
ere, the rigid body is considered to he made up of an infinite number
uous elements. Summations then give way to integration, and so on.
equations of this section accordingly are among the most important
nics.
the homework assignments, we have included, as in Chapter 13,
ery simple rigid-body problems to illustrate the use of the equation
and to give an early introduction to rigid-body mechanics.'h We
trate such a problem.
instructor may wish not to get into rigid-body dynamics at this time. T h i s approach
i
Figure 14.35. Chain after motion of 5 ft.
692
CHAPTER 14 METHODS OT'MOMENTI'M FOR PARTICLES
Example 14.17
A uniform cylinder of radius 400 mm and mass 100 kg is acted on a1 its
center by a force of SO0 N (hee Fig. 14.36).What is the (tiction force.f?
Takefi> = .2.
z
/
Figure 14.36. Rolling cylinder.
We have shown a free-hody diagram of the cylinder in Fig. 14.37. A
reference xy: with origin al C translates with the center 0 1 mass. We first
apply Newton's law relative t o iiiertial reference XYZ.17 Thus. for the X
direction we have for the center of mass C:
N
Figure 14.37. Free hody
Next, we write the moment-of-momentum cquation about the :axis,
which goes through the wnrr,,- of mass. Thus, noting that we have simple
circular motion relative to the ? a x i \ for a11 narticles of thc cvlinder and
I
SECTION 14.8 MOMENT-OF-MOMENTUM EQUATIONS FOR A SYSTEM OF PARTICLES
ple 14.17 (Continued)
the cylinder and I is the thickness of the
and differentiating with respect to time as
.40.f = - ( p l ) ( 2 z ) ( h ) [ q ]
f = -.I 005(p/)d
M = 100 = ( p I ) [ ~ ( . 4 0 ) ~ ]
pl = 198.9kg/m2
1
(.40)0 =
(C)
-x
Ther fore,
I
(.40)W = - X
Subs ituting for pI and r3 in Eq. (b) using Eqs. (c) and (d), we get
,f = (-.1005)(198.9)
[
-~
.fO)
Now solve for X from Eq. (a) and substitute into Eq. (e):
f = ( . 1 0 0 5 ) ( 1 9 58 . 9. O) (l f7' )
I
Solvi g for j,we get
f = 166.6N
We must now check to see whether our no-slipping assumption i s
valid The maximum possible friction force clearly is
fmax
6
= (100)(9.81)(.2) = 196.2 N
whic is greater than the actual friction force, so that the no-slip assumption i consistent with our results.
693
694
CHAPTER 14 METHODS 0 1 ' MOMENTUM FOR PAKTICLES
"14.9
looking AheadBasic Laws of Continua
In the preceding tlircc chapters, we have presented three alternate ;Ipplo;lches.
They were. broadly speaking:
1.
2.
3.
Direct applic;ition of N e w l ~ i n ' slaw.
Energy mcth(ids.
I.inear-momcntuin methods and moiiient~of-miiiiicntumimettiod~
Thesc all conic from a c o n m o i l source (i.e.. Newton'\ law) iind si1 w e can u\e
any one for particle imtl rigid-body pnihlcms.
Later, when you study more cmiiplcx cunlinua such a s a tl(iwing iluid
wilh heat transfer and compression you w i l l have Lo salisfy four hasic laws.
Thcsc hasic laws are:
1. Coiiservati(in 111 mass.
2. Linear momcntum and moment 01 momentum (Ihese are now Newton's
3.
4.
law).
First law of Lhermodyiiamics.
Second law of thermodynamics.
For more general ciintinua. the above nientimicd fmir hasic laws" ire itid?p n i l r . , ~of each other (i.e., they must he separ;itely catisfied) whereas in part i c k and rigid-body mechanics that wc have bccn studying. 2 and 3 directly
ahove. are cquivalenr to each other; I i s satisfied by simply keeping the ma\\
M constant: and 4 i s satisfied by making sure that friction impede\ the relative motion hctwccn two bodies i n contact.
Funhermore, we applied the approaches of the preceding thrcc chapters
to free bodies, For more general continuuni studies. such iis lluid Ilou'. we
can apply thc I.riur hasic law\ to systems (i.c., frcc bodies) and also to socalled coiitrol volumes (fixed voIumes i n space) as discussed i n the I.ooking
Ahead Section 5.4.
Thus. in this b ~ i o k w
, e are conqidering a very simplc phzisc [if coiitinuuni mechanics whcrehy. in effect. we need only consider explicitly
one of the basic laws. Your view w i l l hriiaden a s you move thriiogh thc
curriculum
In some mechanics hooks there i s presented a n elementary preseimtion
lor determining the force develiiped by ii stream of wiitcr or orher fluid on a
SECTION 14.9 LOOKING AHEAD BASIC LAWS OF CONTINUA
ng a similar approach, the other basic laws are developed in the
r the conservation of mass, we equate the net efflux rate of mass
author believes these laws in the above form should be taken up
f the basic laws needed in fluid mechanicsl9 and
IYSek I.H.
Shames. Mechanics ofF/uid.r,McGraw-Hill. 3rd ed., 1992, Chapters 5 and 6 .
695
4.89. A system u r particles i \ shown at time t moving in thc ,r!
#lane.Thc following datir apply:
V, =
m , = I Lg.
m2 = 0.7 Lg.
,>I,= 2 kg.
rnd = I .S hg.
(a)
(h)
(c)
V,
=
I n,
Si + Sj mlscc
-4i + 3,j mlscc
V, = - 4 j mlscc
V, = 3i
4 j mlsec
~
What i \ thc total linear momentum of the \ystem'!
What i s Ihc linear momentum ofthe ccntcr OS m a s s ?
What i s the total inirrnent of mmlentum of the
svstem about the origin i i n d ahout point ( 2 , 6)'!
Figure P.14.90.
14.91.
A system 01 particles :it t i m e f , ha'i ~masscsV I , = 2 Ihm.
I Ihm. in, = 3 lhm and location< and vdocitic\ a\ shown
i n ( a ) . The cnmc system of musscs 15 \houri i n (h) a1 lime I,.What
i q the total linear impulse o n the sysrcm durinp thib time intenal'!
What is the total angular impulse M d i during thih timc i n t r n a l
ahwl thc origin'?
I I Z ~=
VIq
1
(- 2.2
\
"4
Figure P.14.89.
14.90. A \ystem of particles at time
.ies and masse\:
V, = 20 fVsec,
V, = I X ftlsrc.
V, = I S ftlscc.
v, = 5 ftlsec.
f
has the following veloci-
m , = I lhm
mz = 3 Ihm
m , = 2 Ihm
= I Ihm
Determine (a) the total linear momentum of the bystctn, (hJ the
mgular mtrmentum rrf the systcnl ahout the wigin, and IC)
the angular momcntum nf the systrm ahout point (1.
s90
xftkc
14.93. A mechanical system is composed of three identical bod
ies A , B, and C each of mass 3 Ibm moving along frictionless rod
1 2 0 ~apart on a wheel. Each of these bodies is connected with ai
inextensihle cord ti) the freely hanging weight D. The connectioi
of the cords to L) is such that no torque can be transmitted to D
Initially, the three masses A , B , and Care held at a distance of 2 f
from the centerline while the wheel rotates at 3 radlsec. What ii
the angular speed of the wheel and the velocity of descent of I
if, after release of the radial bodies, body D moves I ft? Assumi
that body D is initially stationary &e., is not rotating). Body L
weighs 100 Ib.
45".
-1
m.
I
n
Figure P.14.93.
(b) Time f2
Figure P.14.91-b.
14.92.
I .5 "11.
mass (I
AB whs
that ins
masses slide alvng bar AB at a constant speed 01
Bar AB rotates freely about axis CD. Consider only the
: sliding bodies to determine the angular acceleriltion 01
l e hodies are 1.5 m from CU if the angular velocity 81
is I O radliec.
YO
14.94. Two sets of particles a , b, and L', d (each particle of mass
m) are moving along two shafts AB and CD, which are, in turn
rigidly attached to a crossbar EF. All particles are moving at a
constant speed V, away from EF, and their positions at the
moment of interest are as shown. The system is rotating about G,
and a constant torque of magnitude Tis acting in the plane of the
system. Assume that all masses other than the concentrated
masses are negligible and that the angular velocity of the system
at the instant of discussion is w. Determine the instantaneous
angular acceleration in terms of m, T, w,9,.and s2.
C
D
Figure P.14.92.
w
Figure P.14.94.
697
14.95. A uniform rod with a makc of 7 kglm lies flat on a fiictionless surface. A fhrce of 250 N act? on the rod as shown in
the diagram. What i s the angiilar accclcralion o f the rod'! What is
thc acceleration o f t h c m
14.97. A unifbrm cylinder of cadius I 111 r d l s without dipping
down a 30" incline. What i s tllc angular acccleriltion of thc cylinder i f i t has ii ~nlilss01SO kp''
Figure P.14.97.
5 kg i q acted o n by a
14.98. A cylinder of lungth i m a i d r n
torque T = ( I 1.251 + ? I t ' ) N - m iwhcre I i s in seconds) about its
gcomrtric %xi\. What i s the anpular speed aftcr 10 SCC? The cylinder i\ ill irc\t uhcn the torqur ih ;appllcd.
Figure P.14.95
7~
7
~~~~~~~~
_
*14.96. Consider an aggregate of particles with C i l k thc mass
centcr and m i n t A iiccelcratinz- toward o r :LW:IY f n m C'. Starl wlth
the expression for H ahout 0 given :IS
X P =
I
M,, = € I ,
Figure P.14.96.
'x
~~
~
~
~
,-
,
~
A
mm
I*--
3 111
Figure P.14.98.
xp,,,)x*l'
Formulate Mi, in term, of I; arid use Newton's law to rlirninatr
terms. Next show from the vxiltinp equation that
I
X
14.99. 4 conmiit torque 7 of 800 N-m is :ipplied tu il uniform
cyliiidcr 01mdiur 400 cmn and mass 50 kp. A 1.500-kN weight is
iiitached 10 thc cylindcr with ii liglil cable. What is the accclcration
UI W !
I
14.101. A u m t i i n t t m p e Tu1 500 i w l h i \ applied til a uniform
cylinder of r;itliii\ I it. A light inrxtcnsihlr cable i\ urapped part])
connected to a block W
of W if the cable doe.s not
the hlock. For the cylin-
Figure P.14.103.
14.104. Do Problem 14.103 for the case where a force given as:
& = .1
F = S0i + 7SjN
is applied at point u instead of the 500-N force.
Figure P.14.101.
14.102
a widtt
assume
section
center1
canal with a rectangular cross section is shown having
" and a depth Of
water is
be zero at the banks and 10 V Z Y Parabolically over the
iown in the diagram. If 6 i s the radial distance from the
f the channel, the transverse velocity Vs is given as
V,
What i
water i
axes)?
". The
= $225
~
Of
14.105. A cylinder weighing 50 Ib lies on a frictionless surface.
Two forces are applied simultaneously as shown in the diagram.
What is the angular acceleration of the cylinder'? What is the
of the
center?
a*) ft/sec
angular momentum Ha about 0 at any time f of the
circular portion of the canal (Le., between the x and y
radial component v i s zero.
Y
i
i
Figure P.14.105.
(
14.106. A thin uniform hoop rolls without slipping down a 3V
incline. The hoop material weighs 5 Ib/ft and has a radius R of
4 ft. What is the angular acceleration of the hoop?
Figure P.14.102.
14.103
frictior
the an$
the ma;
hoop with mass per unit length 6.5 kg/m lies flat on a
surface. A 500-N force is suddenly applied. What is
acceleration of the hoop? What is the acceleration of
nter?
Figure P.14.106.
699
700
CHAPTtK I 4 MFTHODS O b M O M t N T U M FOR PAKTICLtb
14.10
Closure
One o l t h e topics studied i n this chapler i s the impact of bodies under certiaiia
restricted conditions. For such problems, we can consider the bodies as particles before and after impact, but during impact tlic bodies act as deformahlc
media for which a particlc model i s n o t meaningful or sufficient. By making
an clcmcntnry piclure of the action. we introduce the coefficient of restitutiun
to yield additional infiirmation we need tii determine velocities after impact.
This i s an empirical approach. s o our analyses are limited to simple prohlems.
T u handle more complex prohlems or to do the siinplc ones more precisely.
we would have ti1 makc a more rational invcstigation o f the deformatioii
actions taking place during impst- that is. i a continuum approach (11 part 111
thc problem would he rcquircd. However, we cannut makc a careful rtudy 01
the deformation aspects i n this text since the suhject 111 h i g h p e e d defornaation of solids i s ii difficult one that i s s t i l l under careful study by engineers
and physicists.
Note in the last two chapterr we started with Newton‘s law F = Mu
and pcrformed the fiillowing operations:
1. ‘Took tlic dot pruduct of hotti sides using piisilion vcclur r .
2. Multiplied both sides by ilr and integr:iicd.
3. Took thc cross product of both sidcs using position Yector r .
These steps permitted a surprisingly large number of very useful t?irmulaliiins
and concepts that have occupied us for some considerable time. These wcrc
thc energy methods. the lincar-momentum methods. and the moment-ofniomentuiii methods. It should now be clear that Newton’s law requires cotisiderable study tu fully explore i t s use.
In our study o f moment of tiiwneiiturn fur a system o f particles. we set
foith one o f the kcy equations nf inechanics, M,, = HA. and we inlroduced i n
the examples several considerations whose mure cai-eful and complete study
w i l l occupy a good purtiim o f the remainder of the tcxt. Thus, i n Example
14.17 we have “in miniature.” as i t were. the major cleinents involved i n the
rludy of much of rigid-body dynamics. Recall khat “c employed Newton’i
law for the mass center and thc n i ~ i n i e n t ~ o f ~ i r ~ o i n eequation
r ~ t u ~ i ~ about the
mass center to reach the desired rcsults. In 50 doing. however. we had to
make use of certain elementary kinematical idcas from our carlicr work i n
physics. Accordingly. ti1 prepare ourselvcs fur rigid-body dynamics in Chap
ters I 6 and 17. we shall dcvote ourselves in Chapter I S to a rather ciareful
examination o f the general kinematics of a rigid body.
Although we shall he much conccrned i n Chapter 15 with the kinematics of rigid budies. we shall nut cczse to consider particles. You will see that
an understanding of rigid-body kinematics w i l l pcrmit us 10 formulate very
powerful rclatiuiis for the gcncral relative motions o f ;I particle involving
references that niuve i n any arbitrary manner with respect to each other.
x
,
Figure P.14.109.
1
Figure P.14.107.
= (100
+
5 0 ~ 'Ib
)
"; = 30 Ib, W, = 60 Ib, and W, = 50 Ib, what is the
force'? The dynamic
14.110. A space ship is in a circular parking orbit around the
earth at 200 miles above the earth's surface. At space headquarters, they wish to get the vehicle to a position 10,ooO miles from the
center of the earth with a velucity at this position of 25,000 miihr.
The command is given to fire rockets directly to the rear for a
specified shon time interval. What is the change in speed needed
for this maneuver? What are the radial and tangential velocity
components of the 25,000 mi/hr velocity vector?
P
Figure P.14.108.
I
,
.,
,,
i
f
I
I
\
\
\
.
-
200 km'!
,
_
_
/
Figure P.14.110.
70 1
14.111. A small clastic ball is dropped from a height o f 5 m onto
a rigid cylindrical body having a radius of 1.5 111.At what position
on thex axis does the hall land after the cullision with thc cylinder?
strikes 1.5 rn above the .r axib and if the c o l l i i i m is pe~ltcllyela\tic, what i \ the maximum height rcached hy the bullet as It iicochets'! Nrglcct air rmihtanoc and rake thc belocity 01 the hullct on
impact as 700 m/scc with a dircctiw thdt I S p ~ r a l l etlo the I axis.
1.
v
I
Figure P.14.115
5m
I
14.116. I n Plohleni 14.1 15. ilsbunic an inela'tic impact with
= .h. A t what position along ,x does the bullet strike the p a r a h h
after the irnpxr'?
t
14.117. A >pace \rehick is in a circular "parking" orhit ( I
aruund the earth 200 k m above the earth's hurfacr. I t i \ lo lransfrr
t u another circular orbit (2) 500 k m ahow the earth's surface. The
transfer t o the second orhit i s dune in two stage,.
x
Figure P.14.111.
I
14.112. Do Problem 14.1 I I for an inelastic impact with
t
= .h.
14.113. A small elastic sphere i s dropped fiom pusitiun (2, 3,
30) f t onto a hard soherical body havine
- a radiub of 5 ft positiuned
so that the I axis of the reference shown i s along a diameter. Far a
perfectly elastic collision, give the speed of the s m d 1 sphere
1. Fire rucheth so the vehicle has an upoper cquill tu the r d i u 0 1
the sccond circular urhit. Whal uhangc of speed i\ required l o i ~
this
2. At u ~ i o p wrockets are (ired again t i l get
into the second circular orhit. Whal i s thi\ srcund change of hpced?
directly after impact.
500 k m
0
I
,, ~-
.
\,',_---..~.
.,.
~~
-
\
\
1
I
I
I
I
I
I
200km-
~.
~~
~
,
,
~~
Figure P.14.117.
*14.118. A tugbwdt weighing 100 tuns is moving toward a statiuiiay buge weighing 200 tom and c u y i n g a load C wcighing 50 ton&.
/
The tug ib moving at 5 knots and its propellers arc devcloping a thrust
of 5,000 Ib when it cwlacts the huge. Ai a result of thc soft padding
Figure P.14.113.
at the nose of the tug. consider that there i s plastic impact. If the load
14.114, D~ problem 14,113 for an inelastic
impact
with = ,6, C I S not tied in any way to the huge and has B dynamic coefticient of
friction o f . I with the slippery deck of the harge, what i\ the sped V
14.115. A bullet hits a smooth, hard, massive two-dimensional of the barge 2 sec after the tug first contacts the barge'! 'The load C
body whose boundary has been shown as a parabola. If ihe bullet slips during a I-see internal starting at the beginning ofthe contact.
x
3i + 1j mlsec. After impingement three droplets are formed moving parallel to the ry plane. We have the following information:
i
D , = .6 mm,
V, = 2 d s e c ,
8, = 45’
D, = 1.2 mm,
V, = I mlsec,
0, = 3W
Find D,, V,, and 8,.
I
Figure P.14.118.
3 mlsec
7
1 m/sec
(a)
Figure P.14.121.
(b)
14.122. If the coefficient of restitution is .8 for the two spheres,
what are the maximum angles from the vertical that the spheres
will reach after the first impact? Neglect the mass of the cables.
I
Figure P.14.119.
14.120. A body A weighing 2 tons is allowed to slide down an
incline o a barge as shown. Body A moves a distance of 25 ft
along the incline before it is stopped at E . If we neglect water
resistanc how far does the baree shift in the horizontal direction?
1
i
14.121.
phere at t
droplet o
The vel
\/
25’
-4
. ..... .
..- -.
2.5 kg
Figure P.14.122.
123. Thin discs A and B slide along a :tior s surface.
Each disc has a radius of 25 mm. Disc ~A has a mass of 85 g,
whereas disc B has a mass of 227 g. What are the speeds of the
discs after collision for c = .7? Assume that the discs slide on a
frictionless surface.
I
. ,
.
,
_____~
Figure P.14.120.
A water droplet of diameter 2 mm is falling in the atmose rate of 2 d s e c . As a result of an updraft, a second water
diameter 1 mm impinges on the aforementioned droplet.
ity of the second droplet just prior to impingement is
I
_
;
Figure P.14.123.
703
14.124. A BB i\ shot at the hard, rigid surface. The speed o l l l i c
pellet is 3110 fisec a\ i t strikrs the suriace. If the directimi of
the velocity for the prllel i s @\en hy thc i d l o w i n g unit vector:
t = -.6i - .Xk
what is thc final velocity \'ectoi of thc pcllet lor ii collision havinz
E
14.127. Cmipnle thc angular cmonicntiim uhout 0 01a uniform
Irrd, of Icngth L = 3 rii iind n i t w pcr unit Icnglh nr (11 7.5 hs/rn.
ill l l i e i n a i i i f Nhcn i t i i \erticitl iind h a \ a n a n g ~ i l aspeed
~ ~ (0 (6
3 i-adlscc.
= .7?
0
7
-
I
k
,/
w
\
A
5'
Figure P. 14.124.
!5. A chain 0 1 wrriught i1on, with length o f 7 111 and il n i a v
of 100 kg, is held so thal i t jus1 touches the \upport AB. I l t l i e ~ h i i i r i
i s rclcased, determine the total impulse driririg 2 sec in thc wrtic:il
directinn cxpcrienced by thc support if the impiicl i\ plastic (i.u.,
the chain does not bounce up) arid i f we m w t : thz support \(I that
thc links land UI the platfimn and 1not 011 each othcr'! IHirit: Notc
that any chain wrsrin*. on AH dclivcrc a vcrtiCal impulsc. Also
check 10 see if thc entire chaiii lands 01, A H hcfiire 2 sec.]
Figure P.14.125.
14.126.
Two trucks x e shown moving up ii I O incliiir. Tmck ,A
weighs 26.7 kN and i s developing il 13.30-kN driving lorce on thc
road. Truck R wzighs 17.8 hN and i \ curinccled with an incxtensihle
cable to truck A. By operating a winch h, truck B approaches truck A
with a conmnt acceleratim oi 3 m/xc.'llill lime f = 0 both r l u c k ~
have a speed of 10 mlsrc. what arc their ipeeil\ a1 timer = 15 scc?
Figure P.14.126.
704
Figure P.14.127.
Fieurc P.14.12X.
i
14.130.
container
stationary
you will I
ity of the
as 101r f
momentu
closed container is full of water. By rotating the
or some time and then suddenly holding the container
we develop a rotational motion of the water, which,
am in fluid mechanics, resembles a vunex. If the velocuid elements is zero in the radial direction and is given
sec in the transverse direction, what is the angular
of the water?
1
14.132. A spacecraft has a burnout velocity Vo of 8,300 mlsec
at an elevation of 80 km above the earth's sulface. The launch
angle a is 1 5 ~What
.
is the maximum elevation h from the earth's
surface for the spacecraft?
Figure P.14.132,
Ill'
Figure P.14.130.
14.131. dentical thin masses A and B slide on a light horimntal
rod that i' attached to a freely turning light vertical shaft. When
the masse; are in the position shown in the diagram, the system
mtates at ;Ispeed w o f 5 radlsec. The masses are released suddenly
from this position and move out toward the identical springs.
which have a spring constant K = 800 Iblin. Set up the equation
for the CO npression 6 of the spring once all motions of the bodie?
relative tu the rod have damped out. The mass of each body ii
1 0 Ibm. Neglect the mass of the rods and coulombic friction
Show that S = .OX361 in. satisfies your equation.
14.133. A set of particles, each having a mass o f 112 slug,
rotates about axis A--A. The masses are moving out radially at a
constant speed of 5 ft/sec at the same time that they are rotating about the A-A, axis. When they are I ft from A-A, the
angular velocity is 5 radlsec and at that instant a torque is
applied i n the direction of motion which varies with time f in
seconds as
torque = ( 6 1 +
~ l o t ) lb-ft
What is the angular velocity when the masses have moved out
radially at constant speed to 2 f t ?
Torque
Figure P.14.133.
14.134. A torpedo hoat weighing 100,000 Ib moves at 40 knots
( I knot = 6,080 ftlhr) away from an engagement. To gv even
faster, all four 50-caliber machine guns are ordered to fire simultaneously toward thc rear. Each weapon fires at a muzzle veloc-
I
I
Figure P.14.131.
ity of 3,000 ftlsec and fires 500 rounds per minute. Each slug
weighs 2 0%.How much is the average force o n the boat
increased by this action'? Neglect the rate of change of the total
mass of boat.
705
14.135 A devicc to be detunatcd with a small charge ir
suspended in space [ser ( a ) ] .Dircctly after detonation. lour fragments are formed moving away from the paint of suspension. The
following information is known ahout thcse fragments:
I Ihm
v, = 200;
m , = 2 lhm
123
=
+
l0O.j f l k c
IXOj
~
m , = 1.6 lhm
V , = 200; + l S O j
~
=
100k Wsec
+
action i T it occurs in I rec'!
*14.138. I n the ,fission procebs i n a iiuclziir reactor, a risU
nucleus first ahsorhs (11 captures a neutron lsee (a)]. A short tinic
later. the "'U nucleu5 hreakr up into lission products plus ncu-
m, =
~
t o the war of 10 Wsec relative tu the initial spccd uf thc octopus.
What horicpowcr is hcing dcvckrped hy the octopus in thc ahme
1X0k fUsec
troiis, which may suhsequently he captured by ulher 23'U nuclei
and inaintain a <.kiiiw r r w l i o n . Energy is rcleased in each fiwion.
111(h) we have shown the results of a possible fission.The fnllnwing information is known fnr this fission:
3.2 Ibm
Kinetic
What is the vclacity <'!
Maw
No.
Energy
(MeV)
Direction of V
.3i - .2j + .YXk
Prr,duct A
I3X
E
Prclduct n
Neutron I
Neutron 2
96
90
IO
eH = l,i
IO
E, =
I
I
E,, =
E,
= .hi
+
+
m8j
+
n,$
.8j
.4i - .hj - .hWk
What i s the energy E of product A in MeV and what is the V C C ~ O ~
E" for the velocity of product B? Assume that before fissiun the
nuclcus of 235U plus captured neutrons is stationary: [Hinr: You
do not h u e to aclually ciinvert MeV to j d c s or atomic number to
kilogram5 to cilrry out the pnihlem.]
Figure P.14.135.
f Y4
14.136. A hawk is a predatory h i d which olten auacks snvallei
birds in flight. A hawk having a inass of 1.3 kg is swooping down
nn a sparrow having a mass of I S 0 g. Just hcfore sei~ingthe rparrow with its claws. the hawk is moving downward with a speed VH
of 20 kmlhr. l h r spnrrow i s moving horizontally at a speed C; of
I S kmlhr. Directly after cei7.ure. what is the speed of the hawk and
its prey'! What i c the loss in kinetic energy in J~xdes'!
14.137. The principal mode of propulsion of an ~ c t o p u ai s tu
take in water through the mouth and then after closing the inlet
to cject the water to the rear. If a 5-lh octopus after taking in I Ih
of water is moving at a spccd of 3 ftlscc, what is its \peed directly
after ejecting the water'? The water is ejcctcd at an average \peed
706
X
lb)
Figure Y.14.138.
Kinematics
of /Rigid Bodies:
Relative Motion
Introduction
ble to analyze complicated motions in a more simple systematic
several references. Second, the motion of a particle is often
set the stage for our main effort in the remaining ponion
I
Translation and Rotation of
15.2
Rigid Bodies
For pu ses of dynamics, a rigid body is considered to be composed of a
continuo s distribution of particles having fixed distances between each
I
707
708
CHAPTER 15
KlNEMAllCS OF RIGID BODIES: RELAI'IVC MOTION
,dl
Figure 15.1. TrJnrlation of il hod?
M
J
Rotation. I1ii ripid hmly IIIOK\ s o that alonx wmc \tlaight line all [he pxticks of thc body. ( I I ~II 1hypothctic;il r x i e m i o ~o ~f t l i c hod). have
velocity relative io w n i e refcrcnce. lhe hody 15 said to he ill roiolron rclatiw to
this relrrencu. 'l'he line o l atation:uy p x l i c l e i i i called thc irris qt m r ~ r r i o , ~ .
\\\
,\
\
\
,\
f
/'
0
We shall now conhider how wc nleahure the rotation of a body. A sillgle revoIution i h defined as the iiiilount of rotiitiun in either a clockwise or a
counterclockwisc direction ahout the n i s of rotation that hrings the hody
back ti1 its original position. Partial revolution\ ciin ciinveniently he rneasurcd
hy obserbing riny line segmenl cuch :IS ,113 i n thc hody (Fig. 15.2) from a
vicwpoint M-M directed along the asis 0 1 rcrlation. I n Fig. 15.3, we have
hhown this vizw of AB at thc bcgititiiiig of the partial rotation as seen along
Ihz axis of rotation. as well iis the \,ien' A'H' iit 1hc end of the paitial rotation.
TIE angle fl that these line\ f<rrni w i l l he the same lirr the initial and final
Firure 15.2. Ilotatim u t a hody.
Figure 15.3. Mrnsurc of a partial wlatiun.
SECTION 15.3
e angle
so formed
, so that infinitesimal rotations d p are vector quantities. Theremagnitude dpldt with
Chasles' Theorem
15.3
e displacement vector for this translation is shown at
ngle A$ about an axis of rotation which is normal to the plane and
ses through point E'.
1
Figure 15.4. Translation and rotation of a rigid body.
CHASLES' THEOREM
709
710
CHAPTER 15
KINEMATICS OF RIGID BODIES RELATIVC MOTlOh
What changes would occur had we chosen some lither point C for such
a procedure? Consider Fig. 15.5, where we have included an alternative
procedure by translating the body so that point C reaches the correct final
position C'. Next. we must rotate the hody an amount A@ ahout an axis of
rotation which is normal to the plane and which passes through C' in order k 1
get to the final orientation of the body. Thus, we have indicated twu routes.
We conclude from the diagram that ihc displacement ARC differs from AR,.
hut there is no difference in the amount of rotation A@, Thus, in general. AR
arid rhe axis of rototion will drpend on rlir poiiit ,,ho.rrn. uliilr rhr r i i n ~ i i i i f01
rotation A@4 1 he the .same.for rill such pobit.r.
!
A R,
Figure 15.5. Tmnslatim and
rotiition of a rigid body
u\inp point\ N ; ~ n dC'.
Consider now the ratios ARlAr and A@IAt. Thcsc quantities can he
regarded as an average translational velociiy and an average rotational speed,
respectively, of the body, which WE could soperpme to get from the initial
position to the final posilion in the timc AI. Thus, ARIA/ and A@lAl rcprescnt
an average measure o f the miitiiin during the time interval At. Ifwe go io rhr
limit by letring A I + 0, use have in.rruntunrous tran.slutional and angular
velocities which, when superposed, give the iristantrineuus motion qfrhr body.
The displacement vector of the chosen point B in the previous discussion represents the translation of the body during the time A/. Furthermore. the chosen
point B undergoes no other n~otionduring At oiher than that occurring during
translation. Thus, we can conclude that, in the limit, the trun.rlntionu1 velor.i/y
used for the hody corresponds to the acrual insrurrrurleous velocity of the chosen point B at time t. The angular velocily w t o be used i n the movement US the
body, as described ahove. is the ~ a m vector
e
for d l puints B chosen Accordingly, w is the instantaneous ungular velocity of the body.
We have thus far considered the movenient of the body along a plane
surface. The same cunclusioiis can he reached for the general motion of an
arbitrary rigid body in space. We can then make the following statements for
the description of the general motion of a rigid body relative to some reference at time t These statements comprise Chasles' theorem.
1. Select any poinl B in the body. Assume that all particles of the hody havc
at the time r a velocity equal to V,, the aciual velocity of the poini B.
2. Superpose a pure rotational velocity w about an axis of rotation going
through point B.
SECTION 15.4 DERIVATIVX OF A VECTOR FIXED IN A MOVING REFERENCE
V , and w, the actual instantaneous motion of the body is deterw will be the same for all points B which might be chosen. Thus,
velocity and the axis of rotation change when different
15.4
Derivative of a Vector Fixed in a
Moving Reference
arbitrarily relative to each other in Fig.
equal to the velocity of 0, onto a rotaaxis of rotation through 0.
Figure 15.6. Vector A fixed in xyz moving relative to XYZ.
suppose that we have a vector A of fixed length and of fixed o n reference xyz. We say that such a vector is "fixed" in
Clearly, the time rate of change of A as seen from reference
can express this statement mathematically as
Howeve), as seen from XYZ,the time rate of change A will not necessarily
be zero. To evaluate (dAldt),, we make use of Chasles' theorem in the
followin ;manner:
1. Consider the translational motion k.This motion does not alter the direction o'A as seen from XYZ. Also, the magnitude of A is fixed; thus, vector
A cannot change as a result of this motion.'
2. We nt:xt consider solely a pure rotation about a stationary axis collinear
with (1) and passing through point 0.
'The line of action of A, however. will change as seen from X Y Z . But a change of line of
action doe, not signify a change in the vector, as pointed out in Chapter I on the discussion
O f equality of "Cctors.
71 1
To best observe this rotation, we shall employ
at
0 a srationury refer-
cncc X’Y’Z’ po\itimed sci lhar %’ coincides with the axis of rotation. This reference i s shown in Fig. 15.7. N o w thc vcctorA i s rotaling ill this instant abiiul
X
/’
I4gwe 15.7. (~‘ylindrica comlxmcnts f i r i \cccor A
h e Z‘ axis. We h a v e \tiown cylindrical coordinates to the end o f A (Le.. iit
point ( 1 ) ; :mil hiivc shown cylindrical components Ai. A,,. and A., In Fif.
15.8. we h a w sh(iun point i i with unil vcctors E?. e a . and e7,, for cylindrical
coiirdinatcs a1 thih point. Wc can accordingly cxpress A as
A
=
A,€,
+ AH€#+ AL.e7
Y’
X
Figure 15.8. Unit vectors fbr cylindriual coordinate
Clcarly, 3s A rotales ah(iu1 z‘.the values of the cylindrical scalar componcncs
o l A for X’Y’Z’. namely Ai, A,. and A,,. do not change. Hence. as seen from
4
SECTION 15.4 DERIVATIVE OF A VECTOR FIXED IN A MOVING REFERENCE
~
I
X'Y'Z,
=
A, = A, = 0. Also, noting that iZ,
= 0, we can say that
1.
We have already evaluated the time derivatives of the unit vectors for cylindrical c rdinates. Hence, using Eqs. I 1 2 8 and 11.32 and noting that '
6
correspo ds to 0,we have
side is simply the cross product of w andA as you can see
cross product with cylindrical components. Thus,
!
We conc ude that
relative to XYZ, we would observe the same time
reference as from the former reference. That is,
and we can conclude that
(15.1)
Thc: foregoing result gives the time rate of change of a vectorA fixed in
reference x.yz moving arbitrarily relative to reference XYZ. From this result,
we see that (dAldi),, depends only on the vectors w and A and not on their
lines of action. Thus, we can conclude that the time rate of change of A fixed
in q z i s lot altered when:
1. The v x t o r A is fixed at some other location in xyz provided the vector
itself s not changed.
2. The a:tual axis of rotation of the xyz system is shifted to a new parallel
position.
We can differentiate the terms in Eq. 15.1 a second time. We thus get
7 13
714
CHAPTER
is
KINEMATICS OF RIGID BODIES: RELATIVE MOTION
Using Eq. 15.1 to replace (dAldrJ,, and using & to replace (dwldt),,,
since the reference being used for this derivative is clear,2 we get
You can compute higher-order derivatives by continuing the process. We
suggest that only Eq. 15.1 be remembered and that all subsequent higherorder derivatives he evaluated when needed.
In this discussion thus far, we have considered a vectorA fixed in a reference hyz. But a reference xyz is a rigid system and can he considered a rigid
body. Thus, the words ‘!fixed in a rsferencr xyz” in the previous discussion
can he replaced by the words ‘ : f w d in a rigid body.” The angular velocity w
used in Eq. 15.1 is then the angular velocity of the rigid body in which A is
fixed. We shall illustrate this condition in the following examples, which you
are urged to study very carefully. An understanding of these examples is vital
for attaining a good working grasp of rigid-body kinematics.
As an aid in carrying out computations involving the triple cross product, we wish to point out that the product
o,k x
(@,k
x cj) = - 0 ; c j
That is, the product is minus the product of the scalars and has a direction
corresponding to the last unit vectnr, j . Remembering this will greatly facilitate our computations.3
Additionally, consider a situation where the angular velocity of body A
relative to body B is given as wI, while the angular velocity of body B relative to the ground is w2. What is the total angular velocity w7 of body A relative to the ground? In such a case, we must rememher that the angular
velocity wI of body A relative to body B is actually the difference between
the total angular velocity w,of hody A as seen from the ground and the angular velocity w2 of body B as seen from the ground. Thus.
w1 =
WT
-w2
Solving for w y we get
w7 = w 1
+
W2
We see from above that to get the total angular velocity w T ,we simply add the
various relative angular velocities just as we would with any pair of vectors.
‘When it is clear from the discussion what reference is involved fur a time derivative. we
shall use the dot to indicate a time derivativc.
’Of course, i f t h e j vector weie a k vector, then clearly we would amiw at a iiull value for
the triple vector product.
SECTION I5.4 DERIVATIVE OF A VECTOR FIXED I N A MOVING REFERENCE
Exal pie 15.1
A dis
with
whicl
const
paral'
veloc
7 is mounted on a shaft AB in Fig. 15.9. The shaft and disc rotate
(d2w
2
ily gi
onstant angular speed wz of 10 radlsec relative to the platform to
earings A and B are attached. Meanwhile, the platform rotates at a
angular speed wIof 5 radlsec relative to the ground in a direction
to the Z axis of the ground reference XYZ. What is the angular
vectorw for the disc C relative to XYZ? What are (dwldt)xrz and
)xyz?
ie total angular velocity w of the disc relative to the ground is eas1 at all times as follows:
w = w,
At th
+ w 2 radlsec
istant of interest as depicted by Fig. 15.9, we have for w :
Disk C
&wJf
I
I
Platforin
Ground
Reference
Figure 15.9. Rotating disc on rotating platform.
To g
alwa:
dot ti
the first time derivative of w, we go back to Eq. (a), which is
valid and hence can be differentiated with respect to time. Using a
:present the time derivative as seen from XYZ, we have
ri, = ri,, + & J 2
Cons
to he
chr
may
since
(b)
'r now the vector w2. Note that this vector is constrained in direction
ways collinear with the axis AB of the bearings of the shaft. This
is a physical requirement. Also, since q is of constant value, we
nk of the vector w2 as $red to the platform along AB. Therefore,
2 platform has an angular velocity of w I relative to XUZ, we can say:
G2 = w , x w 2
(C)
115
7 16
KINEMATK'S or' RIC;I)
C H A P T ~ KI S
Bouifs KEI.ATIVI:
MOTION
Example 15.1 (Continued)
As Cor W,,
namely the other vector i n Eq. (b). wc iiotc Lhal iis s e m frnm
XYZ, w , i s a cnnslant veclur and si) at all times r U , = 0. Hcncc Eq. ( h ) can
be written 21s Si)llows:
& = w l xu,
((1)
This equatinn i s valid at all times and s o can he differentiated again. A I the
instant of intercut as depicted by Fig. 15.9. w c liavc for W:
To get &, we ~ i o wdifferentiate (d) with respect
ij =
to
time. W c llavc
W l x w 2 + w , x &>
+ w I x l w ,x W , i
(1)
=0
where we have wed the Sac1 that h , = 0 at all times a s well a s
interest, we have
ti.( c ) I i r
W,.A t the instant nf
ij = Sk
x ( 5 k x lO.ji =
-2Mjradsec3
Example 15.2
In Example 15.1, consider a position vector p hctween two points nn the
rotating disc (see Fig. 1 5 . 1 0 ) . The length o f p i s 100 rnm and. at the instanl
of interest, i s in the vertical direction. What are the l i n t and hecond time
derivatives o f p at this instan( as seen from the ground referencc'?
It should he ohvious that the vector p i s fixed ti) the disc which has
at all times an angular velocity relatiw ti) X Y Z equal to w , + w,. Hencc.
at all times we can say:
p
= tw, t w ? ) X p
ia)
Figure 15.10. Displaccrnrnl
in disc.
A t the instant o f interest. w e have noting that p = IOOk
ii
=
(Sk
+ lOji x
IO(lk = 1,000imdsec
To get the secnnd derivalivc n t ~ )go
. back (11 Eq. (a) and diSkreiitiate:
p = (&, + & > I x p + ( w , + W 2 ) x p
(hi
V ~ C I Op~
I
I1
SECTlON 15.4 DERIVATIVE OF A VECTOR FEED IN A MOVING REFERENCES
Exa ple 15.2 (Continued)
Notin that cbl = 0 at all times and, as discussed in Example 15.1, that
w2 is ixed in the platform, we can sdy:
p
= (0
+WI
x w 2 ) x p + (wl +w:) x
p
(C)
At th instant of interest we have, on noting Eq. (b):
p
= (Sk x l O j ) X lOOk
+ (5k + I O j )
x 1,000imm/sec2
!
AI hough we shall later formally examine the case of the time derivative of v ctorA as seen from XYZ when A is notfixed in a body or a reference
xy:, we an handle such cases less formally with what we already know. We
illustrate this in the following example.
I
'
Exa pie 15.3
For t
tive
radls
the a
insta
For
e disc in Fig. 15.9, oz = 6 radlsec and d2 = 2 rad/sec2, both relathe platform at the instant of interest. At this instant, wI = 2
c and hl = -3 rad/sec2 for the platform relative to the ground. Find
gular acceleration vector dJ for the disc relative to the ground at the
t of interest.
The angular velocity of the disc relative to the ground at all times is
4
w=01+w2
(a)
dJ = cbl + c b 2
(b)
, we can then say
i
It is pparent on inspecting Fig. 15.1 I that at all times w I is vertical, and
so w , can say:
7 17
71 8
CHAPTER I S
KINEMATK‘S 0 1 : RIGID BODIES: Rlil.ATIVE M U I I O N
Example 15.3 (Continued)
However, w2 is changing direction and, mnst imporlantly. is changing
magnitude. Because of the latter, w2 cannot he considercd lixcd i n ii reference or a rigid body Sor pui-poses of computing b2.To get around this diSficulty, we fix a unit vector j ’ 011f0 tlw plu!fi,rm to he colliiiciir with the
centerline of the shaft A B as shown in Fig. 15.1 I. Wc know the ;In_pular
velocity of this unit vector; it is w , at all Limes. We can then express o2i n
the following manner, which is valid at all time\:
w2 = w2j’
X
(d)
relerenie
Figure 15.11. Unit veclorj’ fixed Lo platform.
We can differentiate the above wilh respect to time as follow\:
b2= [b7j ’
+ w2 J’
Butj’ is fixed to the platlorin which ha\ angular vclocily w , relalive
XYZ at all times. Hence. we have for thc above.
G 2 = (b2j‘ + 0 2 ( w I x j ’ )
10
(e)
Thus, Eq. (b) then can be L’
w e n ns
b = w , k + o i 2 j ’ + m 2 ( w ,X j ’ )
This expression is valid at all Limcs and could he differentiated agnio. At
the instant of interest, we can say. noting thatj’ = j ifit this instant,
b = -3k
+ 2 j + b(2k
2j
xj)
-Yrds
15.1. Is the motion of the cabin of a ferris wheel rotational or
translational if the wheel moves at uniform speed and the occupants cause no disturbances? Why?
i
Z
15.2. A cylinder rolls without slipping down an inclined surface.
What is the actual axis of rotation at any instant? Why? How is
this axis moving?
15.3. A reference xyz is moving such that the origin 0 has at
time t a velocity relative to reference XYZ given as
+
V, = 6 i
12j
+
x
13kftlsec
The xyz reference has an angular velocity w relative to X Y Z at
time t given as
+
o = 1Oi
12j
+
I
X
Figure P.15.4.
2kradIsec
What is the time rate of change relative to XYZ of a directed line
segment p going from position (3,2,-5) to (-2.4.6) in xyz? What is
the time rate of change relative to XYZ of position vectors i’ and k’?
/
15.5. Find the second derivatives as seen from XYZ of the vector
p and the unit vector i’ specified in Problem 15.3. The angular
acceleration of qi relative to XYZ at the instant of interest is
&=5i+2j+3krad/sec2
z
15.6. Find the second derivative as seen from X Y Z of the vector
P , , ~specified in Problem 15.4. Take the angular acceleration of
xyz relative to X Y Z at the instant of interest as
& = 1% - 2k rad/secz
X’
15.7. A platform is rotating with a constant speed 0,of IO
radlsec relative to the ground. A shaft is mounted on the platform
and rotates relative to the platform at a speed o2of 5 radlsec.
What is the angular velocity of the shaft relative to the ground?
What are the first and second time derivatives of the angular
velocity of the shaft relative to the ground!
/
Figure P.15.3.
15.4. A reference xyz is moving relative to X Y Z with a velocity
of the origin given at time I as
V, = 6i
+
4j
+
6kdsec
C;-
The angular velocity of reference xyz relative to XYZ is
w = 3i
+
14j
+
2kradlsec
What is the time rare of changc as SCCCT from XYZ of a directed
~ xyz going from position 1 to position 2 where
line segment P , , in
the position vectors in r y z for these points are, respectively,
p , = 2i’+3j‘m
p7 = 3 i ‘ - 4 j ’ + Z k ’ m
X
Py
Figure P.15.7.
7 1‘
15.8. In I'rohlem 11.7, what are the f i n 1 mil second t i m e
derivati\!ec of il dircctcd line segment p i n the disc at thc instant
[hill the \ystem has thc gewncir) \Iiown? 'The x c t o r p i \ 01
length I O inin
Figure 1'.15.11.
Figure P.15.Y.
X
Figure P.15.12.
ry
Y
Figure P.lS.15.
Figure P.15.13.
15.14. An electric motor M is mounted on a plate A which is
welded to a shaft D. The motor has a constant angular speed w2
relative to plate A of 1,750 rpm. Plate A at the instant of interest is
in a vertical position as shown and i s rotating with an angular
speed w , equal to 100 rpm and a rate of change of angular speed
b, equal to 30 rpmlsec-all relative to the ground. The normal
projection of the centerline of the motor shaft onto the plate A is at
an angle of 45" wilh the edge of the plate FE. Compute the first
and second lime derivatives of w, the angular velocity of the
motor, as seen from thc ground.
15.16. A cone is rolling without slipping such that its centerline
rotates at the rate w , of 5 revolutions per second about the Z axis.
What is the angular velocity w of the body relative to the ground?
What is the angular acceleration vector for the body?
X
Figure P.15.16.
Z
15.17. A small cone A is rolling without slipping inside a large
conical cavity B . What is the angular velocity w of cone A relative
to the large cone cavity B if the centerline of A undergoes an
angular speed w , of 5 rotations per sccund about the Z axis?
C
\
Z
I
0 1
Figure P.15.14.
15.15. A racing car is moving at a constant speed of 200 milhr
when the driver turns his front wheels at an increasing rate, W , , of
.02 radlsec2. If w , = ,0168 radscc at the instant of interest, what
are w and cb of the front wheels at this instant? The diameter of
the tires is 11:l in.
/xi.
Figure P.15.17.
12 1
15.18. An amusement park ride comisis of a staiionary verlical
tower with arms that can swing outward from the Lower and a1 the
same time can rotate about thc tower. At the ends of the arms,
cockpits containing passenger5 can rotate relatiw to ihe arms.
Consider the case where cockpit A rotales at angular speed wz relative to arm BC, which rotates at angular speed m1relative lo the
tower. If % i s fixed at YO", what are the total angular velocity and
the angular acceleration of the cockpit relative til the ground? Use
o, = .2 radisec and w? = .h rad/sec.
~
I
1,
l
Figure P.15.20.
15.21. In PI-rrhlem 15.18, find & of the cockpit A for the case
whcrc 0,= .2 rad/secz and i-l, = .3 rad/iei2.
15.22. In Prohlem 15.13, find & o l bcem AB relative to the
ground it a1 the instant shown the fbllowing data apply:
wl = .3 radlsec
W ,= .2 r a d k e c '
<u2 = .6 rad/sec
W? = - .I rad/sec2
Y
X
Figure P.15.18.
15.19. In Prohlem 15.18. find i, of the cockpit for the case
where 0 = m, = .8 radlsec at the instant that 0 = Yo".
15.20. Mass A is connected 10 an inextensihle wire. Supports C
and D a r e moving as shown.
(a) What is the velocity vector of m a s A?
(b) If cylinder G is free to rotate and there is no slipping, what is
its angular velocity'?
l'he following data apply:
h=2m
L=3m
1=2m
( V $ ) , = .5 t d s
CV, j, = .6 m/s
find the angular acceleration
In Pr~,hlem
for the gun harrel, if, for the instant shown in the diagram. the fnllowing data apply:
I$
= 3 0 radlsec.
4 = .26 r a d / s e c 2 ,
B = 20"
q$ = 30"
6 = . I7 radlsec
8 = -.34 radlsec'
15.24. I n Problem 15.1 1. find V if at the instani shown in the
diagram:
( V ,. 1., = .24 Lm/F
( V , , ) , = 2 1 rn/\
R-lm
a = 45'
l'he last four problems of this SPI w e d w i p d /Or those
dents who huvr studied Example 15.1.
.stu
wI = 5 rad/sec
W. = IO rad/sec2
W, = 2 radlsec
W, = 3 radlsec'
V = IO mlsec
0 = s m/sec2
SECTION 15.5
15.5
APPLICATIONS OF THE FIXED-VECTOR CONCEPT
123
Applications of the Fixed-Vector
Concept
In Section 15.4, we considered the time derivative, as seen from a