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Analytic Mechanics
THE
MACMILLAN COMPANY
NEW YORK + CHICAGO
DALLAS + ATLANTA + SAN FRANCISCO
THE MACMILLAN COMPANY
OF CANADA, LIMITED
TORONTO
Analytic
Mechanics
AWlal 01aID) 15; 1D) AP 1 O)Iyy
VIRGIL
MORING
FAIRES
Professor of Mechanical Engineering
North Carolina State College
SHERMAN
DANIEL
CHAMBERS
Professor Emeritus of Engineering Mechanics
Purdue University
PROBLEMS
IN THIS
EDITION
JAMES
REVISED
HARVEY
BY
CADDESS
Assistant Professor of Mechanical Engineering
Agricultural and Mechanical College of Texas
THE
MACMILLAN
New York
COMPANY
Copyright 1952 by THE
MACMILLAN
COMPANY
All rights reserved—no part of this book may be reproduced in
any form without permission in writing from the publisher, except
by a reviewer who wishes to quote brief passages in connection
with a review written for inclusion in magazine or newspaper.
Printed in the United States of America
First printing
First edition, Mechanics of Engineering, copyright 1934 by
THE
MACMILLAN
COMPANY;
Revised
edition,
Analytic Mechanics, copyright 1943 by THE MACMILLAN
COMPANY.
PREFACE
TO
THIRD
EDITION
While the objectives remain the same as in previous editions, this book
has been rewritten with the idea of making explanations clearer to students.
Statements of important basic ideas, and also warnings of possible pitfalls
are repeated often enough, we hope, to impress students, but not too
often to bore teachers.
Many new illustrative examples have been added.
We believe that students will appreciate these features and will be motivated
by the many problems with an actual engineering flavor.
Some teachers prefer to start with the general case and then derive the
particular cases.
In the authors’ experience it has seemed pedigogically
more effective in presenting this subject to learn the simple particular cases
first, proceeding to the general in easy steps, and then showing how particular
applications may be obtained from the more general equations.
The terms centrifugal force, inertia force, and reversed effective force have
been used in the traditional
way.
However,
we realize that beginning
students are sometimes confused in making free bodies because of the use of
the word force in these names, and we toyed with the notion of using the
name dynamic reaction: it was used occasionally and for this reason the
authors would be glad to learn if the name dynamic reaction helped students
gain the correct conceptions.
The only new mechanics topics, not in the previous edition, are those in
the appendix and the chapter on vibrations.
Thus, as before, this text
should still serve equally well for either one-semester or two-semester
courses.
A new feature consists of numerous historical and biographical sketches
which are included for the purpose of interesting students.
Most of these
sketches came from notes made several years ago by one of the authors.
Much of the original source material is not readily available, but it is believed
that the sketches are substantially correct—except where brevity may give
an incomplete picture.
We wish to acknowledge with gratitude the care with which drawings were
made by Mr. 8. M. Cleland of the A & M College of Texas; the valuable
assistance in checking answers to problems given by J. C. Wilhoit, Jr., J. R.
Ballentine, A. M. Gaddis, L. G. Berryman, P. W. McDaniel, and E. Arnold.
v
vl
PREFACE
TO THIRD
EDITION
The large contribution in revising the problems and in writing many new ones
by Professor J. Harvey Caddess is acknowledged here and on the title page.
We are indebted to several users of the previous edition, some of whom
were not made known to us by the publishers, for numerous valuable suggestions. We should like to thank particularly Professor Eastman Smith,
Dean L. A. Deesz, and Dr. John A. Sauer.
Although we are sorry that all of the suggestions could not be used, we
shall nevertheless look forward to hearing from you who use the book anytime you find an error or whenever you may wish to comment on any part of
its contents.
V. M.‘F:
S2DeG2
PREFACE
TO
REVISED
EDITION
This book follows in a general way the basic plan of organization of
Chambers’ Mechanics of Engineering, but it is a completely rewritten edition.
Because a greater number of problems have been included and because the
explanations and discussions have been made more complete, this book is
longer than its predecessor, although few new topics have been added. An
earnest attempt has been made to lead the student gradually from the elementary conceptions to the more difficult ones, to avoid the discussion of too
many topics in one chapter, to minimize the backtracking that occurs when a
topic mentioned briefly in a previous chapter is to be elaborated, and to make
the explanations complete enough and sufficiently specific, yet not too
rigorous, that the student will have little difficulty in learning by reading.
While no effort has been made to avoid the use of the calculus, the mathematics encountered is usually simple.
Since the calculus so often provides
the simplest approach to a problem, its application should be encouraged in
order to impress the student with its usefulness.
In the dimensions on the illustrations and in many equations, the diameter
of circular members, such as wheels, has been used instead of the radius,
because it is the usual practice in engineering to express the sizes of flywheels,
pulleys, gears, etc., by their diameters.
The student may as well become
accustomed to this practice in the beginning.
For the most part, the answers to the problems are slide-rule answers and
are therefore subject to the inherent errors of the slide rule.
We wish to express our sincere appreciation to Professor Lee P. Thompson
of the A. and M. College of Texas, who shouldered the major responsibility
for the answers to the problems, and to Professors Lloyd G. Berryman of the
A. and M. College of Texas and W. B. Sanders of Purdue University for their
careful work in checking answers.
Also our thanks to Mr. M. B. Gallaway
and Mr. 8S. M. Cleland, of the A. and M. College of Texas, for their care in
drawing the illustrations.
pal GC.
WE a 8
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SYMBOLS
linear acceleration, a dimension.
linear acceleration of the center of gravity.
normal acceleration, tangential acceleration.
Any At
area, a constant.
breadth, a dimension.
a constant.
= iS -)
bet
anos
by©
SS
hay
g =
a
aS
ed
Ct:
Sea
ol
Ny
Dy
FS
M, M,, ete.
M
Mn
MeN
N
Ws
constants, coefficient of damping.
differential of, a dimension.
diameter.
efficiency, coefficient of restitution.
modulus of elasticity, modulus of elasticity in shear.
coefficient of friction, of static friction, of kinetic friction.
forces.
x and y components of a force, ete.
acceleration of gravity (use 32.2 fps?, ordinarily).
standard acceleration of gravity, 32.17... fps?.
height, a dimension.
horsepower.
rectangular moment of inertia, with respect to the x axis, etc.
rectangular moment of inertia with respect to a centroidal axis.
linear impulse.
angular impulse.
polar moment of inertia, with respect to a centroidal axis.
radius of gyration, with respect to a centroidal axis.
seale of a spring, the spring constant.
length, a dimension.
mass, W/g.
moment of a force, about the axis O, etc.
linear momentum.
angular momentum.
angular velocity, revolutions per minute (rpm), revolutions
per second (rps).
a force, a normal force, number of turns.
1x
SYMBOLS
designates an axis, origin, center of rotation.
pressure [lb. per sq. in. (psi) or lb. per sq. ft. (psf)], designates
a point.
:
product of inertia, pitch of threads, sometimes a force, desig-
i)
nates a point.
product of inertia with respect to a centroidal axis.
potential energy.
a force, designates a point.
radius, radius to the center of gravity.
resultant force, reaction force.
components of R, etc.
displacement, distance a body moves.
static deformation caused by a mass on a spring.
thickness, time.
period of vibration, sometimes a tensile force.
work.
velocity or speed, occasionally a dimension (a variable radius).
velocity or speed in ft. per sec. (fps), in ft. per min. (fpm).
velocity or speed of the center of gravity.
volume.
specific weight, lb. per cu. ft. or per cu. in.; density in the
pound unit.
total load, force of gravity.
coordinates of a point.
SR NsWI
aT Re:
a (alpha)
B (beta)
©
S|
>,
y (gamma)
6
A
6
»
(delta)
(delta)
(theta)
(lambda)
coordinates of a centroid or a center of gravity.
dx/dt, dy/dt, dz/dt, ete.
d*x/dt?, d’y/dt?, d®z/dt?, etc.
an angle, angular acceleration.
an angle.
an angle.
an angle.
indicating a difference, a change of, a deflection.
an angle, angular displacement.
lead angle on screw threads.
m (pi)
p (rho)
3.1416...
> (sigma)
summation sign.
an angle, the limiting angle of friction, frequency of a vibration (= 1/T).
¢ (phi)
w (omega)
density in slug units (w/g), variable radius.
angular velocity (usually in radian units), natural circular
frequency of a vibration (also w,).
SYMBOLS
AND
wy
Q (omega)
ABBREVIATIONS
frequency of a forced vibration.
angular velocity of a precession.
ABBREVIATIONS
Oy (le
cf
cfm
cg
cpm
cps
fpm
fps
gpm
hp
kw
mph
psf
psi
rpm
rps
ID
OD
REF
clockwise, counterclockwise.
cubic feet.
cubic feet per minute.
center of gravity.
cycles per minute.
cycles per second.
feet per minute.
feet per second.
gallons per minute.
horsepower.
kilowatt.
miles per hour.
pounds per square foot.
pounds per square inch.
revolutions per minute.
revolutions per second.
inside diameter.
outside diameter.
reversed effective force.
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CONTENTS
PREFACE
TO
THIRD
PREFACE
TO
REVISED
EDITION
EDITION
SYMBOLS
Chapter
ile RESULTANTS AND COMPONENTS
1. Introduction, 2. Force, 3. Vector Addition, 4. Transmissibility of a Force, 5. Subtracting Vectors, 6. Coplanar Components
of a Force, 7. Rectangular Components, 8. Example,
9. Example, 10. Example, 11. Defining a Force, 12. Classification of Force Systems, 13. Closure.
PROBLEMS
(gt, COPLANAR CONCURRENT FORCES
14. Introduction, 15.. Resultant of Coplanar Concurrent
Forces, 16. Example, 17. Equilibrium, 18. Free Body, 19. Equilibrium of Three Forces, 20. Example, 21. Trusses—Joint-toJoint Method, 22. Example, 23. Example, 24. Closure.
PROBLEMS
12
16
33
IIf. MOMENTS AND PARALLEL COPLANAR FORCES
25. Moment of a Force, 26. Varignon’s Theorem, 27. Sign
of a Moment, 28. Example, 29. Resultant of Parallel Coplanar
Forces, 30. Example, 31. Example, 32. Couples, 33. Characteristics of Couples, 34. Example, 35. Resultant of a Coplanar
Parallel Force System in Which 2/'=0, 36. Equilibrium of
Coplanar Parallel Forces, 37. Example, 38. Moment and Shear
at a Section, 39. Example, 40. Torque, 41. Closure.
PROBLEMS
IV. NON-CONCURRENT, NON-PARALLEL, COPLANAR FORCES
57
66
42. Introduction, 43. Resultant of a General Coplanar Force
System, 44. Examples, 45. Equilibrium of a General Coplanar
Force System, 46. Example, 47. Example, 48. Example, 49.
Example, 50. Trusses—Method of Sections, 51. Example,
52. Closure.
PROBLEMS
79
Xill
CONTENTS
XIV
Page
Chapter
92
FLEXIBLE
CORDS
Ve
56.
Example,
55.
Cord,
Parabolic
53. Introduction, 54.
Length of a Parabolic Curve, 57. The Catenary, 58. Example,
59. Example.
99
PROBLEMS
VI. FRICTION
60. Introduction, 61. The Frictional Force, 62. Example,
63. Limiting Frictional Force, 64. Coefficient of Kinetic Friction, 65. Examples, 66. Example, 67. Laws of Friction, 68.
Angle of Friction, 69. Example—Wedge, 70. Example—Axle
Friction, 71. Belt Friction, 72. Example, 73. Pivot Friction,
74. Frictional Torque for Screw with Square Thread, 75. Examples, 76. Friction Wheels, 77. Rolling Resistance, 78. Closure.
PROBLEMS
VII. GRAPHICAL METHODS
79. Introduction, 80. Bow’s Notation, 81. Resultant of Parallel Forces, 82. Resultant of Non-Parallel Forces, 83. Coplanar
Parallel Force System in Equilibrium, 84. Coplanar NonParallel Force System in Equilibrium, 85. Internal Forces
for a Truss.
PROBLEMS
VITT MAXIMUM AND MINIMUM FORCES
102
124
136
145
150
86. Introduction, 87. Example—Body on Inclined Plane,
88. Example—Cantilever Frame, 89. Example—A-Frame, 90.
Example—Compression in Diagonal Brace.
PROBLEMS
IX. NON-COPLANAR FORCES
.
154
156
91. Introduction, 92. Three Rectangular Components of a
Force, 93. Concurrent Forces in Space, 94. Equilibrium of
Concurrent Forces in Space, 95. Example, 96. Example, 97.
Example, 98. Non-Coplanar Parallel Forces, 99. Example, 100.
Equilibrium of Parallel Forces in Space, 101. Example, 102.
Couples in Space, 103. Vectorial Addition of Couples, 104.
Non-Concurrent, Non-Parallel Forces in Space, 105. Equilibrium of General Force System in Space, 106. Example.
PROBLEMS
170
CENTROIDS
178
107. Introduction, 108. Definitions, 109. Center of Gravity,
110. Mass Center and Centroid, 111. Centroids of Areas, 112.
CONTENTS
XV
Chapter
Page
Centroids of Lines, 113. Principle of Symmetry, 114. Estimating Location of Centroid, 115. Integrating for Centroids, 116.
Example—Are of a Circle, 117. Example—Plane Triangle,
118. Example—Sector of a Circle, 119. Example—Area without an Axis of Symmetry, 120. Example—Right Circular Cone,
121. Composite Figures, 122. Example, 123. Example, 124. Example, 125. Example, 126. Theorems of Pappus and Guldinus,
127. Examples, 128. Center of Pressure, 129. Example.
PROBLEMS
XI. MOMENTS OF INERTIA OF AREAS
130. Introduction, 131. Moment of Inertia, 132. Rectangular
Moment of Inertia, 133. Polar Moment of Inertia, 134. Radius
of Gyration, 135. Example—Rectangle, 136. Example—Triangle, 137. Transfer
Formula—Parallel Axes, 138. Example—
Circle, 139. Example—Triangle, 140. Choice of Differential
Element, 141. Example—Circle, 142. Composite Areas, 143.
Example, 144. Example, 145. Graphical Determination of the
Moment of Inertia, 146. Product of Inertia, 147. Transfer
Formula for Products of Inertia—Parallel Axes, 148. Example,
149. Moment of Inertia about Inclined Axes, 150. Maximum
and Minimum Moments of Inertia, 151. Example, 152. Closure.
PROBLEMS
XII. MOMENTS OF INERTIA OF MASSES
196
206
220
227
153. Introduction, 154. Definition, 155. Moments of Inertia
about Planes and Axes, 156. Units, 157. Radius of Gyration,
158. Transfer Formula—Parallel Axes, 159. Integrating for
Moments of Inertia of Masses, 160. Example—Cylinder, 161.
Example—Sphere, 162. Example—Thin Disk, 163. Example—
Cone, 164. Example—Slender Rod, 165. Composite Bodies,
166. Example—Flywheel, 167. Closure.
PROBLEMS
XUWI. PLANE MOTION
168. Introduction, 169. Displacement, 170. Speed and Velocity, 171. Examples, 172. Acceleration, 173. Example, 174.
Example, 175. Example, 176. Constant Acceleration—Rectilinear Motion, 177. Uniform Motion, 178. Example, 179.
Example,
180. Variable Acceleration,
181. Example, 182.
Graphical Representations, 183. Angular Velocity, 184. Relations between Angular and Linear Speeds, 185. Angular Accel-
236
241
CONTENTS
xvi
Page
Chapter
eration, 186. Examples, 187. Constant Angular Acceleration,
188. Example, 189. Curvilinear Motion, 190. Tangential and
Normal Accelerations, 191. Example, 192. Component Motion,
194. Example—Trajectory, 195. Example,
193. Example,
196. Simple Harmonic Motion, 197. Example, 198. Instantaneous Center or Centro, 199. Velocity Ratio, 200. Closure.
PROBLEMS
XIV. RELATIVE MOTION
201. Introduction, 202. Relative Displacement, 203. Example, 204. Relative Velocity, 205. Example, 206. Example, 207.
Example, 208. Relative Motion of Points in a Rigid Body,
209. Example, 210. Example, 211. Relative Acceleration, 212.
Example, 213. Example, 214. Coriolis’ Law, 215. Example,
216. Angular Motion about a Revolving Axis, 217. Closure.
PROBLEMS
X VE FORCE SYSTEMS THAT PRODUCE RECTILINEAR MOTION
303
310
218. Introduction, 219. Sir Isaac Newton, 220. Newton’s
Laws of Motion, 221. Units, 222. Component Forces and
Accelerations, 223. Example, 224. Example, 225. Example,
226. Motion of the Center of Gravity of a Rigid Body, 227.
Location of the Resultant—-Body in Rectilinear Translation,
228. Inertia Force, 229. Methods of Solving Problems, 230.
Example, 231. Variable Forces, 232. Example, 233. Closure.
PROBLEMS
XVI. CURVILINEAR MOTION
234. Introduction, 235. The Simple Pendulum, 236. Conical
Pendulum, 237. Example, 238. Example—Side Rod on Locomotive, 239. Superelevation of Curves, 240. Example, 241.
Example, 242. Closure.
PROBLEMS
XVII.
ROTATION AND PLANE MOTION OF RIGID BODIES
243. Introduction, 244. Rotation of a Rigid Body, 245.
Rotation about an Axis through the Center of Gravity, 246.
Example—Rotation about the Center of Gravity, 247. Example—Rotation about the Center of Gravity, 248. Example
—Rotation about the Center of Gravity, 249. Example—Rotation Not about the Center of Gravity, 250. Line of Action of
the Reversed Effective Force, mfa, 251. Example, 252. Compound Pendulum, 253. Center of Percussion, 254. Example—
343
349
CONTENTS
XVii
Chapter
Page
Bearing Reactions from Unbalanced Inertia Forces, 255. Example—Rod Inclined to Axis of Rotation, 256. Example—
Variable Acceleration, 257. Plane Motion, 258. Reversed Effective Forces and Couples, 259. Example, 260. Example,
261. Example, 262. Combined Sliding and Rolling, 263. Example, 264. Closure.
PROBLEMS
XVIII.
WORK, KINETIC ENERGY, POWER
265. Introduction, 266. Work, 267. Positive and Negative
Work, 268. Work of a System of Forces Acting on a Rigid
Body, 269. Principle of Work and Kinetic Energy, 270. Kinetic
Energy of a Rigid Body in Translation, 271. Units of Work,
272. Example, 273. Potential Energy, 274. Example, 275. Example, 276. Work of a Couple, 277. Kinetic Energy of a Rotating Body, 278. Example, 279. Bodies in Plane Motion, 280.
Frictional Force in Plane Rolling, 281. Example, 282. Example,
283. Example, 284. Variable Forces, 285. Example, 286. Example, 287. Graphical Representation of Work, 288. Example,
289. Power, 290. Conversion Factors, 291. Equations for Horsepower, 292. Efficiency, 293. Example, 294. Example, 295. Example, 296. Closure.
PROBLEMS
XIX. IMPULSE AND MOMENTUM
XX:
377
388
416
430
297. Introduction, 298. Impulse and Momentum, 299. Impulse, 300. Example, 301. Principle of Impulse and Momentum,
302. Example, 303. Example, 304. Example, 305. Angular
Impulse and Angular Momentum, 306. Angular Momentum,
307. Example, 308. Example, 309. Conservation of Linear
Momentum,
310. Example, 311. Conservation of Angular
Momentum, 312. Direct Central Impact, 313. Example, 314.
Oblique Impact, 315. Example, 316. Example, 317. Closure.
PROBLEMS
453
MECHANICAL VIBRATIONS
467
318. Introduction, 319. Harmonic Motion, 320. Kinds of
Vibrations, 321. Displacement, Velocity, and Acceleration in
Harmonic Motion by Rotating Vectors, 322. Example, 323.
Torsion Pendulum, 324. Example, 325. Nodal Point for Two
Bodies on One Shaft, 326. Vibration of a Fluid in a U-Tube,
327. Vibrating Beams, 328. Resultant of Two Harmonic Mo-
CONTENTS
XViii
Page
Chapter
tions of Same Frequency Is a Harmonic Motion, 329. General
Solution of the Differential Equation of Harmonic Motion,
330. Weight Not Attached Directly to Spring, 331. Weights
Attached to More Than One Spring, 332. Damped Free Vibrations, 333. Forced Vibrations, 334. Forced Damped Vibrations, 335. Closure.
O08
APPENDIX
PROBLEMS
497
BALANCING ROTATING ELEMENTS
336. Introduction, 337. Balancing a Single Mass, 338.
Balancing Several Masses in One Plane of Rotation, 339. Example, 340. Masses in the Same Longitudinal Plane but in
Different Planes of Rotation, 341. Masses in Different Longitudinal Planes and Different Planes of Rotation, 342. Example.
PROBLEMS
509
A. GYROSCOPIC
512
ACTION
343. Introduction, 344. Axes for Gyroscopie
Value of the Gyroscopie Couple, 346. Example.
PROBLEMS
APPENDIX
B. UNITS
AND
DIMENSIONAL
Action,
345.
517
ANALYSIS
347. Systems of Units, 348. Density, 349. Mass Moment of
Inertia, 350. Viscosity, 351. Angular Measure, 352. Dimensional Analysis, 353. Conversion Constants, 354. Example,
355. Example, 356. Example.
PROBLEMS
APPENDIX
C. SOME
USEFUL
MATHEMATICAL
APPENDIX
D. SOME
USEFUL
EQUATIONS
INDEX
RELATIONS
OF ANALYTIC
515
MECHANICS
026
527
529
531
‘
Chapter I
RESULTANTS
AND
COMPONENTS
1. Introduction.
Mechanics is that branch of physical science which
treats of the effect of forces upon material bodies.
The general field of
mechanics is further subdivided to include: (1) mechanics of fluids, a phase
of which is called hydraulics; (2) mechanics of materials, more often called
strength of materials, a subject which deals with the internal forces or stresses
in bodies; and (3) analytic mechanics or mechanics of engineering, usually
simply called mechanics, a study of the external forces on bodies, ordinarily
rigid bodies or bodies considered to be rigid, and of the effects of these
forces on the motions of bodies.
This work treats only those phases of the
third subdivision which are of most concern to engineers. Since mechanics
is a foundation subject of use in varying degree to nearly all engineers,
students should realize at the outset that they are now studying
a subject of major importance, no matter what branch of engineering they
may be pursuing.
Analytic mechanics includes a study of (1) statics, which
deals with the forces acting on bodies or structures which are at rest relative
to the earth or which are moving with a constant velocity; (2) kinetics, which
deals with the motion of bodies as it is affected by external forces; (8) some
kinematics, which deals with motion alone without regard to the forces involved; and (4), somewhat by virtue of a traditional arrangement, centroids,
centers of gravity, and moments of inertia. Dynamics is a term usually
understood to include kinematics and kinetics.
The science of analytic mechanics is based on Newton’s laws of motion,
with which you became acquainted in your physics. These laws may be
stated as follows:
I. Every particle remains in a state of rest or moves with a constant
velocity in a straight line unless an unbalanced force acts on it.
II. The acceleration of a particle is directly proportional to the resultant
force acting on it and inversely proportional to its mass, and the sense
of the acceleration is the same as that of the resultant force.
III. To every action, there is an equal and opposite reaction.
1
RESULTANTS AND components
2
[Ch. I
Law I defines the condition of equilibrium and from it we develop the
Law III applies to both statics and kinetics.
first part of this work—statics.
The study of kinetics is developed from Law II.
In mechanics, a force arises out of the interaction of two
2. Force.
If a body (yourself,
or tends to cause motion of the bodies.
causes
and
bodies
for example) is pushed by another body, motion may or may not occur, but
it is important to recognize that the body being pushed also pushes back;
that is, if you are pushed, you will offer resistance to motion, whether you
exert yourself or not, because of the property of inertia which all material
Thus, we observe that forces appear invariably in pairs—a
bodies possess.
force exerted on the body and an equal and opposite reaction (a force exerted
by the body), in accordance with Newton’s Law III. A body which is at
rest or is moving with a constant velocity is said to be in equilibrium (Newton’s Law I). A complete algebraic definition of equilibrium is given in
§105. In a force analysis of a body in equilibrium, we concern ourselves with
the forces acting on the body, but not with the reactions, the forces exerted
by the body.
Force is a vector quantity.
The characteristics of a force vector are that
it has (1) magnitude, (2) sense or direction, and (3) location or line of action.
The magnitude of a force vector is usually expressed in pounds (lb.) in
English-speaking countries.
Large forces may be given in tons (a short ton
of 2000 lb. in the United States or a long ton of 2240 lb. in the British Commonwealth) or in kips.
Iekip = 1000)1b:
The length of a force vector represents its magnitude to some scale, called the
force scale. The force scale is usually specified in pounds per inch, but
it may be in any combination of force and linear units, for example, kips
per inch. An engineers’ scale, which will be very useful in this study, is
conveniently graduated for laying out vectors.
For example, suppose a force has a magnitude of 240 lb.
The length of the vector
which represents this force depends on the scale—as illustrated in the following
tabulation:
Scale of 240-lb. force
Length of vector
30 Ib. per in.
40 Ib. per in.
50 Ib. per in.
60 Ib. per in.
100 Ib. per in,
200 lb. per in.
8 in.
6 in.
4.8 in.
4 in.
2.4 in.
1.2 in.
Examine your engineers’ scale with reference to the foregoing scales.
Observe that the length of the vector is the magnitude of the force divided
by the scale.
Conversely, if the length of a vector and the force scale are
§ 3] VECTOR ADDITION
8
known, the magnitude of the force is determined by multiplying the length
by the scale.
In either case, the linear unit of the scale must be the same
as the linear unit used to measure the vector—inches, in the foregoing example.
Thus, if a vector is 2 in. long and the force scale is 4 kips per inch,
the magnitude of the force is (2)(4) = 8 kips, or 8000 lb.
When the line represents a vector, an arrowhead on one end of the line
defines the sense of the vector.
For example AB in Fig. 1(a) represents a
force F;. The arrowhead shows that the sense of the force is upward and
toward the right. The angle a = 60° and point A locates the vector.
The line of action of a force may be considered a line of indefinite length.
The line representing the force coincides with the line of action, as indicated
in Fig. l(a). As suggested in Fig. 1(a), the beginning point of the vector is
called the tail, and the end with the arrowhead is called the point or head.
F,= 300 lb.
(a)
Line of Action of F,
(b)
(c)
Fig. 1. Adding Vectors.
Observe in (a) that the resultant R of two coplanar forces
passes through the intersection of the lines of action of the two forces F; and F,. This
observation will be useful in problem work.
3. Vector Addition.
Vector quantities, such as force, acceleration, velocity,
and momentum, cannot be added or subtracted as are scalar quantities,
which possess magnitude only. Hight gallons plus two gallons of gasoline
(scalar quantities) are ten gallons of gasoline.
But an 8-lb. force plus a
2-lb. force may or may not be a 10-lb. force, since the magnitude of the sum
depends on the directions of the forces that are to be added.
Suppose we
wish to find the sum of the force /’; = 400 lb. and the force F2 = 300 lb.,
Fig. 1(a), where for convenience the force vectors are in the plane of the paper.
(When all the vectors of a force system lie in the same plane, the forces
See § 12.) The method of adding vectors is shown
are said to be coplanar.
in Fig. 1(b). From any convenient origin O, lay out 1 to some scale, say,
1 unit equal to 100 lb. From the point of 7; at B, lay out the vector /2 in
Observe that the tail of 7 is joined to the
proper magnitude and sense.
point of F;. Now the line AD drawn from the origin to the point of vector F»2
represents the sum of the vectors 7; and /; in magnitude and sense, but not
RESULTANTS AND COMPONENTS
4
[Ch. I
in the location of the line of action, because we have moved the vectors 1
When a vector is placed in a position
and F» for the purpose of this addition.
it is called a free vector.
action,
of
line
its
to,
outside of, but parallel
the addition, Fig. 1(c), laying
making
We may, if we wish, start with #2 in
out AE representing /’, to scale and then adding F by placing the tail of Fy
at the point of F2. This gives AH and HD, the vector sum of which is AD,
the same in magnitude and sense as before. A vector addition, a process
called composition of vectors, is indicated by the symbol +); thus, in Fig. 1,
(Pi
hy
=
TR.
The magnitude of the sum of two or more vectors is equal to the magnitude
of the resultant.
See the next article.
F,=800 lb.
Line of Action of F,
(a)
(b)
Fig. 1.
(c)
Repeated.
4. Transmissibility of a Force.
Experience and experiment show that
the external effect of a force on a rigid body is the same for all points of application that lie on the line of action, the principle of the transmissibility of a
force. Whether the force F; in Fig. 1(a) pushes at point B, pulls at point A,
or acts at C, or any other point along the line AC, the motion imparted to
the body M by F; is always the same.
It should be noted that the internal
effects, that is the stress and strain induced in the body, are affected by
changing the point of application of a force.
To satisfy yourself as to the validity of this law, imagine a heavy block on
a floor with a rope attached to the block.
It makes no difference in the
motion, or lack of motion, of this block if you pull on the rope with a particular
force close to the block or far from the block, provided that the rope makes
the same angle with the horizontal in each instance.
In this illustration,
you are shifting the point of application of the load but maintaining the
same line of action.
Now if we consider that both F, and F, in Fig. 1(a) are applied at the
point C, which is on the lines of actions of both forces, it is evident that
the line of action of the sum R must also pass through C. Thus, using
§ 6 |COPLANAR
COMPONENTS
OF A FORCE
5
the angle 6 found in Fig. 1(b) or (c), we may draw R in Fig. 1(a). When the
vector F is specified in magnitude, sense, and line of action, it is called the
resultant.
Consequently, when a problem asks for the resultant of a force
system, these three characteristics must be specified.
Whenever a body is undergoing an acceleration, there is a resultant force,
in accordance with Newton’s second law, § 1. During the first part of our
study, the finding of the resultant may sometimes appear to be just an exercise. However, in our study of kinetics, the determination of the resultant
is an important part, but only a part, of the solution.
Moreover, it is often
necessary to get the resultant force on a member in order to design it.
In general, it may be stated that if any two or more forces in a plane
intersect at a common point, the resultant of these forces also passes through
the common point. The location of the resultant of other kinds of force
systems will be discussed in subsequent chapters.
5. Subtracting Vectors.
To subtract one vector from another, place the
tail of one vector joining the tail of the other vector, or place the point of one
vector joining the point of the other vector, each vector being laid out to
scale in magnitude and sense.
The line that joins the other ends of the
vectors is the vector difference.
Thus, in Fig. 1(b), #1 is the vector difference
between R and F2; and F» is the vector difference between A and F;. A
vector difference is indicated by the symbol —.
Thus,
R-F,=
Fy,
and
R-F,
=
F..
The algebraic expression for the vector difference, when the vectors are at
right angles, is for example
Py = (BR? — F212,
If the angle between the vectors is other than 90°, an algebraic expression
may be found by the law of cosines (see the next article).
6. Coplanar Components of a Force. The forces F; and F, which added
to give R in Fig. 1 are said to be components of R. However, although
there is only one resultant force for an intersecting force system, there is an
infinite number of pairs of components for any one force. In Fig. 2(a),
AC = F, and AD = F, add vectorially to give R and are therefore components of R. Similarly, in Fig. 2(b), AG = F; and AE = F, are other
components of R. In each case, with these coplanar components as two of
the sides, a parallelogram may be constructed in which & is a diagonal.
This parallelogram, which suggests a method of adding two vectors graphically, is a graphic expression of the parallelogram law. The algebraic relation
between Fj, F'2, and R can be obtained from the trigonometry of Fig. 2(a)
and (b). If the angles between F', and /’2 are designated in general by a,
RESULTANTS AND components
6
[Ch. I
then evidently the angles ACB and AGB would be designated by 180it—"a.
Now in the triangles ACB and AGB, apply the law of cosines; thus
1i92 =
(a)
Be
-|- F,?
QF LF, cos(180
=
—
a).
Since the cos(180 — a) = — cos a, we get from (a)
(1)
R? = Fi + F,? + 2F1F2 cos a,
where a is the angle between the vectors 7; and /2.
we see that
(b)
tan 6
b
Also from Fig. 2(a),
F, sin (e4
—4
Fe
Paice
—_—___-—"o
If a is between 90° and 270°, the cos a@ is a negative number.
Fig. 2. Components of Forces.
Note particularly that the components of a force must
add vectorially to equal the force.
7. Rectangular Components.
For a = 90°, we get the special case of
components which are perpendicular to each other.
Since cos 90° = 0, we
have, from equation (1) and also from the right triangle AKB of Fig. 2(c),
(2)
Feet
Pot
or
R=
(FY? + F.?),
Components of a resultant that are at right angles to each other are called
rectangular components; when they are in the direction of reference axes x
and y, they are usually designated by Ff, and F',. These rectangular components are the most important ones because we find repeated use for them
throughout the study of forces. Observe carefully from Fig. 2(c) that
(3)
F, = F cos 6
and
where @ is the angle between / and F,.
F, = F sin 6,
Using the system of notation Fe oditss
and I’, Fig..2(¢), we call ’, the x component and F’, the y component of :F.
Moreover, when the x and y axes are respectively horizontal and vertical, we
also speak of these components as the horizontal and vertical components. ‘Tt
is seen from Fig. 2(c) that the direction of the resultant of two rectangular
components is defined by
(c)
J
i's
tan 9 = F.
The process of finding components of a force is called resolution.
§ 9] EXAMPLE
”
8. Example.
Find the resultant of a horizontal force of F, = 400 lb., acting
toward the right, and a vertical force of F, = —300 lb., the negative sign indicating
that the force acts in the negative direction, downward.
Sotution.
In accordance with equation (2), we find
R = (F2+ F,?)2 = (4002 + 3002)? = 500 lb.
The slope of the vector & with respect to the horizontal is obtained from (c);
tan’¢7—
ge 400
Emo
F,
d
from which @ = 36.9°. Thus, the resultant passes through the intersection of the
lines of action of F, and F,, pointing downward to the right at. 36.9° below the horizontal.
The student should sketch the relationship of these vectors after the fashion
of Fig. 2(c).
It is frequently convenient to measure the slope of a vector by the smallest
angle it makes with the horizontal.
In this example, the slope may be
stated as above; or we may say that 6 = —36.9°, the minus sign indicating
that the angle is measured from the horizontal in the negative (clockwise)
direction; or we may say that 6 = 323.1° (= 360 — 36.9), where @ is measured
in a positive direction from the x axis, as explained in § 11. In any event,
use at least a free hand sketch (or your imagination after you become expert)
to decide upon the general sense of a vector.
When you give an angle in
the solution of a problem, there must be no doubt
ee
as to what angle you have in mind.
WW
9. Example.
A block with a 1 X 4-+ft. rectangular
section is 20 ft. long and weighs W = 9000 lb. If a
2000-lb. force F acts 8 ft. from the base, as shown in Fig.
3, what is the magnitude of the resultant & of these two
forces and where does the line of action of FR intersect
the base of the block?
GRAPHICAL SotuTion.
First draw the block to scale,
using, say, a space scale of 1/8 in. = 1 ft. (Fig. 3).
Locate accurately the line of action of F. The line of
action of the vector which represents the weight passes
through the center of gravity of the block (the central
point in a homogeneous symmetric body). Remembering
that the resultant of B and W must pass through the
intersection of their lines of action, lay off
W = 9000
lb. to some convenient scale, say, 1 in. = 4000 lb., and
obtain the vector OH.
Now, recalling that the resultant is a vector sum, lay off HG to represent / = 2000
lb. The line OG represents R in magnitude, direction,
ayesae
F = 2000 lb.
Space
Scale
na
acrenaie
1”=4000 Ib.
Ib. %
9000
W
E
Fig. 3.
and line of action, and scales about 9200 lb.; also @ =
1214°. The point of intersection of R and the base is J, such that
about 1 ft. 9.5 in.
ALGEBRAIC SoLuTION.
DJ scales
Since W and F are at right angles to each other,
R = (F? + W?)t? = (81 X 10° + 4 X 10)? = 9200 Ib,
RESULTANTS AND components
8
[Ch. I
From
tan 9 = zie= LO
W
we get 9 = 1214°.
Z
9000
9
Then from similar triangles OFG and ODJ, we have
he
DJ
tte
EE
OD
Of
Es
EG
=
9000
2
2000
~
Disa
aia te
The location of the point J is important in retaining walls and dams, because if J
should fall to the right of C, the wall would topple over unless some other force not
considered here prevented it. Preferably, the resultant should pass through the
middle third of the base of a dam.
10. Example. A force of 5000 lb. acts upward toward the right at an angle of
9 = 30° with the horizontal. What are its horizontal and vertical components?
Sotution.
The horizontal component is
F,, = F cos 6 = (5000)(cos 30) = 4330 lb.
The vertical component is
F, = F sin 6 = (5000)(sin 30) = 2500 lb.
11. Defining a Force. In the absence of a figure which defines a force
vector, we shall specify from time to time the angle which the vector makes
with the horizontal.
In each instance, the angle measured
counterclockwise
from the positive side of the x axis is intended unless otherwise specified, and
the arrow should point away from the
origin. This manner of stating angles
is conventional and often the most convenient.
Thus, in Fig. 4(a), 62 is an
angle greater than 90° and less than
180°, and locates a vector in the second
quadrant. Similarly, 03 locates a vector
Fig. 4.
in the third quadrant, and 64 locates
one in the fourth quadrant.
Unless two vectors intersect at right angles, there are two different angles
between their lines of action, a and 8, Fig. 4(b). Angle 6 is the angle between
a point end of one vector and a tail end of the other.
However, when the
angle between two vectors is stated, the angle a, which is measured between
the point ends of the vectors, is intended.
12. Classification of Force
into two principal groups:
Systems.
Force
systems
may
be classified
(1) Coplanar force systems, in which the force vectors are all in the same
plane, and
(2) Non-coplanar force systems, in which the vectors are not all in the
same
plane. Force systems in this category are often called forces in
space
See Chapter IX.
§ 13] CLOSURE
9
Force systems may also be classified as follows:
(1) Collinear force systems, in which all the forces act along the same line
of action. A collinear system is necessarily coplanar.
(2) Concurrent force systems, in which all lines of action intersect at one
point. A concurrent force system may be either coplanar or non-coplanar,
provided that there are more than two forces. Since a plane can always
be passed through two intersecting lines, a concurrent force system of
two forces is coplanar.
(3) Non-concurrent force systems, in which the lines of action of the force
vectors do not intersect at a point. A non-concurrent system may be
either coplanar or non-coplanar.
(4) Parallel force systems, in which the lines of action of all force vectors are
parallel.
A parallel force system may be either coplanar or non-coplanar.
13. Closure.
The composition of forces as presented in this chapter may
be carried out either by use of the parallelogram law or the triangle law.
PARALLELOGRAM Law.
If two coplanar force vectors are laid out to
scale from their point of intersection, both pointing away from the point of
intersection, and if a parallelogram is completed with these force vectors as
two sides, then the diagonal of the parallelogram that passes through the
point of intersection represents the resultant in magnitude and direction
[Figs. 1(c) and 2].
TRIANGLE Law.
If two coplanar force vectors are laid out to scale with
the tail of one at the point of the other, the third side of a triangle of which
these two vectors are two sides represents the resultant in magnitude with
a sense from the tail of the first vector to the point of the second vector
[Figs. 1(b) and 3].
HISTORICAL
NOTE
Mechanics as a science may be said to have begun with Archimedes (287212 s.c.). To be sure, other civilizations applied the principle of the lever,
but we have no record of any one before Archimedes making a statement of
the principle.
advances
Such a statement is significant when we observe that broad
in science grow
out of important
generalizations
which
become
known as laws or principles. However, in this instance, there was relatively
little progress for some 1800 years, which may be explained in part by the
attitudes of the intellectuals of those days. For instance, Plato said, speaking of Archimedes’ inventions, “... that the machines did utterly currupt
and disgrace the worthiness and excellence of geometry making it descend
from things not comprehensible and without body into things sensible and
material...”
10
RESULTANTS AND components
[ Ch. I
He was born in the
A few words about Archimedes will be of interest.
spent his life,
Greek colony of Syracuse on the island of Sicily, where he
except for some years of schooling in Alexandria, Egypt.
He was primarily
of his great
a mathematician, and a great one, but at the continual urging
mechanical
many
d
invente
he
e,
Syracus
of
and good friend Hiero, the King
ng the
defendi
for
s
machine
war
Many of his inventions were
contrivances.
y, is
geometr
love,
The extent of Archimedes’ real
city against the Romans.
indicated by Plutarch’s statement, “ Archimedes] was so ravished and drunk
with the sweet enticements of this siren [geometry], which as it were lay
continually with him, as he forgot his meat and drink, and was otherwise so
careless of himself, that oftentimes his servants got him against his will to
the baths to wash and anoint him... .”
Among his other achievements, Archimedes is believed to be the first to
understand and use the principle of the compound pulley (block and tackle),
he worked out the relation between the circumference and diameter of a
circle (pi), he invented a screw-thread type of pump (Fig. 5), and, importantly, he defined the laws of buoyancy of a liquid, now familiar to all who
have studied physics.
There is an interesting story about how he happened to discover the laws
of fluid buoyancy,
His King had commissioned a goldsmith to make a
gold crown for a favorite god who had presumably done him a favor.
It was
soon rumored that the goldsmith had cheated, although the weight of the
crown was exactly equal to the weight of the gold furnished for the purpose.
The problem was to determine if the intricate work of art was pure gold
without destroying it. Unfortunately for the goldsmith, Archimedes was at
hand to search for an answer.
For some time, no progress was made.
Then
one day, his bath tub was filled too full and it ran over when he got in.
Archimedes began to speculate on the significance of this phenomenon, and
suddenly, the laws of fluid buoyancy were clear in his mind.
He was so
elated by this revelation that he immediately raced into the street in a condition unsurpassed by Lady Godiva shouting “Eureka,” which freely translated means “T got it.” Knowing that the loss of weight of a body in water
is equal to the weight of the volume of water displaced, it was easy to weigh
the new crown in water and compare this weight with that of an equal
quantity of known pure gold. Since the crown was not pure gold, but part
silver, it displaced more water and therefore lost more weight than the equal
quantity of gold. History does not say exactly what happened to the
dishonest goldsmith.
The modern development of the science of mechanics began to evolve
in
the sixteenth century, largely through the efforts of two
contemporaries,
Galileo (about whom, more later) and Simon Stevinus (or Stevin).
Stevinus
(1548-1620),
a Dutch engineer, born at Bruges, did his work independently
§ 13] CLOSURE
seh
Courtesy the Johnson Publishing Co., Cleveland
Fig. 5.
Artist’s Conception of Archimedes’ and His Spiral Pump.
A comfortable civiliza-
tion cannot be built on man power and animal power alone.
of Galileo and made a number of important contributions to mechanics
(including mechanics of fluids) and other budding sciences.
He is credited
with the discovery of the parallelogram law, but the first clear statements of
the law were probably given independently by Varignon (1654-1722) and
te
prosLEMs [Ch. I
One of Varignon’s important deductions is discussed in § 26. Ase
Newton.
suming that perpetual motion was absurd, Stevinus showed that on inclined
planes of equal vertical heights, uniformly distributed loads would be in
balance (see Fig. 6). Stevinus, being of practical genius, first attracted
B
(b)
2
Fig.6. Inclined Planes.
Stevinus assumed that the flexible member ABCD in (a), whose
mass is uniformly distributed along its length, would not move on the inclined planes
even if there were no friction. He argued that the symmetric portion ADC in (a) could be
removed without upsetting the equilibrium of the remaining part, and that therefore the
shorter and lighter part BC balanced the longer and heavier part AB, as shown in (b).
The consequences of this ingenious process of reasoning were far reaching.
attention by constructing a sort of land boat, a contraption mounted on
wheels and propelled by a sail, which he used between towns along the shores
of Holland.
Stevinus’ work seemed somewhat dim beside the dazzling
brilliance of Galileo’s, but it was extremely important.
Leonardo da Vinci (1452-1519), one of the great geniuses of all times,
knew a good deal about the laws of motion, inertia, and forces, but his
scientific achievements were not published until the end of the eighteenth
century.
In the meantime, his knowledge of mechanics, and more too, had
been developed independently by other scientists.
Problems
1. Two horses A and B are hitched to a
load. If A pulls northward with a force of
300 lb. and B pulls N30°E with a force of
200 lb., find the resultant force on the load,
Solve graphically and algebraically.
Ans. 484 Ib., N11.9°R.
2. A 2000-lb. force acts horizontally toward the left. A 3000-lb. force makes an
angle of 50°, measured clockwise, with the
2000-Ib. force.
Find the resultant force,
solving algebraically and graphically,
3. A 300-lb. force acts horizontally to the
right. A 400-lb. force makes an angle of
140°, measured clockwise, with the 300-lb.
force. Find the resultant force graphically
and algebraically.
Ans. 258 Ib. at — 91.6°.
4, Two persons,
A and B, desire to lift
a 300-lb. object. If A lifts 160 Ib. at an angle
of 18° with the vertical, determine the force
B must exert to achieve the resultant 300-Ib.
vertical lift. Solve algebraically and graphically. Must the force that B exerts be
coplanar with A’s force and the resultant
300-lb. foree? Why?
5. Two 50-lb. forces may have a resultant
of 40 lb. acting at 90° with the plux 2 axis.
Determine the direction in which each of
the 50-lb. forces must
act.
Is there more
than one solution?
Solve graphically and
algebraically.
6. Figure 7 represents the sheave and
cable on the end of a derrick boom.
Let
= 0 and find the resultant force on the
sheave axle. Solve graphically and algebra-
13
PROBLEMS
ically. What
force on the boom is caused
15. In Fig. 9, let F = 3600 lb. and 6 =
by the cable?
Ans. 5.66 tons, 225°.
45°. Assume both pulleys to have no friction so that the tension in the cable CD is
3600 lb. (a) Solve graphically for the force
on the shaft at B. (b) Solve graphically for
7. The same as 6 except let @ = 60°.
4 tons
4 tons
Fig. 7.
Problems 6-8.
8. Figure 7 represents the sheave and
cable on the end of a derrick boom.
If the
4-ton pull in the cable causes a force of 6
5000 Ib.
tons on the sheave axle, determine the value
of 6.
Solve algebraically and graphically.
Ans. 7.2°.
Fig. 9.
Problems 15, 16.
A
3
B
C.
Cc
0
A /,
(d)
Fig. 8.
(e)
Problems 9-14.
9. Figure 8 shows several ways in which
two external forces A and B may act on the
body C. Show algebraically and graphically that each manner of application has the
same resultant external effect.
10. In Fig. 8, let 6 = 60°, A = 2 kips,
B = 5 kips, and find the resultant graphically and algebraically.
Ans. 4.37 kips at 156.5°.
11. In Fig. 8, let 6 = 150°, A = 2 kips,
B = 4 kips, and find the resultant graphically and algebraically.
12. In Fig. 8, let B = 200 lb. and the
resultant force be 85 lb. at 90°. Find A
and 6 graphically and algebraically.
Ans. 218 lb., 23°.
13. In Fig. 8, let B = 200 lb. and the
resultant force be 125 lb. at 270°. Find A
and 6 graphically and algebraically.
(a) If the
14. In Fig. 8, let 6 = 250°.
resultant is 500 lb. at 200°, find A and B.
Solve graphically and algebraically.
3
the force on the bearing at A. Check each
solution algebraically.
Ans. B = 2760 lb.
at 202.5°, A = 7970 lb. at 288.6°.
16. The same as 15 except let 6 = 15°.
17. Let the block in Fig. 3 be made of
concrete which weighs 150 lb. per cu. ft.
If F = 3000 lb. acts horizontally and in the
same plane with W at a distance of 5 ft.
above the base, determine the distance
DJ = « graphically.
Ajis.s leant te
18. The same as 17 except that DJ = x
is to be determined algebraically.
19. If the weight of the block in Fig. 3
is W = 10,000, where is the line of action
of a horizontal coplanar force F = 2000 lb.
when the resultant passes through point C?
Solve algebraically and graphically.
Ans. 10 ft. above base.
20. The same as 19 except that the
resultant is to pass through a point two-
thirds the distance from D to C.
14
PROBLEMS [ Ch. I
21. The
forces
block shown
Fi, F:, and
W
on
in Fig. ‘10 ee coplanar.
the
80°.e
The
and F, algebraically and graphically.
j
Find
the rec tangular components
total vertical vector is therefore Ff; + W.
It is desirable that the resultant FR fall
F,;
Ans. 86.8 lb., 492
lb.
F
fe
WLLL
Fig. 12.
Problems 24-26.
25. In Fig. 12, let F = 18 kips and @ =
150°. Find the rectangular components F,
and F, algebraically and graphically.
26. In Fig. 12 if the rectangular components are fF, = —50 lb. and F, = 400 lb.,
find F and 6 graphically and algebraically.
Fig. 10.
Problem 21.
within the middle third of the base dimension 6, the limiting position being at 1D),
which is one-third the distance C to E.
For this limiting position, show that
_ (Fi + W)
Paes
6nh
where h is the height of the block and n is
a decimal locating
F, as a percentage of
the height. The alee of the block is
talcens atl
Fig. 13. Problems 27, 28.
ees
Peeler”
a7. Figure 13 indicates the foree F that
a cable exerts
on a vertical pole. The force
in the stay wire should cause the resultant
force on the pole to be downward and
collinear with the pole. Let F = 5000 lb.
and @ = 15°, and find (a) the horizontal
and vertical components of the cable pull
F, (b) the horizontal
and vertical
compo-
nents of the reaction in the stay wire, and
(c) the net force on the pole
Ans. (a) 4830, —1290 lb., (b) —4830, —8360,
(ec) —9650 lb.
28. The same as 27 except that F = 7.5
tons.
Fig. 11.
Problems 22, 23.
22. A load of Q = 4000 lb. is acting at
B on the cantilever structure of wie
Te
If the load at A is F = 10,000 Ib., determ
ine
the resultant of F and Q and the distanc
e
from C to the line of action of the resulta
nt
measured
resultant.
23. The
2000 Ib.
along a line perpendicular
same
to the
Ans. 13,600 lb., 15.67 ft.
as 22 except that F =
24. In Fig. 12, let F = 500 lb.
and 8 =
()
og
Fig. 14.
,
0
Problems 29-31.
29. In Fig. 14, a weight W =
200 Ib. is
placed on the platform scales
.
Then a
PROBLEMS
15
force F = 50 lb. is applied at @ = 30°.
What should be the balanced scale reading?
Anise 225) lbs
3C. The same as 29 except that @ = 75°.
31. With the arrangement shown in Fig.
14, if W = 200 lb., F = 100 lb. and the
scales balance to read 275 lb., find @.
Ans. 48.6°.
35. The same as 34 except that a = 30°.
36. The same as 34 except that a = 45°.
37. A 200-lb. force acts at 30° with the
positive «axis. Consider all positive angles
as counterclockwise from this axis and replace this force by its components Ff; and
F, such that: (a) if F: acts at 0° and FP, at
90°, find fF; and F.; (b) if F; acts at —15°
and F» at 75°, find F; and FP; (ec) if F; acts
at 15° and F2 at 165°, find F; and F»; (d)
if F2 = 50 |b.at 20°, find F; and its direction.
AMS, (@) WS2, 100, Co) ial WiLAz (@)
Psexra, MOB, (Gl) iG! Woy, Bis BRL
38. The same as 37 except that the 200Ib. force acts at 45°.
Fig. 15.
Problems 32, 33.
32. A steam engine is diagrammatically
outlined in Fig. 15. If, when the steam is
turned
on,
the
thrust
on
the
connecting
rod is 80,000 lb. when @ = 15°, find the
total steam load F on the piston and the
vertical foree N on the crosshead.
The
engine is not moving.
Ans. 77,280 lb., 20,720 lb.
33. The same as 32 except that 6 = 10°.
34. The force in the connecting rod when
a = 60° in Fig. 16 (see also Fig. 15) is
Fig. 17.
Problems 39-41.
39. A car is stalled (Fig. 17) on a hill the
grade of which is 5% (meaning a 5 ft. rise
in 100 ft. horizontal).
Its weight is W =
4000 lb., the resistance to motion parallel to
the road is F = 20 lb., the normal plane
reaction is N = 3995 lb., and the initial
pull parallel to the road is @ = 60 Ib. (a)
What is the resultant force in magnitude
and direction?
Solve graphically and algebraically.
(b) Doesthecarmove?
Refer
to Newton’s second law.
Ans. (a) 160 Ib. down plane.
Fig. 16.
Problems 34-36.
40. The same as 39 except that Q = 220 lb.
41. The same as 39 except that Q = 350 lb.
20,000 lb., acting in the direction AB.
Replace this force by its rectangular components, one of which is acting in the direction
of tangent 7-7 and the other through the
center of the crankshaft.
Ans. 10,000 lb., 17,320 lb.
42-50. Some people prefer to see fewer
problems in a text.
However, the majority
seems to desire an ample supply.
Those
who wish may use these blank numbers for
their favorite problems.
Other blank numbers will be found at the end of each chapter.
Chapter II
COPLANAR
CONCURRENT
FORCES
14. Introduction.
Our discussions in the first chapter were of a simple
case of coplanar concurrent forces, since two components of a resultant force
necessarily intersect and since any two non-parallel forces in a plane must
necessarily intersect.
However, in many cases, the lines of action of three
or more forces may intersect at a single point, and thus may form a system
of concurrent forces. We shall now discuss concurrent forces which are
also coplanar.
RESULTANTS
15. Resultant of Coplanar Concurrent Forces.
(a) Graphically.
Figure 18(a) shows forces F'1, F'2, F'3, and Fs represented in magnitude and
direction.
As previously stated, the resultant of these forces is their vector
sum.
‘To add these vectors, choose some origin A, Fig. 18(b) and lay out
(b) Foree Polygon
Fig. 18.
Resultant of Concurrent Forces.
PROCEDURE.
Draw AB of indefinite extent, parallel to the vector
scale along AB and mark off the length representing
the ete
point B of AB = F,, drawa
i
i
ahag tity.
line parallel to the vector F, of indefinite extent.
Using the same
scale as before, mark off from B a length representing
the magnitude of F».
This locates
point C, through which a line is then drawn parallel
to vector F;. The fore oing
pr
is repeated until all the forces whose resultant is desired
have been added an thi ee
the final vector point is at E. The line AE which
closes the polygon is the ee: fae
Measure its length with the scale being used
and the angle @ with a protract
i Thi
gives you the magnitude and sense of R.
16
Its line of action passes Segre O a
7
§ 15 ]RESULTANT
OF COPLANAR
CONCURRENT
F, = AB inmagnitude and direction.
FORCES
17
From the point of F1, lay out F2 = BC
in magnitude and direction. Similarly, vectors Ff; and Fy are added
Fy + Fs, which locates point EH, Fig. 18(b).
to
The sum of any number of vectors, added as explained, is the resultant and
as represented by a line drawn from the origin to the point of the final added
vector and the sense 1s always from the origin.
Thus, the line drawn from A to E represents R in magnitude and direction.
However, the line of action must pass through O, Fig. 18(a), the point of
intersection of the original forces.
It does not matter in what order the forces are added, the same vector sum
will be obtained in any event.
Observe too, in Fig. 18(b), that R, is the
resultant of /: and Fs, and that Az is the resultant (or sum) of Ri(= Fi 4 F2)
and F;. Thus, one may think of this process as a repeated application of the
triangle law (§ 13).
Figure 18(b) shows what is called a force polygon.
In an analogous
sense, we often speak of a force triangle when only three vectors are involved.
(b) Algebraically.
In computing a resultant algebraically, we usually
find it convenient first to determine the rectangular components of the
various forces and then to find the resultant of these rectangular components.
It is evident that if the forces A, B, and C, Fig. 19, for example, give a certain
Fig. 19.
sum, this same sum will be obtained by adding the components of A, B,
and C. ‘The procedure then is to find A,, B,, and C,; which, we note, form a
collinear (in same line) group of forces; and also to find A,, B,, and C,,
which are collinear forces too. To find the resultant of a collinear system,
we need only to make an algebraic sum of the forces. By algebraic sum, we
mean that we account for the sense in which each force acts. Thus, in Fig. 19,
force A, acts toward the right, a sense which we ordinarily consider positive.
COPLANAR CONCURRENT
18
FORCES
[Ch. II
If we assume that a force acting towards the right is positive, then a force
Thus, B; and Cs are sages
which acts toward the left is negative.
The algebraic sum of components in the x direction is designated by pak
The symbol =F, is interpreted as “the sum of the forces in the x direction.
Since we usually consider the upward sense as positive, we shall take
Now observe that though we
A, and B, as positive and C, as negative.
the angle with the positive
using
may find it convenient to define a force by
side of the x axis, as for example, 6 in Fig. 19, we find the use of the small
angle between the vector and the « axis, for example, 8 in Fig. 19, more
convenient in determining the =F, and ZF,
(In Fig. 19, note that
B cos 6 = —B cos B.)
Fig. 19.
The actual application
Repeated.
of the principles which
much simpler than the description.
From
and also from equation (3), we see that
A, = A cosa,
B, = B cos B,
Therefore, the sum
immediately as
of the components
(a)
A
wherein
forces.
(b)
cos
have been
the trigonometry
and
described
is
of Fig. 19(a)
G,.= Cocos.
in the x direction may
be written
a — B cos B — C cosy,
the negative signs are used to indicate the leftward
Similarly, we write
sense
of the
2F, = A sina +B sin 6 — C sin Y;
where again the signs indicate the directions in which
the forces act. Knowing the numerical values of A, B, C, a, 8, and y, we
easily find the numerical
values of 2F, and 2F,. F inally, we recall that we
want the vector sum
§ 16 | EXAMPLE
19
of all the components of A, B, andC. We have the sum of the x components
and the sum of the y components.
Thus, the resultant R must equal
R=
=F,H+ =F,.
From the trigonometry of Fig. 19(b), we see that
(4)
Epes
2 a eter glee
Since Fig. 19(b) is drawn approximately to the same scale as Fig. 19(a),
the student may and should check with his dividers the 2/7, and LF, as
shown in (b). In the chosen illustration, the sum of the x components is
negative; that is, ZF, is a force acting towards the left. Also, since =F,
acts upward, the direction of the resultant R is necessarily such that the
vector is in the second quadrant.
Thus
LF,
(c)
iain Q | ie= SF
wherein the signs obtained in the numerical sums of horizontal and vertical
components are used.
To illustrate, suppose 2/7, = +5000 and =F, =
—5000.
Then tan 6’ = (+5000)/(—5000) = —1.
Therefore, 6’ = 135°.
However, the preferred procedure in practice in finding the direction
of R is to
1. Disregard the signs of 2F', and =F, in finding tan @,
2. Note carefully the sense of each sum (The student should sketch each
vector with its proper sense.),
(eX). Note the quadrant in which the vector falls,
4. Find the smaller angle 6, Fig. 19(b), which the resultant makes with
the x axis from
ys
(5)
tan 6 =
a
(both ZF, and ZF, positive),
where the sign obtained for either sum is disregarded and where both sums
are used as positive numbers.
Thus, for values of 2/, and YF, as specified
above, tan 6 = 5000/5000 = +1; and @ = 45°, which of course directs the
resultant as before.
The line of action of R passes through O, the intersection of the vectors A, B, and C. The resultant force replaces the other
forces; it does not act with the other forces.
16. Example.
Find
Sotution.
shall follow the procedure
Fig. 20(a).
We
algebraically
AL, = 0:
the resultant
D, = 0,
All other forces have both horizontal
horizontal forces, we have
of the force system shown in
described
and
in § 15(b).
Observe
that
jt, =O:
and vertical components.
Now
summing
COPLANAR
20
CONCURRENT
FORCES
[ Ch. II
=F, = A+ B cos 60° — C cos 45° — D — E cos 30°
500 + (600)(0.5) — (700)(0.707) — 800 — (900) (0.866)
ae 12743 Ibe
the negative sign showing that it acts toward the left.
Summing vertical components,
we get
=F, = B sin 60° + C sin 45° — FH sin 30° — F
= (600)(0.866) + (700)(0.707) — (900) (0.5) — 1564.5 = —1000 Ib.,
A= 500 lb.
1274 lb. ¢ Ic
the negative sign showing that it acts downward.
Having observed the directions of
<F, and >F,, we know that the vector for R must be in the third quadrant.
The
angle with the horizontal is found from
tan @ =
Zn O00
=F, 1274
0.785.
Thus, @ = 38.1°, where @ is the angle shown in Fig. 20(b).
resultant is
The magnitude of the
R = ((2F,)? + (2F,)?}”
= [(1274)? + (1000)2}2 = 1620 Ib.
as indicated in Fig. 20(b).
point of intersection.
The line of action of R must pass through O, the common
EQUILIBRIUM
17. Equilibrium.
Although we have centered our attention thus far on
external force systems which have a finite resultant, we shall be most
concerned
for a while with external force systems for which the resultant
Ris zero.
A
system of forces in which the resultant is zero is said to be in
equilibrium.
The condition of equilibrium was defined by Newton in
his first law, which
may be stated as follows: Every particle remains in a state
of rest or moves
with a constant velocity in a straight line unless an unbalanc
ed force acts on it.
Obviously, if R = 0, its components, 2/7, and =F, are
also zero. Con-
§ 17 | EQUILIBRIUM
21
sequently, it follows that we have two conditions expressing the relations
of forces in a concurrent system in equilibrium, namely,*
and
Lf = 0
(6)
Bee SO.
These two conditions, equations (6), are independent and may be used to
Moreover, the x and y axes may be taken in any two
find two unknowns.
different directions; that is, the sum of the forces and components in any
However, perpendicular x and y axes are most
direction must be zero.
convenient for our purposes.
Example.
The forces shown in Fig. 21 are in equilibrium.
SoLutTion.
Summing forces in the x direction, we have
What are F and 6?
xf, = 100 cos 30 + F cos o — 200 cos 45° = 0,
from which
(d) F cos 6 = 141.4 — 86.6 = 54.8 lb.
The sum of the forces in the y direction gives
=F, = 100 sin 30 + F sin 6 + 200 sin 45 — 300 = 0,
from which
(e) F sin 6 = 300 — 50 — 141.4 = 108.6 lb.
Fig. 21.
By division from (d) and (e), we get
108.6
tan 6 SS
BAS
i,
1.98,
or
6 == 63 Oe.
Then from (e), we find
108.6
108.6
sin 6
0.893
Most bodies in equilibrium are at rest (not moving relative to the earth),
but, for example, a rigid body moving with constant speed in a straight path
is in equilibrium.
<A rigid body may be any particular mass whose shape
remains unchanged while it is being analyzed for the effects of forces.
For
some purposes in mechanics, this might be a mass of fluid, but most bodies
considered will be solids.
Since no body is truly rigid, what we mean by
“rigid body” is one whose deformation under force is negligible for the
purposes of the problem.
When we know that a body is in equilibrium, we may use this knowledge
to find certain unknown forces from certain known forces. In doing this,
we find it advantageous to use what is called a free body. In sketching a
free body, we need to be able to determine by inspection the directions of
certain forces.
*After a study of moments in the next chapter, we shall see that it is also true that the
sum of the moments of the forces about any point is zero; but this is not an independent
condition in this case.
In a coplanar concurrent system in equilibrium, only two unknowns
may be found by the conditions of equilibrium.
COPLANAR CONCURRENT FoRCcES [Ch. II
22
a
18. Free Body. A free body is a representation of an object, usually
of
law
third
Newton’s
it.
on
acting
rigid body, which shows all the forces
motion states that to every action there is an equal and opposite reaction.
Thus, as explained in §2, if body B exerts a 10-Ib. force on body A, then
body A exerts on B a 10-lb. force. These two forces are equal Sumo
If a free body is made of A, the force exerted by B on A is shown.
in sense.
If a free body is made of B, the force exerted by A on B is shown.
To make a free body of a structure or a member of a structure, sketch
the structure or member isolated in space, then indicate the forces on it,
See
forces produced by other objects when it is in its normal position.
forces
where
recognizing
in
you
assist
will
discussion
following
The
Fig. 22.
occur.
Fig. 22. Free Bodies.
In (b), the forces A and B replace the effects of the pins A and B
in (a) on the member AB.
In (c), the forces C, and C, are the components of the action
of the pin C in (a) on the member CG.
In (e), we have a free body of the post HD, and
in (f), a free body of the entire structure
in which
only the members
CG
and HD
are
considered to have weight.
The student should study these free bodies, noting carefully
their relations with one another.
Observe and compare the directions of the various forces.
(a) Two-Force Members.
Consider the structure shown in Hig 223).
If a force (or load) F is applied at G, certain reactions or forces will be induced at the points A, B, C, D, and H, or else the structure will collapse or
tip over. This observation should be instinctive.
F irst, let us make a free
body of the strut AB.
If this member ig considered to be weightless, an
assumption often made when the weight is inconsiderable as
compared with
the magnitude of the impressed forces, there are only two forces
acting on it,
the force exerted by the pin at A and the force of the pinat B.
These points
A and B are the only contacts of this member with other
members,
In
general, it may be stated that:
§ 18 | FREE BODY
28
There will be a force acting on a member at each point where the member
makes a contact with another member.
Now since there are only two forces acting on AB and since this member
is in equilibrium, that is, 2F (in any direction) = 0, it follows that the forces
at A and B must be equal, opposite, and collinear.
Such a member is called
a two-force member.
‘The locations of the points A and B at which the external forces are applied to AB determine the directions of the forces A and
B. When the forces act to compress the fibers together as in AB, the member
is said to be in compression.
If the forces on a two-force member act away
from each other tending to pull it apart, as in the free body of ED, Fig. 22(d),
the member is said to be in tension.
Observe that a straight piece of flexible
cable, rope, or cord cannot sustain a compressive force along its length. It
must be in tension.
(b) Gravitational Force and Magnetism.
The line of action of the gravitational force or weight of a body (the action of the earth on the body) passes
through the center of gravity (cg) of a body and always acts vertically downward.
Until we learn later (Chapter X) how to find centers of gravity of
bodies in general, we need to know only that the center of gravity of a symmetric, homogeneous body is its geometric center.
The free body of member CG is shown in Fig. 22(c). Observe (1) that
the vector W,, the gravitational force, passes through the central point of the
member, (2) that A is known to act in the direction BA, and (3) that, since
the reaction at C is unknown in magnitude and direction, it is represented by
its x and y components.
In making this free body, we placed a force vector
wherever the member CG made contact with some other member.
For
example, at A, the force exerted by AB on CG is equal and opposite to the
force exerted by CG on AB (action and reaction are equal); thus, we know
the direction of A, Fig. 22(c), since AB is a two-force member.
At C, we could have shown the force C (the resultant of C, and C,) and the
angle @ that C makes with the horizontal instead of the components, but we
shall find later some convenience in using the components in such eases.
Magnetic force may act without two bodies being in physical contact.
Nevertheless, there is an action and a reaction, as in all other applications
of forces.
The magnetic force may act in any direction, depending upon
the location of the magnet with respect to the other body being considered.
(c) Reaction between Surfaces. We shall have occasion to deal for a
while with what we call smooth surfaces.
By a smooth surface, an imaginary
concept, we mean one which offers no frictional resistance to motion.
This
concept is a useful and practical approximation for some bodies at rest.
The
pin connections in Fig. 22 have been tacitly assumed to be frictionless; that
is, the pins are assumed to be smooth in all such structures where there is
no relative motion.
COPLANAR CONCURRENT FoRcES [Ch. II
24
The pins and the members are in contact on a surface. When two smooth
surfaces are theoretically in contact along a line or at a point,
the reaction between the surfaces is normal to the common
plane to the two surfaces at the point of contact.
tangent
As an illustration, consider Fig. 23, which represents a sphere (or cylinder)
resting on an inclined plane and against a stop at B. The body is in equilibrium under the action of the force exerted by the plane at A, the force
(b)
Fig. 23.
Reaction between Smooth Surfaces.
exerted by the stop at B, and the gravitational force. To make a free body,
Fig. 23(b), we remove the plane at A and insert a force F; which is equal to
the reaction between the plane and the sphere.
If the plane is smooth, there
can be no force parallel to the plane, so that the reaction must be normal to
the plane (which is tangent to the sphere).
This observation determines
the direction of F4.
Similarly, we remove the stop and replace the action of the stop by the
vector /’, which must act perpendicular to the tangent at B if the surfaces
are smooth.
Thus, with W and the directions of 7, and F, known,
there
are only two unknowns in this free body, the magnitudes of F; and F2, which
can be found from the conditions of equilibrium, equations (6), or from a
graphical solution (see § 19). Notice particularly that in order to reduce
the unknowns in this problem to two, we had to recognize that the directions
of F; and F, were known.
If a flexible cord in tension passes across a smooth surface, the tension
(force) in the cord is the same on both sides of the surface, Fig. 24. This constancy of force will not exist if friction is present—as we shall see later.
Read the caption for Fig. 24 now.
The student at this point should solve all the problems 130-138.
If you
will take the time now to master these problems, you will material
ly reduce
your future difficulties and improve your understanding of
the subject. These
problems are not numerical, but are exercises in drawing free
bodies.
§ 19 |EQUILIBRIUM OF THREE FORCES
25
19. Equilibrium of Three Forces.
Since the resultant of a force system
in equilibrium is zero, the vector sum of the forces is zero or the force polygon
closes. Now suppose that three non-parallel forces are in equilibrium. Since,
in this instance, the resultant of any two intersecting forces of the three forces
must be equal to, opposite to, and collinear with the third force, it follows that
any three forces in equilibrium must be concurrent and coplanar (or they may
be parallel, as described in the next chapter). This observation is often very
useful (see § 20).
Tyee:
Pin)
Smooth
TT
ie
Ts
cH
cht
|B
(a)
Not
Smooth
|B|
(b)
Fig. 24. Flexible Cord Passing over a Surface.
The illustration represents a flexible
cord from A to B passing over a peg C. If the peg C-has a smooth (frictionless) surface,
the tension JT in the cord on each side of the peg is the same, as shown in (a). Actually,
there is always friction, though sometimes the size of the frictional force is negligible as
compared to other forces involved.
With friction, the tensions in the cord on each side
of the peg are generally different, and may be designated 7, and T:, as in (b). We shall
analyze the case with friction in Chapter VI.
To illustrate a graphical solution, reconsider the sphere of Fig. 23. In
Fig. 25(a), let the vector W represent in magnitude and direction the gravitational force passing through the cg. The vectors F; and Ff, are drawn in
the proper direction but are unknown in magnitude.
To find their magnitudes, lay out to scale the free
veetor Ab ="W im Fig, 25(b).
From point B, draw a line B-b of
indefinite extent and parallel to
vector Fin Fig. 25(a). From point
A, draw a line A-a of indefinite
extent and parallel to the vector
F,. Since it is known that the
sum of these forces is zero, the
intersection of B—b and A-a, point
(b) Force Polygon
C, must determine the magnitude
Fig. 25.
of F, as BC and of F, as CA.
Observe that, when equilibrium exists, the force polygon closes with all
vectors pointing in the same rotary direction.
When any number of forces
are in equilibrium, the force polygon must close.
Making Fig. 25(a) general by calling the angles between the forces a, 8,
and y, we see that the angles of the force triangle in Fig. 25(b) are 7 — a,
COPLANAR CONCURRENT FORCES [Ch. II
26
Applying the law of sines to the triangle in
7 — B, and s — y, as shown.
Fig. 25(b), we have
W
oy
sin(r — a)
Fy
us
sin(a— 8)
Fy,
;
sin(r — 7)’
and since the sin (rt — a) = sin a, etc., we see that
W:
(f)
sina
te
Polls eee
sing
siny’
which is to say that:
Each force in a group of three forces in equilibrium is proportional to
the sine of the angle between the other two forces (Lamz’s* Theorem).
Compare equation (f) with Fig.
25(a).
Lami’s theorem offers
another method of solution when
three concurrent forces are involved.
Thus, from (f),
(a)
(b) Force Polygon
Fig. 25.
Repeated.
a third method of finding F; and F2,
for illustrative purposes, we have
a
(= alu and
sin a
a
(= *)
2=|(—
sin a
The equations of equilibrium,
ZF, =0 and ZF, = 0, present
Using the known angles in Fig. 25(a)
ZF, = F, cos 45° — F, cos 60° = 0,
2F, = F; sin 60° + F2 sin 45° — W = 0,
from which, by a simultaneous solution, we may find F, and F, if W is known.
This third method we find particularly advantageous when more than
three forces are in equilibrium.
An adequate understanding of analytic mechanics can come only through
the solution of a large number of problems.
The subsequent examples will
help the student to learn to analyze engineering problems of
this nature if
he will carefully trace the solution of each one, preferably by
solving each
with paper and pencil.
20. Example. Find graphically the reactions at pins B
and £ in Fig. 26(a), which
diagrammatically represents a derrick, if the load W
= 1000 lb. N eglect the weight
of the members BH and CE.
*Father Bernard Lami (1640-1715) discovered
independently
law. Co-discoverers included Stevinus, Varigno
n, and Newton,
(1687) the parallelogram
§ 21 | TRUSSES—JOINT-TO-JOINT METHOD
27
SotuTion.
Make a free body of the boom.
In Fig. 26(b), the member BH is laid
out to scale to facilitate a graphical solution. The load W is drawn acting vertically
downward through H. Since CE is a two-force member, the direction of the force on
BH at E (Fz) is known.
Therefore, draw Fz through EZ in Fig. 26(b) and parallel
to CH. The reaction at B, F's, caused by the mast on the horizontal boom, is, for the
moment, unknown in direction, although its direction is defined by the intersection
of the other two forces. To indicate that the direction is unknown, we draw a wavy
line of action for Fz.
N
Fig. 26.
Superimposing the solution of this problem on the free body, extend the line of
action of Fz until it intersects the line of action of W atQ. Then lay out QK to represent W to some convenient scale. Since three forces in equilibrium intersect at a
point, the line of action of F's must pass through Q. Therefore, draw BQ, which is
the line of action of Fz. Now through point K, draw KN parallel to QB. The point
of intersection of AN and the line of action of Fz at N is one apex of the force triangle
which defines equilibrium conditions.
Consequently KN represents the magnitude
and direction, but not the line of action, of /’s, which must pass through point B,
The length NQ represents completely the force Fz acting on the boom BH.
21. Trusses—Joint-to-Joint Method.
A determinate structure is one
wherein the internal forces in the various members of the structure may be
obtained by the conditions of equilibrium.
Although trusses for roofs and
bridges, etc., are actually subjected to all manner of loading, engineers
usually assume for purposes of simplicity that the loads W1, We, ete., Fig.
27(a), are applied at pin joints. With a truss loaded in this manner,
all the various members are two-force members and the free body of each
pin is a system of concurrent forces.
Thus, if all the external forces are known, including the reactions at the
supports R; and Re, Fig. 27(b), we may use the principles of this chapter to
determine the internal forces in each member, such as AC and AB, of the
truss. Since the size of each member depends upon the force on it, the internal force is used in the design of the member.
To illustrate the problem in mechanics, consider the truss loaded as shown
in Fig. 27(a).
Figure 27(b) shows the free body of each pin.
Starting with
COPLANAR CONCURRENT FORCES
28
[Ch. II
in
the free body A, we note that 2, is known, that Fy and F, are unknown
Recalling that we have two conditions of equilibrium,
magnitude and sense.
=F, =0 and =F, = 0, and that from these two conditions we can obtain
the value of two unknowns, we know that we can find the magnitudes of F;
and Ff’, by using the two equations
Lh
yt,
hp
ell
OO,
= 1h = 1, COS.) = UW:
A graphical solution may be obtained also, if desired, from the condition
that the force polygon closes.
Smooth
Fig. 27.
Free Bodies of Pins ina Truss.
Trusses are used to support roofs, as suggested
by (c), or bridges, as suggested by (d).
Also they may be built in a number of different
forms for many different purposes.
Now, knowing 2, we may solve the free body at B. Study this free body
until you understand why the horizontal forces are equal and why the vertical
forces are equal.
With the force in BC known to be W, (from the free body B) and with the
force in AC known to be F, (from the free body A), there are now only two
unknowns left in the free body C, the magnitudes of F3 and Fy. Observe
that no solution can be made at C until the unknowns at A and B have been
found, that the solution of the problem must proceed in the foregoing manner.
Again at D, with F3; known from the free body at C, there are only two
unknowns, the magnitudes of F’; and F's.
In making a free body of one of these pins, you must decide in what direction to point the arrows of the unknown forces. Sometimes there is no
difficulty in determining
by inspection
the sense
of each
unknown
force.
§ 22 | EXAMPLE
29
(Study free body A until you have decided that you could have pointed F;
and F, correctly.)
In some instances, however, the correct sense is not so
evident.
In any case, the following procedure may be used.
(1) Indicate the unknown forces as acting away from the pin; for example,
F and F¢, in free body D. This sense is tantamount to assuming the members
DE and DG to be in tension. When the member pulls on the pin, the pin
also pulls on the member, placing the member in tension.
When the member
pushes on the pin, the pin also pushes on the member, placing the member
in compression.
(2) Solve for each of the unknowns, for example, F’; and F's.
A positive answer indicates that the correct sense was assumed.
negative answer indicates that the assumed sense is wrong.
A
Thus, if the numerical value of /’s (in free body D) is determined, it will be
found to be negative.
Therefore, F's should act toward the pin (as shown
in free body G) and the member DG is in compression, not tension as originally
assumed.
Whether the direction is assumed to be toward or away from the
pin, a negative answer indicates that the wrong sense was chosen. This rule
generally holds true in other force analyses, but the known forces must point
correctly.
For instance, in free body D, F’; must be in its correct sense.
observation suggests the importance
F in free body C.
12,565 lb.
1N 16,900 lb.
20°
R= 7250 Ib,
This
of determining correctly the sense of
20’
6000 Ib.
5000 Ib.
R= 9750 Ib.
(a)
Fig. 28(a) and (b).
22. Example. A roof truss is constructed and loaded as shown in Fig. 28. At the
pin N, the following internal forces have been found: NF = 16,900 Ib. (tension),
DN = 7000 lb. (compression), and
NH = 12,565 lb. (tension). For an external load
at the pin of 6000 lb., find the forces in the members NC and NB.
Sotution.
If the structure does not fail, the forces at N are in equilibrium. A
free body of pin N is shown in Fig. 28(b) with 7; and F2, the unknown forces, directed
away from the pin. Proceeding algebraically, we find
=F, = 16,900 — 7000 cos 30° — F»2 cos 30° — 12,565 = 0,
COPLANAR CONCURRENT FORCES
80
[Ch. II
from which Ff, = —2000 Ib., where the negative sign indicates that F» is shown in the
Now
wrong sense and that the member NB is in compression rather than tension.
from =F, = 0, we have
=F, = F, — 7000 sin 30 — 6000 + F2 sin 30 = 0
= F, — (7000)(0.5) — 6000 — (2000)(0.5) = 0,
from which F; = +10,500 lb., where the positive sign shows that the member NC is
in tension. Note particularly that in this sum the force F2 is substituted as —2000,
as previously found.
Y |F,= 10,5000
12,565
Nf} 16,900
~
(b) Free Body of NV
(c) Corrected Free Body of N
Fig. 28(b) and (c).
However, the student may find some advantage in marking out the incorrect
arrowhead on /’, after the correct sense has been determined, inserting the proper
arrowhead, and then using the correct sense of F: in the =F, = 0. In this case, the
free body at N would appear as seen in Fig. 28(c) with #. shown in its true sense.
From Fig. 28(c), we write
=F, = F, — 7000 sin 30 — 6000 — F: sin 30 = 0,
which naturally yields the same result as before for F’;.
GRAPHICAL SoLuTioN.
Graphical solutions are frequently easier and more convenient than algebraic ones.
See Fig. 29 for a graphical solution of this problem.
ER
Fo= 2000 Ib.
—
D
Scale:1”= 8000 Ib.
rey,
Fig. 29. Graphical Solution of Example of §22. Starting at any
convenient origin O, lay
out OA to scale to represent the 6000-Ib. load. From A, lay
out AB = 16,900 lb. From
B, lay out BC = 7000 lb. From C, lay out CD = 12,565 lb.
These vectors are the
heavy ones. Now we know that the addition to these
vectors of the two unknown forces
F, and F, will close the polygon; that is, one of the vectors
F, or F, must pass through O
and the other through point D.
Thus draw a line through D parallel to F,
of indefinite
extent.
Draw a line through O parallel to F 2 of indefinite extent.
The intersection of
these lines at E defines the magnitudes of F, and
Fy», Scaling the lengths DE and EO
should give approximately the same values as those
found algebraically.
§ 23] EXAMPLE
31
The first step is to add vectorially all the known vectors to give a start on the force
polygon.
If the body is in equilibrium, we know that the force polygon must close.
Therefore, unknown forces must act in such a way as to close the polygon.
The
force polygon will determine two unknowns, in this case, the magnitudes of F; and
F.. Their directions are already known.
It does not matter in what order the vectors are added. The reader should
solve this problem by adding the vectors in a different order, say, 6000, 7000,
16,900, 12,565, F2, and Fy. Such an action would measurably help your
understanding of graphical solutions.
20,000 lb.
Tie
Rod
aC
Smooth
Fig. 30(a) and (b).
23. Example.
Two timbers, AB and BC, Fig. 30, are pinned together at B and
joined by a tie rod AC near the ground. If the load at B is 20,000 lb., what is the tension in the tie rod? All surfaces are smooth.
Neglect the weights of AB and BC.
Sotution.
If we are to follow the principles of this chapter, we must choose a
free body with at least one known force and no more than two unknowns.
The pin B
accords with these requirements.
In Fig. 30(b), the free body shows F; and F, unknown in magnitude but known in line of action because BC and AB are two-force
members.
Now we note that the triangle ABC is one that is termed a 3-4-5 triangle
(3:4:5 =9:12:15), so that angle ABC is 90°. Noting from the dimensions of the
sides of the triangle that sin 6 = 0.6 and cos 6 = 0.8, we find from Fig. 30(b)
(g)
(h)
=F,
6 =0
0 —nF, cos
= F,.si
= 0.6F2 — 0.87; = 0.
=F, = — 20,000 + F, sin 9 + F2 cos 6 = 0
= — 20,000 + 0.67, + 0.8F2 = 0.
By a simultaneous solution of (g) and (h), we find F’; = 12,000 lb.
and F2 = 16,000 lb. Now considering the free body of the
pin at C, we see that F,; is known, F; and the reaction Kg
between the ground and the frame are unknown in magnitude.
Thus, from Fig. 30(c), we get
=F, = F, cos 6 — F3 =
= (12,000)(0.8) — F3 = 0,
from which F'; = 9600 lb., the tension in the tie rod.
F,
Swe
F,
Rc
(c)
Fig. 30(c).
COPLANAR CONCURRENT FORCES
82
[Ch. IT
AuTERNATE SotuTion. As previously stated, the x and y axes in 27, and =P, may
By a judicious choice of axes, simultaneous solutions of
be in any two directions.
equations may often be avoided with some simplification of the solution. Whenever
a system of concurrent forces is in equilibrium, the equations of equilibrium can be
Thus, if the magnitudes of two forces are
used to find a maximum of two unknowns.
the unknowns, the axes may sometimes be conveniently taken perpendicular to the
lines of action of the unknown forces. Following this plan, let one axis be along AB
(which is perpendicular to BC, so that BC has no component in the direction AB) and sum forces in the direction
AB. See Fig. 30(b).
=F 4s = F'2 — 20,000 cos 6 = 0,
or F, = (20,000)(0.8) = 16,000 lb. Taking the other axis
along CB, along which AB will have no component, we get
SFac = F, — 20,000 sin a = 0
Fig. 30(b).
Repeated.
or F, = (20,000)(0.6) = 12,000 lb.
24. Closure. Before leaving this chapter, be sure the following principles
and suggestions are understood.
1. When one sense of a vector has been chosen as positive, the opposite
sense must be considered negative.
2. When writing =F, or ZF, it is generally easier to determine the sense
of a component (the sign of a term in the sum) by inspection and to
use the small angle between the vector and the x axis.
jes). A free-body diagram shows the actions of other bodies on the free body.
4. The direction of the force on a pin-connected two-force member lies
along the center line of the pins (smooth pins). The member is in
either tension or compression.
5. In starting the solution of a problem, examine the structure particularly
for two-force members. . Since the direction of the force in such a
member is known, a two-force member can be cut with the simplest
consequences.
6. A flexible cord is always a two-force member and the force
in it is
always tension.
7. The force of gravity always acts vertically downward.
8. The reaction between smooth surfaces is normal to the surfaces
.
9. The lines of action of three non-parallel coplanar forces
in equilibrium
necessarily intersect at a common point.
10. The force polygon must close when equilibrium
exists.
11. When forces are being considered, always make
a free-body diagram.
Here and there throughout this book, the reader
finds reference to weightless members, such as a “weightless cord.”
Since evidently no material
body can be weightless, such expressions are
interpreted to mean “a cord
whose weight is negligible.”’ Sometimes the
weight of a member is neglected
PROBLEMS:
RESULTANTS
Oo
in order to simplify the problem; sometimes the effect of the weight as compared to other forces is actually negligible from an engineering point of view.
Thus the reader should not feel that he is dealing with impractical problems
when such weights are neglected.
Unless the weight of a member in a
problem is defined by the data, its weight is not to be considered.
Problems
Where applicable, always show free-body diagrams in the solution of problems.
Note.
In such problems,
If the problem does not state the method of solution, solve algebrarcally.
a graphical solution for a check is optional.
RESULTANTS
51. Figure 31 shows three concurrent
coplanar forces with @ = 30°. Determine
algebraically and graphically the resultant
of these forces.
Ans. 213 lb., at 344°.
200 Ib.
y
64 = 300°. What force will produce the
same external effects as these four forces?
Solve algebraically and graphically.
Ans. 98 lb., 172.6°.
55. The same as 54 except that M4 =
200 lb.
56. A concurrent force system is composed of the following forces which act in
the compass direction given: 7; = 20 kips,
N30°B;
F.2 =15
kips, North;
Fs; = 10
kips, N75°W; F, = 12 kips, West; Fs =
Fig. 31.
Problem 51.
52. Determine the resultant of the forces
shown in Fig. 32 algebraically and graphically if F = 500 lb.
12 kips, S75°H; and Fs = 18 kips, 845°E.
Find the resultant force, solving (a) algebraically and (b) graphically.
Ans. 22.9 kips at N33.5°E.
57. In Fig. 33, R is the resultant of F,
T, and Q. Let F = 150 lb., 6 = 30°, R =
85 lb., and find Q and a (a) algebraically
and (b) graphically.
IR
|
Fig. 33.
Problems 52, 53.
Fig. 32.
53. Determine the resultant of the forces
shown in Fig. 32 algebraically and graphically if F = 900 lb.
54. A concurrent force system is composed of the following forces, where @ is the
angle with the positive side of the x axis:
F,
=
ED
lps
6,
=
Abas
F,
=
100
lose
6, =
120°; Fs = 150 lb., 63 = 210°; F, = 60 lb.,
Problems
57-59.
58. In Fig. 33, R is the resultant of FP,
T, and Q. Let Q = 800 lb., a = 30°, R =
2kips.
Find F and @ (a) algebraically and
(b) graphically.
Ans. 1650 lb., 335.2°.
59. In Fig. 33, R is the resultant of FP,
T, and Q. Let F = 17 lb., 6 = 40°, and
R = 25 lb.
Find Q and a (a) algebraically
and (b) graphically.
34
PROBLEMS: RESULTANTS
60. In Fig. 34, a boat is being pushed
forward against a resistance of Q = 50 Ib.
If the resultant force in the x direction is
[Ch, II
62. In Fig. 35, the bodies A and B,
connected by a cord and resting on smooth
planes, weigh W4 = 50 Ib. and Wz = 30
The body A is acted on by a force in
such a manner that the tension in the cord
is 30 lb. If the normal reaction between
B and the plane is 25.98 lb., what is the
resulting force on B?
Ans. 15 lb. up the plane.
63. The same as 62 except that the tension in the cord is 20 lb.
lb.
gor
Fig. 34.
Af
Problems 60, 61.
30 lb.
Rk = +100 lb., determine Ff and @ algebraically and graphically.
Ans. 90 lb., 26.4°.
61. The same as 60 except that R =
20 lb.
°
fe
+200 Ib.
Smooth
Fig. 36.
Problems 64, 65.
64. The resultant R = 60 Ib. shown in
Fig. 36, is the resultant of the 20-Ib. and
30-lb. forces and another force F. What
is the force F?
Ans. 54.7 lb., 6 = 233.7°.
Fig. 35.
Problems 62, 63.
2
65. The
5albe
same
as 64 except
69. The
permissible
that
R =
EQUILIBRIUM
66. A weight W = 1000 Ib. is suspended
from a pin B which unites two inclined
timbers AB and CB, Fig. 37. Determine
algebraically and graphically the compressive force on each timber.
Ans. 578 lb.
timber
AB,
Problems 66, 67.
67. In Fig. 37, if the horizontal and
vertical components of the reaction
at Cc
are C, = 500 lb. and Cy = 866 lb., find
W.
HINT: Observe that BC isa two-force membe
r.
68. The permissible internal foree
on
timber AB, Fig. 38, is 4 kips.
If 6 = 30°,
what maximum safe load may be suspe
nded
at B? Solve algebraically and graph
ically,
Ans. 4460 lb.
lb.
force
on
If 6 =
20°, what maximum safe load W may be
suspended at B?
Solve algebraically and
graphically.
Fig. 38.
Fig. 37.
internal
Fig. 38, is 25,000
Problems
68-70.
70. In Fig. 38, if W = 100 lb. and the
reaction at A is 75 lb., find the reaction
at
C and the angle @. Solve algebraically and
graphically.
Ans. 70.9 Ib., 41.6°.
71. A span of telephone wire is suspended
from crossarms of equal elevation and
the
horizontal span is 200 ft. When
a 10-lb.
buzzard alights in the center of the
span,
the sag is observed to be 2 ft. Consi
der
the wire on each side of the bird asa straig
line and neglect the weight of the wire.
ht
(a)
Find the force in the wire caused
by the
PROBLEMS:
bird.
35
EQUILIBRIUM
(b) Reduce
the
sag
to 6 in. and
recalculate your answer. Solve algebraically
and graphically.
72. A 5000-lb. sphere rests on a smooth
plane inclined at an angle 6 = 45° with the
horizontal and against a smooth vertical
wall. What are the reactions at the contact surfaces A and B, Fig. 39?
Ans. 5000 lb., 7070 Ib.
73. The same as 72 except that 6 = 75°.
80. The same as 79 except that the tension in the rope is 400 lb.
Fig. 41.
Fig. 39.
Problems 72-75.
Problems 79, 80.
81. A captive balloon in a balloon barrage exerts a net vertical lifting force of
10,000 lb. Its own weight is 1000 lb. A
horizontal wind causes the cable to which
it is attached to slope at an angle of 30°
with the vertical. What is the force of the
wind on the balloon?
Ans. 5200 lb.
74. In Fig. 39, the greatest force at A or
B caused by the 5000-lb. sphere must not
exceed 10,000 lb. if permanent deformation
of the surfaces is to be avoided.
Find the
largest value 6 may safely have.
Ans. 60°.
75. The same as 74 except that the sphere
weighs only 800 lb.
W= 2000 Ib.
Fig. 42.
Fig. 40.
Problems 76-78.
Problems 82, 83.
82. The derrick shown diagrammatically
in Fig. 42 supports a load of W = 2 kips.
Find the tension in the boom cable and the
compression in the boom when the angle
6 is (a) 30°, (b) 90°, and (c) 150°.
76. A 2500-lb. wheel with a radius of 3 ft.
is acted upon by a force F, Fig. 40, which
tends to pull the wheel over the obstruction
at A. At the instant the wheel is about to
move, the pressure between the wheel and
the ground is zero.
What is the magnitude
of the force F at this instant if @ = 30°?
Ans. 2500 lb.
77. The same as 76 except that @ = 40°.
78. The same as 76 except that the weight
of the wheel is 4000 lb.
79. A 500-lb. cylinder A rests on a
smooth inclined plane, Fig. 41. For a tension in the rope of 250 Ib. find the inclination of the plane and the plane reaction.
Solve algebraically and graphically.
Ans. 30°, 433 Ib.
position produces
boom?
Which
the largest load on
the
Ans. All in kips: (a) 3.5, 2.03; (b)
3.5, 4.03; (¢) 3.5, 5.33.
83. The same as 82 except add the following: (d) Prove that the compressive
load
on
the
boom
is constant.
(e) Does
your proof include the conditions of @ = 0°
or @ = 1807%
84. What are the forces on the members
AB and BC, Fig. 48, if @ = 30°? Solve
graphically and algebraically.
Ans. 7120 lb., 438 lb.
85. The same as 84 except that 6 = 60°.
86. Two spheres are at rest against
smooth surfaces, as shown in Fig. 44.
Sphere A weighs 3200 lb. and sphere B
weighs 400 lb. Let # = 1000 lb. and @ =
PROBLEMS: EQUILIBRIUM
36
90°, and find the reactions at C, D, and
E. Solve algebraically and graphically.
Ans. 850 Ib., 3600 lb., 1850 lb.
[Ch. II
body is subjected to a concurrent force
anon
Ans. 1607 |b.
90. The same as 89 except that Rh =
1800 lb.
2000 lb
7000 Ib.
i
6
Ba |
Fig. 43.
Problems 84, 85.
Fig. 46.
Problems 91, 92.
91. Two weights are suspended from a
flexible cable as shown in Fig. 46. For
6 =
120°, determine
the internal
forces in
the various parts of the cable and the
weight W.
Ans AB — 2/30 lb bey —
1417 lb., CD = 1937 lb., W = 1736 Ib.
92. The same as 91 except that @ = 140°.
Fig. 44.
Problems 86-88.
87. The same as 86 except that / = 0.
88. The same as 86 except that 6 = 75°.
Ans. 760 Ib., 3403 lb., 1734 Ib.
Fig. 45.
Problems 89, 90.
89. In Fig. 45, let Wa = 2000 lb., Wz =
500 lb., We =5000 lb., and force R = 8000
Ib. Neglecting all friction so that the reactions are normal to the surfaces, find the
force Q on top of the wedge.
Solve algebraically and graphically.
Assume that each
Fig. 47.
Problems 93, 94.
93. A continuous string ABCDE, Fig.
47, passes over smooth pegs at B and D,
10 in. on centers.
To the ends of the
string are attached the weights W4 = 7 lb.
and Wz =5 lb. A 10-lb. weight is attached at C and the three bodies are in
equilibrium.
Determine the distance a and
the angle a.
Ans. 3.8 in., 49.5°.
94. The same as 93 except that Wz =
10 lb.
95. A weightless string is 10 ft. long and
is attached to pins A and B, B being 6 ft.
to the right of and 2 ft. below A. A weight
is supported below the string by a frictionless ring. Determine the position of the
ring on the string when the ring slides to
the equilibrium position.
(arnt: Study
carefully the equilibrium condition in the
horizontal direction.)
Ans. 5 ft. below and 3.75 ft. to right of A.
96. The same as 95 except that the string
is 8 ft. long.
PROBLEMS:
57
EQUILIBRIUM
97. A timber ABC, Fig. 48, is supported
by another timber DB.
A load of F =
4.5 kips acts horizontally at C. If @ =
60°, find the reactions at A and B.
Solve
graphically and check by Lami’s theorem.
Ans. Ra = 14.74 kips, 64 = 252.2°, Rp =
14.03 kips, 68 = 90°.
F = 9000 lb. is being exerted in the connecting rod to support a load W on the
cable.
By a graphical solution, find (a) W
and (b) the bearing reaction at A. Check
the result by the law of sines.
Ans. (a) 3690 Ib., (b) 10,300 lb. at 149.4°.
ZB
OH
SAM
oes
OHNO
,
VOLLOLTELT SEE be
Fig. 48.
Problems 97, 98.
Fig. 51.
98. The same as 97 except that 6 = 40°.
C
Smooth
Bell Crank
Weightless
Fig. 49.
Problems 103, 104.
103. The member AB, Fig. 51, weighs
2000 lb. and rests against a smooth wall at
B. Determine graphically the reactions at
Band A when @ = 30°. What are the horizontal and vertical components of the reaction at A?
Ans. B = 578 \b., A = 2084 ]b., 04 = 73.9°.
104. The same as 103 except that @ =
60°.
Problems 99-101.
99. The bellcrank, Fig. 49, has a load
F = 100 lb. If 0 = 30°, solve graphically
for the reactions at B and C.
100. The same as 99 except that @ = 90°.
Ans. B = 379 |b. at 97.6°; C = 289 lb. at
270%
101. The
120°:
same
Fig. 50.
as 99 except
that
@ =
Problem 102.
102. Figure 50 shows a cable drum which
A pull of
is turned by a steam engine.
Fig. 52. Problems 105-108.
105. The bar AB in Fig. 52, with uniform
cross section, is pivoted at pin B and rests
on a smooth plane at A. If the bar weighs
W = 100 lb. and if 6 = 30°, find the pin
reaction at B. Solve graphically and algebraically.
Ans. 70 lb. at 105.7°.
106. The same as 105 except that @ =
60°.
107. In Fig. 52, the uniform bar weighs
100 lb. If the reaction at B is 65 Ib. at
120° with the positive x axis, determine @
and the reaction A graphically and algebraically.
Ams. 54.6 lb. 0.= 36.:6°.
108. The same as 107 except let B =
85 Ib. at 150°.
PROBLEMS: EQUILIBRIUM
58
109. In Fig. 53, the bodies A, weighing
50 lb., and B, weighing 75 lb., are connected
by a cord and rest on smooth inclined
planes.
What is the angle @ if the bodies
are in equilibrium?
Solve algebraically and
graphically.
Ans. 28.15°.
aac
45
°
the wheels and the value of angle @. There
is negligible friction.
Ans. @ = 21.2°.
wi
lig =
200 tons
200 tons
200 tons
200 tons
e
Fig. 53.
[Ch. II
Fig. 56.
Problems 109, 110.
Problems 114, 115.
114. In the bridge truss of Fig. 56, find
the forces in members AB, AF, BF, and FE.
110. The same
weighs 200 Ib.
Fig. 54.
as
109
except
that
B
Ans.
Problems 111, 112.
111. The wheel of Fig. 54 is on the point
of rolling over the block (that is, the reaction
at A is zero, but the wheel has not moved),
If W = 1000 lb., what is the magnitude
and sense of the least force F that. will
produce this condition?
Solve graphically.
Ans. 547 lb., 6 = 33.2°.
112. The same as 111 except that W =
2kips. Solve algebraically and graphically,
Ww
AB
= 283
tons
(C); AF
= 200 tons
(T); BF = 200 tons (T); FE = 200 tons (T).
115. In the bridge truss of Fig. 56, find
the forces in members DC, DE, CE, BC,
BE, and EF.
Specify whether each member is in tension or compression.
Ve orizontal
Smooth
Fig. 57.
Problems 116, 117.
116. (a) Find the forces on all members
of the cantilever truss shown in Fig. 57
when
F = 4 tons.
(b) Using values found
in the foregoing analysis, state the horizon-
tal and vertical components of the pin reaction at H and state the directions in which
they act to maintain the truss in equilibrium.
Ans. (b) Hz = —13.88 tons, Hy = +4 tons.
117. The same as 116 except that F =
2.5 kips.
8”
Fig. 55.
Problem 113,
113. A boy’s wagon is in process of crossing a ditch as shown in Fig. 55. If W
=
100 lb.,
Q = 0, the wheels have a diameter
of 10 in., and if the wagon is at rest, solve
graphically for the normal reactions agains
t
Fig. 58.
Problems 118-120.
118. In Fig. 58, let F = 10 kips and @
=
80°, Find the foree in members AB
and
AD graphically and algebraically.
PROBLEMS:
59
EQUILIBRIUM
119. In Fig. 58, let F = 5 kips and 6 =
60°.
(a) Find the force in each member of
the truss.
(b) Using values found in the
foregoing
analysis,
determine
the compo-
nents of C and the reaction at H, and state
the directions in which they act to maintain
the truss in equilibrium.
120. In Fig. 58, let F = 5 kips and 6 =
30°.
(a) Find the force in each member
of
the truss.
(b) Using values found in the
foregoing analysis, determine the magnitudes and directions of the components of
C and the reaction of EH.
Ans. All in kips: (a) AD
= 5.77 (C), AB
=
DES (GE), iD =) (Dg IDI = B33, (CO),
BIB) = VOY (CO), XC! S 1k (ID), Cle = San)
z
Ay
G
ale fl
Fig. 59.
force against each of the walls is five times
as great as the tension 7 in the turnbuckle.
Avis, libs
124. In Fig. 60, 7 represents a turnbuckle that is used to actuate a toggle press
that pushes walls # and G apart.
Links
AB, BC, CD, and AD are of equal length.
The turnbuckle causes a pull in BD of 1000
Ib. Find the force against each wall (a) if
a, = S30), (lo) i on SS Be
125. In Fig. 61, neglect the weight of
all members and consider all pins smooth.
bey
il.
@ = CO,
eacl
e =
I
F
E
ieee
Problems
12 Fein Hiv 59 Oe—22000) lbs ar— 30.
The reaction at D is 683 lb. upward.
At A
the reaction has components of A, = 1000
Ib. and A, = 1049 lb. Find the forces in
members AB, AG, and BC.
Ans. 1485 lb. (C), 2049 lb. (T), 1866 lb. (C).
1225 nie hies 59) C= 2000 by vas— 30".
The reaction at D is 683 lb. upward.
At
A the reaction has components A, = 1000
lb. and A, = 1049 lb. Find the forces in
members CD and CF.
J = 100)
Using two eoncurrent force systems, solve
graphically for the force in each member of
121, 122.
the assembly.
What are the reactions at
A, D, and EH?
126. The same as 125 except that @ =
120°.
Fig. 61.
Problems 125, 126.
127. In Fig. 62, CD is a rigid, weightless
body,
Fig. 60.
F = 150 lb., the pegs
are
smooth,
Problems 123, 124.
123. In Fig. 60, 7 represents a turnbuckle that is used to actuate a toggle press
that pushes the two walls H and G apart.
If the links AB, BC, CD, and AD are of
equal length, find the value of a when the
Fig. 62.
Problems 127, 128,
PROBLEMS: EQUILIBRIUM [Ch. IT
40
Problem 131.
Bell Crank
B
Weight,
Ws
Problem 132.
Problem 134.
Problem 133.
Problem 135,
Problem 136.
4
A
Weight, Wa
30°
Problem 137.
Problem 138.
PROBLEMS:
and
the
Al
EQUILIBRIUM
cable
is weightless
and
flexible.
force on the piston is 7000 lb., find the
force exerted on the rivet.
Determine the weights of A and B if the
bodies are in equilibrium and CD remains
horizontal.
Ans. 138.6, 75.5 lb.
128. The
same
130-138.
For the figures on p. 40, sketch
a free body of the members designated by
the letters A, B, and C, All plane surfaces
and pins are smooth (frictionless).
Flexible, weightless cords are represented by
single lines.
139-150. These numbers may be used
for other problems.
as 127 except that F =
200 lb.
129. A pneumatic riveter that utilizes a
toggle linkage is shown in Fig. 63. The
diagrammatic drawing, Fig. 63(b), shows
this riveter in an acting position.
If the
Air Piston
G
Frame and
NY
Cylinder
Air Cylinder
Movable Die
Attached to
Plunger
Fixed Die
for Head
(a) Courtesy Watson Stillman Co.
Fig. 63.
Problem 129,
Frame
(b)
Pneumatic Toggle Riveter.
Chapter III
MOMENTS
AND
PARALLEL
COPLANAR
FORCES
25. Moment of a Force.
In making force analyses ot structures and
machines, we often find it convenient to use the moments of the forces.
The moment of a force is the force times a moment arm.
The moment arm
is the distance between the line of action of the force and the point or axis
about which the moment is taken, as measured along a line perpendicular
to the force vector. The point about which the moment is taken, called the
center of moment, may be the trace of some axis on
the plane containing the force vector and the point.
Thus, in Fig. 64, let the vector CD completely define
the force F. Then, the moment of F about the point
A, represented by Ma, is M4 = Fa, where a is the
Na perpendicular distance from A to the vector CD.
If the desired center of moments is located as at B
(Fig. 64) the moment arm is still a perpendicular
Fig. 64. Moments.
distance to the line of action of F, which may be
indefinitely extended; that is, Mz = Fb.
Observe that a force may have an infinite number of moments, since there
are an infinite number of locations of points or axes about which moments
may be taken, and that therefore the
moment of a force is intelligible only
when the center of moment is precisely
designated.
The unit of a moment
is a compound unit, a force unit times
a distance unit, usually inch-pounds
(in-lb.) or foot-pounds (ft-lb.), though
it may be any desired combination
of force and distance units such as
Fig. 65. Moment of Components.
inch-ton.
An interesting characteristic of moments is illustr
ated in Fig. 65, which
shows a crank AB keyed toa shaft A. If a force
F is applied as shown, the
42
§ 26 | VARIGNON’S THEOREM
43
moment of this force about the axis of the shaft is Fh. Now replace F by its
components F, and F,. Then the moment of F, about the axis A is zero
because the line of action of Ff, passes through this axis (moment arm is
zero). The moment of Ff, about A is (F,)(AB).
Since AB = h/sin 6 and
since fF, = F sin 6, we find
(F,)(AB) = (F sin 0( i )= Fh,
sin 6
which is the moment of F about A. Thus, we see that the moment of one of
the components of a force is the same as the moment of the force about a
particular point when the other component is directed through the center of
moments.
This conclusion, which is frequently useful in practice, may also
be derived from Varignon’s theorem, proved in the next article.
26. Varignon’s Theorem.*
Varignon’s theorem states that the algebraic
sum of the moments of two concurrent
forces about any axis perpendicular
to the plane of the forces is equal to
the moment of their resultant about the
same axis.
The vectors A and B represent two
forces with a resultant R (Fig. 66).
Let P be the trace of the axis about
Then
which moments are to be taken.
the moment arms of A, B, and R are
a, b, and r, respectively.
y
Yee
pee
It is required
to prove that Aa+ Bb = Rr. From
the geometry of the figure, it is seen that
(a)
wl
j
9
Fig. 66.
Bess
Varignon’s Theorem.
A cosa+ B cos B = K cos 0.
Multiplying each term of this equation by OP, we get
(b)
A(OP cos a) + B(OP cos 8) = ROP cos 6).
Now, observing from Fig. 66 that a = OP
OP cos 6, we obtain from (b)
(c)
Aa + Bb = Rr.
cos a, b = OP cos 8, Phe y=
QED.
It should be evident, if more than two coplanar forces are involved, that by
successive applications of Varignon’s theorem we may prove that the algebraic
*Pierre Varignon
(1654-1722), a Frenchman,
and a contemporary of Newton, although
not generally counted among the giants of science, made an extremely important contribution to mechanics in his principle of moments. He is also credited by Girwin in his A
Historical Appraisal of Mechanics as having been the first to evolve the differential equations of motion [equations (33), (34), and (35) of Chapter XIII].
4h
MOMENTS
AND PARALLEL COPLANAR FORCES
[Ch. III
sum of the moments of any number of concurrent coplanar forces about any point
in their plane is equal to the moment of their resultant about the same povnt.
27. Sign of a Moment.
When a body is subjected to a moment, the
tendency of the moment is to turn the body. In Fig. 65, p. 42, for example,
the force F tends to turn the shaft A clockwise.
An upward force acting on
the arm AB would tend to turn shaft A counterclockwise.
In evaluating
the effect of the moments of several forces, we must account for the sense of
each moment.
In practice, it is convenient sometimes to take the clockwise
moment as positive, in which case the counterclockwise moment would be
negative. However, it is Just as easy to take the counterclockwise moment
as positive, in which case the clockwise moment would be negative. ahs
latter rule is usually followed in certain more advanced work, but since it
makes little difference in this course, we shall sometimes use one, sometimes
the other rule.
Fig. 67. Transmissibility and Moment of Components.
In Fig. 67, the moment of the force R about point P is Rr. If force R
is
replaced by its rectangular components R, and R,, the moment
of these
components is Rb — R,a, where the negative sign means that
the moment
f,a tends to turn the body about P in a sense opposite to that
produced by
Rk,b.
From Varignon’s theorem,
Rb. — Rye = Rr.
Using the principle of the transmissibility of a force
(§ 4), we may break
up & into two rectangular components at the point
O’, where the line of
action of R,’(= R,) passes through P. Then, Rr
= R,'c, where R,’ is the
same in magnitude as R,. This method of handli
ng the moment of a force
is sometimes convenient when the distance
c is easier to determine than is
the distance r.
§ 29 | RESULTANT
OF PARALLEL COPLANAR
FORCES
45
28. Example.
The crank shown in Fig. 68 has acting on it a 200-lb. and a 300-Ib.
force as shown.
What is the moment of these forces about the axis A and in what
direction will the crank turn?
Sotution.
In order to use the principle
that the moment of a force is equal to
the sum of the moments of its components
(with respect to the same axis), we find
18, = OM sin AG? ss ANT My.
LL
hie 200 Ib.
B
The value of F is not needed because its
Fig. 68.
moment about A is zero. Letting a clockwise moment be positive and summing moments about A, we get
=Ma = (141)(20) — (800)(10) = —180 in-lb.
Since the answer is negative, the net moment is in the counterclockwise sense, the
direction in which the crank will turn if it moves.
29. Resultant of Parallel Coplanar Forces.
Let two parallel vertical
forces Ff; and F, act on the member AB at the points A and B as shown in
Fig. 69.
To find the resultant of these forces 7; and F2 using the principles
already developed,
apply equal and opposite
x
Q
forces J and J’ at A and B
(eH
a)=! —-—--—--—S
i>
ae
x
Fig. 69.
Resultant of Parallel Forces.
respectively. Since these forces J and J’ are also collinear, the equilibrium
or other state of the member AB is not disturbed.
Next, find the resultant
R, of the forces J and F, and the resultant R» of the forces J’ and Fy. Then,
using the principle of the transmissibility of a force ($ 4), move R; and R,
to act at the point C, which is the intersection of their lines of action. With
R, and R, acting at C, replace these forces with their vertical and horizontal
components.
The horizontal components J and J’ at C neutralize each
other as before (as far as the external effects are concerned), and the vertical
46
MOMENTS AND PARALLEL COPLANAR FoRCES [Ch. III
components are collinear through C. Thus, the point C defines the line of
action: of the resultant R = Ff, + F2 of the original forces Fi and Fo. it
the point of application of R on the member AB is somewhere along the line
of action through C, the force R acting alone will produce the same external
BS
effect that the forces /, and I’, produce without R.*
This method may be extended to cover any number of forces by combining
the resultant R just found with some other vertical force /’3 (not shown) to
However, it is too
obtain a new resultant in exactly the same manner.
Fig. 69.
Repeated.
cumbersome as compared with other graphical procedures to be described
later, or as compared with the algebraic method.
Now, considering the similar triangles CKB and BNG (Fig. 69), we have
(d)
Fy _ OK
NG ~ KB
Also, from the similar triangles CK A and AM E, we get
G
F,
CK
ME” KA
Notice that NG = ME
and (e) that
(f)
(Fig. 69).
Then, by division, we
Eee
eek
find from
(d)
ds
Ay
a relation that may be used to establish the locati
on of the point K on the
line of action of R.
*The “external effect?’ refers to the motion
that might be imparted to the member if
it were free to move, or to the bearing reactio
ns if the member were a shaft, for exampl
e.
§ 30 ] EXAMPLE
47
To establish a more general algebraic procedure, consider the point O as a
center of moments.
Then the moment of the resultant R about O is
Mo
—
R(KA
+
21)
=
(ip
=
(Fy, +
F.)(KA
since R = Fi + F, and KA +2, =7, Fig. 69.
taining /'; + F» in this expression, we get
(g)
Mo
oa Fix, aie F\(K A) =F F.(KA)
+
21),
Expanding the term con-
+ Fon.
Then, using
p, = F(KB)
(KA)
from equation (f) for the second F’; in (g), we have
Mo
=
Fix
a
F(KB
+
Foxe,
fb KA
—
21) => Fig
4
F.(BA
+
Lis
which proves that
Rr
=> Fix
where x2 = BA + %, Fig. 69; or in words, it proves that
The algebraic sum of the moments of the forces F, and F, about a center of
moments is equal to the moment of their resultant about the same center.
A similar procedure could be used to show that the algebraic sum of the
moments of any number of coplanar parallel forces about any point in their
plane is equal to the moment of their resultant about the same point. Varignon’s theorem points to the same conclusion.
So we see that to get the magnitude of the resultant of a group of parallel
forces, we make an algebraic sum of the forces,
(h)
VRP
de
and to get the location of the line of action of the resultant, we choose any
convenient center of moments O and equate the moment of the resultant to
the sum of the moments of the individual forces,
(i)
where
Ries,
=Fx = Fix; + Fox. + + + * and the symbols are defined in Fig. 69.
Of course, the sign of a term in the =x defines the sense of the moment.
Four wheels A, B, C, and D apply loads of 3000, 6000, 10,000 and
30. Example.
15,000 lb., respectively, to a beam GH as shown in Fig. 70. Find the resultant of
these loads.
Sotution.
tically; thus
The magnitude of the resultant is obtained by summing forces ver-
R = =F, = —3000 — 6000 — 10,000 — 15,000 = —34,000 lb.,
MOMENTS AND PARALLEL COPLANAR FORCES [Ch. III
48
where the negative sign means only that the force is acting down.
To define R com-
pletely, we locate its line of action by using the principle that the algebraic sum of the
moments of the forces is equal to the moment of their resultant. Choosing any point
for a center of moments, say point A, we get (clockwise positive)
=M 4 = (3000)(0) + (6000)(6) + (10,000) (14) + (15,000) (24)
= 36,000 + 140,000 + 360,000 = 536,000 ft-lb.
3000!Ib.
6000"Ib.
so
!
10,000! Ib, |
|
15,000'Ib,
r=15.76 “4,
——
R=34,000 Ib.
Fig. 70.
Since Rr = 536,000 = (34,000)r, we get r = 15.76 ft. Therefore the resultant
F& = 34,000 Ib. acts vertically downward through a point 15.76 ft. from A, as shown
in Fig. 70.
31. Example. Three parallel forces A, B, and C are acting as shown in Fig. 71.
Determine their resultant.
SoLution.
This example illustrates the significance of the signs of the moments.
As before, the magnitude of the resultant is
R = =F
= 600 + 100 — 300 = 400 lb.,
acting in the same direction as the 600-Ib.
and 100-lb. forces.
Now, taking counterclockwise moments as positive, and choosing
C as the moment center, we get
=Mc = —(300)(15) + (100) (5)
= —4000 in-lb.,
Fig. 71.
where the negative sign indicates that the
resultant must act in such a manner as to
produce a clockwise moment about C.
Then the length of the moment arm r of R
is obtained from
Rr = 4000 = 400r,
orr =
10in.
We now know:
- that the line of action of R is 10 in. from (Bi
. that R acts downward, and
R must produce a clockwise moment about G.
. that
CE
oe
Therefore, the vector R must be located on the right
of point C, as shown in Fig. 71.
Satisfy yourself that R cannot be on the opposite
side of C. The same
location for the line of action of R would be
obtained by using any other
§ 33 | CHARACTERISTICS
OF COUPLES
49
center of moments.
For instance, solve this problem by using B as a center
of moments.
32. Couples.
When two parallel forces are equal but opposite in sense
and not collinear, they form a couple.
The moment of a couple Mc is the
product of one of the forces F and the
perpendicular distance between the
lines of action of the forces 7; or Me
= Fa, where a is the moment arm of
the couple (Fig. 72). An examination
of Fig. 72 shows that a couple turns
or tends to turn a body. For a body
acted on by a certain couple to be in Fig. 72. A Couple. A counterclockwise
couple as shown.
When viewed from the
equilibrium, it is essential that another
other side of this page, it is a clockwise
couple of the same moment but of couple.
opposite sense be acting also.
The properties which define a couple are:
1. the magnitude of its moment, which is usually expressed in in-lb. or
ft-lb.
2. the rotational sense or direction, as clockwise or counterclockwise;
3. the orientation of the plane of the couple in relation to the body upon
which it acts (see Fig. 73).
A couple may be represented by a
vector perpendicular to the plane of the
couple, the length of the vector representing the magnitude of the moment
The sense given to the
of the couple.
Usually the
vector 1s conventional.
The
Fig. 73. Orientation of Couple.
effect of the orientation of the plane of
The
the couple is suggested above.
forces F are acting perpendicular to the
paper to form a couple Fa. The beginning student can probably sense intuitively that the couple in (b) would
produce a greater turning moment about
the x axis than the couple of the same
magnitude in (a). The orientation of
the plane of the couple may be defined
:
by the angles 6 and 0’.
vector points away from that side of
the plane of the couple from which
the couple appears to be counterclockwise. Thus, the vector representing the
couple in Fig. 72 would point toward
the reader, perpendicular to the plane
This rule is often called
of the paper.
the right-hand screw rule, because the
vector for the couple points in the direcsense of
tion that a right-hand screw would move if it were turned in the
Test this notion mentally for Fig. 72.
the couple.
(a) The moment of a couple is the same
33. Characteristics of Couples.
as shown in
about any point in its plane. Let a couple act on the lever AB
MOMENTS AND PARALLEL COPLANAR FORCES
50
Fig. 74(a).
The moment
[Ch. III
of the two forces F and F about any axis Aa
distance x from the nearest force F is
=M,
= F(a +h)
— Fe = Fx + Fh — Fu = Fh,
where Fh is recognized as the moment of the couple.
As a corollary of this
proof, we may state that the couple may be placed in any position in its
Fig. 74.
Moment of a couple is a constant.
plane with no change in the external effects.
action are as shown in Fig. 74(b). Again,
Thus, suppose
the lines of
F(h + 0) — Fo = Fh,
the moment of the couple. Note that the orientation of the plane of the
couple is the same in Fig. 74(b) as in Fig. 74(a).
(b) A couple may be transformed into another couple that produces the same
external effect. At any convenient points a and b on the lines of action of
the forces F’, F forming the couple Fh, Fig. 75, replace each of these forces
Fig. 75.
Transformation of a Couple.
by the components F sin 6 and F cos @. The components
F cos 6 are equal,
opposite, and so taken as to be collinear, and therefo
re equalize one another,
There is left the couple formed by the forces F sin
6 = P, the moment of
which is
(F sin 6)(h/sin 0) = Fh,
the same as the moment of the original couple. Theref
ore, the external effect
of the couple P(h/sin 4) is the same as that of the
couple Fh.
§ 33 ] CHARACTERISTICS OF COUPLES
51
Suppose F = 100 lb. and h = 15 in.
The moment of the couple is then
Mc = (100)(15) = 1500 in-lb.
The conclusion from the foregoing demonstration is that any other couple
in the same plane and in the same sense with the same magnitude will produce the same external effects; for example,
F = 500 Ib.
T= 10 1b,
and
and
h =3in.;
fh = 150. 10%, ete;
Moreover, if the principle of the transmissibility of a force is applied to
Fig. 75, it is seen that the points of application of the forces of a couple may
be located anywhere on a body as long as the magnitude, sense, and the
plane in which the couple acts are not changed.
(See also Fig. 74.)
(a)
Fig. 76.
(b)
Resolving a Force into a Force and a Couple.
(c) A force may be resolved into a force and a couple. Let the force PF
(solid vector) be acting on body A, Fig. 76(a). To replace this force by a
force and a couple, choose any point a and apply equal, opposite, and collinear
forces, each equal to and parallel to the original force F. Since these forces
cancel, there is no change in the external effects. Evidently, we now have
a force F acting upward to the right through point a and a couple Fh. The
forces of the couple may now be changed and the couple shifted as desired,
for example, as in Fig. 76(b), where Pp = Fh. The line of action of F in
Fig. 76(b) must not be changed, but the process may be repeated with a
point other than a, which would result in a different couple.
It follows from this demonstration that a force and a couple may be com-
bined into a single force.
The student should demonstrate this principle to
himself and observe that the resulting force is the same in magnitude and
sense as the original force but with a different line of action. We also deduce
that a single force and a couple cannot balance one another.
For example, in Fig. 76(a), let 7 = 1200 lb. To resolve F into a force
52
MOMENTS AND PARALLEL COPLANAR FORCES [Ch. III
and a couple, choose some convenient point a, whose perpendicular distance
from the line of action of F is h = 6 in., and apply equal and opposite forces
F = 1200 lb. The couple FA is (1200)(6) = 7200 in-lb. So the original
force F in Fig. 76(a) may now be replaced as in Fig. 76(b) by the force
F = 1200 lb. and the couple Pp = 7200 in-lb. clockwise.
(d) A couple may be transferred from one of two parallel planes to the other.
This principle is the outcome of observation.*
For example, the length of
the rear axle of an automobile has nothing to do with the turning moment
delivered to the rear wheels; or, the steering wheel on a car may be on a long
steering column or a short one, with no change in the moment required to
turn the front wheels.
(e) Since couples of the same moment, sense, and orientation plane are
equivalent, paragraphs (b) and (d), the resultant of several couples in the same
plane or in parallel planes has a magnitude equal to the algebraic sum of the
moments of the several couples. ‘Thus, suppose the couples A, B, C, and D
acting on a body are:
M4
Mp
Me
Mp
I 100 in-lb. clockwise;
=
= 250 in-lb. counterclockwise;
= 60 in-lb. clockwise; and
= 30 in-lb. clockwise.
The total clockwise moment is
M4 + Me + Mp = 190 in-lb. The total
counterclockwise moment is M, = 250 in-lb. The resultant couple is
Mr = 250 — 190 = 60 in-lb.
in a counterclockwise
sense because the total counterclockwise
ceeds the total clockwise moment.
34, Example.
Replace the force
the axis and a couple.
moment
ex-
F = 300 lb. in F ig. 77(a) with a force through
C = 4230 in -lb.
Fig. 77.
Sotution.
Introduce at the axis two equal, Opposite, and collinear
forces F =
300 lb., shown dotted, Fig. 77(a). We now have a couple
Fd = (300)(14.1) = 4230
in-lb. and a force F = 300 Ib. acting through the axis
downward toward the left. As
p.
ae may
18 .
be proved however.
See Engineering
Mechanics,
Timoshenko
and Young
§ 36 | EQUILIBRIUM
OF COPLANAR
PARALLEL FORCES
58
suggested by Fig. 77(b), we may say that the original force F produced a force
at the shaft of 300 Ib. and a turning moment of C = 4230 in-lb.
35. Resultant of a Coplanar Parallel Force System in Which 2F = 0. If
a system of parallel coplanar forces is not in equilibrium but =F (in the direction of the forces) = 0, the resultant is a couple.
F,=200 Ib.
F,=100 lb.
Fig. 78.
Resultant isa couple.
F, =150 lb.
F=250 Ib.
A graphical solution for this type of problem is explained
in Chapter VII, Fig. 253.
Consider Fig. 78, in which a member AB is acted upon by the forces shown.
Although the ZF = 0, the moment of the forces about point C is
Mc = (200)(8) — (250)(18) + (150)(30) = +1600 in-lb.,
the positive sign indicating that the resultant moment is in the same sense as
the moment of F2 (counterclockwise about C). Taking moments about
point D, we get (counterclockwise positive)
>Mp = (100)(18) — (200)(10) + (150)(12) = +1600 in-lb.,
These two computations constitute a numerical verifithe same as before.
cation of the statement that the resultant is a couple, in this case, 1600 in-lb.
Since the moment of a couple is the same about any
counterclockwise.
point in its plane, we would obtain the same result in taking moments about
any other point in Fig. 78. Hence, if SF = 0 ina parallel system of forces
and if the resultant is not zero, then the resultant is a couple, the magnitude
and sense of which may be obtained by taking moments about any convenient
point.
Since, as we have learned,
36. Equilibrium of Coplanar Parallel Forces.
couple, two
the resultant of a parallel-force system may be either a force or a
which the
in
conditions are necessary to define the state of equilibrium,
a force,
resultant is zero. In order that the resultant should not be
(7)
Si
0.
In order that the resultant should not be a couple,
(8)
=M = 0,
point. The
where the sum of the moments 2M may be taken about any
used to find
be
may
and
dent
conditions defined by (7) and (8) are indepen
b4
MOMENTS AND PARALLEL COPLANAR FORCES [Ch. III
two unknowns.
It is recalled that any number of moment equations may
be obtained by taking moments about various points. However, no matter
how many moment equations we may have, we shall find that only two
independent equations exist. A moment equation may be substituted for
the equation (7). Thus if two points A and B are chosen as moment centers,
the equations
2Ma=0
and
2M,
=0
will be independent relations when both A and B do not lie on a line parallel
to the force vectors.
37. Example. A beam AB supporting loads as shown in Fig. 79 is resting on end
supports. What are the reactions R; and R» at the supports?
F,=500 Ib.
F,=3001b.
F,= 700 Ib.
Ri
R,
Fig. 79.
Sotution. A free body of the beam is shown in Fig. 79. To get the reaction R,
at the right support, take moments about the left support. This gives
=M.4 = 10F, + 22F, + 30F; — 36R, = 0
= (10)(500) + (22)(300) + (30)(700) — 36R, = 0,
from which R2 = 906 lb. To get the reaction R, at the left support, take moments
about the right support.
=M,p
=
6F,
+ 14F,
+ 26F,
—
36R,
== (()
= (6)(700) + (14)(300) + (26)(500) — 36R, = 0,
from which Ri = 594 1b.
To check these computations, sum the forces vertically:
=F, = Ry — Fi — Fe — F3+ R,
ll 594 — 500 — 300 — 700 + 906 =
0.
The sum is zero, as it should be for equilibrium conditions.
It is advisable in practice
to find the unknown reactions by moment equations, as illustrated
in this example,
and then to check the computations by seeing that 3F = 0.
If SF is not zero, an
error is indicated.
38. Moment and Shear at a Section.
In designing members of machines
and structures, we often need the moment at a particular
section of the member.
Suppose the moment at the section C, Fig. 79, is
desired.
Cut the beam
at this section and make a free body of either part,
say the left section.
This
free body is shown in Fig. 80. To balance the
effects of the forces Ri = 594
lb. and F = 500 lb., observe that a counterclock
wise moment (couple) M
must be applied at C to balance the clockwise moment
produced by R, and Fy
§ 39 | EXAMPLE
55
and that a vertical force S, called the shear, must be applied to balance the
vertical forces. The moment M and the force S, which are necessary for
equilibrium, represent the action that the right part of the beam had on the
(From a vertical
left part of the beam before the right part was detached.
the section in
Since
downward.)
directed
is
sum, it is seen that S actually
Fig. 80 is in equilibrium, the sum of the
moments about any point is zero.
Thus,
taking moments about C, we get (clockwise
positive)
F,= 500 Ib.
DMc = 15k: —5F, — M =0,
R,= 594 lb.
where the moment of the couple M is
necessarily included to produce equilibrium.
A solution for M gives
pig g9,
Section.
15
Bending
Moment
at a
M = 15R: — 5F, = (15)(594) — (5)(500) = 6410 in-lb.,
which is the counterclockwise moment that must be inserted at C to maintain
The magnitude of this moment
this part of the beam in equilibrium.
(6410 in-lb.) is called the bending moment at C. The student should make a
free body of the other part of the beam and find the same result for the
moment at C. From this illustration, we may state a simple rule.
When the forces are perpendicular to the axis of the beam, the bending
moment at any section of a beam is the sum of the moments of all the
forces to the left (or to the right) of the section about any point (axis)
in the section.
The shear is obtained by summing forces vertically, Fig. 80; thus
SF =
Rk, — Fi+
S = 594
— 500+
S = 0,
from which S = —94 Ib., the shear, the negative sign indicating that S was
assumed in the wrong sense.
A simple beam is loaded as shown in Fig. 81. Determine (a) the
39. Example.
and (b) the bending moment at the support B.
R»
reactions R, and
(a) The uniform load of 400 Ib. per ft. may be replaced by a vector W
SotuTion.
Ib. per
acting through the center of gravity (eg) of this distributed weight. At 400
ft. for 16 ft., the total load
W = (400)(16) = 6400 lb.
end
The vector representing this load is shown dotted acting 9 ft. from the left-hand
A,
to
respect
with
forces
all
of
moments
taking
and
vector
Using this
of the beam.
the point of support of Ri, we find (counter clockwise positive)
5M 4 = — (6400)(5) — (4000)(9) — (6000) (26) + 20R2 = 0,
MOMENTS
56
AND PARALLEL COPLANAR FoRCES
[Ch. III
from which R, = +11,200 lb., the plus sign indicating that we correctly assumed the
sense of R» and that it acts up as shown in Fig. 81. The reader should check the
moment arms in the foregoing equation.
Another moment equation should be used to find R,. If B, the point of support of
Rz, is used as the center of moments, the moment of f» will not be involved in the
ees iorW
if
12°
4000 Ib.
6000 lb.
!
Fig. 81.
equation and the answer found for FR, will not depend on the accuracy of the solution
for Ry Accordingly, summing moments about B, we get (clockwise positive)
=M.4 = 20R, — (6400)(15) — (4000)(11) + (6000)(6) = 0,
from which R; = +5200 Ib. The positive sign signifies that R, is shown in its correct sense. It is advisable now to sum the forces in the vertical direction to see if
the sum is zero, as 1t must be if the above solution is correct.
=F, = Ri — W — 4000 + R» — 6000 = 5200 — 6400 — 4000 + 11,200 — 6000 = 0.
Since this sum is equal to zero, we have a check on our solution and R; and R» are
correct as found.
(b) The bending moment at B is the sum of the moments of all the forces to the
left of B or to the right of B. Since there is only one force to the right of B, it is
simpler to use the part BC as the free body for the purpose and find
Ms = (6000)(6) = 36,000 ft-lb.
The student should find the same answer for Ms by using the left-hand part of the
beam.
40. Torque.
Torque is the moment of a force, or forces, about an axis of
rotation.
For example, if A and B (Fig. 82) represent two gears and if F
represents the driving force exerted by gear A on B,
the torque on the shaft for B is F(D/2).
Again, in
Fig. 65, p. 42, the torque on the shaft A produced by
the force F on the lever is Fh = Fa.
If F in Fig. 82 is the force of gear A on gear B, then
the force of gear B on A is equal but in the opposite
sense to ?. Therefore the torque on shaft A is F (d/2).
We see that the torque delivered to shaft A (by a
motor, for instance) is increased by the ratio D/d
on shaft B. Thus, by varying the relative sizes of
gears A and B, we may increase or decrease the torque
Fig. 82.
Torque.
on shaft B, even though the force F remains constant.
PROBLEMS:
57
MOMENTS
Since a shaft, for instance, may have a bending moment and a torque on it,
the special name of torque for a turning moment is useful.
In this chapter, note carefully the following ideas.
41. Closure.
i The moment arm of a force is always perpendicular to the line of action
of the force.
_ The moment of a force has no meaning unless the center of moments
is specified.
It is often convenient to use the moments of the rectangular components
(Do not overlook
of a force in place of the moment of the force itself.
It will often save time.)
_ The resultant of an unbalanced system of parallel forces is either a force
or a couple.
_ The moment of a couple is the same for all locations of the center of
this one.
moments.
_ There are two independent conditions for a system of coplanar parallel
= 0,
forces in equilibrium, ZF = Oand 2M = 0 (or, 2M. = Oand 2Mz
forces).
the
to
parallel
line
same
where both points A and B are not on the
forces in
_ When two unknown forces in a group of coplanar parallel
a
by moment
equilibrium are to be found, it is advisable to find each force
tions.
equation, so that ZF = 0 may be used as a check on the computa
wrong
the
using
from
arising
errors
This procedure does not exclude
force system to start with.
Problems
MOMENTS
151. The hand crank on a tractor engine
is 15 in. long. When idle, the crank hangs
If a man lifts vertically updownward.
crank handle with a force of
the
ward on
d
100 lb., what turning moment is imparte
is (a)
to the crankshaft when the handle
its
from
180°
(d)
and
120°,
90°, (b) 15°, (c)
vertical downward position?
1299
Ans. (a) 1500 in-lb.; (b) 388 in-lb.; (c)
in-lb.; (d) 0.
e
152. The hand crank on a tractor engin
hangs
crank
the
idle,
When
is 11 in. long.
downward.
If a man
lifts vertically up-
of
ward on the crank handle with a force
e) 1s
50 Ib., what turning moment (torqu
imparted to the crankshaft when the handle
is (a) 30°, (b) 60°, (c) 90°, (d) 175° from its
vertical downward position.
’
Fig. 65.
Repeated.
Problems 153-155.
153. In Fig. 65, a = 3 ft. 6 in., 6 = 30°,
and F = 300 lb.
By two different ways,
find the moment about point A.
Ans. 525 ft-lb. ¢.
154. The same as 153 except that 6 =
Ans. 909 ft-lb. c.
120°.
155. The same as 153 except that @ = 60°.
PROBLEMS:
58
156. In Fig. 83, Q = 200 lb. and a =
60°.
Find the moment of Q about point A
by (a) computing the moment arm from A
to the line of action of Q, (b) by breaking
MOMENTS [Ch. [II
is lOO Ib:
(a) Find the moment
(torque) of each force about the shaft center.
(b) Find the resultant of 7; and F,2 and
check its torque on the shaft against the
algebraic sum of the moments of the individual forces.
Ans. (b) 8310 in-lb. c.
160. The same
ome
as 159 except that @ =
161. (a) Two forces act on a crank AB,
Fig. 84, to cause a counterclockwise moment
(torque)
on
the shaft
of 400
in-lb.
If h = 16 in., find 6 and Fy. (b) Is there
more than one solution?
Ans: (a). Hor 6) — 90 sehie— 5448 pasa)
Infinite number.
oe
Fig. 83.
Problems 156-158.
Q into its vertical and horizontal
nents, and
compo-
(c) by breaking Q into rectan-
gular components one of which is along
line AC.
Ans. 4075 in-lb. c.
157. The same as 156 except that a =
iy
158. The same as 156 except that a =
45°.
Fig. 85.
Problems 162, 163.
162. A homogeneous parallelepiped rests
on an inclined plane as shown in Fig. 85.
If @ = 30°, show by numerical computation
that the moment of the weight W = 1000
lb., which acts through the center of gravity,
is the same as the moment of the components
of W (shown dotted) about the point A.
163. A homogeneous parallelepiped rests
on an inclined plane as shown in Fig. 85.
Fy, 600 lb.
Fig. 84.
159. Two
Problems 159-161.
forces act on a crank
Fig. 84, where
h = 16 in,
AB
in
0 = 0°, and
RESULTANT
OF
164. Two parallel forces, 7; = 40 lb. and
F, = 150 lb., act in the same sense with
their lines of action 5 ft. apart.
Determine
the magnitude and line of action of the
resultant.
Solve algebraically and graphleally.
Ans. 190 lb., 3.95 ft. from Fy,
165. Two parallel forces, F; = 18 oz. and
F, = 72 02., act in the same sense with their
lines of action 9 in. apart.
Determine the
magnitude and line of action of the resultant. Solve algebraically and graphically.
166. The resultant of two parallel forces
which act in the same sense is 900 Ib. The
If @ = 60°, show by numerical calculation
that the moment of the weight W = 500
lb., which acts through the center of gravity,
is the same as the moment of the components
of W (shown dotted) about point B.
PARALLEL
FORCES
line perpendicular to the two force vectors
is 9 ft. between the force vectors and divided
into parts in the ratio of 5 to 4 by the result-
ant vector.
Compute the magnitude of the
two parallel forces.
Ans. 500 lb., 400 Ib.
167. The resultant of two parallel forces,
whose vectors are 10 ft. apart and act in
opposite senses, is 700 Ib.
If the resultant
acts 3 ft. from one of the forces, what are
the forces? Show by sketch the relative
positions of these forces.
Ans. 910 Ib., 210 lb.
168. Four soldiers, A, B, C, and D stand
PROBLEMS:
RESULTANT
OF PARALLEL
FORCES
59
8000 lb.
3 ft. apart along and on a 16-ft. long
uniform timber that weighs 500 lb. The
soldiers weigh 150 lb., 175 lb., 200 lb., and
250 lb., respectively, and A stands at one
end of the timber.
Find the location at
which a fulcrum might be placed to balance
the timber.
169. The following forces act vertically
at distances measured along the horizontal
XZ axis as given: fF, = +180 lb., 21 =0;
lb., v3 =
+12 ft.
+7 ft.; and fF, = —110 lb., x =
Determine the resultant.
Ans. 100 lb., x = —6.6 ft.
170. The following forces act vertically
at distances measured along the horizontal
x axis as given: fF; = —3 kips, x1 = —3 ft.;
F. = —4 kips, x2 = 0; Fs; = +5 kips, x3 =
4 ft.; and Fy = —4kips, x4 = 6 ft. Determine the resultant.
6000 Ib.
Fig. 86.
7000 lb. é
Problems 171, 172.
171. A 14ft. lever is acted upon by
forces as shown in Fig. 86. What force F
acting at what distance x will result in a
zero reaction at pin A?
Ans. 12,000 lb., 7.83 ft.
172. The same as 171 except that the
reaction at A is not zero, but the force of
the pin on the lever at A is 1000 lb. downward (parallel to the other forces).
COUPLES
173. Aload W = 500 lb. is suspended by
176. A beam is acted upon by a couple
a cord from a stationary disk as shown in
C (Fy, F2) = 1500 in-lb. and a force Q =
50 Ib., Fig. 89. Replace these effects by a
single force which is specified in sense and
Ans. +50 |b. at —30 in. from Q.
location.
Fig. 87. Replace the load on the disk by
a force through the axis and a couple.
Ans.
—500 lb., 750 ft-lb. ec.
307
Q
Fig. 89.
Fig. 87.
177. The same as 176 except that C =
400 in-lb.
178. The forces shown in Fig. 89 are
Problem 173.
A
Fig. 88.
Problems 176-179.
Problems 174, 175.
174. A load of F = 4 kips acts on an
L =8
ft. cantilever beam, Fig. 88. Replace F by a force at section A-A, where the
beam enters the supporiing wall, and a
couple.
Give sketch showing the couple
and force.
Ans. —4 kips, 32 ft-kips c.
175. The same as 174 except that L =
24 in.
Q = 100 lb., F:1 = 50 lb., and F2 = 150 lb.
Determine the resultant.
Ans. 6000 in-lb. ce.
179. The forces shown in Fig. 89 are
Q = 350 Ib., Fy = 750 lb., and Fz = 1100
lb. Determine the resultant.
180. (a) For W = 8 lb., replace the force
F (Fig. 90) by a force through the axis O
anda couple.
(b) Replace the couple found
in part (a) and the weight W by a single
force.
(c) What reaction occurs at O due
to the single force found in part (b)?
(d)
What single force is acting on the body at
O due to the force found in (a) and to the
force found in (b)? Show a series of free
bodies with the various force arrangements.
Ans. (a) 10 lb. at 30°, 40 ft-lb. c¢; (b) 8 lb.
PROBLEMS: COUPLES [Ch. III
60
down at +4 ft. from O; (c) 8 lb. up and 32
ft-lb. cc; (d) 3 Ib. at 270° and 32 ft-lb. c.
181. The same as 180 except that W =
36 lb.
Fig. 91.
Fig. 90.
Problems 180, 181.
Problem 182.
182. Replace force F = 50 lb. in Fig. 91
by a force through the bolt center and a
couple. How could this torque be applied
so that there would be no lateral force on the
bolt? Ans. 50 lb. at 270° and 600 in-Ib. c.
EQUILIBRIUM
183. On a pair of carpenter pincers, Fig.
92, F = 12 lb. and s = % in. Assuming
coplanar forces, find the force on the cutting
Fig. 92.
spaced 120° apart, Fig. 94. If the wheel is
free to rotate about its axis, what angle @
will the radius to point A make with the
Problem 183.
eS
edges. The pin P is in equilibrium under
the action of two equal forces.
What is the
magnitude of these forces?
Ans. 192 lb., 204 Ib.
Fig. 94.
vertical
when
Problems 185, 186.
the wheel
is at rest (stable
position)?
Ans. 30°.
186. The same as 185 except the weight
of B is 20 lb.
Fig. 93.
Problem 184.
184. A load of W = 2 tons needs to be
raised by a crowbar as shown in F ig. 93.
If a 200-lb. man rests his full weight as
force F, how long must dimension L be to
accomplish this job? Assume all surfaces
smooth.
185. Three weights, A = 5 Ib, B= 10
lb., and C = 15 lb., are attached to the
rim of a uniform wheel at a 2-ft. radius and
Fig. 95.
Problem 187.
187. Three weights are suspended in the
same plane, as suggested by Fig. 95. (a)
PROBLEMS:
61
EQUILIBRIUM
What should be the value of W if 2M,
=
0? (b) Replace each of these weights by a
force through O and a couple.
What is the
resultant couple?
What is the resultant
force at O?
Ansae(a)e
400) lbs
(b) 5G = 0 ftlbs Ry =
1000 Ib.
the axles is F; = 4200 lb., F2 = 5100 lb.,
and F; = 3800 lb. A concentrated load
W = 6 tons must be hauled.
(a) Calculate
the least distance x the load must be moved
forward on the trailer in order to comply
with a law limiting axle loads to 5 tons.
(b) For the location of the load as found in
(a), determine the load on each axle.
Ans.
(a) 11.87 ft.; (b) from front to rear,
5525 Ib., 9970 lb., 10,000 lb.
194. The same as 193 except that the
permissible axle loading is 14,000 lb.
Fig. 96.
Problems
188-190.
188. In Fig. 96, if F = 100 lb. and the
force on A is to be 1 ton, find z and the reaction at roller B. All surfaces are smooth.
189. In Fig. 96,
reaction at roller B
lb.
(a) What must
maximum force on
F = 100 lb. and the
must not exceed 1800
be the distance z for a
A?
(b) What is the
bending moment in the lever at B?
190. The same as 189 except let F =
15 lb.
Ans. (a) 0.05 ft.; (b) 89.25 ft-lb.
Fig. 99.
Problem 195.
195. The arrangement in Fig. 99 may
be balanced on roller A. Let F = 200 lb.,
let the 18-ft. uniform beam weigh 10 lb.
per ft., and neglect the weights of the
pulleys and rope. Calculate the distance x
for this balance.
95
==
iG
;
Fig. 97.
To Atmos
Problems 191, 192.
Cylinder
191. Two weights, A = 30 lb. and B =
15 lb., shown in Fig. 97, are attached to a
bell crank whose weight is neglected.
At
what two values of the angle @ are these
weights in equilibrium?
Which of these
two positions is the more stable state of
equilibrium?
Explain.
Ans. 36.9°, 216.9°.
192 The same as 191 except that B
weighs 70 lb.
193. A trailer-truck is shown in Fig. 98.
Without a “payload” the reaction under
Fig. 98.
Piston
Connected to
Shirly
Boiler
Fig. 100.
Problems 196, 197.
196. One type of safety valve for steam
boilers operates in principle as indicated in
Fig. 100. One end of a small cylinder is
connected by a pipe to the boiler. The
piston in this cylinder is acted upon by
Problems 193, 194.
PROBLEMS: EQUILIBRIUM [Ch. III
62
steam from the boiler. Under normal operation, the moment of the weight W is
sufficient to prevent the piston from moving
upward and allowing the steam to flow from
Let W = 210
the boiler to the atmosphere.
lb. and D = 4 in., and determine the boiler
pressure which would cause the piston to
be on the point of moving, if the friction is
neglected. What is the force acting on the
pin at A for this condition?
Ans. 100 psi, 1050 lb.
197. It is desired that the safety valve
shown in Fig. 100 start opening when the
boiler pressure becomes p = 150 psi. The
diameter, D, of the piston is 6 in. What
should be the weight W if friction is neglected? See problem 195 for a description
of the action of this safety valve.
200. A simply supported beam is loaded
as shown in Fig. 102. Load Q =0 and
xz =6 ft. (a) What are the reactions R,
and R:? Solve by moment equations and
check by >F =0.
(b) Find the bending
moment at the reaction R; Check this
answer by using the alternate part of the
beam.
Ans. (a) 1175 lb., 11,375 lb.; (b) 1300 ft-lb.
201. The same as 200 except that Q =
5000 lb.
202. The same as 200 except that z =
B iit,
203. Figure 103 represents a stringer AB
of a railroad bridge with its wheel loads C,
D, and E, as shown.
What is the distance
x when the left reaction is three-fourths of
the total load?
Is this the maximum value
of R,?
Ans, 13.5) the, m1Os
204. The same as 203 except let Ri =
60% of the load.
Flexible Cord
Fig. 101.
Problems 198, 199.
19ST (a) ine hice Opt Oe—s1 0,000) be
determine the weight W and the reaction
at B if the link AB isin equilibrium.
There
is no friction at the pulley.
(b) What is
the bending moment at the section where
the 6000 lb. load is applied?
Ans. (a) 6000 lb., 0; (b) 20,000 ft-lb.
199. The same as 198 except that Q =
21,000 Ib.
Fig. 104.
Problems 205, 206.
205. What are the reactions at the supports of the bridge truss shown in Fig. 104
if the loads are F; = 25 kips and F, = 35
kips?
Ans. 36.25 kips, 23.75 kips.
206. If the load F; = 10 tons and the
reaction under the smooth
for the bridge
truss
7000 Ib.
Fig. 102,
Problems 200-202.
Fig. 103.
Problems 203, 204,
roller is 15 tons
shown
in Fig.
104,
PROBLEMS:
63
EQUILIBRIUM
find the load
F. and
the reaction
at the
left support.
Fig. 105.
Problems 207, 208.
207. What are the reactions at the supports for the truss loaded as shown in
Fig. 105 when F = 4000 lb.?
Ans. 1000 lb., 15,000 Ib.
208. The same
as 207 except that F =
14,000 lb.
25,000 lb.
50,000 Ib. each
Brandi
rAns: C, —9——250 lbs, Br= 350ilb.
211. The same as 210 except let F =
18 kips.
Fig. 108.
Problem 212.
212. A bicycle pedal and sprocket are
shown in Fig. 108. If the rider weighs
W = 100 lb. and exerts his weight W vertically downward, find the tension 7; in
Wie Gigin i (©) C=] Cr, (bd)
OS a, ©C
@ = 150°, and @) 6)= 180°.
Notice that
32,500 lb each
wi
Fig. 106.
209. A locomotive turntable weighing 60
tons is represented in Fig. 106.
It is desired to run a locomotive and tender with
wheel loads as shown onto the turntable in
such a position that the support reactions
R, and R» are equal.
(a) What should be
the value of x? (b) For this value of 2,
determine the bending moment at the center
of the beam.
Ans. (a) 11.61 ft., (b) 1,400,000 in-lb.
Problem 209.
the tension 7's in the slack side of the chain
1s zero.
— Smooth
vy Bell Crank
90.
Fig. 109.
Problems 213, 214.
213. An ore car B which weighs 15 tons
is balanced by a weight A as shown in
Fig. 107.
Problems 210, 211.
210. On the bell crank in Fig. 107, let F =
100 lb. and 6 = 30°. Find the reactions at
Fig. 109. The angle of the incline is 6 =
60°. What should be the weight A if there
is no friction at any point? Ans. 25.98 tons.
214. The same as 213 except that @ =
207
PROBLEMS: EQUILIBRIUM
64
215. The member AB
upon
(Fig. 110) is acted
by the forces shown, and x = 2 in.
(a) What
are the forces Ff; and PF» on the
lines of action
a-a
and
218. The same
[Ch. III
as 217 except that the
arrangement in Fig. 113 is to be used.
Ans. 200 lb.
b-b, respectively,
500 lb.
AR
x
b
200 Ib.
800 lb.
Fig. 110.
Problem 215.
which will maintain equilibrium?
(b) What
is the bending moment at the 800-lb. force?
Fig. 114.
Fig. 113.
Problem 218.
Problem 219.
At the 200-lb. force?
Ans.
(a) 3750 lb., —3250 lb., (b) 0, ale
in-lb.
4000 Ib.
3500 lb,
Fig. 111.
216. The
in Fig. 111.
beam
AB
is loaded
as shown
the weight W = 4000 lb. suspended from
a pulley.
What should be the diameter of
the pulley?
ALS, WP ith,
Problem 217.
217. In the system of sheaves shown
5000 lb.
220. The principle of operation of a differential hoist is illustrated in Fig. 115, and
a photographic illustration of a chain hoist
is shown in Fig. 116.
As the upper double
Fig. 115.
in
Fig. 112, what force F will hold a weight
of W = 800 lb. in equilibrium?
There are
no frictional losses at the axes.
the
Problem 216.
Equilibrium is maintained by
Fig. 112.
219. The same as 217 except that
arrangement in Fig. 114 is to be used.
Problems 220, 221.
pulley rotates, say clockwise, the cord winds
onto the larger pulley of diameter Dries
115, and unwinds
from the smaller pulley
of diameter d, in which case the weight
W
PROBLEMS:
EQUILIBRIUM
moves up. What force F will hold W =
1000 lb. in equilibrium if D = 18 in. and
d =16in.?
There are no frictional losses.
Ams) 55.0 lb;
65
bell crank are a = 20 in. and b = 12 in.
Neglecting friction, determine the normal
force between the wheels A and B when
=" 60! libs
Fig. 117.
Problem 222.
223. Two horizontal members AC and
CE are supported as shown in Fig. 118.
If a weight of W = 70 lb. is suspended at
E and if a = 20 in., what is the reaction at
Fig. 116.
Fig. 118.
Differential Chain Hoist.
221. Derive an equation which expresses
the force F in terms of W, D, and d for the
differential hoist of Fig. 115. See problem
220 for a description of its operation.
222. A bell crank is used to force wheel
A against B, Fig. 117. The arms of the
A?
Problems 223, 224.
The members
have a uniform weight
of 51b. per ft. of length. Ans. 535 lb. at 90°.
224. The same as 223 except that a =
28 in.
225-240. These numbers
for other problems.
may
be used
Chapter
[V
NON-CONCURRENT,
COPLANAR FORCES
NON-PARALLEL,
42. Introduction.
In this chapter, we shall learn how to solve problems
involving free bodies subjected to more than three coplanar forces which are
neither concurrent nor parallel. We recollect that if a body is in equilibrium
under the action of two forces, these forces must be collinear (§ 18); so that
if the points of application of the forces are known, the line of action is defined.
We also recall from § 19 that if a body is in equilibrium under the action of
three forces, the lines of action of these forces must intersect at a common
point, a principle previously used to solve such problems.
However, the
methods of this chapter are not only applicable for bodies subjected to more
than three forces, but they are often more convenient than any other method
for three-force members in equilibrium.
43. Resultant of a General Coplanar Force System. It is not only necessary for the engineer to be able to make force analyses of members in equllibrium, which includes most structures and many parts of machines, but also
to be able to find the resultant of unbalanced
force systems,
since it is the
resultant of an unbalanced system that causes an acceleration of a body.
Fig. 119.
Resultant of Three Non-Concurrent Forces.
The resultant of a group of coplanar forces is either a force
or a couple.
Consider the forces Fi, F's, and Fy acting on the body A,
Fig. 119(a), with
the lines of action as shown. If these vectors should be
added vectorially to
66
§ 44] EXAMPLES
67
get the resultant, it would be evident that the horizontal component of the
resultant R would be the sum of the horizontal components of the applied
forces Fy, Fs, and F3;thatis, R, = =F,. It would also be evident that the vertical component of the resultant would be the sum of the vertical components
of the applied forces; that is, R, = DF,. (See § 15.) Now with the horizontal
and vertical components of R known, the magnitude and direction of F is
easily found.
Thus, the first step in finding R is to obtain the sums, ZF’,
and =F ,, from which
k= (2h,
(a)
and
tan 6 =_
(b)
Or,
=F,
See _ f,
Although
where @ is the inclination of the vector R with the horizontal.
(a) and
from
resultant
the
of
sense
and
magnitude
the
determine
may
we
unless
action,
of
line
the
of
location
the
(b), we do not know at this stage
an
from
The location is obtained
the forces happen to be concurrent.
application of the theorem that the moment of the resultant about any point
is equal to the moment of its components about the same point (§ 26). In
the force polygon which we mentally constructed to get equations (a) and
(b), we note that F,, Fz, and F; are a set of components of R. Then, choosing
any convenient center of moments, such as O, Fig. 119(a), we may write
(clockwise positive)
2Mo = Rr = Fya — Fb + Fee.
(c)
With the magnitude of the resultant known from equation (a), this expression
may be used to find r, the distance of the vector R from the point O, which
is the information needed to locate the line of action of R, as shown in Fig.
119(b).
That is
_ =Mo
1 es Wes:
(d)
where =M og is the sum of the moments of the forces with respect to Oand R
is the magnitude of the resultant of these forces. If both 2/7, and =F,
equal zero, and 2M is not zero, the resultant is a couple; that is, when
Re S20
Ry = 2F,=0}/
2M #0
Resultant is couple.
The location of the resultant couple may be anywhere in the plane of the
forces.
44, Examples.
by solid vectors.
(a) A member
AB, Fig. 120, is acted upon by the forces shown
What force acting alone will produce the same external effects?
NON-CONCURRENT, NON-PARALLEL, COPLANAR FORCES [Ch. IV
68
Following the procedure outlined above, we get
Sotution.
=F, = 75 + 30 cos 30° + 100 cos 60° = 151 lb.,
acting toward the right.
=F, = 50 + 30 sin 30° — 100 sin 60° =
—21.6 lb.,
acting down. From the directions in which the components act, we conclude that the
resultant acts downward toward the right. Hence, in finding the angle 6, we may drop
the negative sign for =/',.
Thus,
R = [(2F.)? + (2F,)7]4? = (151)? + (21.6)?]4/? = 152.3 Ib.
=
D>
I tan
=
Uaioe
21.6
Ses
(2) = S105
measured in a clockwise direction from the horizontal.
To locate the line of action R,
choose the point O as a convenient center of moments; then
a = 9 ft., b = 5 sin 30° = 2.5 ft., and c = 3 sin 60° = 2.598 ft. = 2.6 ft.,
approximately.
Letting
of the moments as
counterclockwise
be the positive
sense,
we
get the sum
2M o = Rr = (152.3)r = (50)(9) + (30)(2.5) — (100)(2.6) = + 265 ft-lb.,
from which r = 265/152.3 = 1.74 ft. Since =Mo is positive (+265), the resultant
F must act to produce a counterclockwise moment about O. It must also be directed
downward toward the right, as we learned previously. The only location for R that
satisfies these conditions is shown in Fig. 120.
(b) The forces acting on a member AB are as shown in Fig. 121. Determine
the resultant.
SoLuTion.
=f, = 10 cos 30° — 20 cos 45° +548 =0
=F, = 10 sin 30° — 20 sin 45° + 14.14 —5
Fig. 121.
= 0.
Since the components =F, and SF, of the resultant R are equal to zero, R itself is zero, and
the resultant is not a force. The resultant may
or may not be a couple, depending on whether
or not the sum of the moments of the forces
about any convenient axis is finite or zero.
Choosing O as the center of moments, we get
(clockwise positive)
2M o = (20)(8 cos 45°) — (14.14)(18) + (5)(19) = 4 24.3 in lb.
§ 46 | EXAMPLE
69
Therefore the resultant is a couple with a magnitude of 24.3 in-lb. in a clockwise sense.
Any other center of moments would have resulted in the same answer.
45. Equilibrium of a General Coplanar Force System.
A system of forces
is in equilibrium when the resultant is zero.
Therefore we readily conclude
from the foregoing discussion that the following three conditions must be
true when a general coplanar force system is in equilibrium.
(9)
1
0}
=F, = 0,
=M =0.
The x and y axes may be taken in any convenient directions and the sum of
the moments may be taken about any convenient point.
The proof of these conditions is a matter of logic. If ZF, = 0, there is no
component of the resultant in the x direction.
If =F, = 0, there is no component of the resultant in the y direction.
Therefore if lxoula 2/7, = 0 ganel
=F, = 0, the resultant cannot be a force. If 2M = 0, the resultant is not
a couple.
Consequently, when all three conditions are satisfied, the forces
Observe that with these three independent condimust be in equilibrium.
However, as
no more than three, can be found.
and
tions, three unknowns,
adsubstitute
often
explained in connection with parallel forces, we may
vantageously a moment equation for a summation of forces ($46). Accordingly, the three conditions of equilibrium used in the solution of any particular
problem may be any one of the following combinations:
(1)
=F, =0
=F, =0
2M, =0
(2)
“Fr, =0
=M4 =0
=M, =0
(3)
=M, =0
=M4 =0
=Mp =0,
where the subscripts O, A, B, designate any three suitable axes about which
To put it in words, we may use two force sums in any
moments are taken.
two directions and one moment sum to get three independent conditions; or
we may use one force sum and two moment sums; or we may use three moment
In a manner analogous to that suggested for parallel forces, two or
sums.
more moment equations may be used to find unknowns and then one or two
force sums may be used as a check of the solution.
Although the principles are quite simple, the application is not always easy.
The purpose of the numerous examples throughout this book is to teach the
The student should study these examples by solving them,
application.
checking his solution against the one presented here.
46. Example.
of F = 5000 Ib.
are neglected?
The framework of a crane shown in Fig. 122 is supporting a load
What are the reactions at H and D if the weights of the members
A free body is shown in Fig. 122. Choose point H as a center of
Sonution.
moments because the two unknowns H, and H, will not appear in the equation,
NON-CONCURRENT, NON-PARALLEL, COPLANAR FORCES [ Ch. IV
70
The student should always look for such a convenient point because, with three
Thus,
unknowns, there will be only one unknown in the equation obtained.
>My = (5000)(12) — (D cos 30°)(16) = 0,
from which D = 4330 Ib. Now using D as
a center of moments, because the forces
D and H, will not appear in the equation,
we get
>Mp = (5000)(12) — (H,)(16) = 0,
F=
5000 Ib.
from which H, = 3750 lb. ‘The positive sign
indicates that we have chosen the sense of H,
correctly. The sum of the vertical forces is
=F, = D sin 30° + H, — 5000
= (4330) (0.5) + H, — 5000 = 0,
from which H, = 2835 lb.
His
The resultant at
H = (H,? + H,?)'? = (37502 + 2835?)1/2
= 4700 lb.
Hy
at an angle of
Fig. 122.
=
an
te
with the horizontal, acting upward toward the right.
construction of a portable crane.
47. Example.
2835
(=)
="
=
Sila
La
°
Figure 123 shows the actual
An A-frame is supported in a smooth track and loaded on the
horizontal member BH with a 500-lb. load, as shown in Fig. 124(a).
What are the
horizontal and vertical components of the pin reactions at A, B, C, D, and E?
Fig. 124(a).
A-Frame.
, SoLurTion. Taking the whole frame as a free body, we note that
the reactions at
; and # must be vertical, since the guides are smooth.
(The student should imagine
that the track is removed and that the action of the track on
the pins A and # is replaced by the force vectors A and E. The result
is a true free body.)
Now from the
sum of the moments about A, we get
2M 4 = (500)(24) — (£)(20) = 0,
from which ZH, = H = 600 lb. And from
=M x = (500)(4) — (A)(20) = 0,
§ 47 | EXAMPLE
we find A = 100 lb.
71
The positive signs show that the senses of A and H have been
chosen correctly.
If we next examine a free body of any member in this frame, we shall
find that on each member there is one
known force and four unknown forces.
Since there are only three independent
conditions for each member, we shall not
be able to find all the forces on any one
member from the free-body diagram of
that member alone. First make a free
body of BH because the given force is
acting on it [Fig. 124(b)] and because
we can find two unknowns immediately
from moment equations applied to this
member.
If we take moments about B,
we get (clockwise positive)
=M xz = (500)(20) — (D,)(12) = 0,
from which D, = 833 lb.
Then
the equation (clockwise positive)
from
=Mp = (500)(8) — (B,)(12) = 0,
we find B, = 333 lb. Since the equation =F, = D,—B.=0
yields only
that D, = B., these forces cannot be
determined from this free body. However, if we now make a free body of the
member AC, as in Fig. 124(c), we observe
that there are only three unknowns, all
of which can now be found. In making
this free body of AC, note that B, acts
on AC opposite to its direction on BH.
Courtesy Lyon-Raymond Corp., Greene, N. Y.
We know intuitively that this is true,
Fig.
123.
Crane.
This picture shows
because if some body X pushes on
something of the actual construction of a
another body Y, it follows that body Y
special type of crane, designed for picking
must be pushing back on body X with
up an airplane engine and transporting it
an equal force.
(Action and reaction
to the plane for attachment, or for handling
an engine removed for service.
are equal—one of Newton’s laws.) Now
proceeding with the analysis, we may take
moments about the point C (since by using this point we eliminate two unknowns,
C, and C,), and we get (clockwise positive)
>Me = (100)(10) — (333)(6) + (B.)(6) = 9,
Fig. 124(b) and (c).
A-Frame,
NON-CONCURRENT, NON-PARALLEL, COPLANAR FORCES [Ch. IV
72
from which B, = 166 lb.
The sum of the horizontal components yields
oF, = B, —C, = 0,
so that B, = C, = 166 lb.
A sum of the vertical forces gives
>F, = —100 + 333 —C, = 0,
or C, = 233 1b. Finally, from the free body of BH, we note that D, = Bz = 166 lb.
All vertical and horizontal components of all the pin reactions have now been found.
The student should make a free body of the member CE at this point to determine
whether or not the three conditions of equilibrium are satisfied for CH with the forces
as determined above. This step serves as a partial check on the computations for Cz,
CUD, Dip and E
48. Example.
The framework in Fig. 125(a) is loaded by a 1000-lb. force applied
at D. The member AD weighs 1800 lb. and the member AB weighs 1000 lb. What
are the components of the pin reaction at A?
< =
Fig. 125.
30°
1800 Ib.
(b)
The connection indicated at point H is loose enough so that there can be no
vertical reaction at H.
SOLUTION, Some problems require a simultaneous solution of equations obtained
from different free bodies. For example, we see from the free body of member AD;
+ oe 125(b),
OR
P
Fig.
that thererailsare four unknowns,
and that no matter what sum is
i made, at
least ae unknowns appear in the resulting equation. Thus, taking moments about
point C’, we find (clockwise positive)
(e)
2Mc = (1000) (20 cos 30°) + (1800)(7 cos 30°) — (A,)(6 cos 30°)
+ (A,)(6 sin 30°) = 0
=) 17,320' 1-910, 0001-24 wea
ent:
With the member AB as a free body, Fig. 125(¢), we may take moments about point B
and get (clockwise positive)
(f)
=Mxz = (A,)(5.196) — (A,)(9) — (1000)(2.598) = 0,
where the moment arms have been computed from the geometry of F ig.
125(a).
Now
BF a simultaneous solution of equations (e) and (f), we find
A, = 4270 Ib. and
y = 7894 lb.
If the other pin reactions are desired, they may now
be found by the
awe already explained, since properly chosen free bodies need now have no more
an
three unknowns.
The student may complete the solution and find C, = 4270 lb.,
§ 49 | EXAMPLE
GB
C, = 10,694 lb., B, = 4270 lb., and B, = 6894 lb.
that there is no friction at any of the pins.
Observe that we have assumed
49. Example.
A framework is loaded as shown in Fig. 126.
on pins A and D.
Determine the forces
2000 Ib.
o
Fig. 126(a), (b), (c).
(b)
Frame with Three-Force Members.
Sotution.
This example is similar to the preceding one in that at least two free
bodies are considered together in order to arrive at a solution. Moreover, since the
beginner often has difficulty with his free bodies when two or more three-force members are joined at one pin, this example will illustrate some of the fine points for those
who study it carefully. As usual, the reader should make his own complete solution,
following the text as a check.
One might well think first of using the whole frame as a free body, in which case, it
is observed that there are four unknowns—the magnitudes and directions of the
reactions at A and B (or Az, A,, Bz, and B,). Since there are only three independent
equations of equilibrium, all four unknowns cannot be found from this free body.
Although the value of A, may be found from this free body we shall follow another
procedure.
Next, consider the free body of link AD, Fig. 126(b), which shows at D the forces
on this link produced by CD and DB (the force by DB is represented by its compoWe see that there are five unknowns, that there would still be four
nents Q, andQ,).
found from a free body of the entire frame. Thus, another
were
A,
if
unknowns
approach must be found.
We observe that CD is a two-force member and that therefore the line of action of
CD goes through the centerline of the pins C and D. Thus, we decide that CD can
be determined from a tree body of BC, Fig. 126(c), in which we find CD represented
The components of the reaction at B are called P, and P,. (And
by CD, andCD,.
it is important to observe that P, and Py are not the same as the components of the
eround reaction at B of the whole frame, because of the force on pin B exerted by
link DB. This statement will be easier to understand later in the solution.) Proceeding with the free body of CB, Fig. 126(c), we take moments about B and find
[From CB]
CD;-= 1000 Ib.
or
=>M, = (2000)(30) — 60CD, = 0
NON-CONCURRENT, NON-PARALLEL, COPLANAR FORCES
74
The triangle CED is a 3-4-5 triangle; hence the cos 6 = 4/5 = 0.8.
CD,/cos 0, we have
[Ch.
IV
Since CD =
kip) — “== 1250 Ib.
Now returning to the free body of ra Fig. 126(b), we may take moments about
point A and find Q, (clockwise positive) :
[From AD]
=M 4 ll= (1000)(15) + 30Q. — 30CD.,
15,000 — 30Q, — 30,000 = 0,
from which Q, = 500 lIb., the positive sign indicating that Q, is
shown in the correct sense.
(The reader might find some advantage in inserting on his free bodies the numerical values of
the forces as they are found.)
The problem as stated might
be completed now if we had found A, from a free body of the
whole frame. Nevertheless, to emphasize the force arrangement
on pin D, we shall next use the free body of DB.
In Fig. 126(d), take moments about pin B (counterclockwise
| (b)
Fig. 126(b).
peated.
positive)
Re-
[From DB]
=Mes = (1000)(20) + 30Q, — 40Q,
= 20,000 + (30)(500) — 40Q, = 0,
from which Q, = 875 Ib., the positive sign indicating that Q, is shown in the correct
sense.
Sy=1625
DC=1250
(f)
Fig. 126(d), (e), (f).
Frame with Three-Force Members.
Again, from Fig. 126(b), we find
[From AD}
=F, = 1000 — Aj + Q, — DC,
1000 — A, + 500 — 1000 = 0,
or, A, = 500 Ib. and is shown in its correct sense.
fees we get
Taking a sum of the vertical
[From AD]
2=F,= Ay —Q, — DC sine
ll
A, — 875 — (1250) (0.6) = 0,
§ 50 | TRUSSES—METHOD
OF SECTIONS
78
from which A, = 1625 Ib. Now we know the components of the force on pin A.
The free body of Fig. 126(b) shows the forces on AD as produced through the pin D
by DC and DB. The force on pin D produced by AD is equal and opposite to the
resultant of Q,, Q;, and DC.
To study these forces further, let the free body of link AD be as shown in Fig.
126(e). This is a valid free body where S (= S, + S,) is the force on pin D produced
by AD.
Checking back on the values of A, and A, previously found, we see that
[Fig. 126(e)]
Sy = Aly, = G5 Ilo},
and
S. = 1000 — Az = 500 lb.
Now take a look at the free body of pin D, shown in Fig. 126(f). The forces on D
are then S( = 1625+ 500), Q( = 5004 875), and DC( = 1250), which completes
the problem as stated.
Observe that the free body in Fig. 126(e) could have been used in the beginning
and the free body in Fig. 126(b) omitted. [The student would no doubt benefit by
solving the problem without using Fig. 126(b).] Thus, the force S is equal and
opposite to the resultant of Q and DC. Without careful reasoning to establish correct
free bodies, a problem of this kind becomes difficult. For practice, the student
should find the forces on pin B (now) and check them using the whole frame as a free
body.
50. Trusses—Method of Sections.
A truss is made up of a series of
triangular frames—triangular because the triangle results in the most rigid
construction.
In structural design practice, engineers commonly, but not
always, assume that:
1. The weight of the individual members of a structure is negligible as
compared to the loads on the structure.
2. The structure is rigid for the purpose of the force analysis. (No member
is rigid. Any body deflects under the action of a load, no matter how
small the load. The assumption of rigidity is a close enough approximation for most metal parts and their external loads.)
3. All loads are applied at pin joints.
(Hach member is then a two-force
body, and the line of action of the forces produced on the member by
the pins is the line joining the centers of the pins. Moreover, each
member is either in tension or compression, depending on whether the
pins tend to pull the member apart or to compress it. The construction
of roofs laid on trusses is often such that this condition is substantially
met.)
Before the size of the various members of a structure can be determined,
One method of dethe forces acting on these members must be known.
termining these forces is described in § 21, the joint-to-joint method in which
Now that we are
the pins are the free bodies and the forces are concurrent.
will offer some
sections
able to handle non-concurrent forces, the method of
advantages.
76
NON-CONCURRENT, NON-PARALLEL, COPLANAR FORCES [Ch. IV
In applying the method of sections, we find first the reactions at the points
of support. For example, consider the truss of Fig. 127(a). Make a free
body of the entire truss and take moments about the support G to determine
the reaction at H. Then take moments about the support H to determine the
reaction at G and check these computations by using the equation 2/’, gt
Now suppose a section is made across B-B, Fig. 127(a). In order to maintain, say, the left-hand part of the truss in equilibrium, we must apply an
external force to the member JK exactly equal to the internal force formerly
produced in IK by the action of the right-hand part on the left-hand part of
the truss. This force is designated as 7K in Fig. 127(b). Similarly, the external forces applied to the members JK and JZ after they are cut are equal
to the former internal forces. Since this left-hand part of the truss in Fig.
127(b) is acted upon by exactly the same forces as were formerly on it when
LAV WBS
pa
ae (ieee
Fig. 127.
Method of Sections.
it was an integral part of the structure and since the structure is in equilibrium,
then the detached part must also be in equilibrium.
Thus, using the free
body in Fig. 127(b), we may take moments about any point and equate the
sum to zero and we may equate the sum of the forces in any direction to zero.
In making the solution of this problem, observe that it would be convenient
to use J as a center of moments because two unknowns (JK and JL) would
thereby be eliminated.
The >M; = 0 would yield the value of IK. A
vertical sum, ZF, = 0, would result in the value of JK. Then the value of
JL could be determined from 2f, = 0 or from >Mz = 0.
In general, the cut or section may be made across any group of member
s—
but no more than three members subjected to unknown forces
must be cut
in dividing the truss. Three members are the limit because
only three
unknowns may be found from the conditions of equilibrium.
(Only two
unknowns may be found if the three members intersect
at a common point.)
For example, sections A-A, B-B, C-C, and others too
[Fig. 127(a)], may be
made to find all unknown forces in each case.
On the other hand, in a
section D—D, there are four unknowns, and in a section
E-E, there are five
unknowns.
The sections D-D and E-E could be used if
enough of the
internal forces were known go that only three
internal forces were left to be
determined.
See Fig. 128 for a view of a typical steel bridge.
§ 51 | EXAMPLE
77
51. Example. A free body of a truss is shown in Fig. 129(a). Find the reactions
at the points of support and the forces in the members CD, ID, and IH.
Courtesy Texas Highway Department
This illustration shows something of
One End of Typical Highway Bridge.
Fig. 128.
Observe that the members
the actual construction of the joints of a bridge structure.
are riveted to each other, not attached by pins as assumed in the theory. This construction
introduces a rigidity into the structure not allowed for in our force analysis. However,
Even
the force analysis as explained here is commonly used and will give safe results.
if the members are welded together, the force analysis is usually made as it would be for
pin points.
,
20!
20tons
10tons
5 tons
(a)
Fig. 129.
To find the reaction R,, take moments
Sotution.
(clockwise positive), from Fig. 129(a),
about the point 7.
Thus
>My = (R)(100) — (20)(60) — (10)(40) — (5)(20) = 0,
from which R, = 17 tons.
To find R2, take moments about the point A.
This gives
(counterclockwise positive)
>M4 = (R2)(100) — (5)(80) — (10)(60) — (20)(40) = 0,
from which Ry = 18 tons.
have =F, = 0.
Asa check, note that with these values of Ai and Re, we
NON-CONCURRENT, NON-PARALLEL, COPLANAR FORCES [Ch. IV
78
Next, section the truss at y-y, since this section cuts the members whose forces are
desired. Either the right-hand or left-hand parts of the truss may now be used as a
free body. Choosing the left-hand part, we get the free body shown in Fig. 129(b).
In this free body, taking J as the center of moments, we find
=M, = (CD)(20) + (17)(40) = 0,
from which CD = —34 tons. The negative sign means
that the sense of CD has been chosen incorrectly; hence
CD acts toward the left and the member is in compression
instead of in tension (§ 21). Now, taking the point D
17 tons
20 tons
as the center of moments, we find
(b)
Fig. 129(b).
=M p = (17)(60) — (20)(20) — (IH)(20) = 0,
Repeated.
from which JH = +81 tons, where the positive sign
indicates that the sense of JH is correctly shown and the member is in tension. The
final unknown JD may now be found by taking moments about any convenient
point, such as C, but the simplest method is to sum the vertical forces. Thus
ZF, = 17 — 20 + (1D)(cos 45) = 0,
or
ID = +4.2 tons.
Again, the sense is correct and the member JD
is in tension.
In design, it is important in every instance to know whether a member is in tension
or in compression. Hence, the student should always state the kind of load on a twoforcemember.
The plan suggested in § 21 is recommended for solutions by the method
of sections. That is, as in the preceding example, make the free body with the wnknown forces acting away from the adjacent pins, thus asswming a tensile force on the
members.
Then if a positive answer is obtained for the magnitude of the force, the
member 7s in tension, and if a negative answer is obtained, the sense of the force has
been incorrectly assumed and the member 7s in compression (as in the case of member
CD in the foregoing example).
52. Closure.
The reader is reminded again not to overlook the importance
of free-body diagrams.
At every point where the free body formerly made
contact with some other body, there is generally a force. At all such points,
then, a force or the components of a force should be shown on the free body.
The following suggestions and reminders will aid in the solution of problems.
1. In finding resultants, notice that (a) if the force system is collinear or
concurrent, the resultant is necessarily a force; (b) if the force system
is non-concurrent or parallel, the resultant may be either a force or
a couple.
2. Recall that the reaction at a smooth surface is normal to the surface.
3. Look for two-force members, but be sure that a member with
three or
more forces is not handled as a two-force member.
The action at a
section of a two-force member is simply a force. The action
at a section of a three- (or more) force member is generally a force
and a moment.
We do not usually cut a three-force member in
making a free body
(see Fig. 80, p. 55).
PROBLEMS:
RESULTANTS
79
4. Put down the numerical values of the known forces on the free body,
so that the unknowns become apparent at once. If there are more
unknowns than there are conditions of equilibrium, maybe a solution
can be made by choosing other free bodies (see §§ 47, 48, 49). The
beginner sometimes thinks he has too many unknowns because he has
represented the force of a two-force member by two components, whereas
the action of a two-foree member introduces only one unknown—the
magnitude of the force.
. Observe that a free body of an entire structure is frequently necessary
(and nearly always desirable) before a complete solution of a problem
is possible.
It will be helpful for you to notice that all coplanar force systems may be
solved by the equations.
(g)
Be, = 0,
(h)
z
(i)
0,
a
=M =0,
If the forces are concurrent, the
2M = 0 is not needed; only two equations express independent
If the forces are parallel, either 2F, = 0 or ZF, = 0 is not
conditions.
needed.
By noticing the similarities of a subject, you improve your underif there are no more than three unknowns.
condition
standing of it.
Problems
RESULTANTS
241. Three forces act on a body as shown
in Fig. 130. If A = 200 lb., B = 500 lb.,
C = 400 lb., and @ = 30°, find the resultant
force which will have the same effect on the
body (a) using point C as a center of mo-
243. Three forces, A, B, and C, each
equal to 100 lb., act on a body as shown in
Fig. 130, with @ = 40°.
(a) Find a force
which will counteract their combined effect,
describing its location with reference to
point C. (b) Resolve the given force system
into a force at A and a couple.
Describe
each fully.
244. Three forces act on a body as shown
in Fig. 130. If A = 80 lb., B = 80 |b.,
C = 113.2 Ib., and 6 = 45°, find their
resultant.
Ans. 1440 in-lb. ce.
ments in locating the resulting vector, (b)
using point A as a center of moments.
Ans. (a) 154 Ib. at 180° through a point
245. Three forces, A, B, and C, act on a
body as shown in Fig. 130. Their resultant
R = 200 lb. passes through a point which
is 10.4 in. directly above C at an angle of 150°
with the positive x axis. If A = 100 lb.,
23.4 in. above C; (b) 154 lb. at 180° through
a point 18.4 in. above A.
242. The same as 241 except that 0 =
60°.
246. Three forces act on a body as shown
in Fig. 180. Ii A = 10]b. and B = 201b.,
find a force C such that forces A, B, and C
Fig. 130.
Problems 241-246.
and B = 150 lb., find C and @.
PROBLEMS: RESULTANTS
80
may be replaced by a force of 15 lb. acting
at 180° through A and a counterclockwise
couple of 105 in-lb.
rat
Ff,
i,
Fig. 131.
250. If the forces in Fig. 132 are Fy =
30 lb., F2 = 110 lb., F3 = 35. 1b., and Fs =
20 lb., find the resultant.
Ans. 110 lb., 6 = 222:4°, r = 0.
251. If the forces in Fig. 132 are F; =
138:3 [beets
00 Nlb Hee 0s lbs eel
F,4 = 61.3 |b., find the resultant.
2
Problems 247, 248.
247. If the forces in Fig. 131 are Fi =
A00) Ib., Fe = 150 lb.) Fs = 250 Ib. and
F', = 300 lb., find the resultant.
PNR AYR Noa, Ge MSA, Po PAIL tite
248. If the forces in Fig. 131 are Fy, =
200) lbs Ha) lbs Hs)
150 lbssand 47
310 lb., find the resultant.
Fig. 133.
Problems 252-254.
252. A link AB, Fig. 133, is subjected
to the forces F; = 100 lb., F2 = 550 lb., and
W =75lb.
(a) Find the resultant of these
forces.
(b) Will the link move
action of the forces?
Ans.
Fig. 132.
[Ch. IV
Problems 249-251.
249. If the forces in Fig. 132 are Fy =
100 lb., F2 = 60 lb., F3 = 40 lb., and F4
50 lb., find the resultant.
under the
Explain.
(a) 656 Ib. at-816.7°% 7 —0:685) ft;
(b) rotate clockwise.
253. A link AB, Fig. 133, is subjected to
the forces F; = 300 lb., Fe = 494.9, and
W = 100 lb. Find the resultant of these
forces.
Will the link move under the action
of these forces?
Explain.
254. The same as 253 except that F. =
200 lb.
EQUILIBRIUM
255.
wheel
(a) What
force
F will
cause
the
of Fig. 134 to be on the point of
the reaction at B?
At C?
(arnt: Solve
graphically. W, F, and B intersect.)
The
moment arm of F may be determined
graphically if desired.
The surface at B is
rough.
Ans. (a) 195 lb.; (b) 437 Ib., 0.
256. (a) What force F at 6 = 0 results
in a ground reaction at C, Fig. 134, of 100
Smooth
Fig. 134.
Problems 255, 256.
moving over the block A?
weighs 500 lb. and 6 = 30°.
The
wheel
(b) What is
Ib. if the wheel weighs 500 lb.? (b) What
is the reaction at B? The surface at B is
rough.
257. In Fig. 135, let W = 1000 lb., P =
2000 Ib., 7’ = 3000 Ib., and @ = 15°. What
force F will result in the body being on the
point of moving over the obstruction A?
What is the reaction at B?
Ans. 12,000 lb., 0.
258. In Fig. 135, a force F = 300 lb.
acts at an angle 6 = 0° on a wheel whose
weight is W = 800 lb. Let
P=T =O.
If the wheel is in equilibrium, what is the
reaction (force) at C on the wheel?
The
PROBLEMS:
81
EQUILIBRIUM
surfaces are smooth.
Is the reaction at B
equal to zero?
similar to that shown.
If the tumbler is
2.5 in. in diameter and if all surfaces are
smooth, what is the angle 6? Solve graphically and algebraically.
Ans. 19.85°.
262. The same as 261 except the rod is
7 in. long. Note that it is not necessary
to know the weight of the uniform rod.
Fig. 135.
Problems 257, 258.
259. A plate A is attached to a vertical
wall by two rivets B and C as shown in
Fig. 136. If the force F = 250 lb. and
x = 18 in., what are the horizontal and
vertical components of the forces on the
rivets? The following assumptions are nec-
Fig. 136.
Fig. 138.
Problems 263, 264.
263. In Fig. 138, A is a weight of 100
lb., CD is a rigid, uniform body weighing
50 lb., and the pegs are smooth.
If CD
remains in a horizontal position, determine
F and the weight B for equilibrium.
Ans. 77 lb., 64.2 lb.
264. The same as 263 except that CD
weighs 70 lb.
Problems 259, 260.
essary: neglect friction between the plate
and the wall; there is no twisting of the
plate; and the rivets share equally in supporting the load F vertically.
Ans. Bz = Cz = 450 lb., By = Cy = 125 lb.
260. The same as 259 except that 2 =
30 in.
Fig. 139.
Problems 265, 266.
265. A pipe is supported by hangar rods
AE as shown in Fig. 139. The hangar is
pivoted at pin A. If the pipe section
weighs W
Fig. 137.
Problems 261, 262.
261. A glass rod AB, Fig. 137, weighs
5 oz. and is 6 in. long. It is placed in a
glass tumbler C in a position of equilibrium
= 800 lb. and L = 14 in., com-
pute the reactions at A (components), B,
C, and D, neglecting the weight of the
hangar.
Ans, Az =B =D — 655 1b., A;
= C=
800 lb.
266. In Fig. 139, a pipe is supported by
hangar rods AZ of negligible weights.
The
hangar is hinged on the removable pin A.
If the reaction at B should be W/6, what
must be the dimension L?
PROBLEMS: EQUILIBRIUM
&2
[Ch. IV
267. In Fig. 140, let Q = 15 lb. and @ =
45°.
-(a) Find the components of the reac-
tions at pin A andat D.
(b) Find the single
Fig. 142.
Fig. 140.
Problem 267.
force which, if applied to the body (extended
Problem 269.
271. In Fig. 143, find the components of
the external reactions at D and E, of the
pin reaction at A, and the force on the
if necessary), would cause the same reactions
at A and D.
Fig..143.
Fig. 141.
Problem 268.
268. A boy’s wagon with a total load of
W = 100 lb. is being pulled across a ditch
as shown in Fig. 141. The wheels have a
diameter of 10 in. and the wagon has come
to rest.
If Q@ =15]b., find 6.
Ans. 5.8°.
269. It becomes necessary to pull posts
by hand. A vertical pull of 2 tons is
needed to move the post after the initial
set is broken.
The materials at hand are
some timbers and a flexible wire rope.
With the aid of a student of analytic
mechanics, the materials are arranged as
shown in Fig. 142. First, the rope is made
tight; then a horizontal foree F = 100 lb.
is exerted until the post ceases to rise,
moving point A on the rope to B. Then
the rope is made tight again and the process
is repeated.
How far can the post be
raised before the rope is readjusted?
270. Find the reactions at Q and H, Fig.
125
(a), observing
that
the
member
AD
weighs 1800 lb. and that AB weighs 1000
lb. QH weighs 2000 Ib.
Ans. Q = 6020 lb., 09 = 74.6°, H = 1600 lb.
Problems 271, 272.
cable BC when F = 3000 lb.
weight of the members.
Neglect the
Ans.
D; = E = 3000 lb., BC = 4500 lb.,
‘A
COON:
272. The same as 271 except that the
beam AC weighs 100 lb. per ft. of length.
Fig. 144,
Problems 273, 274.
273. On the horizontal boom of Fig. 144
are placed three loads, F, 2F, and 3F, as
shown.
Let 6 = 60°.
(a) What value of
F will produce a thrust of 30,000 lb. on the
PROBLEMS:
strut BC?
(b) What are the vertical and
horizontal components of the resulting pin
load at A?
Ans. (a) 2370 lb., (b) 26,000 lb., 790 Ib.
274. The same as 273 except that 6 =
45°.
Fig. 145.
Problems 275, 276.
275. The boom ACD of Fig. 145 supports
wheel loads of F; = 8000 Ib. and F:; = 6000
ib. whose lines of action remain 6 ft. apart.
What value of x results in a load of 24,000
lb. on the strut BC?
Ans. 15.3 ft.
276.
53
EQUILIBRIUM
The same
as 275 except that Ff; =
2 kips.
Ans. (a) Ey, = 6000 lb.; G = 3000 Ib.; (b)
Cor "3460 br
—1 8000) IbaN(e) mo:
278. The same as 277 except that member BC weighs 40 Ib. per foot of length.
279. In Big. 147, i F = 100 lb: and
@ = 0, find the components of the reaction
at A and at B, and the force in BC.
Fig. 147.
CASO, Ih Vie, ee i () = KOO Mon,
=
30°, and F = 0, find the reactions at A and
B and the forces in BC and BD.
Ans. 63.2 lb. at’ 103:9°, 70 lbs at 21.8°,
70 Ib. (C), 0.
281. In Fig. 147, if Q = 100 lb., @ = 30°,
and F = 200 lb., find the force in member
IBC!
282. In Fig. 147, if Q = 100lb., 9 = 150°,
and F = 200 lb., find the forces in members
IGE, (CID, eave! IED).
Ans. 468 Ib. (C), 200 lb. (T), 0.
Fig. 148.
Fig. 146.
Problems 277, 278.
277. A load of F = 6000 lb. acts on the
(a) Deterboom CB as shown in Fig. 146.
mine the components of the reactions at #
and G. (b) What are the components of
the force exerted on the pin C by the member CB?
(c) If CB is cut, can the action
of one part on the other be replaced by a
force? Explain.
Problems 279-282.
Problems 283-285.
283. In Fig. 148, if F = 200 lb. and Q =
0, find the force in CD and the reaction at A.
284, In Fig. 148, if F = 50 lb., and Q =
150 lb., and @ = 0°, find the components of
the reaction at A and the foree in CD.
285. in Kies 148i 7 = 0) QO = 200 1b:
and 6 = 120°, find the force in CD and CH
and the components of the reactions at A
and B.
Ags, CID = OH illo, (0), Cie) = BO. Mo, (OQ),
A, = —75 lb. A, = —s01b., Bo = 248 Ib:,
B, = —70 lb.
8h
PROBLEMS: EQUILIBRIUM [Ch.
IV
288. In Fig. 151, find the reactions
at
A (components)
and C, and the forces on
the members AD and BD, if F = 10,000
Ib. Neglect the weight of the members.
Ans. Az = —10,000 lb., C = 2890 Ib.,
AD = 8660 lb. (T), BD = 5000 lb. (C).
Fig. 149.
Problem 286.
286. In Fig. 149, if F = 100 lb., Q = 200
lb., and 6 = 30°, find the components of the
pin reactions at all pins.
Q
jase
Fig. 150.
Fig. 152.
Problem 287.
Problems 289, 290.
289. The framework in Fig. 152 is loaded
287. In Fig. 150, if F = 50 lb., and Q =
100 lb., and @ = 30°, find the components
of the pin reactions at all pins.
as shown with F = 5000 lb.
(a) Determine
the components of the reactions at HE and
G. (b) What are the components of the
force between the pin C and the member
BD?
(c) If BC is cut, can the action of
one part on the other be replaced by a
force? Explain.
Ans. (a) HE = 1414 lb., G, = —1414 Ib.,
G, = 5000 lb.; (b) C, = 2500 lb., C, =
5000 lb.; (ec) no.
Fig. 151.
290. The same as 289 except that the
member DCB weighs 30 Ib. per ft. of length.
Problem 288.
TRUSSES—METHOD
OF
291. For the truss of Fig. 153, find the
external reactions and then find the loads
on the members AB, BF, and FG by a single
section.
Ans. ky = 11,5001b., Rz = 16,5001b., AB =
33,000 Ib. (C), BF = 10,000 lb. (T), FG =
28,600 Ib. (T).
292. For the truss in Fig. 153, find the
external reactions and then find the loads
on the members BC, BG, and FG by a
single section.
Check by using the alternate part of the truss.
293. For the truss in Fig. 153, find the
external reactions and then find the loads
on
the members
CD,
single section.
Check
nate part of the truss.
DG,
and
GH
by a
by using the alter-
Ans. CD = 17,000 lb. (C), DG = 6000 lb.
(C), GH = 19,900 lb. (T).
SECTIONS
6000 Ib.
6000 Ib.
“al
ae
15
15/2 15-245:
Ry,
10,000 Ib.
Fig. 153.
Rp
Problems 291-294, 306.
294. For the truss in Fig. 153, find the
external reactions and then find the loads
on the members DE, DH, and GH by a
single section. Check by using the alternate part of the truss,
PROBLEMS:
85
TRUSSES
Zs
Smooth us
Fig. 154.
Problems 295-298.
295. Compute the external reactions
(components at A) and the loads on the
members GH, CG, and GJ, Fig. 154, when
F, = 3000, F2 = 4000, and F; = 5000 lb.
Use the method of sections.
Ans. Az = 5500 Ib., A, = 2440 lb., G =
10,770 lb., HG = 3700 lb. (C), CG = 2510
Ib. (C), GJ = 9920 lb. (C).
296. Compute the external reactions
(components at A) and the loads on the
members CD, CJ, and GJ, Fig. 154, when
F, = 10 kips, F2 =0, and F; = 4 kips.
Use
the method
of sections.
Check
by
using the alternate section.
297. The same as 296 except that PF. =
5 kips.
298. Find the loads on the members
CD, CJ, and GJ in Fig. 154 without finding
the external reactions, if F; = 3000 and
F, = 4000 lb. Use the method of sections.
Check by the method of joints.
Ans. CD = 9200 lb. (T), CJ =0,
GJ =
9940 Ib. (C).
1000 and FP, =
299. In Fig. 155, let PF; =
5000 Ib. Find the external reactions (com-
ponents at HZ) and the loads on the members
Use the method of
CG, CH, and EH.
sections and check by using the alternate
section.
Ans. G,; = E, = 7000 lb., #, = 6000 Ib.,
Fig. 155.
Problems 299-301, 307.
CG = 9910 lb: (1), CH =0,
300. The game as 299 except that Ff, =
2000 Ib.
301. The same as 299 except that PF. =
2000 lb.
302. The Baltimore truss shown in Fig.
156 sustains a uniform load of 4000 Ib.
per ft. on the lower chord.
A and TJ; pin I takes half the load between
A and J; ete. Find the loads on members
BC, WL, and KL.
‘Then find the loads on
BW,JW,andJK.
The method is optional.
PS
Ry
aN
ORE
Span =144 ft or 12 panels @ 12 ft.
Fig. 156.
In making a
force analysis, engineers assume that the
uniform load acts as a series of concentrated
Iayeyoley ey aeloves qovnavsh VIGIL, digs oe on YM Rl
Pin A takes half the uniform load between
B
jain
AE = 7000
Ib. (C).
Problem 302,
1s
PROBLEMS:
GENERAL [Ch.
IV
306. Find the loads on all members
the truss of Fig. 153.
of
56
PROBLEMS
GENERAL
303. The framework in Fig. 157 is loaded
with forces F, = Ff. = F3 = 3000 lb. as
Find the loads on each member.
shown.
307.
Find
the loads
on
all members
of
the truss of Fig. 155.
Rollers
Fig. 159.
Fig. 157.
308. In the truss of Fig. 159, the wind
Problems 303, 304.
loads of F,; = 1000, F2 = 2000,
In the solution, section the members
AD, and ED.
same
as
303
except
that
and F3
1000 lb. are assumed to act at the pin
joints, as shown.
In order to allow for
AB,
Ans. DE = 0, AE = 6300 lb. (C), AB
HOO lo, ©), AUD = S/O is, (CD), 1exC!
38300 lb ©)
D424 0 ele Ce
Ca
DANO) io, (4D).
304. The
Problems 308, 309.
expansion,
the support at G@ is on rollers.
For F, = 5000 lb., find the external
reac-
tions (components at A) and the loads on
members BC, BH, and AH.
Ans, A, = 200021b Ay = 2495 Ib5 G—
5970) lb, BC = 49905lbs (©), BA 0;
AH = 2320 Ib.(2):
309. In the truss of Fig. 159 the wind
Fy
acts at pin A horizontally toward the left.
Ans. DE = 0, AE = 3300 (C), AB = 6600
(O), AUD = S72Q GD), 1D = 2210) (1D),
GD = 320 GD), KC = FO (©), alll to Moy,
loads of F; = 1 kip, F2 = 2 kips, and F3
3 kips are assumed to act at the pin joints,
as shown.
In order to allow for expansion,
the support at Gis on rollers.
For Fy = 4
kips, find the external reactions (components at A) and the load on members CD,
CG, and GH.
305. Find the reactions at A (components)
and # and the loads on all members of the
truss shown in Fig. 158 when F, = 5000
Ib., #2 = 1000 lb., Fs = 4000 lb., and Fy,
1500 lb.
Fig. 160.
1000 lb.
2000 Ib.
Span = 80’.
Fig. 161.
Expansion Rollers
Problem 311.
Problem 310.
PROBLEMS:
&7
GENERAL
310. In the truss
of Fig.
160, F = 60
315. A load of F = 1000 lb. is applied
tons and @ = 25°.
Find the external reac-
to a frame, as shown in Fig. 164.
tions (components
at C) and the force in
horizontal and vertical components of the
reactions at A and H and the forces on the
members BE and CF.
Ans. BE = 1860 lb. (1), CF = 1345 Ib.
(C), — A, = H, = 1600 lb., A, = 1440 ]b.,
H,, = 4438 lb.
all members.
311. If the wind loads on the Fink truss
of Fig. 161 are as shown, determine (a)
the components of the external reactions
and (b) the forces on each member of the
truss when
all other
loads
are
Find the
considered
negligible.
Fig. 162.
Problems 312, 313.
312. If Fi = 5 kips and F, = 8 kips in
the truss of Fig. 162, find the components
of the external reactions and the load in
members 2-3 and 3-6.
Ans. 2-3 = 6 kips (T), 3-6 = 17.25 kips (C).
313. The same as 312 except find the
force in all members.
abl
if
ele
F
Fig. 163.
Problem 314.
314. A Pratt-truss panel consists of four
rigid bars pinned together and stabilized by
two diagonal wires, as shown in Fig. 163.
If F = 40 lb., find the components of the
external reactions and the load in each
Assume no preloadmember of the truss.
Examine this truss for a
ing on the wires.
slight shifting into a parallelogram of the
rigid bars, the short diagonal being the
If the short
wire that will bear no load.
diagonal member were of rigid material, is
it then true that this member would carry
no load when the truss deforms?
Fig. 164.
Problems 315, 316.
316. In the frame of Fig. 164, what is
the value of F if the horizontal
HH, = 3 kaipst
reaction
317. For a load of F; = 1000 lb. in Fig.
165 and for Ff, = 0, find the x and y components of the reactions at A and B and the
loads on members CE and CD.
Ans, B= Ay = 2250lby AZ = 0B, —
SOO I, C2 = SW lo, GD), CD = R000
Ibs (CG):
Fig. 165.
Problems 317-319.
318. For a load of F; = 1000 lb., acting
toward the left, and F, = 6000 lb. in Fig.
165, find the x and y components of the
reactions at A and B, the load on member
CE and the x and y components of the pin
forces C and D on member CD.
Anis. Aj = 0200 be) Bz — 1000) bee Bia
750 lb., Dz; = Dy = 3000 lb., CH = 3750
Ib. (CG):
319. If fF, = 0, the maximum safe load
of the member CD is 28,000 lb. (Fig. 165).
The maximum safe load for member CF is
35,000 lb.
applied?
What
safe
load
Fi; may be
Ans. 9330 lb.
PROBLEMS:
&8
320. With F, = 5000 lb. and F, = 4000
Ib. in Fig. 166, find the x and y components
of the reactions at A and B and the loads
on members BC and CE.
Fig. 166.
GENERAL
[Ch.
1V
crank, let a = 30°, and find (a) the weight
of W and (b) the reaction at C on the bar
BD.
Problems 320, 321.
Ans. Az, = Bz = 16,400 lb., A, = 3340 lb.,
15, = ABO) Mo, JC" = IPF) MN, (CC),
(C10) = ZEN) Moy, (AD),
321. The same as 320 except that Ff; = 0.
Fig. 168.
325. In’ Figs
Problems 325-328.
168), 6) =. 907
sand)
fF — 0:
Find the force in member AC.
As a final
check, make a free body of pin C, showing
the forces in magnitude and sense produced
on it by the members DC, AC, and BC
(components will be all right).
Ans. 100 Ib. (C).
326. The same as 325 except that @ =
60 lb. Sphere
uies
Fig. 167.
Problems 322-324.
270°.
327. In Fig. 168,¢ =QOandF
the force in member AC.
=0.
ANS. Uf Seon
328. In Fig. 168,
Find
ely)
9 = O and F = 200 lb.
Find the force in member AC.
Ans. 973.2 Ib. (T).
322. In Fig. 167, there is shown a bell
crank ABE and a bar BD, each being independently pivoted at B. A 60-Ib. sphere
is attached to the bar at C by a 1-ft. cable
so that the sphere assumes an equilibrium
position
angle a.
at some particular value of the
Neglect the weight of the bar and
the bell crank, let W = 100 lb., and find
(a) the angle a at which the bell crank will
be at rest, and (b) the reaction on the bar
BD at C.
Ans. (2) 52°, (b) 42 Ib.
323. The same as 322 except let W =
60 lb.
324. In Fig. 167, there is shown a bell
crank ABE and a bar BD, each being independently pivoted at B. A 60-lb. sphere is
attached to the bar by a 1-ft. cable so that
the sphere assumes an equilibrium position
at some particular value of the angle a.
Neglect the weight of the bar and the bell
Fig. 169.
Problems 329-331.
329. In Fig. 169, the pipe load is W =
4000 lb. on the vertical center line of pin
PROBLEMS:
&9
GENERAL
C, w = 15°, and D =4 ft. Which pin, A,
C, or #, has the maximum load on it?
How much is this load?
330. The same as 329 except that a =
30°
IMS, Ne = VOD Mod, Aly SS PEO Iss
Solemn hoe OOM letras — 202 .ande
=
4 ft. Ifa weak pin at C will sustain a safe
maximum force of 5000 lb., calculate the
permissible maximum weight W of the pipe
section.
336. In Fig. 172 is shown
Fig. 172.
Fig. 170.
Problems 332, 333.
332. Figure 170 illustrates a three-hinged
arch. Let a = 20 ft., fF: = 0, and Fy =
100,000 lb. Find the z and y components
of the pin reactions at A, B, and C.
Anse A, =—B.a => C,— A, = 5, = 16,670
Ib, @, = 83,3830 Ib.
333. Figure 170 illustrates a three-hinged
archaea Dettam— 22) itp Oe—ell
padite el eel O
kips, and F, = 15 kips. Find the z and y
components of the pin reactions at A, B,
and C.
an inverted
A-frame supported by wheels D and £,
either of which may bear against the top or
bottom rail. The cable passing over the
pulley J supports a load F = 1000 lb. All
Problems 336, 337.
surfaces are frictionless.
Determine the
components of the pin loads at A, B, and C.
is, TNs = 185 = 8) No, 4, = GH lle,
B, = 1000 lb., C, = 827 Ib., C, = 2000 Ib.
337. The same as 336 except that F =
80 lb.
5000 lb.
50001b.
pp Y
5000 1b.
Fig. 173.
Fig. 171.
Problems 334, Boos
334. The three-hinged arch of Fig. 171
(a) Detersupports the loads as shown.
mine the « and y components of the pin
reactions at A, B, and C. (b) Find the
loads on the members DB and EB.
Ans.
(a) Ay = Cy = 7500 lb., Az = Be =
G, = 8660 Ib., By = 2500 lb. <(b) BD =
16,850 lb. (C), BE = 8360 lb. (T).
335. The same as 334 except that the
5000-lb. load at B is removed.
Problems 338, 339.
338. Find the horizontal and vertical
components of the pin loads at D, G, and
C, Fig. 173, when a = 2b and F = 1000 lb.
Ans.
D, = Gz = Cz = 722
|b.,
Dy = 333
Ib., G, = 667 lb., C, = 278 lb.
339. The same as 338 except that a = 46.
340. Find the horizontal and vertical
components of the pin load at C and the
load on the member HK, Fig. 174, when
x = 6 ft. and F = 2000 lb.
Alas, Cl, = NG3X0) toy, Ch, SS SMCO Woy, Lele SS
1730 lb. (T).
PROBLEMS: GENERAL [Ch.
90
a = 10° when Q = 100 lb.
sure (psi) in the cylinder.
IV
Find the presAns. 1444 psi.
343. The same as 342 except that the
piston diameter is 1/2 in.
344. The same as 342 except that a = 4°.
Should a mechanical stop for the lever be
installed to prevent a further reduction of a?
345. An internal combustion engine is
represented diagrammatically in Fig. 176.
Its stroke and bore are each 3 in. (r = 1.5
1)
lt ae
oan
Cio —slomennenmnne
pressure in the cylinder is 400 psi, what is
the torque on the shaft at this instant?
Ans. 1520 in-lb. e.
346. The same as 345 except the crankshaft has rotated so that a = 7° and 6 =
Fig. 174.
Problems 340, 341.
341. The same as 340 except that x =
iL) 10
Alps, Gz = WS) lo, Cp, = 200 ld, 0K =
1730 lb.
Ejaga|
gy
Smooth Surfaces
A
30°
Q
30%
347. Two beams DAS and SCE are supported at D, A, C, and # as shown in Fig.
177. The beams are slightly separated at
the point S. The distances between the
wheels
remain
as
shown,
and
the
loads
imposed by the wheels are H = 12,000,
I = 6000, J = 6000, L = 6000, and K =
1000 lb. As the wheels move along the
beams from D to &, the loads in the struts
AB and BC vary.
For what position of the
wheels will the load on AB be a maximum?
What is the corresponding load on BC?
Ans. AB = 29,800 lb., BC = 1730 Ib.
348. The same as 347 except the position
for maximum loading on BC and the corresponding load on AB are desired.
Fig. 175.
Problems 342-344,
342. Figure 175 shows a hand-operated
hydraulic pressure pump that uses a toggle
linkage and a lever to increase the force on
the piston.
Fig. 176.
The piston diameter is 2 in. and
Problems 345, 346.
349. A diagrammatic representation of
a walking-beam (CAB) pump is shown in
Fig. 178. The rotation of the crank pin
D about E causes the piston J to reciprocate
and pump water.
For the position shown
and with a belt tension on the slack side of
F = 500 lb., what pressure p in psi on J
holds the mechanism in equilibrium?
What
is the pin load at A? Ans. 47.7 psi, 7160 lb.
Fig. 177.
Problems 347, 348.
PROBLEMS:
GENERAL
91
Fig. 178.
Problems 349, 350.
350. For the arrangement described in
349, if a pressure of p = 200 psi is desired,
find the necessary value of F’.
351. For what value of @ will the force in
BD be zero, Fig. 179. Since for this condition, there is no load on BD, would you
advise eliminating this member?
Explain.
352-360. These numbers may be used
for other problems.
Fig. 179.
Problem 351.
Chapter V
FLEXIBLE
CORDS
53. Introduction.
Problems involving flexible cords are encountered in
electrical transmission lines and in suspension bridges. The load is usually
assumed either to be distributed uniformly in a horizontal direction or to be
distributed uniformly along the cord.
54. Parabolic Cord. When the load on the cord (or cable) is uniform
horizontally, the cord is curved to form a parabola.
An approximation of
this situation is suggested by Fig. 180(a), where the uniform load UV may
be considered as the road bed of a suspension bridge supported from the
Fig. 180.
Parabolic Cord.
cable AOB by the hangers.
In order that the curve AOB should be a parabola, an infinite number of hangers would be necessary.
In practice, the
spacing of the hangers is often such that the assumption of a parabolic form
for the curve AOB is a good engineering approximation.
To prove that,
AOB is a parabola, we may make a force analysis.
Choose
as a free body a section of the cord OC, Fig. 180(a), which includes the low
point of the cord O. Thus, in Fig. 180(b), the force Q is the tension in the
cord at O and is the force that must be applied to maintain the
section in
equilibrium.
Similarly, F represents the tension at any other point C, chosen
at random with the coordinates 2 and y, When O is the origin.
The total
load on this section of the cord is wx, because the loading
is w pounds per
foot of horizontal distance x. Since the loading is uniform
horizontally, the
vector representing the resultant of the distributed
load is located at the
92
§ 54] PARABOLIC
CORD
93
distance 2/2 from the origin, as shown in Fig. 180(b). There are no other
forces acting on this section of the cord; hence, it is in equilibrium under the
action of three forces which necessarily intersect at a common
point (§ 19).
From the sum in a horizontal direction, >/, = F cos 6 — Q = 0, we get
(a)
Q = F cos @.
From a vertical sum,
2F, = F sin 6 — wx = 0, we have
(b)
we = F sin @.
Equations (a) and (b) are the two independent conditions for this concurrent
system of forces.
Although a moment equation does not introduce another
independent condition, we may arrive at the equation of the curve easily by
taking moments about point C. Thus
(c)
2
IMo = Qy — a = 0,
or
ee
2
which is seen to be the equation of a parabola with its vertex at O. From a
force triangle of the forces Q, wz, and F, Fig. 181, we note that F is the hypotenuse; so that F? = Q? + (wzx)?, or
F =
=
(Q2 + wx?)2,
the tension in the cord at any abscissa x. Since the
tension increases as x increases, the maximum tension F’ occurs at the supports A and B.
At B,
x = L/2; hence the maximum tension (at the sup-
ie
Q
Fig. 181.
port) is
Bie G * eile
(10)
The value of Q may be found from equation (c) for any simultaneous values
Thus letting y = d, the sag, and x = L/2, half the span, we have
of andy.
we?
wl?
The slope of the cord at any point C is obtained from (a) and (b) by division;
this gives
tan 0 = ame
(d)
where, we recall, tan 0 = dy/dx. At the supports, we find
A
dae:
4d
(Gy)
tan
@
=
30
=
Fis
If the only load on a cable is its own weight and if it is tightly stretched so
that the sag d (Fig. 180) 7s less than 1 O% of the span, the curve of the cable
FLEXIBLE CORDS [Ch. V
I
closely approximates a parabola. In such instances, engineers frequently
use the foregoing force analysis, although the true curve is a catenary (S57)
55. Example. A cable is suspended between two points on the same level with a
span of 500 ft. and a sag of 50 ft. If the load is 2 Ib. per ft., uniformly distributed
horizontally, find (a) the tension in the cable at the low point, (b) the maximum tension, (c) the slope of the cable at the supports.
So.tuTion.
(a) From equation (11), we have
wL? _ (2)(500)?
= 1250 lb.
Sd (8) (50)
(b) The maximum tension is given by equation (10):
Fi
G = (ey]"- [(1250)? + (500)?}/2 = 1346 lb.
(c) From equation (12), we find
Ad
_,(4)(50)
= 21.8°.
6’ = tan = — = tane
L
500
°
56. Length of a Parabolic Curve. In the calculus, it is shown that the
length of any curve is given by the equation
-fb=@)]
Using the value of dy/dx = wx/Q from equation (d), we have the length of
the curve as
L/2
(e)
=a
0
wx?
ji+
1/2
S|
Q?
w2L3
de
=
Deis
24Q?
wtlhs
a
640Q!
’
approximately, where the integrand has been expanded by the binomial
theorem to three terms and integrated.
Using Q = wl?/(8d) from equation
(11), we obtain, from (e),
(13)
sa = b+
Sd*Teaa2d-
Tien WK
where 8,,.x is the length of parabolic cable between supports which are on
the same level. Do not use this expression for any other situation.
In
these expressions, substitute distances in feet, forces in pounds, and w in
pounds per foot.
57. The Catenary.*
When a flexible cord with a uniform weight w per
unit of length is suspended between two points and acted on by the force of
*John (1667-1748) and James (1654-1705) Bernoulli, during a conversatio
nal stroll,
agreed that a chain suspended freely from its ends would assume a position
such that the
center of gravity of the entire body would be in its lowest possible position.
This concept
defines the configuration of the suspended chain.
If the parts of the chain are very small
and of equal weight, the resulting curve is a catenary.
The master thought is the Bernoullis’
application of a basic principle concerning the center of gravity.
vation of equations may proceed in several ways.
The mathematical deri-
§ 57 | THE CATENARY
95
gravity only, the curve of the cord is called a catenary. When the sag d of
such a cord is large as compared to the span L (Fig. 182), the relations for the
parabolic curve are too inaccurate. ‘ To find the relations for the catenary,
take a section of the cord CD as a free body and show the horizontal tension
0
<k
=
a]
x
(b)
Fig. 182.
The Catenary.
Q, the tension F at any random point D, and the force of gravity ws, where s
represents the length of cord (or cable) from C to D and w is the unit weight
of the cord. As before, these three forces must intersect at a common point,
if there is to be equilibrium, but not at a distance of 2/2 from the origin. Since
DH
=)
and
DF, = F sm 0 — ws = 0,
F cos 8 =Q
and
F sin @ = ws,
— i Cop ve)
we have
(f)
from which
ws
tan 6 = —.
Q
Let
Q ==
[hy
Ww
a constant for a particular catenary.
have
Recalling that tan @ = dy/dx, we now
8,
(g)
En
k
dz
Using again the expression
30)
WP
ds = E =f (24) | aa:
da
we get
a
ds
ax
b
E ain (Z) |e = od
ue dx,
ds
(e+ ope
96
FLEXIBLE CORDS [Ch. V
whence, by integration, we find
= logds + (k? + s%)¥"] — log, A,
where — log, A is the constant of integration. Since s = 0 when z = 0, we
find that A = k. Using this value of A in the foregoing equation, we get
(h)
S - lo st (k rass?)!
aD,
/2
Now taking exponentials, we find the equation
ketl® = § + (kh? + 5?)¥/2,
Solving for s after transposing and squaring, we have
(14)
k kK __ p—a/k)
x
eS
2
a5
(e
Ca
r= Gisinh
i
a
Fig. 182.
Repeated.
where s is the length of cord measured from the low point C, Fig. 182.
Notice
that if the cord is symmetric (points of support on the same level),
the total
length of cord is 2s. If we use the value of s = k(dy/dx) from
(g) in (14)
and integrate, we obtain
(i)
Y oF B=
B(ean aE eo FiF))
where B is a constant of integration.
Now if the origin O is at a distance of
k below point C in Fig. 182, then y = k when x
= 0. Substituting these
simultaneous values of x and y in (i), we find B =
0. Hence, for the origin
O as shown in Fig. 182,
(15)
k
y= me + e*!*) = k cosh 2,
k
which is the equation of a catenary in cartesian
coordinates when the low
point of the catenary is a distance k = Q/w above
the origin O. The hyper-
§ 57 | THE CATENARY
97
bolic cosine of equation (15) may be written as a series, a convenience in
some applications.
Equation (15) then becomes
yay?
/2\4
1
fecN*
la
WA
y= bi + ni(z) ” nz) ar uli) +
|
Another relation that is often an aid in the solution of problems is obtained
from the equation
ao= [1+ (2) Y= [14 CV]en
or
dy =
sds
(s? + k?)12?
the integral of which is
yt EB = (st +R,
where E is the constant of integration. For the origin at O (not C), Fig. 182,
y = k when s = 0, in which case H = 0. Thus
y = (8? + ke),
(16)
we know that
From a force triangle of the three forces, F, Q, and ws,
fe =
(j)
Q? + w?s?
F=
or
(Oz ae ws?) 12,
Using Q = wk in (j) and then using equation (16), we find
F
(17)
=
(Q?
—
ws?) i?
—
w(k?
+
Sue
F = wy.
design, occurs
The maximum tension, which the engineer needs to know in
of x and y,
at the supports. Thus, if we substitute the maximum values
(k)
n=
3
and
y =k +d,
of support.
into the foregoing equations, we find values at the points
chapter will be
A résumé of symbols with particular meanings in this
For the parabola:
convenient.
w = pounds per foot of uniform horizontal load.
For either the parabola or the catenary:
are two values of
d = the sag (if supports are at different levels, there
the sag).
is L/2).
L = the span (the maximum value of x
horizontal component
Q = the horizontal force at the low point (also the
of the tension F at any point).
s = length of curve, as defined.
98
FLEXIBLE CORDS [Ch. V
For the catenary:
k = Q/w, the value of y for z = 0.
w = the weight in pounds of 1 ft. of cable.
A 60-ft. (= 2s) cable weighs 360 lb. and is suspended from two
58. Example.
points at the same elevation. For a sag of 15 ft., find (a) the tension at the supports
and (b) the span.
Sonution.
(a) For 2s = 60 ft., w = 360/60 = 6 lb. per ft. Using
y=k+d,
the value of y at the support, in equation (16), we get & = (s? — d?)/(2d), where, it
must be recalled, s is measured from O. Hence (s = 30 ft., d = 15 ft.)
k
#—@
(900 — 225)
2d
30
= 22.5 it.
Then, from (17), for y = k + d = 22.5 + 15 = 37.5, we have
ae
= Oi = (ONG)
= 22S Now.
(b) The span may be obtained from equation (h); thus,
i = roe
*+age
2
s? 1/2
|iE koe. (2+)
s
S0RP 37.
= (22.5)log, (Pee) = 24.7 ft.
which is measured from O.
The span is therefore 2x = L = 49.4 ft.
F,
max,
ee
eee
A
y
w=1 Ib./ft.
Wa
fe
B
a!
——t
Q
eee
k
O
&
Fig. 183.
59. Example. A cable weighing 1 Ib. per ft. is suspended
between two points on the
same level with a span of 300 ft. and a sag of 20 ft. What
are (a) the length of cable,
(b) the tension at the lowest point, and (c) the maximum
tension? See Fig. 183.
Sorurion.
(a) In some instances, as in this example, it is
necessary to make a
trial-and-error solution in order to obtain results.
Since the span L is known, we
naturally consider first those expressions containing
«. If we substitute the vale of
y = k + d from (k) into (15), we notice that
the only unknown will be k&. Thus, for
x = 150 ft. and d = 20 ft., equation (15) become
s
.
y=
k aL 90
=
5(este
+
The solution of this equation is obtained by
trial.
CMI).
Assuming various values of
99
PROBLEMS
until the equation balances, we find k = 566 ft.
Now from equation (14), we have (x/k = 0.265)
(Figure 183 is not drawn to scale.)
_k9 (er/k — e-2lk)
J
566
1
= re (o» _ za) = illed titi,
s=
which is half the length of the cable.
The total length is 303.4 ft.
(b) The tension at the lowest point is
Q = wk
= (1)(566) = 566 lb.
(c) The maximum tension is
Prax = wy = wk + d) = (1)(586) = 586 lb.
(Noting that the sag in this cable is so small that the parabolic formulas should give
a good approximation, the student should check the results by these formulas.)
Problems
361. A cable weighing 6 lb. per ft. is
suspended from two supports on the same
level.
permissible span? (b) What length of wire
Solve as a parabola.
is needed?
Ans. (a) 742 ft., (b) 761 ft.
(a) For a span of 800 ft. and a sag
of 60 ft., determine the maximum
tension.
(b) If six cables as described pass across a
supporting tower and if the span and sag
are the same on each side of the tower,
what vertical load must the tower carry?
Assume the curve to be parabolic.
Ans. (a) 8350 Ib., (b) 28,800 lb.
362. A cable is suspended with its ends
at the same elevation and 200 ft. apart.
The load is uniformly distributed horizontally. When the sag is 5% of the span,
the maximum tension is 2040 lb. What is
the load in pounds per foot?
Ans. 4 Ib. per ft.
363. The
sag is 8%.
same
367. Solve 366 as a catenary.
368. A wire weighing 4 oz. per ft. is
strung between two supports, 300 ft. apart.
One support is 15 ft. higher than the other,
and the sag, measured from the lower sup-
port, is 5 ft. Compute the tension at each
support and the length of wire.
Ans. 250 lb., 254 lb., 302 ft.
369. The same as 368, except that the
supports are 400 ft. apart.
as 362 except that the
364. A cable weighing 2 lb. per ft. is 400
ft. long. The distance between the points
of support, which are on a horizontal, is
398 ft. Compute the sag and the maximum
tension in the cable.
AG
SMe,
CBYAS Mlloy
365. A 3/8-in. copper wire may be safely
loaded to a maximum tension of 12,000 psi
For a sag of 20 ft.,
of cross-sectional area.
what may be the maximum safe span when
the points of support are on the same level?
Copper weighs 556 lb. per cu. ft. Assume
the curve to be parabolic.
366. A wire which weighs 1 lb. per ft. is
The
to have a sag of 10% of the span.
points of support are on the same level.
(a) If the maximum permissible tension in
the wire is 1000 lb., what is the maximum
Fig. 184.
Problems 370-373.
370. In Fig. 184, the cable weighs 2 lb.
per ft., h = 15 ft., y is unknown, and the
smooth pegs at A and B are of negligible
diameter.
Consider initially that the cable
is parabolic and that negligible lengths of
cable hang vertically.
If a = 10 ft. to the
left of the lowest point in the span, find the
weights W, and W2 for equilibrium.
Is the
sag to span ratio such as to justify recalculation as a catenary?
Ans. 534 Ib., 563 lb.
371. The same as 370 except that a = 10
ft. (locating W1) to the right of the lowest
point O.
PROBLEMS
100
372. The same as 370 except that the
location of W, is defined by a = 0.
373. In Fig. 184, the cable weighs 1/2 lb.
[Ch. V
at the lowest point is 1500 lb. Find (a)
the length of cable, (b) the sag, and (c)
the maximum tension in the cable.
(d)
per ft., h = 16 ft, Wi = 125.2 lb., and
W. = 132.8lb.
Thesupport pegs are fixed,
Solve this problem assuming that the cable
forms a parabola and compare answers.
smooth, and of negligible diameter.
Solve
as a parabolic cable to find the sag y and
the distance a of the low point O from the
left peg.
Ans. y = approx. 0.2 ft., a = 10 ft. to left
or to right of O.
374. A power line which weighs 0.25 lb.
377. If a cable, 200 ft. long and weighing
3 lb. per ft., is suspended from points on a
horizontal with a span of 150 ft., find (a)
the minimum tension, (b) the maximum
tension, (c) the sag.
Ans. (a) 164 lb.; (b) 342 lb.; (c) 59 ft.
per ft. is to be subjected to a maximum
tension of 1500 lb. The poles are 200 ft.
apart. If the supports are in a horizontal
plane and if the weight is assumed to be
uniformly distributed horizontally, what is
the corresponding sag? What is the length
of wire between poles? Ans. Sag = 5/6 ft.
375. A suspension foot bridge is supported by two parallel cables, one on each
side of the bridge. The suspension points
are in a horizontal plane. The span is
300 ft., the sag is 40 ft., and the cables
each weigh 3 lb. per ft. The floor of the
bridge is 4 ft. wide and weighs 30 lb. per
ft. of length. For a maximum permissible
tension in a cable of 20,000 lb., what
live
load per square foot of bridge-floor area
can this bridge sustain?
Neglecting stretch
due to load, find the length of each cable.
Ans. 14.8 lb. per ft.2, 309.6 ft.
376. A cable weighing 5 lb. per ft. hangs
as a catenary from points which are 600 ft.
apart and on the same level.
Ans. (a) 705 ft.; (b) 163 ft.; (c) 2315 Ib.
378. A cable in the form of a catenary
is 400 ft. long. How far apart may be the
supports on the same level, if the maximum
tension is not to exceed 400w, where w is
the weight of cable in pounds per foot?
379. The same as 378 except that Pmax =
500w.
Ans. 388 ft.
380. At the point of support of a cable
(catenary), the tension is 25% greater than
the tension at the low point. The cable
weighs 1 lb. per ft., and the sag is 20 ft.
If the points of support are on the same
level, find the span.
What is the length
of cable?
Ans. 111 ft., 120 ft.
381. If the tension at A in Fig. 185 is
10,000 lb., what is the tension at B?
382. (a) For the catenary shown in Fig.
185, what is the maximum permissible span
for a tension of 10,000 lb. at A?
(b) Solve
this problem assuming that the cable forms
a parabola and compare answers.
The tension
Ans. (a) 1656 ft.
g
— Ss
Fig. 185. Problems 381-383,
101
PROBLEMS
383. The tension at A in Fig. 185 is
10,000 lb. What is the elevation of a point
on the cable which is at a horizontal distance
of 500 ft. from the lowest point of the cable?
¢
Fig. 186.
Problems 384, 385.
384. The cable in Fig. 186 weighs 1 lb.
per ft. The lowest point is the point at
which a W = 100-lb. weight is suspended
as shown.
Find the reactions Az, A,, Bz,
and B,, assuming uniform horizontal distribution of the cable’s weight.
Hint: Some
part of the 100-lb. weight is supported by
the right side of the cable and the remainder
by the left side. Show as a free body the
whole cable, the cable section to the right
of the weight, and the cable section to the
left of the weight. You can also prove
that the point of support of the 100 Ib. is
the low point in the span.
Ans. A; = B, = 875 \b., A, = 77.5 Ib.,
1B = 10H.155Io).
385. The same as 384 except the cable
weighs only 0.5 lb. per ft.
386-390. These numbers may be used
for other problems.
Chapter VI
FRICTION
60. Introduction.
Friction is both a boon and a bane to mankind.
For
example, without friction we should not be able to walk; we could not get
our cars in motion, nor could we stop them if they were moving.
On the
other hand, we could well do without friction in such places as bearings and
in cylinders with reciprocating pistons.
In some circumstances, the study of frictional losses requires the use of
higher mathematics.
In this work, however, we shall confine our attention
to simple cases of one body sliding or tending to slide over another with little
or no lubrication.
For such cases, spoken of as dry friction, the laws of
friction have been determined experimentally with sufficient accuracy for
many engineering purposes.
See also § 67.
61. The Frictional Force.
Suppose you are going to pull your desk across
the floor. If you pull lightly, the desk does not move.
The force which
prevents its motion, if there are no obstructions, is a frictional force between
the floor and the desk. The magnitude of the frictional force is directly
dependent upon the magnitude of the pull which you exert.
The harder
you pull in a particular direction, the larger the frictional foree—provided
motion does not occur; that is, provided equilibrium is maintained
.
Of
course, if you pull hard enough, the force that you exert will overcome
the
maximum possible frictional resistance, and the desk moves.
Observe that
the frictional force always acts in a direction to oppose the
motion that
occurs or tends to occur between two surfaces.
While we shall not deal with motion as such in this chapter
, we shall be
concerned with problems where the body moves, since a
body is in equilibrium
when the velocity is constant.
In some cases, we wish to determine whether
or not motion will occur and, in other cases, we wish
to determine the resistance to motion, although most problems will of
course deal with bodies
in equilibrium.
The frictional resistance between two surfaces
which have no relative
motion is called static friction.
The frictional resistance between two sliding surfaces which do move relative to each other
is called kinetic friction.
102
103
§ 63 | LIMITING FRICTIONAL FORCE
A surface which offers frictional resistance is said to be rough, as opposed to
a frictionless surface which is said to be smooth.
A body A, which weighs 100 Ib., rests upon a 30° plane, as shown
62. Example.
in Fig. 187. What frictional force is needed to maintain equilibrium?
Sotution.
In the free body of this block
as shown in Fig. 187(b), we see that the
component of the weight W down the plane
tends to cause A to move downward. The
normal plane reaction N acts perpendicular
to the plane.
Only the frictional force F
tends to prevent motion.
The value of F
necessary for equilibrium may be obtained
from a sum of the forces parallel to the plane,
the direction of the x axis. Thus
=F, = W cos 60° — F = 0,
from which F = (100)(0.5) = 50 Ib., the frictional force necessary to prevent motion.
It may be noted that since only three forces are involved, the lines of action of these
center of
forces must intersect at a common point. Inasmuch as W acts through the
if
determined
be
may
point
common
this
contact,
of
surface
gravity and F at the
desired.
As previously pointed out, the frictional
63. Limiting Frictional Force.
to
force acts in a direction tending to prevent motion and is just enough
experiHowever,
maintain equilibrium as long as the body does not move.
of the frictional
value
lemating
a
is
there
case
ments show that in a particular
F’ is approxiforce
frictional
force, above which it cannot go. This limiting
of proporconstant
The
mately proportional to the normal force, F’«N.
tionality is called the coefficient of static friction f; thus
/
(18)
or
F’ = fN
f= 7
the actual
where F’ is the maximum possible frictional force, but not necessarily
impending.
is
motion
when
acting
frictional force. It is the frictional force
the limiting
Observe that the actual frictional force F may be less than
frictional force F’’, but never greater;
FSF
=jN.
than F’, equiIf F as computed on the assumption of equilibrium is greater
librium does not exist and an accelerating motion occurs.
a particular
The coefficient of static friction is reasonably constant for
. It varies
pair of dry surfaces for the more usual conditions met in practice
ental
experim
Typical
.
widely, however, with the nature of the surfaces
should be observed
values are given in Table I for normally dry surfaces, but it
from the stated
that actual values in particular cases may vary materially
104
FRICTION [Ch. VI
+
values because of different surface conditions.
Experience (or experiment)
is the only suitable guide to the value of the coefficient of friction to use in a
particular application.
TABLE
I.
COEFFICIENTS
OF FRICTION
These are typical values and should not be used in actual practice unless it is known
that they apply. The same statement is generally true of values found in handbooks
and other sources. However, see Marks Mechanical Engineers’ Handbook, 5th ed.,
for a good assortment of values and for numerous references to source material.
Surfaces
f., static
fi, kinetic
SaSc ICON OU Casu lO
inane
ae eee
feather onion ye.
Te. Sere
e
ee
igather‘on wood
-.>.-S, Seo
eee
Steelrunneron snow andice
........
Steel on steel, cleananddry.........
Steel on steel, surface oxidated. . . 2...
Steel on steel, wet withoil
.........,
Steel on steel, lubricated with graphite . . . .
hip on dry asphalt
2
2S 8a.
are
Waeodon wood: dry,
Baia
ein G
ate
Woodlon.caciaron Orye uate oa
oe
0.2
0.5
0.4
0.15
0.3
0.3
0.02
0.8
0.3
0.2
0.06
0.7
0.5
0.6
0.3
0.33
64. Coefficient of Kinetic Friction.* When a body moves over a dry surface,
the relation F = fN again holds true approximately.
In the case of motion,
however, the symbol f is named the coefficient of kinetic friction.
We may
use subscripts to identify these two coefficients of friction ; thus
fs, coefficient of static friction,
fx, coefficient of kinetic friction.
Since, in most circumstances, it will be evident by the context
which coefficient applies, we shall often omit the subscripts s and k.
Experiments show that the coefficient of kinetic friction
for a particular
pair of surfaces is lower than the coefficient of static
friction. Thus, the
locomotive engineer well knows that the locomotive produce
s the maximum
pull when slipping of the drivers is impending, not
when slipping actually
occurs.
He knows too that the maximum braking effect
can be obtained
only if the wheels are nearly but not quite locked
.
See Fig. 188 and read
the caption.
65. Examples.
(a) If the coefficient of static friction in the
example of § 62 is
f = 0.4, will the body A be in equilibrium?
“See Gould, “Determination of the Dynam
ic Coefficient
Conditions,” A.S.M.E, Transactions, vol.
73, p. A6l1,
of Friction
for Transient
§ 65 | EXAMPLES
Sotution.
105
Using the free body of Fig. 187(b), we find the sum of the forces in
y
the y direction as
=F,
from which
= N — W cos 380° = 0,
F.
N = (100)(0.866) = 86.6 lb.
N
“fhNe
The corresponding value of F’ is
Ww
F’ = (0.4)(86.6) = 34.64 lb.,
(b)
the maximum possible (limiting)
In § 62, y we
g frictional force.
Fig. . 187. R
ted.
ee
rd
found that a force of F = 50 Ib. is needed for equilibrium. Since
F’'(= 34.64) is less than the required value of /(= 50), the body is not in equilibrium, but will move down the plane.
Fig. 188. To recapitulate: suppose F’ for the desk shown is F’ = 30 lb. Then your
horizontal pull of @ = 10 Ib. in (a) does not move the desk and the frictional force between
Suppose you
the floor and the desk is also 10 lb., just enough to maintain equilibrium.
increase your effort until you pull with a force Q = 30 lb. = F’ in (b), then motion of the
desk is impending and a slight extra exertion will start it moving, because the frictional
Finally, if you pull steadily with some force
resistance is at its maximum possible value.
Q>F' in (c), say 40 lb., the desk moves and, moreover, it moves faster and faster if the
In this latter case, F = f,N, which is less than f,N,
pull Q = 40 lb. is maintained.
because fi <fs.
(b) If a force Q is applied to the block A of § 62, acting at 45° with the surface of the
plane as shown in Fig. 189, what should its value be in order that the block should be
in impending motion down the plane? As in example (a), f = 0.4.
So,ution.
we get
Choosing x and y axes as shown in Fig. 189,
(a)
(b)
=F, = N — W cos 30° + Q sin 45° = 0,
=F, = Wsin 30° — F’ — Q cos 45° = 0.
From (b), we get
(c)
F’ = (100)(0.5) — 0.707Q = 50 — 0.707Q.
Multiplying each term in (a) by f and noting that F’ = fN,
we find
fN = F' = fW cos 30° — fQ sin 45°
I (0.4) (100) (0.866) — (0.4)(0.707)Q.
(d)
F’ = 34.64 — 0.28289.
Using the two values of F’ in (c) and (d), we get
50 — 0.707Q = 34.64 — 0.2828Q,
Fig. 189.
Motion Im-
pending Down.
106
FRICTION [Ch. VI
from which Q = 36.2 lb. Observe particularly that the limiting frictional force is not
the same as it is in example (a), even though the coefficient of friction is the same.
The change is due to the effect of Q in changing the normal force VN. The student
should complete the problem by solving for F'’ and N.
It is possible for the force Q to be so large that motion impends up the plane, in
which case the limiting frictional force acts down the plane, as in (c).
(c) The same as (b) except that motion is impending up
the plane.
SoLtution.
The arrangement of forces for this condition
is shown in the free body of Fig. 190. Using x and y axes
as before, we get
Fig. 190. Motion Impending Up.
(e)
=F, = N + Q sin 45° — W cos 30° = 0,
S3r(f)
=F, = Q cos 45° — W sin 30° — F’ = 0,
where, in (f), the upward direction along the plane is taken
as positive. Following the method of the preceding solution,
we find Q = 85.6 lb. for
W =
100 lb.
66. Example.
If a body is relatively tall and slender, it may tip over instead of
sliding. Suppose W = 100 lb., f = 0.4, and the inclination of the plane is 30°, as
before. But let the body be as shown in Fig. 191 with a gradually increasing force
Q applied to it.
Will the body slide or tip over?
SotutioN.
To solve this problem, we find two values of Q:; that which would
cause the body to slide if it did not tip over, and that which would cause it to tip over
if it did not slide. In the beginning, we do not know the location of the line of action
of N, nor whether or not the frictional force is the limiting frictional foree. For the
first part of the solution, we assume that F’ is acting and determine the force
Q to
cause impending sliding.
Summing forces perpendicular to the plane, we
find
zF, = N — W cos 30°= N — (100)(0.866) = 0,
or N = 86.6 lb. Then
F’ = fN = (0.4)(86.6) = 34.64 Ilo.
as found in § 65(a).
in the x direction,
Q
m
/>
=P, = Q — F’— W sin 30°= Q — 34.64 — 50 = 0,
Wising
s
1m
|
Weos 30°
From the sum of the forces
we find Q = 84.64 Ib., the force needed to cause
the body to be on the point of sliding up the plane.
Ie
N
N
Fig. 191.
iy
=
ahs iff. ;
Body Tipping Over.
When tipping over is impending, the total plane
reaction will be at the corner O
a point on the line of action of N for this conditi
on.
The line of action of W goes
through the
center of gravity (eg).
Thinking then of O as the center of moment
s, we
see that if the body turns over about O,
the moment of @ with respect to O anus
be
greater than the moment of W about O.
When turning is impending, these two
moments are equal. This is one of those
situations where it may be easier
e to find the
moments of the components of W rather than
of W itself. With this approach,
=M, = 10Q — (6)(W sin 30°) — (2)(W cos
30°) = 109 — 300 — 173.2 = 0,
or Q = 47.32 lb., the force necessary to
cause impending tipping.
Since Q = 47.32
§ 67 | LAWS OF FRICTION
107
(tipping) is less than Q = 84.64 (sliding), the body will tip over before it slides. We
shall investigate this type of problem more thoroughly during our study of kinetics.
67. Laws of Friction. Situations involving friction may be divided into
three classes, dry friction, fluid friction, and the in-between class, usually
designated as imperfect lubrication.
Fluid friction exists when a fluid
(lubricant) separates the sliding surfaces completely.
In the case of fluid
friction, the friction is due to the shearing action on the lubricant.
The
shearing force for the lubricant depends on its viscosity and other significant
factors, such as the rate of shear. In the twilight zone between fluid friction
and dry friction, we find, as expected, all conditions from nearly dry to
nearly fluid friction.
Further discussion of fluid friction and imperfect lubrication is beyond the scope of this book,* except as problems may be solved
by a given coefficient of friction.
The phenomenon of dry (sliding) friction is not too well understood in
all its details.
It has been variously ascribed to the interlocking of the inevitable irregularities of an actual surface, to molecular forces at the contact
points or surface, and to a plastic deformation of the solid surface because
of the load.** Probably in each circumstance, one or more of these explanations apply.
In any case, experiments show that the coefficient of
friction of dry surfaces varies with the materials of the contact surfaces, the
load (significantly for extremes of load), the velocity (significantly for extremes of velocity), and no doubt with other factors.
Nevertheless, the laws of dry friction stated by Morin in 1831, and applicable to many engineering problems, will be used in this book.
These
laws are:
1. The coefficient of friction and the value of the frictional force are
independent of the area of contact.
[Doubtless, this is untrue for point contact. See (2).]
2. The limiting frictional force for two bodies in contact is directly proportional to the normal force; that is, the coefficient of friction 1s constant
for all values of the normal force.
[Some experiments at large intensity of pressure show that the coefficient
On
of friction may increase almost 50% for a fourfold increase in pressure.
the other hand, under certain other conditions, there is no change of the
coefficient of friction for this same pressure increase. |
3. The coefficient of friction is independent of the velocity, although static
friction is greater than kinetic friction.
*For applications of the laws of fluid friction to lubricated bearings, see Faires, Design
of Machine Elements, wherein references to more fundamental discussions are found.
For a more general treatment of friction, see Germant, Frictional Phenomena.
**See Dokos, “Sliding Friction under Extreme Pressures,” A.S.M.E. Transactions,
vol. 68, p. A-148.
65, p. 317.
Also, W. Claypoole,
“Static Friction,’
A.S.M.E.
Transactions,
vol.
FRICTION [Ch. VI
108
[This law is very much in error for wide variations of speed, but is substantially true for ordinary pressures and speeds less than about 600 fpm,
except for extremely slow speeds of the order of a few thousandths of a foot
per minute, and for speed variations of the order of 500 fpm.
speeds, f increases, approaching the values for static friction.
At the lowest
As the speed
increases, f decreases, but not in any well-ordered pattern for various situations. For instance, it changes from 0.2 at 2640 fpm to 0.06 at 5280 fpm
for iron on steel. In another example, friction wheels transmitting power
(§ 76), it is found from experience that f varies with the slip between the
wheels, the coefficient of friction at 100% slip (one wheel stationary) being
some one-fourth to one-third of the value at 2% slip (the difference in the
surface velocities of the wheels is 2%).|
In engineering practice, one needs to decide upon the coefficient of friction
to use with thoughtful care.
It is certain that the actual value of f will not
be the same as the value used in calculations.
Moreover, actual values of f
in a particular part of a machine will vary from moment to moment.
If it
is a matter of estimating the heat produced because of friction in a bearing,
for example, it will often be better to err on the high side, because if the
actual value of f is lower than the one used in computations, the bearing will
not “heat up” so much and there will be less danger of overheating.
On
the other hand, there are situations where a lowering of the coefficient of
friction below that expected may adversely affect operation.
For example,
the coefficient of friction between concrete pavement and rubber tires may
approach 1 under certain conditions, but become about 0.3 or less on some
wet pavements.
Low coefficients of friction have resulted in many highway
accidents.
Also a dangerous situation may develop if the coefficient of
friction of a brake decreases markedly.
68. Angle of Friction. When motion is impending, or when motion is
taking place, it is often convenient to replace the limiting frictional force
F’ and the normal force N by their resultant R, which is called the total
plane reaction. Figure 192 shows three cases in which the resultant of F’
and N is used, F’ and N being shown dotted in order to make clear that
Rf is being substituted for F’ and N.
In Fig. 192(a), the block A is resting in impending motion on an inclined
plane. Since the block is in equilibrium under the action of two
forces, R
and W, these forces must be collinear. The angle that the total
plane reaction & makes with the line of action of N is called the angle of static
friction
and is designated by ¢ (phi). Since the tan ¢ = F’/N (Fig. 192)
and since
f = F'/N, it follows that
(19)
f=
§ 69 | EXAMPLE—WEDGE
109
that is, the coefficient of static friction is equal to the tangent of the angle
of static friction. We also observe in Fig. 192(a) that the angle @ is equal
to the angle ¢. Evidently, if the inclination of the plane is increased slightly,
the downward component of W would be great enough to overcome the
limiting frictional force /’’ and the block would move.
Fig. 192.
Angle of Friction.
The angle of inclination of the plane when motion of a body under the
action of gravity is impending is called the angle of repose.
We observe that
the angle of repose is equal to the limiting angle of static friction. ‘This observation suggests a simple method of experimentally determining the coefficient
of friction.
Place a body on a plane and adjust the inclination of the plane
Then find the tangent of the
until the body is on the point of moving.
angle of inclination.
In Fig. 192(b), the force Q is so large that motzon 7s impending up the
Observe that, as before, ¢ is the angle between the lines of action of
plane.
Moreover, the lines of action of Q,
N and R, and that f = tan ¢ = F’/N.
The angle a is not in general
point.
common
a
at
R, and W must intersect
the angle of repose and the vector R slopes away from the line of action of
N in a direction such that it will have a component (/’) which opposes the
motion or the tendency for motion.
Again, in Fig. 192(c), a force Q is such as to induce the limiting frictional
force F'’, so that f = tan ¢ = F’/N.
If the body is moving, the same general relationships exist between the
The only
frictional force, the normal force, and the total plane reaction.
kinetic
of
coefficient
the
f;,
becomes
f
motion
of
difference is that in the case
F is
where
For this circumstance, we have f, = F/N = tan ¢,
friction.
the kinetic frictional force and ¢ is the angle of kznetic friction.
The use of the foregoing principles may be made clearer
69. Example—Wedge.
In Fig. 193(a), a resistance of
by applying them to common engineering problems.
the wedge B. Neglect
against
presses
turn
in
1000 lb. is exerted on the block A, which
all surfaces. What
for
1/3
=
f
that
assume
and
B
and
A
blocks
both
the weight of
force Q will cause the wedge B to be in impending motion downward?
FRICTION [Ch. VI
110
So.turion. First make a free body of A. There will be some advantage in this
problem in using the total plane reaction R in place of the components F’ and N.
However, we may consider first the components which are shown dotted in Fig. 193(b).
When the wedge
The forces N, and N» act of course normal to the surfaces as shown.
B moves downward, A must move toward the left; hence /; must act toward the right
to oppose the motion. With Ff, and N; acting as shown, 2; must point upward toward
the right at an angle of ¢ with the normal force N;._ To determine the direction of F2,
note that if the wedge B moves downward, this is the same relatively as A moving up
along B; that is, #20on A must act downward to oppose the relative motion of A upward
along B. Another point of view is that if B moves down, it will evidently exert a
frictional force fF, on A tending to push A down (and the free body shows the forces
Fig. 193.
Wedge.
acting on the body). Thus, R», the resultant of F2and N», must act downward toward
the left on A at an angle of ¢ with the normal No. The frictional angle ¢ is the same as
before because the coefficient of friction is the same.
Since this free body involves a known force and two unknowns, R; and R 2, these
For
unknowns may be found.
ii =
tan o=
1
3”
we find
ous
26);
Now by summing forces horizontally and vertically, we find
(g)
=F, l|= 1000 + Risin ¢ — Ry cos(¢ + 15°) = 0,
(h)
=F, = Ri cos ¢ — Rysin(¢ + 15°) = 0.
By a simultaneous solution of (g) and (h), we get Ro = 1536 lb.
Next make a free body of B.
We may place N» and F» (and R;) on this free body
immediately, because we recall that Newton’s laws provide that the action
of A upon
B shall be equal and opposite to the reaction of B upon A.
Then, we note that Rs must
have an upward component to oppose the downward motion of B and that it acts
at
an angle of ¢ = tan“(1/3) with the normal N3, since f and ¢ are the
same
surfaces. The equation
=F, = Rzcos(¢ + 15) — Rs cos(¢ + 15) = 0,
shows that Rs = R. = 1536 lb. From the vertical sum,
=P, = 2R.sin(¢ + 15) —Q = 0,
we find Q = 1690 lb.
for all
§ 71] BELT FRICTION
ALi
70. Example—Axle Friction. A 3-in. shaft is supported by bearings on 20-in.
centers A and B, Fig. 194. The loading on the shaft is equivalent to two forces,
C = 2000 lb. and D = 1200 lb., located as shown, as far as the bearing reactions are
affected. What is the frictional force at each bearing if f = 0.008?
Total
Resistance
(a)
(b)
Fig. 194.
Journal Friction.
Sotution.
The first step is to determine the magnitude of the bearing reactions
Aand B. Taking moments about A we have
=Ma = 10C — 12D — 20B = (10)(2000) — (12)(1200) — 20B = 0,
from which
B = 280 1b.
Taking moments about B we have
=Mez = 30C + 8D — 20A = (30)(2000) + (8)(1200) — 20A = 0,
from which A = 3480 1b. A check by a sum of the vertical forces shows that there is
no error in the solution of these equations. Now the coefficient of friction for a journal
bearing is defined as the frictional resistance at the periphery of the shaft (the total
amount of which is represented by F’) divided by the bearing reaction.
If the bearing
reaction is represented by N, we then have f = F/N, as before. Thus the frictional
force at B is
Fx = (0.008)(280) = 2.24 lb.;
and that at A is
F'4 = (0.008) (3480) = 27.84 lb.
The frictional moment at bearing A is My = Far = (27.84)(1.5) = 41.76 in-lb.
71. Belt Friction. The engineer often needs to
estimate the frictional effects of a flexible band
wrapped around a cylindrical body, as in belt drives
and band brakes.
In Fig. 195(a), consider the
cylinder A as stationary, although this is not
necessary, and note that the belt BC wraps about
the cylinder, subtending an angle 0. Let a pull 7;
be exerted, tending to pull the belt about the pulley
against a resistance at the other end of the belt of
Length of
eeuians -dL5
(AN
3
C
7
D
T,
(a)
T2, such that slipping of the belt on the cylinder is — Fig. 195(a). Belt Friction.
Evidently 7 is necessarily larger than
impending.
T,. In the conventional analysis of this situation, a free body of an elementary length of belt dZ is used and the coefficient of friction between the belt
and the cylinder is assumed to be constant.
FRICTION [Ch. VI
112
The free body is shown in Fig. 195(b). We find a normal force dN and
the corresponding limiting frictional force dF’ = f dN, which opposes the
tendency to slip. A pull P is shown on the
right end of the element, and, because the
tension increases in a counterclockwise direction, a pull of P + dP is shown on the left
end of the element.
Since the angle subtended by the element is dé, the angle
between the forces P and P + dP and the
chosen x axis is d0/2. Now, summing forces
in the x direction [Fig. 195(b)], we get
LF gam)
(b)
Fig. 195(b).
bot OP) cos F — P cos
— aF’
Belt Friction.
= dP cos
— dF’ = (0)
Since the angle dé is very small, we may let cos(d@/2) = 1 and find
dha =EdP,
In the y direction, we have
2F, = dN — (P + dP) sin
(i)
=
dN — 2P sin
Y— P sin @
— aP sin
= 0,
The sine of a small angle is very nearly equal to the angle itself; hence, we
may use d6/2 for sin(d@/2). Also the product of two infinitesimal quantities,
dP and d6/2, may be neglected.
With these changes, equation (i) becomes
dN = P dé.
Multiplying both sides of this equation by f, noting that f dN = dF’ = ar.
and separating the variables, we get
oy
/
T2
dP.
0
Dey, = p / dé,
0
where the limits of the integration with respect to P are from the smaller
T: to the larger T, and where dé is integrated for the entire angle of contact,
from 0 to 6. The integration gives
(20)
loge
ie
= fo,
2
or
iL
Frees ise
2
or
T, = Tre",
This equation gives the relation between the tight tension 7’, and
the slack
tension 7, when the angle @ is expressed in radians and when
slipping is
impending.
Moreover, this equation entirely neglects the effect of the
113
§ 72 |EXAMPLE
centrifugal force, which is a significant factor in high-speed belts (speeds
greater than about 3000 fpm).
Evidently the total frictional force between the belt and the cylinder is
B15 Jie.
In work with brakes, we usually express the effectiveness of the brake as the
moment of the frictional force about the center of rotation of the brake drum;
this moment is called the braking torque. ‘Thus, the braking torque or
frictional moment is M,; = FD/2 (Fig. 195(a), where D is the diameter of
the brake drum.
pyc
(a)
Fig. 196.
Band Brake.
72. Example.
A band brake, as shown in Fig. 196, consists of a leather band on a
14-in., cast-iron drum, for which f = 0.3. The data are: a = 18 in., b =4
in,
D = 14 in., and 6 = 270°. Determine the frictional torque for a clockwise rotation
of the drum and a force Q = 100 lb.
SoLution.
First make a free body of the lever AB [Fig. 196(b)] and find the pull
on the band at B from a sum of the moments about the pivot point C. For a clockwise
rotation of the drum, the pull at B is smaller than that at C; that is, the force at B is
T, and that at C is 71, if the notation of equation (20) is used. We have
=M¢ = (100)(18) — 4 T2 = 0,
from which 7, = 450 Ib.
Then, from (20), we find
Ty = Tet? = 450 -8%8r/2 = 1850 Ib.,
The frictional force, as seen in Fig. 196(c), is
where 6 = 270° = 3x/2 radians.
‘The frictional torque is
lb.
1400
=
450
—
1850
=
T2
—
F = 7,
M, = (1400)(7) = 9800 in-Ib. = 817 ft-lb.
The reaction R, which would be useful in designing the pin C, may now be found by
principles previously explained.
Figure 197 shows two applications of belt friction.
FRICTION [Ch. VI
114
gzoronemneng
15
(a) Differential Back Stop.
(b) Car Spotter.
Fig. 197. Applications of Belt Friction. Belt friction is useful in many ways.
(a) Is
an illustration of a differential band brake used as a back stop; that is, it is a self-locking
brake for one direction of rotation.
By self-locking, we mean in this instance that once
the brake is applied, it locks the wheel or brake drum.
This property is useful as a safety
precaution in hoists and conveyors, since the back stop prevents the load from going
down in case of power stoppage.
The self-actuating property is provided by the way
in which the band is attached to the actuating member (the lever AB, Fig. 196). A differential band brake is not necessarily self-locking.
See problems 507 and 508.
The capstan driven by a small motor in (b) does the work in spotting freight cars. This
same machine may be used for a variety of other moving jobs.
73. Pivot Friction. In collar bearings and end bearings, in disk clutches
and other circumstances, there is relative motion or a tendency for motion
between surfaces of an annular shape where a normal pressure exists between
the surfaces.
In a step bearing or end bearing, Fig. 198(a), the greatest wear
occurs at the largest radii because the corresponding rubbing speeds are
greatest. Since the outside portion of the bearing wears away faster, undesirably high pressures soon occur near the axis of the bearing.
For this
reason, an annular thrust disk is used or the shaft may be bored somewhat
as suggested at A, Fig. 198(a), in order to remove entirely the surface on
which the pressure is likely to be excessive.
(a)
End Bearing
(b)
Disk Clutch
Fig. 198.
Pivot Friction.
§ 73 | PIVOT FRICTION
115
The disk clutch presents a similar problem.
A diagrammatic representa-
tion of a typical disk or plate clutch is shown in Fig. 198(b). The member
C may slide along the shaft D but rotates with the shaft. If a force N,
usually exerted by springs in automotive clutches, is applied to the parts B,
the arrangement is such that these parts B move toward the left and clamp
the disk C between B and G. The capacity of these contacting surfaces to
resist slipping permits the transmission of power from shaft E to the shaft D.
Compare Fig. 198(b) with Fig. 199, which gives some photographic details
of actual clutches.
Either one of two assumptions is usually made in computing the frictional
torques for such cases:
1. The wear may be assumed to be at a constant rate for all points on the
contacting surfaces, or
2. The pressure is assumed to be uniformly distributed over the contacting
surfaces.
Uesciite Plate
»
Sprin J\
Bey
Pressure
Operating
Levers
7) —~Pressure
|
Plate
saa
Plate
L
” ~Flywheel
Courtesy Borg & Beck, Chicago.
Courtesy Link-Belt Co., Chicago.
(a)
(b)
In this type of clutch, the spring
Single Disk Automobile Clutch.
Fig. 199(a).
applies pressure to the pressure plate which in turn presses the disk against the face of
When the clutch pedal is pushed
This is the “engaged” position.
the flywheel.
down, a release bearing engages the lever A and results in a force at B which moves
Notice that the disk engages the driven
the pressure plate away from the disk.
shaft through splines. When the clutch is engaged, the splined shaft turns with the
flywheel. There are two annular surfaces, each transmitting a frictional torque.
Fig. 199(b). Industrial Twin-Disk Clutch. This clutch has two disks C and four annular
The spring designated in this picture pushes the pressure plate
friction surfaces.
away from the disks, and it serves the purpose of separating
pressure exerted through the operating levers is released.
the
faces
when
the
116
FRICTION [Ch. VI
In either case, the coefficient of friction is also taken as constant for all points
in contact. Although neither of these assumptions is in accordance with
the facts, the results for either assumption are generally nearly enough correct
for most engineering purposes.
To illustrate the procedure, let us find the frictional torque for two annular
surfaces in contact for the case of uniform wear.
The wear in general is
proportional to the work of friction, which in turn is proportional to the
intensity of the pressure p in psi, and to the distance from the center of
rotation p (because of the change in speed). Thus, it is assumed that the
wear is proportioned to pp = C, a constant.
Now, in Fig. 200, consider
a differential
area
dA =p dé dp, subjected to a normal force dN.
Since force is unit pressure p times the area,
dN
=p dA
for uniform wear.
force N gives
=
pp d@dp
= C de dp
Integration for the total normal
To
20
N = fdN =C / / dp d0 = 2nC(r, — r,),
Ti
Fig. 200. Friction
0
from which
Surface.
N
Gj)
C= Qn(r, — ri)
The frictional torque is the moment of the frictional force about the axis,
M, = pdF’, and
dF’ = fdN = fC dé dp,
where p is the distance of dA from the axis (moment arm of dF ’).
then that the frictional torque for the differential area dA is
We see
dM, = p dF’ = Cfo dp dé,
or
M,=
cf | [Vode do = ont (nt = re),
rs
0
Using the value of C from (j), we find
(21)
M,
=
Nrf(r,?
2a(1o
ae 1?)
—
=
fN
("a a)
11)
where r, is the outside radius and r;, is the inside radius
of the annular area.
If r, and r; are in inches, the torque is in inch-pounds;
if in feet, the torque
is in foot-pounds.
The student should now be able to find the equation for
the frictional torque if the pressure is assumed to be
uniform. See problem
475. Observe that equation (21) gives the frictional
torque for one pair of
surfaces in contact and that in the typical automotive
plate clutch, Fig. 199(a),
there are two pairs of surfaces transmitting the power.
§ 74] FRICTIONAL TORQUE FOR SCREW WITH SQUARE THREAD
117
74. Frictional Torque for Screw with Square Thread.
Square screw
threads are often used on jacks, shaft straighteners, and screw presses of
various kinds. By virtue of the screw thread, a large mechanical advantage
may be obtained, so that a relatively small force applied at the end of a lever,
which tends to turn the screw, may produce a large force along the axis of
the screw.
Although the engineer can never be too sure of the true value of
the coefficient of friction, he often needs to estimate the frictional torque
necessary to turn the screw against an axial load by using an approximate
value of the coefficient of friction.
First, we note that the load W produces
a pressure between the threads in contact
in the base of the jack, Fig. 201. If the
screw is turned, the threads rub on each
other, so that the applied effort must be
ER
_ Load
ese
great enough not only to raise the load
W, but also to overcome the frictional
resistance.
If we observe that a thread
is simply equivalent to an inclined plane
wrapped around a cylinder and that raisFig. 201.
Screw Jack.
ing a load W on a jack is equivalent to
pulling a load up an inclined plane, the force analysis is easy to understand.
Imagine the load in the form W of Fig. 202 being pulled up the incline of
the thread by a force P perpendicular to the axis
“si
Dias De
of the screw and always acting tangent to the
mean circumference of the thread. The reduction
of the area supporting the load does not affect the
mechanics problem, because the coefficient of
friction is practically independent of the area
(§ 67). Thus, considering the thread to be unSSS*>
wrapped from the screw, we may picture the
Load W on aThread
forces as shown in Fig. 203, where FR is the total
Fig. 202. Plan View of plane reaction, the resultant of 7’ and N. For
Thread,
Load
Concentrated
these forces in
on Small Area.
oreo?
De
equilibrium,
we may write, from the equation ZF’, = 0,
(k)
R
P =RsinQ + ¢).
Similarly, from DF, = 0, we get
(1)
y
W = Recos(\ + 4).
Ane akan
_W= Load
ean Circumference
From equations (k) and (1), by division,
[ton Dy,
we find
Fig. 203.
Raising Load.
is perpendicular to the
(m)
P = W tan(a + ¢).
The force P
axis of the
screw. \ = tan—[lead/(7D,,)].
FRICTION [Ch. VI
118
Recall that the force P is perpendicular to the axis of the screw, so that the
Thus, multiplying
torque (moment of P about the axis) is M = PD,/2.
both sides of (m) by D,,/2, we find
199
(22)
M
PDA 9See WD
ie
2
2
aay poh
a
tan (A + ¢)
=
WD.»m f tand + tan ¢d
2
1 — tan A tan ¢/’
where Jf = the turning moment or torque
Cap Does
in inch-pounds (D,, in inches)
ie =
:
Pivot Friction
that must be applied to the
Here
screwto raise the load W. (In
= 5 Handle @
“ the force
aes
ane
ao
case
Q on the han
biter
EB
osLoad
on These
<=
Friction Here
dle is desired, note that IZ =
Bor
at
a
ae
| oP
Threads
Qa, where a is
the radius
Pitch
ste
Base
from the axis of the screw to
BbaeHity3
the point of application of
Q, shown in Fig. 201),
Fig. 201. Repeated.
D., = the mean diameter of the
square thread in inches, D, = (D, + D;)/2 (Fig. 202),
tan ¢ = f, the coefficient of friction,
@ = the angle of friction, and
\ = tan “{[lead/(7D,,)], an angle called the lead angle.
=f
Screws may be single-threaded screws, double-threaded, three-threaded,
four-threaded, or more.
The lead of a screw is the distance a nut would
move, measured along the axis, in one turn. The pitch is the axial distance
from a point on a screw thread to a corresponding point on the adjacent
thread (see Fig. 204).
1
Number of threads per inch
Pitch in inches = <
Fig. 204. Pitch and Lead. In multiple-threaded
screws, the word thread as commonly used has two meanings.
We say, for example, that a triple-threaded screw
has
three threads running parallel and we trace a “thread
” and find that we trace from A to
Bin one turn. On the other hand, in calling the pitch
the axial distance between corresponding points on adjacent “threads,” the word
has another meaning.
Lead and pitch
are the same in a single-threaded screw which has
only one “thread ” En eneecke
know what pitch and lead mean have no difficult
y with the words
The ied
ictured
is not a square thread; it is known as an Acme
thread.
;
pes
§ 74 | FRICTIONAL TORQUE
FOR SCREW
WITH
SQUARE THREAD
119
For example, if there are 4 threads per inch, the pitch is 1/4 in. (Read the
caption to Fig. 204.) In a single-threaded screw, the lead is equal to the
pitch. In a double-threaded screw, in which two threads run parallel about
the axis, the lead is equal to twice the pitch. In a triple-threaded screw, in
which three threads run parallel, the lead is equal to three times the pitch;
etc. Single-threaded screws are nearly always used when the principal purpose is to obtain a good mechanical advantage (mechanical advantage is the
ratio of the load moved divided by the effort required to move the load, W/Q
in Fig. 201, including the mechanical advantage of the lever or handle).
Multiple-threaded (double, triple, etc.) screws are used generally for power
transmission because, as shown below, the larger lead angle results in a
greater efficiency.
Thus, single-threaded screws are generally found on jacks
and presses, and multiple-threaded screws on worm drives where efficiency
is important.
The larger diameter of the screw D, is called the outside diameter or prefer-
ably the major diameter; and the smaller diameter D; is called the inside
diameter, preferable the minor diameter.
Observe that equation (22) does not account for the pivot friction which
occurs at the cap in Fig. 201. Although only a small error is caused by
neglecting this source of friction, it may be partially accounted for by adding
the frictional torque for the pivot friction as obtained from, say, (21) to the
torque obtained from (22).
The efficiency of a screw may be defined as the torque required with no
friction to turn the screw with a load W on it divided by the torque required
If there is no friction, the angle of friction ¢ = 0 and equation
with friction.
(22) becomes
(n)
(ee
WDn tan X,
2
Therefore, the efficiency is,
the torque to raise W, Fig. 201, with no friction.
from (22) and (n),
M’' _ tan) (1 — tan ¢ tan d)
Met 7
tan \ + tan¢
(0)
_
om
tand(1 — ftand).
tanr +f
The value of ¢ increases when \ increases—up to \ ~ 45°, roughly.
The student should now make an analysis to show the situation when the
load is lowered, that is, for example, when the car is let down with a screw
left
jack. Make a new free body, Fig. 205(a), with P acting toward the
and with R in its proper new direction.
foregoing one, find
(23)
With a procedure analogous to the
WDn
i
M = —5* tan (¢ — 2).
120
FRICTION [Ch. VI
If M as obtained from (23) is positive (that is, if ¢ > X), it is indicated that
an effort must be made to lower the load and the screw is said to be self-locking. If M is negative, the torque must be exerted to keep the load from
moving down of its own accord. In this case, \ > ¢, and the screw is not
self-locking. It is desirable that such screws as jack screws be self-locking.
(a) Self-Locking
(b) Not Self-Locking
Fig. 205. Condition for Self-Locking Screw.
If $<) as in (b), the component of R up
the plane is smaller than the component of W down the plane.
If, for example, this
condition existed in a jack,"then the load would move down (the jack screw would move
down) of its own weight. This would be a bad situation in a jack, but in a worm drive,
it may very well be an advantage, because then it would be possible for the gear to drive
the worm
(that is, the drive could be reversed).
If \ were only a little smaller than ¢,
a screw might work itself down if it were subject to vibration, although it would not do
so in a static situation.
Since efficiency increases with \ and since we may be interested in the
condition of self-locking with maximum efficiency, we note that the corresponding maximum value of \ isA = ¢. From (0), we see that the corresponding efficiency is (tan \ = tan ¢ = f)
io)
2)
en 2
oer
Oe
which shows that if the self-locking feature is desired, the maximum
possible
efficiency is something less than 50%.
Incidentally, if self-locking were
important, one would not dare try to design for the exact condition
of \ = g,
because the coefficient of friction and $ cannot be predicted
with accuracy.
75. Examples.
(a) A 3/4-in. Square-thread screw used in a jack has
5 threads per
inch and a minor diameter of 0.575 in. Allowing for
some lubrication, the coefficient of friction should be about f = 0.12 after motion
begins. If this is a singlethreaded screw, determine its efficiency and find out
if it is self-locking.
Sotution.
The size (3/4 in.) of a screw is its major diamete
r.
Therefore, the
mean diameter ig
wae Do+ Di _ 0.75 + 0.575
5
9
= 0.6625 in.
Since the pitch (lead) is 1/5 = 0.2 in. (the recipr
ocal of 5 threads per inch), we find
0.2
Lead
tan \ =
tDm
=
(0.6625)
= 0.0961.
§ 76 | FRICTION WHEELS
121
Now using equation (0), we get
S
tan \(1 —ftanr)
tanxi+f
0.0961[1 — (0.12)(0.0961)]
= 44%,
0.0961 + 0.12
=
After observing that tan ¢ = f = 0.12 is greater than tan \ = 0.0961, we know that
¢> and that therefore the screw is self-locking. Read the caption to Fig. 205.
(b) Determine the same items if the foregoing screw is a five-threaded screw, other
conditions remaining the same.
in.
Soutution.
Then
For a five-threaded screw and a pitch of 1/5 in., the lead is (5)(1/5) = 1
tan >’ =
and
Lead
tDm
S
1
(0.6625)
= 0.48,
_ 0.48[1 — (0.12)(0.48)] = 15.4%.
r
0.48 + 0.12
For this screw, tan \ = 0.48 is greater than tan ¢ = 0.12, and the screw is not selflocking. Since the coefficient of friction given is the kinetic coefficient of friction, the
torque required to start the load will be greater than that obtained by substituting
f = 0.12 into equation (22) by some 25-30%.
76. Friction Wheels. Friction wheels
are often used to transmit a small
amount of power, especially where the
slipping of the wheels on one another
is not objectional or is even advantageous.
In Fig. 206, either wheel A or
wheel B would be so mounted that it
could be moved toward the other and be
pressed against it, thus inducing some
Fig. 206. Friction Wheels.
normal force N between the wheels.
The maximum possible frictional force is then F’ = fN. The corresponding
turning moment or torque on the wheel B is then F’Dzg/2. The torque on
the shaft of A is F’D4/2, which, it is observed, is not the same as that on B.
Photographic illustrations of friction drives are shown in Figs. 207, 208,
Courtesy Rockwood
Fig. 207.
Spur Friction Wheels.
Manufacturing Co., Inc., Indianapolis.
Fig. 208.
Bevel Friction Wheels.
FRICTION [Ch. VI
122
manuand 209. Asan example of a recommended coefficient of friction, the
and
207
Figs.
facturer gives f = 0.280 as a working value for the drives in
209 when the materials are tarred fiber and cast iron. The smaller (driving)
For
wheel has the fiber (softer) surface.
the disk drive, Fig. 209, designed for use
in a tractor,
.
.
the
manufacturer
recom-
mends the use of f = 0.323 for a tarred
The
fiber wheel and a cast iron disk.
speed of the driven disk in Fig. 209
may be varied by setting the wheel at
different distances from the center of
rotation of the disk.
77. Rolling Resistance.
Since no material is perfectly rigid, any force, no
matter how small, will cause a deformation at the point of application of the
Cos Into® “force.
Courtesy Rockuod Manufacturing
FOr example, al tigid avieel walk
| Press into the surface on which it rests,
or a rigid surface will deform the wheel,
as suggested by Fig. 210. Since neither material can be rigid, both bodies
deform and the actual situation will be something between those shown in
Fig. 210(a) and (b). In any event, deformation is the source of so-called
rolling resistance, Inasmuch as the wheel must be moving against a component of the reaction R between the ground and wheel at all times.
Let the wheel, Fig. 210, be on the point of moving or be moving with a
constant velocity (in equilibrium).
Then the only forces acting on the wheel
are its weight W, the motivating force Q, and the resultant reaction R with
the surface CH. Since three forces are in equilibrium, their lines of action
intersect at a common point. Thus, if Q acts through O, the center of the
wheel, Ff necessarily acts also through O (which is assumed to be the center
of gravity of the wheel).
Fig. 209.
Friction Disk and Wheel.
Fig. 210.
Rolling Resistance.
128
§ 78 | CLOSURE
If the point A is taken as the center of moments, we find Wa = Q(BQ).
Since the indentation of the surface is ordinarily quite small in relation to the
size of the wheel, BO is practically equal to D/2, the radius of the wheel.
Therefore, we may write
Wa=
ee
or
Q = a
Some experimental values of a, called the coefficient of rolling resistance
(not a coefficient of friction), are available, but ordinarily little reliance can
However, we note from this
be placed on the results from this equation.
analysis that the softer the material, the greater is the indentation, the value
of a, and the resistance to rolling. Moreover, we understand why a heavy
load offers a greater resistance to rolling, than a light load, other conditions
The factor 2a/D can be determined experimentally
remaining the same.
for particular conditions and used as a coefficient modifying the weight on
For example, a typical value for an automobile tire on pavement
the wheel.
If a car weighs 3600 lb., its resistance to rolling would be
0.025.
=
is 2a/D
about Q = 0.025W = 90 lb. for the specified conditions.
Resistance to rolling is much less than that to sliding. For this reason,
in order to reduce the required starting effort, ball and roller bearings are
widely used and are especially advantageous where a moving part is stopped
and started frequently.
In problems involving rough surfaces, the usual care in
78. Closure.
The only difference in the prosetting up the free body should be taken.
cedure is the addition of the frictional force which is absent for “smooth”
Hence, if the free-body idea has been mastered, the student will
surfaces.
quickly learn to apply the principles of this chapter.
A body subjected to one or more frictional forces may be:
1. In motion, either in equilibrium or not, in which case F=f,N,
2. In equilibrium and with motion not impending, in which case F < Ff’ =
fsN, or
3. In equilibrium with motion impending, in which case F = F’’ = f.N,
situations,
where in each instance F is the actual frictional force. In some
may be
§ 66, the dimensions of the body and the application of the forces
being on the
such that the body turns over instead of sliding or instead of
because
mind,
in
kept
be
to
need
ies
point of sliding. These various possibilit
far
carried
is
it
unless
the solution to a problem is not known to be correct
nonor
um
enough to determine these details of the state of its equilibri
equilibrium.
n which
For a body in limiting friction, we now have an additional equatio
are
apply
which
s
The four equation
makes it possible to find four unknowns.
XM =0.
2F, = 0,
ya ED)
F’ = fN,
PROBLEMS: FRICTION, PLANE SURFACES
124
[Ch. VI
For
We are not always interested in using all four of these equations.
example, the moment equation may be used to determine the location of
the line of action of the normal force N, yet oftentimes, as in § 65 and others,
we are unconcerned about the exact location of N.
Each of the special cases discussed in this chapter is of practical importance, but note especially the case of belt friction, 71/72 = e/, because it
will be repeatedly useful in other parts of this book.
Problems
FRICTION,
PLANE
391. A 2000-lb. body is resting on a
rough horizontal surface for which f = 0.2.
A force of 100 lb. acts upward toward the
right at an angle of 30° above the horizontal.
What is the frictional force? What is the
limiting frictional force?
SURFACES
horizontal.
(a) If f, = 0.6 and f, = 0.4,
will the block remain where it stops sliding
up the plane, or will it slide back down?
Show why fully. (b) Same as (a) except
that f, = 1/2.
Ans. 86.6 lb., 390 lb.
Q
~
6
Fig. 212.
Problems 397-399.
397. If f; = 0.4, W = 100 lb., and the
Fig. 211,
Problems 392-395.
392. A 1000-lb. body is resting on a
rough horizontal surface for which f =
0.25. A compressive member CD transmits a 2000-lb. load to the body, as shown
in Fig. 211.
(a) If 6 =0, what force Q
will result in impending motion toward the
right?
(b) If Q were equal to zero what
would happen?
Ans. (a) 1683 lb., (b) moves leftward.
393. The same as 392 except that motion
is to be impending toward the left.
394. The same as 392 except that 9 =
horizontal foree Q = 100 Ib. in Fig. 212,
find (a) the largest angle @ for equilibrium
and (b) the smallest angle @ for equilibrium.
Ans. (a) 66.8°, (b) 23.2°.
398. The same
as 397 except that Q =
20 Ib.
399. A 108-Ib. block rests upon an inclined ramp which is directed upward toward the right at an angle of 30° with the
horizontal.
A 30-Ib. force acts horizontally
toward
the right against the block which
is then in impending slipping.
Determine
the frictional force and the coefficient of
friction (Fig. 212).
Ans. 28 lb., 0.258.
ie?
395. For the body in Fig. 211, what is
the value of @ and Q such that force Q is
the least value that will cause impending
motion toward the right when f = 0.25?
SUGGESTION: Construct a force polygon,
noting that the line of action of the total
plane reaction R must make the angle
¢ =tan'f with
the vertical
direction.
The shortest vector Q that will close the
polygon is the minimum value.
Ans. 14°, 1630 lb.
396. A block of weight W is projected
up a plane that is inclined 30° with the
Fig. 213.
Problems 400, 401.
400. A block A, weighing 300 lb., is
acted upon by a force Q whose direction
is at @ = 75° with the vertical, as shown
in Fig. 213. A weightless cable connects
A to another block B, which weighs 500 lb.
If fa =f
= 1/3,
(a) what force Q will
PROBLEMS:
cause
FRICTION,
impending
PLANE
motion
125
SURFACES
toward
the left?
(b) What is the tension in the cable?
Ans. (a) 303 Ib., (b) 167 Ib.
401. The same as 400 except that fz =
0.6.
407. Let the disk A described in 405
have a weight of W = 200 lb. and let
Q = 50 lb. (Fig. 215). Find the distance
z at which motion is impending.
408. For the disk A described in 405,
let Q = 500 lb. acting at a distance x =
9 in. above the top surface of the shaft
(Fig.
215).
Another
horizontal
force
P
acting toward the right is applied to the
disk at a distance of 6 in. below the aais of
the shaft.
What least value of this force
P will cause motion to impend?
Ans. 117.7 lb.
Fig. 214.
Problems 402-404.
402. A horizontal beam is supported by
a fixed pin A and a pin B which is connected
to a block C that weighs’ 4000 Ib. (Fig. 214).
A load of P = 20,000 lb. acts on this beam
as shown.
The block C rests upon a 6000Ib. block D which in turn is on a fixed
surface H. Between C and D, foo = 0.2,
and between D and EH, foz = 0.3. For
6 = 0, what force Q produces impending
motion of the block D toward the right?
Ans. 9800 lb.
403. The same as 402 except that P =
5000 lb.
404. The same as 402 except that 0 =
30°.
Fig. 215.
Fig. 216.
Problem 409.
409. A telephone-cable workman improvises a working platform that locks onto a
pole by friction alone as shown in Fig. 216.
The platform weight is W = 40 lb. and
there is ample clearance between the platform and the pole. If f; =0.38 and the
workman and his tools exert a load of Q =
250 lb., find the least distance x from the
pole the force Q may be without disaster.
The workman uses a‘safety belt.
Problems 405-408.
405. A disk A, 4 in. thick, is supported
on a 6-in. shaft, as shown in Fig. 215. A
force Q, acting at some distance x from the
surface of the shaft, tends to slide the disk
toward the right. If x is large enough, no
amount of force Q could cause the disk to
slide. What is this limiting value of x?
There is a slight clearance between the
hole in the disk and the shaft, and the
weight of the disk is small enough to be
negligible. Let f = 0.25.
Ans. 5 in.
406. A force Q = 1000 lb. is acting on
the disk A described in 405 at a distance
zx = 10 in. from the top surface of the
6-in. shaft (Fig. 215). For a value of f =
0.25, what is the limiting frictional force?
What is the actual frictional force?
Ans. 1625 lb., 1000 lb.
Fig. 217.
Problem 410.
410. Figure 217 represents a cage for a
mine hoist which, with its load, weighs
W = 5000 lb. However, the load is not
centered in the cage, with the result that
the line of action of W is 3 ft. from the
center line. The clearance between the
guides A, B, C, D, and the track is exaggerated for clearness.
For a value of f =
0.2, find Q for impending motion up.
Ans. 5750 lb.
124
PROBLEMS: FRICTION, PLANE SURFACES [Ch. VI
We are not always interested in using all four of these equations.
For
example, the moment equation may be used to determine the location of
the line of action of the normal force NV, yet oftentimes, as in § 65 and others,
we are unconcerned about the exact location of NV.
Each of the special cases discussed in this chapter is of practical importance, but note especially the case of belt friction, 71/T2, = e*, because it
will be repeatedly useful in other parts of this book.
Problems
FRICTION,
PLANE
391. A 2000-lb. body is resting on a
rough horizontal surface for which f = 0.2.
A force of 100 lb. acts upward toward the
right at an angle of 30° above the horizontal.
What is the frictional force? What is the
limiting frictional force?
Ans. 86.6 lb., 390 Ib.
SURFACES
horizontal.
(a) If f, = 0.6 and f; = 0.4,
will the block remain where it stops sliding
up the plane, or will it slide back
Show why fully.
that f. = 1/2.
Q
Problems 392-395.
392. A 1000-lb. body is resting on a
rough horizontal surface for which f =
0.25. A compressive member CD transmits a 2000-lb. load to the body, as shown
in Fig, 211.
(a) If @ =0, what force Q
will result in impending motion toward the
right?
(b) If @ were equal to zero what
would happen?
Ans. (a) 1683 lb., (b) moves leftward.
393. The same as 392 except that motion
is to be impending toward the left.
394. The same as 392 except that 6 =
down?
as (a) except
~
6
Pane
Fig. 211.
(b) Same
Fig. 212.
Problems 397-399.
397. If f, = 0.4, W = 100 Ib., and) the
horizontal force @ = 100 Ib. in Fig. 212,
find (a) the largest angle 6 for equilibrium
and (b) the smallest angle @ for equilibrium.
Ans. (a) 66.8°, (b) 23.2°.
398. The same as 397 except that Q =
20 Ib.
399. A 108-Ib. block rests upon an inclined ramp which is directed upward toward the right at an angle of 30° with the
horizontal.
A 30-lb. force acts horizontally
toward the right against the block which
is then in impending slipping.
Determine
the frictional force and the coefficient of
friction (Fig. 212).
Ans. 28 lb., 0.258.
ice
395. For the body in Fig. 211, what is
the value of 6 and Q such that force Q is
the least value that will cause impending
motion toward the right when f = 0.25?
suaaEsTIoN: Construct a force polygon,
noting that the line of action of the total
plane reaction R must make the angle
@ =tan ff with the vertical direction.
The shortest vector Q that will close the
polygon is the minimum value.
Ans. 14°, 1630 Ib.
396. A block of weight W is projected
up a plane that is inclined 30° with the
Fig. 213.
Problems 400, 401.
400. A block A, weighing 300 Ib., is
acted upon by a force Q whose direction
is at
06= 75° with
the vertical,
as shown
in Fig. 213. A weightless cable connects
A to another block B, which weighs 500 Ib.
If fa =fe =1/3, (a) what force Q will
PROBLEMS:
FRICTION,
PLANE
127
SURFACES
424. Figure 221 shows a body A, whose
weight is 6000 lb., resting on a 30° plane
for which fa = 0.15. The tendency of A
to move down the plane results in a thrust
on the body C through the member AB
with a = 15°, Between C and the plane,
fo = 0.8. If motion is impending, what is
the weight of C?
Fig. 221.
Problems
430. In Fig. 223, let Q = 200 Ib. and let
other data be as given in 429,
Will the bodies move?
What are the frictional forces?
Aiaiyy yy = Whaylay Woy, Lice =! 1afay Moy.
424, 425.
Fig. 224.
Problems 431, 432.
425. The conditions of this problem are
the same as in 424 except that a = 0° and
the body C weighs 10,000 Ib. What are
the frictional forces /'4 and F¢ if the bodies
are in equilibrium?
Ans. 957 lb., 2360 Ib.
1/3, and fg = 0.15.
For impending clockwise motion of the pulley, what is We?
The bearing at Cis smooth.
Ans. 6810 lb.
Smooth -
432. If We = 5000 lb. in Fig. 224 and
other data remain as in 431, find the frictional forces F'4 and Fz.
431. In Fig. 224, Wa = 2000 lb., fa =
Fig. 225.
426.
Fig. 222.
Problems 426-428.
In
222,
Fig.
let
Wa
= 400
Ilb.,
fa = 1/8, and fs =1/4. If the body B is
on the point of moving downward, deter-
mine the tension in the cable and the weight
of B.
Ans. 315.6 lb., 700 lb.
427. The same as 426 except that B is
about to move upward.
428. In Fig. 222, let Wz = 2 kips and let
Problems 433, 434.
433. A 3-ft. diameter oil drum is pulled
by a force Q as shown in Fig. 225. The
drum weighs 300 lb., 6 = 90°, and f4 =
fe =1/3.
Find Q when motion impends,
Is this impending motion to be spinning or
rolling?
434.
Ans. 150 lb., spinning.
The
same
as 4383 except
that
9 =
AD
fa =fs =0.2.
What is the weight of A
if motion of B impends upward?
ps
Cuma
45°
Fig. 223.
0)
>
Fig. 226.
60°
Problems 429, 430.
429. In Fig. 223, body A weighs 400 lb.,
body B weighs 200 lb., a = 15°, fa =
fs = 0.3. What force Q causes impending
motion toward the left?
Problems 435, 436.
435. A driver tries to back his car over
a high curb, as shown in Fig. 226. The
28-in. (= D) wheel supports a load of W =
800 lb., f(all surfaces) = 1/2, and @ = 60°.
Would the wheel spin or roll first? What
torque must be applied to cause motion to
impend?
Ans. 5770 in-lb.
436. The same as 435 except that f = 0.1.
PROBLEMS: FRICTION, PLANE SURFACES [Ch. VI
128
487. A ladder AB with a load of W =
500 lb., as shown in Fig. 227, is held in
impending motion toward the right by the
horizontal force Q. If fa = 0.2 and fz =
Ans. 439 lb.
0.3, what is the value of Q?
Fig. 227.
point B rests against a vertical wall.
fa =fc = 0.3, what
If
value of Q results in
impending motion of the end A toward the
left?
Ans. 58 |b.
441. The lower end of a 15-ft. ladder is
9 ft. from the vertical wall against which it
leans.
What minimum coefficient of friction between the ladder and the wall and
floor is necessary for equilibrium?
Problems 437, 438.
438. If the force Q on the ladder described
in 437 is 100 lb., what are the frictional
forces ’4 and Fg when there is impending
motion at A?
439. A 50-lb. ladder rests on a horizontal
floor for which f = 0.2 and against a vertical
wall for which f = 0.3. If the length of
the ladder is 12 ft., what is the angle that
it makes with the wall when it is on the
point of slidmg down and what are the
frictional forces at this point?
Ans. 23.1°, 9.44 Ib., 2.83 lb.
440. A member ABC, Fig. 228, weighing
900 lb. and with its center of gravity at the
BEARING
Fig. 228.
442. A uniform 20-ft. long ladder weighs
100 lb. and rests against a wall so that it
makes an angle of 60° with the ground.
When the coefficient of friction is the same
at the ground and at the wall, how far up
the ladder can a 200-lb. man go before the
ladder slides down, (a) if f = 1/2 and (b) if
i = ie
Ans. (a) to top, (b) 12.4 ft.
FRICTION
443. A 2-in. shaft rotates slowly clockwise as it transfers a gravity load of 500 lb.
to a bearing as shown in Fig. 229, in which
the clearance is exaggerated.
Assume that
the laws of dry friction apply and let f, =
shaft on the free body diagram and observe
that the horizontal displacement of the shaft
center is r sin ¢.
Fig. 230.
Fig. 229.
Problem 440.
Problem 443.
0.01. Find the reaction of the bearing on
the shaft. What is the frictional torque?
Be sure to include the couple applied to the
Ans. 5 in-lb., approx.
Problems 444, 445.
444, That part of the weight of the shaft
and band wheel B which is carried by the
bearing A, Fig. 230, is 2000 lb. The belt
pulls are 7, = 1000 lb. and 7, = 400 lb.
when the shaft rotates at a uniform speed.
If the coefficient of friction at the axle is
f = 0.015 and if 6 = 0°, what is the fric-
PROBLEMS:
BEARING
FRICTION
129
tional force and what is the frictional torque?
Arise ololion 153) inl.
445. The 2000-lb. pulley B in Fig. 230
is subjected to belt pulls of 7; = 800 lb.
and T, = 600 lb. when 6 = 30°. If f =
0.18 for the axle, what is the limiting
Total Resistance
(or
=F
kinetic) frictional force at the axle? Will
rotation occur?
Ans. 578 |b., yes.
446. A 6-ft. flywheel weighing 6000 lb.
is keyed to a 6-in. shaft weighing 1000 lb.
3999
A load W is suspended from a rope that
wraps about the flywheel.
Iff=0.015for
What is the frictional force at each bearing?
What is the total frictional torque?
The
the axle in the bearings, what
produces impending motion?
load
W
447, In Fig. 194, C = 12,000 lb., D =
THE
Fig. 194.
Ib., and f = 0.02
ghaft is turning.
Ans. Fa = 384
lb.,
(b)
Problem 447.
at the
Fz = 84
bearings.
lb.,
M; =
702 in-lb.
WEDGE
448. A body A rests in a 30° groove, Fig.
231.
Iff = 0.3, at what angle @ will motion
of the body impend?
Ans. 49.3°.
Fig. 231.
(ay
Repeated.
For all slipping surfaces, let f = 1/3. If
there is a horizontal resistance acting on
C of R = 8000 lb., what force Q will impose
impending motion of C?
Ans. 11,200 lb.
Problems 448, 449.
449. The same as 448 except that the
angle of the groove is 12° instead of 30°.
Fig. 233.
Problems 452, 453.
453. The same as 452 except that R =
3000 lb, and f = 0.5.
Fig. 232.
Problems 450, 451.
450. In Fig. 232, let 6 = 15° and f = 1/3
for all slipping surfaces.
If A is weightless,
what load Wz may be raised by a force
On
FEE
2000M oa
Fig. 234.
451. In Fig. 232) let 6 = 30° and
weight of
B be Wz
= 5000 lb.
Problems 454, 455.
the
If f =
1/3
454.
If A, B, and C, Fig. 234, each weighs
for all slipping surfaces, what value of @
causes impending motion of A (considered
500 1b., andf, = 1/4 for all surfaces, find the
force @ that is needed if counterclockwise
weightless) toward the right?
rotation of the lever is impending.
Ans. 605 lb.
452. A wedge B is inserted between a
fixed
surface
which
weighs
A
and
a movable
We = 5000
block
Jb., Fig.
C
233.
Ans. 357 lb.
455. The
same
as
454
surfaces of C are smooth.
except
that
all
PROBLEMS: BELT FRICTION [Ch. VI
180
BELT
FRICTION
456. A leather belt passes around a 6-ft.
band wheel, where f = 1/3, with an angle
If the tight tension
of wrap of 0 = 120°.
is 3000 lb. and if slipping is impending,
what is the value of the slack tension?
Ans. 1493 Ib.
457. If a belt passes around a 6-ft. pulley
with a coefficient of friction of f = 0.3 and
F, = 3000 lb. and F, = 1000 lb., what is
the angle of wrap?
Slipping impends.
458. If the coefficient of friction between
hemp and yellow pine is 0.25, how many
times will a ship’s hawser have to be wound
around a snubbing post if a man exerting
a pull of 100 lb. is to hold a force of 10,000
lb. at the other end of the hawser?
459. Two weights W and Q are suspended
one from each end of a rope which passes
Fig. 236.
Problems 464, 465.
464. In Fig. 236, fe = 1/7, fo = 9,
fs =0.2, 6 = 30°, and Wz = 100 lb. If
B is to have motion down the plane impending, find the weight of A.
Ans. 19.8 lb.
465. The same as 464 except that the
impending motion of B is up the plane.
about a stationary drum, where f = 1/3.
If Q is about to move downward, what is
the value of Q/W?
Ans. 2.85.
460. A cable is wrapped around a 12-in.
stationary drum 3.5 times.
A load W,
suspended from one end of the cable, is
about to move downward but is prevented
from doing so by a load Q = 1000 lb. on the
other end of the cable. If f = 0.25, what
is the value of W and what is the twisting
moment on the drum?
Ans. 243,000 Ib., 121,000 ft-lb.
461. A cable wraps about a stationary
drum 5.5 times. The pull on one end of
the cable is 8000 lb., on the other end, 300
Ib. If motion is not to occur, what must
be the minimum value of the coefficient of
friction?
Fig. 237.
Problems 466-470.
466. The band brake of Fig. 237 keeps a
drum from turning when a weight W is
suspended by a cable from the 4-ft. drum.
For the band brake, let a = 30° and f = 1/3.
For Q = 150 lb., find the maximum weight
W that can be supported without movement.
The pivot point of the lever is a
fixed point.
Ans. 1344 lb.
467. The same as 466 except that a =
90°.
468. The same as 466 except that the
load W is suspended from the right-hand
side of the drum, instead of from the lefthand side as shown.
For which position of
Fig. 235.
Problems 462, 463.
462. ‘Two fixed cylinders, Fig. 235, have
radii r4 = 1 ft. and rg = 2 ft. and 6 = 1
radian.
The load W is increased until W =
130 lb., when it is on the verge of moving
downward.
Find f.
Ans. 0.0835.
463. The same as 462 except that f =
0.3 and W is to be found.
the load is the brake more effective?
469. In Fig. 237, let W = 6000 lb., a =
30°, and f= 1/3. If motion is impending,
what is the value of Q?
Ans. 670 lb.
470. The same as 469 except that the
load W is suspended from the right-hand
side of the drum, instead of from the left-
hand side as shown.
471. A load W is suspended from a cable,
as shown in Fig. 238. The block brake A
prevents W from moving down.
If f = 1/3
for the brake and Q = 150 lb., what is the
maximum weight that can be supported
without movement?
Assume that the pres-
PROBLEMS:
BELT
Fig. 238.
131
FRICTION
Fig. 239.
Problems 471, 472.
sure is uniformly distributed over the brake
shoe and e = 6 in.
Ans. 630 lb.
472. Solve 466 and 471 and tell which
brake is the more effective.
473. In Fig. 239, a force of Q = 150 lb.
is applied to the lever; for the band brake,
f = 1/3; for the bearings, f = 0.12; the 6-in.
PA VOPR
Re CeieL
ON)
475. Show that the frictional torque for
a pivot bearing is [see Fig. 198(a), p. 114]
on
_ SENS
Problems 473, 474.
shaft weighs 800 lb.; the 6-ft. drum weighs
3000 lb.; the 8-ft. brake wheel weighs 2400
lb. Considering the friction in the bearings
and the brake friction, determine the load
W for impending motion.
Ans. 1395 lb.
474, The same as 473 except that the
load W is suspended from the right-hand
side of the drum.
Se)
3a a)
when it is assumed that the pressure is
uniformly distributed over the pivot.
476. A pivot bearing, such as that in
Fig. 198(a), supports a load of N = 4600
lb. The diameter of the bearing is 15 in.
and of the counterbore at
Ais5in.
Iff =
0.12, what is the frictional torque based on
the assumption of the uniform wear?
477. The same as 476 except that the
computation is to be based on the assumption of uniform pressure distribution.
Solve
475 first.
478. A plate clutch for a truck, similar to
that in Fig. 199(a), is held in engagement
by springs that exert a total normal force of
Fig. 240.
1600 Ib. The outer diameter of the facing
is 13 in. and the inner diameter is 714 in.
For the asbestos lining, f = 0.35. What
frictional torque in foot-pounds may this
clutch transmit when computed on the
basis of uniform wear?
Ans. 236 ft-lb.
479. All the dimensions and requirements
are the same as in 478 except that clutch
has two disks, as in Fig. 199(b).
480. The same as 478 except that the
computation is to be based on the assumption of uniform pressure distribution.
Solve
475 first.
481. In the hydraulically operated brake
of Fig. 240, the disk J is prevented from
rotating, but it may be moved axially to
engage and disengage the member K. A
fluid pressure of p psi in the cylinder A
applies the brake.
An electric motor con-
Problems 481, 482.
PROBLEMS: PIVOT FRICTION [Ch. VI
132
nected to the shaft GH exerts a torque of
60,000 ft-lb. tending to turn the disk K,
where the static coefficient of friction is 0.3.
What force on the piston rod B and what
fluid pressure p in the cylinder is necessary
Assume the condition
to prevent motion?
Ans. 80,000 Ib., 255 psi.
of uniform wear.
482. The same as 481 except that it is to
be assumed that the pressure between J
and K is uniformly distributed on the
braking surface. Solve 475 first.
Ans. p = 251 psi.
THE
483. The propeller shaft of a steamer has
a thrust bearing (for the thrust of the propeller) with 6 collars, each with an outside
radius of 15 in. and an inside radius of 10 in.
If f = 0.09 and if the average normal pressure on the collars is 80 psi, compute the
frictional torque.
Assume the condition of
uniform wear.
484. The same as 483 except that it is
to be assumed that the pressure is uniformly
distributed.
Solve 475 first.
Ans. 215,000 in-lb.
SCREW
485. A screw jack with a square thread
whose outside diameter is 3 in. and whose
inside diameter is 214 in. is designed to
support a load of 60,000 lb. The screw has
134 threads per inch and the coefficient of
friction should be about 0.12.
(a) What
torque is necessary to raise the specified
load? What force must be exerted at the
end of a 4-ft. lever to produce this torque?
(b) The same as (a) except that the load
is to be lowered.
(c) If there were no friction on this jack, what torque would be
required to raise the load?
(d) If the
efficiency is taken as the torque required
for a frictionless jack divided by the torque
required for the jack with friction, compute
the efficiency.
ROLLING
Ans. (a) 15,500 in-lb., 323 lb.; (b) 4400
in-lb., 91.7 lb.; (c) 5450 in-lb.; (d) 35.1%.
486. The same as 485 except that f =
0.25.
487. The same as 485 except that the
jack screw is double threaded.
488. A square-threaded screw, with a
major diameter of 114 in., a minor diameter
of 1 in., and 31% threads per inch, is used
in a jack for which f = 0.1 A force of 50
Ib. acts on a handle 20 in. from and perpendicular to the axis of the screw.
(a)
What load will this force raise?
(b) What
force on the handle is necessary to lower the
load found in (a), other conditions remaining
the same?
Ans. (a) 9710 lb., (b) 5.24 lb.
RESISTANCE
489. A 48-in. iron wheel, with a load of
10,000 Ib. and rolling on a steel track, has a
coefficient of rolling resistance of a = 0.03 in.
What horizontal force passing through the
center of the wheel will maintain uniform
velocity, other resistances being zero?
Ans. 12.5 Ib.
490. The same as 489 except that the
line of action of the force Q is tangent to the
top of the wheel.
tires is about a = 0.02in.
With the ‘‘supercushion” tires, a = 0.03 in., approximately.
Find the percentage increase in the rolling
resistance.
Is this increase a serious concern for a 3000-lb. automobile with 28-in.
OD wheels?
Ans. 50%, no.
492. If a diesel locomotive with 48-in.
drive wheels is found to have a rolling
resistance of 1.25 lb. per ton on rails that
weigh 140 lb. per linear yard, find the
coefficient of rolling resistance as applied
to the locomotive.
491. The rolling coefficient for an automobile on a good road with regular balloon
FRICTION
DRIVES
493. In a friction drive such as indicated
in Fig. 206, p. 121, the driven wheel B is
stalled by an overload.
The face of the
wheels is 8 in. wide and the normal pressure
between the wheels is 100 lb. per in. of
face. The diameters of the wheels A and
B are 12 in. and 30 in., respectively.
If
the kinetic coefficient of friction is 0.14,
what is the torque on each shaft?
Ans. Ma = 672 in-lb., Mz = 1680 in-lb.
494. The allowable running pressure between
bevel
friction
wheels,
Fig.
208,
is
250 Ib. per in. of face for a tarred fiber
driving surface, The mean diameters of
PROBLEMS:
FRICTION
183
DRIVES
the wheels are 5 in. and 10 in., and the face
is 3 in. What are the frictional force and
the frictional torque on each wheel for
f =9.28?
(The face is the length of the
line of contact when the wheels are engaged.)
Ans. 210 lb., 525 in-lb., 1050 in-lb.
495. The hoisting machinery for a light
derrick is shown in Fig. 241. The driving
friction gear A has a compressed-paper face
in contact with a cast-iron friction gear B,
for which f should be about 0.22 when the
normal force between
per in. of face.
the gears is 180 lb.
The face (or width) of the
gearsis 12in.
On the shaft with the driven
gear is a 12-in. drum on which winds a cable
which supports a load W, as shown.
What
load W may be raised at constant velocity?
@ =@,
Ans. 5060 lb.
496. The same as 495 except that it is
desired to know what normal force per inch
of face is necessary to raise a load W = 6200
lb.
cen
Fig. 241.
Problems
Fig. 242.
495-497.
Problem 498.
GENERAL
497. In Fig. 241, the brake is used to
hold the load in position after it has been
raised.
What force Q is necessary to hold
a load of W = 6200 lb. when f = 1/3 for
the brake?
The wheels A and B are not
in contact.
Ans. 1748 lb.
498. Figure 242 shows a pair of friction
similar
to that
shown.
The
diameter
the tumbler is 2.5 in.,fa = 0.01, andfz
(smooth).
minimum
What
values
are
of
the maximum and
6 for equilibrium?
See also problem 261.
Alo, PAD, 169) 59°,
tongs used for lifting marble slabs.
(Chains
used for this purpose may mar the surface.)
(a) For f = 1/3 at the shoes # and D, what
is the total limiting frictional force? What
is the
actual
total
frictional
force?
(b)
Determine the forces acting on the pin at A.
499. A glass rod AB, Fig. 243, weighs
5 oz. and is 6 in. long. It is placed in a
glass tumbler C in position of equilibrium
of
= 0
Fig. 243.
Problem 499.
PROBLEMS: GENERAL [Ch. VI
134
500. A steel bar, weighing 3 lb. per ft., is
placed upon 1-in. pins A and B as shown
in Fig. 244. Let the bar C be 6 ft. long,
a = 4 in., and let the coefficient of friction
Fig. 246.
Fig. 244.
Problem 500.
at the pins be f = 0.2. What is the angle
6 when the bar is on the point of sliding
from the pins?
501. A long, 12-in. wide, uniform plank
of weight W is placed inside a 24-in. diameter culvert, Fig. 245. As the culvert is
slowly rolled it is found that the plank first
slips when its broad side is tipped to an
angle of 42° with the horizontal.
Find the
coeficient of static friction.
Neglect the
thickness of the board; use a very carefully
drawn free body.
What is
though the guides were smooth.
this minimum angle 6? Consider the weight
of the follower to be negligible.
Ansy 133%.
504. In Fig. 247, if f = 0.3 for both the
band
brake
Problems 501, 502.
502. A long, 12-in. wide, uniform plank
of weight W is placed inside a 24-in. diameter culvert, Fig. 245. If f, = 1/3 between
the surfaces, find the angle with the horizontal that the board will make at the time
shipping first occurs as the culvert is slowly
rolled. Neglect the thickness of the board;
use a very carefully drawn free body diagram.
503. Figure 246 represents a cam with
its follower.
The follower slides on the
surface of the cam where f = 0.2. Visualizing the action of the cam and follower,
you will see that the angle @ may be so small
that no amount of turning moment on the
camshaft could force the follower up even
and the incline
and
6 = 60°,
determine the force Q that will permit the
weight W = 3000 lb. to be lowered with
uniform speed. The force Q on the lever
pulls the band tight on the wheel, braking
it.
Ans. 236 |b.
Fig. 247.
CEA.
Fig. 245.
Problem 503.
505. The same
Problems 504-506.
as 504 except that @ =
45°.
506. The same as 504 except that W =
5 kips.
507. Make a free body of the lever in Fig.
248 and use 7; = T2e/" to derive the expression for the frictional force
pL Waakelie 1)
Lor
m — nele
for clockwise rotation.
Recall that the
braking
(frictional)
force
is
F = T; — T».
What is the relation between m and n
when an applied load W less than nothing
(negative W) produces a positive F. This
is the condition for a self-locking brake.
See Fig. 197(a), p. 114.
PROBLEMS:
GENERAL
508. For the differential band brake
shown m Fis, 248, D = 18 in.) 7 = 2) in.,
m = 12 in., and the angle of contact 0 =
195°.
The band is lined with asbestos
which gives f = 1/3. Let W = 30 lb. and
Fig. 248.
Problems 507, 508.
Fig. 249.
a = 26 in. (a) Determine the frictional
force and frictional torque for clockwise
rotation of the drum.
(b) Same as (a)
except that rotation is counterclockwise.
See problem 507.
509. A force of Q = 200 lb., acting on
the lever AC, Fig. 249, applies a block
brake at H, through the link BD.
At £,
the coefficient of friction is 0.2 and the
normal reaction is acting along a line OH
at
45°
with
the
vertical.
The
load
must slip around G, where the coefficient of
friction is 0.35.
Ans. 3392 lb.
511. In Fig. 250, the angle @ is gradually
increased until motion of A impends.
Let
GO =]Mim, 0 = Sim, aol =O,
Deeside
body tip over or slide?
oe.
“a
W,
supported as shown by a cable wrapped
about the 4-ft. drum, is prevented by this
brake from moving down.
If motion is
impending, what is the value of W? All
bearings are smooth.
Ans. W = 1696 lb.
510. The same as 509 except that the
pulley G does not turn.
Thus, the cable
Problems 509, 510.
b
0
Fig. 250.
Problems 511, 512.
512. The same as 511 except that f = 0.3.
513-520. These numbers may be used for
other problems.
Chapter VII
GRAPHICAL
79. Introduction.
We
have
METHODS
already
learned
how
to construct
a force
polygon, which is all that is needed for the graphical solution of problems
involving a concurrent system of coplanar forces, inasmuch as the polygon
will serve to determine two unknowns.
We recall that for a concurrent
system of coplanar forces in equilibrium, there are two independent conditions,
>F, = 0 and =F, = 0, from which two unknowns may be found.
If the
system is not in equilibrium, the magnitude and direction (the two unknowns)
of the resultant can be found.
(The line of action of the resultant is known
to pass through the same point as the other vectors.)
However, in a system of non-concurrent coplanar forces, there are three
independent conditions of equilibrium, =F, = 0, ZF, =0, and 2M = 0.
If the system is not in equilibrium, the three unknowns may be the magnitude,
the sense, and the location of the line of action of the resultant.
We shall
now develop graphical methods for the solution of non-concurrent coplanar
force systems involving three unknowns and graphical methods for the
solution of parallel coplanar force systems involving two unknowns.
In order to do accurate graphical work, use a hard pencil with a sharp
point and work carefully. Also make your layouts as large as practicable,
although for instructional purposes, the use of 8! x 11-in. paper should
usually be satisfactory for the problems of this course.
Fig. 251. Bow’s Notation. Thevector
A-Bin (b) represents the magnitude
and
direction
of ab
in
(a); B-C in
(cyiirepresents the magnitadenand
direction of bc in (a).
186
80. Bow’s Notation. The graphical solution of the types of problems to be discussed
in this chapter is greatly aided by a systematic method of designating forces and
vectors—a method known as Bow’s notation. In this system of notation, the vector
which represents the true line of action of
® force is designated by two lower-case
letters, as ab and be.
The free vector
8
:
:
($3) which represents the magnitude
and
sense (but not the line of action) of a force
§ 81 | RESULTANT OF PARALLEL FORCES
187
is designated by capital letters, one at each end of the vector.
Thus, a free
vector for force ab is named AB (Fig. 251) and is laid out to scale in
graphical solutions.
The student should study carefully the application of
Bow’s notation in the examples which follow and be sure that he uses it
correctly in all of his own solutions.
81. Resultant of Parallel Forces.
The determination of the resultant of
a group of parallel forces is shown in Fig. 252. The lines of action of the
forces must be accurately located to some convenient space scale. The forces
are named ab, bc, etc. The name of the resultant of any group of vectors so
named will be the combination of the first and last letters, in this case ae.
The next step is to draw the force polygon, which in this case is a straight
line because the forces are parallel. Thus, starting at some point A (Fig. 252),
lay out AB to scale representing the force ab. Observe that B is above A,
which indicates that the vector AB points upward.
Then from the point
of AB, lay out BC. The location of point C is below B because the vector is
Now
Similarly, to BC is added CD and DE.
known to point downward.
Force
Scale,
7
1= 200 lb.
Space Scale, ie tts
49"
a
225
b
167
==—
20°—>
bic
.
i
200
100
125
(a) Funicular Polygon
Fig. 252.
Ba
(b) Force Polygon
Resultant by Force and Funicular Polygons.
PROCEDURE.
(1) The lines of action of
the forces ab, bc, etc., are located in their
proper relative positions in accordance with
some convenient space scale, say, 3/8 in. =
1 ft. Name the forces and note that the
resultant will have the name ae. Next,
==
draw the force polygon.
(3) From any
DRAW FUNICULAR POLYGON.
convenient point on the line of action of ab,
say, point m, draw the lines oa and ob
PROCEDURE.
(2) Lay out a force polygon
as usual.
Choosing any starting point A
and some convenient scale, lay out AB
( = 225 lb. = ab), then BC ( = 200 lb.);
The
CD( = 100lb.),and DE( = 125lb.).
distance from A to E represents the magni{ude and sense of the resultant (AE scales
200 lb.). From any convenient pole
O,
:
To find the line of
draw OA, OB,:-- OE.
action of ae, draw the funicular polygon.
parallel respectively to OA and OB in the
<—_—"
force polygon.
From the intersection of
ob and bc, draw oc parallel to OC; from the intersection of oc and cd, draw od parallel
to OD; from the intersection of od and de, draw ce parallel to OE. Since oe and oa are
both components of ae, the intersection of these components defines the line of action
of ae as shown.
The force ae = 200 lb. in (a) is the resultant of the forces shown.
GRAPHICAL METHODS
138
[Ch. VII
noting that the polygon was started at point A and that the last point is Z,
we know that the polygon does not close and that the resultant is a force AE,
which may be scaled. Moreover, we know that the resultant points downward because its name A# reads downward.
(If the point # had been
above A, the resultant would have pointed upward.)
In order to locate the line of action of ae, a string or funicular polygon
must be drawn.
In doing this, first choose some pole O, Fig. 252(b), at any
convenient point. Because more accurate results are possible with a large
than with a small pole distance (Fig. 252), the pole O should be as far from
B
Space Scale, ae
3=1
149" +=
16
=——
Force Scale,
ft.
1"= 200 lb.
20>
akb
bic
cid
dje
225
200
100
125
(a) Funicular Polygon
Eko
Fig. 252.
(b) Force Polygon
Repeated.
the force polygon as practicable—a greater relative distance than that used
in Fig. 252(b) is much to be preferred.
Having chosen the pole O, draw the
lines or rays OA, OB, ++ + OE.
Consider now the free vector AB.
The
lines AO and OB are components of AB (§ 6), inasmuch as their vector sum
is AB.
Recalling that the components of a force must intersect on the line
of action of the force, we may choose some point m on the line of action of
ab, Fig. 252(a), and through this point draw the lines or strings oa and ob of
indefinite extent parallel respectively to the rays OA and OB.
From a like point of view, BO and OC are components of BC; and in Fig.
252(a), strings ob and oc are components of bc. Since any two components
intersect on the line of action of their resultant, we note that oc should be
drawn through the point n which was determined by the intersection of ob
and be, and that it should be drawn parallel to OC.
In a like manner, od
and oe are drawn.
Now remembering again that the intersection of the components lies on
the line of action of their resultant and noting that the resultant is AH with
components of AO and OZ, we readily conclude that the intersection of oa
and oe in Fig. 252(a) determines the line of action of ae, as shown.
§ 82] RESULTANT
OF NON-PARALLEL
FORCES
139
When the resultant of the forces is a couple, the strings oa and oe will be
parallel, as shown in Fig. 253, in which is a graphical solution of the example
of § 35 and Fig. 78, p. 53. Since =F = 0, the force polygon closes (points
E and A coincide), showing that the resultant is either zero or a couple.
(If
Since the funicular polygon does not close, the resultant is a couple.
Study
zero.)
been
have
would
resultant
the
closed,
the funicular polygon had
the caption to Fig. 253 for more detail.
Space Scale, a
8"
10"
ft.
12”
|
‘el
Force Scale,
1'= 200 lb.
alb
bc
cid
dhe
100
200
250
150
EA
Pole Distance 4|
(b) Force Polygon
(a) Funicular Polygon
Fig. 253.
Resultant Couple.
Given the forces shown in (a) spaced to scale, name them with consecutive
PROCEDURE.
(An error in the naming of the
lower case letters ab, - + - de in the manner indicated.
forces may result in an error in solution, although the method can be applied without
Lay out the force polygon, in (b),
naming the forces if the principles are understood.)
Select a pole O and draw OA, OB,
and note that it closes, points E and A coinciding.
Then construct the funicular polygon in (a). From any point m on ab and
+++ OE.
parallel respectively to OA and OB, draw oa and ob, components of ab; from the inter-
section of ob and bc, draw oc parallel to OC; and so on until oe is drawn.
The resultant
couple consists of forces oe and ao with a moment arm h. To the space scale given,
The magnitude of the forces is AO in (b), which scales 300 lb. ( = EO
h measures 5.3in.
also). Thus, the resultant couple is (5.3)(300) = 1590 in-lb., which checks the value
found in § 35.
The resultant of three coplanar
82. Resultant of Non-Parallel Forces.
before, if the forces are named
As
254.
Fig.
of
example
the
in
forces is found
first and last letters, ad. A
the
by
ab, bc, and cd, the resultant is designated
foree polygon (ABCD) is drawn and the pole O is chosen as far from the
The closing line AD of this force polygon defines
polygon as practicable.
The string or funicular polygon
of the resultant.
direction
and
the magnitude
that AO and OD are comknown
is
it
locates the line of action, because
ponents of 4D and that the true lines of action (ao and od) intersect on the
See the vector & in Fig. 254(a).
line of action of ad, the resultant.
graphically using Bow’s notation, name
resultant
a
finding
in
Remember:
all the known forces ab, bc, etc., without repeating any of the letters and the
name of the resultant will be the combination of the first and last letters
used (except see Fig. 253).
GRAPHICAL METHODS
140
83. Coplanar Parallel Force System in Equilibrium.
[Ch. VII
Figure 255 shows a
typical problem in finding the reactions at the supports of a beam subjected
In general, it is better to name
to parallel forces.
all of the known
forces
Then name the unknown forces, being sure that the
first, starting with ab.
last letter used is also the first, a. This plan of naming the forces reminds
us that, since the forces are in equilibrium, both the string polygon and the
force polygon must close, so that if we start at the point A, we return to
point A.
F=150 Ib.
#=175 Ib.
oc
B
as
aNd
Ne
Ta
Ae
24”
b\ec
c
a
Force Scale,
[
1= 200 lb.
ae
lb
Space Scale, Vay ft.
(a) Funicular Polygon
Fig. 254.
(b) Force Polygon
Resultant of Non-Parallel Forces.
PROCEDURE.
(1) Assuming that the lines of action
of the forces are defined by the points m, n, and p
PROCEDURE.
(2) Choosing any
origin A, lay out AB = 150 lb.,
and the angles a, 8, and ¥, lay out vectors of indefinite
extent for ab, bc, and cd.
Then draw the force
polygon.
—
DRAW FUNICULAR POLYGON.
(3) From any convenient point s on the line of action of ab, draw the components oa and ob parallel respectively to OA and OB.
Through the intersection of ob and bc, draw oc parallel
BC’= 175'lb., and CD = 200 Ib.
to scale. This locates point D.
The resultant in magnitude and
to OC.
Through the intersection of oc and cd, draw
od parallel to OD.
The interection of oa and od is a
point on the line of action of ad, the resultant of the
system of forces.
AD.
The vector ad is drawn parallel to
direction is AD, which scales
150 lb. at the angle 6 = 104.5°
with the horizontal.
From any
convenient pole O, draw OA,
OB, OC, and OD.
Now to find
the location of the line of action
of AD, draw the funicular
polygon.
<——==
Figure 255(a) represents a simple beam with three known loads, named
ab, be, and cd. The unknown reactions at the points of support are named
de and ea. Observe that the location of the lines of action of de and ea
are known.
After laying out the known forces in the force polygon, we see that there
are two other forces which, when added (vectorially) to CD (Fig. 255), will
close the polygon.
While the free vectors DE and EA are evidently collinear with the other vectors in the force polygon, the location of the point
E cannot be determined until the funicular polygon ts drawn.
After the
strings oa, ob, oc, and od have been drawn, we observe that oe is a component
of both ea and de. But one component of ea is oa, which locates the point m
on ea through which the other component oe must pass; and one component
§ 84] COPLANAR NON-PARALLEL FORCE SYSTEM IN EQUILIBRUIM
141
of de is od, which locates the point n on de through which the other component
oe must pass.
Hence, with two points on oe known, this line may be drawn,
as shown dotted in Fig. 255. Now, since the component OH of the free
vectors
HA and DE must be parallel to oe, we may draw ray OE, thus locating
point # and defining the magnitude of HA and DE.
The point # does not
necessarily lie between the points A and D in the general case.
Force Scale,
1= 300 lb.
Space Scale, fr ft.
(b) Force Polygon’
(a) Funicular Polygon
Fig. 255.
Parallel Forces in Equilibrium.
PROCEDURE.
(1) Having laid out the space
diagram to scale, start the force polygon.
(3) From
any convenient
point s on the
vector of the first known force ab, draw its
string components oa and ob parallel respectively to OA and OB. Through the intersection of ob and bc, draw oc parallel to
OC.
Through the intersection of oc and
cd, draw od parallel to OD.
Through the
intersection of od and de and through the
intersection of oa and ea, draw a line oe,
shown dotted.
This line defines the direction of OE, which may now be drawn. ==>
PROCEDURE.
(2) Starting at any origin A,
lay out to scale AB = 225 lb., BC = 75 lb.,
and CD = 200 lb. Choose a pole O and
draw OA, OB, OC, and OD.
Now draw
the funicular polygon.
<Gamxs
(4) Draw OE parallel to oe, locating point
E. Then DE, reading upward and acting
upward, represents the magnitude and sense
of de; EA, reading upward and therefore
acting upward, represents the magnitude
and sense of ea. By scaling DE and EA,
we find de = 325 lb., and ea = 175 lb.
Another ex84. Coplanar Non-Parallel Force System in Equilibrium.
reprewhich
256,
Fig.
in
shown
is
parallel,
not
are
forces
the
which
in
ample,
160
weighs
which
arm
an
of
end
the
at
lb.
sents a crane with a load of 6000
held
are
forces
These
Ib. per ft. The vertical column weighs 200 Ib. per ft.
in equilibrium by the ground reaction, whose direction and magnitude are
unknown but whose point of application is known (at G@), and by the tie
rod at H which exerts a force de. Since only the magnitude of de is unknown,
there are only three unknowns and the problem can be solved by the forceand-funicular-polygon method.
For the purpose of finding reactions (that is, the forces de and ea in this
case), a distributed load may be represented by a vector passing through
the center of gravity of the load. This rule results in the vectors ab and bc
GRAPHICAL METHODS
148
[Ch. VII
as shown in Fig. 256. Using a vector for the distributed load is the equivalent of using the resultant of a multitude of parallel forces. It should be
emphasized that while this practice does not affect the external balance of
forces, it does change the internal stress pattern of the member involved.
As before, the lines of action of the various forces are carefully located in
the space diagram [Mig. 256(a)] in which the funicular polygon will be drawn.
Then the known forces are added in the force polygon and point D is located.
However, one other known condition must be added to the force polygon.
A line DE of indefinite extent is drawn from D in the known direction
After the rays OA, OB, OC, and OD are drawn, the funicular polygon
of de.
may be completed,
In drawing the funicular polygon, observe that % must be started at the
Space Seale, $=1 £
1920
Weight
200 Ib. /ft.
Non-Parallel Forces in Equilibrium.
PROCEDURE.
(1) As before, the vectors
are first located in space in their proper
relative positions and the known forces
are named first, ab, be, ete.
—(3) Since
only one
point on the line of
action of ge is known, point G, the funicular
polygon must be started at this point. Thus,
through G, draw oa, one component of ae.
Through the intersection of @b and oa,
draw the other component of ab, which is
ob. Continue as previously described until
point J on de is reached.
Noting that oe
is a component of both de and ge and noting
that two points (G and J) on oe are now
known, draw oe, shown dotted.
plete force polygon.
Then com—
PROCEDURE.
(2) The force polygon is
started by adding the known vectors AB,
BC, and CD.
A pole O is chosen and the
rays OA, OB, OC, and OD are drawn.
Since the direction of DE is known, a line
DE of indefinite extent is drawn.
Then
the funicular polygon may be completed.
—
(4) Draw
components
OF parallel to oe.
Since the
DO and OF must add to give
the vector DE, the intersection of OF and
DE locates the point EF which defines the
magnitude of DE.
Since the force polygon must close, the closing line FA repreSents the ground reaction ge in magnitude
and sense.
The answers found are shown
on the figure.
$85 | INTERNAL FORCES FOR A TRUSS
143
point G, because the two components oa and oe of ae must intersect on the
line of action of ae, and since G is the only known point on this line of action,
The point J is
it must be the point through which the string oa is drawn.
located by drawing the strings ob, oc, and od, parallel respectively to the rays
This point J is the point through which one component
OB, OC, and OD.
od of de passes and through which therefore the other component oe must
pass. Also, the point G is the point through which one component oa of ae
passes and through which the other component o¢ must pass. Thus, the
string o¢ is defined by G and J and is added to the solution.
Now since the ray OF is parallel to oe and a component (with DO) of DE
and since the direction of DE and the magnitude and direction of DO are
known, the force triangle DOE may be completed.
The point £ is thus
defined by the intersection of OZ, drawn parallel to oc, and ED previously
drawn of indefinite extent parallel to de.
Inasmuch as the crane is in equilibrium, the force polygon must close;
therefore the closing vector HA must represent the ground reaction ea in
magnitude and sense. The magnitudes of DE and HLA and the direction
of HA may be scaled from Fig. 256(b).
Remember: in solving equilibrium problems using Bow’s notation, name
the known forces first ab, be, etc. and let the name of the last-named force
contain the letter a. With the first and last letter used being a, it is indicated
that the system is in equilibrium.
85. Internal Forces for a Truss. Graphical procedures may be applied
for finding the forces acting on the members of a truss, as shown in Fig. 257.
Usually, of course, the external reactions (in this case, the tension of 26,000
Ib. in the tie rod and the supporting reaction PR = 28,500 lb. at the pin
joint 7) are unknown.
If so, the force and funicular polygons may be used
to determine these forces, as previously explained.
In order that the illustration be kept as simple as possible, the solution for these external forces
is not shown in Fig. 257.
In applying Bow’s notation to trusses, we find it convenient to name the
spaces between the lines of action of the forces in the manner shown in Fig.
257. Thus, the region between the reaction /? and the tie rod is termed a;
between the tie rod and the 15,000-lb. force, b; between the 15,000- and
25,000-lb. forces, c; etc. Then the triangular regions bounded by the members
of the truss are also named, in this problem, e¢, f, g, h, and 7. Any force is
now designated by the combination of the letters on each side of its line of
action.
Thus the reaction / is called da (reading in a clockwise sense about
pin 7); the force in the tie rod is called ab, reading in a clockwise sense about
pin 1. Similarly, the internal force in the member 1-2 is be or eb; in the
member 3-5, fg or gf; ete.
GRAPHICAL METHODS
144
[C h. VII
ure by the joint-to-joint
The method of solution is a graphical proced
of the truss are two-force
system. We recall that when all the members
a system of concurrent
members, each pin is a body in equilibrium under
We must remember, too, that only two
forces whose directions are known.
by the conditions
unknowns in a system of concurrent forces can be found
at a pin where
d
starte
be
must
n
Hence, the force polygo
of equilibrium.
a pin.
such
is
1
Point
there are no more than two unknown forces.
the force in a
of
In applying Bow’s notation at a pin, we read the name
free body.
clockwise direction about the pin which is being considered as the
ea.
and
be,
ab,
forces
the
under
ium
For example, pin 1 is in equilibr
15,000 lb.
| 25,000 Ib.
Force Scale,
2B
1’= 40,000 lb.
ab=26,000 Ib.,
Tension in
Tie Rod
reat)
Space Scale, £ =1
ft.
Fig. 257.
da=28,500 Ib.
Be
Finding Internal Forces—Two Force Members.
Starting at pin 1, lay out AB, Fig. 257(b), in magnitude and sense from
some convenient origin A. Then, noting that the directions of be and ea
are known, draw a line BE of indefinite extent from point B and parallel
to be, and a line AF of indefinite extent from A parallel to ea. The intersection of these lines locates point HE. The force polygon for pin 1 is then
Since BE reads
ABEA, in which BE represents be in magnitude and sense.
which means
1,
pin
the
on
pushes
downward toward the left, the force be
Also the line HA represents force
that the member 1-2 is in compression.
ea, and since HA reads from left to right, it is indicated that the force ea
acts toward the right and pulls on the pin. Thus member 1-3 is in fenszon.
With be known, there are now only two unknowns at pin 2. Reading
clockwise about pin 2, we have eb and be (= 15,000 Ib.) known, ef and fe unknown.
The vector HB is already drawn from the free body of pin 1 (equal
and opposite to BE). From 3, lay out BC to represent bc. From C, draw
a line CF of indefinite extent parallel to cf, and from # draw a line HF parallel
to fe; the intersection of these lines locates the point F. The force polygon
for pin 2 is then composed of the vectors HB, BC, CF, and FE, acting on the
pin 2 in the directions in which the letters read in a clockwise direction about
pin 2. For example, since FH reads downward, the member 2-3 pulls on
pin 2, and is therefore in tension.
At 3, we note that ae and ef (reading clockwise about 3) have already
PROBLEMS: RESULTANTS
145
So from Ff, draw a line F/G of indefinite extent parallel to fg,
and from A draw a line AG parallel to ga. The intersection of these lines
locates G, thus completing the force polygon AEHFGA for pin 3.
This procedure is continued until the forces on all members of the truss
have been found.
Observe that in this truss gh = 0; therefore, ag = ah,
so that the points G and H coincide in Fig. 257(b).
Moreover, since the
members 5-6 and 6-7 can obviously have no stress, points D and I coincide.
If the graphical solution is accurate, these coincident points will automatically
appear in the construction of the force polygon.
The dotted portion of the
force polygon, Fig. 257(b), is the equilibrium diagram for the external forces
only.
been drawn.
Problems
Solve all the assigned problems of this chapter graphically.
Norte.
RESULTANTS
521. In Fig. 258, Q = 200 lb. and Ff =
100 lb. Find the resultant.
Ans. —450 lb., 12.45 in. to right of 225-lb.
force.
r Ears 16" car
20
u"
a
is,
lb.
lin lie, BE,
Q
Fig. 258.
F
125 Ib.
Problems 521-525.
522. In Fig. 258, Q = 600 lb. and F =
600 Ib. Find the resultant.
523. In Fig. 258,
Find the resultant.
Ans.
amel
17 = Bail)
526. To the lever shown in Fig. 259 are
applied the forces fF; = 300 lb., F2 = 200
lb., and F; = 100 lb. Determine their
resultant.
527. What is the resultant of the forces
Fy = 980) lb: Fa — 620) lb. and Fs = 3860
Ib. (Hig. 259)?
225 Ib.
@ =O)
Find the resultant.
“Ans.
C = 13,400) 1n-lb. ce.
528. -Ine-hic. 260) A = 200) Ibs Bb = 500
lbp uC==—s400 sb and s@)— 302) Find eibe
resultant.
See problem 241.
529. The same as 528 except that 6 =
60°. See problem 242.
Q = 0 and F = 125 lb.
—225 lb., 11.1 m. to right of 225-lb.
force.
524. In Fig. 258, Q = 100 lb. and Ff =
200 Ib.
Find the resultant.
Fig. 260.
Problems 528-532.
530. In Fig. 260, A = B = C = 100 lb.
and 6 = 40°. Find the resultant.
See
problem 243.
531. In Fig. 260, A = 3 kips, B = 4 kips,
C= & teas, eine! @) = OO",
Iiuarl iplae
resultant.
Ans. 54 in-kips ce.
532. In Fig. 260, A = 12 lb., B = 9 lb.,
Problems 526, 527.
6) = iS o,,
resultant.
aime)
@ =
HR,
lial
tla
PROBLEMS: RESULTANTS [Ch. VII
146
167 lb. at 289.6°
533. A horizontal steel I-beam has a
clear span of 30 ft. between two simple
The vertical loads on the beam
-supports.
Ans.
acting respectively at distances of 4 ft.,
10 ft., 18 ft., and 25 ft. from the left support.
Determine the resultant of these loads and
its distance from the right support.
resultant.
are 1000 lb., 2500 Ib., 3800 Ib., and 4600 lb.,
F= 5b.
Ans. 11,900 lb., 12.14 ft.
F,=101b.
| R= 30 1b.
A force of 1000 lb. acts vertically
upward at a distance of 24 in. from ab;
another force of 1250 lb. acts vertically
downward at a distance of 54 in. from ab;
and another force of 750 lb. acts vertically
upward at a distance of 90 in. from ab, all
to the right of ab. Solve for the resultant.
Ans. C = 2000 ft-lb. ce.
535. In Fig. 261, Q = 100 lb., a = 30°,
Ee
r = 3.76 in.
536. In Fig. 261, Q = 100 lb., a = 30°,
Find the
fe 2268) lb and) 6 —"/9=
534. A horizontal bar is acted upon by a
force ab of 500 lb. pointing vertically downward.
with
from 100-lb. force.
150 lb. and
6) — 45...)
Hind
the
resultant.
ro
F,
Fig. 262.
Problems 537, 538.
537. Find the resultant of the force system in Fig. 262 when Fy, = 15 |b.
Ans. 26.2 Ib. at 54.6°, 3.6 units from origin.
538. Find the resultant of the force system in Fig. 262 when F's = 35 lb.
geese ret 12*
A
Q
100 tb.
B
]Ae
Fig. 261.
eye
Problems 535, 536.
EQUILIBRIUM
539. In the framework of Fig. 263, F =
5000 lb. Ifa = 6ft., solve for the reactions
at A and B.
540. Find the reactions R; and
Fig. 264. See problem 291.
R» in
Ans. 16,500 Ib., 11,500 lb.
See problem 340.
Ans. 1750 lb., 6750 lb.
6000 Ib.
6000 Ib.
Ri
10,0001b.
Fig. 264.
Fig. 263.
Problem 539.
Rp
Problem 540.
541. A horizontal shaft is supported by
bearings on 50-in. centers. The vertical
downward loads on this shaft are: Fy =
3600 lb., 10 in. to the right of the left-hand
bearing; #, = 7600 lb., 40 in. to the right
of the left-hand bearing. What are the
bearing reactions?
PROBLEMS:
EQUILIBRIUM
147
542. The same as 541 except that the
bearings are on 30 in. centers.
Ans. —133 lb., 11,330 lb.
Fig. 265.
547. In Fig. 268, let F = 2000 lb., a =
90°, and let the member BD weigh 1800 Ib.
Find the pin reactions at B and C using
force and funicular polygons.
Ans. 7180 Ib. at 182.7°, AC = 8290 Ib. (C).
Problem 543.
543. In Fig. 265, if s = 1/2 in. and F =
Fig. 268.
20 lb., find the force at the cutting edges.
Problems 547-549.
See problem 183.
548. The
7 1000 lb.
1500 Ib.
same
as 547 except that the
reactions at H and K are desired.
549. The same as 547 except that a =
GO,
Alas. F535 Ilo. On = 1RRAS, FER lly.
Fig. 266.
Problems 544, 545.
544. In Fig. 266, find graphically the
reactions at 1 and 2 when x = 10 in. and
y =4in.
Ans. 1505 Ib. at 175.7°, 889 |b.
545. The same as 544 except that y =
1/2 ae
Fig. 269.
Problems 550-552.
550. In Fig. 269, 6 = 90° and F = 0.
Find the ground reactions at A and B.
See problem 325.
551. The same as 550 except that 6 =
270°. Ans. 390 Ib. at 262.6°, 300 Ib. at 90°.
Fig. 267.
Problem 546.
552. The same as 550 except that the
force in member AC is desired.
Take as
many
546. A pipe is supported by hangar rods
AE as shown in Fig. 267. The hangar is
pivoted at A. If the section of pipe weighs
W = 800 lb., and £ = 14 in., find the
reactions (a) at A and B and (b) at C and
D. See problem 265.
steps as necessary,
solving the force
systems graphically for successive free body
diagrams.
553. Determine the reactions at pins 6
and 7, Fig. 270, when F; = 20,000 lb. and
F; = 30,000 lb.
Ans. 115,300 lb., 65 = 154.3°, 103,900 Ib.
PROBLEMS: INTERNAL FORCES IN A TRUSS [Ch. VII
148
INTERNAL
FORCES
IN
A
TRUSS
554. Determine the load on each member
of the framework in Fig. 270 when Fi =
20,000 Ib. and F, = 30,000 lb.
See problem
Fig. 273.
Fig. 270.
Problems 553, 554.
555. The loads F; = 1000 lb., F2 = 2000
Ib., and F; = 1000 lb. in Fig. 271 represent
the action of the wind on a roof truss.
(a)
What are the forces at the supports 1 and
Problem 558.
558. In Fig. 273, if the load at C is 2000
Ib., find the reactions at A and F and the
forces in all members of the truss.
Ans. Ra = 1188 lb., Rr = 1290 lb. at 141°,
AB = BC = CE = 1980 lb.(C), BD =0,
CD = 380 lb.(T), DE = 630 lb.(C), DF =
2090 lb.(T), HF = 1355 lb. (C).
5? (b) Determine the force on each member of the truss.
Fig. 274.
Fig. 271.
Problems 559-561.
Problems 555, 556.
556. If the wind loads on the truss of
Fig. 271 are taken as FP; = 1500, F2 = 2500,
and F; = 2000 lb., determine (a) the external reactions and (b) the force on each
member of the truss when all other loads
are considered to be negligible.
Ans. 2-6 = 2885 lb.(C), 3-4 = 3750 Ib.(C),
5-6 = 250 lb.(T).
559. In the truss of Fig. 274, the wind
loads of F; = 1000 lb., F2 = 2000 Ib., and
F;, = 1000 lb. are assumed to act at the
pin joints as shown.
In order to allow for
expansion, the support G is on rollers. For
F4 = 5000 lb., find (a) the external reactions
and (b) the force in each member.
Ans. (a) 5970 |b., 3200 lb. at 51.2°; (b)
AMIE) = 1510) = 2) Wo (©), Teves
(5 Ue =
23201b.(1), CD = DE = 17321b.@), EG =
DG = 2000 lb.(C), CG = 4890 1b.(C), CH =
5/0 Mb) GA = 560 Ibs(@):
560. The same as 559 except that 74 =
2500 Ib.
561. The same as 559 except that Fy =
F, =0.
Fig. 272.
Problem 557.
557. In the truss of Fig. 272, determine
(a) the load on the tie rod and the reaction
at 4, and (b) the load on each member, all
by graphical solutions.
Let F, = 40,000
Ib. and F2 = 30,000 lb.
562. In Fig. 275, F; = 1 kip and F, = 5
kips. Find (a) the external reactions and
(b) the force on each member.
Ans. In kips, (a) 9.23 at 49.4°, 7; (b) AB =
CH = 1.414(7), AD =1(), BC = 2(T),
BD = 1.414(C), CD =8.49(T), CH =0,
CGl— 9191 (ee D ii—18 (©) anG: ewe
7(C).
PROBLEMS:
INTERNAL
Fig. 275.
563. The
FORCES
149
IN A TRUSS
Problems 562, 563.
same
as
562
except
that
a
force of 3 kips acts horizontally to the left
at pin H.
Fig. 277.
Problems 566, 567.
566. The framework in Fig. 277 is loaded
with forces Ff, = F, = F3 = 3000 lb. as
shown.
Find the load on each member.
(See problem 303.)
567. The same as 566 except that F'; acts
at pin A horizontally toward the left. (See
problem 304.)
568. In Fig. 278, when fF; = 3000 lb.,
F, = 4000 lb., and F; = 5000 lb., find (a)
the external reactions and (b) the loads on
the members of the truss.
(See problem
Fig. 276.
Problems
564, 565.
295.)
569. The same as 568 except F’; acts at H.
Ans. In pounds,
564. Find (a) the reactions at A and H
and (b) the loads on all members of the
truss shown in Fig. 276 when F; = 5000 lb.,
Hip — "OOO lioness — 4000 lon and le —
1500 lb. (See problem 305.)
565. Same as 564 except that F; acts
down at G.
Ans. In pounds, (a) 3265 at 40°, 6900; (b)
AB = BO = 4190(©), AG = 1140(1), BE =
0, CG = 6650(T), CD = 5260(C), DH =
7970(C), DG
= 3300(C), GH
= 3960.
(a) 5990 at 203.6°, 10,770;
(b) AB = BC = 4880(T), AH = 1280(T),
BH =CJ =DJ =0,CH = 5000(1),, CG =
DNC), CD = 1D = GAD), kel = CW =
9920(C), GH = 3710(C).
570. If the wind loads on the Fink truss
of Fig. 279 are as shown, determine the
force on each member of the truss when all
other loads are considered negligible.
See
problem 311.
571-580. These numbers may be used
for other problems.
Smooth =
Fig. 278.
Problems 568, 569.
Span = 80’
Fig. 279.
shi
10
Expansion
Rollers
Problem 570.
AR
a
Chapter VIII
MAXIMUM
AND
FORCES
MINIMUM
Quite frequently, a structure or a machine is subjected
86. Introduction.
to a varying load. In designing the elements of such a structure or machine,
the engineer often needs to know or to estimate the range of variation of the
load. In any case, he needs to know the maximum load in order to be able
to design a part that will not fail in service when this maximum load is imposed. In other circumstances, the minimum force that will accomplish a
A general principle which should be concertain purpose may be desired.
sidered for all engineering projects is: obtain the maximum effectiveness
for the minimum cost. Although many problems involving maximums and
minimums are highly specialized and too advanced for consideration here,
we can suggest by examples the nature of some of the more elementary
cases. Perhaps you have already solved some similar problems in preceding chapters.
87. Example —Body on Inclined Plane.
A body of weight W lies on a rough
plane inclined at an angle of a degrees with the horizontal (Fig. 280). What are @
and Q for the minimum value of Q that will produce
impending motion up the plane?
ALGEBRAIC SotuTION.
The forces acting on the
body are W, Q, N, and the limiting frictional force F’.
Replacing F’ and
N by their resultant, the total
plane reaction R, we then use the weight W as a
free body acted upon by the forces W, Q, and R.
Note that the vector
R makes
an angle ¢ with the
normal, where ¢ = tanf is the limiting frictional
angle. Summing the forces in the x and y directions,
we get
(a)
(b)
=F,
=F,
=Q cose — Rsing — Wsina = 0,
= Qsine + Reos ¢ — W cosa = 0.
Eliminating R from (a) and (b), we find
Qcos@ —Wsine
sin ¢
_W cosa
-
— Qsin@
cos ¢
)
whence
Q(cos 6 cos ¢ + sin @ sin ¢) = W(cos asin ¢ + sin « Cos ¢).
150
§ 88 | EXAMPLE—CANTILEVER FRAME
161
Using the trigonometric relations for the functions of the sum and difference of two
angles and solving for Q (see appendix C), we have
iy;
(c)
sin(a + ¢)
Q
cos(@ — ¢)
Observe from this equation that Q will be a minimum when the cos(@ — ¢) is a
maximum (the terms W, a, and ¢ having heen given). The maximum value of the
cosine is 1, for an angle of 0°. Thus, the minimum value of Q is W sin (a + ¢)
when 6 = ¢.
GRAPHICAL SoLution.
Assume that the load W, the inclination of the plane a,
and the coefficient of friction f are known.
Then ¢ = tan “f/f, from which ¢ is determined.
Observe now that W is known completely and that the direction of RF is
known.
In constructing the force polygon, choose some origin
6
O, Fig. 281, and from it lay out W. From one end of the vector
R
W, say O, draw a line Oa of indefinite extent parallel to the
known direction of R. Since the vector Q is the only other
force involved, this vector Q, when drawn through the other end
of the vector for W, must close the polygon.
The shortest
vector Q which will close the polygon is evidently one which
is perpendicular to Oa. Therefore, this vector Q shown in Fig.
281 represents the least force that will result in impending
Fig. 281.
motion; it evidently makes an angle of « + ¢ with the horizontal. By comparison
with Fig. 280, we see that @ must be equal to ¢ for this minimum value of Q.
r at
K
7
(b)
Fig. 282.
88, Example—Cantilever Frame.
A cantilever frame, composed of four pinconnected members, is supported from a vertical wall by pin joints at # and F, Fig.
282. <A rolling wheel J carries a load of W = 6000 1b. The safe capacity of the strut
AB is 16,000 lb. in compression, and the safe capacity of DC is 30,000 Ib. in tension.
What is the maximum distance x from the point # that the wheel may be rolled without exceeding the safe loads specified when a = 4 ft.?
Sotution.
The horizontal component I’, of the reaction at / may be found by
treating the whole frame as a free body, as shown in Fig. 282(a), and by taking moments at the point H. From the equation
I (F.)(4) — (6000) (x) = 0,
=M xz =
we have
(d)
1H, = WA
ay
Since AB and CD are two-force members,
they may be cut, and the member
BP
MAXIMUM AND minimum Forces
152
[Ch. VI II
A convenient center of moments is the
.
drawn as a free body, as shown in Fig. 282(b)
AB and CD.
point G, the intersection of the lines of action of
From the equation
2M ¢ = (Fy)(20) — (Fz)(6) = 0,
and using F, = 1500 x from equation (d), we get
Using D as
d.
Since Fy came out positive, it is correctly shown acting downwar
the center of moments, Fig. 282(b), we find
=M p = (AB)(6) — (F,)(14) = (AB)(6) — (450 x)(14) = 0,
We have chosen correctly the sense of AB, as a compresfrom which AB = 1050x.
Inasmuch as the allowable
is a positive number.
obtained
answer
the
since
force,
sive
of x to be
load for AB was specified as 16,000 Ib., we find the corresponding value
16,000
TE
(e)
15.2 ft.
Next, summing forces horizontally in Fig. 282(b), we find
=F, = (CD)(cos 45°) — Fz = (CD)(cos 45°) — 1500 x = 0,
from which CD = 2120z, which is shown correctly as a tensile force.
sible load on CD of 30,000 lb., the safe distance is
30,000 _
For a permis-
Since the value of x from (f) is less than that from (e), the strength of CD is the determining factor and the maximum distance of the load from the point £ should be
14.14 ft.
Fig. 283.
89. Example—A-Frame.
An A-frame is mounted on wheels which in turn rest
upon a smooth plane, Fig. 283. There is a track on the horizontal member BD, upon
which a wheel with a load W may rollfrom B to D. What position of W results in the
maximum force on the pin C? The members ABC and CDE are continuous.
Sotution.
Let x measure the distance of W from the point D.
frame as a free body, and taking moments about H, we find
IMez = 12k, — Wa + 2) = 0,7
R
1=W
—
(ae
2)
in
Using the whole
§ 90 |EXAMPLE—COMPRESSION
IN DIAGONAL BRACE
158
Considering BD as a free body (not shown) and taking moments about D, we find
ese8
Now, making a free body of AC, as shown in Fig. 283(b), and taking moments about
point C, we have
2Mc = (R1)(6) — (B,)(4) — (Bz) (4) = 0
We +2)(6)
(W2)(4)
=
So
SE
= BD) = 0,
or B, = W/4. From a sum of the horizontal forces, we note that C, = B, = W/4.
Since the distance z is not involved in this expression for C,, we conclude that C,
remains constant for all positions of the load W. Next, taking the sum of the vertical
forces in order to find Cy, we get
Wh, = Ih = 18, = CS =
Wa@t+2)
2D
Waza
ee
from which
1
Cy =W
(g)
ae
=
(;
a)
Since C, is constant, the reaction at C isa maximum when C, isa maximum. Examination of (g) shows that C, is a maximum when z = 0, that C, is zero when x = 4 (the
midpoint of BD), that it changes direction for values of x greater than 4. As long as
the movement of W is limited by the points B and D, the maximum value of C, is
/6, and the total maximum reaction at C is
W
n [+(e
In designing the frame, the engineer must make the size of the pin at C large enough
to withstand safely this maximum force.
Fig. 284.
A horizontal boom DC with a
90. Example—Compression in Diagonal Brace.
constant load F is supported by a brace AB, Fig. 284. If the length L of the brace
remains constant, what should be the angle @ for the minimum load on the brace?
Since the brace AB is a two-force member, the force at the point B is
Sotution.
AB (here shown at A). Considering DC as a free body, we have
=Mp = Fd — (AB)(asin 6) = 0,
where
h = xsing.
Thus
A
=
Fd
-
x sin @
=
Fd
;
L cos
6 sin 6
PROBLEMS
154
(h)
2Fd
=
AB
sin 8, we find
Since sin 26 = 2 cos
where we have used « = L cos @.
[Ch. VIII
~ Lsin 26
is, when the
By observation it is seen that the force AB is least when 9 = 45°; that
respect to
with
AB
iating
different
by
also
result
this
sin 20 is unity. We may reach
obtain
we
zero;
to
result
the
equating
6 in (h) and
d(AB) _
do
4Fdcos20 _
= 0.
L sin? 26
This expression may be zero only when cos 26 = 0, whence 26 = 90°, and 6 = 45°.
Problems
581. A body weighing 500 Ib. rests on a
rough plane inclined at an angle of 30°. If
f = 1/3, what least force will result in impending motion up the plane? Solve graphically, and also algebraically.
Ans. 374 lb., 0 = 18.4°.
582. Derive an expression for the least
force necessary to move a body down a
rough inclined plane. What angle @ does
the force make with the plane? (¢>a,
Fig. 280.)
583. A 1500-lb. wheel, 4 ft. in diameter,
is to be pulled over a 1-ft. obstruction which
is on a 15° incline, Fig. 285. What least
force Q will cause the wheel to be just on
the point of moving over the obstruction?
Ans. 1449 lb., @ = 60°.
585. Figure 286 shows schematically a
15,000-lb. tractor with a side-boom.
The
center of gravity shown is for the tractor
without a boom load. The cables are power
operated over sheaves to raise or lower the
boom or its load.
If the parts of the
machine are strong enough to raise any
weight that will not overturn the tractor,
find the most advantageous angle a at
which the boom may raise a maximum
homogeneous 4-ft. diameter weight W.
What is that maximum W?
Ans. 60°, 30,000 lb.
586. The same as 585 except that 3 men
of the plane.
weighing 200 lb. each stand on the left
track, 4 ft. to the left of the tractor’s center
of gravity, to act as an added counterweight.
587. Figure 287 represents a bridged section in the track for a special overhead
crane.
The track to the left of A and to
the right of B is independently supported,
leaving the bridged section supported only
Fig. 285.
Fig. 286.
584. The same as 583 except that the
wheel is on the left side of the obstruction,
about to move
in the downward
Problems 583, 584.
direction
Problems 585, 586.
PROBLEMS
155
Fig. 287.
Problem 587.
by reactions R, and R». What is the distance x when (a) R; is a maximum and (b)
R, is a maximum?
Determine Ri in (a)
and R» in (b).
Ans. (a) 15,200 lb. at
= 24 ft.
in impending motion of A toward the right?
Ans. 7370 lb., 0 = 18.42°.
591. The same as 590 except that the
least force F for impending motion of A
toward the left is desired.
6
Q
Cable
belerern aad
Sh
Fig. 288.
Fig. 290.
Problems 588, 589.
fp
Problem 592.
:
ae
588. The coefficient of friction between
a 200-lb. man’s feet and the floor is 0.5.
592. In Fig.8 290, lieWa = 300, ge We = 500
least
lb., fa = 0.3, and fs = 0.4. Whatmotion
floor. See Fig. 288. What is the heaviest
toward the left?
For
a large box, f, = 0.3 with
the same
force
Q will result in impending
Cc
box that the man can pull by means of a
rope? What angle does the rope make
with the horizontal?
Ans. 428 lb., 16.7°.
589. The same as 588 except that the
coefficients of friction are exchanged for the
man
and the block; fan = 0.3 and fpox =
0.5.
Fig. 291.
Problem 593.
593. In the A-frame of Fig. 291, what
position of the wheel load W will cause the
maximum force on the pin at C, when the
limits of motion of the wheel are the points
Band D? What is the value of « when the
force on C is a minimum?
Aes,
Fig. 289.
Problems 590, 591.
590. In Fig. 289, the body B weighs 6000
Ib. and A weighs 3000 lb.
For fa = fa =
1/3, what is the least force F that will result
Chin
ORLA
ule
Gt
ys ito,
Ghae = 0481W when a) = 10 tt:
594. In Fig. 284, DC is an I-beam, 10 ft.
long and weighing 50 lb. per ft. The safe
load for an 8-ft. ( = L) brace AB is 90,000
Ib. What is the maximum safe load F that
™ay be supported at C?
595-600.
These
for other problems.
numbers
Ans. 35,750 Ib.
may
be used
Chapter
IX
NON-COPLANAR
91. Introduction.
FORCES
It is evident from Fig. 292 that any vector may be
considered to be a vector in space and that any three reference axes, Oz,
Oy, and Oz, may be used to define or locate the vector.
This vector may then
be resolved into three components, one of which is parallel to or coincident
with each of the three reference axes. As in coplanar systems, it is convenient to use components and axes which are mutually perpendicular.
Thus, we usually find rectangular components when components are needed.
Moreover, we often find it advantageous to reduce a system of forces in
space to two or more systems of coplanar forces.
ee
(eee
She
C
D
Owe
a
WLR e_
1
er rye
i
eee 7
Fig. 292.
ok
a
“2
e
Force in Space.
92. Three Rectangular Components of a Force. Let OC, Fig. 292, be a
vector representing a force whose line of action makes the angles a, 8, and
y with the x, y, and z axes respectively.
If a rectangular parallelepiped is
constructed such that the vector OC = F is its diagonal, we see from Fig. 292
that the sides (OG = F,, OA = F., and OE = F’,) of this parallelepiped are
three rectangular components of the vector.
Thus we may write the expressions for the components as
(a) F, = F cosa,
F, = F cos B,
and
F, = F cos y.
If the x, y, and z components are known, the resultant F
may be obtained.
Squaring each term in the equations (a) and adding, we get
F2+ F,? + F, = F2(cos? a + eos? B+ cos*'+).
156
§ 93 |CONCURRENT FORCES IN SPACE
157
Since the cos? a + cos?8 + cos?y = 1, we find the magnitude
sultant in terms of three rectangular components to be
(b)
Pee
el
of the re-
gee
93. Concurrent Forces in Space.
If the resultant of a number of concurrent forces in space is desired, the x, y, and z components are found for
each force, with the point of concurrence as the origin O. Then from the
sum of the x components, designated by 2F',, the sum of the y components
>F,, and the sum of the z components 2/,, we find the resultant
(24)
R= ((2F,)? + (ZF ,)* te (2).
If R makes the angle 6, with the x axis, 6, with the y axis, and 6, with the
z axis, we have
(25)
cos 6; =
Sk
ae
Sie,
WF
cos 0, = Ta
from which the angles 6,, 6,, and 6, may be found.
same respectively as a, 6, and y, Fig. 292.)
(These angles are the
Fig. 293.
The direction of a force in the practice of statics is often determined by
the location of some member of a machine or structure, such as a strut,
cable, or link. Hence, in practice, the lines of action of at least some of the
forces are generally determined from the geometry of the structure, or are
given as known applied loads. We may locate a line of action by determining the location of two points on the line. Thus, the vector F in Fig. 293
may be located by the points 0, 0, 0 (which is the origin) and 8, 6, 4 (w = 8,
y = 6,andz = 4). From the analytic geometry, the length of the diagonal is
(c)
p(n? + y? + 27)M?,
with the unit the same as used for z, y, and z. Compare this expression with
equation (b) and note that angles a, 6, and y may be obtained from equations analogous to (25) by using dimensions instead of forces:
x
Z
cosy = —
cos 6 = Z,
cos a = —
(d)
p
p
p
NON-COPLANAR FORCES [Ch. IX
Os
For convenience, we shall sometimes
Dermnrrion oF A Forcr tN Space.
x, y, 2. Thus,
define a force by a sequence of four numbers in the order of F,
through
the numbers (500, —6, 4, 8) mean that ei force F = 500 Ib. passes
of action
the origin and the point «= —6, y = 4, and z = 8. If the line
numbers
to define
Example. ‘Tho lines of action of the following forces pass through
A = (35, 6, 12,~y 15), B = (70, 6, —12, 8), C = (120, —4, —S8, —3).
the origin:
Determine
does not pass through the origin, it w ould take seven
a force according to this scheme (Fy a1, Y, 21, Xa Y2y 22).
tho resultant,
‘To get the functions of the angles which the forces make with the axes,
Sonuvion.
use equations (¢) and (d),
Kor force A:
= (6+ 12 + 15)" = 20.1
COS a
As
A eos a
6
=
30.17
Ba
6
12
cos 8 =
zB
= 35 (0) =
A, =
1048 lb.
15
cos y =
A, =
Acos@
5
sa
ent
Aocosy
—
201°
30.1.
on { 12
= (05)
= 20.9 lb.
= 35 (§) = 26.1 lb.
Similarly, for foree A:
ep = (+
p, =70(—°_)
dy»
q
i
15.6
=26.91
FS
S00
)
et
15.6
Bp, =70( ="
):
¥
=
ans
a9
15.6
‘
=
—53.8
oo.
lb.
15.€
= 42+ 8 + 32 = 9.43
Y
Q 43
U ote
=
Ss
70585) = $5.9:1b.
B, =
C, x = 120
-
122+ 8)
eert ih .
20(—8
i)
0.43
=
=
102
lb.
0
'
:
Cc, =
—3
‘
Behe
= —38.2 lb.
Summing the components in the coordinate directions, we have
SF, =
1048
+ 26.9 — 51 =
—18.67 Ib.
=F, = 26.1 + 35.9
ii = 20.9 — 53.8 — 102 =
— 38.2 =
—134.9 lb
+23.8 lb.
Using equations (24) and (25), we define the resultant as follows:
R = (18.678 + 134.9" + 23.8) = 187.7 Ib.
@
’
=
cos™
—13.67
137.7
=
OF
7
hy
Bet
2
—134.9
@ = cos | So}
sae SE
RN
= 168.2°,
aad
23.8
® = cos" | 137
—_—.f
] = 80.1°,
Where @,, @,, and @ are respectively equivalent to a, 8, and » when F = Rin Fig. 293
169
§ 95 | EXAMPLE
94. Equilibrium of Concurrent Forces in Space. As in a system of coplanar
concurrent forces, k = 0 for the case of equilibrium.
Letting R = 0 in (24),
we see that the three conditions of equilibrium for a concurrent group of
forces in space are
(e)
=F, = 0,
LF, = 0;
=F, =0
from which three unknowns may be found. Observe that if 4 group of
concurrent forces are in equilibrium, the projections (components) of these
forces on each of the three coordinate planes yields a system of coplanar
forces in equilibrium in each plane.
| {|WY &s
i
“
~
Boom Timber
Fig. 294.
95. Example.
Figure 294 represents a boom constructed of two timbers AF and
CE which are fastened together at #. The cable DE is holding the timbers in 4 horizontal plane and supporting the vertical load of W = 1000 lb. The horizontal line
AC is the intersection of the plane of the boom with the vertical wall. The construction is such that the boom members can turn about axes at A andC.
What are the
loads on the timbers and the cable?
So.tution.
The point EH, at which the lines of action of the forces intersect, is
taken as the free body in Fig. 294(b). The compressive forces in the boom timbers are
Q» and Q;, and the tensile force in the cable is 7. Summing forces vertically, we have
=F, = T sina — 1000 = 0,
or T = 1000/(0.6) = 1667 lb.
From the equation
=F, = Q2sin B — Q3s8inB = 0,
we note that Qo = Q; (as may be seen also from the symmetry of the members).
the x direction, we get
(f)
ZF,
= Q.cos 8 +Q3 cos 8 —
In
T cosa =
2Q.(0.707) — (1667)(0.8) = 0,
from which Q. = Q; = 9421b.
Observe that in the solution of this problem, we might
have substituted a sum of the moments about the axis AC for ZF, in determining T.
NON-COPLANAR FORCES
160
[Ch. 1X
Two spars, AC and AB, Fig. 295, are set up as shear legs and sup96. Example.
on the
ported by a guy AQ. «For a vertical load W = 3000 lb., what are the forces
spars and the cable?
(b)
Fig. 295.
SoLuTion.
Since a is not equal to 6, Fig. 295, the forces C and B on the spars
AC and AB respectively are not equal. .Moreover, these forces are not in a plane normal to the load line of W, as in the previous example. Many problems of forces in
space can be satisfactorily resolved into two or more coplanar systems.
Using this
plan of attack, we may find the components of C and of B in the xz plane and in the
y direction. These components are
Can = C COS BH.
Cy = C gia Bs,
(Ba — bb COs oe.
Joy, = 13a) 2B.
Then the free body of the forces acting at A may be drawn as shown in Fig. 295(b).
Considering only those forces in the xz plane (W, 7’, Cz:, and B,:), we have
(g)
(h)
=F, = T cos 20°— (Cz. + Bzz) cos 65° = 0,
ZF, = (Cz. + Bz) sin 65°— T sin 20°— W = 0.
Solving equations (g) and (!) simultaneously for the unknowns,
tension 7’ in the cable is
we find that the
fF = 1790 Ib,
Cz + Biz = 3980 lb. = C cos 35° + B cos 25°.
From =F, = 0, we find
(j)
On = 15,
or
C sin 35° = B sin 25°.
(i)
Using C = B sin 25°/(sin 35°) from (j) in (i), we have
B gin 25° cos 35°
sin 35°
@)
+ B cos ORO
25° =«€ 3980;
from which B = 2630 Ib., the compression in the member AB.
in (j) gives C = 1940 lb., the compression in the member AC.
This value of B
97. Example.
The safe loads for the members of the shear legs of Fig. 295 are:
for the cable, 7 = 3000 lb.; for AB and AC, 4000 lb. each. What dead
load W may
be supported safely?
SoLutTion.
Of course, having solved the problem of § 96, we know the
relative
forces on the members for some load W; therefore, we are in a position
to determine
§ 97 | EXAMPLE
161
quickly which member will be the one to limit the load. (Study the solution of § 96
and see if you can decide now which member this is.) However, we shall proceed in
this solution as though we knew nothing of the previous solution.
From the free body, Fig. 295, we see that the y components of B and C are necessarily equal, By = C,, so that any difference in the forces B and C on the legs will be
indicated by the components C,, and B,.1n the xz plane. From the sum of the forces
in the y direction, we get
C sin 35° = B sin 25°,
or
sin 35°
1.320,
B = aa 550 (©) = 1.32C
That is, for any load W, B is larger than C. Therefore, since AB and AC have the
same limiting strength, the safe load W depends on the load B in AB—unless it is
limited by T = 3000 lb. Taking the force in B = 4000, we find the corresponding
force C as
C
_
B _ 4000
= 3030 lb.
(i. Fee
The values of C,. and B,. corresponding to the limiting value of B = 4000 are
B,z = B cos 25° = (4000) (0.906) = 3620 lb.
Cz = C cos 35° = (3030) (0.819) = 2480 lb.
Biz + Cuz = 3620 + 2480 = 6100 lb.
Proceeding with a graphical solution, select a convenient origin O, Fig. 296, and
1. Draw a line OH making an angle of 65° with
the horizontal, which is the inclination of the
plane ABC, Fig. 295.
2. From the same origin draw another line OK at
an angle of 20° with the origin, the inclination of the cable.
3. Lay out B.. + Cz. = 61C0 along OH and from
la
Buz +Cxz= 6100
S
the point end of this vector, drop a vertical
line, which is the direction of the line of action
of W, until it intersects OK at point N.
4. Then NO is the magnitude of 7 (about 2740 lb.)
when B,, + Cz, = 6100.
5. Since this value of 7 = 2740 is less than the
permissible value of 7 = 3000, this force
triangle, shown heavy in Fig. 296, is the solution of the problem. The safe value of W is
scaled as W = 4590 lb.
fig
1'=2000 Ib.
x
is
3
Buz
W.
If we had begun by laying out the value of 7’ =
3000, locating point K, Fig. 296, we would then
7'=3000
have constructed the vertical line from K and found
point H. The corresponding value of B,. + Cz, for
equilibrium is measured by the length OH. It is
Fig. 296.
seen, Fig. 296, that the safe value of B..+ C2
= 6100 is exceeded and that therefore this solution does not solve the problem.
NON-COPLANAR FORCES
162
[Ch. IX
3000
To put it another way, W = 5020 lb. from Fig. 296 when the safe load T =
=
W
is
answer
the
therefore,
lb.;
4000
=
B
load
safe
lb.; W = 4590 Ib. when the
4590 lb.
In the case of non-coplanar, parallel
98. Non-Coplanar Parallel Forces.
forces, it is convenient to use coordinate planes such that one of them is
perpendicular to the lines of action of the forces. For example, consider the
parallel forces F'1, f2, F'3, and F's, Fig. 297. The plane M NPQ is perpendicular
to these vectors and should contain, in accordance with conventional usage,
These axes may be placed in this plane in any convenient
the x and y axes.
position. Here, we have arbitrarily drawn them as shown, although it is
often preferable to place these axes so that all of the forces involved are on
the same side of a particular axis. Then the locations of the lines of action
are defined by the coordinates x and y which mark the points where these
lines of action pierce the zy plane.
Fig. 297.
Resultant of Non-Coplanar Parallel Forces.
Suppose it is desired to find the resultant of the forces shown in Fig. 297.
Recalling that the moment of the resultant about any axis is equal to the
sum of the moments of the components about that axis, we may take moments
about the y axis to find 7, and moments about the x axis to find 7. These
coordinates, = and J, locate the line of action of the resultant R. The
magnitude of FR is found from a sum of forces in the z direction.
As a coordinate, 2; in Fig. 297 is a negative number; but in taking moments
about an axis, it is well to think of all the coordinates Va
ee
as
positive numbers and to determine the sign of the moment solely from the
direction in which the force will apparently turn the plane about the axis of
moments.
Thus, if all the x’s are taken as positive numbers in Fig. 297,
the sum of the moments about the y axis is
(k)
2M,
=) yay
—
Fox
+
F 3x3 == Fx,
=
Dra
=
deg
where the counterclockwise direction as you look toward the origin along
the y axis is taken as positive. Compare the sign of each term in (k) with
Fig. 297 to be sure that positive terms have a counterclockwise moment
§ 100 |EQUILIBRIUM
OF PARALLEL FORCES IN SPACE
168
about Oy and negative terms a clockwise moment.
axis we have
(1)
2M,
= Fiyit Foye — Fsy3 — Pays =
Similarly, about the x
UFy = Ry.
The magnitude of the resultant is then
(m)
Rk = =F, =F;
— Fi — Fy, — Fs.
With R determined from (m), find 7 and 7 from (k) and (1).
99. Example.
Let the data for Fig. 297 be as follows (in the order F, 2, y):
F,(100, — 10, 15), F2(300, 10, 25), F3(200, 30, 20), and F.(150, 25, —5), where the
coordinates are in inches.
Determine the resultant.
SoLution.
The magnitude of the resultant is
R =
—100 — 300 + 200 — 150 =
where the negative sign shows that R points downward.
wise moments be positive, we find
—350 lb.,
Then, letting counterclock-
=M,, = (100)(10) — (300)(10) + (200)(30) — (150)(25) = + 250 in-lb.
The positive sign shows that R must be so located with respect to the y axis that it produces a counterclockwise moment about the axis. The distance from the y axis is
obtained from Rz = 250, whence
Sl
OE
350
Tie
In order to act downward and produce a counterclockwise moment about Oy, R must
be located to the left of Oy (looking toward the origin); that is, x is really —0.714 in.
Now,
However, the safest manner of locating R is by inspection, as explained.
positive)
moments about the x axis give (counterclockwise
=M, = (100)(15) + (300)(25) — (200)(20) — (150)(5) = +4250 in-Ib.,
where the positive sign shows that R must be so located as to produce a counterclockwise turning moment about the x axis (looking in the direction x-O, Fig. 297).
Since R is known to act downward, this means that it must lie in the positive y direction. The distance from the z axis is found from Ry = 4250; whence 7 = 4250/350 =
12.1 in. In Fig. 297, R is shown in the correct quadrant for this example, but = in
particular is exaggerated.
100. Equilibrium of Parallel Forces in Space. If a group of non-coplanar
parallel forces are in equilibrium, it is evident from (m) that, if the forces
are parallel to the z axis,
(n)
Si = 0
in order that there should be no resultant force.
In order that there should
be no resultant moment, we see from (k) and (1) that
(0)
=M,=0
and
=M, = 0.
Equations (n) and (0) give us three conditions of equilibrium.
NON-COPLANAR FORCES
164
[Ch. IX
A card table with one broken leg is supporting books weighing
101. Example.
20 Ib., as shown in Fig. 298. What vertical force F is applied at the point indicated
if the table is on the point of turning over?
Sotution.
If the table is on the point of turning over, the reaction
(Fig. 298). The sum of the vertical forces is
=F
=
Rk, +
RR,
—
20
—
F=
R3; = 0
0.
Choosing the x and y axes arbitrarily as shown, we find
=M, = 30R,; — (10)(20) — 26F = 0,
=M, = 30R, — (8)(20) — 15F = 0.
The solution of these equations for F gives F = 21.8 lb.
102. Couples in Space. Any number of couples in space may be combined
into a resultant couple.
Considering only two couples for simplicity, we
have the couples C; = Pp and C: = Qq in random planes (Fig. 299). No
matter what the location of these planes,
they will intersect if extended far enough,
unless they are parallel.
This line of
intersection AB is shown vertical for
H convenience only. Now, recalling that a
couple may be moved to any position in its
plane without changing its external effect
on the body on which it acts [§ 33(a)], we
may move the couple Pp until one of the
forces P lies along AB and the other
a force P lies in the plane ABDE.
Next,
Fig. 299.
Couple Qq is turned until
the forces are vertical and then con-
verted to an equal couple Pq’.
one Ppa
Ito
an
iy
weet
oe ste
equal
couple
Pq’
such
that
C. = Q¢ = Pa’ [§ 33(b)]. With the forces
of the couple Cz equal to P, we may
move the couple C2 until one of the forces P lies on the intersection AB
but points in the opposite sense to the force P of the couple C; first placed
§ 103 ]VECTORIAL ADDITION OF COUPLES
165
on the intersection of the planes. In this case, the other force P of the couple
C2 lies to the right of the intersection, and the forces P on the intersection
cancel one another.
With the forces on the intersection equal and opposite,
there is left a force P in the plane ABDE and a force P (with opposite sense)
in the plane ABGH.
Since these remaining forces are parallel, they form
a couple whose moment arm is p’ (Fig. 299) and whose moment is therefore
Pp’. This couple Pp’ is the resultant couple Cz. The moment arm p’
may be found from the triangle whose sides are p, q’, and p’ by the law of
cosines; thus
(p)
De a
et
PUR COs:
Evidently, if there is another couple in another plane to be combined with
C; and C», it may now be combined with the couple Pp’ = Cz in the manner
just described.
By a continuation of this process, any number of couples in
any number of planes may be combined into a single resultant couple. That
is to say, the resultant of any group of couples is a couple (or zero).
For the case of parallel planes, we may recall at this time that a couple
acting on a rigid body may be moved from one plane to a parallel plane
without changing the external effect on the body [§ 33(d)]._ Hence, couples
in parallel planes may be added algebraically.
103. Vectorial Addition of Couples.
As stated before, a couple may be
represented by a vector which is perpendicular to the plane of the couple.
The length of the vector represents the magnitude of the moment of the
couple, usually either in inch-pounds or foot-pounds.
The sense of the
vector is customarily such that it points away from that side of the plane of
the couple on which the couple appears to have a counterclockwise moment.
D
(a)
Fig. 300. If a right-hand screw thread were turned counterclockwise, it would move
toward the right from the plane ABHD; therefore, couple Pp is represented by a vector
C; pointing as shown.
If a right-hand screw were turned clockwise, it would move through
the plane ABEG away from the reader; therefore, couple Qg is represented by a vector
C2 pointing into the plane ABEG.
If the planes intersect at an angle ¢, the perpendiculars
(C; and C,) will also intersect at an angle ¢, as shown in (b).
166
NON-COPLANAR FORCES [Ch. IX
is, the
This usage conforms to the right-hand screw-thread rule. That
move
would
screw
and
right-h
a
vector points in the direction in which
Thus,
couple.
the
of
when turned in the direction indicated by the forces
rd direcin Fig. 300, the vector for couple Ci = Pp would point in a rightwa
point
tion from the plane ABHD; and the vector for couple C. = Qq would
ABEG.
plane
away from the reader through the
These two couples may be added vectorially as shown in Fig. 300(b).
This simple case of two couples may be solved algebraically by use of the
law of cosines:
(q)
Cr? = C2 + C2? — 2C1C'2 cos ¢.
To test equation (q) to see that it yields the same results as equation (p),
let Ce = Pp’, C1 = Pp, and C2, = Pq’. Substituting these values into
(q), we get
P%p'? = P%p? + Pq’? — 2P?pq’ cos ¢,
Recall that the
which reduces to (p) when P? is canceled from each term.
Ce.
vector
the
to
plane of the resultant couple is perpendicular
As we know, any force
104. Non-Current, Non-Parallel Forces in Space.
may be replaced by a force and a couple. Thus, in Fig. 301, let /’ represent
a force vector whose butt end has the
coordinates 2, y, and z in relation to any
chosen reference axes. At the origin of
these axes, introduce equal and opposite
forces F'’ whose magnitudes are equal to
F and whose line of action is parallel to
the line of action of F. Now the force F
and the force F’ represented by a solid
line form a couple, which may be moved
anywhere in its plane or into parallel
planes, and there is left at O the force F’
Fig. 301.
represented by a dotted line. In a similar
manner, any number of other forces Q1, Qo,
etc., may be transformed into a force through O anda couple. Thus, if the
resultant of a group of forces F’, Q:, Qe, etc., is desired, we may first reduce
this group of forces to a system of concurrent forces at O and a group of
couples; second, find the resultant R of the concurrent forces at O (§ 93);
and third, find the resultant couple (§ 103). Thus, we see that the resultant
of a force system in space may be a force and a couple. The resultant R
will in general have components in the x, y, and z directions whose magnitudes
are ZF',, 2F,, and =F, respectively.
The magnitude of the resultant force
is therefore
(r)
R = [(2F.)? + (2F,)? + (2F)}?.
167
§ 106 |EXAMPLE
In general, the plane of the resultant couple will be oblique to the reference
axes.
Let the vector representing this resultant couple be M, Fig. 302. ‘This
vector has the rectangular components M,, M,, and M,. Observe that the
plane of the couple represented by 1/7,
is parallel to the yz plane and that
therefore M, is the moment of the
resultant couple about the «w axis.
Likewise, M, is the moment of the
resultant couple about the y axis,
and M, about the z axis. It follows
that if the sum of the moments of
the forces in a system of forces in
:
‘
Fig. 302.
Resultant Moment from Compospace is taken about the x axis, the nents,
:
:
result is 2M,.
Also by taking moHaving
ments about the y and z axes, we get 2M, and =M,, respectively.
found these component moments, we get
(s)
M = [(2M,)?+ (2M,)?+ (2M)?
in accordance with Fig. 302, where M is the magnitude of the resultant
If & #0, the
If R = 0, but M ~ 0, the resultant is a couple.
couple.
a force and
by
recall,
we
represented,
be
resultant is a force, which may
a couple.
105. Equilibrium of a General Force System in Space. If a general force
system in space is in equilibrium, there must be neither a resultant force
It follows
nor a resultant couple; that is, in (r), R = 0 and in (s), M =0.
of
conditions
six
from equations (r) and (s) that there are the following
equilibrium:
Zien),
ZF, = 0,
my, i)
=M, = 0,
SMe, 0)
2M,
= 0.
In solving problems involving forces in space, examine carefully the arrangement of forces in order to reduce the system to two or more coplanar systems,
if possible.
106. Example. Figure 303(a) represents a drum used in a mine hoist. A cable winds
The
on the drum, which is directly driven by a steam engine whose crank pinisatC.
horizontal
the
above
60°
of
angle
an
at
view)
(plan
D
position
at
cable leaves the drum
(as shown in the end view), passes over a pulley above the mouth of the mine, thence
downward to the suspended mine cage. The line of action of the weight of the crank
and the shaft (1700 Ib.) is shown at H. The line of action of the weight of the drum
(7000 lb.) is shown at Ff. The cage is at rest and the force on the connecting rod is
C = 21,000 Ib. What is the weight of the cage and its load if it is being balanced by
the force C in the connecting rod? Find also the bearing reactions.
NON-COPLANAR FORCES
168
[Ch. IX
The end view shows the projection of all forces acting on the shaft,
Souurion.
etc., including the bearing reactions K4 and Ra, indicated with a wavy line because
they are unknown in magnitude and direction. Although the projection of a balanced
system of forces on this plane is a system of forces in equilibrium, the problem cannot
be solved completely from this projection because there are five unknowns and only
three conditions of equilibrium can be obtained for the projected coplanar forces.
However, we may take moments about the axis in this end view to determine D,
LLSLLL
7.
Zh
LLZZLIEES
8225
(b).
20,650
Horizontal Plane—Looking
(c).
Vertical Plane.
Down on Plane.
(d). The couple Cc is equal in magnitude to the cou ple Dd, but is
i of opposite
i
sense
Therefore, the forces D’
(= D) and C’ ( = C) are left actin at th
i :
;
Cc and Dd are not drawn to scale.
ie
Sits
se
Fig. 303.
§ 106 | EXAMPLE
169
since both Ra and Rez pass through the axis.
(1.5) (cos 35°) ‘= 1.23 ft.
The moment
arm of the force C is
The sum of the moments about the axis yields
4D = 1.23C = (1.23) (21,000)
or
D = 6450 lb.,
the weight of the cage and its load.
To get the bearing reactions, first imagine the forces C and D each being replaced
by a parallel force through the axis and a couple. See Fig. 303(d). The couples cancel
one another and may now be disregarded, because =M = 0 for equilibrium (Ce = Dd).
Second, resolve forces C and D at the axis into horizontal and vertical components.
Third, note that the system of forces in space is now replaced by two systems of
coplanar forces, and draw the free bodies for the horizontal and vertical planes.
Figure 303(b) shows the forces acting on the shaft (represented by a line) in the
horizontal plane, where A, and B, are horizontal components of R4 and Rz» respectively. The forces on the shaft in the vertical plane are shown in Fig. 303(c), where
A, and B, are respectively the vertical components of Ra and Rg. The components
needed in these free bodies are:
D, = D cos 60 = (6450)(0.5) = 3225 lb.
D, = D sin 60 = (6450) (0.866) = 5580 lb.
Cz = C cos 10° = (21,000)(0.985) = 20,650 lb.
C, = C sin 10° = (21,000) (0.1736) = 3640 lb.
F, ° = F = 7000
lb.
F,
= 0.
& » = EH =1700
|b.
EH,
= 0.
To solve for the horizontal bearing reactions [Fig. 303(b)], let us first take moments
about bearing A:
=M 4 = (3225)(6) — 8B, + (20,650)(9) = 0,
from which B, = 25,700 lb.
From the sum of moments about B,
M2 = (20,650)(1) — (3225)(2) — 84, = 0,
we find A, = 1775lb. The positive answers obtained for A, and B, show that correct
senses were assumed for these components in Fig. 303(b). From the vertical plane,
we have
=M a. = (7000)(4) + (1700)(5) — (5580) (6) + 8B, — (3640)(9) = 0,
from which B, = 3720 lb.
=Ms
Similarly, from
= 8A, — (7000)(4) — (1700)(3) + (5580)(2) — (8640)(1) = 0,
we get A, = 3200 lb.
The reactions at the bearings are:
Ra = [(A.)? + (A,)2? = [775)2 + (8200)2]4? = 3660 lb.
64 = tant =stalie seas = 61° (with horizontal).
Re = [(B-)? + ,)*P? = [(25,700)? + (3720)2]! = 26,000 lb.
6z— tans Be = tae
-
Bs
ot
25,700
= 8.3° (with horizontal).
The sense of Ra and of Reg should be determined from an inspection of the results
obtained in Fig. 303(b) and (c). From the directions of A, and A,, we see that R4 is
in the first quadrant, looking at the shaft from the right end. Similarly, Rs is found
to be in the third quadrant. The positions of R4 and Rs are shown roughly in the
end view of Fig. 303(a).
PROBLEMS: CONCURRENT FORCES [Ch. IX
170
Problems
CONCURRENT
FORCES
601. Each group of numbers given below
defines a force in space (see Fig. 293) that
passes through the origin. The order of the
numbers is F, x, y, 2. (200, 10, 5, 4), (500,
8 3), (300, 4, 8, 8)} (200, —8, —2, 6),
Determine the resultant.
Ans. 868 Ib., 02 = 96.6°, 6, = 39.6°, 0. =
LAP.
602. The same as 601 except that the
forces are: (1000, —5, 6, —2), (2000, 6,
=§ =), COO, &, 4, =).
603. The same as 601 except that the
forces are: (20, —5, —3, 2), (50, —6, 5,
4
Forces in pounds.
Fig. 305.
Problems 607, 608.
—8), (80, 5, 5, 5).
6, = 43.4°,
Oe = ORG.
604. The same as 601 except that the
forces are: (50, 3, 3, 3), (60, —6, —6, —6),
(40, 4, —6, 5).
Ans.
Ib.,
41.3
62 = 128.5°,
608. The same as 607 except that timber
AB is only 16 ft. long.
609. Figure 306 represents a tripod BA,
BC, and BE, an arrangement set up to
start the digging of a mine shaft.
The
diameter of the sheave at B is 2 ft., F =
5000 Ib., and the dimensions given are the
Fig. 306.
Fig. 304.
Problems 605, 606, 644.
605. Figure 304 represents a boom that
supports a load W = 1000 lb. If AB =
BC
= BD
= 6 ft. and BE
= 8 ft., find the
tension in the cable and the force in each
timber.
Ans. DE = 1667 lb., AH = CH = 883 |b.
606. The same as 605 except that BC =
10 ft.
607. Three timbers, AB, BC, and BD,
each 20 ft. long, form a tripod, Fig. 305.
The ends of the timbers on the ground
form an equilateral triangle ACD, the sides
of which are each 20 ft. long. If the safe
compressive load for each timber is 20,000
lb., what safe load
from point B?
W
may
be suspended
Ans. 49,000 lb.
Problems 609, 610.
distances between the lettered points. What
is the load on each member of the tripod?
Points A, B, D, and the forces F are in the
same vertical plane.
ANS AB es 89 5ilbai((C) 03 Hes
Ot
4570 lb.
610. The same as 609 except that the
cable strand BH is pointed downward and
parallel to the line BD,
Ans. AB = 4490 lb., BC = BE = 5410 lb.
611. A 12-ft. gear, weighing 6000 lb., is
to be mounted on a vertical 8-in. shaft.
During erection, the gear is supported on a
platform ACB, Fig. 307.
Assuming that
the three supporting tripod legs (AD, BD,
CD) come to a point at D, determine the
load on each leg.
612. A derrick
mast
(Fig. 308),
30 ft.
tall, is supported by two stiff legs, CB and
CD, each 60 ft. long. These legs are in
PROBLEMS:
CONCURRENT
171
FORCES
vertical planes intersecting at the centerline
of the mast and making an angle of 75°
with each other. The load at the end of
the 60-ft. boom AF is W = 5000 lb. The
vertical plane of the boom bisects the angle
between the planes ACB and ACD.
For
a = 90°, find the load on the mast.
Ans. 12, 270 lb. (C).
613. The same as 612 except that the
plane of the boom is at right angles to the
plane ACD, counterclockwise.
|
9660 lb.
Fig. 309.
Problems 614, 645.
614. A uniform 300-lb. disk is supported
in a horizontal position by 3 cords attached
symmetrically as shown in Fig. 309. A
slip-ring at D may be pushed downward to
increase the angle that the cords make
with the vertical.
If the cords have a
safe strength of 150 lb. each, find the minimum safe value of h.
Fig. 310.
Fig. 307.
Problem 611.
Plane ACE Bisects
Angle BAD
Problems 615, 616.
615. The shear legs AB, BC, and the
cable HB support a load of W = 30,000
lb., Fig. 310. The plane of the legs makes
a 75° angle with the horizontal.
The points
A, D, C, and E are in a horizontal plane,
and the points HZ, B, D, and
are all in a vertical plane.
on the legs and the cable.
the load
W
Find the loads
Ans. AB = 20,900 (C), BC = 13,270 (C),
BE = 8480 (T), in lb.
Fig. 308.
Problems 612, 613.
616. The cable, for the shear legs described
in 615, can support a safe load of 33,000 lb.
The safe load for leg AB is 70,000 lb.; for
leg BC is 60,000 lb. What may be the
safe load W?
Ans. 100,000 lb.
PROBLEMS: PARALLEL FORCES [Ch. IX
172
PARALLEL
FORCES
617. In Fig. 311, the parallel forces are
A = 50 Ib., B = 60 lb., Q = 100 lb., and
D = E=0.
Determine their resultant.
Ans. —90 Ib. through (—1.333, —0.566).
618. In Fig. 311, the parallel forces are
13 =O; @ = ih lho, 10) = il) No. save!
4 1b.
Determine their resultant.
622. The same as 621 except that one
cable is attached to a point 2 ft. from an
apex on a median line. The other cables
remain at the other apexes.
A
30”
L
B
Se
10”
C
Fig. 312.
Problems 623, 624.
623. A table whose top is triangular in
shape (as shown in Fig. 312) has a load of
W = 100 lb. as shown.
What is the force
on each leg at A, B, and C?
AMOS PA eS ACTS) ily 15) = IES Io (Cl =
37.8 lb.
624. The same as 623 except that the
weight of the triangular top is 30 lb. The
Fig. 311.
Problems 617-620.
center of gravity, which locates the line of
action of the weight, lies at the intersection
of the medians of the triangle.
619. In Fig. 311, the parallel forces are
A=0,B=Q=5
kips, D=0;) and
#=
10 kips.
Determine their resultant.
Ans. +10 kips through (4.5, —3.5).
620. In Fig. 311, parallel forces A = B =
100’ Ib., Q:= 50 Ib., D = # =0, and an
additional force F (not shown) is such that
the resultant is 75 lb. upward through point
(—2.67, 3). Determine force F.
Ans. F = +125 lb. through (—4.0, 4.6).
621. A horizontal steel plate in the shape
of an equilateral triangle, 12 ft. on a side
and 6 in. thick, is supported by three
vertical cables, one attached at each vertex.
The center of gravity, which locates the line
of action of the weight, lies at the intersection of the medians of the triangle. What is
the tension in each cable? Steel weighs
490 lb. per cu. ft.
Ans. 5100 lb.
Fig. 313.
Problem 625.
625. A circular table top, 8 ft. in diameter
and weighing 600 lb., is supported by three
legs at A, B, and C, as shown in Fig. 313.
A load of E = 400 Jb. is to be placed on
the circumference of a circle of diameter
D = 6 ft. in such a position that the force
on the leg A is 375 lb. Determine the
angle that a radial line through EZ makes
with the x axis.
Ans.
+66.7°.
COUPLES
626. In Fig. 314, the spaces are in inches
and the parallel forces are A = B = 5 |lb.,
@=0,
and D=E=10
|b. Find the
resultant couple.
Ans. 37.7 in-lb., ce. about an axis through
(0,0,0) and (3,7,0).
627. In Fig. 314, the spaces
represent
inches and the forces are A = D = EF = 0,
and B = Q = 10lb.
If there is also in the
horizontal plane a clockwise torque, as
viewed from above, of 50 in-lb., find the
resultant couple.
PROBLEMS:
COUPLES
Ans. 65.6 in-lb. cc. about an axis through
(0,0,0) and (—3,
—3, —5).
628. The same as 627 except that the
torque is counterclockwise.
629. In Fig. 314, A = H = 10 lb,
B=
Q = D =0, and the spaces are in feet.
If a clockwise
torque
(looking
100 ft-lb. is also applied
find the resultant.
down)
of
to the Z axis,
GENERAL
630. A welded wall bracket as shown in
Fig. 315 is to support a central load of
W = 500 lb. at a distance e = 20 in. from
the wall. The bracket is bolted to the
wall, the bolt holes being located by a =
15 in., b = 10 in., and c =2 in. Assume
Fig. 316.
Fig. 315.
Problem 630.
Problems 631-633.
633. For the windlass described in 631,
what is the position of the crank for the
maximum bearing reaction at B? What
is this reaction? The force F acts at all
times perpendicular to the centerline of the
crank.
Ans. a = 0, Rg = 720 lb.
634. The shaft in Fig. 317, supported by
the bearings A and B, is driven by the
that F; = 5 Fo, where F; is the force exerted
by each upper bolt and F2 by each lower
bolt, and that the reaction R between the
wall and the bracket may be concentrated
along the bottom edge of the bracket in the
extreme case or in case the bolts become
What maximum load is induced in
loose.
the bolts?
631. The windlass represented in Fig.
316 is in equilibrium when the load W =
600 Ib. and a = 60°. What are the horizontal and vertical components of the bearing reactions at A and B?
Ans. Az = 52 lb., Ay = 210 Ib., Be = 312
Ib. By = 540 Ib.
that a =
except
631
as
same
632. The
150°. The force F continues to act normal
to the centerline of the crank.
Fig. 317.
Problems 634, 635.
PROBLEMS: GENERAL
174
30-in. pulley C, which receives a horizontal
belt. The 15-in. gear D delivers the power
to another gear H which is so located that
the tangential force /’p between the teeth
makes an angle of a = 30° with the horizontal as shown.
Let fF; = 600 lb. and
F, = 200 lb., and find the bearing reactions
Ans. Ai, =
Problem 636.
636. The resultant force on a worm in a
worm-gear reduction is R = 6800 lb. The
axis of the worm shaft is represented as
coinciding with the y axis in Fig. 318. The
coordinates of a point on the line of action
of R arex = 4, y = —12,andz=5.
The
centerline distance between the bearings A
and B is 11 in. The bearing A takes the
—375|b., Br =
416.3 lb., By = 1375 lb.
638. The same as 637 except that 6 = 60°.
Raand Rg. The dimensions shown parallel
to the shaft are centerline distances.
Ans. Az = 768 lb., Ay = 120 lb., B, = 725
WWey.,, 13%, =) PAXO) Io}.
635. The same as 634 except that a =
120%
Fig. 318.
—50.3 lb., Ay =
[Ch. IX
Fig. 320.
Problem 639.
639. Figure 320 represents the rear gate
of a conventional pickup truck with the
gate held in a horizontal plane by chain AC.
The dimensions are as follows: AB = BC =
DE = EG = 16 in. and BE = 72 in. with
hinges at the corners B and HZ. The truck
goes straight forward until it is stopped by
a force of F = 200 lb. in the chain which
is hung on a fixed object. The taut chain
at G makes a 6 = 60° angle with the vertical and it is in a vertical plane containing
points D and #. Find the tension in chain
AC and the force on each hinge. Neglect
the deflection and friction.
Ans. AC
= 141.4 lb.(T),
B =
141.4 Ib. at
315°, # = 200! Ib. at 150°.
Helical Gear
entire thrust load (the axial component of
R). Determine the thrust load on A and
the radial bearing reactions on both A and B.
Ais, Zk, = COO Moy, Aly = OO te, 4b, =
1513 Glo
a— OOO bas
sil SOA tos
637. A rotating shaft, Fig. 319, has three
unbalanced weights attached to it. During
rotation, these weights exert centrifugal
forces @1, Qo, and Qs; acting always in a
radial direction.
See the end view.
The
angles between the lines of action remain
as shown; that is, the weights rotate together.
At a certain
speed, Q: = 500 lb.,
Q2 = 1000 lb., and Q; = 1500 Ib. Determine the horizontal and vertical components
of the bearing reactions at A and B when
6= 30%
Fig. 321.
640. Figure 321 shows a helical gear that
receives power from a mating gear and
transmits that power through the shaft to
a roller chain sprocket C. The gear and
sprocket have respective pitch diameters of
8.64 in. and 6.68 in. The components of
the force on the gear teeth are Fy = F; =
Q» (not vertical)
Fig. 319.
Problems 640-643.
Problems 637, 638.
PROBLEMS:
GENERAL
175
768 lb. (transmitted tangential force), Fz =
443 lb., and F, = 323 lb. Bearings at A
and B support the shaft, A taking all the
end thrust.
When the tight side of the
chain makes an angle of a = 30° with the
horizontal, find the x, y, and z components
The shaft is in
of the bearing reactions.
equilibrium.
Ans. In pounds, A, = 443, A, = 27, A, =
527. B, = 0, By, = 1600, B, = —700.
641. The same as 640 except that a =
180°, with 7; tangent at the top of the
sprocket.
642. The same as 640 except that a may
be varied at the will of the operator, and
the designer desires the maximum reaction
on bearing B. (HINT: It is simpler to find
the reaction at B caused by the fixed loads
F,, F,, and F, and to add vectorially to it
the reaction at B caused by 7, whose
direction is variable.
(BSverinble
which
becomes
Thatis B = Bixea DP
Biized ae [Brrsserte
for
the maximum reaction possible.)
Ans. 1788 lb.
643. The same as 640 except that a may
be varied by the operator, and the designer
desires to know the maximum radial reaction
possible at bearing A. See hint in 642.
Ans. 914 lb.
644. The same as 605 except that each
of the uniform timbers AE and CE weighs
500 lb.
645. The same as 614 except that an
additional weight of 10 lb. is attached at B.
Note that the plate will no longer remain
level; neglect the effect of the thickness of
the plate.
Ans. 0.54 ft. with symmetric cords.
646. Two cables A and B terminate on a
pole as shown in Fig. 322 and exert forces in
Fig. 322.
the horizontal plane at C. The guy cable
CD makes an angle with the pole of 45°
and the anchor at D is to be so located
that the pole will have only a compressive
load (that is, 2F, =O and =F, = 0). Let
@ = 30°, A = 5000 Ib., B = 8000 Ib., and
CE = 25 ft. Find the value of angle a and
the tension in the cable CD.
Ans. 115.8°, 6280 lb.
647. ‘The same as 646 except that 9 = 60°.
648. The same as 646 except that the
cables sag so that the cable forces at the
pole make an angle of 10° with the horizontal, sloping downward.
Ans. 115.8°, 6190 lb.
649. A 1000-lb. radio tower is 100 ft. high
and is held in place by a vertical insulated
base support and by 4 guy wires attached
40 ft. above the base. Each guy makes a 45°
angle with the pole, and they are placed on
the north, west, south, and east sides of the
pole. If an initial tension of 50 lb. is
placed in each guy, find (a) the load on the
base insulator if no wind is blowing and (b)
the force in each wire and on the base
insulator if a south wind exerts a force of
200 lb., the center of pressure being 60 ft.
above the base. Assume that the wires do
not stretch.
Ans. (a) 1141 lb.; (b) wire tensions are
E=N=W
=50lb.,S8 = 474 Ib.; insulator force = 1441 lb. up and 100 lb. north.
650. The same as 649 except that the
wind is from the southeast.
651. Figure 323 represents a crane on
which a monorail dolly supports a weight
W = 500 lb. Guys AB and BD are horizontal and along the y axis as shown.
Guy
wire BI is in the zz plane, making a 45°
angle with the vertical.
When the hori-
Problems 646-648.
PROBLEMS:
176
zontal boom EG is rotated so that @ = 60°
(in horizontal plane) and S = 12 ft., find the
tension in each guy and the a, y, and 2
components
of the socket
reaction
at H.
Neglect friction, the weight of the members,
and the initial tension in the guys.
Ans. In pounds, AB = 100(T), BD = 0,
IB = 245(T), H, = 178, H, = —100, H.
= 673.
[Ch. 1X
GENERAL
652. The same as 651 except that @ = 30°.
653. A pivoted-motor
belt drive
The dimensions are h = 175% in., e = 834
Socket
Support
pad
=
—
x.
Courtesy The Rockwood Mfg. Co., Indianapolis, Ind.
Fig. 324.
to a
blower is shown in Fig. 324. A schematic
diagram of a similar drive is shown in Fig.
325. A 50-hp, 1140-rpm motor, with its
base and pulley weighs W = 1037 lb. The
belt pulls are 7; = 691 Ib. and 7, = 230 |b.
The Rockwood Pivoted-Motor Drive.
PROBLEMS:
GENERAL
lags
in., a = 434 in., b = 2514 in., anda = 26.8°.
The end view of Fig. 325 gives the significant
longitudinal dimensions.
The belt pulls are
assumed to be centered in the center plane
of the pulley, 5 in. to the left of bearing A.
In designing bearings A and B in the pivot
base, the designer needs to know the forces
on them.
Find the horizontal and vertical
components of these bearing reactions, assuming Maz = 0.
Ans. In pounds, A; = 1017, A, = 850,
of member ABCD.
When W = 50 lb. and
0 = 30°, find the pin reactions at A and B,
assuming that they are tangent to the
surfaces of member ACDB.
Forces H =
F =0.
The vertical forces on ACDB are
at pins B and C.
AMS Ale = [ety = G10 Il, Ay = 1B, = 10
0.» Ala = ©, 1B, = HO Mloy,
B, = 196, B, = 396.
657. The same as 655 except that there
also is a clockwise couple of C = HF = 200
ft-lb., the plane of the couple being perpendicular to plane CDEF.
656. The same as 655 except that 9 =
120°.
654. The same as 653 except that the
motor is not running and 7; = 7.2, their
magnitude being the reaction to the weight
of the pivoted assembly.
655. Figure 326 represents a mechanism
that may support a weight W in various
positions.
Pinsat A, B, C, and D area part
ANS,
AG — Ben (00 Mlb. eA p=) Boe
889 lb., A; = 0; B, = 50 Ib.
658-670. These numbers
for other problems.
Motor Pulley
5 “we
A
B
Pivot Axis
Fig. 326.
Problems 655-657.
may
be used
Chapter X
CENTROIDS
The processes of this chapter are analogous to those
107. Introduction.
In order to
involved in finding the resultant of a system of parallel forces.
locate the line of action of a resultant force, moments of the component
In a similar manner, we shall take
forces are taken about one or two axes.
moments in this chapter, not only of forces, but also of lines, areas, and
volumes.
The student should be on the alert for this similarity of procedures.
108. Definitions. A particle may be said to be a point which has mass.
It may be thought of as an infinitesimal portion of matter whose dimensions
are negligible. A finite body is considered to be made up of particles.
Any material body or particle is subjected to the gravitational pull of the
earth. When the body or particle is on or near the earth’s surface, this force
of gravity is called the weight of the body.
Since the weight of a body as
measured by the force of gravity varies slightly from place to place on earth,
a certain location (Greenwich, England) is taken as having standard gravity,
where g = 32.17 «+= fps? (use 32.2)
The density of a body is the quantity of matter per unit volume.
It may
be measured in pounds per unit volume, where the pound is a standard pound
(the force of gravity on a pound mass at standard gravity).
When density
is measured in standard pounds per unit volume, it is often called the specific
weight or weight density.
Without entering into a discussion of weight and mass at this time (see
§§ 221 and 347), we may point out that a body weighing W pounds on a
spring balance has a mass of W/g slugs, where g is the value of the acceleration
of gravity in feet per second-second (fps”) at the point where the body weighs
W pounds on the spring balance.
If the specific weight or pound density of some material is w pounds per
cubic foot, the slug density, represented by p (rho), is p = w/g slugs per cubic
foot, where g is the standard acceleration of gravity.
The density of a solid body of reasonably homogeneous material is usually
assumed as constant, although the density of a hypothetical body may be
178
§ 109 |CENTER OF GRAVITY
179
assumed to vary according to some rule. Actual bodies, too, may have a
density varying in accordance with some more or less clearly defined rule; for
example, the great mass of gas which composes the atmosphere evidently
varies in density in a manner not difficult to describe approximately.
:
(a)
(b)
Fig. 327.
Center of Gravity.
109. Center of Gravity. Consider the body A in Fig. 327 as a homogeneous solid. It is made up of an infinite number of particles or differential
masses with weights dWi, dW2, dWs, + + +, ete.
The force-of-gravity vectors
of these weights are of course virtually parallel to each other and to the chosen
zaxis. The points at which the lines of action of these vectors pierce the zy
plane are designated by a, b, c, +++.
Let W be the total weight of the
body, that is, the resultant of all the differential weights dW1, dW2, dW3, °° °.
Then, taking moments about the yz plane, we obtain
Wz = (dWi)a1 + (dW2)a, + (dWs)ast+ ++ +, = 2(dW)z,
where Z is the x coordinate of the line of action of the resultant W, and where
the sum of the moments of the differential weights about the yz plane, >(dW)z,
may be written in the form fz dW, the limits of the integral being determined
by the physical limits of the body. Thus
(a)
~. joa
basalt
i
In an analogous manner, with moments
from the equation WY = (dW
)y that
(b)
about
the xz plane, we conclude
JyadW
eet
Now imagine Fig. 327(a) turned through 90° about the y axis to give Fig.
327(b). This rotation results in a horizontal z axis, a vertical x axis, and the
forces of gravity acting at right angles to the direction shown in Fig. 327(a).
CENTROIDS
180
[Ch. X
If moments are taken about the zy plane for this position (b) of the body
and the foregoing procedure applied, we conclude that
Pp a ae
(c)
These coordinates Z, 7, and Z, which refer to the reference planes, locate a
point in the body through which the resultant force of gravity W for the
This point
whole body passes, no matter what the position of the body.*
is called the center of gravity of the body.
110. Mass Center and Centroid. The center of mass of a body is that
point at which the mass of the body may be conceived as being concentrated,
so that the moment of the concentrated mass about any axis or plane is the
same as the sum of the moments of all the elements of the mass about the
same axis or plane. This statement is seen to parallel the mathematical
expressions (a), (b), and (c), the only difference being that the mass is substituted for the weight.
Recalling the statement in § 108 that the mass m in slugs is equal to the
weight W in pounds divided by the acceleration of gravity g, we may transform equation (a) as follows,
ey
a fe fa dW/g ew Sx dm.
W/g
m
By analogy, we may also write
(e)
eenm
and
gag ledm,
m
Equations (d) and (e) define the coordinates of the center of mass with respect to the three conventional reference planes.
In finding the center of
mass by these equations, one would multiply the volume by the specific
weight and divide by g; that is
(f)
m=fan=[2av =|oar,
where w is in pounds per cubic foot and p is the density in slugs per cubic
foot. For practical purposes, the center of mass is the same point as the
center of gravity (see the footnote on this page). The term center of gravity,
which is used more often than center of mass or centroid, is denoted by the
abbreviation eg. Its location is very important in two popular machines,
the airplane and the automobile.
Analogous to center of gravity and center of mass is center of volume.
The
*This statement is slightly in error because the lines of action of the forces are not
precisely parallel and because the force of gravity on an element dW varies minutely as the
distance of the element from the earth changes (say, during a 90° rotation).
However,
the error is exceedingly small and is unimportant in engineering calculations.
§ 111 |CENTROIDS OF AREAS
181
center of volume is generally spoken of as the centroid of the volume.
If
the whole volume is thought of as consisting of many minute parts dV, corresponding to the infinitesimals dW and dm, we may sum the moments of
these infinitesimal volumes dV with respect to any coordinate plane (to get,
for instance, fz dV) and divide this sum by the total volume to find a coordinate of the centroid; thus
2 a fad.
V
,
(26)
t=
mee iia a
ma
icay
Uae
Le
cae
vy?
The various centroids given by equations (a), (b),* ° *(e) and (26) may be
Referring to equation (f), we see
coincident under certain circumstances.
If the
that if the material is homogeneous, the specific weight is constant.
body is not tremendously large, the acceleration of gravity is the same for
all points in the body. If both of these conditions are true, we have
w = faw =wfav,
m= fam =" fav =pfav
That is, w, g, and p may be taken outside of the integral sign. In the ratios
(a) «+ +(e), consequently, these terms would cancel. So for the conditions
stated, the centroid of the volume is the same point as the center of gravity
and the center of mass. But it must not be forgotten that if the density
does vary, the basic equations (d) and (e) must be used to find the center of
mass.
The centroid of a volume is not influenced by the density.
Fig. 328.
The moment of dA about the y axis is x dA; about the x axis, y dA.
In designing beams and columns, we need to
111. Centroids of Areas.
The centroid of
locate the centroid of the sectional area of these members.
an area is about the same as the center of gravity of a very thin sheet of
Suppose the sheet in Fig. 328 has a total area A and a thickness
material.
of t. The moment of a differential volume (dV = ¢dA) about plane zOy is
at dA, and the sum of the moments of all the differential volumes may be
indicated as
(g)
Vz = tA% = atdAi+ rtdA,+
+++
= Zaid.
cENTROIDS [Ch. X
182
(26), we see that they
Comparing this expression with that for Z in equation
es with
Also, we recall that the centroid of the volume coincid
are the same.
sheet
this
if
Hence,
neous.
the center of gravity, if the material is homoge
7m
lying
as
it
of
think
of material has an infinitesimal thickness ¢, we could
2,
(21,
dA’s
the plane «Oy, Fig. 328. In this event, the moment arms of the
be canceled
Yi, Y2 * * *) are distances to the y and x axes and the term ¢ may
have
we
from each side of equation (g). Then
Az =, dApy-+4jdAo
+ +: = 2204,
or, for an area A in the «zy plane and in integral form, the coordinates of the
centroid are
(h)
ed =
peas
and
TT] =
He
where the x and y axes are any two convenient axes in the plane of the area.
In applying equations (h), we may have to find the total area A from an
For example, dA = dx dy, and an integration may be made if
integration.
the boundaries of the area are defined in terms of x and y. This principle is
Problems involving the center
explained in detail in subsequent examples.
for engineering purposes by
solved
be
often
may
plates
thin
of gravity of
finding the centroid of the corresponding area.
112. Centroids of Lines.
Suppose we
have around wire bent in the form ABD of
a semicircle, Fig. 329. Let C be the location
of the center of gravity of this wire.
Of
course, C’ lies in the plane passing through
A, B, and D which slices the wire into two
Fig. 329. The moment ofdZ about
equal parts (see $113).
Now suppose the
the y axis is x dL; about the x axis,
:
?
:
pau
centroid C is connected to the wire by
weightless, rigid members AC, BC, and DC.
Then the wire can be balanced in a horizontal position by a point support
at C.
If the wire is imagined to decrease in sectional area, it approaches a line,
which has only one dimension (length).
Then point C would be the centroid
of the semicircular line. Approaching the defining equations from the
viewpoint of center of gravity, as we did for areas, we have
dV = AdL,
where A is the cross-sectional area of the wire.
If the wire is made of homogeneous material, the center of volume coincides with the center of gravity.
For volumes, we may write
Vz = ALZ = XxA aL.
§ 114] ESTIMATING LOCATION
OF CENTROID
183
Canceling the A term as being inapplicable to a line and including the expression for the y coordinate by analogy, we get the coordinates of the centroid
of a line in integral form as
(i)
oa
ap
Sedb
ZL
and
=
SyL ah )
where L is the total length of the line. The line may be straight or curved,
or made up of a series of straight or curved elements.
The actual location
of the center of gravity of a wire of relatively small section may often be
found by considering the wire to be a line.
113. Principle of Symmetry.
Consider the plane area, in the form of an
H-section in Fig. 330, which is symmetric about the x and y axes.
In taking
the sum of the moments of the differential areas about the x axis (fy dA),
we observe that for every dA on the positive side of the x axis, there is another
dA on the negative side with precisely the same moment arm y, but of
opposite sign.
Therefore
fydA = 0, and the centroid of this area lies on
the x axis. Similarly, for every dA on the right (positive) side of the y axis,
there is a dA on the left side with an opposite moment, so that the fe dA = 0,
and the centroid must lie also on the y axis. Consequently, the centroid is
at C, the intersection of the x and y axes.
Fig. 331.
For similar reasons, the centroid of the three lines in Fig. 331, which form
a symmetric figure, must be at the intersection of the x and y axes, since in
each instance, an element of the line dL on one side of one of these axes is
“balanced” by a like element on the other side. Thus, from equations (h)
and (i), we conclude that:
If a plane figure has a line (or lines) of symmetry, its centroid lies on that
line (or lines).
In a like manner, we conclude that:
A plane of symmetry of a solid body contains the centroid.
The
For example, any diametral plane of a sphere is a plane of symmetry.
of
center
common point of the intersection of any three such planes is the
the sphere, which is its centroid.
In the examples which follow and
114. Estimating Location of Centroid.
in your own solutions of problems, it is advisable to take a practical look
CENTROIDS [Ch. X
184
For
at the answer and see if the centroid appears in its correct position.
plane,
al
horizont
a
in
located
D
and
B,
example, if there are three weights A,
of
whose magnitudes are shown in Fig. 332, you should locate the center
and
eravity of these weights at point C with little or no thought of the rules
(Be sure you do this.)
equations.
Similarly, in locating the centroids of a line or plane area, we may think
of the line as being a curved or bent wire in a horizontal plane, and of the
area as a thin sheet of metal in a horizontal position. Then the centroid is
that point which would support the line or area in the horizontal position.
The centroid may or may not be located on the line or within the area.
10
10
Fig. 333.
Fig. 334.
To estimate by eye the location of the centroid of an area, draw at least
two lines, each of which seems to divide the area into two parts, each of
which appears to have the same moment about the line. The intersection
of these lines should locate the centroid C approximately, as suggested in
Fig. 333. To test your first estimate, draw another line to see if it passes
close to your first estimate of C.
The same idea applies to a curved line, as shown in Fig. 334. Remember
that the moments of the areas and the moments of the curves should look as if
they balance.
This process is purely qualitative and you may make a bad
guess. But it should keep you from being satisfied with a silly answer.
115. Integrating for Centroids.
Unless a plane figure has two axes of
symmetry or unless a solid figure has three planes of symmetry with a common point of intersection, it is frequently necessary to use the integral calculus to locate the centroid.
To set up the integral, it is essential to choose
a differential element
(1) all of whose parts are at the same distance from the axis of moments,
or to choose an element
(2) whose own centroid ts known in location.
The application of these principles is best taught by examples.
116, Example—Arc of a Circle.
r which subtends an angle of 28.
Locate the centroid of an are of a circle of radius
§ 117 | EXAMPLE—PLANE
TRIANGLE
185
Sotution.
In Fig. 335, we have chosen the x axis as the bisector of the angle 28,
because this line is a line of symmetry and therefore contains the centroid of the are AB.
Since the centroid lies on line Oz, it remains only to find the distance =.
Choosing an element dL = r dé as shown in Fig. 335, and taking moments about
the y axis, we find
+B
iEi
|
+8
(r cos 6)(r dé) = ef cos 6d@ = 2r? sin B,
— 6
u=
Thus,
:
—6
+6
+6
[ rao = rf 278"
—B
—B
5
fed
:
2esnp
rsin Bp
where
is expressed in radians and is half the total angle subtended, and where the
unit of Z is the same as used for r.
[eX
MLLLLLLLLLLLLLLLL
Fig. 335.
117. Example—Plane
plane triangle.
Fig. 336.
Triangle.
Determine
the location of the centroid of a
Choose the x and y axes, as shown in Fig. 336, with the x axis coinSo.ution.
ciding with a base of the triangle. Then choose a differential area, all parts of which
are the same distance from the z axis. This area is dA = (a2 — 2,1) dy, as seen from
Fig. 336. Therefore,
Aj -[y dA = [ve — a1) dy.
(k)
From the similar triangles ABD and EBG, we find the proportion
t=
i
ey
b
from which 22 — 21 = (b/h)(h — y).
we get
hey’
Substituting this value of x2 — 2, into (k),
A
ab
i
)
rete
|
ee
is
Since the area of the triangle is bh/2, the distance to the centroid from the x axis
fydA
bh , bh_ he
YS tA
6. 2a 8
CENTROIDS [Ch. X
186
base AD.
Thus, the centroid is located at a distance of one-third of the altitude from the
, we
By taking, say, AB as the base and 2 axis, and repeating the foregoing procedure
interThe
AB.
base
this
again find that the centroid is one-third of the altitude from
the
section C of the lines which are parallel to the bases AD and AB and which satisfy
medians.
the
of
on
intersecti
of
point
the
being
also
as
d
recognize
above findings is
With this observation, it is easy to remember the location of the centroid of a triangle.
Find the centroid of a sector of a circle of
118. Example—Sector of a Circle.
radius r, whose central angle is 28.
BSS \
L=
dL=rd9@
Ss
Cc
Fig. 337.
Shaded area = dA.
Fig. 338.
Soxrurion.
If we choose the axis of symmetry as the x axis (the y axis would do as
well), only one integration is necessary because 7 = 0. We shall use the elementary
area OEF (Fig. 337), because by doing so we must apply the results of § 117 and use
the element (2) defined in §115.
For an infinitesimal angle de, this area becomes a
triangle with a base dL = r dé and an altitude of r. Thus,
dA.
r? de
=
Now observe that all points of the area are not at the same distance from the y axis.
However, we now know the location of the centroid of a triangle at one-third of the
altitude from the base and on a median.
Thus, if points Z and F, which coincide in
the limit, are a distance of r cos @ from Oy, then the point C, the centroid of OFF, isa
distance of x = (2/3)r cos @ from Oy. Moreover, since the moment of an area about
an axis ts equal to the area times the distance from the axis to the centroid of the
area, we have
2
r? do
xrdA = =(r cos 9)
2
0
and
9
2
Saas
Az = iEdA = [ie cos —
;
l| ch
3
= f°
+ DW
a.B >
a
a
™
I
to és
cos 6 dé
Bp es
mM—_
The area of the sector may be found by integrating dA or it may be taken as a proportion of the area of a circle.
-=
In either case,
JedA
AS
@rsing
>
Wer
A = 6r?, so that
— 2rsing
Spe?
§ 119 |EXAMPLE—AREA
WITHOUT AN AXIS OF SYMMETRY
187
where £ is half the central angle of the sector.
4\/2r
For a 90° sector, 7 =
For a 180° sector,
or
+
= 0.6r, approximately.
= — = 0.4247, approximately.
It is recommended that in your solutions you find the fadA (or fa dL, etc.) as a
separate operation; then divide by A (or L, etc.), showing the fdA as a separate
operation when it is necessary to integrate for A; which is the usual procedure in
these examples. If you want to learn this material with the least effort, be sure to
work through all examples with pencil and paper, filling in omitted details.
ALTERNATE SoLuTION.
Usually, there is more than one elementary area dA which
may be used in finding centroids.
For this sector of a circle, we may just as conveniently use the dA indicated in Fig. 338. In terms of polar coordinates, one side of
this area is dp and the other side is p d@. Therefore, from Fig. 338,
dA
=
p dé dp.
The moment arm of the area is x = p cos 6. Since there are two differentials, da and
We observe that the limits of the variable
do, a double integration is necessary.
radius p are from 0 to r in order to include the whole sector, and that the limits for
gare from —@to+s.
Thus
r
+e
r
(+6
0 dé
Ar, = iEdA -| | (p cos 6)(p dé dp) -| (! p? dp cos
o J-6
0 J-8
lis
P
ais (2 sin B) =
the same as the value of Az previously found.
Dips
97 sin B,
It is possible to solve this problem
using rectangular coordinates, but it would be more troublesome.
Determine the location of
119. Example —Area without an Axis of Symmetry.
the centroid of the area enclosed by ares of the parabolas y? = 9x and a? = 4y, where
the linear units are inches.
The area involved is that enSontution.
heavy lines in Fig. 339. One
the
by
closed
point of intersection of the curves is evidently
at the origin O; the other point is found by a
simultaneous solution of the equations of the
Thus, from one equation, we find
curves.
= 2/4. Substituting this value of y into
(5.24,6.87)
y
1
the other equation, we get
4
ey
16
or
a
f=
ees
5.24 in.
This value of x used in one of the equations gives
y? = 9x = (9)(5.24) = 47.16,
Fig. 339.
or
47 = 6.87, in.
Consequently, the coordinates of the second point of intersection (in addition to the
CENTROIDS [Ch. X
188
Let the differential area be
origin) are « = 5.24, y = 6.87, as shown in Fig. 339.
parallel to the x axis. We see then that
dA = (Goss) ay;
curve
where 2» is the value of x on the curve a? = 4y and 2; is the value of z on the
y? = 9x; that is,
py PANE?
and
gia iy
Using these values of 71 and x2, we find A from
y?
6.87
.
2
te
At
a
ayal2
Ee ie Aea
LE
Hee
ee
;
ME bk Bs De
= foa~for-siv= f(a) =[Ba wr jones
To get fy dA, use the value of dA above and find
Aj =
y! he= 37.3
3/.9 m. in’
y\ 2y?? — —“)}dy = 25 — ——__
| dAA = /wif
Og
a
aT) Aa
FtiNC
Es
Since each of the foregoing integrals are with respect to y, the limits are naturally from
the smallest value of y in the area (= 0) to the largest value of y (= 6.87). The value
of 7 is now found to be
~ al fydAy 368
= ont bin.
P= ar
Tp)
In determining 7, we may use the same element of area as before. In this case,
however, not all points in the differential area are equidistant from the y axis. Since
in the limit, the differential area is a rectangle, the centroid of dA is known by the
principle of symmetry.
That is, the distance of C, Fig. 339, from the y axis is
x = (to + %)/2 = (Qy/? + y?/9) /2.
Using this distance in fx dA, we get
ps
AZ —=
1
y
y
ay
lz
—
[ras =- [3(x1/2 ar BN ay
1
==
[%8?
ys
4y —
1
[49
ly =
ys
6.87
=
=
in.$
ALTERNATE Soututions.
It will be worth while to analyze this problem using other
elementary areas. An element of area parallel to the y axis may be used, as shown
in Fig. 340. In this case,
2
dA = (y2— yi)dx = (so = =)dex,
where
ve
oy = Bae
and
are obtained from the equations of the curves.
Bare
For moments
about the y axis
*
d
we have
5.24
At =/e dA -|
0
2
xv (so — =)dx = 28.15 in
as before. Observe that this set up is somewhat simpler than that for Az by the
previous elementary area.
§ 120 |EXAMPLE—RIGHT
CIRCULAR CONE
189
To get the moment about the x axis, we use the moment arm (Fig. 340)
yt
y
ys _— dal?+ 22/4
2
7
2
:
where y is the distance to the centroid C of the elementary area in Fig. 340.
5.24
Ay = JydA = i ( eC
3: -1/2
2
Then
i =)dx = 37.3 ins
4
2
0
as before.
(5.24,6.87)
Fig. 340.
Fig. 341.
Another element of area which may be used is shown in Fig. 341, anareadA = dz dy.
Thus, if the first integration is with
For this area, a double integration is necessary.
have, from J fa dx dy,
we
y,
to
respect
with
second
the
respect to « and
6.87
(2, = 22
1
(%8
nA
raedy = 5f (4 — 2)au.
az=fras=|
0
Z = 92/9
=J
0
81
[FIRST INTEGRATION WITH RESPECT TO 2]
Note that the first integration must be from one curve to the other and that therefore
the limits must be expressed in terms of y. The limits of the second integration then
are the extreme values of y, from 0 to 6.87.
If the first integration of ffx dx dy is with respect to y, the limits would be from
yi = 22/4 to y2 = 3217, In particular
5.24
ar=fras=|
0
[FIRST INTEGRATION
(302
‘ x dx dy,
WITH
2/4
RESPECT
TO y]
where it is noted that the limits of the second integral are from 0 to the maximum
on the
value of x = 5.24. Observe that the second limit of the integral should be
positive side of the first limit.
The student should complete the preceding integral
to see that it gives the same result for Z.
you
If you study all phases of this example until you understand it completely,
areas.
should have little trouble with finding centroids of finite areas from differential
120.
Example —Right Circular Cone.
Find the centroid of a right circular cone,
whose altitude is h and whose base has a radius rf.
CENTROIDS
190
[Ch. X
Sorurion. Arrange the cone so that its axis coincides with one of the coordinate
axes, say the x axis, Fig. 342. Now since the xz and zy planes are planes of symmetry,
their intersection (the zx axis) contains
the centroid. It remains only to determine &.
Choose a differential volume dV contained between two planes perpendicular
to the x axis and a distance dx apart.
The volume of the thin disk corresponding to this differential volume is the area
of the base times the thickness:
,
dV
=
rz’ dz.
By proportionality from similar triangles
in Fig. 342, we find
ee
—=2
Dp
te
Z= —
or
h
h
Substituting this value of z in the expression for
V
=
awe
/dV 7 = /
dx
rr
=
dV, we have
:
@
ah
|, x? 2 dx
—
3
=
°
Using the same value of dV in fx dV, we get
Vz =| adV = | oretde =|
I?
p
Jo
[Pe
ode
2f2
=~ —
4
Therefore
I)
pv
a
CN
ee
Cael
ee
Ue
that is, the centroid of a right circular cone is on the axis at a distance of 3h/4 from
the vertex (h/4 from the base). We should recall that if the density is constant, the
centroid and the center of mass are at the same point.
121. Composite Figures. A composite figure is one made up of two or
more basic figures, such as a triangle and a rectangle, a circular are and a
straight line. The location of the centroid of a composite figure is found by
applying the now familiar principle that the moment of an area (or curve)
about an axis is equal to the product of the magnitude of the area (or length
of curve) and the distance from the axis to the centroid of the area (or curve).
Thus, suppose that areas A’;, A’s, A’s, etc., have centroids which are at the
distances x’, 2's, x's, ete., respectively, from a particular axis. The centroid
of the composite area is then located by the principle
AZ = (A
+A2+
AR +
°°)B = Ale
+ Alex's + A'sx’s +
o2*,
In short,
(27a)
Az = DA’x’,
where A is the total net area for which the centroid is desired, and A’ symbolizes a component part of the total area whose centroid is known to be a
§ 122 | EXAMPLE
191
distance x’ from the axis of reference.
Lacking an axis of symmetry, we can
find the other coordinate of the centroid of the composite area from the
equation
(27b)
BR
aye
By analogy, it should be evident that the coordinates of a composite line are
given by
(k)
LE = XL'x’
and
Ly = =L'y’,
where L is the total length of line, and L’ is the length of a part of the line
whose centroid is a distance x’ from the reference axes.
For a composite solid body, the centroid is located by
(1)
Vie
Vz),
Vol=, 2247
and
Ve=
>V'2’,
my = =m'y’,
and
Ui = PN
and the center of mass by
(m)))
ma =" 2's",
We remember that when there are lines and planes of symmetry, the computations for locating centroids are reduced accordingly.
122. Example.
Locate the centroid of
the section of an angle iron whose dimensions are as shown in Fig. 343. (In finding
x
7
z
stand a specified load, the location of the
centroid is sometimes needed.)
>
SoLtution.
Divide this composite area
into any convenient areas, the centroids of
which are known; for example, the rec-
trarily,as shown.
B
z
the dimensions of an angle iron to with-
tangles A’; = KBGD and A’; = KEFO.
The reference axes have been placed arbi-
3”
y
YD
EERIE:
LA
a
Xo
--——H«/_ ----_-_-__-_---—- ari
Other axes may be used.
Fig. 343.
The coordinates of the centroids of these
areas are evidently
fy —= 4p in.,
y=
0.6 ine
A’, = 78q. in.,
and
y', = 0,5 10.,
Also,
A = 13 sq. in.,
and
y’5 = 3 in.
A’, = 6 sq. in.
Applying the principle of equation (27), we have
Az = Aya’: + A’ ors,
from which = = 34.5/13 = 2.65 in.
Ay = Aly’ + A’y’s,
that is,
13% = (7)(4.5) + (6)(0.5),
From the equation
that is,
1397 = (7)(0.5) + (6)(3),
we get 7 = 21.5/13 = 1.65 in. In obtaining 7, we have, in one sense, assumed the
downward direction as the positive direction of y. On the other hand, we may consider
these areas as we considered forces and arbitrarily take either direction of turning
about an axis as positive.
In any case, the engineer should estimate mentally (or
CENTROIDS
192
[Ch. X
“by eye”) the approximate location of the centroid and satisfy himself that his
answers are reasonable.
ALTERNATE SoxtuTion. In solving this problem, we may take the rectangle
OBHF, Fig. 343, as the basis and, from the moment of this area, subtract the moment
which is not part of the figure. Thus, about the y axis, we have
of the area DGHE
13% = (48)(4) '— (85)(4.5) = 34.5,
from which 7 = 34.5/13 = 2.65 in., as before.
123. Example. A homogeneous rod and a hemispherical part of the same material
are welded together, as shown in Fig. 344. Where is the center of gravity of this composite body?
Fig. 345.
Sotution.
We choose the x axis as shown because the body is symmetric with
respect to this axis, which therefore contains the center of gravity. The y axis is
chosen arbitrarily at the junction of the two parts. Since all points in this body have
the same density, the center of gravity is virtually coincident with the centroid.
Hence, consider volumes only. The location of the centroid of the hemisphere is
found by integration, by methods previously explained, as
ve’) =—r=—in.
8
4
(See problem 705 at the end of this chapter.)
, _ (mr
aS
V=-
2n(2)3 = 16.76 in,
Dae
D?
The volumes are
3
a(1)?
iad
(8) = 6.28 ins,
V = 16.76 + 6.28 = 23.04 in.
Now, since the parts of the composite body are on opposite sides of the y axis, their
moments will have opposite signs. Letting the moment of V’; be positive, we get
from equation (1)
23.04% = (16.76)(0.75) — (6.28)(4) = —12.55,
from which
= —0.545 in., the negative sign showing that the centroid C is on the
opposite side of the y axis from the volume V’;, as shown in Fig. 344. Putting it
another way, the moment of V’; was taken as positive. The resultant moment
(—12.55) is negative. Therefore, C must be located on the opposite side of the y
axis from volume V’; to accord with the negative moment.
§ 125 | EXAMPLE
193
124. Example.
The part is shaped as explained in the previous example, but
the hemispherical head is made of bronze, which weighs 550 lb. per cu. ft., and the
shank is made of aluminum, which weighs 165 lb. per cu. ft. Locate the center of
gravity.
SotuTion.
Since the parts are not now of the same density, equations (m) must
be used, except that we may use weight instead of mass.
The weight of volume
V’, of bronze is (1728 cu. in. per cu. ft.)
ay
(850)(16.76)
_SS hs) lb.
=
wV 1=
1728
W
The weight of volume V’> of aluminum is
Sees
a(165) (6.28) 0e
W = W’,+
W’. = 5.33 + 0.6 = 5.93 lb.
Taking moments about the y axis with the moment of W’; being positive, we have
Wz = 5.932 = (5.33)(0.75) — (0.6)(4) = +1.94,
where the positive sign indicates that the c.g. should be located so that the moment
of the total weight W is positive; that is, on the same side of the y axisas W’;. Then
ie
to the right of the y axis.
on
Compare the answer of this example with that of § 128.
125. Example. A right circular cone of homogeneous material (Fig. 345) has a
diameter D = 20 in., an altitude h = 30in. A cylindrical hole, whose geometric axis
is coincident with that of the cone, is bored in the base. This hole has a diameter
d = 8in.andadepthb = 6in. Locate the center of gravity of the cone with the hole.
Sotution.
Since the density is constant, use the volumes.
1xD?,
Vi=V
ec
cone
=
eT
V's =
Vhole
a
a4
V =
V’, —
h =
—_(20)2(30)
(3)
=
3141.63141. in.,
i 3
2
4
b —
AU)
4
=
301.6
WA
V’. = 3141.6 — 301.6 = 2840 in?
Also
£', = 0.752 = (0.75)(30) = 22.5 in.,
from the example of § 120; moreover,
vs =h—
from Fig. 345.
5= 30-3
= 27 in.
Taking moments about Oy, we find
Va = =V'r’
2840 &
(3141.6) (22.5) —
(301.6)(27)
=
62,470,
where we observe that the moment of the material that would fill the hole is subtracted
from the moment of the whole. Thus, 7 = 62,470/2840 = 22 in.
CENTROIDS
194
[Ch. X
(a) Theorem I. If a surface
126. Theorems of Pappus and Guldinus.*
non-interof revolution is generated by a plane curve revolving about any
to the
equal
7s
surface
d
generate
the
of
secting axis in its plane, the area
curve
the
of
centroid
length of the curve times the distance traveled by the
while the surface is being generated.
Consider any differential element dL of some curve MN, Fig. 346,
Proor.
which is revolved through an angle of 6 radians (S 27 radians) about the x
axis. The differential area generated by the element dL is then dy dL, where
the length of the are swept out by the element dL is 6y. The area generated
by the curve MN
is therefore
A=
|oat
=of vat
for a particular value of 6. But since from
equation (i), § 112, fy dL = Ly, we have the
area of the surface generated by ZN as
Fig. 346.
Generating an Area.
i=
afydL =
6yL
Po]
where 67 is the distance traveled by the centroid of the curve MN.
(b) Theorem II. If a volume of revolution is generated by a plane area
revolving about any non-intersecting axis in the plane of the area, the generated volume is equal to the magnitude of the area t2mes the distance traveled
by the centroid of the area while the volume is being generated.
Proor.
The plane area MN is revolved about the 2 axis, Fig. 347(a),
through an angle of @ radians (S 27 radians).
An element of this area dA
moves through a distance of 0y and generates a differential volume of dy dA.
Thus the total volume is
v=/oyaa =o
[yaa
for a particular value of 6. However, from equation (h), § 111, we note that
JydA = Aj. Therefore, the volume generated by revolving MW
is
V = ofydA = 6A,
where 67 is the distance traveled by the centroid of the area.
*Pappus of Alexandria, who lived sometime about A.p. 290, wrote extensively, some
eight volumes, on mathematics, and held forth at length on the properties of the centroid,
including Theorem II. However, he wrote at a time when interest in geometry had
waned, so that much of his work was lost and not much attention was paid to the rest.
Some years later, about 1300 years, Paul Guldinus, or Guldin (1577-1643), published a
voluminous book dealing with the centroids of curves, surfaces, and solids.
In this work,
which covered the subject more thoroughly than it had been covered before, he referred
to the work of Pappus and employed these theorems.
Thus, the theorems have come to
be identified by either name.
§ 128 |CENTER OF PRESSURE
195
(b)
Fig. 347.
Generating a Volume.
The figure (b) suggests the application of the theorem
to a “doughnut,’”? whose volume is A7@, where 6 = 27 for the whole doughnut.
127. Examples.
(a) Determine the area generated by revolving the semicircular
arc MPN of Fig. 348 through 270° about the x axis.
SoLtution.
From
§ 116, we find the centroid C of the are to be at a distance of
rein 6’ (2)(1) - 4
8
(7/2)
from the line MN.
length of the arc is
Thus,
=
ft.
7 = 1.5+4/r
y
ft.
The
C
He
tect
=r
L
=
27 ft.
“ie
if A
Therefore, for 6 = 270° = 37/2 radians, we have
A
T
4
2
Tv
aeptiee (154 *)Qn = 82 sq. ft.
(b) Determine the volume
360° about its diameter.
Soztution.
ay
x
Fig. 348.
2
generated by revolving a semicircular area through
Let the radius of the semicircle be r.
Then, the centroid of the area is
located at a distance of 7 = 2rsin 6/(3 8) = 4r/(8r) from the diametral line of the
semicircle (§ 118). For a value of ¢@= 360° = 27 radians and a semicircular area of
A = m1 /2, we have
A4r\(
a7?
4
=
pel
= ((On (2) 5 )= 3
— ars,7
V=o07A
which we recognize as the proper expression for the volume of a sphere, the volume
generated by revolving the semicircular area about its diameter through 360°.
The theorems may also be used, conversely, to determine centroids of
Thus, in the solution of example (b) above, it is reasoncurves or of areas.
able to suppose that A = mr?/2 and V = 47r3/3 are known, and that 7 is
The equation V = 6yA may then be used to find 7, that is, to
unknown.
The student who remembers the formula
find the centroid of a semicircle.
easy means for remembering the formula
an
as
this
for A and V may regard
for 7.
128. Center of Pressure.
If an area is subjected to a constant or variable
pressure, there is a point in this area through which the vector representing
PROBLEMS
196
[Ch. X
the total resultant force on the area passes. This point is termed the center
If the pressure is uniform, the center of pressure evidently
of pressure.
If the
coincides with the centroid of the area over which the pressure acts.
pressure varies from point to point on the surface, the resultant force does
In this event,
not in general pass through the centroid of the surface area.
the center of
if
known
be
must
pressure
the
of
variation
of
the manner
pressure is to be found.
One of the most common applications is the case of a vessel open at the
top to atmospheric pressure and containing a liquid. Since the gage pressure*
of the liquid at any point on the sides of the container is directly proportional
to the depth of the point from the surface of the liquid, this pressure varies
according to a straight-line law, being zero at the surface and a maximum at
the bottom of the container.
At a depth of ZL feet, the volume of liquid
above a square foot of area is AL = (1)(L) cubic feet. For a specific weight
of w pounds per cubic foot, the pressure is wl pounds per square foot (psf)
or wh /144 psi. An example will show the method of finding the center
of pressure.
129. Example.
In Fig. 349, OC represents a beam with a load varying from zero
at O to a maximum of p psiat C. Let the beam have unit depth perpendicular to the
plane of the paper with no change in the loading in this direction at a particular
section of the beam.
Locate the center of pressure.
SoLution.
Ata distance x from O, the pressure p, on a differential length of the
beam dz is virtually constant. Hence, the differential load on this length is dW =
p:dA = p, dx, and the total load on the beam is fp, dz. From similar triangles,
Px/p
y
wi
=
ef Ly Oe
=
px /L.
Thus,
the
total load is
|
Pp
W
ae,
sara
Fig. 349.
Center of Pressure.
ee
Px
pL
a
9
for a beam of unit depth.
From the
relation Wz = fx dW (§ 109), we get
xe
W
pL
ee
2
which locates the line of action of the resultant load.
This conclusion might have been
reached directly by noting the triangular distribution of the load and recalling the
location of the centroid of a triangle (§ 117).
Problems
671. What is the location of the centroid of a right circular cylinder? Of a sphere?
Of a rectangular parallelepiped?
Of a box whose sides are each rectangular?
What
principle did you use in arriving at your answers?
*“Gage pressure”’ signifies pressure above (or below) atmospheric pressure,
PROBLEMS: LINES (BY INTEGRATION)
LINES
(BY
197
INTEGRATION)
672. Find by integration the centroid of a circular are which subtends an
angle of 150°, using the y axis as an axis of symmetry.
Ansmy — Onan
673. The same as 672 except that the arc subtends 180°.
674. The same as 672 except that the arc subtends 90°.
Ans. 0, 0.901r.
675. The same as 672 except that the are subtends 2 radians.
676. Locate the centroid of a quadrant of a circular are by integration
when the z and y axes mark the boundaries of the quadrant.
ANS. B=
= OLB
677. Locate by integration the centroid of a circular arc which subtends
a central angle of 60°, measured counterclockwise from the positive y axis.
678. Locate by integration the center of gravity of a slender wire bent
into a parabolic curve whose equation is y? = 4x and whose ends are defined
by points (0,0) and (4,4in.).
Ans. L = 5.92 in., Z = 1.64 in., 7 = 2.3 in.
679. The same as 678 except that the end points are (—4, 4) and (V8, 2).
AREAS
(BY
INTEGRATION)
680. An area is bounded by the parabola y? = 8z and the straight line
z= 5in.
Determine by integration the centroid of this area.
Ans GF =O, B= Bim,
681. The equation of an ellipse is x?/25 + y?/16 = 1. Locate by integration the centroid of that part of the area of this ellipse which falls in the
first quadrant.
The linear unit is a foot.
INOS, GE SPANK
SN sive
682. Locate the centroid of the area included between the parabola
xz? = 4y, the line x = 4, and the z axis.
683. The same as 682 except that the right-hand boundary is the line
iy = Oita
Ans) G = 6.10, 9 = 6.08 1n684. I.ocate the centroid of the area bounded
by the parabola x? = 9y,
the x axis, and the line e = 38. The unit is the inch.
685. Locate the centroid of the area included between the parabola
y? = 8a, the y axis, and the line y = 8 in.
Ansaa = 2.4, 7 = 610:
686. Find the coordinates of the centroid of the area bounded
by the
curves y? = 16x and y = 2z.
687. Find the coordinates of the centroid of the area bounded by the
Ans. 6.4, 8.
curves y2 = 16x and y = z.
688. Find the coordinates of the centroid of the area bounded by the
curves y? = 16x and x? = 16y.
Ans. & = 9 = 7.2.
689. Determine
by integration
the centroid
of a triangular
area
the
corners of which are (0,0), (0,4) and (6,0).
690. Determine % and 7 for the area between the curve y = sin x and
the x axis between the limits x = 0 and x = m. Table of integrals may be
used.
Ans. 1.57, 0.392.
691. The same as 690 except that the upper limit is
= 2 in.
PTNAP
KE
PAS
pI
SRRA
KBD
PROBLEMS: AREAS (BY INTEGRATION)
198
692. Integrate
for the centroid
of the
sector of a circle which is located in reference
to the « and y axes as shown in Fig. 350.
Ans. 19.1, 93.1 in.
Let 6 = 60°.
Fig. 350.
[Ch. X
696. Find the centroid of the shaded
area A2 of Fig. 352 when a = 1/9 and the
coordinates of P are (3,3 ft.).
Problems 692, 693.
693. The same as 692 except that 6 = 1
radian.
Fig. 353.
Problems 697, 698.
y
697. Find the centroid of the shaded area
A, of Fig. 353 when a = 1 and the coordi-
D
—
Fig. 351.
6
3:
nates of point B are (4,8 in.). Ans. 1.48, 5in.
698. The same as 697 except that area
A: Is to be used.
Problem 694.
694. Integrate directly for % and 7 of the
segment shown crosshatched in Fig. 351.
Let D = 10 in. and 6 = 60°.
Fig. 354.
Fig. 352.
Problems 695, 696, 741, 742.
695. Find the centroid of the shaded
area A, of Fig. 8352 when a = 1/3 and the
coordinates of P are (3, 9 in.).
Ans. 1.19, 5.14 in.
Problems 699, 700.
699. In Fig. 354, a = 1 and area Aris
bounded by lines y = 2 in, y = 4 in., x =
1 in., and y? = ax’. Find the centroid of
Ai.
Ans. 1.56, 3.16 in.
700. In Fig. 354, a = 1/4 and area A; is
bounded by lines
= 4 ft., x = 6ft., y =1
ft., and y? = ax’. Find the centroid of Ao.
701. A right circular cone has an altitude
h, and the radius of its base is r.
Determine
the location of the centroid of the lateral
surface of this cone.
uint:
Use
the
dif-
ferential area dA between two planes, each
PROBLEMS:
AREAS
199
(BY INTEGRATION)
parallel to the base of the cone, which are
a differential distance apart.
Ans. 2h/3 from apex.
702. Integrate for the location of the
center of gravity of a very thin hemispherical shell (taken as a surface), whose radius
is r. HINT: Use the differential area dA
between
two
planes,
each
parallel to the
SOLIDS
(BY
base of the hemisphere, which are a differential distance apart.
Ans. r/2 from base.
703. The same as 702 except that the
surface is considered to have a variable
density such that the density at any point
on the surface of the hemisphere is proportional to the distance of that point from the
plane of the base.
Ans. 2r/3 from base.
INTEGRATION)
704. Locate the centroid of a right circular cone by integration, using the origin of
the coordinate axes at the center of the
base, point a, Fig. 342, p. 190.
Ellipse
705. Locate the centroid of a hemispherical volume whose radius is r. Let the
origin of the axes be at the center of the
base and use the differential volume between
two planes parallel to the base and a differential distance apart.
Ans. 3r/8 from base.
Fig. 356.
Problems 708, 709.
710. A solid of revolution is generated by
the shaded area of Fig. 357 being revolved
about the y axis. Determine the coordinates of the centroid of this volume.
Ans. ¥ = 6.67.
Fig. 355.
Problems 706, 707.
706. Find Z for the paraboloid shown in
Fig. 355. The equation of the curve bounding the paraboloid in the xz plane is 2? =
4px. The radius of the base is b and the
height of the paraboloid is a.
Ans. © = 2a/3.
707. The same as 706 except that the
equation of the curve in the xz plane is
z2 = 8x and the plane forming the base
passes through the focus of the curve.
The
focus is at « = 2. Solve this problem by
integration.
708. A semi-ellipsoid is bounded in the
xy plane by the curve 2?/a? + y?/b? = 1
(Fig. 356). Determine % for this volume
of revolution.
Show details of integrations.
Ans. 3a/8.
709. The same as 708 except that the
bounding curve is «2/64 + y?/49 = 1. Solve
this problem by integration.
Fig. 357.
Problems 710, 711.
711. The same as 710 except that the
solid of revolution is formed by revolving
the shaded area about the w axis.
TANS. Gl =" 200g
712. A right circular cone has a diameter
of base of 20 in. and is 24 in. high. The
density p of any point varies directly
as its distance from a plane through the
apex and parallel to the base.
Find the
center of gravity.
Ans. 19.2 in. from vertex.
713. The projection of a segment of a
sphere ABC is shown in Fig. 358. Using
such lettered dimensions as are necessary,
derive an expression for 7 for this volume
in terms of the radius of the sphere r.
714. Determine the centroid of a frustum
PROBLEMS: SOLIDS (BY INTEGRATION)
200
‘IB
of a right cone whose bases are 20 in. and
10 in. in diameter and whose altitude 1s
10 in. Integrate for the result.
715. The length of a hypothetical rod of
constant cross section A is designated by h.
The density p of this rod varies from zero
at one end O to kh at the other end according
a
=
in
to the law p = kx, where x is measured
the direction of the length from O. Locate
the center of gravity.
Ans. @ = 2/8.
716. A
20-in.
COMPOSITE
LINES
wire,
as
considered
[Ch. X
ai
E
Fig. 358.
Problem 713.
Fig. 360.
Problem 719.
z
a
weighted line, is bent to form the radii and
the are of a circular sector whose radius is 5
in. Letting one axis be a line of symmetry,
locate the center of gravity of the bent wire.
cate the vector representing the weight of the
truss. Ans. With origin at D: 7.22, 2.12 ft.
Fig. 359.
717. A wire,
Problems 717, 718.
considered
as a weighted
line, is to be bent to form a 10-in. semicircle
joined to the straight part OA shown in
Fig. 359. It is desired that this bent wire
balance about the x and y axes at O.
Where should be the centroid C of the
section OA, and what should be the length
of OA?
Ans. QAb—sllorGonenGn—no2e0 a.
718. The same as 717 except that the
straight part OA is twice as heavy per unit
length as the semicircular portion.
719. All the members of the truss shown
in Fig. 360, are the same size and weigh the
same per foot of length. As an engineering
approximation for some purposes, the members of this truss are small enough in relation
to the space occupied by the truss that they
may be considered as weighted lines. Lo-
COMPOSITE
Fig. 361.
Problem 720.
720. If the members of the truss shown
in Fig. 361 are small enough in cross section
to be considered as weighted lines, and if
all members have a uniform weight per
foot, find the center of gravity of the truss.
The external loads are not to be included.
AREAS
721. Determine % and 7 for the area
Ans. 2.2, 6.2 in.
shown in Fig. 362.
722. Determine % for the segment of the
circle ABC shown in Fig. 363.
723. Determine & for the shaded area
shown in Fig. 364.
Ans. 11.4 in.
724. Determine % for the shaded area
shown in Fig. 365.
725.
Find the centroid of the area shown
in Fig. 366.
726. A built-up beam consists of two
3x5x9/16-in. angle irons and a flat plate of
t = 3/4 in. and b = 12 in. section, Fig. 367.
A handbook gives the location of the centroids Ci and C2 of the angles as shown, and
also the area of each angle as 4.18 sq. in.
PROBLEMS:
COMPOSITE
201
AREAS
A,=4.18 in?
Each
Fig. 367.
Problems 726, 727.
In computing the strength of this beam, we
need to know the location of the centroid
of the built-up section.
Find its location.
Ans. 1.41 in. from top.
727. The same
10 in.
as 726 except that b =
Web Plates
git
3!
4
4
ie)
Ne
=
pelle
Fig. 368.
Fig. 365.
Problem 724.
nie
|
i 8”
Problem 728,
728. The section of a chord of a steel
bridge truss is built up as shown in Fig.
368. The location of the centroid of this
section is needed in computing the strength
of the chord.
Locate the centroid.
The
centroid of the four angles is 12.81 in. below
the top of the cover plate.
Ans. 9.91 in. from top surface of cover plate.
PROBLEMS: COMPOSITE VOLUMES AND soLips [Ch. X
202
COMPOSITE
729. Figure 369 shows the section of a
The six steel
reinforced concrete beam.
rods are each 1 in. in diameter and are
arranged as shown.
How far from the top
surface is the centroid of the sectional ares
of the rods?
Fig. 369.
Problems
729, 730.
730. In Fig. 369, the steel rods weigh
490 lb. per cu. ft. and the concrete weighs
150 lb. per cu. ft. For a unit depth (perpendicular to the paper) and a 1-ft. width, locate
the center of gravity of this beam from the
top surface.
Ans. 2.03 ft.
731. A piece of homogeneous material is
turned on a lathe to the shape shown in
Fig. 370.
Find the distance from the small
end to the center of gravity.
732. A parallelepiped, with a square sec-
tion 12 in. on a side, is 8 ft. long. From
one of its ends and along its geometric axis
is bored a cylindrical hole 6 in. in diameter
Fig. 370.
GENERATED
AND
VOLUMES
and 6 ft. long.
How far is the center of
gravity of the remaining solid from the
unbored end?
Ans. 3.82 ft.
733. A homogeneous right circular cylinder of radius r and height h has a symmetric conical hole cut in its top end.
If
the cone has radius r/2 and height //2,
find the center of gravity of the remaining
solid with respect to its base.
Ans. 0.484h from base.
734. The same as 733 except that the
geometric altitude of the cone is 1.5, resulting in an opening in the lower base of the
eylinder.
735. A right circular cone with a base 40
in. in diameter and an altitude of 60 in. is
divided into a cone and a frustum of a cone
by passing a plane parallel to the base and
20 in. below the apex.
Determine the distance of the center of gravity of the frustum
of the cone from the larger base.
Ans. 13.83 in.
736. From a solid hemisphere, with a
radius of 21 in., is cut a right cone whose
base has a radius of 12 in. and whose
altitude is 1S in. The plane of the base of
the cone coincides with the plane of the base
of the hemisphere, and the axis of the cone
falls on the axis of the hemisphere.
How
far is the center of gravity of the remaining
solid from the base?
Problem
AREAS
737. The surface of a right cone is generated by the hypotenuse of a right triangle
when the triangle is turned about one of
its bases as an axis. Using the theorems
of § 126, determine the surface area and
volume of a cone in terms of the altitude
and the radius of the base.
738. Using the principles of § 126, find
the surface area of a hemisphere.
Let the
generating line be a 90° are of a circle.
Also find the volume of a hemisphere, using
a quadrant as the generating surface.
Ans, 2ar?, 2rr3/3,
SOLIDS
731.
AND
SOLIDS
739. The section of the plastic steering
wheel of an automobile is approximately
round and 7/8 in. in diameter.
The mean
diameter of the steering wheel is 1716 in.
Determine the surface area of the wheel and
the volume of plastic material needed, neg-
lecting consideration of the spokes.
i
Ans. 151 sq. in., 33 cu. in.
740. A l-in. diameter circular area is
revolved about an axis 6 in. from and parallel
to the vertical diameter of the circle, Find
the surface area and the volume of the
torus generated.
SOLIDS
203
741. The area A, of Fig. 352 is revolved
360° about the y axis to generate a homogeneous solid. Find the center of gravity
of that solid. See problem 695.
Ans, & =. aj = Gh my
742. The same as 741 except that area
A, is to be used. See problem 696.
as shown in Fig. 372. If cast iron weighs
450 lb. per cu. ft., what is the weight of the
rim?
PROBLEMS:
GENERATED
AREAS
AND
Fig. 373.
745. (a) Find by integration Z and 7 for
the shaded area shown in Fig. 373.
(b)
Section
A-A
Find the volume generated if this area is
rotated 360° about the z axis.
Ans. (a) 6, 1.2 in., (b) 80.5 cu. in.
(b)
(a)
Fig. 371.
Problems 745, 746.
Problem 743.
746. The
same
as 745 except that the
shaded area is rotated 270° about the y axis.
743. In designing the rim of a flywheel,
we may need the linear speed of the centroid
of the section of the rim. The rim of the
30-in. flywheel in Fig. 371 is sectioned in
detail in Fig. 371(b).
How
¢10,4v3)
far is the cen-
troid of a rim section from the axis of
rotation?
Find the weight of the rim if
cast iron weighs 450 lb. per cu. ft.
Ans. 13.9 in., 113.5 lb.
al
ae
el
Fig. 374.
Problem 744,
744. The rim of a cast-iron flywheel has
an outside diameter of 48 in. and a section
CENTER
747.
747. The shaded area in Fig. 374 is
bounded by the two branches of the hyperbola 22/25 — y?/16 = 1, the x axis, and the
48"D
Fig. 372.
Problem
OF
line y = 4\/3.
Find (a) J, and (b) the
volume generated by revolving this area
through 360° about the z axis.
Ans. (a) 3.91 in., (b) 2350 cu. in.
PRESSURE
748. Figure 161, p. 86, shows five forces
acting on a Fink truss.
Find the center of
pressure of these forces.
749. A concrete wall, 6 ft. thick and 18
ft. high, is acted upon by water (Fig. 375)
which tends to overturn the wall. The
weight of the wall tends to prevent overturning.
The concrete weighs 150 lb. per
cu. ft. and the water weighs 62.5 lb. per cu.
ft. Neglecting the effects of the end restraints, find the ratio of the moment resist-
ing turning to the applied turning moment
(of the water).
For convenience,
consider
Fig. 375.
Problem 749.
PROBLEMS: CENTER OF PRESSURE [Ch. X
204
a unit length of the wall perpendicular to
the paper.
Ans. 1.14.
750. A concrete dam in the form of a
rectangular parallelepiped is 15 ft. high and
6 ft. thick.
At the ends of the dam, the
forces tending to hold the dam upright are
to be considered negligible.
How high is
the water behind the dam when the moment
of the weight of the dam resisting overturning is three times the moment of the
water pressure tending to overturn it? Let
the weights of concrete and water be 150
and 62.5 lb. per cu. ft., respectively.
Consider a unit length of thedam.
Ans. 10.9 ft.
Fig. 377.
Problem 752.
AB represents an intensity of load of 100
lb. per ft. of length of AB, compute Az, Ay,
and CD.
Ans. 222.5 lb., 814 Ib., 445 lb.
Al4-13
ra
teal
Flow
.
oS
N
Fig. 376.
Problem 751.
751. A horizontal member AC, Fig. 376,
has on it a load distributed in such a manner
that the plan of distribution is closely
respresented by the ordinate of the triangle
ABC.
The scale of the ordinate is: 1 ft.
represents an intensity of load of 300 lb.
per ft. of length of AC. Determine the
FL eas
Fig. 378.
Problem 753.
reactions A,, A,, and CD.
752. A horizontal member AC, Fig. 377,
is subjected to a load which is distributed so
that the zntensity of load at a particular
753. The section of a concrete dam is
shown in Fig. 378. The water is considered
to be always at the top, AB.
What is the
value of x when the moment of the weight
of the dam resisting overturning is twice the
moment of the water pressure which tends
to overturn the dam?
Let the weights of
point is proportional to the ordinate of the
parabola OHB, whose equation is 2? =
—12y for the x and y axes shown.
If 1 ft.
of ordinate to the curve from the member
cu. ft., respectively, and neglect the effect of
the restraints at the ends of the dam.
Ans. 0.88 ft., approx.
concrete and water be 150 and 62.5 lb. per
GENERAL
754. The shaded area of Fig. 379 is
revolved 360° about the y axis. What is
the surface area and the volume generated?
Use the principles of § 126.
cal when it is on the point of toppling over?
755. The same as 754 except that the
shaded area is revolved through 136°.
Ans. 872 in.?, 3222 in.3,
:SS
756. The section of a 200-ft. smokestack
is shown in Fig. 380.
(a) Considered as a
solid of revolution, what volume of concrete
does the stack require?
(b) Where is the
eg of the stack?
(c) If the stack should
lean over due to an undermined foundation,
what is its angle of inclination with the verti-
| "”
”"6
10
Fig. 379.
Problems 754, 755.
PROBLEMS:
205
GENERAL
758. The
same
as problem
757 except
that the sphere is removed.
y
Fig. 382.
ores
x
Problem 759.
759. The frustum of a right circular cylinder is shown in Fig. 382. The base is
perpendicular to the geometric axis. The
plane of the top makes an angle @ with the
geometric axis.
(a) Determine a general
expression for the volume of the frustum.
(D>) te O SOO, Pp SA ith, ahavel i SO), inal
the centroid of the frustum.
HinT: A convenient element to use is dV = 2yz dz,
where the projection of the top plane forms
the line z = A + (r + 2)/tan 0.
Ans.
(a) mr2(h + r/tan @), (b)
sl
a
Problem 756.
Fig. 383.
Problem 760.
760. A 4 x 6-ft. cylindrical
383,
Fig. 381.
757.
A uniform
Problems
slender
757, 758.
bar
BCD,
Fig.
381, weighs 25 Ib. Welded to its end is a
sphere weighing 50 lb. Consider the wall
smooth and find @.
Ans, 44.2°,
has
f=
3 it. of water
drum,
Fig.
im it,
Whe
vessel is gradually tipped to an angle of
B = 30°, Fig. 383(b).
(a) Will any of the
water spill?
(b) If it is released in this
position, will it turn over or return to the
original position (neglect the weight of the
vessel)? See problem 759.
761-770. These numbers
for other problems.
may
be used
Chapter XI
MOMENTS
OF
INERTIA
OF
AREAS
130. Introduction.
Since only mass has inertia, the term moment of inertia
applied to an area is a misnomer.
However, the mathematical form of the
expression for the moment of inertia of an area is identical, except for symbols,
with the mathematical form of the expression for the moment of inertia of a
mass.
These analogous forms have led to the common usage of the term
moment of inertia applied to an area.
Nevertheless, we should consider the
moment of inertia of an area to be merely a mathematical expression.
This
mathematical expression is important because it appears in the theoretical
equations for the strengths of beams, shafts, and columns.
Hence, in the
general work of designing structures and machines, the engineer must be
ready to find the moment of inertia of an area, sometimes called the second
moment of an area.
131. Moment
of Inertia.
The definition of moment
of inertia is mathe-
matical:
(28)
Moment of inertia of an area = /a’ dA,
where a is the distance of an element dA of an area from an axis about
which
the moment of inertia is desired.
Generally,
The axis is in the plane of the area, in which case it is called a rectangular
moment of inertia I, or
The axis is perpendicular to the area, in which case it is called a polar
moment of inertia ./.
The limits of the integral in (28) are of course such as to include
the entire
area whose moment of inertia is needed.
If we remember that in finding centroids we used the form Ja
dA, which is
sometimes termed the first moment of the area, we may
call fa?dA the
second moment of the area.
Observation of the similarities of the procedures
of this chapter with those of the preceding chapter will
be helpful.
The
first important difference that might be noted is that moment
of inertia is
always positive, involving as it does a’, which is positive
whether a is positive
or negative.
Observe too that if the linear unit used is inches, the moment
206
§ 134] RADIUS OF GYRATION
207
of inertia has the units inches to the fourth power (in.‘). If the linear unit
is feet, the moment of inertia is in feet to the fourth power (ft.*‘).
132. Rectangular Moment of Inertia.
Consider the area A, Fig. 384,
lying in the zy plane. The differential element dA is at a distance y from
the x axis. Hence, the moment of inertia of
the area about the wz axis is
(a)
(ers /yd;
and, similarly, that about the y axis is
(b)
is -/e dA,
Fig. 384.
These are rectangular moments of inertia.
133. Polar Moment of Inertia.
Since the element dA, in Fig. 384, is a
distance r from the z axis and since the z axis is normal to the plane of the
area, the polar moment of inertia about this particular axis is
(c)
J - [7dA,
Now note in Fig. 384 that r? = 7? + y®.
if -/r dA
Using this relation in (c), we get
foe + y*)dA - [2 dA +f dA,
or
(d)
Joc
Ro
This expression (d) is an important relation which is often useful.
In words,
it says that the polar moment of inertia of a plane area about a particular
axis perpendicular to the plane of the area is equal to the sum of two rectangular moments of inertia about axes in the plane of the area, provided that all
three axes are mutually perpendicular and intersect at a common point.
Since
there are an infinite number of axes about which moments may be taken,
there are an infinite number of moments of inertia for a particular area.
134. Radius of Gyration.
The radius of gyration k of an area is also a
mathematical conception, defined by the relation
(29)
a=
4)
A
or
k=
(4)
A
a rectangular and a polar radius of gyration, respectively.
T=
7A.
or
Since, from (29),
J == h2A.
we may say that if the entire area were concentrated at a point whose distance
from the axis (or pole) is equal to the radius of gyration k, the moment of
OF INERTIA OF AREAS [Ch. XI
MOMENTS
208
original area.
inertia of the concentrated area would be equal to that of the
metal plate
This idea becomes more real if we conceive of the area as a thin
the axis
from
distance
a
at
which can be melted down into a single particle
equal to the radius of gyration.
In computing the strength of columns, we find it necessary to use a radius
of gyration, usually the minimum for the area of a section of the column.
Since there is an infinite number of axes about which moments of inertia
may be taken, it follows that there is an infinite number of radii of gyration.
Therefore, we should begin at onceto think of both conceptions in relation
to a particular axis; for example, k, refers to the radius of gyration with
respect to the z axis and would be obtained from
Derive the expression for the moment of inertia of a
135. Example—Rectangle.
rectangular area about an axis through the centroid and parallel to a side. Also
find the radius of gyration with respect to this axis and the polar moment of inertia
with respect to an axis through the centroid.
Sotution.
In the rectangle shown in Fig. 385, choose
a differential element of area dA such that all points in
the element are the same distance from the x axis. Since
all points of this element, whose area is 5 dy, have the
same moment arm y about the x axis, we get
°
+h/2
rh/2
bys
I
POW
{
ip OW) = 2 | y? dy = ——
Dae Sail oe e
TTL LTT LT TTT RTT
J
—h/2
/0
12
We may find the value of J, in the same way by integration, using the element of area of width dz, Fig. 385;
or, by analogy with the result for 7,, we may write
Fig. 385.
hbs
nae pS
on account of the symmetry of the figure.
the centroidal axis from equation (d) is
Jc-=1,
bhi + hbi _ hb
1, =
136. Example—Triangle.
a base.
‘Thus, the polar moment of inertia about
eae
12
aD
ip i
ON
ee
A
= 1b"):
eas Chi
[ind the moment of inertia of a triangular area about
Sotution.
As before, choose an element dA, all parts of which are the same distance
from the x axis (Fig. 386). The length of the element is x. — 2; = x;s0 thatdA = xdy.
By similar triangles,
eae
b
hie
or
=
Noh
—
eae
h
§ 137 | TRANSFER
FORMULA—PARALLEL
AXES
209
Using this value of x in the expression for dA, we get
dA
= ean =
bh — y)dy
h
d
and
ae
l= | pdd G2)
pola
fu
[ad y)dy
b| hy
ys |"
Al?
t|
anit bh nel
| Dare
Pipa
OG
The corresponding radius of gyration is
Fig. 386.
Fig. 387.
137. Transfer Formula—Parallel Axes. A simple and very useful relationship exists between the moment of inertia of an area about a given axis and
the moment of inertia of the same area about a centroidal axis parallel to the
given axis. Let C in Fig. 387 be the centroid of the area A, so that the z
and y axes are centroidal axes.
The x’ and y’ axes are any other axes in the
plane of the area and parallel to the chosen centroidal axes. The moment
of inertia of the differential area dA about the y’ axis, for example, is
dl, = (d — x)?dA, and, with the proper limits of integration, the moment
of inertia of the entire area A about the y’ axis
hey = |i — x)2dA
= [ead e 2a | adA+@ faa,
where the terms have been rearranged after squaring
(d — x).
In this ex-
pression, [x dA (which is equal to AZ) is zero, since the y axis passes through
the centroid; also, {dA = A. Moreover, by comparison with Fig. 387, we
note that fx?dA is equal to the moment of inertia of the area about the
centroidal y axis, I,. Hence, we find
Dopasel
yt Ad
or, in general terms,
(30)
I
I=14
Ad?,
MOMENTS
210
OF INERTIA OF AREAS [Ch. XI
the parallelwhich is the transfer formula for parallel axes, sometimes called
inertia I of
of
moment
the
In words, equation (30) says that
axes theorem.
a
an area about any axis is equal to the moment of inertia of the area about
of the
parallel centroidal axis I plus the product of the area A and the square
that
arly
particul
distance between the axes d®. Note
One of the axes involved in equation (30) must be a centroidal axis.
The linear units are usually inches, although of course any other linear
units, such as feet, may be used.
By a similar procedure, we may show that the transfer formula for polar
moments of inertia takes an analogous form,
(e)
J =J + Ad’,
where J is with respect to a centroidal axis and J is with respect to some
other parallel axis at a distance d from the centroidal axis. By using the
defining equation (29) for the radius of gyration, we have J = k?A and
I =f?A.
Substituting these values into equation (30), we find
WA = kA + Ad?,
ke =
+ a,
(f)
where k and k are taken with respect to the same axes as J and J, respectively.
Equation (f) applies also to radii of gyration obtained from parallel polar
axes.
Observe that the symbols with the bar, J, J, and k, represent values
with respect to a centroidal axis.
138. Example—Circle.
(a) Find the moment
an axis normal to the area and tangent
moment of inertia about a ciametral axis.
Fig. 388.
SoLution.
of inertia of a circular area about
to the circumference.
(b) Also
find the
Fig. 389,
(a) The axis 2’, Fig. 388, normal to the paper and parallel to the cen-
troidal axis z through point C, is such an axis as that described in the statement of the
problem.
The simplest method of solution is to find J and then apply the transfer
§ 140 |CHOICE OF DIFFERENTIAL ELEMENT
OTT
formula. Thus choose a differential area dA = 2 xp dp, all parts of which are at the
same distance from the z axis, Fig. 388. The moment arm of this area is p; therefore,
D/2
=
(g)
J=Jo=
[rad =2f
D4
1
i
AT?
ee
Using the transfer formula, we get
J = Ja
= J 25 Ad
i
aD?
dD?
4
4
aD! ot
32
(b) For a circle,
J, = Z,.
=
37D!
aa
Sart
Sy
Then since
it follows that
139. Example—Triangle.
Using the results of the example of § 136, find the
moment of inertia of a triangular area about a line through a vertex and parallel to
the opposite base.
Sotution.
This problem may be solved by the use of the transfer formula, but
we cannot transfer directly from the x axis to the x’ axis, Fig. 389, since neither axis is
a centroidal axis. However, we can first find J from J, and then find J,, from J.
We learned in § 117 that the centroid of a triangle is one-third of the altitude from
the base; so the axis through C is a distance h/3 from the x axis. From § 136, we get
I, = bh3/12.
From
the transfer formula, we find
:
bh3
=
ea
reel
2
=
bh
fh
12
ee
bh\f
4h?
bh
=
36’
whence
2
I,
,
pote
bh3
/2 —
m+
(S\)
bh3
—
4
.
140. Choice of Differential Element.
In each of the preceding examples
in which integration was involved, we chose a differential area whose every
part was at the same distance from the axis about which the moment of
inertia was desired.
In accordance with this same principle, we could have
used a differential area dA = dx dy, or, in polar coordinates, dA = p dé dp,
in which event a double integration would have been necessary.
Such a
choice is direct and satisfactory, but it is not always the most convenient.
A more convenient choice may be a differential element of area whose
parts are not all at the same distance from the axis in question, provided we
know the moment of inertia of the element about the axis with respect to
which the moment of inertia of the area is desired.
The use of these elements
is best shown by an example.
MOMENTS OF INERTIA OF AREAS [Ch. XI
212
Find the moment of inertia I, of a circular area about
141. Example—Circle.
a diameter when the differential element is (a) dx dy (Fig. 390), and (b) 2y dx (Fig.
391). (c) Find the corresponding radius of gyration.
If we integrate first
(a) For dA = dx dy, we have I, = ffy?dxdy.
So.tution.
with respect to 2, the limits are from —zx to +a. This integration in effect sums up
the second moments of the area dx dy along the horizontal dotted strip in Fig. 390.
Thus
cia
(h)
iy
+2
T
zt
/ / y? dx. dy = afi y? dx dy.
=
—&
0
0
In the foregoing expression, the integral from a minus limit to a positive limit of the
same value is equal to twice the integral from zero to the positive limit. This condition applies when the exponent of x in the following expression, for example, is zero
or an even whole number (do not use this idea blindly);
+a
a
/ ede
= af AOR
=
U
The limit z in (h) is variable, being from one side of the curve boundary to the other
side (Fig. 390). It is thus obtained from the equation of the circle x? + y? = 7? as
ah ==
(7? =
y? yt,
Fig. 390.
Fig. 391.
If one of the limits in a double integration is variable, the integration for the variable
limit must be made first. If the limits of a double integration are both constant,
it makes no difference which integration is made first. In this instance, we must
integrate (h) first with respect to a.
tema
fata a]
a= (r—y)W?
:
0
0
mafne —aay
r
0
The differential area in this integral, 2(r? — y?)"? dy, is represented by the horizontal
dotted strip in Fig. 390. The second integration may be made by substituting
y =r
sin 8, or by direct use of a table of integrals.
We find
2
1,=4]
zs
yO? —
|Z
4
y)2 dy = AIS An
Sinaia
Compare this result with that in § 138(b).
Dt
Ar
page
Sanh
§ 143 |EXAMPLE
PQS:
(b) The differential area shown in Fig. 391 is rectangular in shape. In the example of § 135, we found the moment of inertia of a rectangular area about its centroidal axis to be bh3/12. Applying this knowledge to the moment of inertia of the
rectangular strip GH in Fig. 391, we let b = dx and h = 2y, and we find*
Then using y? = (r? — z?)3/ from the equation of the circle, and letting x = r sin 0
(or by direct use of a table of integrals), we get
8
les a=, ORK
a
oe
(ae
3 da ee =f
ANP ke a
(CEOe, AVG
are
A
A
nD:
64?
which of course is the same as the result in part (a).
(c) The radius of gyration about a diametral axis is
:
hays a art :
2 pian = 5
(qo (rr?)
thes otk (4)
142. Composite Areas.
When a composite area ($121) can be divided
into a group of simple areas, such as rectangles, triangles, and circles, the
moment of inertia of the composite area about a particular axis is the sum
of the moments of inertia of the simple areas, each about this same axis. If
a composite area has a “hole’’ in it, that is, if there is a blank surface within
the boundaries of the composite area, the moment of inertia of the net composite area 7s equal to the moment of inertia of the gross area including the
“hole” minus the moment of inertia of the “hole,’’ each moment
being taken about the same axis.
of inertia
143. Example.
In designing a beam, the moment of inertia of a section of the
Find the moment of inertia of the T-section
beam about a centroidal axis is needed.
shown in Fig. 392 about a centroidal axis parallel to
MN.
Sotution.
The first step is to locate the centroid of the area.
Let the T-section
be divided into two rectangular areas A’; and A’, whose centroids are at C; and C2,
respectively.
Taking moments of these areas about the line MN we get
Ay =
Aly’: + A’ oy’ »,
169 = (10)(0.5) + (6)(4),
from which 7 = 29/16 = 1.81 in. This dimension locates the centroid C of the
composite area and determines the distances C\C = 1.31 in. and@,C = 2.19in. Now
use the transfer formula and the expression for the moment of inertia of a rectangular
area about a centroidal axis from § 135, ie., bh?/12, for each area A’; and A’s, The
moment of inertia of the area A’; about the centroidal axis Cx is
=
1
Top = Tone + AGP? = oe
‘
+ (10)(1.31)? = 17.99 in.4
*We would have found this same answer in part (a) above if we had integrated first
with respect to y.
MOMENTS
214
OF INERTIA OF AREAS [Ch. XI
The moment of inertia of the area A’» about the centroidal axis Cz is
Teo = [s+
(16)
A’od? =
12
+ (6)(2.19)? = 46.78 in.4
Thus the moment of inertia of the composite area about the centroidal z axis is
I, = Int + Ing = 17.99 + 46.78 = 64.77 in.4
ALTERNATE Sotution.
As a check and as an alternate solution, we may find the
moment of inertia of the rectangle
MNPQ and subtract from this the moment of
inertia of the areas outside of the T-section but within MNPQ.
Call the area 11 N PQ
area As; then A = Az —(A; + Az). The centroid of A; is at O and the distance
OC is seen to be 2.19 — 0.5 = 1.69in. Therefore, for area MNPQ, we find its moment
of inertia about the centroidal axis Cz as
Iz, = 1, + AOC)=
oe
+ (10)(7)(1.69)?= 485.76 in.4
The moment of inertia J’, of the two blank areas on each side of the stem of the T
with respect to the x axis is
[=f
(9)a
(6) + (9) (6) (2.19)?= 420.99 in.4
AG
The value of /, for the T-section is ae
I, =
In3 —
I’, = 485.76 —
420.99
=
64.77 in.!,
which checks with the previous solution.
Fig. 392.
144, Example.
A semicircular area is removed from a triangular area as shown
in Fig. 393. Find the moment of inertia of the net composite area about the z axis.
Sotution.
Using the result of § 136, we find the moment of inertia of the triangular
area | about the x axis, Fig. 393, to be
In =
bhs
Ti
(30) (20)3
IMGpUET
;
:
20,000 in.‘
We have also found the moment of inertia of a circle about a diameter to be a D*/64
(§ 141). Thus, for a semicircle,
This gives
i
area 2, Iz. = D*/128,
Teme L2)s
LIS eas
2S
or half that for a circle.
= 509 in.4
The moment of inertia of the net composite area is then
I, = Ie: — Ing = 20,000 — 509 = 19,491 in.!
§ 147 |TRANSFER FORMULA FOR PRODUCTS
OF INERTIA—PARALLEL AXES
216
145. Graphical Determination of the Moment of Inertia.
One simple
and direct method of approximating the moment of inertia of an area when
the boundary curve is irregular and cannot be expressed in equation form is
suggested in Fig. 394. Divide the area into narrow strips parallel to the
axis about which the moment of inertia is desired—the narrower the strips,
the more accurate the results.
Compute the
areas Ai, Ao, As3, etc., by using the average
Estimate the location
height times the width.
of the centroid of each area (approximately half
way across a strip bounded by two straight
lines), and measure the distances 21, 2, 23, etc.,
Then the moment of inertia of the
Fig. 394.
irregular area about the y axis will approximate
(i)
i
=
Ayr?
+
Asx,”
+
As23?
+
-- ,
where the sum includes all parts of the total area.
Fig. 394.
146. Product of Inertia.
Another mathematical expression that appears
in certain theoretical analyses is frydA.
This integral too has been given a
name—the product of inertia P,,.
(j)
jo ie |ee
The method of procedure in finding the product of inertia with respect to
any two axes is similar to that explained for finding moments of inertia.
However, unlike moments of inertia, the product of inertia may be positive
or negative, or it may be zero.
It is zero when one, or both, of the axes is
an axis of symmetry; for, in this event, each element of area with a moment
arm of +2 will be balanced by a like element with a moment arm of —a, if
the y axis, for example, is an axis of symmetry.
(The principle is the same
as in taking first moments of areas in finding centroids.)
The product of
inertia is also zero with respect to any principal axes ($150).
The usual
unit for product of inertia is inches to the fourth power (in.*), though of
course other linear units may be used.
147. Transfer Formula for Products of Inertia—Parallel Axes.
The differential element dA = dx dy of the area A, Fig. 395, is distance d, — x from
the y’ axis and d, — y from the x’ axis. Thus, the product of inertia of the
area A about the x’ and y’ axes is
poe |ee
eal
= dud, [aA -d,fxaa — defyaa + fevaa.
MOMENTS
216
OF INERTIA OF AREAS [Ch. XI
Since the x and y axes are centroidal axes, the first moment of the area about
Therefore, the second and third integrals in the foregoing
either axis is zero.
expression are zero, fedA = 0 and JydA = 0. Moreover, fdA =A, and
we recognize fry Dale P) as the product of inertia about the centroidal
We have then
axes parallel to the v’ and y’ axes.
(31)
Pirpo=eP
Ada,
which states that the product of inertia P,,, about any two rectangular axes
Q'-x’ and O’-y’ is equal to the product of inertia P about parallel centroidal
axes plus the product of the area A, the distance d, from the centroidal x
axis to the x’ axis, and the distance d, from the centroidal y axis to the y’ axis.
Fig. 395.
148. Example.
Jind the product of inertia of an 8 x 5 x 1-in. Z-section about the
x and y axes shown in Fig. 396.
Sotution.
Divide the Z-section into three rectangular areas, Ai, A2, and A3 as
shown.
Using a differential area dx dy in the area Aj, we have
4.5
feos =| ibfe OLE ld) =
Bl Bi
“
=
+35 in.4,
J0.5
where the limits of integration for x are from 0.5 to 4.5, and for y, from 3 to 4 (Fig.
396). For the area A», we find
—3
(-0.5
Np -| mi
—4
42
RUG
—4.5
1737
og 7-0-5
— E| E|
a
=A
“
= (—3.5)(—10)
= + 35 in4,
= Ale
where we observe that the limits are taken from left to right and from lower border to
upper border, the positive direction in each case.
For the area A3,
ed f +0.5
Pave -| / nyde dy = 0}
ah Ifis
a result which is immediately obvious if we note that the « and y axes are axes of
symmetry.
Hence, the product of inertia of the Z-section with respect to ay is
Poy = Poy tt ayn
Pays = Oo + 3o = (0 ins
§ 149] MOMENT
OF INERTIA ABOUT INCLINED AXES
ailiy
In making a sum of the products of inertia for a composite area, use an algebraic
sum; that is, if a value of P,, is negative, this value should be subtracted from the
sum of the positive values.
ALTERNATE SotutTion.
The transfer formula may be used conveniently in this
problem, because the product of inertia of each of the rectangular areas A, and As is
zero with respect to centroidal axes of these areas. The coordinates of the centroid
of area A; with respect to the x and y axes are d, = 2.5in. and d, = 3.5in. Thus, for
area A,, we see that
Pry: = Py + (Aded,)s = 0 + (4)(2.5)(3.5) = 35 in.!
Since the coordinates of the centroid of area Az are d, = —2.5 and d, = —3.5, the
product of inertia of this area is
Pry2 = P+ (Adzdy)s = 0 + (4)(—2.5)(—3.5) = 35 in.!
These results are of course the same as those found above.
149. Moment of Inertia about Inclined Axes.
Suppose the moment of
inertia of an area about some axis O-v’ (Fig. 397), which makes an angle 6
with the x axis, is desired when the values
of J, and J, are known.
(This situation
may arise, for example, in connection
with the Z-section of Fig. 396 where the
moment of inertia about some xv’ axis may
be needed.)
Sometimes the values of J
about certain axes will be relatively
easy to find, whereas about some inclined axis, a direct integration may be
difficult. In this event, a transfer formula
for intersecting axes will be convenient.
Fig. 397.
The differential area dA, Fig. 397, has
coordinates (x,y) and (x’,y’), respectively, for the two pairs of axes as
shown.
From the geometry of the figure we see that y’ = y cos 6 — x sin 0.
Hence, the moment of inertia of the area A about the 2’ axis is
1 = [ura = |(y cos 0 — sin 0)? dA
cost ofytd
= 2sin dos 0 [rydd +sinto fata,
whence
(k)
I-=T,
cos* 0 — P,, sin 26 + £,sin*:é.
Thus, in order to find 7, about an inclined axis, J,, Z,, and P,, with respect
to some other coordinate axes must be known.
An expression for J, may be found in a similar manner, or by substituting 90° + @ for @ in the preceding expressions.
The value of J, in terms of
the angle 6 between the x and x’ axes is then found to be
(1)
Iy = I, sin? 6+ P,, sin 26 + I, cos? @.
MOMENTS OF INERTIA OF AREAS [ Ch. XI
218
(1) to the sum
Equating the sum of the left-hand sides of equations (R) and
of the right-hand sides, we get
lo
(m)
ly
=
Tale
of inertia
From this equation, we may conclude that the sum of the moments
the
about two rectangular axes is a constant for all rectangular axes with
same origin. Recalling the relation between rectangular and polar moments
of inertia (§ 133), we see that this constant is the polar moment of inertia,
because each side of (m) is equal to J,, Fig. 397.
150. Maximum and Minimum Moments of Inertia. Since we need the
minimum radius of gyration, for example, in the design of columns, we
should determine the conditions which define the corresponding axis. In
accordance with the calculus, the dI,-/d6 set equal to zero will yield the
Differentiating equation
value of 6 for which J, isa maximum or minimum.
(kk) with respect to 6, and equating the derivative to zero, we have
dls.
= — 27, sin 6 cos 6 — 2P,, cos 20 + 21, cos # sin 6 = 0
dé
=— 7, sin 26 — 2P,, cos 26 + I, sin 20 = 0,
from which
(n)
reane2Ge—
2P oy
From equation (n), we find two values of 6 differing by 90°, and thus, we
have two axes defined.
These axes are called the principal axes for the
origin in question. About one of these axes, the moment of inertia is a
maximum; about the other, a minimum. A second differentiation will tell
which axis gives the maximum /, in accordance with mathematical rules;
but the simplest way is to sketch the two axes on the figure and recall that
the closer the elements of the area are to the axis, the smaller the moment
of inertia; or, use one of the values of @ in equation (k), find J,,, and compare
with 7, + I, [see equation (m)].
Referring to equation (n), we see that if the product of inertia P,, is zero,
6=0 and @= 90°. That is, the x and y axes are the principal axes when
P,, = 0. Conversely, we may state that the product of inertia with respect
to the principal axes is zero.
Moreover, since the product of inertia is zero
when one axis is an axis of symmetry, an axis of symmetry is a principal
axis; in other words, if an area has an axis of symmetry, we may conclude
that the moment of inertia about this axis is either a maximum or a minimum
for all axes that pass through a particular origin. The minimum centroidal
moment of inertia is the smallest of all moments of inertia; the maximum
centroidal moment of inertia is exceeded by that about any other parallel
axis and by the moment of inertia about an infinite number of other axes at
some distance from the centroid.
§ 151 | EXAMPLE
219
Notice that P = 0 only when the centroidal axes of reference are principal
axes (which may or may not be axes of symmetry).
151. Example.
example of § 148.
Find the minimum
radius of gyration of the Z-section of the
SoLution.
Figure 398 shows this Z-section with certain appropriate dimensions.
We shall first find 7, and J, for the section.
(For a rectangle, J = bh3/12).
Notice
the use of the transfer formula.
(1)a
Area Ai:
I, = ——
+ (4)(1)(2.5)? = 30.33 in.4
Area Ag:
i
nu + (4)(1)(2.5)?= 30.33 in.4
Area A3:
log =
oi = (0.677in4
Therefore
I, = 30.33 + 30.33 + 0.67 = 61.33 in.!
Area Ay:
In. =
Area Ao:
hon = Oley
f + (4)(1)(3.5)?= 49.33 in.4
Area As:
I,3 =
Therefore
I, = 49.33 + 49.33 + 42.67 = 141.33 in.4
1)(4
3
4)(1
( ' ° + (4)(1)(8.5)?
= 49.33 in.4
4)(1
1
= 42.67 in.4
Using P,, =+-70 in.‘ from the example of § 148 and the equation (mn), we find
DD Poe
eer
(2)(70)
re ris
14133
hs
from which 26 = —60.25° or — 240.25°.
Thus, 6 = —30.12° or —120.12°, in which the
—30.12° is measured of course clockwise, Fig.
398 (angle @ was taken as positive when
measured counterclockwise, Fig. 397). The
corresponding positive answers for @ are
+59.88° and +149.88°.
For the axes as
named in Fig. 398, an angle of @ = 59.88°
(By
used in equation (k) will give I,,.
examination
of Fig.
398,
we
can
see
Fig. 398.
that
I,, < 1,,, since the area is crowded much closer to 0-2’ than to O-y’.)
(o)
I,, = I, cos? @ — Pry sin 26+ I, sin?6
= (141.33) (0.252) — (+70)(0.868) + (61.33)(0.748) = 20.6 in.4
By comparing this answer for J,, = 20.6 in.‘ with 7, + I,, we are now sure that
this is the minimum value. Moreover, by finding J,, from equation (1), we may
check the calculations by seeing if J,, + I,, = I, + I,. Observe that if Pz, had
been negative, the negative sign would have been included in the above equation (0).
The minimum radius of gyration is
_\12
kmin = (43)
on
e\12
= (2°)
= 1.136 in.
PROBLEMS: BY DIRECT INTEGRATION
220
[Ch. XI
152. Closure. The reader is urged to study the examples thoughtfully,
as they illustrate many of the possible variations in determining moments of
Remember that the elementary area may be:
inertia of areas.
1. A dz dy or a p dé dp (Fig. 390),
2. A narrow strip of thickness dx or dy, all parts of the strip being at the
same distance from the axis of moments (Figs. 385, 386, and 388), or
3. A differential element of area whose moment of inertia about the axis
of moments is known, even though all parts of the element are not the
same distance from the axis (Fig. 391).
Since most shapes in engineering practice will be rectangles or circles or
combinations of these, it is good to know the values of J for centroidal axes
The parallel axis theorem is often a great time saver in
of these shapes.
Some useful values of moments of inertia include:
practice.
:
For circle,
74
[= a?
a
a D4
39)
f=
hb?
72?
3
For rectangle, J =
:
For triangle,
=
—
or
=)
SDA
J = 36°
where the symbols have the usual meanings.
Problems
BY
DIRECT
INTEGRATION
771. (a) Derive an expression for the
moment of inertia of a rectangular area
about an axis coinciding with a long side.
(b) What is the polar moment of inertia
about an axis through a corner of the
rectangle?
(c) What is the radius of gyration with
respect to this polar axis?
The rectangle is 8 x 10 in.
(d)
What are the
numerical values for (a), (b), and (ec)?
Ans.
(a) bh3/2, (b) bh(h? + b2)/3, (ec)
[(h? -- 6?) /3)}/2> (d) 1707 in.4, 4378 in.4
7.4 in.
772. For a triangle of base b and altitude
h, find (a) a general expression for the
moment of inertia with respect to a centroidal axis parallel to the base, (b) the
expression for the corresponding radius of
gyration, and (c) their numerical values if
b =3 in. and h=4 in.
(Avoid using a
right triangle, making the solution general.)
773. The same as 772 except that the
reference axis is through the vertex and
parallel to base b.
774. Find the moment of inertia of an
equilateral triangle, whose sides are 6 in.
long, with respect to (a) a side and (b) a
centroidal axis parallel to a side.
Ans. (a) 70.2 in.4, (b) 23.4 in.4
775. (a) Using polar coordinates (dA =
p dé dp), determine
the expressions
for the
moment of inertia and radius of gyration
of a semicircular area about its bounding
diameter.
(b) If the diameter of the circle
is 20 in., find their numerical values.
Ans. (a) rr4/8, r/2; (b) 3920 in.4,°5 in.
776. (a) Using polar coordinates (dA =
pdédp), determine the expressions for the
moment of inertia and radius of gyration
of a quadrant of a circular area about the
axis of symmetry of the quadrant. (b) If
the circle has a 4-in. radius, find their
numerical values.
Ans. (a).0.0712r4%, 0.3017; (b), 18.2: m4
1.204 in.
777. An annular area is bounded by two
circles of radii ro and r;, where ro is for the
PROBLEMS:
BY DIRECT
221
INTEGRATION
outside circle. Determine by integration
the centroidal polar moment of inertia and
the corresponding radius of gyration.
Use
dA
=
y=cos x
p dé dp.
Ans. (a/2) (rot —rs*), [(ro? + 1s?) /2]"/.
778. The same as 777 except that the
reference axis is to be a diameter.
Ans. (1/4) (704 —ri*), (02
+ 732) ¥/2/2,
a7
Fig. 401.
0
acs
5
Problems 783-785, 799, 800.
Fig. 401, between the limits of « = —7/2
in. and « = 7/2 in.
Ans. 0.444 in.4, 0.471 in.
784. The same as 783 except that the
upper limit is « = 1 in.
785. The same as 783 except that J, is
desired.
y’
Fig. 399.
Problems 779, 780, 795-798, 820.
779. If the coordinates
of P, Fig. 399,
are (3 in., 27 in.), find the moment of inertia
with respect to the y axis of (a) A1 and
(b) Ae.
Ans. (a) 121.5 in.*; (b) 121.5 in.4
780. If the coordinates of P, Fig. 399, are
(3 in., 9 in.), find the moment of inertia with
respect to the z axis of (a) A: and (b) A».
Fig. 402.
Problems 786, 787, 818, 819.
786. For the shaded area in Fig. 402, by
direct integration find (a) Jz, (b) I,, and
(c) I,’.. The linear unit is the inch.
Ans. (a) 68.2 in.4, (b) 146.1 1n.4, (c) 77.7 in.
787. The
same
as 786 except that the
equation of the curve is y? = 9x.
Ans. (a) 230.3 in.4; (b) 219.5 in.4; (c) 117 in.4
Jy
Fig. 400.
a
Problems 781, 782.
781. Find J, for the shaded area in Fig.
400 between the curve y? = bx and the line
x
y = az, (a) using the element dd shown
and (b) using a similar element dA that is
vertical.
I, for
(c) Find the numerical
value of
O
Fig. 403.
x=4—|
Problems 788, 789, 821.
a = 1 and b = 4 when @ and y are
in inches.
Ans. (a) b4/(20a5), (c) 12.8 in.4
782. The same as 781 except that J, is to
be found.
783. Find the moment of inertia and the
radius of gyration with respect to the z
axis for the area under the cosine curve,
788. For the shaded area in Fig. 408, by
direct integration find (a) J,, (b) Z,, and
(@) fii
Ans. (a) 12:2 in.4; (b) 51.2 in.4; (c) 8.53 in.4
789. The same as 788 except
equation of the curve is 2? = Qy.
that the
PROBLEMS: TRANSFER METHOD
222
TRANSFER
[Ch. XI
METHOD
790. For the semicircle of radius r, Fig.
404, find the moment of inertia and radius
of gyration with respect to (a) the centroidal
axis, %, (b) the x’ axis, and (c) the y axis.
Use I for a semicircle with respect to a
diameter as a starting point. See problem
775 and § 138.
(d) Also find J..
Ans. (a) 0.1092r4, (b) 0.629r4, (c) mr4/8,
(d) mr4/4.
mine
4.4
Problems
Problems 793, 794.
the
moment
polar
of inertia
with
respect to an axis through B.
794. For the annular area of Fig. 406,
derive the expression for J,’. By using the
result of problem 778 and the transfer
formula, no integration is needed.
O
Fig. 404.
Fig. 406.
790, 791.
Ans. 1(5ro4/4 — ro2ri? — ri4/4).
791. The same as 799 except that only
the right quadrant A is to be used.
Fig. 405.
792. (a) Determine an expression for the
polar moment of inertia for the quadrant
(6 = 90°) of a circular area about the axis
The centroid
of this quadrant is at Z = 0.601r.
(b) For
a 5-in. radius, find this polar moment of
inertia and the corresponding radius of
gyration.
Using information in the text,
no integration is necessary.
of P, Fig. 399,
are (3 in., 27 in.), find the moment of inertia
for area A, with respect to (a) the y axis,
(b) the centroidal y axis, and (c) the line
xz =4.
The centroid of A; is at (1.2 in.,
15.48 in.).
Ans. (a) 121.5 in.4, (b) 34 in.4, (c) 510 in.4
798. The same as 795 except that the
area A» is to be used.
The centroid of A»
797. If the coordinates of P, Fig. 399,
are (3 in., 27 in.), find the radius of gyration
of area
Az with
(b) the centroidal
= 5.
respect
to (a) the x axis
x axis, and
(c) the line
The centroid of A» is at (2.4 in.,
oti. iis):
Ans. (a) 9.86 in., (b) 6.12 in., (ce) 6.7 in.
798. The same as 797 except that area
A, is to be used.
The centroid of A, is at
(EZ ne Osa sau.)e
Ans. (a) 0.233r4, (b) 145.4 in.4, 2.72 in.
793. For the annular area of Fig. 406,
derive the expression for the polar moment
of inertia with respect to an axis through
A and normal to the zy plane. By using
the results of problem 777 and the transfer
formula, no integration is necessary. Deter-
COMPOSITE
If the coordinates
isiat (2:4 ms 7-08 ims):
Problem 792,
through point A, Fig. 405.
795.
799. Find
the moment
of inertia
and
radius of gyration of the shaded area under
the cosine curve, Fig. 401, with respect
to the line y = —4 in. The centroid is at
(0, 7/8 in.).
Ans. 38.7 in.4, 4.4 in.
800. The same as 799 except
reference line is to be x = 4 in.
that the
AREAS
801. In the standard structural steel
angle section of Fig. 407, a = 1 in., b = 2
in., and ¢ =5 in. Find the centroidal moments of inertia J, and I,.
Ans. 13.83 in.4, 1.333 in.4
802. In the standard angle section of
Fig. 407, let a = 1/4 in., b = 3 in., andc =4
in.
Find the centroidal moments of inertia
IL, Savels Thy.
803. In the standard structural steel Zsection of Fig. 408, let a = 3/4 in., b =3144
in, ande =6in.
Find J, and f,.
Ans. 42.1 in.4, 15.44 in.4
804. In the standard Z-section of Fig.
PROBLEMS:
COMPOSITE
Ee
=|
223
AREAS
ally
S
x
—
Fig. 407.
<I:
Probiems 801, 802.
greet
3
za
3
Pe
Fig. 408.
NA
ae
Fig. 411.
Problems 803, 804, 828.
Problems 807-810.
gyration about an axis parallel to AA and
tangent to the circle.
809. For the shaded area in Fig. 411,
find the polar moment of inertia and radius
of gyration about an axis perpendicular to
the paper through point (0,4).
Ans. 907 in.4, 5.14 in.
810. The same as 809 except
axis is through point (2,2).
3
SP) (39
oe -1
Fig. 409.
that the
Problem 805.
408, let a = 1/2 in., b = 34 in, and c = 5
in. Find /, and I,.
805. In the channel section of Fig. 409,
let a
=1 in, b =4 in, and
Find 7, and J,. nove: This
c = 10 in.
channel is
nearly a standard size.
Ams. 205.3 in.4, 20.3 in.‘
806. In Fig. 410, let b = 10 in., h = 6
in., and the diameter of the hole D = 2 in.
Find the moment of inertia and radius of
gyration of the shaded area about axis AB.
807. For the shaded area in Fig. 411,
find I, and I44.
Ans. 728 im.4, 179.6 in:4
808. For the shaded area in Fig. 411,
find the moment
of inertia and radius
of
Fig. 412.
Problems 811, 812.
811. For the composite area of Fig. 412,
find the moments of inertia J; and I,.
The centroid of the semi-ellipse is at (1.7
ey Oe
Ans. 101 in.4, 314 in.4
812. For the composite area of Fig. 412,
find I,, [y, Jzy, and the corresponding polar
radius of gyration.
The values from 811
may be used.
PROBLEMS: COMPOSITE AREAS [Ch. XI
224
813. Determine /, and J, for the shaded
area shown in Fig. 413. Also find the polar
moment of inertia and the radius of gyration
of this area about an axis through point A.
Wipes, dye thy =] ZOD
ales din =
WOES
ka = 4.77 in.
atoll
b
Fig. 413.
Problem 813.
814. (a) For the shaded area of Fig. 414,
compute J,. (b) Determine the moment
of inertia about a centroidal axis parallel to
the x axis.
(c) Determine J, and the polar
moment of inertia for an axis through O.
Ans. (a) 688 in.4; (b) 164 in.*; (c) 928.6 in.4,
1616.6 in.4
Fig. 415.
Problems 815, 816.
The dimensions and other
from a handbook are:
data
as taken
For the channel, L = 12 in., d =3 in.
(very nearly), area of section = 6.03 in.2,
weight per foot = 20.7 lb., the centroid
is at
(0, 0.7 in.),
and
the
centroidal
moments of inertia and radii of gyration
are, respectively, J,,, = 3.9 in.4, kz, =
0.81 m.,
4.61 in.
J,-= 128.1
m4
and
k, =
For each angle, f= 3 in., e = 5 in., area of
section = 3.75 in.?, weight per foot =
12.8 lb., the centroids are at b = 1.75
in.,
Fig. 414.
Problem 814.
815. The cross section of a built-up Ibeam to be used in an auditorium span is
shown in Fig, 415. The dimensions are
=1 in, b =8 in, and
d= 50 in. All
four angles are identical.
Find the moments of inertia and the radii of gyration
about the horizontal and the vertical centriodal axes, neglecting the alteration for
rivet holes or welding.
i
c = 0.75
in.,
and
the
centroidal
moments of inertia and radii of gyration
are, respectively, Jz, = 9.5 in.4, ke,
1.59 in., I, =2.6 int and k,,
0.83 in.
For the composite figure, find (a) the cen-
troid, (b) the centroidal 7, and kz, and (ce)
the centroidal Z, and ky. The added weld
metal is neglected.
Ans. (a) 0, —0.656 in., (b) 43 in.4, 1.78 in.,
(ce) 340 in.4, 5.01 in.
Ans, I, = 63,600 in.4, &, = 21 in., I, =
1675 in.4, ky = 3.26 in.
816. Solve problem 815 first.
Now
change the beam by using 2-in. thick plates
on the top and bottom.
Find the ratio of
the percentage increase of moment of inertia
about the horizontal centroidal axis to the
percentage increase of area of the cross
section.
817. Two standard angles are welded to
a standard
channel as shown in Fig. 416.
Fig. 416.
Problem 817.
PROBLEMS:
PRODUCTS
OF INERTIA
AND
PRODUCTS
AXES
OF
PRINCIPAL
INERTIA
818. What is the product of inertia of
the upper half of the shaded area of Fig.
226
AXES
AND
PRINCIPAL
==
ee
402 about the « and y axes?
Of the lower
half?
Ans. +42.7 in.4, —42.7 in.4
ieee Sion
Lg" S
ro)
i
lees
Fig. 418.
i
||
819. For the area of Fig. 402 find the
product of inertia about centroidal x and y
axes 1f
= 2.4in. and 7 = 0.
820. For the area A» of Fig. 399, point P
istatn(2 ides
ine) andeai—. l/ 20 hind 2,
821. What is the product of inertia of the
shaded area of Fig. 403 about the x and y
axes?
Ans. 21.33 in.4
Problems 826, 827.
828. For a standard Z-section, Fig. 408,
the dimensions are a = 1/2 in., b = 344 in.,
¢ = 5 in., 1419.1
>
=
in and J, = 9.1 in.4
Find the principal axes and the least radius
of gyration.
829. A rectangular area is 10 x 20 in. in
~
oO
rE
Fig. 417.
Problems 822-824.
822. For the angle section shown in Fig.
417, [, = 30.8in.4, I, = 108in.4, and Pry =
—10 in.4. What is the moment of inertia
about an axis through C making an angle
of +30° with the x axis?
Ans. 34.5 in.4
823. The same as 822 except that the
axis makes an angle of +150° with the
x axis.
824. An angle section is shown in Fig.
A4i’7for which f, = 30.8 in.4, 7, = 10.8 in.4,
and P,, = —10 in.t
Determine the max-
imum and minimum centroidal moments of
inertia and the minimum radius of gyration.
Ans. 34.95 in.4, 6.65 in.4, 0.859 in.
825. A right triangular area has legs of
6 in. and 8 in. For axes passing through
the point of intersection of the legs, find
(a) the polar moment of inertia and (b) the
least radius of gyration.
826. The section of a column has the
dimensions shown in Fig. 418. What is
the least radius of gyration of this section?
Ans. 1.8 in.
827. What is the least radius of gyration
of the shaded area of Fig. 418 for axes
passing through point A?
Ans. 1.96 in.
size.
(a) What
is the moment
of inertia
about a centroidal axis which passes through
a corner?
(b) What is the least radius of
gyration for the origin of axes at a corner
of the rectangle?
Ans. (a) 2670 in.4, (b) 3.58 in.
Ny
oo
Fig. 419.
J.
Problems 830, 831.
830. For a standard structural steel angle,
igi
Ode —sleineOl—n On tne ch—=— Sime
% = 2.65 in, 9 = 1.65 im, I, = 38.8504
and J, = 80.8 in.*. Locate the principal
axes and determine the least radius of
gyration.
PA TiS) Crnrm
alloc Gulia
831. For astandard structural steel angle,
Fig. 419, a = 1/2 in., b = 3 in., c = 5 in,
% = 1.75in., andy = 0.75 in., [, = 2.6 in.4,
and J, = 9.5 in. Find the principal axes
and the least radius of gyration.
Ans. kmin = 0.65 in.
832. Calculate the minimum radius of
gyration for the combination of two channel
sections shown in Fig. 420.
Ans. 6.25 in.
PROBLEMS: PRODUCTS OF INERTIA AND PRINCIPAL AXES [Ch. XI
226
~
or
+.
|—— 18” —-|
Fig. 420.
Problem 832.
|Y
YY
Ys)
Fig. 422.
Problems 834, 835.
834. In Fig. 422, the area A, is bounded
by the y axis, the curve, and the line y = 5 in.
For a = 0.2, find the following: the area, ,
lay
Fig. 421.
Problem 833.
833. An extruded aluminum tube is shown
in cross section in Fig, 421, on which A =
2.000 in. and B = 3.375 in. Find (a) Jz,
(b) Ly, (c) Pry, (d) the principal axes, and
(e) the least radius of gyration.
Observe
that this section has three pairs of principal
axes.
Clay
uhem
prineipaleaNes,
elena
Taras ANd lenin
835. In Fig. 422, the area A» is bounded
by the zx axis, the curve, and the line x = 4
in.
Fora
TAG,
lc la eee
= 1, find the following: the area,
1Ree
mine
UOem
prin clpalleaxeswelenass
836-840. These numbers
for other problems.
may
be used
Chapter XII
MOMENTS
OF
INERTIA
OF MASSES
153. Introduction.
Any material body of mass m has inertia. It is
sufficient for the present to say that inertia is the property of a body which
As Newton said, it is because of
makes the body resist a change of motion.
a mass’s inertia that an unbalanced force must act on the mass to give it an
The moment of inertia* is a mathematical property which has
acceleration.
Thus, in chapters which
to do with change of motion that is not rectilinear.
bodies in order to deal
certain
of
inertia
of
moment
follow, we shall need the
with their motions.
154. Definition. The mathematical definition of the moment of inertia of
a mass is the same in form as that for an area; namely,
(32)
ie =| dm,
where dm is a differential mass, all parts of which are at a distance r from
some axis or plane about which the moment of inertia J is to be determined.
The limits of the integral must be such as to include all differential elements
dm that make up the body.
155. Moments of Inertia about Planes and Axes. The moment of inertia
of a mass m may be with respect to either a plane or an axis. In Fig. 423,
let a differential element of mass have the dimensions dz, dy, and c, with the
c dimension parallel to the z axis. Since all parts of this element are then
at the same distance y from the xz plane, and at the same distance x from
the yz plane, the moments of inertia with respect to these planes are
(a)
et =| dm
and
Lan - {xdm,
*The name moment of inertia for fr2dm was first used by Leonhard Euler (1707-1783),
a Swiss mathematician who made many important contributions to astronomy, hydrodynamics, and optics, and who developed the transfer formula, §158. As a professor of
mathematics, he spent some eleven years in St. Petersburg (now Leningrad) and twenty
five years in Berlin. He was a man of broad culture. You will no doubt become familiar
with Euler’s column formula during your study of strength of materials.
227
228
MOMENTS
OF INERTIA OF MASSES
[Ch. XII
respectively.
(Observe that while all parts of this differential element are
the same distance from the z axis, all parts are not at the same distance from
the x and y axes.) Moreover, with all parts of the element dm at the same
distance r from the z axis, the moment of inertia with respect to that azis is
(b)
iG aff dm.
Since, from the geometry of the figure, r? = v2 + y?, we see that
(c)
I=
fredm = [at + y") dm
= Tye + Tae
that is, the sum of the moments of inertia of a mass with respect to two
planes at right angles to each other is equal to the moment of inertia of the
mass with respect to the axis formed by the intersection of the planes.
This
rule is sometimes helpful in determining a moment of inertia about an axis.
Fig. 423. For convenience, the
mass m is assumed to have a uni-
form thickness c.
Fig.
424.
Parallel
Axes.
The
Zand Z’ are a distance d apart.
axes
156. Units.
The units of the moment of inertia of a mass evidently involve
the units of mass.
In scientific work, the English unit of mass is usually
the pound, in which case, as we shall see more clearly
later ($§ 221 and 347),
the unit of force is a derived unit (not a pound).
However, engineers are
accustomed to using the pound as a unit of force.
As a result, the mass,
the units of which are then derived, is found to
be W /g, where W is the
spring-balance weight of the body (the force of gravity
) and g is the acceleration of gravity at the point on the earth’s surface
where the weight is determined.
For most engineering purposes, it is sufficie
ntly accurate to
consider g constant, with a value of g = 32.2
ft. per sec.? (fps?). Thus,
with the weight expressed in pounds, the corresp
onding derived units of
mass are obtained as
W
gq’
lb.
ft./sec.2
Ib-sec.?
ft.”
a unit that is called a slug (a slug = lb-sec
.? per ft.).
Now since the linear
§ 158 |TRANSFER FORMULA—PARALLEL
AXES
229
unit foot is involved in our conception of mass when we use g in feet per
second-second (32.2), the moment arm r must also be expressed in feet in
order to keep the units consistent.
Consequently, the units of moment of
inertia (fr? dm) are
Ly
» ., lb-sec.?
ft
= ft-lb-sec.? = slug-ft.?
Recall that we frequently use the specific weight w of materials in pounds
per cubic foot and that the mass density p is then w/g slugs per cubic foot.
Since we are dealing with masses in subsequent chapters, the density should
be in mass units, slugs per cubic foot.
157. Radius of Gyration.
The radius of gyration k is defined by
i
(d)
k=
1/2
(4)
or
m
LT = mk?.
In the form I = mk?, we see that the radius of gyration k is the distance from
an axis to a point at which the entire mass may be imagined as concentrated
with the result that the concentrated mass has the same moment of inertia
about the axis as the actual distributed mass.
158. Transfer Formula—Parallel Axes. The mass m, Fig. 424, has a
The 2’
center of gravity* at C through which pass the x, y, and z axes.
for
axis,
x
axis is parallel to the z axis, with an origin at O’ on the original
convenience only. Choosing a differential element dm, all of whose parts
From the
are the same distance r from the 2’ axis, we have J,, =fr?dm.
figure, we see that
ye
AD
0)
ee
Thus
Ip -[r dm = fie — d)? + y?] dm,
(e)
ty = [tam — 2a [dm + aefam + [yam
In this group of integrals, we observe that
Je dme=
Tyz
and
Jv ia
Hence the sum of these two J,,+ 1,, = I, [equation (c)]. Moreover,
fxdm = 0, since the yz plane passes through the center of mass; and {dm = m.
Therefore equation (e) takes the form
i
2=11, + md,
or, in general symbols,
(830A)
I=I1+éd,
*Strictly, C is the mass center ($§ 109 and 110).
MOMENTS
230
OF INERTIA OF MASSES [Ch. XII
which we observe is the same form as in the case of areas, equation (30), p. 209.
As before, the moment of inertia J of a mass about any axis is equal to the
moment of inertia of the mass about a parallel gravity (centroidal) axis I
plus the product of the mass m and the square of the distance d between the
axes. Now let I = mk? and J = mk?, and substitute these values in (30A).
This procedure gives
(f)
k? = k? + d?,
the relation of the radii of gyration for parallel axes, where & is with respect
to the gravity axis.
159. Integrating for Moments of Inertia of Masses. A safe choice of the
differential mass is one whose parts are all at the same distance from the
designated axis. This may be one whose finite length is parallel to the axis,
as in our discussion so far in this chapter (Fig. 423), or it may be of the form
dx dy dz (or similarly in polar coordinates).
However, such a choice does
not always give the easiest solution.
A second choice is a differential element whose moment of inertia about
the designated axis is known.
In this event, a direct sum (integral) of all
these elements for the entire mass gives the desired result.
A third differential element that may be used is one whose moment of
inertia about its own centroidal (gravity) axis parallel to the designated axis
is known.
In this case, the transfer formula (830A) may be applied to the
differential mass and the integral made for the entire mass.
Examples will
serve best to show the methods.
160. Example—Cylinder.
(a) Derive an expression for the moment of inertia
of a homogeneous right circular cylinder about the geometric axis.
(b) If a cast-iron
cylinder is 24 in. in diameter and 48 in. long, what is its
moment of inertia about the same axis? Cast iron weighs
450 lb. per cu. ft. (¢) What is the corresponding radius
of gyration?
SoLuTION.
(a) The axis referred to is the z axis, Fig.
425. An element is chosen with a differential sectional
area v dé dv, where v is the variable radius, and with a
length h, as shown, whose parts are all a distance v from
the z axis. The differential volume is
dV =hdA
= hv dodo;
the differential weight is
dW =wdV
= whv de dv;
and the differential mass is
Fig. 425,
Cylinder.
ae
dW = sew
hv do dv = phv de do,
i = 4
§ 161 | EXAMPLE—SPHERE
231
where p is the density (w/g).
Using fr? dm, where r = v, we get
If =f fom do dv = nf diy,v8 dv do
19)
D
=
=
()
pha D*
ds
mD?
= mr?
4
8
8
Oe
ph oau
64
where r = D/2, the radius of the cylinder.
Compare mr?/2 with the answer obtained
in § 138 for the centroidal polar moment of inertia of a circular area (Ar?/2).
(b) The mass of this cast-iron cylinder is
10%
m = ph——
—
450
r4
(5 )o()
= 175.6
175.6 slugsslugs.
The moment of inertia J, is
Limeals
mr?
(175.6) (1)?
a
2
= 87.8 slug-ft
(c) The radius of gyration is
ge
mr
1% (4) - & 04
ge
r
il
toariaenE ce inet
pee
161. Example—Sphere.
(a) Derive the expression for the moment of inertia of
a sphere about a diameter.
The density of the material is constant.
(b) Find the
moment of inertia about a line tangent to the sphere.
(c) Find J about an axis
which is a distance of 10r from the cg.
Sotution.
Figure 426 shows the projection of a sphere on the zy plane. Use a
differential element defined by two planes perpendicular to the xy plane and at a distance dz apart. This element is a cylinder of height dx and radius y, with the geometric
axis C-a (Fig. 426). From § 160, we know that J for a cylinder about a geometric
axis is mr?/2, where, in this example, the m is
oe
2.
dm = p dV = pry? dz.
(x,y)
Thus, for the element dm,
io
_dmr _
os
(pry? dx)(y?)
2
6
Summing all the d/,’s from « = —r to x = +1, we get
=
fae]
2 — 4)? da,
¢
where y? = r? — 2* from the equation of a circle.
Vee
8
=
7
= om
—arrp
Fig. 426.
Sphere.
ferential
mass
used
The difis de-
scribed in the second paragraph
of §159, which you should re-
2/4
—prr® = —(
Isa
(r4 — 2 ra? + x4) dx
|r? =
hats (;mc )
= mr
read now.
where m = pV = 4nr*p/3. Note particularly that this example uses a differential
element described in the second paragraph of § 159.
(b) The moment of inertia about the x’ axis, which is parallel to the z axis, is
I, = 1+ md
= =mo + mr? = Tm
OF INERTIA OF MASSES [Ch. XII
MOMENTS
232
(c) If the axis is a distance of 10r from C, the center of the sphere, the moment of
inertia is
l=If+md@=
=mr? + m(10r)? =
100.4mr?.
It is of very practical importance to notice that for an axis this far away
(10r), the moment of inertia is virtually the same as though the sphere were
a particle. In other words, if the sphere had been considered as a particle,
all parts the same distance from the axis, its moment of inertia would have
been md? = 100mr?, a value in error only about 0.4%. The closer the sphere
to the axis, the greater is the error in this approximation; the further away
from the axis, the less the error.
ya
-
:
>
X adi
nf
x
a
x
Fig. 427
Thin Disk.
Fig. 428.
Cone.
This example illustrates the use of
the element defined in the
third paragraph of §159.
162. Example—Thin Disk. Derive an expression for the moment of inertia of a
thin homogeneous disk about a diametral axis through its centroid.
SotutTion. Choose an element with a sectional area of v do dv, where v is the variable radius, and with a length equal to the thickness of the disk Ah, Fig. 427. Let I,
be required. In the lower view of Fig. 427, we see that the bottom of this element is a
distance aC (projected) from the y axis, while the midpoint of the element is at the
distance v cos @. Thus since all points in this element are not at the same distance
from the y axis, this element theoretically should not be used. However, if the
thickness Ah is small enough, much smaller relatively than shown in F ig. 427, this
element will result in a useful approximation; and if the thickness Ah becomes infinitesimal, the results will be precise.
The differential volume is
dV = vdeo dv Ah;
§ 164 ]EXAMPLE—SLENDER ROD
233
then
dm ll= p(v de dv Ah)
and
rv? dm = (v? cos? @)(p)(v da dv Ah)
with respect to the y axis.
Therefore
ji
oh
2
I, =I, = p anf / cos?
6 do v3 dv = p Ah - = ‘ approx.,
0
0
where m = pV = pAhm”.
Compare mr?/4 with the answer obtained in § 141 for the centroidal rectangular moment of inertia of a circular area (Ar?/4). Observe that mr?/4 is
the exact value of the moment of inertia with respect to the yz plane; that is,
a diametral plane of a cylinder.
If h = 0.1r, the error in using mr?/4 for
I, is less than half of one per cent. As h/r becomes less, the error in using
the approximate formula becomes less.
163. Example—Cone.
Derive the expressions for the moment of inertia of a
right circular homogeneous cone (a) about an axis through its apex and parallel to the
plane of the base, and (b) about a parallel gravity (centroidal) axis.
SoLuTION.
(a) Using the result obtained in the previous example (§ 162), we are
now prepared to choose, with advantage to ourselves, an elementary mass in accordance with the third paragraph of § 159. In Fig. 428, let the differential mass be a
circular disk of thickness dy and radius x. The mass of this disk is
dm = padV = pra? dy;
and the moment of inertia of the disk about its own gravity axis (from I = mr?/4,
§ 162) is
nig
dl,
=
(prix? dy) ee
Since the x, and x axes are parallel, we may use the transfer formula (380A), and find
n2
dI, = dl,,
+ dm@
=
(pr2? dy) = + (pra? dy)y?.
From similar triangles, Fig. 428, we get «/r = y/h, or
= ry/h. Using this value
of x in the SM equation and integrating, we get
*h
ai
ies
fa \hs 3
r2
rh
3
ey
art
pa
i
Re pay + EE yay =3(, 3
ce
5m
sete
(b) The value of J about a gravity axis of the cone may be obtained, as in part (a),
by integration.
However, with the answer in part (a), we may easily use the transfer
formula.
The centroid of a right circular cone is at the distance 3/4 from the vertex
(§ 120). Therefore, the moment of inertia about a gravity axis parallel to the plane
of the base is
=
= 4
3
Pe
ee
2
thy Ee= oden5 m (J
ae i)
Qh?
m 16
3m
¥) 4 h? )
:
30 (
164. Example—Slender Rod. Derive an expression for the moment of inertia
of a long slender rod (a) about an axis through an end and normal to the long dimension of the rod, and (b) about a parallel centroidal axis.
Sotution.
the sectional
OF INERTIA OF MASSES [Ch. XII
MOMENTS
284
(a) The y axis, Fig. 429, is such an axis as defined in the problem.
area
dm = pdV = pAdx.
shown,
for the element
Then,
A be constant.
Let
dV = A dz and
The moment arm of this element to the yz plane is x; hence
L
Tye -[ 2 Hi
—
=
edx
raf
=
AL’
3)
0
=
L
shea
3
If the element A dx is at some distance from the y axis, its various parts, while they
are not at the same distance from the y axis, have nearly the same moment arm.
Hence, if the rod is long enough and slender enough, the foregoing expression gives
approximately the moment of inertia about the y axis, since:
The elements farthest from the axis are the most significant in determining the
moment of inertia.
Fig. 429.
Slender Rod.
If the rod is a slender circular cylinder with r/h = 0.1, the error in using
I, = mL?/3 is less than 1%. The smaller the value of r/h, the smaller
the error.
(b) Using the transfer formula, we find the approximate moment of inertia about
a parallel gravity axis to be
I=I1,-—md@=
mL?
3
mL?
4
mL?
12
165. Composite Bodies.
Just as for areas, the moments of inertia of
masses are always positive.
Hence the moment of inertia of a composite
body is the sum of the moments of inertia of the parts of the body, the
same axis of reference being used for each part. Consequently, if a body
has a hole drilled in it, the moment of inertia of the drilled body zs equal
to the moment of inertia of the original solid body minus the moment of
inertia of the removed material, each moment of inertia being about the
same axis.
166. Example—Flywheel.
In the design of a flywheel, its moment of inertia
about its axis of rotation is needed.
Determine the moment of inertia about the axis
of rotation of the cast-iron flywheel shown in Fig. 430. Cast iron weighs 450 Ib. per
cu. ft.
So.ution.
(a) Rim,
The moment of inertia of the rim is equal to the moment
of inertia of a solid disk 96 in. (8 ft.) in diameter and 18 in. (1.5 ft.) thick minus that
of a disk 96 — 18 = 78 in. = 6.5 ft. in diameter, and 18 in. thick. Using
mi?
I
=
2
pVr?
=
2
§ 166 | EXAMPLE—FLYWHEEL
236
(§ 160), we get the moment of inertia of the rim about the axis of rotation as
450 \ [18?
|
ABO \ f16.52
| een]
(eeees
| (Perec
5) (3.25)
(3.25)?= 4840 slug-ft.?
Jug-ft.2
I, mre 6)
5)
i ib te)(4) Pips Blea
i 5)
(b) Arms.
As an engineering approximation, we may consider the arms as slender
rods. First, we find the moment of inertia of an arm about an approximately located
center of gravity C. (The taper of the arms and the fillets are neglected.)
We then
add to this the product md* to make the transfer to the axis of rotation.
For an
Fig. 430.
Flywheel.
average area of cross section of A = 18 sq. in. = 18/144 sq. ft. (Fig. 430), the volume
of material in one arm is AL, where L (= 2.5 ft.) is the length of an arm.
For the
arm, I = mL?/12 (§ 164). Hence, for 6 arms, the moment of inertia about the axis
of rotation is
f
2
IE
AGOE 23 7a) — (ne + mt)
IT,
1
450
18
450
= 0)(6) (BRS)
2.5)
)(2.5)?
(2.5)? ++ (
—— )(—18 2.5)s)(2) 2)? |
|= 118.618.6 slug-ft
slug-ft.2
ee
(c) Hub.
The moment of inertia of the hub is equal to the moment of inertia of
a solid cylinder 18 in. (1.5 ft.) in diameter and 24 in. (2 ft.) long, minus that of a
cylinder 9 in. (0.75 ft.) in diameter and 24 in. long. Using IJ = mr?/2 as in part (a),
we have the moment of inertia of the hub about the axis of rotation as
(1\f 450 \(AE
(1 2)corsenye-(2)
(1\( S222)
450 \ 10.75" coazsy Gen’
t=_ (2S)
= r2staet ane
Therefore, the total moment of inertia of the flywheel is
T=1,+ 1.4 1; = 4840 + 118.6 + 13 = 4971.6 slug-ft.2, say, 4970 slug-ft
Comparing the values of the moments of inertia of rim, arms, and hub, we readily see
why the approximations involved in getting J. for the arms result in only a minor error
in the total J. Asa matter of fact, engineers generally neglect the effect on I of the
hub entirely and frequently neglect the effect of the arms too. In this example, J; is
97.5% of the total I.
PROBLEMS: DIRECT INTEGRATION
236
[Ch. XII
167. Closure. Observe carefully in the preceding examples how the
elementary mass was chosen and used to find the moment of inertia of the
whole body. In § 160, the element was chosen so that all parts of it were
In § 161, an element was chosen
the same distance from the reference axis.
whose moment of inertia about the reference axis was known (from § 160),
although all of its parts were not’ the same distance from this axis. Then
in § 163, an element was chosen whose parts were not all the same distance
from the reference axis and whose moment of inertia about the reference
axis was unknown, except as we applied the parallel axis theorem, knowing
the moment of inertia of the element about 7zts own gravity axis (from § 162).
The moments of inertia which will be the most useful include (constant
density) :
For a cylinder (about its geometric axis, $160),
L= 7 ‘
For a sphere (about a diameter, $161),
l= :mr?;
For a slender rod (about a centroidal axis perpendicular
i
to its geometric axis, $164),
mL*.
12%
For a thin disk (about a centroidal axis in its “plane,’’ §162),
J = mr
If the mass is relatively concentrated at some distance d from
the axis [§ 161(c)],
Too
Problems
DIRECT
INTEGRATION
841. For the rectangular parallelepiped in
Fig. 431, determine in terms of mass m an
expression for I, if the element of mass
used is (a) pt dx dy, (b) pb dx dz, (c) ph dy dz,
(d) pdx dy dz, (e) pbtdz, (f) pthdy, and
(g) phb dz.
Mine (he + b2),
3
842. The
as 841 except
(b) Find
he
dz
J, for the prism
whose base is in the xz plane.
Express
answers in terms of the mass of the prism.
Is the sum of your answers equal to the
answer to 841?
Ans.
Ghee
that the
to the base in the yz plane.
Ans. mh?/3.
843. The same as 841 except that the
moment of inertia is to be found with
Tespect to the centroidal x axis.
Ans. (m/12)(b? + #2),
844. Two triangular prisms are formed
by passing a vertical plane ABCD through
the rectangular parallelepiped of Fig. 431.
(a) Find J, for the prism whose base is in
the yz plane.
=
same
moment of inertia is to be found with respect
(a) (m/12)(h? + 3b),
(b) (m/12) (3h? + b2),
845. A rectangular parallelepiped has a
rectangular base of dimensions b x ¢ and a
height h. (a) Derive expressions for the
een an ae inertia of this homogeneous
Fig. 431.
Problems 841-844,
ody with respect to the ‘planes
adjacent, eae (b) Using the naien
PROBLEMS:
DIRECT
INTEGRATION
(a), find the moment
line of intersection
287
of inertia about the
of these
(ec) Then find the moment
two
planes.
of inertia about
the geometric axis.
Ans. (a) mb?/3, mt?/3; (c) m(b2 + #2) /12.
846. A right circular cylinder of steel is
18 in. in diameter and 30 in. long.
(a)
What is its moment of inertia about an
element
(line) of the surface parallel to the
geometric axis? Steel weighs 490 Ib. per
cu. ft. (b) The same as (a) except that the
axis 1s a distance of 5 ft. from the geometric
axis. Use the parallel axis theorem.
Ans. (a) 56.7 slug-ft.2, (b) 1699 slug-ft.?.
847. The same as 846 (a) except that the
moment of inertia about a plane containing
the geometric axis is desired.
848. (a) By direct integration, determine
the moment of inertia of a homogeneous,
right circular cylinder about a diameter of
its base. The height is h and the radius is
r. Let the differential element be a thin
disk parallel to a base.
(b) What is its
moment of inertia about a gravity axis
which is parallel to the plane of a base?
Ans. (a) m(3r?2 + 4h?) /12; (b) m(r2 + h2/3) /4.
849. A brass sphere, 6 in. in diameter, is
attached to a link which turns about an
axis 30 in. from the center of the sphere.
If brass weighs 530 lb. per cu. ft., what are
the sphere’s moment of inertia and radius
of gyration about the axis?
Ans. 6.76 slug-ft.?, 2.505 ft.
850. Using the element of the sphere
indicated in Fig. 426, find I, by direct
integration and check the answer for J, in
§ 161. Also find the corresponding radius
of gyration.
851. Derive the equation for the moment
of inertia of a sphere about a diametral
plane.
Ans. mr?/5.
852. An aluminum rod, 1 in. in diameter
and 2 ft. long, weighs 169 lb. per cu. ft.
(a)
Derive
an
exact
expression
for
the
moment of inertia of this rod about a
diameter at one end.
(b) Using the result
in (a), compute the values of this moment
of inertia and the radius of gyration.
(c)
Using the approximate expression of § 164,
compute this moment of inertia and radius
of gyration; compare these answers with the
exact values.
853. The same as 852 except that the
rod is made of magnesium which weighs
112 lb. per cu. ft.
854. The same as 852 except that the
rod is made of Monel which weighs 555 lb.
per cu. ft.
Ans. (a) m(3r?2 + 4L?)/12; (b) 0.251 slug-ft.?
855. Using the element indicated in Fig.
428, derive the expression for the moment
of inertia of the homogeneous cone about
its geometric axis by direct integration.
Whatis the corresponding radius of gyration?
Ans. 3mr2/10, 0.548r.
856. By direct integration, determine the
moment of inertia of the homogeneous cone
of Fig. 428 about a gravity axis parallel to
the plane of the base. Check the answer
given in § 163(b).
857. (a) By direct integration, determine
the moment of inertia of the homogeneous
cone of Fig. 428 about an axis which coincides with a diameter of the base.
(b) If
the diameter of the base is 12 in., the height
is 24 in., and the material is cast iron which
weighs 450 lb. per cu. ft., what is its moment
of inertia about a diameter of the base?
Fig. 432.
Problem 858.
858. A semi-ellipsoid, whose mass is m,
is generated by rotating a quadrant of an
elliptical area about the x axis.
(a) Find
the moment of inertia of this mass with
respect to the y axis, using the vertical
element AB of mass indicated in Fig. 432.
(b) Then find the mass m of this body and
express J, in terms of mass.
(c) The semiellipsoid is made of mortar which weighs
100 lbs per cul it,, a= 3 it, and b= 2 ft:
Determine J, and ky.
Ans. (a) 2prab?(a? + b?/4)/15;
(b) m(a? + 62/4) /5.
y
Fig. 433.
Salle
Problems 859, 860.
859. To balance the reciprocating parts
of a certain engine, a 6 = 45° sector of a
cast-iron disk, Fig. 433, is used. When
PROBLEMS: DIRECT INTEGRATION [Ch. XII
288
the effect of this counterweight is determined, the moment of inertia about the z
axis is needed.
Find J. Showing the differential mass used and its moment arm,
integrate for the equation needed.
Cast
iron weighs 450 lb. per cu. ft.
Ans. 3.66 slug-ft.?
860. The same as 859 except that 0 = 60°.
861. A solid is generated by revolving
the plane of the parabola y? = 9x about the
x axis.
The length of the solid in the x
direction is 3 ft. Find its moment of inertia and the radius of gyration about the x
axis, if it is made of steel weighing 490 Ib.
per cu. ft.
Ans. 17,460 slug-ft.?
x
y2= ax3
Fig. 434.
862. The same as 861 except that J, is
desired.
863. The rim of a cast-iron wheel has a
round 1-in. diameter cross section and the
mean diameter of the rim is 12 in. If cast
iron weighs 450 Ib. per cu. ft., find the
moment of inertia with respect to the axis
of rotation.
Make your computation a reasonable approximation for ordinary engineering purposes.
864. A part
of a certain
projectile
Problem 864.
ordinates (2,2 ft.). Let a = 14 and find J,
and k,. The material is steel that weighs
490 lb. per cu. ft.
867. The same as 866 except that area
Az» is to be used.
is
defined by revolving the area Aj, Fig. 434,
about the y axis. Area A; is bounded by
ines =] 4%, = s = 4 ame) oF = cae
where a = 1. The unit is the inch. Find
Thy Baral [er
865. The area A, of Fig. 435 is revolved
about the y axis to generate a homogeneous
sold.
(a) Determine an expression for J,
and k, when the upper boundary is line
y = 8. (b) Check your answer by selecting
an alternate element.
Ans. Iy = 86.2pa*/3,
866. The area A, of Fig. 435 is revolved
about the x axis to generate a homogeneous
solid, when point P is defined by the co-
VARIABLE
y= ax?
Fig. 435.
Problems 865-867.
DENSITY
868. The rod in Fig. 429, instead of being
homogeneous, has a density that varies
directly as x; that is, p = Cz, where C is a
constant.
(a) Show that the approximate
moment of inertia about the y axis is I, =
mL?/2.
(b) What is the moment of inertia
about an axis parallel to the y axis and
passing through the center of mass?
Ans. mL2/18.
869. The density of the cone in Fig. 428,
instead of being constant, varies as the
distance of the point from a plane through
the vertex and parallel to the base. What
is its moment of inertia J,?
Ans. (m/6)(r? + 4h?2).
870. The density of the cylinder in Fig. 425,
instead of being constant, varies as the
distance of the the point from the z axis.
What are its moment of inertia J, and the
corresponding radius of gyration.
PROBLEMS:
COMPOSITE
239
SOLIDS
Fig. 436.
COMPOSITE
Problems 872, 873.
SOLIDS
871. An 8-in. Monel (w = 555 lb. per cu.
ft.) disk with a 3-in. coaxial hole is used as
If the disk is 1 in. thick, what
a flywheel.
Ans. 3.02 in.
is its radius of gyration?
872. Determine the moment of inertia of
the steel spool shown in Fig. 436 about the
Steel weighs 490 lb.
geometric axis AB.
Ans. 866 slug-ft.?
per cu. ft.
that there
except
872
as
same
873. The
is a cylindrical hole 2 ft. in diameter whose
The hole passes
axis coincides with AB.
entirely through the spool.
874. (a) Derive an equation for the moment of inertia of a right circular cylinder
879. The same as 878 except that k, 1s
desired.
880. The same as 878 except that kz is
Would any likely size of handle
desired.
pe
a
of radius r,, which has a coaxial cylindrical
hole of radius r;. Express the answer in
terms of the mass of the hollow cylinder.
(b) Using this result, determine the moment
of inertia of the rim of a steel flywheel with
an outside diameter of 5 ft., an inside
diameter of 4 ft., and a rim width of 8 in.
Ans. (a) m(ro? + 112) /2; (b) 367 slug-ft.?
875. What is the moment of inertia about
a diameter of a hollow cast-iron sphere with
an outside radius of 4 ft. and an inside
radius of 2 ft.? Cast iron weighs 450 lb.
Ans. 23,300 slug-ft.?
per cu. ft.
876. Find the moment of inertia about
m
the geometric axis of the cast-iron frustu
of a cone shown in Fig. 437, which has a
short 4 x 2-ft. cylindrical hole in the base.
Ans. 19,900 slug-ft.?
except that the
876
as
same
877. The
height h = 3 ft. instead of.6 ft. The base
diameters remain the same.
of
878. The wood handle, of the mallet
Fig. 438, is L = 3 ft. long, weighs 3.14 lb.,
The head,
and has a uniform cross section.
of
weighing 16.1 lb., is a wood cylinder
of
diameter D =6 in. Find the radius
the
gyration of the body with respect to
handle
the
with
head
mallet
The
y axis.
Ans. 2.59 ft.
inclusion is homogeneous.
Fig. 437.
Problems 876, 877.
Fig. 438.
Problems 878-880.
PROBLEMS: COMPOSITE SOLIDs [Ch. XII
240
diameter
Explain.
affect
the answer
significantly?
Ans. 0.315 ft.
rotation occurs about AA.
What is the
moment of inertia of the links and spheres
about AA?
For the links, an approximation similar to that in § 164 is acceptable.
ber
Fig. 439.
Problem 881.
881. A steel link (w = 490 Ib. per cu.
ft.), as shown in Fig. 439, rotates about the
axis of the 3-in. hole. Allowing for the
hole, compute the moment of inertia and
radius of gyration of the link about the
axis of rotation.
(Do not make the approximation of a slender rod.) Compare this
answer with that obtained by using the
approximation for a slender rod and neglecting the effect of the hole.
Ans. 12.2 slug-ft.2, 1.46 ft.
Fig. 441.
Problems 883, 884.
883. A section of cast-iron (w = 450 lb.
per cu. ft.) flywheel is shown in Fig. 441.
Considering the web and the hub, determine
the moment of inertia about the axis of
rotation.
If the web and hub were neglected, what would be the percentage error
in I?
Ans. 8.7 slug-ft.2, 13.5%.
884. The same as 883 except that the
web is replaced by 6 arms.
Each arm has
a round, 2-in. diameter section.
Age
Fig. 442.
Problem 885.
885. A short steel cylinder (w = 490 Ib.
per cu. ft.) is 10 in. in diameter and 10 in.
long. To one end is attached an aluminum
hemisphere
Fig. 440.
Problem 882.
882. A flyball governor is shown diagrammatically in Fig. 440. The 8-in. eastiron spheres are attached to steel links
which are 18 in. long and have a sectional
area of 1 sq. in. each. The center lines of
the links and spheres and the axis AA are
in the same vertical plane. In operation,
(w =
170 lb. per cu. ft.) 9Do
the other end is attached a brass hemisphere
(w = 530 Ib. per cu. ft.). See Fig. 442.
(a) Locate the mass center of this composite
body.
(b) Find the moment of inertia of
this body about a gravity axis parallel to
the x axis shown.
Ans. (a) ¥ = 8.85 in., (b) 1.85 slug-ft.?
886-890. These numbers may be used for
other problems.
Chapter XIII
PLANE
MOTION
This chapter and the next one deal briefly with the
168. Introduction.
As defined in § 1,
subdivision of mechanics which is called kinematics.
are
kinematics is the study of the motion of particles (whose dimensions
infinitesimal) and bodies without regard to the forces which affect the motion.
because
Some study of this phase of mechanics is necessary at this time,
always
tion,
considera
for
unbalanced force systems, which is the next subject
act.
produce motion or a change of motion of the bodies on which they
The laws of motion were investigated scientifically first by Galileo Galilei
family. Al(1564-1642), who was born at Pisa to an impoverished noble
physical
the
to
shifted
though he was educated for medicine, his interest
life,
later
in
d
Most of his experiments on motion were conducte
sciences.
of Pisa was
although the famous one of two weights released from the Tower
performed when he was 26.
bodies, a
Aristotle had reasoned that heavy bodies fall faster than light
s day
Galileo’
of
one
logical and plausible conclusion, and nearly every
idea
e’s
No one thought to question Aristotl
believed that this was true.
neously a halfuntil, some 2000 years later, Galileo decided to drop simulta
Tower of Pisa.
Leaning
the
from
pound weight and a 100-lb. cannonball
ground at the
the
struck
When the spectators observed that both weights
he was pracsame time, they roundly hissed Galileo, thinking that perhaps
People always “Inow’’ so many things which are not so,
ticing witchcraft.
and such a result was contrary to what the people ‘‘knew”’.
tative the
In passing, it will be interesting to suggest further how authori
a monk
spots,
When Galileo discovered sun
writings of Aristotle became.
e which had just then
heard of his discovery and, having access to a telescop
The monk then reported
been invented, he verified Galileo’s observation.
superior searched his
The
the spots to his bored and dubious superior.
told the monk, ‘Be
Aristotle in vain. Finding no mention of sun spots, he
or of your telescope.”
assured therefore that it is a deception of your senses
ental verification.
experim
is
science
of
Today, we know that the very essence
we may evolve a
or
We may theorize first and then verify by experiment,
241
242
PLANE MOTION
[Ch. XIII
theory from experimental data, but it is not science if the theory (philosophieal conclusion) is not backed up by observable facts and measurements.
To return to the story, Galileo asked himself how bodies fell. His first
assumption, that the velocity of a falling body was proportional to the distance traversed
(v = Ch) was in error.
However,
he reasoned himself out
of this notion, and decided that the velocity would be proportional to the
time of descent (v = at). Finding no fallacy in this assumption, he proceeded
to experimental proof. First, he decided to reduce the gravitational acceleration (and velocities within a particular time) by using an inclined plane.
Another problem was the matter of timing.
The pendulum clock had not
been invented; sand and water clocks were traditionally used. Galileo used
a large vessel of water with a very small orifice at the bottom over which he
could hold his finger. The amount of water to run out upon the removal of
his finger was practically proportional to the elapsed time because the drop
in the water level was negligible during his experiments.
From his assumption (v = aft) and using the definition of v (v = 's/i), he
derived the relation s = at?/2. From this latter equation, he learned that
the spaces traversed in 1, 2, 3, 4, + + units of time were proportional to 1,
4,9, 16,
+ +, ete. Marking these spaces on the inclined plane and using a
rolling ball, Galileo soon had experimental verification of his law, thus introducing the conception of acceleration a. The discrepancies between his
experimental results and the algebraic relations he attributed to friction and
air resistance, although air pressure had not been established.
The ramifications of the deductions that Galileo made from his experiments on the
inclined plane are too numerous to give here (see Mach, Science of Mechanics).
Galileo evolved the conception of inertia (Newton’s first law), establis
hed
the forms of beams of uniform strength, conceived the pendulu
m clock,
rediscovered Archimedes’ law of fluid buoyancy (which had been
lost during
the centuries), and made many astronomical discoveries.
When Galileo heard that some one had invented a telescope,
he immediately
constructed one for himself without waiting for details.
(Compare this
event with what scientists are saying today—that
if the scientists of one
large country know that something has been developed in another
country,
they themselves, knowing that it can be done, can also successf
ully develop
the same something, perchance a better version of it.)
It was Galileo’s
astronomical work that invoked the displeasure of many
of his contemporaries. To suggest the nature of the difficulties of scientis
ts in the beginning
of the modern era, we shall quote one outburst after
Galileo announced his
discovery of the satellites of Jupiter: “... From this
and many other similarities in nature such as the seven metals, etc.,
... we gather that the
number of planets is necessarily seven.
Moreover, these satellites of Jupiter
are invisible to the naked eye, and therefore can exercise
no influence on the
243
§ 169 |]DISPLACEMENT
earth, and therefore would be useless, and therefore do not exist.
the Jews
and other ancient
nations,
as well as modern
Besides
Europeans,
have
adopted the division of the week into seven days, and have named them after
the seven planets. Now if we increase the number of planets, this whole
and beautiful system falls to the ground.” Such are the pitfalls of logic.
The best that Aristotle could do was to classify motion as natural (such
as the motion of the stars and planets) or unnatural (such as the thrown
stone, which would naturally soon come to rest—with respect to the earth).
We shall study general (Newton’s) laws of motion later, which apply to
earthly bodies and roughly to heavenly bodies, and in the meantime learn
the names of a few kinds of motion of which Aristotle never heard.
169. Displacement.
Displacement is a directed distance, and as such it
In common with force vectors, it has the properties of
is a vector quantity.
magnitude, sense, and location (line of action). Thus, suppose a person
for 2
starts at some point designated as the origin O, Fig. 443, and walks
The
miles in a straight line directed N 30° E (read this, north 30° east).
iMileeB
A
corresponding displacement is represented by the vector OA laid
out in magnitude and pointing in
the direction of the displacement.
Then if this person walks one mile
due east, the vector
representing
this displacement is AB. It is supposed, of course, that the terrain
is perfectly level, so that these disFig. 443. Displacement.
placements are in the same plane.
starting
At the point B in space, the pedestrian is not 3 miles from the
pedestrian’s
point (the origin O), but some shorter distance OB. The
swm) of the
(vector
t
resultan
the
y
resultant displacement from O is evidentl
:
obtained
be
may
This resultant vector OB
vectors OA and AB.
1. By graphical solution,
2. From the law of cosines (§ 6), or
displacements
3. By summing horizontal and vertical components of the
Dsy)
and
Lsz
and then finding the resultant
Sx = [(282)? + (Z8,)7.
an exact counterWe recognize that each one of these methods of solution is
had been force
part of a method that we might have used if these vectors
vectors.
443, other than
Suppose that the point of reference is some point O’, Fig.
24h
PLANE MOTION
[Ch. XIII
the point O from which the displacement occurred.
The simplest way of
expressing the resultant displacement is in terms of radius vectors and
angles (polar coordinates).
Relative to O’, the point O is defined by the
radius vector p; and the angle a, which is, of course, a negative angle as
measured in Fig. 443. The point B is defined by pz and 8. Since the vector
sum of p; and OB is equal to ps, it is evident that the vector difference is
Ale.
1
nA)
ae
ae Sie
Fig. 443.
Repeated.
Mile
lerB
—>p1=OB=s
LOA,
Ry
that is, the resultant displacement.
This method of finding a resultant
displacement tells nothing of the
path that may have been followed
between O and B, as the path may
have been any one, such as OCB.
Observe in the foregoing illustration that the pedestrian walked 3
miles, which is the distance that he
from the starting point is something less
traversed, but his displacement
than 3 miles.
Displacement problems, such as the preceding, will not occupy much of
our time here; but the significant point, that displacement is a vector quantity, should be fixed firmly in mind.
Soon we shall be dealing simultaneously
with velocities, accelerations, and forces, as well as displacements, all of
which are represented by vectors which may or may not point in the same
direction.
In general, we shall follow the usual convention; component
displacements in the upward direction and component displacements toward
the right are positive.
In many problems, it may be convenient to change
this convention, but consistency in the use of algebraic signs must be maintained in any particular problem.
170. Speed and Velocity. The speed of a particle at any instant is the
time rate with which it is traversing distance.
Technically, although the
words speed and velocity are often used as though their meanings were identical,
velocity is a vector quantity, and therefore possesses both properties of
magnitude and sense; whereas speed is a scalar quantity.
In other words,
the speed is only the magnitude of the velocity.
This distinction is important
because changes of velocity are important; and a change in the sense (direction)
of a velocity is as truly a change of velocity as is a change in the magnitude
of a velocity.
Consider an airplane, whose engine is warming up at a constant angular speed, in revolutions per minute, before the plane takes
off.
A point on the tip of the propeller is moving at constant speed in the path
of
a circle, but its velocity is continuously varying, since the direction
of the
velocity changes through 360° with every revolution of the propeller.
245
§ 171 ] EXAMPLES
Let a point traverse a distance As during some time interval At. Then
the time
the average speed of the point is v = As/At. As the distance and
interval become smaller and smaller, we may write as the limit
ee
eeAcmee
ee
Cs
aa,
of the time
Thus we see that when s is expressed as a continuous function
at any
speed
instantaneous
the
t, the derivative of s with respect to é gives
the
and
speed
time t. If a constant speed is maintained, the instantaneous
any
in
Speed may be expressed
average speed are of course the same.
In engineering work, the most
time.
and
convenient units of distance
feet per minute vm (fpm), and
(fps),
common units are feet per second v,;
miles per hour (mph).
its simplicity
A method of notation which is becoming popular because of
the symbol.
is to indicate differentials with respect to time by a dot over
For example,
z stands for dx/dt =|
y stands for dy/dt I
2 stands for dz/dt =
§ stands for ds/dt =
S§
SS
vz,
v or 0s,
in the x, y, and z direcwhere v;, V,, and v, may be considered as components
ly, two dots over the
tions respectively of the resultant velocity v. Similar
to time; thus
respect
with
n
symbol represent the second differentiatio
% stands for d2a/dt? = dv,/dt = Qz,
gy stands for d*y/dt? = dv,/dt
llsa IS
2 stands for d?z/dt? = dv,/dt
% stands for d?s/dt? = dv;/dt = as,
some acceleration vector a.
where a;, a,, and a, may be components of
use both the differential and
Acceleration is discussed in § 172. We shall
It follows from
one he prefers.
dotted forms, and the reader may adopt the
the defined meaning of the dotted symbol that
i= Mais a
dt
ute opens) is
(a) If the motion of a parachutist (before the parach
171. Examples.
speed 2 sec.
his
is
what
t,
constan
a
is
C
expressed by the relation s = Ct ft., where
15?
=
after he leaves the plane, if C
expression with respect
Since v = ds/dt, we first differentiate the given
Sotution.
to t, and find
26:
Soe
which is equal to (60) (3600) /5280 = 41
Thus, after 2 sec., 0. = (2)(15)(2) = 60 fps,
mph, approximately.
246
PLANE MOTION
[Ch. XIII
(b) A car is moving in such a manner that s = 88t, where s is in feet and ¢ in
seconds.
What is the car’s speed after 10 sec.?
So.tution.
By differentiation, we get
’
§
ds
=
Vs
=
—
dt
= 88, a constant.
Therefore, after 10 sec., the car is moving with a speed of 88 fps (= 60 mph) and it
continues to move with this speed as long as the car’s movement conforms to the
given rule that distance is proportional to the time.
172. Acceleration.
Acceleration a is the time rate of change of velocity.
The velocity may change in either magnitude or direction (or both); hence
the acceleration of a point (or particle) may be due to a change in the magnitude or in the direction, or in both the magnitude and direction, of the velocity.
Moreover, the acceleration too is a vector quantity, since it has both magnitude and sense.
The acceleration a may be constant or variable.
Suppose a point is moving in a straight line (rectilinear motion), so that
there is no change in the sense of the velocity. Then the acceleration is due
to a change in the magnitude of the velocity only. Thus let Av be the change
in speed during a time interval At. The average acceleration during rectilinear motion is a = Av/At. Letting the increments Av and At decrease
until they become infinitesimal, we get the instantaneous acceleration as
eA
yISdy
©)
¢ Sy
~ ae?
which gives the total acceleration only for the case of rectilinear motion.
Observe that since
Se
we find
a
dt’
dim
medL-)
thus, for rectilinear motion,
dy
wed?s
Another useful differential equation for the acceleration
of a point is obtained
by multiplying the numerator and denominator of the
right-hand side of
(a) by ds; that is, for rectilinear motion,
(35)
dvds
Tdi ds
ds dv
dv
di ds =” ds’
me
Vita Ges
If the displacement were represented by some other
symbol, for instance, x
instead of s, (85) would take the form v dv =
a dx. In rectilinear motion,
the resultant acceleration is in the direction of the
straight-line path. However, the acceleration may be either in the sense in
which the point is moving
or in the opposite sense.
When the sense of the acceleration is the same
ag
247
§ 174 ] EXAMPLE
that of the velocity, the velocity is increasing. When the sense of the acceleration is opposite to that of the velocity, the velocity is decreasing. When
a moving point is slowing down, its acceleration is often termed deceleration
or retardation.
Keep in mind that the sign of the acceleration designates the sense of the
acceleration vector. Since we often choose the direction of motion as the positive
direction, a negative acceleration in this event means a slowing down or
retardation.
The most common unit of acceleration in engineering is feet per secondsecond (fps?). Other units sometimes used include feet per minute-minute
(f{pm2), miles per hour-second.
What is the acceleration of the parachutist of § 171(a) 2 sec. after
173. Example.
Assume that he moves in a vertical path.
plane?
the
he leaves
Using the equation given, s = Cé, we obtain
SoruTion.
f=y=—airy =f Ct,
al
ay
DT
tS
@s
(2)(15) =:30iips?,
20
se
and is
when C = 15, as given in §171(a). Note that this is a constant acceleration,
true.
the same for any time ¢ as long as the rule s = C? holds
ac174. Example. A particle is moving with rectilinear motion such that the
(initial
fps
10
is
speed
the
celeration a = 4t fps’. When we start counting time,
traversed?
speed). After 4 sec., what are the speed and the distance
Using equation (34), a = dv/dt, we may write
So.ution.
v
4
/ dv = af t dt,
10
0
where the limits of the integrals are set in accordance
problem.
with the statement of the
:
Integration gives
Ob Sere
LO
2e| =a 32,
0
or
V2 = 42 fps.
indefinite integral and find
To find the distance, first integrate dv = 4t dt as an
p= $s = 20+,
C may be found from the simulwhere C is the constant of integration. The value of
subscript o for v, and t, indicate
The
10.
=
C
or
0;
=
t,.
when
10
taneous values v. =
v = ds/dt in the preceding
values at the origin. Using this value of C = 10 and using
equation, we get
8y
4
4
i n=
2] ra+
10 dt,
0
0
3
0
4
So = = ae oe = 82./.{.
0
248
PLANE MOTION
[Ch. XIII
175. Example. A particle is moving in a straight line so that § = a = 2s. If it
starts from rest, what is its speed after it has moved 10 ft.? What time has elapsed?
Sotution.
Using equation (35), v dv = ads, and a = 2s we get
ja= 2
ie
=
s+
sas,
(,
where C is the constant of integration.
Since the problem states that the particle
starts from rest, »» = 0 when s, = 0. These values in the foregoing equation give
C= 0. The speed after any distance s is traversed is then
OP = OA
After s = 10 ft., the speed is
vist Sx 2)1 1004/9 = 14 foe:
Using v = ds/dt in the previous equation, we find a relation between distance and
time; thus
ds
=>
or
—
dt
ey,
=
Is
"1
10
t
i ee / dt,
Ou
from which
t=
—-
0
= 1.68 sec.,
the time for the particle to move 10 ft. with the motion as defined bya = 2s:
176. Constant Acceleration—Rectilinear Motion.
When we know the
law that governs the acceleration of a particle which moves on
a straight
line, we may use equations (34) and (35) to derive une detailed
relations
between a, v, s, and ¢, as suggested by the examples of §§ 174
and 175. A
special case of importance is that in which the acceleration a is
constant or is
assumed to be constant.
Thus from (34), we have
b
(b)
=>
do
—_—-
or (
=
dv —= adt.
Integrating (b), we find
(c)
$= v= ai -- C3,
where C; is a constant of integration.
Let» = v, When t = 0 and substitute
these values in (c): This substitution gives C, =
v,; hence equation (c)
becomes
(d)
v=v,+
at
or
a=
=,
§ 176 |CONSTANT
ACCELERATION—RECTILINEAR
MOTION
249
where Av is the change in speed (second speed minus the first speed, v — vp)
during the time ¢. Since v = ds/dt, we have, from (d),
fas
(e)
» fart afeat
s =f +S + Ce
If we start measuring space and counting time at the same instant, that is,
if s = 0 when ¢ = 0, we find C2 = 0; in which case, (e) becomes
(f)
+S
s=o¢
Now, if the acceleration a is constant, we find from (35) a useful relation;
[oa =afas
whence
y
3g = as + C3.
space,
If the particle is moving with a velocity , when we start measuring
C3 in
of
value
this
then v = v, when s = 0, from which C3 = »,”/2. Using
the preceding equation, we get
vy?
vee
y2 = 9," - 2as.
or
Gor == ii
(g)
acceleration
It should be kept in mind that these relations hold only if the
If the body starts from rest, v. = 0, and equation (g) becomes
ais constant.
iP SSP
= Las
, v? = 29h,
You are probably familiar with the special case of this equation
h is the
and
gravity
of
tion
that for a falling body, where g is the accelera
other
or
air
height through which the body has fallen from rest (without
resistance) when the speed is v.
instant 4, s is
In the equations (d) and (f), v is the speed of a point at any
speed (the
initial
the
is
0»
¢,
instant
the distance the point has moved at the
the point.
of
tion
accelera
speed at the origin of time), and a is the constant
a after
tion
t accelera
In equation (g), » is the speed of a point with constan
speed of the point is 0.
it has moved a distance s from an origin where the
the displacement, all of
If the acceleration is opposite to the velocity and to
when the sense of the
these equations hold good, but the value of a is negative
velocity is taken as positive.
he must be sure that
If the student uses these equations in any problem,
that is, that the
article,
this
of
the problem falls within the limitations
acceleration is constant,
a =C
=a
constant.
PLANE motion [Ch. XIII
250
177. Uniform Motion. The term uniform motion is applied to a particle
or body which moves with a constant speed, v = C. If v is constant,
dv a
a =
0,
or
a = 0,
the acceleration is zero. This is recognized as the special case of statics, as
far as forces are concerned.
When a = 0, we have v = s/t.
178. Example.
The velocity of a freight train increases at a constant rate from
15 mph to 45 mph in half an hour.
Sotution.
What is its acceleration (fps*)?
Convert mph to fps.
4
5 mph
(45) (5280)
3600
—
fps == 66 fps it
_ (15)(5280)
,
3600
fps = 22 fps.
15 mph =
Therefore, by (d), § 176, the acceleration is
2
where t = 0.5 hr. =
179.
Av
;
66 — 22
1800
=
»
0.0244
fps?,2
1800 sec.
Example.
A body A is projected vertically downward from a 500-ft. cliff
with an initial velocity of 10 fps. One second later, a body B is projected vertically
upward from the bottom of the cliff with an initial velocity of 70 fps. (a) When do
these bodies pass one another?
Time is to be measured from the beginning of the
motion of A. (b) How far above the bottom of the cliff are the bodies when they
pass?
(c) What is the direction of motion and speed of B when they pass?
air resistance.
ta=0
yi
Sotution.
picture
(a) The
of what
N eglect
first step is to get a clear mental
is taking
To do this, make
place.
a
Uownd
sketch to show the pertinent information
Direction
Positive
may choose either sense as positive. In this solution, we
shall let the upward sense be positive. We notice that the
(Fig. 444).
We
bodies, considered as particles, are acted upon by gravity,
so that the acceleration of the bodies is the acceleration of
gravity g = 32.2 fps’. However, since the acceleration of
500’
gravity is downward, a =— g in equations (34) and (35),
if the upward sense is positive. Also, let the origin of space
be at the level of B, which places A at an initial position of
+500 ft. By (34), the equation of motion for either body
Fig. 444,
A or Bis 6 = dv/dt = — g, or dv =—
gdt.
By integration,
we may get the particular relations of s, v, t, and a for each body as follows:
Bopy A
VA
Bopy B
t
/ in=
+10
vB
ofa
,
0
Dh
ae
of a
70
(Initial velocity = — 10)
va +10 =— gi
ess
t
/ v=
10
—gt
1
(Initial », =+ 70, initial t = 1)
ve —-10 =—gt+g
ds
Oe aie
| Oa
La
251
§ 180 |VARIABLE ACCELERATION
i ina tof a—of tuto
1
1
0
(Initial s, = 0)
=
500
—
10
—
a
ll 70t — 70
SB
&
1
Ba pale sas EDues gy g
2
sa
t
t
t
SB
t
t
SA
| iv= —10 fa —o fea
0
0
500
(Initial s, = +500 ft.)
2
102.2 ¢ — r — 86.1
that is,
When the bodies pass, they will be at that instant at the same elevation;
get
we
sp,
and
s.4
of
values
preceding
the
Equating
Sa = Sg.
2
2
500 — 108: — r = Ny
= e = 86.1)
.
from which t = 5.22 sec., the elapsed time after A is projected downward
the expresfrom
found
be
may
B
of
level
the
(b) The distance of the bodies from
find
we
Thus
sec.
5.22
=
¢
ing
substitut
by
sion for either s4 or Sz
- (32.2) (5.22)?
=O,
sa = 500 — (10)(5.22)
2
for sa is positive,
which is also equal to sz at this instant. Note that, since the answer
the same origin
have
B
and
A
(Both
the distance 9.2 ft. is measured upward from B.
meaning besame
the
have
ment
displace
and
of distance. In this instance, distance
.)
direction
vertical
the
in
all
cause the motion is
the body B is
(c) From the preceding expression for vz, the velocity of
ve = 10 —gt
+g = 70 —
(32.2) (5.22) + 32.2 =—
65.9 fps.
downward; B is returning
The negative answer in this case shows that B is moving
moving down, we may
is
B
that
ion
verificat
another
For
from its highest point.
our knowledge that
using
find the time it takes for B to reach its highest point by
vs = 0, we have
Setting
zero.
rily
at the peak of its travel its speed is momenta
vm = 0 = 70 —gi-kg,;
or t = 3.18 sec.
body
Since this time is less than ¢ = 5.22 sec. obtained in part (a),
same height.
B must be moving down when it and A are at the
In the event of variable acceleration, it is
180. Variable Acceleration.
or ¢, as some of the prein general necessary to express @ as a function of s
Such a relation would be of the form
ceding examples have shown.
C= fle
or
a = f(s)
ion v dv = a ds, equation
In the first form, we may use the fundamental express
variables, as
(35), and we obtain a differential equation with two
v dv = f(s) ds.
Ifo = /(0),. the fundamental
take the form
expression dv = a dt, equation
(34), will
dv = f(t) dt.
of acceleration, the solution
Even with a simple law expressing the variation
difficult.
of the resulting differential equation may be
252
PLANE MOTION
[Ch. XIII
181. Example. A particle whose acceleration is a = 3¢ — 12 fps®, is moving at a
certain instant in a straight line with an initial velocity of 15 fps in the same sense
as the initial acceleration.
(a) At the end of 10 sec., what is the velocity of the particle? (b) At the end of 14 sec., what is its displacement from the origin?
SoLution.
(a) At
{= 0, a =— 12 fps. Therefore, the given initial velocity is
v. =—10 fps, that is, it is negative if it is to have the same sense as the acceleration.
Using a = 6 = dv/dt, we get
v
10
/ dv =| (8t — 12)dt,
—15
0
r+15=
38
“an
a
0
[3-21
v=
300
eget
120;
+15 fps.
(b) To find the displacement in a second integration, make the first integration
indefinite. An indefinite integration gives
dx
ro
im
_ 3
Bee
TS -- C,
12t
where the constant of integration C is determined by the conditions that v = v, = — 15
when t= 0. Using these simultaneous values, we find C = —15.
Substituting this
value of C’ in the preceding equation and integrating, we find
x
14
i dx =} (1.542 — 12t — 15)dt
0
0
ee
1.58
122
‘
pee tgRent
5)
ade Sane te
3
;
Assuming that the positive sense is toward the right, the negative sign for
x shows
that the displacement is to the left from the origin.
O
t
(b)
Fig. 445.
182. Graphical
s-t Diagrams.
Representations.
The slope represents velocity.
The
variables
which
appear
in the
motion equations, s, ¢, v, and a, may be combined to
plot curves representing
the relation between any two of them.
However, certain pairs of these var-
lables produce curves of particular usefulness.
advantageous groupings briefly.
(a) Displacement Diagram (s-t curve).
We shall now consider these
A curve
with
the coordinates
s-t, Fig. 445, is sometimes called a displacement diagra
m, inasmuch as the
ordinates s represent displacements.
In some circumstances, it may be con-
§ 182 | GRAPHICAL REPRESENTATIONS
258
venient to measure the distances a body has moved after various times #.
If so, a series of points such as B and C, Fig. 445 (a), may be plotted for
simultaneous values of s and ¢ and a smooth curve drawn through these
The slope of the curve at any point D 1s
points to obtain some curve OA.
ds/dt, which is recognized as the instantaneous speed, v = 8 = ds/dt. Hence,
the steeper the curve, the greater is the speed. This curve near A is horizontal (zero slope), indicating that the velocity is zero (no change in s).
Suppose that the tangent drawn at D, Fig. 445(a), has a slope of
Sl
=
oe
2 in. of distance
:
1 in. of time
The actual velocity is determined from the scales used for s and ¢ in drawing the
curve.
Suppose the scales were
1 in. = 4 sec. (for ¢),
lit, = ZO is, Gor 8).
Then the slope expressed in feet per second (velocity) units is
, = #20) = 2S iOS
(4)
Keeping in mind the significant characteristic of this diagram, namely,
v = ds/dt = slope of the curve, we may easily interpret such a curve as
0-1-2-3, Fig. 445(b). We see that the speed (slope) increases to a maximum
at a point about midway between O and 1, that the speed reaches zero at
the point 1, that no motion occurs from | to 2, and that the speed (slope) is
constant from 2 to 3. The negative slope from 2 to 3 indicates that the
velocity vector is pointing in the negative sense.
On occasion, we may be able to plot an s-t curve, or one of the other
curves, when a mathematical relationship is impossible or very difficult to
In cases of this nature, the graphical procedure readily yields
obtain.
important information on the motion of a point.
(b) The Velocity-Time Curve. ‘The curve OB, Fig. 446, defines a relation
the
between v and t, v = f(é). At any point D, we observe that the slope of
Knowdv/dt.
=
6
=
a
on
accelerati
curve is dv/dt, which is recognized as the
on.
ing the scales used for v and t, we may convert the actual slope to accelerati
curve are
Suppose the slope at D is 0.8 and the scales used in plotting the
1 in. = 20 fps (for v),
in’ ="5 sec(for't);
then
20 ft.
Speed scale\
QS
=
seat)
sec-iny)
_
pLobe
5 sec.
in.
\80(0.8)(20)
5
oe
pote
eo
254
PLANE MOTION
[Ch. XIII
When the slope is negative, the speed is decreasing and the body or particle
is decelerating.
1
Another significant characteristic of the v-t
B_
diagram is that the area ‘under’ the curve
represents to scale the distance traversed by
the point in question.
Observe that the area
t of the differential element at H, Fig. 446, is
oO
eo:
v dt; but since v = ds/dt, we have v dt = ds.
Fig.
446.
slope
represents
The
area
v-t
Diagram.
The
Thus
acceleration.
represents
displace-
/[pen
=
ik
v dt,
ment.
which is the area “under” the curve when the correct limits of the integral
are used. Suppose we wish to know the distance traversed between 1 and 2,
Fig. 446, then the integration is made from ¢; to tz. If the curve OB is obtained from experimental data and v dt is difficult to integrate, the area under
the curve between 1 and 2 may be found by a planimeter or other means.
This area in square inches, say, may be converted to distance when the scales
are known.
Thus, if the area under 1-2 is 0.6 sq. in. and the scales are as
before, we have
(Area in sq. in.)(Speed scale)(Time scale) = Distance
(0.6 in2) x (oC )x =sec.)
sec-1n.
Fig. 447.
v-t Diagram,
tion Constant.
Accelera-
Ta
60 ft.
Fig. 448.
a-t Diagram.
represents speed.
The area
In Fig. 447, the curve n2 represents a motion with constant acceleration
(constant slope). When ¢ = 0, the ordinate is some finite value Vo, Which
is the initial speed. At the point 2, the ordinate is v2, the speed of the
point
after the elapsed time ¢. The slope of the line n2, which is equal
to the
acceleration, is
Ve — Uo
eT ea
Compare
this expression
or
with equation
v2 =v, + at.
(d), § 176, and observe
how
the
§ 182 ] GRAPHICAL REPRESENTATIONS
255
equations of § 176 may be obtained from graphical representations.
To get
equation (f), § 176 (the expression for s the distance moved), find the area
under n2, Fig. 447. We see that this area is equal to a rectangular area v,t
plus a triangular area ($)(v2 — v,)t. The whole area and the distance is
given by
s =v
—
a
2
¢ + (95,5 Sea
at Volt + easeeetayes
ma ;
aaa
2
where we have used a = (v2 — v,)/t. Compare with equation (f).
(c) The Acceleration-Time Curve.
The significant characteristic of the
a-t curve, Fig. 448, is that areas “under’’ the curve represent speed. To
prove this, set up the shaded differential element, whose area is a dt. Since
a = dv/dt, we have adt = dv. Hence, fadt, which represents the area, is
equal to f dv, which is the change of speed. For example, the area under
the curve from point 7 to point 8, Fig. 448, represents the change of speed
between the times ¢7 and ts. The scale of the area is the product of the
acceleration scale and the time scale.
0
O
j
G
Fig. 449.
(d) Velocity Diagrams.
B
s
v-s (Velocity) Diagram.
In studying the motion
of machine
parts, we
For
often construct a so-called velocity diagram, which is a v-s curve.
example, the curve OpA of Fig. 449 represents the relation between the
The
velocity and displacement of a shaper cutting tool on the return stroke.
the
of
stroke
the
of
length
the
represent
to
scale
to
out
laid
distance OA is
are
curve)
the
is,
(that
stroke
the
of
tool. The velocities at various points
found easiest by graphical means (as explained in books on kinematics and
This diagram is used principally to study the variation of
mechanism).
velocity, but the acceleration at any point p may be found from it by using
the known relation v dv = ads.
Tangent and vertical lines drawn at p form the triangle pBC. Since this
triangle is similar to the one whose legs are dv/ds, Fig. 449, we may find
Knowing this ratio, we may multiply it by the actual
dv/ds from pB/BC.
velocity at p to get the acceleration at p. Thus suppose v, = 100 fpm and
PLANE MOTION
256
[Ch. XIII
Then (9 in. = 0.75 ft.
BC = Qin. (These are actual values for the machine.)
and dv/ds = 100/0.75)
i
(a) = 13,333 fpm?,
or converting the time unit to seconds, a, = 3.7 fps?. Observe that at the
highest point of the curve, the slope (acceleration, too) is zero. To the
right of the highest point, the slope (acceleration) is negative and the velocity
is decreasing.
In connection with the foregoing diagrams, it is worth while noting that,
if the displacement s-¢ diagram of the rectilinear motion of a point is known,
the other diagrams, for v-t and a-t, may be drawn.
The procedure is to
measure carefully the slope (speed) of the s-t curve at a series of points, and
plot this slope against the corresponding time. A smooth curve through the
points obtained gives the v-t curve.
Then plotting slopes (accelerations) of
the v-t curve against time will give the a-¢ diagram.
183. Angular Velocity. If a body is rotating about either a fixed or a
moving axis, it is said to have angular velocity. The angular velocity is
measured by the time rate of change of angular displacement of a line in the
body.
Suppose the body B (not a point or a particle), Fig. 450, turns about
an axis O so that the line OA takes the position OA’. The resulting angular
displacement of the body is the angle A@ between the two positions of the
line, OA and OA’.
Now suppose that this rotation through
the angle A@ occurred during a time interval of At. Then the average angular
velocity w of the body B is 6=w=
Aé/At.
Letting the increments A@ and
At become progressively smaller, we find
;
Fig.
0
36
450.
.
0 =
>
een
=
PACTS
41710
ee ee
a
arr
the instantaneous angular velocity of the body. The usual units of angular
velocity are: radians per second, radians per minute, revolutions per minute
(rpm), and revolutions per second (rps).
184. Relations between Angular and Linear Speeds.
Observe in Fig. 450
that the distance moved by point A along the are AA’ iss =r Aé, where A@
is measured in radians. In terms of differentials, we have ds = r dé.
Using
this value of ds in equation (33), we get
(87A)
Y Sygate
)
18 fps
or
v= Tw,
§ 185 |]ANGULAR ACCELERATION
257
where r is in feet and w is in radians per second in order to give v = 2 fps.
Since there are 27 radians in one revolution, w = 27s, where n; is in
revolutions per second; or w = 27n» radians per minute, where n» is in
revolutions per minute.
(37B)
(87C)
Up = 2a elPI,
Um = TDnrm fpm,
and
and
Vs = 2mrn, fps
v, = aDn, {ps
where r and D are expressed in feet. Equations (37) are important relations
between the linear and angular speeds. As we see from these equations,
the linear speeds of any two points in a rotating rigid body are directly
proportional to their distances from the center of rotation of the body,
that is
Yap sb
UB
rR
where vy is the speed of point A, vz is the speed of point B, ra is the radius
of A, and rz is the radius of B.
Using a definition analogous to that given
185. Angular Acceleration.
say that angular acceleration a is the time
may
we
*
for linear acceleration,
During a time interval At, let the change
rate of change of angular velocity.
of angular velocity be Aw. Then, as before, the average angular acceleration
is Aw/At; and the instantaneous angular acceleration is (a = w= 6)
= tim 22
ea 0Ab a
Since w = d6/dt, we have dw/dt = d0/dt?, and therefore
x
6=a=
(88)
dw
di
d’0
de
By multiplying the numerator and denominator of the middle term of (38)
by dé, we get,
(39)
dwd@
CEST
dédw _ dw
TITY Baan
or
a dé = w dw.
a
The equations (38) and (39) are basic for analyzing the angular motion of
or
body which turns about an axis, whether the angular acceleration is constant
on
accelerati
angular
an
or
velocity
angular
variable. The positive sense of an
When
may be chosen arbitrarily to be either clockwise or counterclockwise.
w, the
the angular acceleration a is in the same sense as the angular velocity
When the sense of a is opposite to that of w,
angular velocity is increasing.
Angular acceleration is usually expressed in radians
then w is decreasing.
per second-second (rad. per sec.”).
on or linear velocity
*When the words acceleration or velocity appear alone, linear accelerati
is used.
angular
adjective
the
to,
referred
is
motion
angular
When
is understood.
258
PLANE MOTION
[Ch XIII
186. Examples.
(a) A body starting from rest rotates counterclockwise according
to the law o = 0.148 — 0.3 + 0.8¢. After 6 sec., what is (a) the angular displacement,
(b) the angular velocity, and (c) the angular acceleration?
So.tution.
(a) The angular displacement is obtained directly from the given law
by substituting ¢ = 6:
6 = (0.1)(6)* — (0.3)(6)? + (0.8)(6) = 15.6 rad. = 2°-8 rev. = 2.48 rev.
T
(b) Differentiating 6 with respect to t [equation (36)], we get
ap
ee
SE
Ee oar So eenoe:
After 6 sec., the instantaneous angular velocity is
w = (0.3)(6)? — (0.6)(6) + 0.8 = 8 rad. per sec.,
which is equivalent to (8)(60)/(27) = 76.4 rpm.
(c) Differentiating w with respect to ¢ [equation (38)], we get
d(0.3 — 0.6¢ + 0.8)
q = dw
— = —____
=
dt
dt
Observe that the acceleration is not constant.
0 6 — 0.6.
.
After 6 sec., the instantaneous angular
acceleration is
a = » = (0.6)(6) — 0.6 = 3 rad. per sec.
The reader should note the characteristics of this motion as revealed by the equations.
For example, the origin of displacement is the same as the origin of time (@ = 0 when
t = 0). Moreover, the value of w never becomes zero, yet the angular acceleration
has a negative value at first, is zero at t = 1 sec., and is positive for all values of ¢
greater than 1.
(b) A body is rotating about a fixed axis so that its angular acceleration is a = 6 =
4? —t-+ 4 rad. per sec.” If the initial angular velocity is 10 rad. per sec., what are
the angular velocity and angular displacement after 2 sec.?
Soturion.
Using (38), a = dw/dt, we get
feo=fae-e4 oa
48
(Os
3
= at
4t +
Ge
The constant of integration C is found from the simultaneous values w = w
= 10
when ¢ = 0; which gives C = 10. Then after 2 sec.,
= oe - ;+ (4)(2) + 10 = 26.7 rad. per sec.
Letting » = dé/dt, we may integrate for ¢@ as follows,
feBER
cnereide
\
=
E + ie + ior= 32 1g rad6
ma Cio
which is equivalent to 32/(2r) = 5.1 rev.
:
a
§ 188 |EXAMPLE
259
187. Constant Angular Acceleration.
When the law of variation of the
angular acceleration can be expressed as a function of ¢ or 6, equations (38)
or (39) may be used to derive detailed relations between a, w, 6, and t, as
suggested by § 186. In the special case of constant angular acceleration, we
find forms of the equations analogous to those obtained for constant linear
In order to benefit by this analogy, the reader should review
acceleration.
§ 176 while studying the following derivations and observe the similarities.
With a constant, we get, from (88)
[os = a |
at,
whence
wo=at+Ci,
Ifw = w, whent = 0, we find Ci = a,,
where C; is a constant of integration.
and the preceding equation becomes (w = 6)
(Compare with v = v, + at)
wo = w, + at.
(h)
Since w = d6/dt, we find, from (h),
2
whence
[ao =cofartafea
O=
ot
+S +r
If 6 = 0 when ¢ = 0, then C, = 0, and this equation becomes
2
6=a¢
(i)
+ “
2
(Compare with s = v,¢ + x )
If a is a constant in (39), we have
iEdw = a
ae,
whence
2
s = a + C3.
If the body is moving with an angular velocity #, when 6 = 0, we get C3
= w,2/2. This value of C; in the previous equation gives
(j)
w? = w,?
+ 2a6.
(Compare with v? = v,” + 2as)
In this equation, the origin of time and angular displacement occur simulAt this same instant, however, the
@=0.
taneously, that is, when ¢ = 0,
Ifw, = 0, this term simply drops from the equations.
angular velocity is w.
If the sense of a is opposite to that of w, its value is negative if the sense of w
is taken as positive. A quick and easy procedure is to set up the differential
equation, (38) or (39), to fit a particular problem and integrate for the
desired results.
A wheel which is rotating 300 rpm is slowing down at the rate of
188. Example.
2 rad. per sec.2 (a) What time will elapse before the wheel stops? (b) At what
rate in rpm is the wheel revolving after 10 sec.? (c) Through how many revolutions
has it turned during the first 10 sec.? (d) What is the total angular displacement?
(e) Compute the number of revolutions from the time t = 10sec. until the wheel stops.
260
PLANE MOTION
[Ch. XIII
Sontution.
(a) As sketched in Fig. 451, the sense of a is opposite to that of w;
therefore, if the direction of motion of the wheel is taken as positive, a = — 2 rad.
per sec.2. Moreover, w. = (300)(2r)/60 = 107 rad. per sec. Using equation (38),
we get
0
_&~ 2 rad./sec?
if
whence
Ax
-——(@)}-
t
iHdw = — 2fdt,
==]
ort6, Sa;
or
t = 15.7 sec.,
the elapsed time when the wheel has stopped.
\\
(b) In making an integration, the limits are always
from the first condition to the second condition.
When
we begin counting time in this example (when ¢t = 0),
w, = 107 rad. per sec. The second limits are then » = w»
Fig. 451.
and t2 = 10 sec.
w2
10
/ day =
—
2f dt,
10 7
we —
Thus, we find w2 from (38) as follows:
0
107 =
(— 2)(10) =
— 20,
we = 31.4 — 20 = 11.4 rad. per sec.,
which is equivalent to (11.4)(60)/(2r) = 109 rpm, approximately,
the speed after
10 sec.
(c) To get the angular displacement after 10 sec., we may integrate
11.4
P)
‘i ody
=
—
2f dé,
107
0
where the second limit for w (= 11.4) is the angular velocity after 10 sec.
method of obtaining @ is to integrate
equation (h), § 187.
Another
Using the latter method, we have
6
10
"10
/ do ll 10 | dian 2 t dl,
0
0
ll
0
100
(10 r)(10) — ca/ )= 214 rad.
2
In order to convert to revolutions, divide by 2m rad. per rev., or 6 = 214/(27r)
=
34.06 rev.
(d) The total angular displacement may be obtained either from equation
(h) or
from w dw = ado.
Since we found in part (a) that it took 15.7 sec. for the wheel to
come to rest, equation (h) integrated from 0 to 15.7 sec. with respect
to time will
give the desired 0; thus
0
i do
0
15.7
aoe |
0
15.7
dt
rE
af
0
t dt.
§ 190 |TANGENTIAL AND NORMAL ACCELERATIONS
261
The other procedure, to use equation (39), is somewhat simpler. The limits of w
are from 10z, the original angular velocity, to 0; the corresponding limits for @ are
0 to 6; hence
0
/
0
w da
2 af dé,
10 7
_ 1007°
2
0
=
— 29
or
() = May ial.
(e) The angular displacement from ¢ = 10 sec. until the wheel stops is found as
247 — 214 = 33 rad., or (33)/(27) = 5.25 rev. during 5.7 sec. This result may be
obtained also, within slide-rule error, by the integration of the equation
0
i
6
oda
11.4
=
—
[ de
0
for the value of 6, where 11.4 = w at the instant when ¢ = 10 sec.
189. Curvilinear Motion.
A point moving in the path of a plane curve
which is not a straight line is said to have curvilinear motion.
Since the
velocity of the point at any instant is in the direction of the displacement at
that instant and since the direction of the displacement at a point on a curve
is the direction of the tangent at that point, the velocity of a point in a
curved path is in the direction of a tangent to the path at the instantaneous
position of the point. In short, the velocity vector is always tangent to the
path of the point.
A point may move in a curved path with a constant speed, but inevitably
its velocity varies because the direction in which the point travels varies.
This variation of the direction of the velocity results in an acceleration, even
with constant speed—an acceleration which we term the normal acceleration.
190. Tangential and Normal Accelerations.
Let a point move in the
curved path shown in Fig. 452.
Let the two positions A and B, defined by
radius vectors pa and pz and the angles 6 and 6 + A@, be a distance As apart.
Let the speed of the point at A be v, and at B, v + Av. And let the time
taken by the point to move from A to B be At. Remembering that the
change of velocity is that vector which when added to the first velocity gives
the second velocity, we may construct the vector diagram in Fig. 452(b).
Thus, with a change of velocity equal to the vector HF, the acceleration is
EF /At, since this change HF occurs in the time At.
acceleration is
Hence, the instantaneous
However, to evaluate this limit, we find it convenient to use the components
EG and GF of the vector EF. Substituting HG 4 GF for EF above, we get
(k)
EO
iy me
i =
Yh
peal
Cae
aay
PLANE MOTION
262
[Ch. XIII
In Fig. 452(b), we observe that, in the limit with A¢ = 0, the component
EG is (v + Av) — v = Av, whieh is the change in the speed. Therefore, the
first term of equation (k) becomes
Ge Avo
times
At
eat?
which is recognized as the time rate of change of the speed, an acceleration
which is directed tangent to the curved path. This acceleration due to a
change of the magnitude of the velocity is called the tangential acceleration
a,, and it is a component of the resultant or absolute acceleration.
(b)
Fig. 452. Normal Acceleration.
Point C is the center of curvature of an arcAs.
limit as Af approaches 0, point C is the center of curvature at point A.
The other rectangular component,
be transformed as follows.
We find
In the
the second term in equation (k), may
GF = (v + Av)sin Ad
from the triangle DGF.
Thus, the average acceleration due to the component GF of the velocity change EF is
(1)
(v + Av)sin Ad
Ai
The following conditions apply as At approaches zero:
The point B approaches point A.
The direction of GF is normal to the curve at A (giving rise to the name
normal acceleration a, for this component).
Av approaches zero, so that v + Av approaches v.
The sin Ag approaches A¢.
§ 191 | EXAMPLE
268
With these conditions in mind, multiply and divide expression (1) by r = AC,
the radius of curvature at the point A (Fig. 452) and then transform the
expression as follows:
Oi,
i
=
\hien
r(v + Av)sin Ad _
At— 0
r At
2 TR ee ee
AtSO) AE
kim 2 Ad
at>o
atedl.
7 At
er”
since ds/dt = v, which is the speed of the point at the position and instant
in question.
If either the speed v of the point or the radius of curvature r
of the path of the point varies, this value of the normal acceleration a, is an
instantaneous value.
Since, by (387), v = rw, we may write a, in the form
y?
(40)
lore
or rw? = vw,
in which w is the instantaneous value of the angular velocity of a line connecting the position of the moving point with the center of curvature at that
position. For the case of Fig. 452, » is the instantaneous angular velocity
of the line CA. As previously mentioned, the vector for a, is directed
normal to the curve toward the center of curvature.
For a point in curvilinear motion, then, the magnitude of the total acceleration (resultant acceleration) may be found from its two rectangular com-
ponents, in the form (curved path is stationary)
a= (a2 + 4,2)",
(m)
in which a; is zero if the speed of the point is constant and a, is zero only for
When the
rectilinear motion or as the velocity passes through a zero value.
radius of curvative r is constant (circle), we get, from a, = dv/dt and v = ra,
(41)
dv _
d(rw) _
dw
OeeceaiegyMERE Yippee
= ra
or
a, = ra,
where a is the instantaneous angular acceleration of a radius to the point
whose tangential acceleration is a; The resultant or total acceleration may
also be defined by the vector sum
a=
aHQ,
where a is the resultant of the vectors for a; and a,.
In the wheel of § 188, a point A is located on the horizontal center191. Example.
What is the
line (position A1, Fig. 453) at the instant the wheel begins to slow down.
sec.?
10
after
point
this
of
total acceleration
The angular displacement after 10 sec. as obtained from § 188 is
Sotution.
9 = 214 rad. = 34.06 rev. Therefore, after 10 sec., the wheel has made 34 complete
[Ch. XIII
PLANE MOTION
264
revolutions plus 0.06 of a revolution. This puts the second position of A at A»,
Fig. 453, which is (0.06)(360°) = 21.6° above position A1. Since the radius of the
wheel is 15 in. = 1.25 ft., we have
OS
fee = (UA),
SS 725) ips
directed oppositely to the rotation of the wheel, because the wheel is slowing down
and we are considering the direction of motion as positive. At this instant, » = 11.4
rad. per sec. (§ 188); hence,
Cpe
— (led) (li)
directed toward the center of the wheel.
2a Oe
pups.
The total acceleration is
a = (2.5)? -F 162.5)72" = 162.52 fps?,
a value which is of course very nearly equal to ap, since a, is very large compared to a,
in this instance. The angle that the resultant vector makes with the vector a, is
2) == tan -1 “
-
=
tan -1 2.5
162.5
—
she)
0.88°.
The total acceleration is therefore directed downward toward the left at an angle of
@ = 21.6° + 0.88° = 22.48° with the horizontal, as shown exaggerated in Fig. 453.
. ale
oc=2 rad./sec?
&
v*=2gh (free fall)
Fig. 453:
v2=2gh
Fig. 454,
192. Component Motion.
Since displacement, velocity, and acceleration
are vector quantities, they may be resolved into any convenient components.
The
components
a, and
a, are examples
of such
convenient
components.
Frequently, rectangular components in certain x and y directions are desired.
Thus, let a, represent the component of the acceleration, v; the component
of the velocity, and let x represent the component of the displacement, each
in the x direction.
Since the x axis is a straight line, this component of the
motion is necessarily rectilinear, no matter what is happening to the actual
point. Hence, the equations for rectilinear motion apply. We have
GES
x
ce
dt?
eve triete
era a ORE Es
tet
dt
rae
Oe
dt?’
Vy
d
Ve
=,
=
A,
AX.
Similarly, the equations for the motion in the y-direction are
i
ney
ase Tap
yas
t
Ue
dv
ad
Tt re ae
vy dvy = a, dy.
§ 193 |EXAMPLE
265
The magnitude of the resultant velocity at a particular instant is obtained
from the relation
v= (vet + 0/2),
where v, and v, are simultaneous values.
vector v and the vector v; 1s
6 = tan
The angle between the resultant
5)
me
-!—
= tan ne
ve
x
after the manner of getting the sense of a resultant force vector, equation
(c), p. 19. The magnitude of the resultant acceleration is
a=
(a,?> + Op es
where a, and a, are simultaneous values.
is found from
data
The direction of the vector for a
1 dG,
Dy
The total acceleration is also defined by the vector sum
@ =a,
ay
193. Example.
A body M, considered as a particle, moves from rest down a
smooth inclined plane (Fig. 454). If it moves solely under the influence of gravity and
the normal reaction on the plane, what is its speed after it has moved from A to B?
So.tution.
If the body M fell freely under the action of gravity alone, its acceleration would be g = 32.2 fps? vertically downward.
Therefore, if it is constrained to
move along a smooth plane, its acceleration would be that component of the acceleration of gravity in the direction of the plane; in this case a, = g sin 6, where the x axis
is parallel to the plane. Use the equation v,dvz = a, dx and integrate:
v,
L
ibVz dv; = g sin 6 i dx,
0
0
Bae
:
ail g(sin 6)L;
or, since sin 9 = h/L,
v2 = gh,
(n)
from which the speed at B may be determined.
If this body should fall freely through
a vertical distance h from point A, the speed would be obtained from the equation*
vy
h
ifvy dvy =
0
of dy,
0
whence
(0)
v,? = 2gh.
Since the values of v, and v, in equations (n) and (0) are identical, we conclude that the
speed attained by a body sliding a distance L down an inclined plane without friction of
any kind is the same as the speed attained by the body in falling without friction (and
*Note that in this example the z and y axes are not perpendicular to each other.
PLANE MOTION
266
[Ch. XIII
without air resistance) through the vertical component h of that distance. It is also
true that the speed attained by a body moving downward along any other smooth
path is the same as that which would be attained in a free fall through the vertical
distance defined by the path. However, the velocity is not necessarily the same in
each case. Ina free fall, the velocity is directed vertically downward; the velocity
of a particle as it leaves a sloping or curved path is in the direction of the tangent to
the path at the point B of departure (Fig. 454).
194, Example—Trajectory.
Investigate the motion of a projectile (considered
(This analysis neglects the effects of the rifling of the
as a particle) in a vacuum.
The initial velocity of the
gun barrel, wind movement, and the wind resistance.)
projectile is v,, directed upward at an angle @ with the horizontal (Fig. 455).
Sotution.
(a) The origin of coordinates is taken at the point where the projectile
leaves the gun, at which point the time tis zero. Since the projectile is subjected only
to the acceleration of gravity, the acceleration in the horizontal direction isa, =
= 0,
Fig. 455.
Trajectory.
and that in the vertical direction is a, = y = — g, the negative sign indicating that
the acceleration is downward (the upward sense being chosen as positive). Investigating the horizontal motion first, we have, using equation (34),
_
dds
cs
dt
Since the differential of v, is zero, vz is a constant which is equal to the horizontal
component of the initial velocity v.. See Fig. 456. Thus
(p)
Vz = Voz = Vo COS 8,
where @ is a constant for a particular firing of the gun.
dx
Since v, = dx/dt, we have
Vo COS 6,
[aw= Uo COS ofa
R |=
(v, cos 6)t + Cj,
where C; is the constant of integration.
Since
x = 0 when t = 0, then C; = 0 ; thus
(q)
Gi" (VS COSLA)E
This equation shows that the horizontal displacement is directly proportional to the
time ¢. For the motion in the vertical direction, we have
pe Plante
MaDeAine
fou=-ofa,
y=
—gt+C,,
267
§ 194 ] EXAMPLE—TRAJECTORY
where C2 is the constant of integration. When t = 0, vy = Vy = % sin 6, the vertical
component of the initial velocity; so C2. = v, sin @. Thus
=
— gt + v, sin 6,
a
Up
(r)
=| = oftai +e sins fa
=
C=
(apr
iar + (vo sin 6)t + C3.
Since y = 0 when ¢ = 0, the constant of integration C3 = 0; therefore,
gt?
:
bes + (v, sin 6).
ee
(s)
Equation (r) gives the vertical component
of the velocity of the projectile at any
At first, ». sin @ is
time ¢ after firing.
this period the
During
ereater than gt.
value of v, is positive, showing that the
vector points upward. Later, gt becomes
ereater than v, sin 6, in which event, the
value of v, is negative, showing that the
Evidently,
vector points downward.
then, the path of the projectile (the trajectory) moves upward from the origin
(when 0 < @ < 90°) and then turns downward (Fig. 456). In equation (s), when
gt?/2 becomes greater than (v. sin @)t, the
value of y is negative and the projectile
has passed the point B.
(b) The equation (t) of the trajectory
OAB, Fig. 455, in terms of x and y is obtained by eliminating ¢ from equations
(q) and (s); this gives
hp
(t)
Joe
Ry Secret anieame:
2 v,? cos? 6
Reproduced by permission from Mechanics,
Fig. 456. Multiflash Photograph of Projectile. The white dots are photographs of
a projectile (a small ball) taken by high-
speed flash photography. The time interval
between flashes is constant. Notice that the
horizontal distances between adjacent positions of the projectile are all the same. The
vertical distances between successive positions decreases as the projectile moves upward, increases as it moves downward—
the acceleration is constant downward. Even
though the target may be at E, where y has
a negative value, the range is defined by L.
The maximum value of y is Ymax = A
This is an equation for a parabola.
(c) The time of flight t; is by definition
the time for the projectile to return to the
is that
level of the origin. The condition here (at B, Fig. 456)
get
we
zero,
to
equal
(s)
n
in equatio
y
=e
by
F. W. Sears,-published by Addison-Wesley Press.
y = 0.
Letting y
g? + (v, sin 6)t = 0,
9
whence the time of flight is
(u)
healee
2v, sin 6
Figs. 455 and 456,
(d) The range L is by definition the horizontal distance OB,
beyond point B.
or
of
ahead
although the projectile might actually strike a target
PLANE MOTION
268
[Ch. XIII
The range (x = L) may be determined by letting y = 0 in equation (t); or by substituting the value of t; from (u) in (q). The latter procedure gives
=I
= On COO
2v,2 sin 6 COs 6
=
g
For a particular initial velocity v,, the maximum
sin 20 is a maximum.
(e) The maximum
Vo? sin 26
range JL,,,, 1s obtained when the
The corresponding value of 6 is 45°.
height is the maximum value of y. This value of y may be
obtained by letting x = L/2 in equation (t), or by letting ¢ = t;/2 = v, sin 6/g in
equation (s). Using the latter plan, we find
g(Uo sin 6)?
Daa
=
(v. sin 6)? _
Vo? Sin? 6
29
29?
See also Fig. 457.
195. Example.
A moving point
follows the path of the hyperbola
zy = 16.
The y component of the
velocity is constant at the value
vy = dy/dt = 6 fps. At the point
(8, 2) on this curve, determine (a)
the tangential (absolute) speed, and
(b) the x component of the acceleration a;.
Reproduced by permission from Mechanics, by F. W. Sears,
published by Addison- Wesley Press.
Fig. 457.
Multiflash Photograph of a Ball.
The ball starts from rest at A and flash pictures
are taken at constant time intervals. The ball
accelerates as it rolls down the incline AB,
moves with constant speed on the horizontal
from B to C (notice that the distance between
positions remains constant), then it moves as
a projectile from C to D. The path from C to D
is a parabola for a projectile given an initial
velocity in a horizontal
direction
(0 =0, Fig.
456), and the equations for a projectile apply to
SoLution.
(a) Differentiate
= 16 with respect to ¢.
ry
Since y = dy/dt = 6, this equation
becomes
(v)
y= cb Ge = 0.
When x = 8 and y = 2, we get
this part of the movement.
Ye
le:
Ae
eS
dt
y
24
fg
0 = (v2 + v,7)N® = [(—24)? + (6)2]"/2 = 24.75 fps,
the speed at the point (8, 2). With v, = —24 and », = 6, the vector v ev idently
points upward toward the left at an angle ¢ = tan“(v,/vz) with the horizontal
.
(b) Differentiating equation (v) with respect to ¢, we find
dy \f da
Px
i
(2\(«) ee
6
dx
dt
dy
tT Ue
dt
dx
ae6 it
§ 196 |SIMPLE HARMONIC MOTION
269
Since dz/dt = — 6 2x/y, we get
Ax
me eR
ices)
ay B72) (8)
:
wine
od 144 fps?,
directed toward the right, since the sign is positive.
Since y = 6 fps is a constant,
y =a, = 0:
196. Simple Harmonic Motion.
One of the most important examples of
variable acceleration in engineering problems is simple harmonic motion,
because this motion, which is vibratory or periodic, is the starting point in
the study of vibrations (Chapter XX). Since the modern tendency is to
run machines at higher and higher speeds, the study of vibrations was never
more important than it is now.
A point describes simple harmonic motion
when its acceleration varies directly as the displacement of the point from
an origin, but is in the sense opposite to that of the displacement.
That is,
when the displacement is positive, the acceleration is negative, and vice
versa.
In mathematical form, the definition is
(42)
a=
eerie
shi ina
—Cx
or
d*x
dé
=
—Cx.
The physical significance of this equation is shown in Fig. 458. As the
point P moves along Ox away from the origin O toward the right, it is slowing
down, because the acceleration is toward
the left. It will come to rest at some
position A and then begin to move
toward the left, gaining speed until it
passes the origin O. On the left side of
O, the acceleration is toward the right;
so that the particle again slows down
until it stops at some point B. Since
the acceleration is still toward the right,
the point P now moves toward the right
with increasing speed until the origin O
is reached, after which it is again reFig. 458.
Harmonic Motion.
tarded.
Thus, the motion occurs enEquations for harmonic
tirely between some definite limits A and.
motion may be found by integrating the equation v dy = —Cux dz, followed
However, we can
by another integration of the expression for v = dz/dt.
find by simpler means a solution to the differential equation (42).
In Fig. 458, imagine a point Q moving at constant speed on the circum-
We
ference of the circle whose center is at O and whose radius r is OA.
of
diameter
a
on
Q
pot
this
shall find that the motion of the projection of
Let ¢ = 0 when the point P (and Q)
the circle is simple harmonic motion.
isat A. The point A then is the origin of téme, a condition not to be over-
PLANE MOTION
270
When
looked.
the radius to the point Q makes
displacement of the point P is x =r
cos 6.
[Ch. XIII
an angle @ with Oz, the
If w is the constant angular
velocity of the radial line to the point Q, and if t = 0 at A, then 6 = ot.
Thus we have
(w)
«x =rcosat
(x)
p=
= SF= ~orsinat
4. .
wie
ahs
Catia
aioe
or
(Q)
4
— wr Cos wt,
dx
Sy
= er,5
where v and a are, respectively, the
vélocity and acceleration of the point
P. Since, by successive differentiations
Fig. 458.
Repeated.
of x =r cos wt, we arrived at the differential equation d?x/dt?? = —w*x, it follows that x =r cos wf is a solution of this equation.
Moreover, comparing equations (42) and (y), we
observe that they are the same in form.
Hence, if C = w*, equation (42)
defines the motion of the projection P of the point Q, so that this projected
motion must be harmonic.
The maximum distance reached by the oscillating point P from the origin
O is called the amplitude (distance r = OA, Fig. 458). The period T of
harmonic motion is the time taken by the point P to complete a cycle or
oscillation; for example, the time to make the two strokes AB and BA,
Fig. 458. If » is the angular velocity of the radius vector OQ in radians per
second, the period is
(z)
(PS
rad. per
rev. (cycl
den recy ae)
rad. per sec.
or
2
T = =" see. per cycle.
w
Since w = C'/?, the period is also J’ = 27/C'/?, where C is defined by equa-
tion (42).
The frequency ¢ is the number of cycles per second (or other unit of time)
and is therefore seen to be the reciprocal of the period; or
Wie
re
Ce
20
Observe that the acceleration is a maximum when the velocity is zero; that
is, although the velocity is zero at points A and B, Fig. 458, the time rate of
change of velocity at these points is a maximum.
Chapter XX gives a
more extended discussion of harmonic motion.
197. Example.
A point moves according to the law a = —16 2 with an amplitude
of 3 in. (a) Find the period and frequency.
(b) Determine the displacement,
velocity, and acceleration after 10 sec.
§ 198 |INSTANTANEOUS
CENTER
271
OR CENTRO
(a) From the law of motion, C = w? = 16, whence w = 4 rad. per sec.
Sotution.
Thus the period and frequency are
»)
if = 7 = ilo Sees,
¢=
ee = 0.637 cycle per sec.
iL530/
(b) After 10 sec., the number of cycles is (10)(0.637) = 6.37 cycles. Hence, the
point has completed 0.37 of a cycle from point A, the origin of time. The corresponding time is (0.37)(1.57) = 0.581 sec. In order that the angle ot be less than 360°, we
thus use tf= 0.581 sec., instead of 10 sec., in the equations of harmonic motion.
angle wt is then (4)(0.581) = 2.32 rad. = 133°.
The
The displacement is
xz = 7 cos wt = 3 cos 133° = — 2.045 in.,
where the negative sign shows that the point is on the left side of the origin of displacement.
The velocity is
y = — rw sin wt = — (3)(4)sin 133° = — 8.78 in. per sec.,
where the negative sign shows that the point is moving toward the left. The acceleration is
@ = — rw? cos wt = — (3)(16)cos 188° = + 32.75 in. per sec.”,
where the positive sign shows that the acceleration is directed toward the right.
The student should practice the derivation of equations (x) and (y) in solving
this example.
If a body moves so that each
198. Instantaneous Center or Centro.
particle of the body remains in a particular plane, the body is said to have
No matter how complicated the plane motion of a body
plane motion.
may be, the body is rotating at any particular
instant about some axis. Imagine that a thin plate
M, Fig. 459, is tossed vertically into the air. ==Lhe
plate will have a vertical translation as a whole,
but it is likely to be spinning about some axis perThus, a
pendicular to the plate at the same time.
particular point in the plate will have an undefinable curvilinear motion that in general is not the
same as the curvilinear motion of some other point.
Suppose that Fig. 459 is the position of the plate
Fig. 459. Centro.
at a particular instant. Suppose that by some means
we had learned that a point A was moving in
the direction indicated by the arrow at A; and that the point B was moving
in the direction shown by the arrow at B. Recalling that the velocity vector
the curve
is tangent to the curve described by a point and that a normal to
line
passes through the center of curvature and is also normal to a tangent
perand
extent
indefinite
of
BO
and
AO
(see Fig. 452), we draw the lines
of
pendicular, respectively, to the vectors v4 and vg. The intersection O
these lines is the instantaneous center of rotation or the centro; because,
line BO and
since the point B moves about a center of curvature along the
PLANE moTION [Ch. XIII
272
since the point A moves about a center of curvature along the line AO, the
intersection of these lines must represent the center of rotation of the whole
body, provided, of course, that M is a rigid body,
so that there is no relative motion of the points A
and B, for example.
The point O is momentarily
stationary when the velocities used in this construction are absolute velocities. By absolute velocity, we
mean the velocity of a particle relative to the earth.
Thus this point O is a point of zero velocity (as is
any center of a rotating body); that is, it is a point
without velocity when the earth is the reference
body. Although the body M is rotating about O
oY
Fig. 459.
Repeated.
at this particular instant, it will probably be rotating about some other point in the next instant, and
still another in the next, etc. If we say the center of rotation is moving, the
reader tries to imagine a moving point that has no velocity and is naturally
confused.
The fact is that this body M rotates about a series of fixed points
and about each one for an instant.
A smooth curve drawn through these
various fixed centers of rotation is the focus of the centros for the body, a
curve called a centrode.
An illustration should help. Consider a wheel.
If the wheel is slipping,
as in the case of a sharply braked, screaming automobile wheel, there is
rubbing of the bottom of the wheel on the road.
If the wheel is rolling, i.e., if it is not skidding
on the road, the point on the bottom of the
of oe
wheel in contact with the road is momentarily
BE
stationary.
This point of contact O, Fig. 460,
is the point of zero velocity and is therefore
the instantaneous center or the centro of the
rolling wheel.
As the wheel rolls along, the
O' No Slipping
centro is continuously moving ahead at the
Fig. 460. There are applicaspeed aay which the wheel advances. See Fig.
Rig eee
aK a eae
461 for evidence that the point of contact between the ground and the wheel is stationary.
If the wheel zs slipping,
there is a point of zero velocity relative to the ground (a centro), but it is
not the point of contact with the ground.
Suppose the wheel, Fig. 460, is moving with a velocity of ve. This means
that the axis (and the car) is moving at a velocity of vc. The speed of any
point in a rotating body is proportional to its distance from the center of
rotation. Thus, if the radius of the wheel is 7, we have va/vc = 2t/ ox
va = 2v¢; that is, the top point of the wheel moves at twice the speed of the
axis. This same result is obtained by graphical construction.
Let the
§ 198 |INSTANTANEOUS CENTER OR CENTRO
273
vector vc, Fig. 460, be laid out to scale. Draw the line OD through the end
of vector vc. Draw a line AD perpendicular to OCA.
The intercept va
represents the speed of the point A, as we see from the similar triangles OCE
and OAD.
Since the speed of any point B is proportional to the instantaneous radius OB, vg may be found from similar triangles by swinging an
are of radius OB to locate the point B’ and then erecting a perpendicular to
OCA at the point B’. It is seen from Fig. 460 that vg/vc = OB/OC, and
that the vector vg therefore represents the speed of point B.
ls
This photograph provides visual evidence that the
Rolling Automobile Wheel.
Fig. 461.
with the ground.
center of rotation or centro of a rolling wheel is at the point of contact
in near-circular
Observe how the curved streaks on the wheel bend about the contact point
but blurred at
arcs. Notice too that the markings on the tire are clear cut at the bottom,
mph when the
25
at
traveling
was
car
This
greater.
were
velocities
the
where
other points
picture was taken.
It should be said in passing that while the point O on the wheel is the point
Consequently, this
of zero velocity, it is not a point of zero acceleration.
method of handling velocity vectors cannot be applied to finding absolute
Acceleration problems of this type are discussed in the next
accelerations.
Considerably more detail on instantaneous centers is found in
chapter.
books on kinematics and mechanism.
PLANE MoTION [Ch. XIII
27h
a
199. Velocity Ratio. The velocity ratio VR between two members of
the
machine, both of which are undergoing plane curvilinear motion, is
angular velocity of the driving member divided by the angular velocity of
This is illustrated by two rolling wheels or two meshing
the driven member.
gears, Fig. 462. Suppose that the wheel A drives the wheel B. Then the
velocity ratio is
OR
WA
wp
NE
where the angular velocities may be expressed in any convenient units, say
radians per second or rpm, but the units must evidently be the same for
both parts in a particular ratio.
Now if these
wheels roll on each other at the point of contact D,
that is, if there is no slipping at D, the tangential
velocities of any points P and Q on the surfaces of
the wheels are the same, that is, vp = vg. Since
Dawa
;
2
and
10.
Doe
2.
— Dawp
he
wa
Dp
WB
D4
Vp = ro =
we find
Dawa
2
2
:
which is the correct ratio when there is no slipping
at the point of contact. For this case, then,
an
Swe
WON.
ee
or the velocity ratio is the diameter of the driven wheel divided by the diameter of the driving wheel. The definition of velocity ratio may be applied
to any two members in a machine whose angular motions are related to one
another; for example, the statement that the velocity ratio between the
engine and rear wheels of an automobile is 3, means that the engine turns
through three times as many revolutions as the wheels in a unit of time.
200. Closure.
The principal objectives of this chapter are to acquaint
you with the conception of acceleration, especially normal acceleration, with
which you may not be too familiar, and with the usefulness and some applications of the basic kinematic equations of motion:
v = § = ds/dt,
a=
vdv
= du/dt,
= ads,
w = 0 = d6/dt,
a=
= dw/dt,
wdw
= adé.
Of course, there are some other related details that you need to know, but
observe that these equations are as basic to kinematics as =F = 0 and
PROBLEMS:
>M
RECTILINEAR
= O are to statics.
275
MOTION
You will make frequent use of the relations v = rw
and a = ra.
In passing, it may be interesting and instructive to note that a point may
have zero velocity but not zero acceleration, or zero acceleration but not
zero velocity.
Problems
In solving problems involving constant acceleration, start
TO STUDENT.
equations (for example, a = dv/dt and vdv =a ds) and
basic
the
with
case
in each
proceed through the integration (or differentiation) in order to build wp your confidence
This practice will tax your memory the least
in your ability to use the basic equations.
excellent tools. You gain so little when
some
with
e
and give you a good acquaintanc
equation.
an
in
numbers
substitute
to
is
do
all you
SUGGESTION
RECTILINEAR
MOTION
891. A passenger on a train counts the
clicks of the wheels as they pass over the
joints of the rails, noting 62 clicks in 20 sec.
(a) If the rails are 29 ft. long, what is the
approximate speed of the train in mph?
(b) If the passenger counts the clicks for
only one minute, what round number when
divided into the total clicks per minute
would give the approximate speed in mph?
Ans. (a) 61.3 mph; (b) 3.
892. A train moving at 45 mph is decelWhat is the
erated to 15 mph in 22 sec.
In
average acceleration in fps?? in fpm2?
miles per hr-sec.? The direction of motion
is taken as positive.
893. A fighter plane pilot uses his wing
flaps to decelerate the plane at an average
rate of g/3, or 10.7 fps’. (a) How long
must he apply the flap controls to decrease
(b) What
the speed from 400 to 250 mph?
is the speed of the plane 4 sec. after the
flap controls are applied in fps, in fpm, and
in mph?
Ans. (a) 20.6 sec.; (b) 543.3, 32,600, 371.
894. An automobile is retarded uniformly
at the rate of 15 fps? from a speed of 60 mph
What is the elapsed time?
to 9 yd. per sec.
How far does the car travel in this time?
895. How far does a car travel in changing its speed from 10 fps to 60 fps when
a= Sitps2e
896. A car made a test stop from 60 mph
in 300 ft. on a certain type road. Should
the driver see a roadblock 250 ft. ahead and
should it take him 0.6 sec. to apply the
brakes,
find the speed (in fps, fpm, and
mph) with which he would hit the roadblock, assuming uniform deceleration at
the test rate.
Ans. 51.5 fps, 3090 fpm, 35.1 mph.
897. A car made a test stop from 60 mph
in 300 ft. If the deceleration was uniform,
find the speed in mph (a) 1 sec. after the
brakes were applied, (b) after the car has
traveled 100 ft. (convert to fps and to fpm),
and (c) find the time required for the stop.
898. A ball is thrown vertically upward
and observed to go level with the top of a
161-ft. tree. Find (a) the initial velocity,
(b) the distance traveled during the first
second
of flight,
during
the
third
(c) the distance
second
traveled
of flight
it = 2 sec. to t = 3 sec.), and
(from
(d) the time
required for the ball to return to the starting
point.
Ans. (a) 101.9 fps, (b) 85.8 ft., (c) 21.4 ft.,
(d) 6.32 sec.
899. A stone is thrown downward from
a 100-ft.-high tower with an initial velocity
of 20 fps. (a) With what speed did it hit
(b) What is the velocity when
the ground?
if =2
flight?
sec.?
(c)
What
was
the
time
of
(d) What initial speed would reduce
the time of flight to 50% of that found in
part (c)? (Errors are less likely in this
problem if one assumes the downward direction as positive.)
Ans. (a) 82.75 fps, (b) 84.4 fps, (c) 1.95 sec.,
(d) 51.35 fps.
900. The same as 899 except that the
initial velocity is 10 fps upward.
PROBLEMS: RECTILINEAR MOTION [Ch. XIII
276
901. As a balloon is rising at a speed of 10
fps, a sandbag is pushed overboard.
Four
seconds later the sandbag hit the ground.
Find the maximum altitude reached by the
sandbag and the altitude when it was pushed
overboard.
Ans, ZNO tte, 21 7.0) tbe
902. A balloon is ascending vertically
with an acceleration of 2 fps?. At an altitude of 200 ft., while the balloon is moving
at 20 fps, a rock is released.
With what
velocity does the rock strike the ground?
What is the elapsed time? What total
distance did the rock move?
Neglect air
resistance.
903. Two elevators in adjoining shafts
approach one another after starting simultaneously from rest when they are 500 ft.
apart. The top elevator A travels down
with a constant acceleration of 1 fps?. The
other elevator B has a uniform acceleration
of 2 fps? upward.
After what time are
they opposite each other? How far has
each traveled at this instant?
Ans. 18:25 sec., S4 = 166.7 ft., sp = 333.3 ft.
904. The same as 903 except that the
lower car has a uniform acceleration of 0.5
fps? upward.
905. The
same
as 903 except
that the
top elevator starts 2 sec. ahead of the bottom elevator.
906. Car A, traveling at a constant speed
of 60 mph on a straight road, passed a
parked police car P. Car P passed A 51 sec.
later. If P delayed 1 sec. in getting started
and its continuous acceleration is considered
constant, find ap and vp at the instant P
passes A.
Ans. 3.59 fps?, 179.5 fps.
907. Two automobiles, A and B, are
traveling in line on a straight highway at
the same speed of 100 fps. At time t = 0,
A’s brakes are applied to give a constant
acceleration of a = —15 fps?. One second
later, B’s brakes are applied to decelerate
RECTILINEAR
913.
(a) When
expression for the velocity v. Assume the
initial conditions to be v, and s, when t = 0.
The motion isin astraight line.
(b) Ifv, =
5 fps and s, = 10 ft., find the velocity when
+[v, + 2(s? — s,2)]1/2
914. The linear acceleration of a particle
is defined
fps’.
by the equation
a = (t3 + 3¢2)
If it starts from rest, find the dis-
placement, velocity, and acceleration when
t =4sec.
cars
Could such motion long continue?
Ans. 115.2 ft., 128 fps, 112 fps?.
is reduced
If the distance between the
to zero
without
collision,
what is the least distance at which B could
have been following A?
Find the time ¢
and velocities v4 and vg when this bumperto-bumper condition occurs.
The initial
position of A is suggested for the s = 0
position.
Ans. 30 ft., 4 sec., va = vp = 40 fps.
908. Two automobiles are traveling in
line along a straight highway at 100 fps.
The front driver A applies his brakes and
decelerates his car uniformly at 15 fps?.
One second later, the rear driver B applies
his brakes.
At what minimum uniform
rate must B decelerate in order to avoid a
collision if the initial distance between cars
is 160 ft.2 (Note: At the crucial moment,
the cars have the same velocity and are at
the same point, pratically.)
909. A stone is dropped into a vertical
mine shaft.
It is heard to strike the bottom
of the shaft after 6 sec.
Estimate the depth
of shaft. Use the speed of sound as 1130
fps and neglect the effect of air resistance.
Ans. 497.2 ft.
910. The same as 909 except that the
stone is given an initial downward velocity
of 30 fps.
911. Two mortar shells are fired vertically into a vacuum.
The muzzle velocity of
the first shell A fired is 600 fps, of the second
shell B, 900 fps. If these two shells are to
be together when the first shell is at its
highest point, what would be the time
interval between the firings? Ans. 11.5 sec.
912. A rock is dropped from a bridge
across a canyon 1000 ft. deep. At the same
instant another rock is projected vertically
upward from the bottom of the canyon.
If they pass at a point 300 ft. from the
bottom, what is the initial velocity of the
rock that is projected upward?
MOTION
a = 2s, derive a general
t = 2 sec.
Ans. (a) v =
it at 20 fps?.
=
VARIABLE
ACCELERATION
915. The motion of a point moving along
a horizontal straight line varies according
to the equation a = 12\/s. When ¢ = 2
sec., the point is 16 ft. to the right of
the origin, has a velocity of 32 fps to
the right, and has an acceleration of 48
fps? to the right. Determine the velocity
and acceleration when the ¢ = 3 sec.
916. A particle moves according to the
equation a = 5t"/8,_ If the initial conditions
are t = 0, v. = 6 fps, and s, = 0, find the
displacement during the interval between
t = 4 sec. and ¢ = 10 sec.
Ans. 340 ft.
PROBLEMS:
RECTILINEAR
MOTION
= VARIABLE
917. The same as 916 except that a =
2801/3.
918. The acceleration of a particle moving with rectilinear motion is a = Cx, where
C is a constant.
At a certain instant, its
speed is 15 fps and its displacement from
an origin is 30 ft. Somewhat later, its
speed is 60 fps and its displacement is 130 ft.
What is the value of C? What are the units
Of
Ans. 0.211 rad. per sec.?
919. The acceleration of a particle moving with rectilinear motion is a = 5t'/?.
If the initial velocity is 6 fps, what space
is traversed in 16 sec.? What is the space
traversed during the last 4 sec. of this
16 sec.?
Ans. 1461 ft., 724 ft.
920. A particle starting from rest and
moving in a straight line has a displacement
s = 303 — 8t2 +5.
What is the acceleration after 5 sec.? What is the change of
acceleration during the 10th sec.?
921. A particle moves with rectilinear
motion so that its displacement is given by
s =
(2 — 2t + 9)1/? ft., where
sense
924. Newton’s law for the motion of
bodies falling toward the earth from great
distances (millions of miles) is a = —C/s?,
where the constant of proportionality C =
gr2.
Neglect
all
resistances.
(a)
Let
a
body start from rest at a distance h from
the surface of the earth
(radius r), choose
the center of the earth as the origin, and
show that v2 = 2ghr/(h +r), where v is the
velocity after falling to the earth’s surface.
(b) Let r = 4000 miles. If h is small, say,
a few thousand feet, compared to r, to
what form does this equation reduce?
925. The general expression obtained in
accordance with Newton’s law (problem
924) for the velocity after a fall from a distance s, to some distance s from the center
of the earth is v2 = 2C(1/s — 1/s.). Using
v = ds/dt, show that
pees
sii)
1/2
E cos7! (<)
1/2
+ (ss — sue]
¢ is in sec-
onds. Compute its velocity and its acceleration at the end of 3sec. What is its arithmetic average acceleration during the 3d
sec.? Ans. 0.578 fps, 0.193 fps, 0.245 fps?.
922. A particle whose acceleration a =
3t — 12 fps? is moving at a certain instant
in a straight line with an initial velocity of
15 fps in the same
277
ACCELERATION
as the initial
acceleration.
At the end of ¢ =3
sec.,
what are the velocity and displacement of
the particle?
Ans. —37.5 fps, —85.5 ft"
923. The same as 922 except that t =
30 sec.
GRAPHICAL
926. A body falling in air actually undergoes a diminishing acceleration, and a point
is reached at which the acceleration is zero.
At the instant the body starts falling, the
acceleration is g. For such bodies as rocks,
bullets, and bombs, the acceleration a =
g — gv?/v'? = (g/v’?)(v’2 — v?), where v’ is
the limiting speed. Show that
ts of
v to
t= om log. G = -).
SOLUTIONS
927. The v-t diagram of a point is shown
in Fig. 463. Let the area of this diagram
be 0.58 in.2, the time scale 1 in. = 0.12 sec.,
and the velocity scale 1 in. = 30 fps. Find
the acceleration of the point at (a) position
B, (b) position A, (¢) position Gi (Gl)
Determine the displacement corresponding
to OD.
The integration may be made by substituting S = So cos? 6.
Ans.
(a) 150 fps?,
(b) 0, (c) —267
fps?,
(d) 2.09 ft.
928. The v-t diagram for the rectilinear
motion of an automobile between two points
shown in Fig. 464.
(a) What is the acceleration during the first 10 sec.? during the
30
ze
B
20
&
210
(@
12
0.16D
oe
10
t, sec,
Fig. 463.
Problem 927.
Fig. 464,
Problem 928.
20
PROBLEMS: GRAPHICAL soLuTions [Ch. XIII
278
next
10 sec.? during the last 3 sec.?
using the positions of the crank pin numbered
in Fig. 465.
(b) From the s-t diagram construct the v-t diagram.
(c) From the v-t diagram, construct the a-t diagram.
(d) From
these diagrams, determine the maximum
velocity and the maximum acceleration.
Ans. (d) Approximately —5.23 fps, —27.4
fps?.
933. High-speed photography is now used
as a means of measuring the displacement
of moving objects.
Thereby As is measured
by comparing successive pictures taken At
time apart (see Fig. 457 and several other
flash photographs in this text). Then by
graphical analysis the accelerations can be
quite accurately found.
The exposure rate
is 120 frames per second.
Let As; = 1 in.,
ANSoe—alep te AS sae
NS
ye nT
Ass = 4 im. Ase = 41n., As; —s3 10. As3.—
2. in., Ass = 1 in., and Asiop = 0.5 im: Blot
the s-t curve.
Then plot the v-t curve and
the a-t curve.
What are Umax and Qmax
during this 10-frame exposure?
(b)
What distance is traveled by the car?
929.
(a) A body falls freely for a time t.
Sketch the s-t, v-t, and a-t diagrams.
What
does the area under each curve represent?
What does the slope of each curve represent?
(b) A body is projected upward with an
initial velocity v and permitted to fall back
to the initial point. Neglecting air resistance, sketch the s-f, v-t, and a-t diagrams.
930. Sketch the s-t, v-t, and a-t diagrams
for the
stone
in 900.
Numerical
values
are not required.
931. A ball is dropped from a height h
above a floor. If the ball is perfectly elastic
and there are no losses, it will distort elastically as it comes to rest after initially
touching the floor. It then rebounds and
theoretically returns to the initial point
(see Fig. 691). Assume the accelerations
during the distortion phase (while it is in
contact with the floor) to be constant and
sketch the a-t, v-t, and s-t diagrams
two complete bounces of the ball.
for
ANS. Umax = 1 fpS, Gmax = —1.05 fps?.
934. The displacement of a particle is
given by the equation s = ¢ — 0.2/? ft.,
where ¢ is in seconds.
(a) Plot an s-t diagram for an interval of 5 sec.
What is the
932. A reciprocating steam-engine mechanism, with a 1-ft. crank and a 4-ft. connecting rod, is represented in Fig. 465.
The crank turns at a uniform speed of 50
rpm.
Thus, each of the divisions 0-1, 1-2,
etc., on the crank circle represents a certain
time unit. Let 1in. = 0.2 sec. of time and
lin. = 1 ft. of displacement.
displacement after 5 sec.?
(b) Derive a
v-t diagram from the s-t curve, and check
by differentiation the velocities obtained
graphically for t = 1 sec. and t = 38 see.
(c) Derive an a-t diagram from the v-t curve,
and check the acceleration by differentiation.
(a) Construct
a displacement s-t diagram for the piston,
Crosshead
A’ Co
ne
Fig. 465.
ANGULAR
cting Rod
Problem 932,
MOTION
935. A 12-ft. flywheel attains a speed of
120 rpm from rest in 60 sec.
(a) What is
the average angular acceleration? (b) What
is the final velocity in fps of a point on the
rim?
Ans. (a) 0.2095 rad. per sec.2, (b) 75.4 fps.
936. A flywheel attains a speed of 120
rpm from rest after turning through 360
revolutions.
What is the average angular
acceleration?
937. (a) Through how many revolutions
will a drum turn in 40 see. when the initial
angular velocity is 5 rad. per sec. and the
acceleration is constant at 2 rad. per sec.2?
(b) After this 40 sec., what angular acceleration will bring the drum to rest in 3 min.?
Ans. (a) 287 rev., (b) 0.472 rad. per sec.2
* 938. The speed of an electric motor is
changed from 10 rpm to 1800 rpm in 4 see.
For this interval, find (a) the arithmetic
average angular acceleration and (b) the
angular displacement in revolutions.
939. (a) A turbine turning at 1200 rpm
is brought to rest at the uniform rate of 100
PROBLEMS:
ANGULAR
&79
MOTION
944, The same
rpm?. What are the number of revolutions
turned and the elapsed time?
(b) If this
turbine slows to 300 rpm from 1200 rpm
during 7500 rev., what is the corresponding
uniform acceleration in rad. per sec.??
945. A rotating
6 = 213 + 442
Ans. (a) 7200 rev., 12 min.; (b) 0.157 rad.
perssec.-
940. From
a speed of 2000 rpm,
as 943 except that @ =
1 — 0.10?.
a fly-
wheel will come to rest with constant acceleration in 3 min. after the power is cut off.
(a) When t = 1 min., find w and @. (b)
When 6 = 1200 rev., find w andt.
Time is
measured from the instant that deceleration
begins.
941. A flywheel may be coupled to a
driving shaft by a jaw clutch.
The flywheel initially has a speed of 200 rad. per
sec. and is decelerating at a constant rate of
2 rad. per sec.2.. The shaft is started from
rest in the same sense at the same time and
body
follows
+ 10 rad., when
the law
¢ is in sec-
onds. Determine the angular displacement,
velocity, and acceleration after 4 sec. Is
the acceleration constant?
Ans. 202 rad., 128 rad. per sec., 56 rad.
per sec.?
946. The same as 945 except that @ =
3t? — 2t + 4.
947. If a = 2t, derive general expressions
for w and 6 when the initial angular velocity
is w, and when the initial angular displacement is 6.
948. If a = 26, what is the general expression for w when the initial angular velocity
is w, and when 6, = 0?
Ans, @ = + (wo? + 26?) 1/2.
ee
2
2.: 2%
6)?
rcos @9<——(L*-r“sin*
5(ani rcos
Fig. 466.
6+ (P- r’ sin’ 6) nie
Problems 950-952, 1001.
is accelerated at a uniform rate of 10 rad.
per sec.?.. The operator desires to make a
“flying” engagement of the clutch, which is
possible when their velocities are the same.
At what time t should he engage the clutch?
942. An automobile engine turning at
2600 rpm is driving the car through high
The wheels are 30 in. in diameter.
gear.
If the velocity ratio to the rear wheels is
3.42 in high gear and 6.5 in second gear,
what is the speed in mph when the car is
(a) in high gear and (b) in second gear?
Ans. (a) 67.8 mph, (b) 35.7 mph.
angular displacement of a roThe
943.
tating body follows the law @ = t — 0.10?
Determine the
rad., where ¢ is in seconds.
angular displacement, velocity, and acceleration (a) after 3 sec., (b) after 10 sec. (e)
What is the maximum displacement during
the first 10 sec.?
Ans. (a) 2.1 rad., 0.4 rad. per sec., —0.2 rad.
per sec.?; (b) Orad., —1 rad. per sec., —0.2
rad. per sec.?; (¢) 2.5 rad.
949. The same as 948. except that a =
6 + 36?.
950. Figure 466 represents the reciprocating mechanism in an airplane engine,
with a crank OA, a connecting rod AB, and
a piston B. The displacement of the piston
as measured from the center of the crank
shaft O is seen to be
s =r cosé + (L? — rsin?@)"/2,
Show that the expressions for the velocity
and the acceleration of the piston B are
;
To,
(a)
< ll
_ ro(sin6+ a7, Sin 20),
(b)
a=
ra*(cos 6+ a cos 20).
pint: Expand the second term of the equation for s to two terms, and obtain
ee
ec
s@- L Boe eae
s=rcosé+
5, Sin’é +
:
then differentiate.
If the second term in
the equations (a) and (b) were zero, what
PROBLEMS: ANGULAR MOTION [Ch. XIII
280
kind of motion would the piston describe?
See § 196.
951. If r = 2 in., L = 10 in., and if the
engine of 950 turns 2000 rpm, determine
the velocity and acceleration of the piston
when 6 = 45°.
952. In the airplane engine mechanism,
Fig. 466, the point C is the center of mass
of the connecting rod. With the origin at
the center of the crankshaft O, the coordinates of C are
NORMAL
AND
954. A train moves at constant speed
around a railroad curve of 2000-ft. radius.
If the normal acceleration is 3.872 fps?,
what is the speed of the train in mph?
955. A point on a rotating body changes
its speed uniformly from 10 fps to 20 fps
while it moves 120 ft. If the radius of the
point is 6 ft., what is its absolute acceleration at the instant its speed is 20 fps?
Ans. 66.7 fps?.
956. A point moves in the circumference
of a 20-ft. circle through an are distance of
320 ft. in 10 sec.
Its initial speed is 10
fps and the tangential acceleration is constant.
At the instant that it has moved
200 ft., what are the values of a, and a,?
957. A point at a radius of 2 ft. in a
rotating body has an initial speed of 160
fps. During a period of 6 sec., the angular
deceleration of the body is 5 rpm each 1.5
sec. Compute the tangential and normal
components of the acceleration (a) after
3 sec.; (b) after 6 sec.
, (L? — r? sin?6) 1/2,
y = 5 sin 8.
Determine the horizontal and the vertical
components of the velocity and of the
acceleration of the center of mass C, when
b= 60%)
= Grint 7a— sop ineand a7 —
1500 rpm.
See the hint in 950.
Ans. v; = —18.1 fps, az, = —17.15 fps?,
vy = 4.8 fps, ay = —8.5 fps’.
TANGENTIAL
953. A motorcycle goes around a circular
track whose radius is 200 ft. at 80 mph.
What is its normal acceleration?
Ans. 68.8 fps?.
Ans.
z=rcos@+
ACCELERATION
960. The same as 958 except that a4 =
g/2 — t/12 and v, = 10 fps.
961. The angular displacement of a rotating body follows the law 6 = ¢ — 0.1#?
rad. when ¢ is in seconds.
What are the
tangential and normal accelerations of a
point whose radius is 2 ft. after (a) 3 sec.;
(b) 5 see.?
Ans. (a) —0.4
fps?, 0.32
fps?;
(b)
962. A rotating body whose motion follows the law a = 0.1¢ has an initial angular
velocity of 5 rad. per sec. After 10 sec.,
what are the tangential and normal accelerations of a point whose radius is 6 in.?
963. A rotating body whose motion follows the law a = —46/2 has an initial
angular velocity of 50 rad. per sec.
After
a rotation of 7 revolutions, what are the
tangential and normal accelerations of a
point whose radius is 18 in.?
Ans. —39.8 fps?, 1410 fps?.
964. At a particular instant, a point is
on the horizontal centerline of a wheel
whose angular velocity is 300 rpm counterclockwise and whose acceleration is 2 rad.
per sec.” clockwise.
When t = 2 sec., find
(a) 0.698 fps?, 12,450 fps?; (b) 0.698
fps’, 12,120 fps?.
958. A body A, Fig. 467, is suspended
from a cable wound around a 5-ft. drum and
is moving down with a constant velocity
of 10 fps. When ¢ = 3 sec., (a) determine
the angular and linear velocities of point P
which is on the flywheel that turns with
the drum and (b) the normal and tangential
accelerations of point P.
959. The same as 958 except that a4 =
2 fps’. Let v, = 10 fps.
Ans. (a) 6.4 rad. per sec., 25.6 fps, (b)
164 fps?, 3.2 fps,
—0.4
fps?, 0.
Fig. 467.
Problems 958-960.
PROBLEMS:
NORMAL
AND
TANGENTIAL
281
ACCELERATION
965. An automobile starts from rest and
moves around a circular path whose radius
and tangential accelerations for a point on
this body at a radius r.
Ans. a, = Cr sin 0, Qn = 2 Cr(1 — cos 8).
967. A crank moves with an acceleration
of a = (1 — sin 6)?/3.
Its initial speed
is 600
is 20 rpm.
the position of the point and the magnitude
of its absolute acceleration.
The radius to
the point is 6 in.
ft.
Its tangential
a, = (s + 6).
acceleration
Determine
is
the tangential
and normal accelerations after the car has
gone 100 ft.
966. A body has an angular motion such
that a = C sin 6, where C is a constant.
Determine
970. An automobile starts down a 7%
grade at a speed of 10 mph.
After 20 sec.,
its speed is 50 mph. At the moment its
speed is 50 mph, what are the horizontal
and vertical components of the velocity v
and acceleration a, if a is constant?
How
far has the car traveled?
971. A body slides down a smooth 15°
incline with an initial velocity of 6 fps.
How far does it slide in 7 sec.? How far
does it slide during the seventh second?
Ans. 246 ft., 60 ft.
incline which is 10 ft. long.
a smooth
60°
If its initial
velocity is 5 fps, what is its velocity at the
bottom of the incline?
973. A. body moves freely up a smooth
After traveling 70 ft., its veloc30° incline.
ity is 10 fps. What was the initial velocity?
Ans. 48.5 fps.
974. A body moving freely up a smooth
incline slows down from 60 fps to 20 fps
in a distance of 100 ft. What is the inclination of the plane?
975. A good golfer imparts an initial
velocity of 136.5 mph to a good golf ball.
If the direction of this velocity is 30° with
the horizontal,
find
(a)
Ans. (a) 20 rpm; (ib)! 13:17 fps?,,0.
MOTION
969. If an automobile travels with the
uniform speed of 40 mph up a 6% grade,
what are the vertical and horizontal component velocities?
Ans. 2.4 mph, 39.9 mph.
972. A body slides down
10 revolutions,
(1 — cos 6)#/2/3.
the expressions for the normal
COMPONENT
After
the angular velocity in rpm, (b) the magnitudes of the normal and tangential accelerations of a point whose radius is 3 ft.
968. The same as 967 except that a =
find (a) the horizontal dis-
tance traveled in flight, (b) the time of
flight, (c) the maximum height reached by
the ball. Assume that the ground is level,
and neglect air resistance.
Ans. (a) 1077 ft.; (b) 6.22 sec.; (c) 155.6 Tie
976. The same as 975 except that the
direction of the velocity is 45° above the
horizontal.
977. A mortar projectile has a muzzle
velocity of 900 fps. What must be the
angle of elevation if the shell is to hit a
target 2000 ft. away on a level with the
gun? Neglect air resistance.
ATS 2229798 fells
978. A shell has just sufficient muzzle
velocity to clear a hill, the crest of which
is 2 miles away horizontally and 1500 ft.
above the cannon.
With the elevation of
the gun set for maximum range, what is this
muzzle velocity? Neglect air resistance.
Ans. 630 fps.
979. A bombing plane in level flight at
an altitude of 40,000 ft. is moving 400 mph.
How
far ahead
of the target, as measured
horizontally, must a bomb be released in
order to strike the target? What is the
time of flight? Neglect air resistance.
Ans. 29,300 ft., 49.8 sec.
980. Assume that sound travels Q fps,
and that t sec. elapse between the time of
discharge of a field-artillery gun and the
time at which the gunner hears the sound
of the shell burst. Neglecting air resistance
to the shell, show that the range is given by
Q(gt — 2v, sin @)/g, where v 1s the initial
velocity of the shell and @ is the elevation
of the gun.
981. A body A slides 10 ft. down a smooth
plane that is inclined at 45° with the horThe end of the plane is 40 ft.
izontal.
At the instant that the
above the ground.
body A leaves the inclined plane, a body B
is projected vertically upwards from the
ground with an initial velocity of 30 fps.
(a) With
what
velocity does A leave the
plane? (b) Where does the body A strike
the ground, as measured horizontally from
the end of the incline? (c) As measured
from the instant the body A leaves the
incline, what time has elapsed when the
bodies A and B are the same distance from
the ground?
(d) How far above the ground
are the bodies when they are at the same
PROBLEMS: COMPONENT MoTION
282
level? (e) What is the magnitude
sense of the velocity of body B?
Ans.
and
(a) 21.35 fps; (b) 17.72 ft.; (ce) 0.88
sec.; (d) 13.9 ft.; (e) 1.4 fps up.
982. The same as 981 except that the
inclination of the plane is 30°.
that v, = a constant and ay = —g.
Determine the magnitude and direction of its
velocity when the point is at the position
(—100, 0).
988. A point P moves in the path of a
curve defined by the equation y = e”. Its
tangential velocity is constant and is equal
to 12 fps. Ata position defined by y = 10
ft., what are the components
Fig. 468.
[Ch. XIII
v, and v, of
the velocity?
Ans. 1.195 fps, 11.95 fps.
989. A point moves along the parabola
x? = 36y. At the position where x = 40,
the tangential velocity is v = 12 fps. At
this position, what are the components v;
and v, of the velocity.
Problems 983, 984.
983. A motorcycle stunt rider passes
point A, Fig. 468, at a speed of 75 mph.
What is the maximum value of h if the
motorcycle (considered as a particle) is to
jump the 20-ft. ditch? Neglect the air
resistance and somersaults.
Ans. 10.84 ft.
984. The same as 983 except the rider
passes A with a velocity of 80 fps.
985. A particle A slides down a parabolic
chute whose equation is x? = 4y. It starts
ata value of y = h. Show that the speed at
point (0,0) is »v= V/2gh, friction neglected.
986. A point P moves in the path of the
hyperbola 22/36 — y2/16 = 1. The x component of the velocity is constant at v, = 9
fps. At the instant that P is at the position
(12, 4\/3), what is the acceleration a, in the
y direction and what is the tangential
velocity?
Ans. —1.73 fps?, 11.36 fps.
Fig. 470.
990. The sliding members A and B, Fig.
470, are constrained to move at all times
in
the
y and
zx directions,
respectively.
They are connected by the rod whose length
is L = 10 ft. At the instant when xz = 8
ft., vs = 20 fps toward the right and ag =
—15 fps? toward the left. Determine the
velocity and acceleration of A at this
instant.
Ans. —26.67 fps, —165.1 fps?.
991. A point moves so that
x=-100
x = 5 cos Ct and y = 5 sin Ct,
x*+ 50y -10,000=0
Fig. 469.
Problems 990, 1002.
Problem 987.
987. A point P moves in the path of the
parabola x? + 50y = 10,000
(Fig. 469), so
HARMONIC
when ¢ is in seconds. Determine the equations of the path of the point, its velocity,
and its acceleration.
In general, how is
the resultant acceleration directed?
Ans, x? + y? = 25 ft.2; + 5C fps;
+ 5C? at tan—1(180° + Ct).
MOTION
992. A point moves according to the law
a = —16x with an amplitude of 3 in. (a)
Find the period and frequency.
(b) Determine the displacement, velocity, and acceleration when ¢ = 3 sec.
993. A particle moves with a harmonic
motion whose amplitude is 4 ft. The maxi-
mum acceleration is 20 fps?. Determine
the maximum velocity of the particle and
the period of the motion.
Ans. 8.96 fps, 2.81 sec.
994. In Fig. 458, p. 269, show that the
acceleration of P at the position B is equal
to the normal acceleration of Q when it is
PROBLEMS:
HARMONIC
283
MOTION
at B. The point P is moving with harmonic
Also show that at any position of
motion.
The follower makes one oscillation for each
revolution of the cam.
P, its acceleration
999. A body oscillates twice a second
with an amplitude of 3in. Use the relation
a =vdv/ds = —C*s and integrate to find
is the horizontal
com-
ponent of the normal acceleration of Q.
995. In Fig. 458, p. 269, the point Q
moves in the circle with a constant tan-
gential speed of 10 fps. The radius of the
circle is 4 ft. Five seconds after Q, going
counterclockwise, passes the point D, what
is the velocity of P, the projection of Q on
Ans. 0.697 fps, 0.279 ft.
the diameter?
996. The radius of the circle in Fig. 458,
p. 269, is 20 in.
The period of P is 3 sec.
Determine for P (a) its maximum
(a) the maximum
velocity, (b) the displace-
ment when v = 10 in. per sec., and (c) the
velocity when s = 2 in.
Ans. (a) 37.7 in. per sec.; (b) 2.89 aes
(c) 28.1 in. per sec.
1000. The same as 999 except that the
frequency is such that 7’ = 1 sec.
velocity,
(b) its maximum acceleration, (c) its acceleration when @ = 30°.
997. The cam shown in Fig. 471 raises
and lowers the follower a distance s = 3 in.
If the cam makes
with harmonie motion.
60 rpm, determine
the maximum
velocity
and maximum acceleration of the follower.
The follower makes a stroke in each half
revolution of the cam.
Ans. 9.42 in. per sec., 59.2 in. per sec.?
998. The cam shown in Fig. 471 raises
im.
and lowers the follower a distance s = >
with
is 2
velocity and
If the period
motion.
harmonic
oscillation
sec.,
find
acceleration
the
of
maximum
Fig. 471.
of the follower.
Problems 997, 998
CENTROS
n
1001. Note that the direction of motio
Fig.
in
rod
ting
connec
the
of
end
of each
6 =
466 is known and locate its centro for
cen45°,r =2in., L = 10 in. Using this
ty
tro, determine the instantaneous veloci
at
turns
engine
the
when
B
of the piston
(See problem 951.)
2000 rpm.
1002. Locate the centro of the rod AB
d in
in Fig. 470 for the position define
Problem 990.
1004. A 3-ft. wheel, Fig. 472, rolls toward
the right. If the velocity of the center of
the wheel is v = 30 fps, what are the absolute velocities of the points J and K?
1005. The
v = 10 fps.
Ans. 55.5 fps, 22.95 fps.
as 1004 except that
Using this centro, check the
velocity of A as found in 990.
al
1003. A ladder 25 ft. long is in a vertic
the
position against a vertical wall. If
outward
bottom of the ladder is dragged
speed
on a horizontal surface at the constant
top of the
of 4 fps, what is the velocity of the
ake les Allee
ladder when the bottom point
—3 fps.
Ans.
wall?
from the
J
Fig. 473.
Y
K
Problems 1004, 1005.
Problems 1006, 1007.
1006. (a) The pin B, Fig. 473, has an
absolute velocity of 10 fps. What is the
(b) What
absolute velocity of the pin C?
is the velocity ratio between the links AB
and
Fig. 4/2.
same
DC
for
the
instant
defined
in
(a)?
The link AB is the driving lnk.
Ans. (a) 11.2 fps; (b) 1.485.
1007. The same as 1006 except that the
link AB is 2 in. long.
284
PROBLEMS: CENTROS [Ch. XIII
1008. The wheel in Fig. 474 is rolling so
that the absolute velocity of the pin A is
10 fps when 6 = 60°. What is the absolute
velocity of the slider B? The pin at B is
level with the bottom of the wheel.
Ans. 7.4 fps.
1009. The same as 1008 except that
For the position shown, let w4 = 40 rpm
counterclockwise and find the speed of B
and the angular velocity of link AB.
@ = IOP,
1010. Figure 475 shows a sliding block B
that is driven by a crankpin A through
link AB.
The block B is constrained to
move on a circular are whose center is at C.
NS
Fig. 474.
Problems 1008, 1009.
Solve
graphically.
Ans. wag = 2.23 rad. per sec.
1011-1020. These numbers may be used
for other problems.
Fig. 475.
Problem
1010.
Chapter XIV
RELATIVE
MOTION*
201. Introduction.
In the previous chapter, we were concerned mostly
In this chapter, we shall study the
in a plane.
point
a
of
with the motion
relative motions of two points, an approach which leads us to knowledge of
When a body moves so that
the kinematics of rigid bodies in plane motion.
each particular point in the body remains in the same plane, it is said to have
plane motion.
All motion is relative. Although we speak of the absolute speed or acceleration of a particle, we mean the speed or acceleration with respect to a point
So far as we know, no object in the
on the earth, which is itself moving.
However,
fixed point of reference.
no
is
there
that
so
stationary,
is
universe
it quite
find
who
engineers,
to
handicap
no
the absence of such a point is
motion
all
consider
and
sufficient to let the earth be the reference body
relative to the earth as absolute motion.
As an illustration of relative motion, consider a point on the tip of the
Relative to the axis of the shaft of the engine,
propeller of a moving airplane.
the point has a certain speed. This certain speed is the absolute speed of
the tip when the plane (and shaft) is standing still and the propeller is turning.
Since the shaft is moving through space, this speed is not the absolute speed
of the tip. Probably the reader understands intuitively that the absolute
speed of the tip will be some combination of the speed through space of the
shaft and the speed of the tip of the propeller relative to the shaft.
Suppose two men A and B start at some
202. Relative Displacement.
and walk in different directions. If A
476,
Fig.
point represented by O,
walks a distance s4 in a direction N 20° E and if B walks a distance sp
in a direction E30°N, the relative displacement is represented by the
distance AB, Fig. 476. In speaking of the relative displacement, we say
*The material in this chapter is not essential to an understanding of succeeding chapters
Some of the problems, however, require the use of
and may be omitted in short courses.
omitted, particular care should be used in choosis
chapter
this
if
that
so
these principles,
Some teachers may prefer to postpone the study
ing problems for assignment to students.
of this chapter until after Chapter XVI and before XVII.
285
286
RELATIVE MOTION
[Ch.
XIV
the displacement of A relative to B, or the displacement of B relative to A,
in order to distinguish the sense of the first relative displacement from
that of the second.
The displacement of B relative to A, designated by
8z/A, 18 the straight-line vector which A must traverse to get to the location
of B (provided both displacements s4 and sz originated at the same point).
See Fig. 476(b). Thus we may write
(a)
Sp
=
Sab
SB/a.
Similarly, we see from Fig. 476(c) that
(b)
SA =
SB+
Sa/B;
that is, the displacement of A is equal to the displacement of B vectorially
plus the displacement of A relative to B. The senses of the relative dis-
(b)
Fig. 476.
(c)
Relative Displacement.
placements sz; and s4,p are defined by the equations (a) and (b).
equations may be written:
SBj/A
=
SB—
Sa,
SA/B
=
Si =
Sips
These
that is, the vector difference of the absolute displacements is the relative
displacement.
If the displacements originate at different origins O and O’ ;
Fig. 476(a), the relative displacement is not affected, since the vector difference of the absolute displacements is the same in either case.
Recall that displacement is not necessarily the same as distance traveled.
Both men A and B may weave over the countryside, but if they end up at
points A and B from a start at O, the relative displacement is defined by
equations (a) and (b) and by Fig. 476.
203. Example. An airplane heads due north with an airspeed of 90 mph. A
60-mph wind is blowing toward the northeast.
What is the distance traveled by the
plane in 1 hr.?
SotuTion.
In this problem, we know that the distance moved by the air in 1 hr.
is 60 mi, NE = s4. We know the displacement of the plane relative to the air,
sp,a, is
90 mi. north. The displacement of the plane relative to the ground is given by
Sp =
Sab
Spya.
287
§ 204 |RELATIVE VELOCITY
The displacement of the airplane sp may be found eraphically as shown in Fig. 477, or
it may be found algebraically by the law of cosines. We have
sp = [sa2 + spa? — 2 Sasp,a cos 135°}?
= [3600 + 8100 — (2)(60)(90)(— 0.707)? = 139 mi.
The direction of the displacement may be found as
9 = tan“(Zs,)/(Zsz),
or it may be scaled from the solution of Fig. 477 as N 18° E, approximately (@ = 72°).
204. Relative Velocity. To find the relation of the velocities of two
points, consider Fig. 478 which represents a stick on which a nut slides freely.
With the nut close to your hand near the end A, suppose you swing the
Fig. 477.
Fig. 478.
Relative Motion.
stick through a horizontal curved path. The nut moves outward, as it
We shall concentrate our attention on two points, P
naturally would.
These points are coincident at the beginning of motion, point P
and M.
on the nut and point M on the stick. It is usually convenient in practice to
consider one of the bodies as a reference body and the other body as the
relatively moving body. The nut of Fig. 478 could be taken as the reference
body, but it is somewhat less of a tax on the imagination to let the stick be
on
the reference body, because we seem to think naturally of the nut sliding
and
the stick rather than of the stick sliding in the nut. The absolute
you
If
taken.
is
view
of
point
what
relative motions are the same no matter
stick.
the
on
sliding
nut
are an observer standing on the stick, you see the
If you are an observer standing on the nut, you see the stick sliding through
the nut.
Let the stick move from position A1B; to A2B2. The initial position
At the second position of the
of points P and M are at P; and M, as shown.
stick, point M is at M2 and point P is at Po, Fig. 478. The absolute displace-
288
RELATIVE MOTION [Ch. XIV
ment of M is designated as As,,, the absolute displacement of P is As,, the
relative displacement of P along the stick is Asj/m. We see from the vectors
of Fig. 478 that
As, = A&m, + ASp/m-
Suppose these displacements occur during the time interval At.
terms of the foregoing equation by At and get
Divide the
Asp — A8m ., AS8p/m_
Ai inte
Sekt
If the time interval is infinitesimal
(At
0), the corresponding displacements are so small as to be straight
lines coincident with the paths, even
if the paths are curved, and the various
ratios As/At become instantaneous
velocities.
Therefore, we may write
(c)
Vp =
Um +> Up/m,
which is to say that the absolute velocity of point P is equal to the absolute
Fig. 478. Repeated.
velocity of point M vectorially plus the
velocity of P relative to M.
Thus,
we see that the relative velocity of two points is the vector difference of
the absolute velocities of the points. Having observed in equation (c) that
similar relations exist for velocities as for displacements, we may write
the various equations relating the velocities of points P and M in general
as follows:
(d)
Vp =
Um +
Up/my
or
Vpim
=
Vp
(e)
Um
Vp
Um/p,
or
Um/[p
=
Vm
=
Um;
—
Up.
If the stick were curved instead of straight, we would obtain
the same relations by a similar analysis.
In applying equations (d) and (e), remember
that they are for an infinitesimal movement.
The stick has moved so little,
it is still in the same place. The significance of this warning
in the case of
the stick and nut is suggested by Fig. 479, where we see
that the relative
velocity vector vp» is drawn parallel to AB, since the directio
n of motion of
the nut must be along AB.
However, if the velocities are constant (this includes
direction) during the
displacements being considered, the equations (d) and
(e) hold for finite as
well as infinitesimal time intervals.
205. Example. A ship is moving directly northeast
at 10 knots (vs) in still water.
A man walks across the deck facing due east at 4
mph (Umjs). What is the man’s
absolute velocity (vn)? Note: 1 knot = 1.152 mph.
§ 207 | EXAMPLE
289
So.ution.
Since the velocity of the man is equal to the velocity of the ship vectorially plus the velocity of the man relative to the ship, we have
Um = Us
Umis = (10)(1.152) + 4 = 14.6 mph.
The method of graphical solution is shown in Fig. 480, where we observe that the law
of cosines will afford an algebraic solution. The angle 6 = 33.9°.
206. Example. A cruiser is traveling at 26 knots due north. Ata certain instant,
a shot is fired with a muzzle velocity of 2000 fps at a stationary target which is on a
line E 18° 26’ S from the position of the gun. What should be the bearing of the gun
to make a hit? [i knot = 1.152 mph.]
SoLutTion.
The speed of the cruiser is
Ve
Up
_ (26)(1.152)(6280)
3600
= 44 fps.
\N
320
{
Um/s
Usfc
108°26’
B
Fig. 481.
Fig. 480.
Fig. 479.
In this problem, we know the direction of the absolute velocity of the shell, 18° 26’
south of east, represented by a line OA, Fig. 481; we know the magnitude only of the
relative velocity of the shell v,,. = 2000 fps; and we know the direction and
magnitude of the velocity of the crusier v.. Hence a triangle OAB, Fig. 481, may
be constructed from which, by the law of sines, we find
Ve
Us/e
Ai fo aN
7,
ie ey ee ae
esi
108-43")
2000
toe
Vi2
The bearing of the gun then must be 18° 26’ + @ = 18° 26’ + 1° 12’ = 19° 38" south
(The value of 6, shown exaggerated in Fig. 481, is too small for graphical
of east.
a very large scale for the vectors is used.)
unless
solution
Fig. 482 represents a mechanism with four links, A (a part of the
207. Example.
link), B (a crank which rotates about the fixed pin ab), C (a slider
ground
frame or the
which slides along link D), and D (a link which moves with variable angular velocity
This mechanism, called a Whitworth
when B moves with constant angular velocity).
provide a slow-cutting stroke for a
to
quick-return mechanism, is used as a device
no metal cutting is being done.
when
stroke
quick-return
a
then
machine tool, and
In problems of this kind we generally know the constant angular velocity of the link b,
from which the absolute linear velocity of the pin P at the end of link B can be computed. Find the absolute velocity of point # in link D.
290
Sotution.
RELATIVE MOTION [Ch. XIV
We first observe that there is a point in link D coincident with P in B.
Call this coincident point m (in link D).
The known data are generally:
(1) and (2), the magnitude and sense of the velocity v, of P (in B). The magnitude
of vp may be computed from vp = 2rrn [equation (37)] and the vector must be
perpendicular to the radius P-ab, Fig. 482,
(3) the direction of the velocity v,, of the coincident point m (in D), which must
be perpendicular to the radius P-ad,
(4) the direction of the relative velocity ¥p/m, which must be parallel to the lmk D
since the slider C is constrained to move along link D.
Thus, in finding graphically the velocity of m (in D):
1. Lay out v, to scale in magnitude and sense, Fig. 482.
2. Draw a line PQ perpendicular to the radius P-ad of indefinite extent.
3. Draw a line parallel to the radius P-ad (parallel to the direction of motion
of C) from the end of the vector for v,, point h in Fig. 482.
Fig. 482.
Whitworth QuickReturn Mechanism.
Point m
is in link D directly under P.
Fig. 483.
Body.
Relative
Velocity, Points in Rigid
Then the intercept on the line PQ is the velocity v,,, inasmuch as the equation
Vp =
Um =
Vp/m
is satisfied. Having found the absolute velocity of m, a point in the link D, the
velocity of any other point in the link D may be found because the velocities of any
two points in a rotating body are proportional to their distances from the center of
rotation ad (§ 184). That is, the velocity v, of point #, for example, may be found
from the equation
Ye
r,
E-ad
Des
hn
1g
where r, and 7, are measured from the center of rotation ad of the link D.
208. Relative Motion of Points in a Rigid Body.
The types of problems
involving relative motion with which we deal include:
1. One body in a moving stream, as an airplane in a wind or a boat crossing
a river.
2. One body moving relative to a rigid body, in which the motion of the
first body is constrained (or defined) by the construction of the second
(or reference) body or by the method of attaching the bodies together,
as in Fig. 482.
3. Two points in a rigid body.
291
§ 209 |EXAMPLE
The equations (d) and (e) may be applied to each of these types of problems.
However, there is an aspect of the type (3) that may be useful.
If a body M is rigid, any two points A and B, Fig. 483, remain at a fixed
Theredistance apart, no matter what motion the body itself undergoes.
is one
A,
to
relative
B
of
or
B,
to
fore, the only possible motion of A relative
Thus, in Fig. 483, if the body M is in curvilinear motion with
of rotation.
7
some clockwise angular velocity w, then vg/a = rw and va/z = 70, where
is the distance between the points A and B. This knowledge of the relative
motion of two such points is useful in finding the absolute velocity of any
point B in a body when the absolute velocity of some other point A in the
body is known, the relation being
Vp =
VahHvea
=
VAY
To,
velocity of
where r is the distance between the points and w is the angular
the body.
s how vz
For example, the construction at point B, Fig. 483, indicate
The vector Bd is laid out parallel
may be found from a known va and vga.
Bd. The velocity
to and equal to Ac. Then de = vaja = Tw is added to
Be.
by
of B is represented in magnitude and sense
B in a rigid body
Suppose we let the line joining any two points A and
es of the points
be called the x axis. Then the components of the velociti
A and B along the « axis are Vaz and vp,
Now observe that
va, Taust be equal to vzz,
apart or closer together,
otherwise the points A and B would move further
the body is rigid.
but their distance from each other does not change if
velocity of a
In the previous chapter, we learned how to find the
209. Example.
(Fig. 460). Using the rolling
point in a body when the location of its centro is known
vee
wheel as an example again, we shall find the absolut
the absolute
given
having
484,
Fig.
B,
point
the
of
locity
UB/c
velocity of the center C of the wheel.
If the wheel is rolling with an angular veSotution.
ty of the center of the wheel is
veloci
locity w, the
Ve = Tw.
[ROLLING WHEEL]
the periphery of
Moreover, the velocity of any point on
varc = Tw = Ve
hence
Tw;
is
No Slipping
center
the
to
e
the wheel relativ
any particular
in
g
dealin
t
withou
Thus,
tude.
in magni
Fig. 484.
s/c perpendicular
magnitudes, we may lay out a vector
then add to this
to the line r joining the points B and C,
representing vc. The sum of these two
length
equal
of
vector
vector a horizontal
ty of point B, since
vectors is recognized as the absolute veloci
\ [///
Va =VUcH UBie.
292
RELATIVE MOTION
[Ch. XIV
The vectors ve and vz,c are equal in length only in case the wheel is rolling without
slipping and the point B is on the periphery of the wheel. In any event, however,
UBIC
=
To,
[ROLLING OR SLIPPING]
where r is the radius of the point B and w is the angular velocity of the wheel.
210. Example.
The lengths of the various links A, B, C, and D in the mechanism
of Fig. 485 are known, as well as the angular velocity of the link B. For the position of the links shown, find the absolute velocity of the point cd, the center of the
pin which connects the links C and D.
Sotution.
The center of the pin bc has a velocity rw in the direction perpendicular to the link B, where r is the distance ab-bc, Fig. 485. Since the center of this pin
bc is a point in both links B and C, we now know the absolute velocity of a point
in
Ved/be
Fig. 485.
Four-Link Mechanism.
the link C. Moreover, the point cd is a point in the link
C.
we may write
(f)
Vcd
=
For these two points,
Vbe => Ved /be+
The velocity of the point cd relative to bc, Vcaise,
Must be in a direction m—n or n—m
perpendicular to the line joining bc and cd, since this
link C is assumed to be
rigid.
Similarly, the absolute velocity of point cd must
be in a direction P-q or q-p perpendicular to the link D, because cd is a point in the link
D and because all points in the link
D rotate about the fixed center ad. Thus, of the
three vectors involved, we know the
magnitude and sense of one of them, and the lines
of direction of the other two vectors.
Consequently, we may lay out v, to scale from
any convenient origin O, Fig. 485(b);
then, from one end of the vector draw a line
parallel to mn, and from the other end of
de draw a line parallel to pg. The intersection
of these two lines defines the limits of
the vectors v.¢- and v-¢. The sense of each of
these vectors must be such as to satisfy
the equation (f); therefore the velocities are
directed as shown in F ig. 485(b). The
magnitude of v.¢ may be scaled from the diagra
m when the scale is known.
211. Relative Acceleration.
Relative accelerations may become some
what
more involved than relative velocities.
However, take a simple case first.
In Fig. 486, the x and y axes are fixed
relative to the earth, so that velocities
and accelerations relative to these axes
are absolute values. The x’ and y’
axes are attached to the rigid body
A, so that velocities and accelerations
related
to these axes are relative values.
The point P represents a particle
§
211 | RELATIVE ACCELERATION
293
moving relative to body A (relative to x’y’), and body A also has motion
which can be defined in terms of the coordinates x, and ym and the angle 6,
where M is any convenient reference point or pole in body A.
It is simpler to analyze the situation when @ equals a constant (@ = C),
which is to say that body A does not rotate, although any particular point
might move in a curved path; that is MZ may move along some path cd with
The absolute
the various positions of the body parallel to one another.
coordinates of P are x and y, or
2 = Xm + x'cosd — y’sing,
Y = Ym + 2’sind + y'cosd,
(g)
(h)
Knowing
from the geometry of Fig. 486.
differentiate equations (g) and (h) with
respect to time ¢, holding @ constant
Using the dot noas a special case.
tation (z, etc.) for convenience, we
find the x component of the velocity
of P as
Gs,
that dx/dt =v, = 4, we may
— 2, + cos) — Y sind,
and the y component of P’s velocity
as
Ge
p=
Ga
@ sind + ¥’cos?.
To digress from accelerations a moment, we might study equations (i) and
(j) to become acquainted, in a case with
:
Particle P is moving relative t
4, Say, in the path ab.
O
Fig. 486.
which we are already familiar, with the
reasoning to be used. We know that
the vector addition of the two components
that is,
(i) and (j) gives the velocity of P, vp;
Vp = Lp +> Yp-
Substituting the values from (i) and (j), we find
(k)
Vp = (dm +> Ym) 1H (4’cos 6 — y/’sin 6) + (#’sin 6 + y’cos 6).
velocity of
We observe that ¢, and y» are the x and y components of the absolute
that
and
M
point
reference
or
the pole
bien a
Vp
=
Um.
y’ components of the
The last two terms in equation (Kk) are, respectively, the x’ and
of P relative
velocity
the
of
velocity of P; that is, they are rectangular components
to M; thus,
(z’cos 6 — y’sin 0) + (a’sin 0 + y'c0s 0) = Ypim,
and equation (k) becomes
Vp =
Un +> Upim)
obtained.
which is the same except for subscripts as equation (d) previously
294
RELATIVE MOTION
[Ch,
XIV
Returning to equations (i) and (j), we recall that da/dt = dv,/dt = a, = &.
Therefore, differentiating these equations with respect to time to get the x
and y components of the acceleration of P, we find (for 6 = C)
Ln = Lm + (Cos 6 — ¥’sin 6),
(1)
Yin = Yin + (&’sin 6 + ¥’cos 8).
(m)
As in the case for velocities, the absolute acceleration of the pole M in Fig.
486 is the vector sum
Om = Ln + Ym-
The term in the parentheses in equation (1) is the 2’ component of the
acceleration of P, that is, a component of the acceleration of P relative to M.
And the term in the parentheses of (m) is seen to be the y’ component of the
acceleration of P relative to M.
The vector sum of these components is
Ap/m=
(#’cosd — y’sin#) + (#’sind + 7y’cos6)
Therefore, adding the right- and left-hand sides of (1) and (m), we get
Ap
=
Lp te Yp =
Am +
Ap/my
a form of equation exactly analogous to relative motion
equations for dis-
placements and velocities. However, it must be recalled that in the case of
accelerations, this equation was obtained when the body was asswmed to have
no rotation (6 =C).
If the body should rotate while a particle moves relative to it, @ would vary and the result
would be somewhat longer than that
obtained with 6 constant.
For the
special case under discussion then, we
may summarize our findings as
(a)
Ap
=
Am
>
Anim
Ap/m
=
Ap
>
Am.
This same relation holds between
the accelerations of any two points P
and M in a rigid body, even when the
body has a motion including rotation.
Fig. 487. P and M attached to body A. If both P and M are attache
d to
the moving body, point M may be
brought to a state of zero acceleration by subtracting the vector a as shown
at M in Fig. 487. If this same acceleration is subtracted from that of some
other point P in the body, then the acceleration that P would have would be
the relative acceleration of P with respect to M. That is, Fig. 487 at P,
Ap
Ca.)
SO)
lh, =
haa
By comparison, we see that this equation is the same as (n). If
points
P and M both have linear acceleration only, equation (n) is easily applicabl
e
995
§ 212 |EXAMPLE
as it stands.
If either or both points P and M have curvilinear motion, it
is generally going to be convenient to use the normal and tangential comIn terms of these components, we have
ponents of the accelerations.
Ap =
Anp
Arp,
Am
Anm
+ Aim,
=
Ap/m = Anp/m +> Gip/m)
or
(0)
Ap = Anp YPA1rp = Anm VY Aim YP Anp/m 1 Ap/m,
where a, stands for the normal acceleration of P, dim stands for the tangential
acceleration of M, etc. The normal acceleration (rw?) of P relative to M
when these points are each attached to a rigid body (Fig. 487) is
Onp/m
=
lw’,
where I is the distance between the points P and M and w is the angular
velocity of the body A. The tangential acceleration (ra) of P relative
to M is
Qip/m
=
la,
t in
where a is the angular acceleration of the body. In case a componen
equation (0) is zero, it simply drops out of the equation.
TTITIVTITITTT
TT
No Slipping
(b)
(a)
Fig. 488.
of 5 fps?.
An automobile moving at 3 mph has an acceleration
212. Example.
tion of a
accelera
e
absolut
ne the
The outside diameter of the tires is 30 in. Determi
point A on the periphery of the tire and on a line which
as shown in Fig. 488.
makes 45° with the horizontal
The wheel does not slip.
C is the same as that
Since the acceleration of the center of the wheel
So,utTion.
angular acceleration
the
,
no slipping
of the car, we have ac = 5 fps’. Therefore, with
of the wheel is
a=
= = ae = 4 rad. per sec.”
296
RELATIVE MOTION [Ch. XIV
The tangential acceleration of point A relative to C is
Giarc = Ta = (1.25)(4) = 5 fps?,
the same as ac.
With a linear velocity of the center C
as vc = 3 mph = 4.4 fps, the angular velocity of the
wheel is
;
SS.
SS
125
,
=
. 2 rad. . per is
3.52
sec .
Hence the normal acceleration of A relative to C is
Gnajc = Tw” = (distance AC)w? =
(1.25)(3.52)? = 15.5 fps’.
These relative accelerations are directed as shown in
Fig. 488(a). Now in the equation
(b)
Fig. 488(b).
Repeated.
Qa = Ac +> Gaic = Ac
HD Ataie
HD Anaie,
all terms on the right-hand side are known in magnitude and sense. The magnitude
of their vector sum (= 14.7 fps), as shown in Fig. 488(b), is the magnitude of the
total acceleration of A. The acceleration of A may also be obtained algebraically by
summing the x and y components of the vectors, as in the case of force vectors.
213. Example.
A 12x 14-in. steam engine turns steadily at 100 rpm.
The length
of the connecting rod is 28 in. When the crank is in a position 30° past head-end dead
center, as shown in Fig. 489, determine (a) the velocity of the piston (C), (b) the angular velocity of the connecting rod BC, and (c) the absolute acceleration of C and
the
acceleration of C relative to B.
SoLuTion.
(a) With a 14-in. stroke, the radius of the crank-pin circle is 7 in.
The angular velocity of the crank is
was = 100rpm
_ (100)60 (27)
= 10.47 rad. per sec.
The linear velocity of the crank pin is
ey
v
— (5)(1047) = 6.1 fps.
Now lay out the mechanism to scale in the proper position (Fig.
489), and draw the
vector vz to represent 6.1 fps. The solution of the velocity
problem may be completed at the point B. We note that
(p)
Vol
VE?
Voss:
and that vc;s is perpendicular to the line BC ; hence draw a line BE through
B perpendicular to BC of indefinite extent. We also note that Yc must be in
a horizontal
direction ;hence draw a horizontal line through the other end H of
the vector vz. The
intersection of these two lines at the point Z determines
the limits for the vectors ve
and vc;s. In order to satisfy equation (p), the vectors
must be directed as shown in
Fig. 489(a), where it is seen that vc/p- vs = vc.
scaled as
The velocity of the piston is
vc = 3.71 fps.
(b) To get the angular velocity of BC, we
can use the instantaneous center or
centro (§ 198) of BC. To locate this centro, we
note that the direction of motion of
the
point B is known and that the direction of motion of
the point C is known.
Per-
$213 |EXAMPLE
297
pendiculars to these directions of motion at the points B and C intersect at O, the
centro of BC ($198).
Now we know the magnitudes of the velocities of two points
B and C in the link BC, from either of which the angular velocity may be found.
Scaling the drawing we find OB = 32.3 in. and OC = 19.7 in. Therefore
Bc
9
UB
6.1
= OR = 353/12 7 2.265 rad. per sec.,
;
Bill
Ve
one == Oa =
19.7/12
= 2, 26 rad. i per sec ;
Also, since v¢/p = Tw = (BC)wac, we may get still another computation for wz¢ from
eet
MPS
ae CB
HG 04 Bay aW:
One calculation is a check on the other and also a check on the accuracy of the graphical
work.
a,=63 fps?
Fig. 489.
Slider-Crank Mechanism.
the relative acceleration equation
(c) To get the acceleration of C, we examine
<>
ac/B = GB +> AnciB + AtciB.
(q)
Qc = GBH
The acceleration az
stant, and is
is entirely a normal acceleration, since AB is moving with w con7
ap
=
GnB
=
r(waps)?
=
63.8 fps®.
(5)aoane =
The normal acceleration of C relative to B is
28
Gncjp = T(wec)? = (3)e20 SSO)
’
yee,
we
SRR
Also the
The acceleration ac must be in either the sense m—n or n—m, Fig. 489.
are
there
Thus,
direction (p-q or q-p) of adicyn must be perpendicular to the line BC.
found.
be
can
which
ons,
accelerati
two
only two unknowns, the magnitudes of these
ag = 63.8 fps?
Starting at any convenient origin D, Fig. 489(b), add the vectors
and dnc/z = 11.9 fps?.
Through the end of this latter vector, point G, draw a line
The interperpendicular to BC. Through the origin D, draw a line parallel to m—n.
g to the
Referrin
adc.
and
aic/s
section of these lines defines the limits of the vectors
ion of
accelerat
The
shown.
as
directed
be
must
equation (q), observe that the vectors
the piston is measured as
ac = 63 fps?.
Shown dotted in
The acceleration of C relative to B is the vector sum dnc/s +4 dics,
fps’.
33.2
=
ao/z
scales
sum
This
Fig. 489(b).
298
RELATIVE MOTION [Ch. XIV
214. Coriolis’ Law.*
The general case of relative motion would occur if
the body A, Fig. 486, had some rotational motion, as well as translation;
that is, if @ varied while zp, Ym, x’, and y’ varied.
In this event, the components of the absolute acceleration of particle P may be obtained by two
differentiations with respect to time starting with equations (g) and (h),
p. 293.
However, the physical significance of the difference between the
equation (u) obtained below and equation (n) will be clearer from the following derivation.
Although it is based on a particle moving in a path in a
rotating body, the result is general and complete.
First, we shall familiarize ourselves with the representations of Fig. 490.
The rotating body MW is shown in two positions M,; and Mo», separated by an
angular displacement Aé@. There is a path AB and a particle P which
moves in this path.
In the position
)
Up /ps
t
Af,, the initial position of the particle
is designated as P;. The point A, isa
(a)
. Up'/z,
point A in the body M which coincides
with the initial position of the parM7
ticle. (The points P; and A, are
Ys:
R
shown separated only to help to disaS
tinguish between them.) If the path
AEOTATS
(body J) were stationary, the parne os
ticle would move from the position
P,; to a position P’, coinciding with
a point B in the body M, during a
Fig. 490.
Coriolis’ Law.
time interval At. However, during
this time A¢, the body takes the position 1/2 shown dotted, so that the particle is in some position P» coinciding with the second position B, of point B.
The initial velocity of point A in the body M is designated as va,; the
initial velocity of P relative to point A as vp,4,; and so on for other
velocity symbols.
The body is rotating with an angular velocity wi and
an angular acceleration ay,.**
The initial absolute velocity of the particle is evidently
Up,
=
VA, bd UP\/A}-
In the second position, the absolute velocity of P is
VP, = UBy + UPy/ Bo.
The change of the absolute velocity of P is
(r)
Avp
=
Ups
VP,
=
UB» pH UP»/
Bo —
VA,
UPy/Ay-
*This article may be omitted in shorter courses.
**The following development
and Kinetics of Machinery.
is similar to that given by Dent and Harper,
:
Kinematics
299
§ 214 | CORIOLIS’ LAW
In the second position M2, we see that
VBo = VAg 1 UBo/ Ao:
get
Using this value of vz, in (r) and letting va, > va, = Ava, we
Avp = Ava 4H UBo/Ag 1 UPo/Bg — UPVAL
(s)
path after it has
Now vp//z, is the velocity of the particle relative to the
vp./s,, is the
velocity
The
ry.
moved to the point B with the path stationa
vector difThis
on.
same in magnitude as vp’/z., differing only in directi
ference is seen from Fig. 490(a) to be
st = 2vp,/py sin (46/2),
may be written in the
which, for very small values of A@, where sin Ad ~A@,
then
form st = vps, 46. From Fig. 490(a), we have
—
UP»/By
UP'/By, a
UP9/Bo Ae.
Using this value of vp./s, in (s), we get
Avp
(t)
=
Ava
UBo/ Ao +
)
UP'/ By, HD UP5/Be A@—
VPy/Ay
of velocity of the particle
Now vp-/p, — P14, is seen to be the change
Let this change
stationary).
relative to the path (that is, with the path
that the point
ty
veloci
the only
be designated by Aveyaw. We observe that
B may
A of magnitude
about
have relative to A is one of rotation
As wa
in M. As the time interval
(§ 208), since A and B are particular points
If these points
ach coincidence.
approaches zero, points A and B appro
along the path,
ce
distan
lly the
are very close together so that As is virtua
At, ~Thus
the value of As will be As = v At = vpjm
UBo/Ag — UP/MOM At.
Moreover,
as
At
approaches
Zero,
UP2/B»
approaches
the
instantaneous
we may designate by vpjw. Using
velocity of P relative to the path, which
raph in equation (t), we find
the equivalents discussed in this parag
Avp
=
Ava)
Avpjm 1
Vp/MOM
At }
vpjm Aé.
At, we have
Dividing each term of this equation by
Avp
At
Ava ,, Avpjm + 2
VP/MWMy,
iN
iG
the time interval has become inor, in terms of different symbols when
expressing Coriolis’ law,
finitesimal, we obtain the equation
(u)
ap =
Ou 1D Apjm + 2Up/mMoM,
300
RELATIVE MOTION [Ch. XIV
where we recognize
the term
Avp/At (= ap) as the absolute
acceleration
of particle
P,
the term Av4/At (= ay) as the acceleration of a point on the body M
coinciding with the position of the particle,
the term Avp;y/At (= apm)
relative to the path.
as the total acceleration
of the particle
The final term, 2vp;wy, is a supplementary acceleration arising from two
sources: one part of it is due to the change in the direction of the relative
velocity vector that results from the rotation of the path (vp: to Up2/B2),
and the other part is due to the relative velocity of the two points of reference
A and B. We can combine these components into the one term 20P/mMoeu
because they both are in the same direction.
This direction is evidently
normal to the path at the position of the particle.
To decide upon the sense of the acceleration vector 2vp;mwu, proceed as
follows:
1. Draw a vector representing vp/i7;
2. Note the direction of the angular velocity of the body waz;
3. Imagine a force applied to the point of the vector Vpym In such a manner
as to turn the vector in the same sense as wu;
4. Then the sense of 2vp;ywy is the same as this imaginary force.
The other components of the total acceleration of P are conveniently
broken up into normal and tangential accelerations.
Thus, if the path is
rotating about some center O, the absolute acceleration of the
point in the
path which coincides with P is
Gu
where
=
AnmM HD Qim
ry is the radius from
=
rywy?
the center
9
Db ruam,
of rotation
O to the point in the
path. The component rywy?, since it is a normal
acceleration, will be
directed toward the center of rotation, and the sense
of ryay is determined
by the sense of ay.
If we let wp and ap, respectively, represent the angular velocity and the angular acceleration of a radial
line joining the particle
and the center of curvature of the path, and let rp be
the radius of curvature
of the path at the position in question, the compon
ents of Qpjm May then
be expressed as
Apjm
=
Qnpjm 1
Aipjm
=
Ppwp? DH rpap.
As usual, rpwp? must be directed toward the
center of curvature of the
path, and the sense of rpap is determined
by that of ap. The equation
(u) may thus be written in the form
(v)
Gp = Gnu +> Gim 1 Gnpjm 1 Qipj_m D 2vpjmom,
801
§ 215 | EXAMPLE
where any one of these terms may be zero in particular instances. The
Coriolis component 2vpjmoa = 0 only when the motion of the path is a
translation.
215. Example. A shaft governor P, Fig. 491, moves in an are m-n as the speed of
The flywheel is turning at 100 rpm and its angular acceleration
the flywheel changes.
is momentarily 12 rad. per sec.2_ The velocity of P with respect to its path is 4 fps and
the governor has an angular acceleration in the direction shown of ap = 18 rad. per
sec.2 with respect to the flywheel. What is the absolute acceleration of the center
P of the governor weight?
Qip/m
Flywheel Governor.
Fig. 491.
tion ap in
We shall compute the various components of this accelera
Sotution.
arrow the
short
a
with
indicate
shall
we
and
(v),
equation
in
terms
the order of the
to the student.
direction of each component, a procedure which is recommended
angular velocity of the flywheel is
/60 = 10.47 rad. per sec.
ww = (100)(27)
with P is 10/12 ft. Hence
The radius of the point M on the flywheel coincident
Anum = to? = (73)
= 91 2%ps5) 7
the direction PO.
directed toward the center O of the flywheel in
Cte
10
ie = ke
IX
= NOsifoeey
The angular velocity of O’P relative to the flywheel
perpendicular to OP.
12
Anpim
= To? =
(So
=
24 fps’,
\
in the direction PO’;
8
Qupjm = Ta = (S)os =
2 iDS
is
The
302
RELATIVE MOTION
perpendicular to PO’.
[Ch. XIV
With vp;m = 4 fps and wy = 10.47 rad. per sec., we get
2vepj)mom
=
(2)(4) (10.47)
=
133},1/ fps?,
By
in the direction PO’, since a force applied at the point of the vector for vp,y to turn
this vector counterclockwise (see wi) would act parallel to the direction PO’.
The
vector sum (ap = 146 fps?) of these components may be obtained graphically as
shown in Fig. 491(b).
216. Angular Motion about a Revolving Axis. Let the angular velocity
of wheel B, Fig. 492, relative to a stationary axis Q be w2; then let the link A,
carrying the axis Q, rotate with an angular velocity w:, as shown.
By the
principle of relative velocities, we may say that
(w)
UPR = UP,
UPZ/P,;
where Pg is a point P in the wheel B, and P4 isa coincident point in the link A. (Imagine the link A extended to include point P.)
IS T3 = 11
12, we have
UP,
=
£3W,
=
Since
(ri b
the radius of P,
12)W4.
The value of vpz/p4 = Tow2, because we is the angular
velocity of B with respect to A.
Using these expressions for the velocities in (w) we get
(x)
UPR
=
(ry a
UP,
=
1101
12)
DP lowe
=
1T1w1 HH T2W1
Hb
T9W2,
ro(w1 pb We).
In this equation, we see that riw: = vo.
equal to vp,/a, Iasmuch as the relation
UP,
=
Vo 1
Therefore,
r2(w: 4 w2) must
be
UP p/Q
must hold true. Thus, since the velocity of P in wheel B relative
to the
axis Q is T2(w: +> w2), the rotations of the wheel about the axis @ must be
the vector sum of the angular velocity of the link A and the angular velocity
of the wheel B relative to@. Since these axes Q and O, Fig. 492, are parallel,
the vector sum is simply the algebraic sum.
For example, if w; = 100 rpm
and w, = 50 rpm, each in a clockwise direction, the velocity of Px relative
to Q is (100 + 50)re. If w: = 100 rpm clockwise and w. = 50 rpm
counterclockwise, vp,/@ = (100 — 50)ro in a clockwise sense, since
w; > wo. This
principle will sometimes be helpful in solving relative velocity
and relative
acceleration problems.
The conclusion to be reached from the foregoing discuss
ion is that if a
body B is rotating on an axis with a relative angular velocit
y of w2(= ws/4)
and then if the axis is in a body A which rotates at a
velocity of wi(= wa),
the absolute angular velocity of B is the algebraic
sum (rotations in the
PROBLEMS:
DISPLACEMENTS
3803
AND VELOCITIES
same or parallel planes) of these angular velocities,
op
= 01 tw.
= wa
+ wB/A
It follows from a differentiation of w, = w1 + 2 with respect to time that
ap = a+ a = a4 + apy
(y)
where a2(= aga) is the angular acceleration of a body B with respect to
another body A whose angular acceleration is oi(= aa), and ag is the absolute angular acceleration of B. As for the case for angular velocities, the
sum a4 + aga in equation (y) is an algebraic sum when the bodies rotate in
the same or parallel planes.
This chapter gives a brief coverage of relative motion
217. Closure.
because we shall apply the knowledge gained only occasionally in succeeding
However, the study of velocities and accelerations is Important
chapters.
in its own right. Any engineer may find himself confronted with a problem
on motion and some background knowledge, however limited, will be a
welcomed beginning.
The simple relations, say,
Vp = Um
Vpim
and
Ap = Am +> Apim
are most frequently useful, although there are many instances when the
and fluid torque
Coriolis law can be applied—for example, fluid couplings:
converters in automobiles, steam and water flow in the blades of turbines,
and certain quick-return mechanisms used in machine tools. Be sure to
As an illustration, va;s is
remember the significance of the subscripts.
the vector v4 minus the vector vg; and vzya is the vector vg minus the vector
VA; etc.
kind
All problems involving two points in a rigid body, no matter what
given
relations
simple
the
by
solved
be
of plane motion the body has, may
above; the Coriolis component is zero.
Remember that the center of rotation
of zero
(centro) is a point of zero velocity, but it is not necessarily a point
acceleration (unless it is a fixed point).
Problems
Include a properly lettered diagram for each solution, even when
NOTE TO STUDENT.
principles.
the problem is solved algebraically, and use relative motion
DISPLACEMENTS
1021. From a train traveling at 60 mph
is thrown a parcel whose speed relative to
the train is 30 fps in a direction
making
AND
VELOCITIES
30° with the forward end of the train.
What is the absolute speed of the parcel?
Ans. 115.1 fps, @ = 7.49°.
804
PROBLEMS: DISPLACEMENTS AND VELocitTiEs [Ch. XIV
1022. A ship has an absolute velocity of
12 mph ona course N 30° E. A man moves
on deck with an absolute velocity of 13.9
mph in a direction N 60° E. What is the
velocity of the man relative to the ship?
1023. A battle cruiser moving in still
water at 26 knots on a course N 38° E is
struck by a torpedo whose velocity is 13
knots due west.
With what velocity did
the torpedo strike the cruiser?
Ans. 35.5 knots, 8 54.73° W.
1024. A river flows north at the rate of
4 knots. A ferry whose water speed is
8 knots is to land at a point due east of the
starting point. (a) What course must the
pilot set? (b) If the crossing requires 12
min., how wide is the river?
[NoTE: 1
knot = 1.152 mph.]
1025. An airplane pilot is to go from
field A to a field B which is 100 miles away in
a direction S 40° E from A. A 30-mph wind
is blowing from S 20° W.
‘The airspeed of
the plane is 200 mph.
What course should
the pilot set and how long is required for
the trip?
Ans. § 32.54° EK, 32.8 min.
1026. The same as 1025 except that the
wind is blowing from N 80° E.
1027. Two freight trains approach each
other on a double-track railway.
Train A
moves at 40 mph and is 3/4 mile long.
Train B moves 30 mph and is 1/2 mile long.
From the moment these trains are a certain
distance apart, it 1s 5 min. before their
rear flags are abreast.
What was the initial
distance between the fronts of these trains?
Ans. 4.58 mile.
1028. A one-mile long freight train A is
going north at 45 mph and, on a parallel
track, a 1/4-mile long passenger train is
going north at 60 mph.
(a) What distance
is required for passing?
(b) With what
speed should the passenger train move in
order to reduce the passing distance by
one-third?
1029. A motor boat with a water speed
of 13 knots is to cross a river 3/4 mile wide
which is flowing at 3 mph due west. The
boat is to land at a point 1 mile up the river
and on the north side.
What should be the
course of the boat and how long does the
trip take?
Ans. N 60° E, 6 min.
1030. An amateaur airplane pilot measures the map distance from Dallas to St.
Louis to be 520 miles N 50° EH. He heads
his airplane N 50° E and flies for 4 hours
with an airspeed of 130 mph.
Upon seeing
a city below, he lands to discover that he is
actually in Kansas City which is 440 miles
N 15° E from Dallas.
(a) Find the velocity
of the wind that blew him off course.
(b)
In what direction must he head his plane
and how long should it take him to fly
from Kansas City to St. Louis at the same
airspeed and encountering the same wind
velocity?
Ans. (a) 76.2 mph from § 74° E; (b) S 74° E
for 5 hr. 40 min.
1031. The same as 1030 except the pilot
flew only 3 hours under the same conditions,
landing at some field A instead of Kansas
City.
1032. A plane cruises at an airspeed of
120 mph.
The pilot wishes to go from A
to B, where B is 240 miles from A on a line
10° north of east. The wind is blowing
15 mph in the direction 30° west of south.
How long does it take him to go from A to B?
Ans. 2 hr, 11 min:
1033. A duck is flying south at 60 mph
and at a height of 60 ft. A hunter is on
the ground 80 ft. (horizontally) due east of
the bird when he shoots at the duck with
a gun whose muzzle velocity is 1500 fps.
Neglecting air resistance and assuming a
straight line trajectory, find the angle of
elevation and compass direction the hunter
desires to use.
1034.
Ans. 36.8°, 8 84.64° W.
The same as 1033 except the hunter
fires 1/2 sec. later.
Ans. 32.5°, S 62.5° W.
1035. A jet plane flies directly over an
anti-aircraft gun position at 600 mph and
an altitude of 1000 ft.
As the plane flies
away at this speed and altitude, the gun
fires at the instant when the line-of-sight to
the plane 1s 60° above the horizontal.
If
the projectile is assumed to travel a straight
line at the muzzle
velocity of 3200 fps, at
what elevation should
hit the plane?
the gun be fired to
Ans. 46.2°.
1036. An airplane A is spotted at a certain point A traveling at 400 mph due west.
At the same time, a plane B leaves a field
B and sets a course N 20° W.
The field B
is 150 miles 8 80° W from the point A. At
what speed must the plane B travel to intercept the plane A?
How long does it take?
The planes are on the same level and the
air is still.
Ans. 452 mph, 18.3 min.
1037. The positions of the points A and
B and the motion of the plane A are as
defined in 1036.
However, the plane B
travels at an airspeed of 460 mph.
What
must be the course of the plane B in order
to intercept the plane A, and how long does
it take?
PROBLEMS:
DISPLACEMENTS
AND
805
VELOCITIES
1038. A plane A leaves a landing field A
and moves with an absolute velocity of 400
mph due east.
Ten minutes later, a plane
B leaves a field that is 100 miles directly
southeast from the field A and travels at an
airspeed of 500 mph.
What must be the
course of the plane B in order to intercept
the plane A, and how long does it take?
No wind is blowing.
piston is 18 in. and the crank turns 60 rpm.
For 6 = 60°, what is the velocity of the
piston?
Ans. —4.08 fps.
1043. The same as 1042 except that
6 = 240°.
Ans. N 49.8° E, 13.15 min.
1039.
This problem is the same
as 1038,
with plane A moving with an absolute
ground speed of 400 mph in an eastward
direction, except that a steady wind is
blowing from the northwest at 40 mph.
Ans. N 44° EH, 12.9 min.
1040. An airplane A is traveling due
south at a constant speed of 300 mph.
At
a certain instant, another plane B, located
400 mi. due south of A, is moving east at a
constant speed of 200 mph.
(a) When will
the two planes be closest together?
(b)
What is this minimum distance?
Ans. (a) 0.923 hr.; (b) 222 miles.
1041. An ice boat is headed due north.
A 40-mph wind blows in a direction N 30° W.
The plane of the sail is at an angle of 30°,
as measured clockwise, with the direction of
motion
of the boat.
Fig. 494.
Slider-Crank Mechanism.
Prob-
lems 1044, 1045, 1068.
1044. A steam engine mechanism is represented in Fig. 494. When the engine turns
120 rpm ( = ws), what is the velocity of
the piston D for
= 10 in.? What is the
maximum velocity of the piston?
Ans. 15.8 fps, 16.05 fps.
1045. In Fig. 494 when x = 10 in. and
the piston speed is 25 fps, find the angular
velocity of the flywheel in rpm.
What is
the angular velocity of link C?
(a) If the velocity of
the wind relative to the boat is parallel to
the plane of the sail, what is the speed of
Assume that the frictional resistthe boat?
ance is negligible.
(b) If the relative veloc-
ity of the wind to the boat is perpendicular
Which
to the sail, what is the boat’s speed?
is the better way to trim the sail when
maximum speed is desired?
(c) Should the
plane of the sail make a larger or smaller
angle with the direction of the boat in
order to increase the speed of the boat?
Justify your answers.
Ans. (a) 69.3 mph; (b) 23.1 mph; (c) Smaller.
Fig. 495.
1046. In Fig. 495, AB = 9 in., BE = 9
in., and EC =12 in. The lines EC and
AD are horizontal, link CD is vertical.
Link AB is rotating clockwise at 60 rpm.
What is the absolute velocity of point # on
the rigid link BEC?
Fig.
496.
Mechanism.
Scotch
493.
1042, 1043, 1067.
Fig.
Crosshead.
Problems
1042. In Fig. 493, the pin attached to the
crank slides in the slotted member as the
crank rotates, thus giving reciprocating motion to the pump piston. The stroke of the
Problem 1046.
Ans. 3.83 fps at 338°.
Oscillating-Arm
Problems
Quick-Return
1047,
1048,
1090.
1047. A quick-return mechanism, as used
on shapers, is depicted in Fig. 496. As the
crank B rotates (at 200 rpm), the slider C
moves along the link D, which oscillates.
For a crank radius of 10 in. and a value of
= 30°, find the absolute velocity of point
E.
Ans. 20.7 fps at 82°.
PROBLEMS: DISPLACEMENTS AND VELociTIES [Ch. XIV
506
1048. The
=
same
as
1047
except
that
Bi
ine
Ca
Gring Ds —. 9 Rin
ese
in. If B turns with a constant angular
velocity of 1000 rpm, what is the velocity of
point cd when 6 = 45°?
Ans. 39 fps at 347°.
Nee,
Rolling Wheel
vv
7/
Fig. 497.
Problems 1049, 1060-1062.
1049. A 72-in. driving wheel on a locomotive with a crank circle 30 in. in diameter
is represented in Fig. 497. If the train is
moving at 45 mph, find the absolute veloccities of the crank pin when it is in the
positions A and B, The wheel rolls without
slipping.
Ans. va = 87.6 fps at 12.8°, vp = 57.4 fps
Mi Sana
Fig. 499.
Problems 1054-1056.
1054. The wheel in Fig. 499 is rolling so
velocity of pin A is 10
fps when 6 = 60°. Find the absolute velocity of slider B. See problem 1008.
Ans. 7.4 fps at 0°.
that the absolute
1055. The same as 1054 except that
6 = 150°. The distance from the point of
contact to A remains at 2 in.
1056. The same as 1054 except that
ve = 10 fps to the right and v4 is to be
found.
J,
K
Fig. 498.
Problems 1050-1052.
1050. In Fig. 498, if the wheel is rolling
toward the right at 10 fps, find the absolute
velocities of points J and K.
1051.
In Fig. 498, if the wheel
center is
moving toward the right at 10 fps, but an
excessive torque causes the wheel to slip so
that its angular velocity is 6 rad. per sec.,
(a) find the absolute velocity of the point
in contact with the ground.
(b) Find the
absolute velocity of J.
Ans. (a) 1 fps at 0°; (b) 17.5 fps at 21.3°.
1052. The same as 1051 except that » =
12 rad. per sec.
1053. In a mechanism similar to Fig.
485, p. 292, the lengths of the links are:
RELATIVE
Fig. 500.
Problems 1057-1059, 1079, 1080.
1057. The pin B, Fig. 500, has an absolute
velocity of 10 fps. Find the absolute veloeity of pin C. See problem 1006(a).
Ans. 11.2 fps at 210°.
1058. In Fig. 500, if ve = 20 fps at 240°,
find the angular velocity of driving link
AB.
Ans. 88 rad. per sec., ec.
1059.
In Fig. 500, if wag
sec., find the absolute
point of link BC.
5.62 in.
is 10 rad. per
velocity of the mid-
The
length of BC
is
ACCELERATIONS
1060. A train moving at 5 mph has an
acceleration of 15 mph per min.
What is
the acceleration of a point A, Fig. 497, on
the 30-in. crank circle of the 72-in. drivers?
1061. The same as 1060 except that the
acceleration of the crank pin when it is in
the position B, Fig. 497, is desired.
Ans. 2.75 fps? at 149.6°.
1062. The same as 1060 except that the
accelerations of the points D and E on the
wheel are desired.
1063. If the center of the rolling spool
of Fig. 501 has a velocity of 5 fps toward
the right and an acceleration of 10 fps?
toward the left, determine the velocities
and accelerations of the points D and B.
PROBLEMS:
RELATIVE
507
ACCELERATIONS
Ans. For point D, 11.2 fps at 296.6°, 72.8
fps? at 164°; for B, 8.66 fps at 90°, 53.6 fps?
Ey Wess
is 18 in. long; also wap = 2 rad. per sec.,
and a4p = —2 rad. per sec.?, where clockwise is positive.
(a) What is the absolute
velocity
of C?
(b) What
is the absolute
acceleration of C?
Ans. (a) 3.45 fps; (b) 3.1 fps”.
1070. The same as 1069 except that
wap = 20 rad. per sec.
1071. The same as 1069 except that
6 = 180°. The slope of BC remains at 30°.
Fig. 501.
Problems 1063, 1064.
1064. The same as 1063 except that the
velocities and accelerations of the points
A and C are desired.
Fig. 504.
1072.
In Fig. 504, the wheel moves
with
a velocity of 10 fps and an acceleration of
10 fps?, both toward the left. For the
position shown, determine the absolute
accelerations of the pins A and B.
Ans. 26.7 fps? at 198°; 27 fps? at 180°.
Ve
Rolling
Fig. 502.
Problem 1072.
Problems 1065, 1066.
1065. In the rolling wheel of Fig. 502,
(a) Determine
v, = 10fpsanda, = 5fps®.
the absolute velocity and acceleration of
B.
(b) Determine the velocity and acceler-
ation of the point B relative to a point A
on the wheel.
Ans. (a) 12 fps at 34°, 27.5 fps? at 7°; (b)
6.7 fps at 90°, 40.5 fps? at 312°.
1066. The same as 1065 except that the
acceleration a, is toward the left.
1067. The same as 1042 (Fig. 493), except
that the acceleration of the piston is also
Ans. 14.8 fps? at 180°.
desired.
1068. The same as 1044 (Fig. 494), except
that the acceleration of the piston D is
also desired.
Fig. 503.
1069.
In
the
Problems 1069-1071.
slider-crank
mechanism
shown in Fig. 503, @ = 30°, the crank AB
Fig. 505.
Problem 1073.
1073. The sliding members A and B,
Fig. 505, are constrained to move at all
times in the y and z directions, respectively.
They are connected by the rod whose length
is L = 10 ft. At the instant when x = 8
ft., the velocity of B is vg = 20 fps toward
the right and ag = —15 fps’ toward the
the left. Determine the velocity and acSee problem
celeration of A at this instant.
Ans. —26.67 fps, —165.1 fps?.
990.
1074. A 12-ft. wheel is rolling toward the
At a certain instant, the velocity
right.
and acceleration of the center are +30 fps
Determine for
and +6 fps2, respectively.
this instant the absolute velocity and ac-
PROBLEMS: RELATIVE ACCELERATIONS
508
celeration of a point P on the wheel, whose
are
coordinates in relation to the center
= Sih, Aacl p= —ahsie
[Ch. XIV
C when 6 = 45°. The value of wec may
be found from vez.
1079. The link AB of the quadric chain
in Fig. 500, has a counterclockwise angular
velocity of w = 10 rad. per sec. and, in the
same sense, an angular acceleration of a = 3
rad. per sec.*. For the position shown,
find the absolute acceleration of pin C.
HINT: Find first the normal and tangential
components dnc and ay.
1080.
Fig. 506.
Ans. 13.9 fps? at 253°.
The same as 1079 except that the
acceleration of the midpoint of BC is desired.
Link BC is 5.62 in. long. urint: Problem
1079 must be solved first. Why?
Problems 1075, 1076.
1075. In Fig. 506, the weight W is in
motion such that v, = 2 fps and a, = 4
fps’, both downward.
The rolling wheel
is pulled along by a cord attached to its
axle O. Find the velocity and acceleration
of (a) point A, (b) point B. Show all
vector equations and solve graphically.
Ans. (a) 0, 2 fps? at 90°; (b) 3.46 fps at 30°,
8.48 fps? at 24.1°.
1076. In Fig. 506, the velocity and acceleration of W are, respectively, 4 fps and
8 fps’, both downward.
The wheel slips
as it rolls with w = 1.5 rad. per sec. and
a = 3 rad. per sec.2, both clockwise.
Find
the absolute velocity and acceleration of A.
aby:
1 2
[ A!
10!
Problems 1077, 1078.
1077. In Fig. 507, as B rotates around
A, point C slides along the inclined plane.
At
the
position
shown,
where
@ = 45°,
wpo = lrad. per sec. clockwise and ago = 0.
Find w4, and a4, at this instant.
Ans. 4.48 rad. per sec., cc.; 2.9 rad. per sec.2,
CG,
1078. In Fig. 507, as AB rotates at a
constant speed of 4 rad. per see. clockwise,
point C slides along the inclined plane.
Find the velocity and acceleration of point
CORIOLIS’
2-ft.
arm,
Fig.
508,
rotates
§ 216).
By the principle of relative motion,
find the absolute velocity and the absolute
acceleration of point D. (See problem 1085).
Ans. 31.4 fps at 30.65°, 495 fps? at 286.6°.
1082. The same as 1081 except that as
is in the opposite sense to that shown.
1083. The same as 1081 except that the
velocity and acceleration of point F are
desired.
1084. The same as 1081 except that the
and acceleration of point @ are
(See problem 1086).
Ans. 19.4 fps at 350.8°; 400 fps? at 238.8°.
velocity
desired.
LAW
1085. Solve 1081 by using the Coriolis
law for accelerations.
1086. Solve 1084 by using the Coriolis
law for accelerations.
1081. The
Problems 1081-1086.
about a fixed center and carries a 3-ft.
wheel B which rotates about a spindle in
the end of the arm.
For the arm £,
a, = 6 rad. per sec.? and w; =8 rad. per
sec. The wheel B has a motion such that
a2 = 3 rad. per sec.? and w: = 10 rad. per
sec. with respect to its own center (see
Horizontal
Fig. 507.
Fig. 508.
1087. Epicyclic gear trains are used in
gear reduction units of various kinds; for
example,
in automoble
transmissions.
Figure 509 represents part of such a gear
PROBLEMS:
CORIOLIS’
309
LAW
Fig. 510.
Fig. 509.
train.
The
Problems 1087, 1088.
i6-in. gear A is a stationary
internal gear and the 4-in. gear C rolls on A
as it moves with the rotating arm B.
Suppose the arm B is suddenly speeded up
from rest. The designer may wish to investigate for the maximum accelerations in
order to estimate the maximum forces
involved.
Problem 1089.
1090. The crank B of the quick-return
mechanism, described in problem 1047 and
Fig. 496, rotates at constant speed. The
crank B, whose radius is 10 in., rotates at
200 rpm.
At the position where @ = 30°,
determine (a) the velocity of the slider C
relative to D, (b) the angular velocity of
D, (c) the total acceleration of the pin
connecting B and C, and (d) the tangential
and radial components of the absolute acceleration of point EZ.
Ans. (a) 10.7 fps; (b) 4.51 rad. per sec.;
(c) 365 fps?; (d) 611 fps? and 93.5 fps?.
For the arm B, let w; = 10 rad.
per sec. and a = 20 rad. per sec.” for the
position shown.
For the gear C, w2 = 40
rad. per sec. and a2 = 80 rad. per sec.?
with respect to its own center.
Determine
the acceleration of point H (a) by using
the Coriolis law, (b) by using relative
accelerations.
Ans. 200 fps? at 0°.
1088. The same as 1087 except that the
acceleration of point G is desired.
1089. In Fig. 510 is represented the vane
of a centrifugal water pump with a particle
of water P traveling outward along the
Relative to the vane, the tangential
vane.
velocity of P is 15 fps and the tangential
acceleration is 15 fps. The angular velocity of the pump is uniform at w = 20 rad.
per sec.
of P?
What is the absolute acceleration
Ans. 1198 fps? at 278°.
Fig. 511.
Problem 1091.
1091. A body B slides on a link A which
rotates about a fixed center, Fig. 511.
If
aa = 10 rad. per sec.?, wa = 6 rad. per sec.,
Veja = 2 fps, and ag/a = 25 fps? outward,
what is the absolute acceleration of B?
Ans. 35.2 fps? at 149,85°.
1092-1100. These numbers may be used
for other problems.
Chapter XV
FORCE SYSTEMS THAT
RECTILINEAR MOTION
PRODUCE
218. Introduction.
We are now prepared to return to the consideration
of the effects of forces on particles and bodies.
In this study, we shall be
dealing with external unbalanced force systems which invariably cause
changes in the motion of the bodies on which they act. This branch of study
is called kinetics.
In this chapter, we shall deal with coplanar force systems
(or systems which can be reduced to or considered as coplanar systems) for
which the resultant is a force.
In these systems of forces, there is no resultant
couple. Renewing this study of forces, we recall that a free body carefully
sketched with the external force vectors in their proper relationships is a
most effective aid to clear thinking.
219. Sir Isaac Newton.
Even the briefest mention of Newton (1642-1727)
and his contributions to science is too long for a footnote.
Hence, a few
words here about the man from whose laws (§ 220) the science of analytic
mechanics may be derived.
Many people credit Newton (born on Christmas in the year of Galileo’s
death) with being the greatest scientist of all times.
It would certainly be
difficult to prove any one else greater.
He was born at Woolsthorpe, England, of parents who were farmers of moderate means.
His mechanical talent
and his lack of interest in farming became apparent early. Among his grammar school achievements were a water clock and sun dials. Within two
years after receiving his degree from Trinity College, Cambridge, he discovered the binomial theorem, took the first steps toward the invention of
calculus, started experiments on color, and had begun to speculate on eravitation. His achievements were so varied and numerous that only a few will
be mentioned: the reflecting telescope, the composite nature of sunlight, a
science of optics, the invention of a thermometer, invention of fluxions
(forerunner of modern calculus), and, of course the most monumental, the
law of universal gravitation.
The idea that the sun and not the earth was the center of our universe
was generally accepted at the time of Newton, and some of Newton’s scientific
310
§ 220 |NEWTON’S
LAWS OF MOTION
311
predecessors had speculated upon the notion of universal gravitation, but no
Newton’s first calculation in 1666 to establish the
proof had been devised.
concept of universal gravitation was based on an erroneous estimate of the
size of the earth. Had not M. Picard made a more accurate estimate of the
earth’s size in the nick of time, there is no telling how much the development
of the science of mechanics might have been delayed. Having learned of
Picard’s measurement in 1682, Newton returned to this problem with reIt showed
newed hope. The calculation was a comparatively simple one.
the moon
prevented
also
stone
a
of
falling
the
that the acceleration governing
has a
moon
the
orbit,
closed
a
in
from moving in a straight line. Moving
the
of
because
normal or centrifugal (§ 236) acceleration toward the earth
With the calculation partially completed,
attractive force between masses.
be verified. He became so jittery that a
to
going
was
he saw that his theory
friend had to complete the calculation for him.
In 1695 he was made
Newton was active in public life in many ways.
He
1705.
Warden of the Mint of England and he received a knighthood in
ities
laid no claim to unusual sagacity and possessed none of those eccentric
eccentric.
as
classed
be
ss
mindedne
absent
sometimes found in genius, unless
He was
He attributed his successes to application and patient thought.
ce.
intoleran
very religious and very tolerant of everything except
As we have learned, Galileo established
220. Newton’s Laws of Motion.
tion, and time for
the kinematic relations of displacement, velocity, accelera
It was evident from
bodies moving under the influence of a constant force.
its motion, but
his work that a force was necessary to cause a body to change
ted the
formula
Newton
that
later
years
it was not until some one hundred
stated
be
may
laws
s
Newton’
laws which related the force to the motion.
as follows:
with a constant
I. Every particle remains in a state of rest or moves
acts on it. (We
velocity in a straight lme unless an unbalanced force
brium (R.=,0),
equili
of
on
recognized this statement as defining the conditi
a special case.)
tional to the resultant
II. The acceleration of a particle is directly propor
its mass, and the sense of
force acting on it and inversely proportional to
(This is the
resultant force.
the acceleration is the same as that of the
basic condition of kinetics.)
te reaction.
Ill. To every action, there is an equal and opposi
ant force is necessary to
Considering the first law, we see that a result
move in other than a straight
cause a body to change its speed or to make it
causes it to resist a change
line. This property of a body or particle which
unbalanced (resultant) force
in its motion is called its inertia. Thus, if an
speed of the particle will be
acts upon a particle, it may be that only the
812
FORCE SYSTEMS THAT PRODUCE RECTILINEAR MOTION
[Ch. XV
changed (if the line of action of the resultant force is in the direction of
motion), or only the direction of the velocity may be changed, or the velocity
may be changed both in magnitude and in direction.
The second law, which gives quantitative expression to the first law,
may be expressed in mathematical form. Keeping in mind that a “particle”
is a hypothetical something which has weight and mass but occupies no
space, let A represent the resultant force, a the acceleration, and m the
mass of a particle; then Newton’s second law says
(a)
R
UR ers
or
a =C—,
|
3
where C' is a constant of proportionality.
In statics, the action and reaction of the third law refers, for example,
to the action of a body A on a body B and the equal and opposite reaction
(force) of the body B on the body A. In kinetics, the reaction may be not
only in the form of an equal and opposite force, but also in the form of a
change in the motion of the body or particle being acted upon.
221. Units.
Mechanics is a rational subject.
In any equation of mechanics the units must balance; that is, if one side of the equation is in
pounds, the other side must be in pounds also. Thus, if we write Newton’s
second law in the form R = C’ma, where C’ is the reciprocal of the constant
C’ in equation (a), the unit of R must be the same as the unit of C’ma.
Asa
matter of fact, we may drop the constant C (or C’) altogether, and write
the equation in the form
(b)
R'='ma,
in which case, the units of any two of the quantities R, m, and a must define
the units of the other quantity.
From equation (b), we may say that a unit
force is that force which gives to a unit mass a unit acceleration.
In fact,
such a relation of units exists in the centimeter-gram-second system, where a
unit of mass is a gram, a unit of acceleration is a centimeter per secondsecond, and the corresponding unit of force is a dyne. This system of units,
which is used for all engineering and scientifie work in continental Europe,
is
commonly used in scientific work and in electrical measurements in
this
country.
However,
the English system of units is not so simple.
Some groups,
the engineers among them, generally take the unit of force as a
pound and
the unit of acceleration as feet per second-second.
With these two units
defined, the unit of mass must be pound-second-second per foot,
or a slug
(§ 156), if the constant C is to be unity. The scientists often use
the unit
of mass as a pound, in which event, the unit of force must be
Ib-ft.
Rk = ma
=,
[Not used in this text.]
sec.”
§ 222 |COMPONENT
313
FORCES AND ACCELERATIONS
We shall of course continue to use the pound as
a unit called a poundal.
the unit of force.
Suppose that a single force Ri acts on a particle of mass m and produces
an acceleration of ai, in accordance with Newton’s second law. If another
single force Ry acting on the same particle, produces an acceleration @2, we
may write (from Newton’s second law)
i leah
R.
da’
where R; and R, are any two resultant forces. We know that if only the
force of gravity acts on a particle, its acceleration is the acceleration of
gravity, g. The force of gravity W is the weight of the particle. Thus,
we know (or can easily measure) the value of one force W and the value of
the corresponding acceleration g. Then with any force R producing an
acceleration a, we get the proportion
3
NERS
(c)
a
g
k= isa.
or
g
If a consistent system of units is used, so that, in equation (a), C = 1, we
may compare equations (c) and (b) and conclude that if the pound is the
unit of force, the mass is
(d)
Wa
od beg
lbp
wilb-see7
ft./sec.2. © ft. (slug).
involvThus, the equation that we shall use in solving problems of motion
ing force, mass, and acceleration is
(43)
R=ma
= fe
quantities’
where the resultant force R and the acceleration a, both vector
tion is
accelera
the
,
constant
is
force
If the resultant
have the same sense.
varies.*
tion
If the resultant force varies, the accelera
constant.
In the solution of problems,
222. Component Forces and Accelerations.
te axes. If neither
coordina
certain
we often find it convenient to choose
the relations
of these axes is in the direction of the resultant acceleration,
forces are
between the component accelerations and the component
Ww
“FF, = —; Gz,
(e)
Ww
(f)
ZF, = vi Ay;
where
(CADPR CE ie arte
and
(a,? + 4,7)!? = a.
Appendix B.
*A somewhat longer, but still brief, discussion of units is found in
814
FORCE SYSTEMS THAT PRODUCE RECTILINEAR MOTION
[Ch. XV
If it happens, for example, that a, = a, the absolute acceleration, then
a, = Oand SF, = 0. In the analysis of the motion of a particle in a plane,
there are therefore two independent conditions (the force system is a concurrent system).
These conditions are expressed by equations (e) and (f),
where either a, or a, may be zero. The values of YF, and ZF, are found
just as explained in the chapters on statics from a free body of the member
being studied.
Not only should the student use the principles of the free
body, but he should also be prepared to use the kinematic relations previously developed, because many problems in kinetics cannot be solved
without the application of the principles of kinematics.
223. Example.
A body A weighing 20 lb. is resting on a 45° incline for which
f = 0.2 (kinetic friction). A horizontal force Q = 10 lb. acts on the body as shown in
Fig. 512. If the body starts from rest, what is its velocity after 5 sec.?
Sotution.
In this type of problem, the body may be considered as a particle.
First, we note that in order to obtain v 4, we must know the acceleration of A. Second,
we expect that a may be found from the equation R = ma.
Therefore, draw a free
body. Weare sure that W acts vertically downward, that N acts normal to the plane.
The direction of Q is specified.
Since we are not sure in what direction the body will
move, we are not sure in what sense to point the frictional force. However, the
component of W down the plane is (20)(0.707) = 14.14 lb., which tends to cause the
body to move downward.
The component of Q up the plane is (10)(0.707) = 7.07 lb.
Hence, if any motion occurs, it will be down the plane and therefore F points upward.
Noting that the motion is entirely parallel to the plane and that consequently =F, = 0,
we find
=F, = N — Qsin 45° — W cos 45° = 0,
from which
N = (10)(0.707) + (20)(0.707) = 21.21 Ib.
Therefore, if the bodies move, F = fN = (0.2)(21.21 ) =4 .242 lb.
we get (upward direction along the plane is positive)
Using =F. = ma,
=
=F, = Q cos 45° + F — W sin 45° = —a
M
g
20
= 7.07 + 4.24 — 14.14 = —a,
g
from which a = —4.56 fps*, a constant acceleration as long as the forces remain
constant. Substituting a in the sum =F, with a positive sign is tantamount to assuming
that the acceleration is wp the plane. The negative sign for the answer shows that
it
is down the plane. With this acceleration, the velocity after 5 sec. is
v = at = (—4.56)(5) = —22.8 fps
down the plane.
224, Example.
SotuTion.
we get
This example is the same as that in § 223 except that Q = 25 lb.
Summing forces in the y direction and solving for N as before, Fig. 512
N = (25)(0.707) + (20)(0.707) = 31.8 lb.
315
§ 225 | EXAMPLE
The maximum frictional force for f = 0.2 is
Pe
—(02)(318)
= 6:36 lbs
If there is any doubt as to whether motion will occur, a sum of the forces parallel to
the plane, omitting the frictional force, compared to the frictional force, will tell.
With the upward direction as positive, we find
Q cos 45 — W cos 45 = (25)(0.707) — (20)(0.707) = 3.5 lb.
friction
This is the net motivating force. Since it (3.5 lb.) is less than the kinetic
ee
=
(6.36 Ib.) to be overcome, motion does not occur. _
‘i
=
aa
Af
Q
10 lb.
x
&
FOESN
Spal
\
Fig. 512.
-225. Example. _ A bedyA weighing W4 = 50 lb. is ona plane inclined at afi angle
plane isfa = 0:3. A cable
of 6 = 30° (Fig. 513). The coefficient of friction on this
for which the coefficient of
C,
member
attached to this body passes over a stationary
a body B weighing
suspended
is
cable
the
of
end
friction is fe = 0.2. From the other
what is the speed
and
cable
the
in
P,
and
Pi
tensions
the
are
Wz = 100lb. What
is negligible.
cable
the
of
of the bodies after they move 20 ft. from rest? The weight
rd, pulling the
Evidently, if motion occurs, the body B moves downwa
Sonution.
into three free
system
the
up
body A up the plane. To solve this problem, break
get
bodies. Starting with the free body of A, we
one
(Ng
or
=F, = N — 50 cos 30° = 0
the direction of motion as posiTherefore, F = fN = (0.3)(43.3) = 12.99 Ib. Taking
find
we
plane,
the
to
tive and summing the forces parallel
Wa
oF, = P, — F — Wasin 30° = pd
(g)
oF, = P; —37.99 = — a.
another equation is needed for a
In this equation, there are two unknowns. Hence
the cable slides across the member C is
solution. The relation between P1 and P» as
we fnd-(e~=_120° = 2.09 rad.)
given by equation (20), p. 112, from—which
(h)
N8 1 52P a
Put?aePes
Dice
pest
so that still another equation is needed.
Now there are three unknowns, P1, P», and a,
of B.
forces ZV on the free body
This equation is obtained from a sum of the vertical
316
FORCE SYSTEMS THAT PRODUCE RECTILINEAR MOTION
[Ch. XV
The positive direction, which has been taken in the direction of motion, is now downward.
100
(i)
zV = 100 — P, = i a.
Observing that the acceleration of B is the same as that of A, provided the cable
is inextensible, we may write a4 = ag = a. To solve the equations (g), (h), and (i),
we substitute the value of P; from (h) into (i), and obtain
100
Equations (g) and (j) may be solved simultaneously for P2 and a.
12) = 0 Ilo.
and
Doing so, we find
a = 7.73 fps’.
The value of P; = 76 lb. may be found from either (&) or (i). Since the forces are
constant, the acceleration is constant and the speed of the bodies after they have
moved 20 ft. is found from the equation
v = 2as = (2)(7.73)(20) = 309.2,
whence » = 17.6 fps.
226. Motion of the Center of Gravity of a Rigid Body.
Consider a body
of mass m (Fig. 514) in plane motion.
If this body is composed of particles
whose masses are dm, dino, +++ dm,, then
m = dm, +dm2.+---dm,
Fig. 514.
=
Xdm.
No matter what the nature of the motion
of a particular particle may be at a
certain instant, the resultant force dR on
the particle is equal to its mass times its
acceleration, that is,
dR = (dm)a
in accordance with Newton’s second law. This force
dR = (dm)a is called
the effective force for the particle. Referring the motion
to any fixed x
and y axes, Fig. 514, and using components in these directi
ons we may say
for any particle dm in general, that
dk, = (dm)a,z
and
dR, = (dm)a,,
where dR, is the x component of the effective force
dR on the particle and
a, is the x component of the acceleration a of the
particle, and where the
y subscript refers to similar components in the
y direction. Considering
the x direction only for the moment, we may
write for the individual particles
dR = dmian,
AR 2 = dm20z2,
dRon = dm
nQen-
Then by addition, we find
dR
t+ dRye +--+
bing = dma
AM2One + +>. GiyGak:
§ 226 |MOTION
817
OF THE CENTER OF GRAVITY OF A RIGID BODY
The sum of the terms on the left side of this equation is seen to be the sum
of the x components of the resultant forces for all the particles of the body;
that is, this side of the equation may be represented symbolically by 2 dR;.
But the sum of these components for all the particles is the « component
R, of the resultant force on the body, dR, = R,. Thus, we have
(k)
ive =
dmiGr1
== AM2A22
mF Betas
OMe.
Now the zx coordinate of the center of mass of the body is defined by the
relation (§ 110)
mE
= dmyt, + dmor,.
+ +++ + dm 1Xp.
Successive differentiations of this equation with respect to time give
aco paclt
Te
a +.--+ dm, ain
°r + dme ue)
may = dm
mM ap = Ohi qe + dmz qe +..-+ dmy
(1)
m a
=
aM 1471
+ AM2Az2
qe
ap oce a5 Divide
nt of
where we recognize that, for example, d?xi/dt? = az, the x compone
the
be
to
seen
is
the acceleration of particle 1. The acceleration d?x/dt®
which
body,
a component of the acceleration of the center of mass of the
of (k) and
sides
nd
right-ha
the
that
see
also
We
we may designate by d;.
is,
that
equal;
are
sides
d
Therefore, the left-han
(1) are identical.
(m)
Oe
2
R,=m ap 7 Mae:
In a like manner, using the y components, we find
2 ly,
ay
(n)
Ry, =m
:
acceleration @ of
Then, in terms of the resultant force R and the resultant
the mass center, we get
(0)
Rk = ma.
, the resultant force
It has been shown that, for any body in plane motion
ation of tts center of mass
on it is equal to the mass of the body times the acceler
The resultant force on a particle of a body may
(or its center of gravity).
the resultant of an external
be either an internal force alone, or it may be
interior of the body, it is
and an internal force. If the particle is in the
actions of adjacent
acted upon only by the adjacent particles and the
body is rigid, that is,
particles are considered as internal forces. If the
fixed, these internal forces
if the relative positions of the particles remain
Some of the particles on the
remain in equilibrium among themselves.
which may or may
surface of the body are subjected to external forces,
318
FORCE SYSTEMS THAT PRODUCE RECTILINEAR MOTION
[Ch. XV
not be in equilibrium. Thus, the resultant of the effective forces acting
on all the particles of a rigid body is simply the resultant of the external
force system.
This principle is called D’Alembert’s principle.*
Therefore,
we find # in equation (0), or #, or R,, in the usual manner from a free-body
diagram showing the eaternal forces. It is important to remember that the
acceleration in equation (0) is the acceleration of the center of mass and
that the vectors for R and a have the same sense.
227. Location of the Resultant—Body in Rectilinear Translation.
It is
often convenient to know the location, relative to the body, of the line of
i
—
Yo
sa
° dmyz
(%2, Yo)
action of the resultant.
Taking moments
of the x components of the effective forces
dR, = (dm)a, on the particles of Fig. 514
about the wz plane, where the ¢ axis is
perpendicular to the page, we have
(dmiaz1)y1 + (dmodz2)y2 + +++ + (dren)Yn
Fig. 514.
For the special
Repeated.
since all particles
then becomes
case
of rectilinear
trans-
lation, we know that a@z1 = a;2 =Gzn = Be}
have the same acceleration.
The preceding expression
ar(dmiyr + dmay2 + ++» + dmnYn) = azd(dm)y.
We recognize that Z(dm)y = m7; hence
(p)
a,2(dm)y = a,my = Ry,
where a,m = R,, by equation (m).
yz plane, we may write
(q)
Similarly, by taking moments about the
(dimidyi)t1 + (dmoay2)t2 + --- + (dmydyn)tn = lige
Since the components R, and R, intersect on the line of action of the
resultant R, equations (p) and (q) show that the x and y coordinates of
a point on the line of action of the resultant are the same as the coordinates
of the center of mass; therefore, the line of action of the resultant force
on
a body in translation passes through the body’s center of mass (eg). Note
carefully that this statement holds for a body in translation (no angular
motion).
*Jean le Rond D’Alembert
(1717-1783), a Paris-born mathematician and philosopher,
studied law and medicine, but his principal contributions to posterity were
in mathematics,
especially integral caleulus.
His book, Traité de Dynamique, is important not only for
the presentation of the principle which goes by his name, but also
for its use of the caleulus
in the field of mechanics.
Not having had calculus as a tool, Newton
relied on geometry
in his Principia.
D’Alembert was a contemporary and good friend of Voltaire.
He
was an excellent musician.
It should be mentioned that the job of developing analytic
mechanics (the mathematical approach) was practically completed
1813) only a few years after the death of D’Alembert.
by Lagrange
(1736-
§ 229 |METHODS
OF SOLVING
PROBLEMS
319
228. Inertia Force. If the right-hand term in the equation Rk = ma is
transposed to the left-hand side, we have
(r)
R — ma = 0.
From this equation, we see that 7f a force equal to md, collinear and opposite in sense to the resultant R, is added to a free body, the result is a system
of forces in equilibrium (since 2/7 in any direction would then be equal
to zero). This statement applies when the resultant is not a couple, a case
we shall discuss later.
The resultant force is called the effective force; hence, this opposite force
ma is called a reversed effective force (REF) or an inertia force. The
inertia force is a dynamic reaction and it should not be included in a freebody diagram unless the method of solving a problem, as explained in the
next article, calls for its addition.
If desired, the components of the inertia
force, md, and md,, may be used, though this procedure is seldom advantangeous for bodies in rectilinear motion.
The line of action of the reversed effective force (inertia force) for a body
in translation passes through the center of gravity (eg). Also notice the
statement that the inertia force is a dynamic reaction (Newton’s third law)
Its
and that it is not a force in the same sense as the other external forces.
chapters.
succeeding
through
proceed
we
as
significance will become clearer
From the preceding discussion we
229. Methods of Solving Problems.
One
conclude that either of two plans may be used in solving problems.
the
then
forces;
external
actual
the
only
of
plan is to make a free body
conditions which apply are
(s)
=M., = 0,
LF, = ma,
ng
[INERTIA FORCE NOT IN FREE BODY]
where we note that the sum of the moments (2M., = 0) is taken about
the cg. The cg is used as a center of moments because it is known to be on
Other points may be used, but we must
the line of action of the resultant.
remember
that the
=M #0
for an unbalanced
system
of forces except
when the center of moments is on the line of action of the resultant.
The other plan for solving problems is to add the inertia force ma to
the free body and then use the conditions of equilibrium.
(t)
Sia,
LF, = 0,
DM = 0,
[INERTIA FORCE INCLUDED IN FREE BODY]
where the x and y directions are any two convenient directions and where
>=M represents the sum of the moments about any convenient point. However, the line of action of the inertia force must be properly located; through
the center of mass (for translation) and its sense is opposite to that of the
acceleration (inertia force = dynamic reaction).
FORCE SYSTEMS THAT PRODUCE RECTILINEAR MOTION
820
[Ch. XV
A ladle of molten metal is supported by a carriage whose wheels
230. Example.
A and B are 12 ft. apart (Fig. 515). The total weight of the ladle and carriage is
1000 Ib., the center of gravity of which is 8 ft. below the track. A force P = 150 lb.
is exerted horizontally at a distance d = 4 ft. below the surface of the track.
If the
frictional force at each wheel is 0.05 of the corresponding normal force, find the
acceleration, and find the normal and frictional forces at the wheels. Assume that
the ladle and the carriage are rigidly connected.
First Sotution.
From the free body of Fig. 515, which shows all of the external
forces, we sum forces in the y direction where the acceleration is zero and get
Shee NN
A sum in the 2 direction gives _—
ee 1000r==s0.
ae
=
—
“SF, = P — Fa — Es =mai-——_—__
nay
es
“150 — F, — F, — 10008.
ae
A
eae
The relations between the frictional and normal forces are
Fa =fNa = 0.05Na,
\
Fs =f{Nz = 0.05Nzp.
.
y
*
Fig. 515.
Fig. 516.
76
mg moments about the cg, (we recall that the center of moments‘nust be on the
line of
action of R if =M = 0), we find
Pa
=Meg
~~
=
6Na
—
6Ne
—
8F4
—
8Fz3 +
4P
=(()
= 6Na — 6Nee— 8F 4— 8F 3+600 = 0.
We now have five equations-and_five unknowns “The solution of these equations for
the unknowns yields (the student should make this solution)
Ne = 517\b.,
Fe = 25.85lb,
Na=483lb,
Fa = 24.15lb.,
a = 3.22 fps.
Second Sotution.
In this solution, the inertia foree m&@ = 1000 G/g is added to
the free body (lig. 516). By this act, we place the body in a simulated equilibrium
so that the sum of the forces in any direction and the sum of the moments about ony
point 1s each equal to zero. Summing forces in the vertical direction, we get
ZF, = Na+ Ne
as before.
— 1000 = 0,
A sum in the horizontal direction (Fig. 516) gives
SFPeay
i
ll 150
Te
ep
— Fu — Fz —
g
10004a _9
§ 231 | VARIABLE FORCES
821
The relations between the frictional and normal forces are as before,
Fa=fNa = 0.05Na,
Fp gNe
0.05oe
Now choosing point A as the center of moments, we find
2M 4 =4P
+ 12Ny
— 6W
—8—* =0
80000 _ 0.
= 600 + 12Nz — 6000 —
Again, we have five unknowns and five equations, from which we should find the same
values of the unknowns as we did in the first solution. The student should be sure to
do all the mathematical work in making each of these solutions. It is worth noting
that, since f = 0.05 for both points A and B,
Fa+Fs=fNa+fNe =f(Na+ Ns)
= (0.05)(1000) = 50 Ib.
This value of F4 + Fs in the equation for =F, yields @. The other details are left
to the student.
There are times when one or the other of the preceding methods is advantageous
in solving problems.
The reader is advised to learn both methods; however, at the
discretion of the instructor, he may concentrate on one method to the exclusion of
the other where time is short.
7
Al
ee
231. ‘Variable
ere-are a number of instances where the effective
force is not constant, one of the most common being the force exerted by a
spring. The action of helical and leaf springs usually follows Hooke’s law*
with reasonable accuracy, at least as long as the material of the spring is
not stressed beyond a certain point (called the elastic limit). As applied
to springs, Hooke’s law states that the force exerted by a spring is directly
*Robert Hooke (1635-1703), a brilliant contemporary of Newton, born on the Isle of
Wight, was the son of a minister.
Being a sickly youth and unable to attend school
reguarly, he was left much to his own devices.
There was soon no doubt but that he
had a marvelous mechanical aptitude, an interesting and usually disagreeable personality,
and also a prodigious mind.
He mastered Euclid’s six books of geometry in one week.
Probably
because
of an
inferiority
complex,
originator, but lacked that characteristic
he drove
himself
(which Newton
hard.
He
was
a great
had) of carrying through his
ideas to a logical conclusion.
While attending Christ Church, Oxford, he aided Boyle in
his experiments with air. Among his prolific endeavors were: a study of the role of air
in respiration and combustion; a statement that indicated his belief in universal gravitation
(about which he quarreled with Newton); the relation between changes in barometric
pressure and weather; the determination
of the frequency of vibrations for each musical
tone; origination of the short wave theory of light; first application of the spiral spring to
time pieces (the center of a violent dispute with Huygens who had independently done
the same thing); pointed out the rotation of Jupiter.
He thought of many things first,
but usually left his notions unproved so that eventually someone else got the credit.
Physically unattractive, somewhat deformed, always in ill health, and irascible, Hooke
was involved in many unpleasant controversies with his contemporaries.
His discovery
of Hooke’s law was first written in cryptic form in 1660 as follows:ceiiinosssttuu,
which rearranges to ut tensio sic vis, free translation of which is “force is proportional to
elongation.”’
The cryptic form was for the purpose of keeping his discovery a secret so
that he might profit from it by a blanket patent, which he never got. He lived in near
poverty, but left a good sum of money at his death in a locked chest.
$22
FORCE SYSTEMS THAT PRODUCE RECTILINEAR MOTION
proportional to the deflection of the spring.
of a spring; then the corresponding force
Fas
or
where K is a constant for a particular spring.
[Ch. X V
Let s represent the deformation
F = Ks,
This constant K has various
names—the spring constant, the rate of the spring, the modulus of the
spring, and most common, the scale of the spring. Since K = F'/s, the unit
of K is seen to be force unit per length unit. We ordinarily state the scale
However, to be consistent in the units, we
of a spring in pounds per inch.
generally convert the inch unit to a foot. Thus, a scale of 50 lb. per in. 1s
equivalent to (12)(50) = 600 lb. per ft.; that is, 50 Ib. will compress the
spring 1 in., and 600 Ib. will compress it 12 in. or 1 ft. (if such a compression
The free length
were possible within the elastic limit of the material).
(Fig. 517) of a spring is its length when no load is imposed on it.
In general, if a force is variable, it is shown on the free body in terms
of some function of s, ¢, v, etc. Then, the sum of the forces is made in the
usual manner and equated to ma (ZF = ma). Then we may be able to use
one of the equations a = vdv/ds, a = dv/dt, or a = d’s/dt? and integrate
for some desired result.
232. Example.
A body (W = 75 lb.) is resting on a smooth plane which inclines
6 = 30° with the horizontal. It is held in position by being attached to a spring
Center of
Vibration \
whose weight is negligible and
whose scale is A = 50 lb. per in.
If the body is pulled down the
plane from its equilibrium position
for a distance of r = 2 in. and
then released, determine (a) its
i)
velocity when it has returned to
sy
sewn
or
er
“Gory
the equilibrium position, (b) the
acceleration at the instant it is
I cede
sor,
:
:
:
eal
released, (c) the time of vibration.
we
Sotution.
(a) The situation
Fig. 517.
described in the problem is pictured in a general way in Fig.
517. If the body is at rest, the extension of the spring is s,. In this equilibrium
position, a sum of the forces parallel to the plane, placed equal to zero, shows that
ae,
j
ois
Ks, = W sin 0.
Now let the body be pulled down a distance r from the equilibrium position and
released.
It will vibrate with an amplitude r = 2 in. When the body is a distance
from the equilibrium position, its total deflection is s. At this instant the force up
the plane is
ae
:
‘
.
Ks = K(s. + 2) = Ks, + Kz =:W sin 6+ Kz.
Taking as positive the direction down the plane (as assumed in the foregoing displacements s, and x), we get
=F, = Wsing
— Ks = Wsine — W sine
— Kz = Ho
g
§ 232 | EXAMPLE
823
or
ge
(u)
Kg as
This equation gives the acceleration a at any displacement z. Compare with equation
(42), p. 269. Substituting the foregoing value of a in v dv = a dz, we have
v
0
0
;
==i=
—| —
I
W
save
W
tive
Ww
72
a
a
;2 |
fe
a
where the limits are from zero velocity at the instant of release to the desired velocity
v, and from a displacement of x = r = 2 in. = 1/6 ft. to zero displacement (the position of equilibrium).
Integration gives
edz =
vdv = —
ee
Dy
—
Ome \
Ue.
Ww
(600) (32.2)
CANCE
from which v = 2.67 fps for W = 75 lb. and K = 600 lb. per ft.
(b) At the instant of release, the displacement is ¢ = r = 2 in. = 1/6 ft.
(600)(32.2)(8)
the acceleration is
gx
0)(382.2)(@
=
= — 42.9 fps?.
W
75
Hence
ESS
(c) By comparing equation (u) of this example with equation (42), p. 269, we see
that the motion of this body is simple harmonic, where
=e KayWe
CS
Thus, the equations for simple harmonic motion may be used to solve parts (a) and
(b), and the student should check the preceding results by the harmonic equations.
From
Fin
(v)
py Ae
we find w = 16.03 rad. per sec.
Kg =
(600)(32.2)
OA
75
=
5)
BS;
Then by equation (z), p. 270, we have the period as
Qa
Qa
da = —ae 08 = ().391 1 sec.,
the time for the body to return to the position from which it was released.
On a
will continue
frictionless plane, this vibration, called an undamped free vibration,
vibration is
a
Such
rest.
to
indefinitely. With friction, the body will soon come
called a damped free vibration.
We may derive various equations, if we desire, for a body vibrating on
an incline as in Fig. 517.
example, we find
W
For instance,
Ks
from
the first equation
sin 0
Using this in the expression for w*, we get
weet Kg Lai sin 6
Ss
W
The frequency ¢ is the reciprocal of the period,
ety
pets
1
Ve ihGee) Wa
gsin 6\"”
ncpe
Ufo
in this
FORCE SYSTEMS THAT PRODUCE RECTILINEAR MOTION
824
[Ch. XV
where g sin @ is observed to be the component of the gravity acceleration
vector parallel to the plane. However, the reader is urged to solve problems
at this stage by making a free body and reasoning his way through as illustrated in the foregoing example.
Referring to equation (v), we see that the period (and the frequency)
depend on the weight W and the scale K. In the days of the automobile
without shock absorbers, it was always noticeable how much easier riding the
car was when loaded than with only the driver in it. The heavier load
reduced the frequency of vibration.
If the scale K of the spring is increased,
the spring is ‘‘stiffer’’ and the frequency is increased.
In an automobile,
increasing K makes the ride more jarring (same road conditions).
A “‘soft”
spring (smaller AK) will result in your bouncing farther but not so jarringly.
It might be said that the ride you get in a modern automobile is the result
of a compromise between stiffness (A) and the action of shock absorbers,
based largely on experience.
See Chapter XX for additional material
on vibrations.
233. Closure.
This chapter has dealt with the application of Newton’s
second law only to bodies in translation.
Whenever possible, determine
the direction of motion (the sense of the velocity vector) and choose this
direction as positive.
Then in making sums, remember that acceleration,
as well as force, is a vector quantity and that, therefore, we must be vigilant
about giving the proper sign to a. If the direction of motion is taken as
positive, a@ is negative for a deceleration.
If the numerical value of a is
known, put it into the equations with its correct sign or the solution will be
wrong.
If nothing is known about a, it may be put into the equations with
a positive sign, equivalent to assuming a positive acceleration; then if the
answer is negative, we know we chose the wrong direction for a.
By all means, always make a complete free body before beginning the
algebraic solution, and somewhere on this sketch include a vector representing the acceleration—to help you to remember about its sign. In the process
of doing this, consider whether you want to include the inertia force ma
(reversed effective force, abbreviated REF) in the free body and use
2F = 0,
[Free body should match this equation, include REF]
where ma is included in SF; or to omit the inertia force and use
LF = ma,
[Free body should match this equation, omit REF.]
where ma is not included in =F. If the inertia force is included, make it
a dotted vector to distinguish it from the real forces. If you must take
moments, there is often some advantage in including the inertia force on
the free body because this allows you to take moments about any con-
PROBLEMS:
HORIZONTAL
OR VERTICAL
MOTION
OF PARTICLES,
525
a CONSTANT
venient point. If moments are taken without the inertia force as part of
the free body, the center of moments must be on the line of action of the
resultant; such a point for a body in translation is the cg.
Problems
NOTE.
Always show complete free bodies in each solution.
HORIZONTAL
PARTICLES,
OR
a
VERTICAL
MOTION
OF
CONSTANT
1101. The weight of the reciprocating
parts of the steam engine of § 213 is 322 lb.
For the conditions of this example, the
acceleration of the piston was found to be
a. = 63 fps? (p. 297).
What is the resultant force on the reciprocating parts at this
instant?
1102. A body slides down a 60° incline
forwhichf = 1/4. (a) Ifit starts from rest,
how long does it take to slide 60 ft.? (b)
What should be the inclination of the plane if
the body slides down with constant speed?
Ans. (a) 2.24 see.; (b) 14.03°.
1103. A 644-lb. loaded sled is towed
behind a truck by a rope that is inclined
upward from the sled at a 20° angle with the
horizontal.
If f = 1/5 for the sled and if
the tension in the rope is 200 lb., find (a)
the acceleration and (b) the time required
for the speed to change from 12 fps to 40 fps.
1104.
A
3000-lb.
automobile
is to
be
brought to rest with a constant from 60 mph
in a distance of 160 ft. What is the total
frictional force between the tires and the
ground?
Ans. 2250 lb.
1105. A book is thrown across the room
and found to slide 17 ft. as it comes to rest.
If the kinetic coefficient of friction is 0.2,
compute the horizontal speed with which
the book is thrown.
1106. A serious student wishes to perform
an experiment in determining the speed of
a book.
Using a blackboard eraser, he
coats one side of the book with chalk dust,
so that the book will mark a trail as it
slides. Then he places the book on a long
table whose inclination is varied until the
book will slide down at a uniform speed.
This angle of inclination is observed to be
26.6° with the horizontal.
Now with the
table top horizontal, he throws the book
onto the table, and measures a 7.5 ft. trail
left by the book as it slid to rest.
Calculate
the horizontal speed with which the book
struck the table.
Ans. 15.5 fps.
Fig. 518.
Problems 1107, 1108.
1107. Two bodies A and B, connected
by a rod C, Fig. 518, have an initial speed
of 30 fps. A force of P = 150 Ib. acts at
6=0°.
Let Wa = 322 lb., We = 644 lb.,
fa = 0.04, and fg = 0.15.
Determine
the
force in the rod C and the velocity of the
bodies after 15 sec.
1108. Two bodies A and B, connected
by a rod C, Fig. 518, have an initial speed
of 6 fps and move 300 ft. in 30 sec.
Let
Wa = 966 lb., We = 1288 lb., fa = 0.04,
fs = 0:15, and @)= 15°. For constant acceleration, determine the force P and the
final velocity.
Ans. 270 lb., 14 fps.
1109. A hydrogen-filled balloon of weight
W is falling vertically downward with a
constant acceleration a. What amount of
ballast Q must be thrown out in order to
give the balloon an equal upward acceleration a? Neglect air resistance.
Pile Driver
g cg
822 lb.
Fig. 519.
1110.
A
Problem
pile-driver
1110.
hammer,
Fig.
519,
weighing 322 lb., is to be raised 30 ft. with
a constant
acceleration
in 2.5
sec.
The
226
a CONSTANT
PROBLEMS: HORIZONTAL OR VERTICAL MOTION OF PARTICLES,
total friction in the guides is constant at
200 lb. What is the required pull P in the
Ans. 618 lb.
cable?
Cam
Fig. 520.
Problems 1112, 1113.
PARTICLES
ON
a CONSTANT
1111. A man in an elevator has a weight
If the elevator
of 100 lb. on his shoulders.
moves upward with an acceleration of 3 fps?,
what load does he support?
1112. The reciprocating follower A, Fig.
520, which together with its attached parts
weighs 6 lb., is moved upward by the cam
B a distance of 4 in. during a 75° turn of
the cam with constant acceleration.
If the
cam turns at a constant speed of 120 rpm,
what is the force between the cam and the
follower?
If the permissible load is 100 lb.
per in. of face of the cam, how wide should
the face be, if the computed force only is
considered?
Ans. 11.45 |lb., 0.1145 in.
1113. The same as 1112 except that the
speed of the cam is 400 rpm.
AN
INCLINED
1121.
Fig. 521.
In Fig. 522, the bodies
amcl
@ = 20) ls,
(@) Ih
the block is initially at rest, what
is the
maximum value of the coefficient of static
friction if motion is to occur?
(b) What
is the velocity of the body A after it has
traveled 10 ft.?
Ans. (a) 0.1517, (b) 6.43 fps.
1116. In Fig. 521, 6 = 40°,f, = 1/3, and
force Q is removed while the body A is at
rest. Find the velocity of body A after
5 sec.
Find the limiting value of f, if A
is to move
1117.
(Q = 0).
In Fig. 521, body A weighs 32.2 lb.,
fe = 1/4, and force Q = 30 lb.
Find
the
angle 6 for which the body will have an acceleration of 3 fps? up the plane.
1118. This problem is the same as the
example of § 225 except that 6 = 60°.
1119. A toboggan slides 100 yd. from
rest down a 20% grade in 15 see.
Find
the average coefficient of friction.
Ans. 0.115.
1120. A toboggan slides 100 yd. from
rest down a slope in 10 sec.
Find the
angle the runway makes with the horizontal
f the coefficient of friction is 0.1.
A and
B
The rigid
If W4 =
10 lbs. Wie = 208 be i
ie,
of the
02
and 6 = 30°, find the acceleration
bodies and the force in bar AB.
Problems 1114-1117.
fp ]Oul,
PLANE,
are in motion down the plane.
bar AB is of negligible weight.
Ans. 8.04 fps?, 0.77 lb. (T).
1114. In Fig. 521, body A weighs 50 lb.,
6 = 30°, f, = 0.1, and force Q = 40 lb. If
the body starts from rest, what is its velocity
after 4 sec.?
Ans. 8.5 fps.
1115. In Fig. 521, body A weighs 50 lb.,
0 =802,
[Ch. XJ vf
Fig. 522.
Problems 1121-1125.
1122. The same as 1121 except that bar
AB weighs 5 lb., being supported by the
pins as shown.
1123. The same as 1121 except that the
bodies are in motion wp the plane.
1124.
In
Fig.
522,
the
blocks
are
in
motion down the plane.
If W4 = We =
16:1 Ibs 6) = 305 fate) and fiz — 1/4
find the acceleration of the blocks and the
force in the weightless bar AB.
Ans. 7.97 fps?, 0.58 Ib. (C).
1125. In Fig. 522, the blocks
are in motion down the plane.
We
= 10 lb., 6 = 30°, a =
A and B
If W4 =
—1.61 fps?, and
the compressive force in the weightless rod
AB is 1 lb., find the coefficients of friction
fa and fp.
Ans: 0.70, 0252:
1126. While moving up a 1% grade, a
freight locomotive exerts a constant drawbar
pull of 60,000 lb. The train resistance is
15 lb. per ton.
(The train resistance con-
PROBLEMS:
PARTICLES
ON
AN INCLINED
PLANE,
527
a CONSTANT
sists of the friction in the bearings, the
rolling resistance of the wheels, air resistance, etc., and is generally estimated as a
force of so many pounds opposed to the
direction of motion.)
If the speed of the
train changes from 5 mph to 30 mph in a
drawbar pull of 135,200 lb. produced by a
Virginian steam locomotive.
To reach the
distance of 8 miles, what is the weight of
the train?
Ans. 1650 tons.
1127. A 1000-ton train has a resistance
1128. The same as 1127 except that the
train is double-headed with 2 Virginian
locomotives, each producing the same drawbar pull.
(see note in 1126) of 10 lb. per ton and a
TWO
DIRECTIONS
1129. A freight car starts down a 1.5%
grade from rest under the action of gravity.
The constant frictional resistance to motion
is 10 lb. per ton of weight.
After 5000 ft.,
it passes onto level track where the resistHow far does it
ance remains the same.
go before it stops. There are no cows or
other obstructions on the track.
top of a 1-mile long, 144%
grade with a
speed of 20 fps, (a) find the speed with
which the train must “hit” the grade.
Ans, 126 fps.
OF
MOTION
(PARTICLES)
1133. With the initial conditons described
in problem 1181, find the displacement. of
A after 6 sec.
Ans. +149 ft.
12
Weightless
Ans. 15,000 ft.
1130. The same as 1129 except that the
grade is 1%.
Fig. 524.
Fig. 523.
1134. In Fig. 524, P = 0, Wa = 600 lb.,
We = 225 lb., and fa = 1/3. Neglecting
the weight of the cable and pulley and the
friction at the pulley, compute (a) the
distance A moves in 20 sec., (b) the tension
Problems 1131-1133.
in the cable, and (c) the speed of the bodies
1131. In Fig. 523, bodies A and C are
connected by a weightless flexible cord over
a smooth
surface B (fg = 0).
The
coefhi-
cient of frictionfa = 1/3. If Wa = 64.4]b.,
We = 96.6 lb., and the initial velocity of
A is 30 fps toward the left, find the time in
seconds for A to travel 10 ft. Ans. 0.39 sec.
1132. The same as 1131 except that
pe
0.2:
INVOLVING
Problems 1134, 1135.
after 20 sec. The bodies start from rest.
1135. The bodies A and B, Fig. 524, are
moving toward the right at 20 fps. Let
Wa. = 600lb., We = 225 1b.,fa = 1/3, and
neglect the weight of the cable and pulley
What conand the friction at the pulley.
stant force P will bring the bodies to rest in
a distance of 30 ft.? What is the tension
Ans. 189.4 lb., 271.6 lb.
in the cable?
PULLEYS
(PARTICLES)
a
1136. In Fig. 525, let Wa = 966 lb. and
1/3. The speed of A changes from
jo
va, = 10 fps to v42 = 35 fps during 25 sec.
disDetermine (a) the weight W, (b) the
tance moved by W during 25 sec., and (c)
the tension in the cable.
that
1137. The same as 1136 except
v4, = 60 fps and va = 10 fps.
Ans. (a) 508 lb.; (b) 437.5 ft.; (e) 262 Ib.
A
Weightless Cable.”
Weightiess and
Frictionless Pulleys.
Fig. 525. Problems 1136, 1137.
PROBLEMS: INVOLVING PULLEYS
528
(PARTICLES)
[Ch. XV
Weightless Cable.
fh
Weightless and
Frictionless Pulleys.
Fig. 526.
Problems 1138, 1139.
30°
1138. In Fig. 526, Wa = 200 lb., We =
LOOM Dial — b/A and fiz —s/3
eHow tar
and in what direction does A travel from
rest during 30 sec.?
in the cable C?
1139. The
Wa
What is the tension
in the cable D?
Ans. 326 ft., 52.2 lb., 104.4 Ib.
same as 1138 except that
= 30 1b:
1141. Let the weight of the bodies in
Fig. 528 be represented by Wa4 and We.
Derive an expression for their acceleration
and the equation giving the tension in the
weightless cord. Let
W4>W es and neglect
the inertia and friction of the pulley.
1142. In Fig. 529, an elevator A is supported by a cable which passes over a
Weightless Cable.
Weightless and
Frictionless Pulleys.
Fig. 527.
Problem 1140.
1140. A mine cage A, Fig. 527, weighs
9660 lb. and has a speed of v, = 30 fps
downward.
The total friction in the guides
is constant at 100 1b. The cable supporting
the cage wraps around a 2-ft. drum which
is connected to a 10-ft. brake wheel.
Between the brake shoe and the wheel, f = 1/4.
Consider the mass of the wheel and drum
to be negligible.
If the cage comes to rest
in a distance of 60 ft., compute the tension
in the cable and the force P on the braking
lever.
Driver, B
Ans. 11,810 lb., 945 lb.
Fig. 529.
Fig. 528,
Problem 1141,
Problems 1142, 1143,
driving sheave B, thence half around an
idling sheave, thence over the driving sheave
again, and down to a counterweight CW.
In this type of drive, the motor turns
the sheave B, which, by virtue of the
friction
between
the cable
and
the
PROBLEMS:
INVOLVING
PULLEYS
829
(PARTICLES)
sheave, causes the elevator to move up or
down.
Let Wa = 5000 lb., a4 = 3 fps?,
and let the coefficient of friction of the
cable over sheave B be fg = 0.1. Neglecting the inertia effects of the sheaves and
assuming that sheaves B and C are the
same size, determine the necessary minimum
weight of the counterweight CW if the
cable is not to slip on the driving sheave.
Ans. 3220 lb.
1143. In the traction drive described in
1142, Fig. 529, let Wa = 7500, fz = 0.1,
and let the weight of the counterweight be
6000 Ib. What maximum upward acceleration may be given the elevator A without
slipping of the cable on the driving sheave?
the acceleration of each body and the
tensions in the cords P; and Pp».
Ans. a4 = 2.93 fps?, 131 Ib., 65.5 Ib.
1145. The same
Wa = 240 lb.
Fig. 531.
as
1144
except
that
Problems 1146, 1147.
1146. In Fig. 531, the masses of A, B, and
C are, respectively, 1 slug, 2 slugs, and
4 slugs. The cords are weightless and
flexible. Sheaves D and £E are considered
Fig. 530.
1144.
In
Problems 1144, 1145.
Fig.
530,
Wa
= 120
lb. and
W» = 80lb. Consider the cord and pulleys
weightless, and neglect friction. Determine
RIGID
In 2 sec. after all
free and weightless.
elements are simultaneously released from
rest, find the absolute acceleration, velocity,
and displacement of body A.
Ans. 19.25 fps?, 38.5 fps, 38.5 ft.
1147. The same as 1146 except that the
mass of C is 2 slugs.
BODIES
of the following
Unless the instructor specifies otherwise, make two solutions
and one with tt.
body
free
the
on
(REF)
force
problems, one without a reversed effective
s of each method.
advantage
the
clear
vake
This
procedure
will
n
Make two free bodies.
NOTE.
1148. Body A, Fig. 532, is accelerated
from rest to 40 fps in 20 ft. Attached to
it is the 50-Ib. vertical homogeneous bar B
that is pivoted at pin C. Find the forces
at C and D, (a) not using the REF, (b)
using the REF.
Ans. C = 52.4 lb. at 72.7°,
D = 46.6 lb.
1149. This problem is the same as the
example of § 230 except that the line of
action of P is a distance d = 10 ft. below
the
track.
Solve
(a) without
REF,
(b)
with REF.
1150. A 128.8-Ib. garage door is supported on frictionless rollers as shown in
Fig. 533. For a force F = 20 |b. applied
y = 3 ft. from the bottom of the door, what
are the reactions at A and B? Solve (a)
without REF, (b) with REF.
Ans. 66.9 lb., 61.9 Ib.
1151. A 128.8-lb. garage door is supported on frictionless rollers as shown in
(a) What force F applied at a
Fig. 533.
distance
reaction
Fig. 532.
Problem 1148.
y = 2 ft. would result in a zero
(b) What force F
at roller B?
applied at a distance y = 6 ft. would result
PROBLEMS: RIGID BODIES [Ch. XV
330
Fig. 533.
in a zero reaction
at roller A?
Problems 1150, 1151.
Solve
(a)
without REF, (b) with REF.
1152. An overhead crane in a foundry
moves a 20-ton ladle, which is suspended
from cables.
From the point of tangency
of the cables on the crane drum to the center
of gravity of the ladle is 30 ft. Ina distance
1157. In Fig. 534, the weight of the bar
is W = 100lb.
Neglect the size and weight
of the wheel and of the block.
If the
reaction at B is 10 lb. upward, and f, = 0.2,
find @ and
Q (a) without
with REF.
REF,
and
(b)
Ans. 34.3 fps?, 124.7 lb.
of 60 ft., the speed of the crane steadily
increases from 5 to 15 fps. What is the
total pull in the cables and the inclination
of the cables with the vertical?
Ans. 40,100 lb., 2.96°.
1153. A truck, with a gross weight of
6440 lb., has a wheel base of 12 ft. Its
center of gravity is 4 ft. above the pavement
and 5 ft. ahead of the rear axle.
If the
truck is brought to rest in a straight line
from a speed of 70 fps in a distance of
90 ft., (a) what are the vertical reactions at
the front and rear wheels, and (b) what is
the total frictional force between tires and
pavement?
1154.
The
truck
described
in
1153
is
going down a 10% grade at a speed of
50 mph.
The coefficient of kinetic friction
between
the wheels
and
the road
is 0.4.
If the driver locks all four wheels, determine
the normal reactions on the front and rear
wheels.
The truck moves at all times in a
straight path.
Ans. 3520 lb., 2885 Ib.
1155. A driver of a 3000-lb. automobile
brakes his car so that each wheel is about
to slip on the pavement, where f = 0.8.
The wheel base of the car is 120 in., and
its center of gravity is 2 ft. above the ground
and
6 ft. back
of the front axles.
If the
car is going 60 mph when the brakes are
applied, how far does it go before it comes
torest?
What are the frictional and normal
forces at the front and at the rear wheels?
Ans. 150 ft., Fy = 1344 lb., F, = 1056 lb.
1156. The same as 1155 except that the
car is on a gravel road, where f = 1/4.
Fig. 534.
Problems 1157-1160.
1158. The same as 1157 except that
lis =U
1159. The same as 1157 except let the
reaction at B be 10 lb. downward.
1160. In Fig. 534, the weight of the bar
is W = 100 lb. Let f, = 0 and neglect the
size and weight of the wheel and of the
block.
If Q = 500 lb., find @ and the
reaction at B. Solve (a) without REF,
(b) with REF.
1161. A 2 x 2 x 6-ft. box B containing
cork and weighing 322 lb. is placed on a
car A, as shown
in Fig. 535.
At C is a
fixed peg for which fo = 0. The car A is
given an acceleration of a = 3.22 fps? leftward.
(a) What is the weight W if the box
is about to turn over?
Solve without and
with REF.
(b) If the coefficient of friction
between A and B is fs = 0.1, would the
box turn over or slide?
1162. The
fig =O.
Ans. (a) 51.2 lb.; (b) slide.
same as 1161 except that
PROBLEMS:
RIGID
331
BODIES
b4
is 40 fps. In what least distance may the
car be brought to rest if the block neither
tips or slides? Give a complete solution.
Problems 1165, 1166.
Fig. 537.
1165. In Fig. 537, the homogeneous body
A weighs 1200 lb. Let @ = 30° and fa =
0.3.
Motion
Determine
1166. The
6 = 45°.
Fig. 536.
(a) the weight
W
when
body A is on the point of turning over, and
(b) the corresponding tension in the weightless cord. The pulleys are weightless and
Ans. (a) 4860 Ib.; (b) 1670 lb.
frictionless.
SS
same
as
1165
except
that
Problems 1163, 1164.
1163. A 1932-lb. car C, Fig. 536, is loaded
with a block A of material weighing 2576 lb.
Fig. 538.
(a) What is the acceleration of the car when
the block is on the point of tipping over
about the edge B? Solve without and with
(b) Iff = 0.4 between the block and
REF.
the car, what is the acceleration when the
(c) What is the
block is about to slide?
maximum value of the force P if the block
Problem 1167.
1167. A homogeneous body A, weighing
200 Ib., is loaded on a truck on an inclined
surface as shown in Fig. 538. Let f = 0.4.
(a) Will the body tip over or slide if the
Exacceleration a is gradually increased?
reactions at the wheels?
tered on the car.
The load is cen-
plain. (b) What is the maximum acceleration
if the body is to maintain its position?
Ans. (a) slide; (b) 1.018 fps?.
Ans. (a) 16.1 fps?, (b) 12.88 fps’, (c) 1802
Ib., (d) 1787 lb., 2720 lb.
1164. A 1932-lb. car C, Fig. 536, is loaded
with block A of material weighing 2576 lb.
= 0.4.
The initial speed
gasoline relative to the horizontal?
Ans. 5.74°.
neither slides nor tips? Neglect friction at
(d) What are the vertical
the wheels.
Let
P = Oandfac
SPRINGS,
1168. An automobile has a constant horizontal acceleration of 3.22 fps?. If the
gasoline in its tank is at rest relative to the
car, what is the slope of the surface of the
VARIABLE
FORCE
SYSTEMS
time, may be found at the end of
Additional spring problems, if desired at this
NOTE.
Chapter XX.
1169. A helical spring with a scale of
24,000 Ib. per in. is compressed 2 in. between
a fixed surface and a block B which weighs
161 lb. The block rests on a 30° inclined
plane for which f = 1/3, and the axis of the
spring is parallel to the plane. If the spring
PROBLEMS: SPRINGS, VARIABLE FORCE SYSTEMS
332
and block B are released, how far up the
plane does the block go? The spring (considered weightless) acts on the block only
for the 2 in. which it was compressed.
Ans. 31.47 ft.
1170. A weight of 50 Ib. is suspended
from a weightless spring in a vertical position. The scale of the spring is 20 lb. per
in. If the weight is pulled down 21% in.
and then released, determine (a) the maximum acceleration, (b) the maximum veloc-
ity, and (c) the frequency of the motion.
1171. A coil spring used for the front
suspension of an automobile has a scale of
400 lb. per in. If the weight on this spring
is 1000 lb., determine the frequency of the
vibration when the wheel hits a bump, assuming that no shock absorbers are used.
Ans. 1.98 cycles per sec.
1172. A 10-lb. body falls a distance of
3 ft. and strikes a spring whose scale is
K = 30 lb. perin. How much is the spring
compressed when the body has come to rest?
1173. The problem is the same as the
example of § 232 except that W = 30 lb.
Ans. (a) 4.24 fps; (b) —107.3 fps?, (c)
0.247 sec.
1174. A loaded freight car, weighing 70
tons and moving at 6 mph, strikes a spring
bumper, consisting of a nest of springs, at a
point 4 ft. above the top of the rails. The
nest of springs as a group has a scale of
40,000 Ib. per in. and may be compressed
as much as 12 in. The distance between
the front and rear trucks of the car is 30 ft.
and the cg of the whole is 6 ft. above the
top of the rails. Neglect rolling resistance
and determine (a) the maximum compression
of the springs, (b) the maximum acceleration,
and (¢c) the maximum vertical reactions at
the front and rear trucks.
Ans. (a) 10.04 in.; (b) 92.4 fps?; (ce) 96,784
lb.; 43,216 lb.
1175. The same as 1174 except that the
rolling resistance is constant at 15 lb. per
ton and is assumed to act at the surface
of the rails.
1176. A 322-lb. weight is suspended from
a spring whose axis is vertical.
It is pulled
down a certain distance and released. When
the displacement is —2 in., the velocity is
+14.1 fps. When the displacement is +3
in., the velocity is —10.6 fps. Determine
the scale of the spring, the period, the
maximum
acceleration,
and maximum
ve-
locity of the weight.
Ans. 8000 lb. per ft., 0.209 sec., 450 fps?,
15 fps.
[Ch. XV
1177. The resistance to motion of a 3220lb. boat in still water is approximately
3v Ib., where v is in fps. If the speed of
the boat is 9 fps when the engine is stopped,
(a) how far will the boat go and (b) how
long will it take before it comes to rest.
Ans. (a) 300 ft.; (b) infinite.
1178. The resistance to motion of a 1610Ib. boat in still water is approximately 30 lb.,
where v is in fps. If the speed of the boat
is 6 fps when the engine is stopped, find
the speed after it goes 100 ft. and the time
required for the displacement.
1179. A 3220-lb. boat is propelled from
rest by a constant thrust of 50 lb. The
resistance to motion is 3v, where » is in fps.
(a) How long does it take for the boat to
attain a speed of 6 mph?
(b) For a thrust
of 50 lb.. what is the maximum
the boat?
(c) How
far does
speed of
the boat
go
in attaining a speed of 6 mph?
Ans. (a) 25 sec.; (b) 16.6 fps.; (c) 125 ft.
1180. A man and parachute weighing
196 lb. reach a speed of 80 fps before the
parachute opens.
At this instant the parachute opens, and the resistance to falling
is 0.81v2, where v is in fps. (a) What is
the limiting speed with the parachute open?
(b) How
far does the man
move
while his
speed drops to 20 fps? (c) How long does
it take to reach this speed?
1181. Aresultant force acting on a 64.4-lb.
body in the direction of motion is F = 4¢2.
The initial velocity of the body is 4 fps.
What is its velocity after 2 sec.? What is
the corresponding displacement?
Ans. 9.33 fps, 10.67 ft.
1182. A motive force of F = 4¢2 acts on
a 64.4-lb. body. There is a constant force
of 30 lb. acting opposite to the sense of the
initial motion.
The initial speed is 4 fps.
Determine the velocity of the body after
(a) 0.2 sec., (b) 2 sec., and (c) 5 see.
1183. A particle moves in a straight line
according to the law s = t8 — 75t ft., where
t is in sec.
(a) What is its arithmetic
average velocity during the fourth second?
(b) When does it come to rest?
What is
its acceleration at the instant it is at rest?
(c) If the rightward direction is positive,
in what direction is it moving after 4 sec.?
Where
is it located
at this instant?
(d)
If the particle weighs 6.44 lb., what is the’
resultant force on it at 4 sec.?
Ans.
(a)
1184.
—37.5
fps; (b) 5 sec., 30 fps?;
(c) left, —236 ft.; (d) 4.8 lb.
The connection between the crank
pin and piston rod shown in Fig. 539 is
PROBLEMS:
SPRINGS, VARIABLE
FORCE
833
SYSTEMS
Crank Pin A
Piston Rod
Ye
Fig. 539.
Problem 1184.
called a Scotch crosshead.
This connection
is sometimes used in driving pumps.
The
stroke is 20 in. and the crank turns uniformly at 120 rpm.
(a) If the weight of the
reciprocating parts is 644 lb., what is the
corresponding inertia force when the piston
is 4 in. from its head end _ position?
What is the maximum
inertia force? (b) If
the diameter of the cylinder is 18 in. and
the net fluid pressure on the piston is 60 psi
for a complete stroke, what is the maximum
force on the crank pin?
Ans. (a) 1580 lb., 2630 Ib.; (b) 17,870 Ib.
1185. A flexible chain weighs 20 lb. per
ft. and is 30 ft. long. It is placed on a
horizontal surface, for which f = 1/3, with
10 ft. of it hanging vertically over the edge.
If the chain is released in this position,
how long does it take for the remainder of
the chain to leave the horizontal surface?
Ans. 2.42 see.
1186. A flexible chain weighs 10 lb. per
ft. and is 20 ft. long. It is placed on a 30°
inclined plane, for which f = 0.4, with 5-ft.
of the chain hanging over the lower end of
If the chain is released in this
the incline.
position, what is its velocity at the instant
the chain leaves the incline? How long does
it take for the chain to leave the incline?
Ans. 25.7 fps, 1.425 sec.
The maximum horizontal recoil
1187.
of a 966-lb. gun is to be 1.5 ft. with an
initial recoil speed of 12 fps. The reaction
is to be handled by compressing air in a
cylinder. The volume of air in the cylinder
at any point is vA cu. ft., where A is the
area of a section of the cylinder and 2 is
the distance from the piston to the cylinder
head. At the beginning of the reaction,
the value of x = 2.5 ft. giving a volume of
air of 2.5A, and the pressure of the air at
this point is p1 = 14.7 psi. At the end of
the reaction, the value of xis 2.6 — 1.5 = 1
ft. During the compression of the air, the
relation pV = C, a constant, is assumed to
be true, where p is the variable pressure
of the air. The variable resisting force is
then F = pA. Determine the constant C
in terms of A, and compute the necessary
diameter of the recoil cylinder.
Ans. 9.03 in.
Fig. 540.
1188.
A constant
Problem 1188.
force
of Q = 100 lb.
acts on the 360-lb. body A, Fig. 540, but
the angle @ varies according to the law
@ = xt/30, where 6, measured counterclockwise, is in radians and ¢ is in seconds.
Motion of A starts when ¢ = 0 and @ = 0.
Neglecting friction, find the velocity and
displacement of A after 15 sec.
1189-1200. These numbers may be used
for other problems.
Chapter XVI
CURVILINEAR
MOTION
A body or particle is constrained to move in a curved
234. Introduction.
path by the action of external forces which produce a change in the direction
Of course, as we learned in kinematics, the change of the
of the velocity.
direction of the velocity results in an acceleration, called the normal acceleration, expressed by the relations (§ 190)
Oy = RH? =
= = MH.
ip
In this chapter, we shall deal with cases of motion of particles or of bodies
in which
body.
the resultant force passes
Aside from
the introduction
through
the center
of the normal
of gravity
acceleration
of the
and the
corresponding dynamic reaction (ma,), this chapter presents no principles
of kinetics not covered in the previous chapter.
235. The Simple Pendulum.
of weight
W
a
a:
Fig. 541.
A simple pendulum consists of a particle A
(Fig. 541) attached to a weightless, inextensible string B,
the other end of which is attached to a fixed
point C. Equilibrium exists if the particle is
at rest on a vertical line through C.
If the
particle is displaced so that the string B makes
some angle 6 with the vertical and is then
released, it will swing in the are m-—n of a
vertical circle. If there is no friction of any
kind, the particle will oscillate indefinitely between the same two points on the path m-—n.
The only forces acting on the particle are P,
the tension in the string, and W, as shown in
Simple Pendulum.
the free-body diagram (Fig. 541).
The acceleration of A may be represented by its components a, and a,. Choosing the rightward direction from O as positive,
we get, from the sum of the forces in a tangential direction (assuming a,
unknown),
334
§ 235 |THE SIMPLE PENDULUM
ie
835
Ww
or
a;
:
=
— gsin 8,
that is, the acceleration varies as sin 6, and is directed toward
Since a = dw/dt = d?6/dt? = a,/r (by using a, = ra), we get
d?6
ay
—
or
(a)
a
dt.
ih, ——-
=
=
Fp
g sin 6,
—S
d?0
a=
the left.
a]
Gee
—
7, Sin 8,
—
—_
where the radius of the path m-—n is r = L, the length of the simple penEquation (a) may be solved for the angular velocity of the particle
dulum.
However, for the special case of very
by using the relation a = wdw/dé.
small ares of swing, sin 6 is very nearly equal to @; hence
(b)
d’6
g
eal: 0,
ines
[when sin 6 = 6]
a form of differential equation for which a solution has already been found.
Comparing this equation (b) with the equation (42), p. 269, we see that the
solution of (b) must be in the same form as the solution to (42). Noting
that the quantity w in equation (y), p. 270, is replaced here by (g/L)*!?, we
have, for the period 7’, from (z), p. 270,
Dy.
be
ips
jh We
= Gime = (5)
This expression gives the time of a complete vibration, from one extreme
point of the movement to the other extreme, and back again to the original
Equation (c), however, is reasonably accurate only when the
position.
maximum value of 6 (= 8, Fig. 541) is not greater than some 4° to Gee Lon
such angular displacements, the period of the pendulum 7’ is independent
of the amplitude 8. The smaller the value of 8, the more accurate is the
period as obtained from (c).
To find the speed of the particle A, we first substitute a =
This gives
equation (b), and integrate.
BY
fod =ae -~£ [ode
Semler
a]
sit sof
whence
where the constant of integration is determined
Thus
when 6 = 8; that is, C = g6?/(2L).
2
=
so that
(4)
Lin, Tetie
Te at
dw/dé in
fll
on t &
by the conditions w = 0
lig
—©),
0" l| Lia? = gb(6?
CURVILINEAR MOTION
336
[Ch. XVI
(radians)
from which the tangential speed v at any angular displacement
is small.
@
may be found. These expressions too are suitable only when
However, we should recall that the speed attained by a particle moving
down a frictionless path depends only on the vertical displacement through
which the particle moves (§ 193); that is, the maximum speed of a simple
pendulum may be found in any case from
y? = 2gh,
where h is the maximum vertical displacement of the pendulum from its
position of rest. For example, the maximum fall of the pendulum of Fig.
541 is
h=L-—
LecosB.
The tension P in the string is obtained from a sum of the forces in the
direction AC. In this direction, the acceleration is the normal acceleration
a, = v?/L; therefore, from 2 = ma,
(e)
Lr =P
—
Weoos 6 = ma,
=
We?
gL’
where v is the instantaneous speed at the displacement #6.
correct for any angle 6, since no approximation is involved.
Fig, 541.
Repeated.
Fig. 542.
Equation (e) is
Conical Pendulum.
236. Conical Pendulum.
In a conical pendulum, the particle A moves
with constant speed v in a horizontal circle, so that the string B, Fig. 542,
generates the surface of a cone. In this case, the only acceleration is the
normal acceleration a, directed toward the center of the circle O. In analyzing this motion, we may include the inertia force ma, = Wv?/(gr), where'r
is the radius of the circular path of A. This reversed effective force ma, is
often called the “centrifugal force.”
Inclusion of the inertia force (the
dynamic reaction) places A, as a free body, in equilibrium.
A vertical sum
of the forces gives
837
§ 237 |EXAMPLE
(f)
Pcos6 -W=0
A horizontal sum gives
Wv?
;
(g)
r — Psné=0
or
P cos 6 = W.
or
eee
2
gr
From these two equations we obtain, by division,
(h)
v?
tan. 0.= ==
gr
Oh
=;
g
From this equation we see that the inclination @ of the string and the size
r of the circle increase as the speed of the particle increases. This property
of a conical pendulum is used in the form
of a flyball governor to regulate the speed
For example, if a steam
of an engine.
engine speeds up, the flyballs (particle A)
This action,
move outward and upward.
closes a
partially
through connecting links,
valve, thus reducing the amount of steam
admitted to the cylinder, which causes the
With such a goverengine to slow down.
may be held
engine
nor, the speed of an
8
ee
Equanearly (but never exactly) uniform.
for Governor.
Flyballs
543.
Fig.
tions (f) and (g) may be used to determine
details of
some
This figure shows
Observe also that tan @ =
two unknowns.
r/h, Fig. 542, See Fig, 548
atioe
Ea eee
steam engine.
The force which is equal and opposite to
the centrifugal force (the dynamic reaction)
force (or
is called the centripetal force. The centripetal force is an actual
component; in this case the component of P in the direction AQ)
which causes the particle to move
in a curved path.
237. Example. A vertical shaft AD,
Fig. 544, rotates about a fixed axis and
has attached to it an arm BF anda supporting link CH. It is thus held rigidly
On the end
in position relative to AD.
of the arm isa 75-lb. ball F. Assuming
that the weights of the arm and the link
CE are negligible, determine the angular
velocity of the shaft (rpm) when the
compressive load on the link CE is half
the static compressive load.
SotutTion. Figure 544(b) shows a free
body including the inertia (centrifugal)
CURVILINEAR MOTION
838
[Ch. XVI
force man. However, we first let man = 0 (static condition) in order to find the static
load on the link CE. We observe that the link CH is a two-force member and may
therefore be cut.
Hence, taking moments about B with ma, = 0, we get
>Mz = W(60 sin 30°) — H(36 sin 75°) = 0,
R=
(75)(60)(0.5) _Oem: da
OIG
The shaft is now rotated so that H = 64.7/2 = 32.35
lb. Inasmuch as the weight of BF is being neglected,
the only inertia force is the one for the ball.
With
this inertia force included in the free body, a condition of equilibrium is simulated.
Hence, we may
again equate 2M, to zero. Using the equation
Man
Wet | Wre?
(gr)
G8
we get
Wr?
=M,; = ( a (oocos 30°)
+ (32.35) (36 sin 75°) — (75)(60 sin 30°) = 0.
» Fig. 544(b).
Repeated.
.«-
In this equation,
W = 75 lb., and r = 30 in. = 2.5 ft.,
as is seen from Fig. 544.
Wrw?/g, we must use r in feet.
If g is taken as 32.2 fps? in
Solving for w, we get w = 1.93 rad. per sec., or
_ (1.93) (60)
(27)
= 18.4 rpm.
If desired, the horizontal and vertical components of the reaction at B may now be
found from the horizontal and vertical sums of forces, because the inertia force Wrw?/g
can be evaluated now that » is known.
238. Example—Side Rod on Locomotive.
Determine expressions for the reactions at the pins of a side rod connecting the driving wheels A and B (Fig. 545) of a
locomotive, when the side rod is in its lowest and highest positions.
Sotution.
Observe that the side rod, Fig. 545, moves so that each of its positions
is parallel to all other positions; that is, so that the rod itself has no angular motion
although each point in the rod undergoes curvilinear motion. The rod is said to BS
moving with curvilinear translation.
If the drivers are rolling with a speed of v fps, the angular velocity of the wheels
and hence that of the pins attached to the wheels, is» = v/r,, rad. per sec., where i
(b)
Fig. 545.
§ 239 |SUPERELEVATION
OF CURVES
339
is the radius of a driving wheel. With each of its ends attached to a wheel, all points
in the rod have the same acceleration relative to the frame of the engine, 7w*, where
7 is both the radius of the circles described by the pins and the distance of the center
of gravity of the rod from its center of rotation relative to the frame; and where w is
the angular velocity of the wheels.
In its lowest position, the side rod has an acceleration vertically upward; hence the inertia (centrifugal) force acts downward
and through the center of gravity. Its value is mfw? = Wrw/g, where W is the
weight of the rod. With the center of gravity at the midpoint of the rod, the reactions
Q at the ends are equal, by virtue of symmetry.
Summing forces vertically [Fig.
545(a)], we get
mF, = 2Q — nifo® — W = 0,
W (ie?
which is the maximum force on the pins. Observe that increases as w (or v) increases.
When the side rod is in its highest position [Fig. 545(b)], the inertia force acts upward.
Again summing forces vertically, we have
aM ('2 1)
SF, = mrw? — W — 2Q = 0,
W [ Fo
If 7w2/g is less than unity, Q is negative, which means simply that the forces Q act
Notice that, as the wheels roll, the value
Unless the centrifugal force mfw? is
limits.
these
between
of Q varies continually
balanced, there results a repeated pounding on the rails with forces that may be very
large at high speeds [position (a), Fig. 545]. Since such pounding is undesirable, it is
Observe too that m7u* would become
the customary practice to use counterbalances.
the track [position (b), Fig. 545].
from
wheels
the
lift
to
speeds
large enough at high
upward instead of downward as assumed.
On highways and race tracks, it is com239. Superelevation of Curves.
in order to reduce the chances that a
curve,
the
“bank”
to
practice
mon
it goes around the curve. On railway
as
vehicle may slide off the highway
curves, the outer rail is elevated above the inner rail in order to reduce
(or eliminate) the flange pressure between the wheel flanges and the rails.
Moreover, in either case, the use of superelevated curves adds to the comfort
of the passengers, since the sensation of being thrown away from the center
of curvature (by the ‘centrifugal force”) is decreased or eliminated.
Consider a car going around a curve of radius r at constant speed. In
At right angles to the
the fore-and-aft direction, the car is in equilibrium.
directed toward the
n
acceleratio
normal
a
is
direction of motion, there
forces at the
frictional
If the curve is level, only the
center of curvature.
tires cause the direction of the velocity to change (keep the car from skidding
off the road). The motion of the car is considered to be curvilinear trans-
It is similar to that of the side rod of § 238.
lation.
In this
In Fig. 546(a), imagine the car stopped on a banked curve.
Fo,
and
case, the dynamic reaction ma, = 0, and the frictional forces Ff;
acting upward parallel to the bank, prevent the car from sliding inward.
CURVILINEAR MOTION [Ch. XVI
840
Now imagine the car moving slowly at first, and then faster and faster.
The inertia force (dynamic reaction) ma, is added to the free body to simuAs the speed increases, ma, increases, since
late equilibrium conditions.
a, = v2/r. We see then that the frictional forces wp the bank will decrease
with increase in speed, since the reversed effective force is taking over the
job previously done by friction.
A point will be reached where the frictional forces are zero.
In this
event, the free body is as shown in Fig. 546(b).
The normal reactions are
equal, N; = No, where formerly Ni; > Ne. From the standpoint of comfort
to passengers, this curve has an ideal bank for this particular speed, which
makes F, = F2 = 0.
(a)
Tends to slide inward.
No sliding tendency.
Tends to slide outward.
Fig. 546.
Now if the speed is increased beyond the point pictured in Fig. 546(b),
the frictional forces act down the bank to prevent the car from sliding outward and upward, and the free body is as shown in Fig. 546(c).
If the
speed is increased until the frictional forces are the limiting frictional forces,
the car is about to skid and will do so if the speed is increased still more.
In the meantime, however, another event may have occurred.
The
car may have turned over. When the car is on the point of turning over,
the reaction N,; = 0, the car is riding on two wheels, and the moment of
ma, about the bottom of the right wheels is exactly balanced by the moment
of W.
A further increase of ma, results in overturning.
The reversed effective (inertia) force ma, is shown
center of gravity.
acting through
the
In Fig. 546(b), where Ni = No, their resultant N = N, + Np passes
through the center of gravity.
Replacing N, + N» by N and summing
forces horizontally, we get
b
LF, =
Wr?
7a
gr
where ma, = Wv?/(gr).
:
Nisin
a0
or
Nigimesese
gr
Summing vertically, we find
ZF, = Ncos@-W
=0
or
N cos 6 = W.
§ 241 | EXAMPLE
841
By division, we then get
mts
Zoe
cal (o) = gr
(H
[ 1=F,=
0],
where @ is the angle of bank for a speed v fps and a radius of curvature of
r ft., when there is no tendency for the body to slide either away from or
Except for the special case of Fig. 546(b),
toward the center of curvature.
N, # No, and N does not pass through the center of gravity.
Superelevation is the difference between the elevations of the outer and
inner edges of a road or of the outer and inner rails of a railroad.
A concrete highway, 30 ft. wide, is curved to a radius of 500 ft.
240. Example.
and superelevated 2 ft. If f = 0.2 for wet, muddy concrete, what is the maximum
speed of the car if it does not skid?
Sotution. A free body showing all the forces
acting on the car, and the inertia force Wv"/(gr)
to place it in equilibrium, is shown in Fig. 547.
The simplest solution to this problem is obtained
by recognizing that the resultant F of the forces
Fi, Ni, F2, and N2 must necessarily pass through
the center of gravity of the car, since this resultant R together with the weight W and the
inertia force Wv?/(gr) must be in equilibrium,
and since three forces must intersect at a point
Moreover, if
in order to be in equilibrium.
Fig. 547.
slipping impends, the inclination of R with a
normal to the incline must be the limiting angle
of friction ¢. Thus, the vector R makes an angle 6 + ¢ with the vertical. Now,
find
from horizontal and vertical sums of the three forces R, W, and Wo?/ (gr), we
Rsin(eo + ¢) =
By division, we obtain
Wr?
mx
R cos(@ + ¢) = W.
2
tan(@ + ¢) = a
gr
¢ = 15.14°.
Using ¢ = tan 0.2 = 11.32° and @ = sin-(2/30) = 3.82°, we get 6 +
Then
(0.2705)|"? = 66 fps = 45 mph.
(500).2)
v = (gr tan 15.14°)!? = [(32
For the conditions in the preceding example, what are the normal
941, Example.
that is, what are
and frictional forces on the left wheels and on the right wheels,
the ground, the
above
ft.
Fy, Ni,-F2, and Nz», Fig. 547? The center of gravity is 2
lb.
3200
is
wheel tread is 5 ft., and the weight of the car
of this probSorution. At least one moment equation is necessary for a solution
A, is taken
point
the
by
ted
lem. If the line of contact at the right wheels, represen
in the
appear
would
N,
unknown
one
only
as the center of moments, we observe that
equation.
ft. and
The moment arms of W and Wv*/(gr) are found to be a = 2.63
842
CURVILINEAR MOTION
[Ch.
XVI
b = 1.83 ft. (Fig. 548). Using Fig. 547, but omitting R in favor of its normal and
frictional components, we get
W 2
2M = Nib) = Wat
- b= 0
= 5N1 — (3200)(2.63)
(3200) (66)2(1.83) _ 0
(32.2) (500)
ahaa
from
;
which
N; =
13865
(0.2)(1365) = 273 lb.
Fig. 548.
Not a Free Body.
The
moment arms a and b of the forces W
2
f
é
"28
tact (A) of the right wheels is given
0ad, we have
in detail for the assistance
the student.
The angle
gerated for clearness.
cal
!orces perpendicular
Ni+
Ne —
Wcosé
Ff, =
es
—
Sumf
to the plane of the
of
@ is exag-
Thus,
js taken from the previous example.)
and Wv?/(gr) about the line of conhere
lb.
(The speed » = 66 fps
We
2
sing=0
Cas Z2tanoe—m(2)(2). 30) e— Ossi ts
Ae is ee
=
4/008 0 =
2:
ieprcw ft.
2
1365 + N2 — (3200) (0.998)
DE = (2.5 +c) sin @ = 0.1755 ft.
b = BE — DE = 1.8295, say, 1.83 ft.
__ (8200) (66)?(2/30) _
(32.2) (500)
:
from which N2 = 1883 Ib. Then F2 = (0.2)(1883) = 376 lb. A sum of the forces
parallel to the plane may be used as a check on the preceding solution.
242. Closure.
By this time the reader should have perceived the advantages of a systematic approach to the solution of problems in kinetics.
First, a free-body diagram should be made showing all the external forces.
Then a decision should be made as to whether or not to add the inertia force
or forces (the dynamic reactions).
If the inertia force is not included in the
free body, we must take moments, if we take moments, about a point on
the line of action of the resultant of the external force system.
Such a point
is the center of gravity for bodies in translation and for the cases discussed
in this chapter. If the inertia force 7s included in the free body, moments
may be taken about any convenient point, but the inertia force
must be
located with its proper line of action.
For the eases considered thus far,
the line of action of the inertia force passes through the center
of gravity
and its sense is opposite to that of the acceleration.
Something else not to
be overlooked is the matter of positive and negative senses.
Choose the
senses for, say, positive component displacements and then
be sure that
accelerations, velocities, and forces are used with the same
sign convention.
Probably in most instances, it will be good to choose the sense
of the velocity
as positive.
Habitual consideration of these factors, together with a
handy
knowledge of the principles learned in the study of statics
and kinematics,
should soon provide the student with a good facility
in solving problems
in kinetics.
PROBLEMS:
MOTION
IN VERTICAL
343
PLANE
Problems
MOTION
IN
VERTICAL
1201. A “‘seconds” pendulum is one which
swings from one extreme position to the
other in one second.
What should be the
length of a simple “seconds” pendulum at
a place where g = 32.2 fps’? The angular
PLANE
(b) his elevation
at the highest point if his elevation at the
lowest point is 2 ft., and (c) the tension in
the rope at the highest point?
Ans. (a) 132.4 lb., (b) 3.55 ft., (c) 113.9 lb.
passes the lowest point,
displacement is less that 5°.
Ans. 39.19 ---in.
1202. A pendulum is adjusted to oscillate once a second at town A. It is then
moved to town B where it is found to oscillate
37,132 times in 10 hours.
Compute g 4/QB-
turning the pendulum
bob up one turn.
Which town has the higher elevation?
1203. A simple pendulum has a length
of 45 in. Its bob (weight) may be screwed
up or down on an American standard fine
screw thread, size No. 10, with 32 threads
per inch. Find the change in the number
of oscillations per 24-hour day caused by
(Satisfactory accuracy may be obtained
Ans. +28.
using 6-place logarithms.)
length L
of
um
pendul
a
1204. (a) If
vibrates once a second, find the period of
vibration if the length is changed to L/2,
(b) Ifa pendulum of length
2L, and 4L.
L vibrates once in time ¢, find the length
required for the period to become ¢/2, 2t,
and 3¢.
1205. A simple pendulum is 30 in. long
and has a maximum angular displacement
of 5°. The weight of the bob (particle A,
Fig. 541) is 4 lb. If g = 32.2 fps?, find
Fig. 549.
Problems 1209, 1210.
1209. A particle of weight W slides on a
smooth, circular, vertical ring of radius 1,
Derive an expression for its veFig. 549.
locity at the lowest point if the starting
position is defined by the angle @ = 60°.
1210. The same as 1209 except that the
answer is to be obtained in terms of any
Ans. (2gr (1 — cos 6)]"?.
angle 0.
(a) the maximum tangential acceleration,
(b) the maximum normal acceleration, (c)
the period of vibration, and (d) the max-
imum tension in the cord.
Ans. (a) 2.8 fps?; (b) 0.245 fps?; (c) ie/(osecs:
(d) 4.03 lb.
1206. A simple pendulum is 75 ft. long
and has a maximum angular displacement
A,
of 6°. The weight of the bob (particle
Fig.
541)
is 392
Ib.
For
g = 32.2
fps?
De.
and for an angular displacement of
normal
and
tial
tangen
the
(a)
determine
the
components of the acceleration, (b)
tension in the cord, and (c) the period.
Ans. (a) 1.122 fps?; 0.313 fps’; (py na2eoulbe:
(c) 9.58 sec.
1207. The same as 1206 except that the
length of the pendulum is 40 ft. and the
bob weighs 64.4 Ib.
1208. A boy weighing 120 lb. swings on
a single strand of rope 30 ft. long and passes
the lowest point at a speed of 10 fps. What
he
are (a) the tension in the rope when
Fig. 550.
Problems 1211, 1212.
1211. A particle A of weight W is free
to slide in a smooth parabolic curve, Fig.
550, whose equation is x? = l6y (units in
fect). If the reaction of the path on the
particle at the lowest point O is 3W, find
the coordinates of the point from which
Ans. 11.3, 8 ft.
the particle was released.
1212. Same as 1211 except that x? = 64y.
1213. A load W is suspended by a weightless, inextensible cord. For what displacement 0, Fig. 551, is the tension in the cord
PROBLEMS: MOTION IN VERTICAL PLANE [Ch. XVI
S44
with the load at the lowest point twice the
tension at the instant of release?
Ans. 41.4°.
1214. The string AB, Fig. 551, passes
through 360°. What should be the speed
of a weight W at the lowest point if it is
to pass the top point without falling into
the circle? There is no friction and AB =
12 in.
Ans. 12.68 fps.
1218. In the “loop-the-loop”’ represented
in Fig. 552, the car starts from a height
h = 60 ft. and the diameter of the loop is
D = 40 ft. Determine the reactions of the
car on the track at points # and G, neglecting the friction.
Lf(80, 100)
|a
ei
ie
Fig. 551.
Ww
Problems 1213-1216.
1215. If the weight W in Fig. 551 goes
through 360° and produces a tension in
the cord of 3W when it is at its highest
point, what is its speed at the lowest point?
What is the tension in the cord at the lowest
point?
AB
= 12 in.
Fig. 553.
Problem 1219.
1219. A toboggan starts at B, Fig. 553,
on a smooth parabolic track, whose curve
is expressed by x? = 64y. The toboggan
enters a vertical circle at O with a velocity
which is just enough to get it past the top
of the circle at A without falling.
(a)
What is the diameter of the vertical circle?
(b) What is the reaction between the tobog-
gan and the track at the point C?
Ans. (a) 80 ft.; (b) 3W.
1216. In Fig. 551, AB is a weightless,
rigid rod. What should be the speed of
weight W at the lowest point in order to
clear the highest point? What is the force
in the rod at the highest point? At the
lowest point? Neglect frictional effects.
F\ioy —= fe} al
tas
A
Fig. 554.
Problems 1220-1222.
1220. A particle of weight W = 50 Ib.
starts from rest (v. = 0) on a circular path
at the position A, Fig. 554, and moves to
the position B. The radius of the path is
40 ft., and
Fig. 552.
Problems 1217, 1218.
1217. The diameter of a “loop-the-loop”
is D = 40 ft. From what minimum height
h, Fig. 552, should a car of weight W = 966
Ib. start in order to stay on the track at
the top of the loop £, friction neglected?
What is the pressure of the car on the track
at the position C?
Ans. 50 ft., 2900 lb.
the resistance
is constant
at
10 lb. and is always in the line of motion.
What is the speed at the position B?
Ans, 42 fps.
1221. The same as 1220 except that
Vo = 27 fps.
1222. In Fig. 554, a particle starts with
¥ = 10 fps at the position A, moves to the
position B against a constant frictional force
in the direction of motion of 10 lb. The
radius of the path is 40 ft. What is the
weight of the particle if it comes to rest at B?
Ans. 15.1 |b.
PROBLEMS:
MOTION
IN VERTICAL
845
PLANE
1223. Airplane pilots can withstand an
acceleration of 9g for short periods of time.
For an acceleration of 9g, what should be
the minimum radius of the path of a test
pilot as he pulls his plane out of a 600-mph
dive? With what maximum force would
a 150-b. pilot press on his seat in this
maneuver?
Ans. 2675 ft., 1500 lb.
1224. A weight is suspended from an
18-in. string attached inside to the top of
an automobile which is on a level road.
If this car
accelerates
gradually,
what
is
the inclination of the string with the vertical
when the acceleration is 5 fps??
1225. The steam engine on a locomotive
has a stroke of 30 in., giving a crank radius
highway profile so that at 60 mph the
effects of centrifugal force will not increase
or decrease the load on the tires of vehicles
by more than 50%.
Find the least radius
of curvature (vertical) that he may specify
(a) for a hilltop and
(b) for a dip in the
road.
Ans. (a) 472 ft., (b) 472 ft.
1227. If a roadway has been designed
with a minimum radius of curvature for
hilltops and dips as described in 1226,
calculate the per cent decrease or increase
in tire loads when a car goes 90 mph (a)
over a hill and (b) over a dip. Does the
curve have to be the are of a circle? Is
this speed safe? Explain.
# = 15 in., Fig..555. The side rod AB
weighs 3220 lb. The inertia effect of the
side rod may be balanced by weights placed
on the driving wheels, which are 6 ft. in
diameter in this case.
The weight is usually
in the form of a segment, shown crosshatched in Fig. 555. If @ = 45°, what
should be the thickness of the steel segmental counterweight in order to balance
exactly the inertia force mfw? for the side
rod?
Ans. 0.645 ft.
1226. An engineer considers designing a
MOTION
IN
Fig. 555.
HORIZONTAL
1228. An automobile travels at a constant
speed of 60 mph around a 400-ft. radius
A 2-lb. weight is suspended
highway curve.
from its top on a 15-in. string. Find the
angle the string makes with the vertical.
6”
Problem 1225.
PLANE
1230. Certain parts of a projectile fuse
are to withstand 30,000g of acceleration.
To test the parts, a device similar to Fig.
556 is considered, where the parts are to be
attached to the arms and represented by W.
The test assembly weighs W = 3.22 lb.
The weight of the arms is negligible.
For
an acceleration of 30,000g on the test assembly W, find (a) the angular velocity
necessary in rpm, (b) the force in the arms,
and (c) the angle 0.
Fig. 556.
Problems 1229, 1230.
1229. In the flyball governor represented
in Fig. 556, the weight of each flyball is
W =10 lb. If the weight of the arms is
neglected, at what speed (rpm) will the
angle @ be 30°? What is the tension in an
Ans. 31 rpm, 11.53 lb.
arm at this speed?
Fig. 557.
Problems 1231, 1232.
1231. Figure 557 represents a weighted
flyball governor, The circular disk A, which
346
PROBLEMS:
weighs 100 Ib., rests upon the flyballs (W =
10 lb.) in such a manner that each flyball
supports half the weight of the disk. The
inertia forces of the disk are evidently in
Within limits, the flyballs
equilibrium.
may move outward, raising the disk someNegwhat and thereby operating a valve.
lecting both the frictional effects and the
weight of the arms, determine the speed
of the governor
(rpm)
when
@ = 30°.
At
this speed, what is the tension in the arm?
Ans. 76.3 rpm, 69.4 lb.
1232. The
0 = 45°:
same
as
1231
except
MOTION IN HORIZONTAL PLANE [Ch.
XVI
and weigh 1 lb. each.
What bending
moment is induced at the base H# of the
pins?
Ans. 122.6 in-lb.
1236.
A horizontal disk, 4 ft. in diameter,
rotates about a vertical axis. A body W
rests on the disk at a radius of 1.5 ft. If
f = 1/3, at what speed (rpm) is the body
about to slip? Assume that this speed is
attained with small (negligible) tangential
acceleration.
that
Fig. 560.
Problem 1237.
1237. A body B, Fig. 560, in the form
of a parallelepiped stands on end on a disk
A with its eg at a radius of r = 8 in. from
the axis of rotation.
What is the speed
(rpm) of the disk if B is on the point of
Fig. 558.
1233.
tipping over?
Assume that this speed is
attained with small (negligible) tangential
acceleration.
What minimum coefficient of
friction is necessary for tipping to occur?
Ans? 42 rpm, f = 0.42
Problems 1233, 1234,
Figure 558 shows a more
common
method of weighting a governor (compare
with problem 1231). The weight W’ =
100 lb. moves upward along the spindle
AD as the flyballs (W =
10 Ib.) move
out-
ward.
The pins at C and # each carry
half the weight W’.
For @ = 30°, determine
the corresponding speed (rpm) of the governor.
Neglect the weights of the arms
and the effect of friction.
Also determine
the forces on the arms AB and BC.
Ans. 157.7 rpm, 69.3 lb., 57.8 lb.
1234. The same as 1283 except that
P= ANS,
Fig. 559.
Problem 1235.
1235. A symmetric
rotates at a constant
arm 4, Fig. 559,
speed of 600 rpm.
Attached to the arm at a radius of 6 in.
are cylindrical pins P which are 4 in, long
Fig. 561.
Problems 1238, 1239.
1238. A 32.2-lb. body B, Fig. 561, is at a
radius of 2 ft. on a conical disk which rotates
about a vertical axis.
If f = 1/3, at what
speed (rpm) is the body B on the point of
moving toward the center of rotation?
The
tangential acceleration is negligible.
1239. The same as 1238 except that the
speed at which B is in impending motion
away from the center of rotation is desired.
Ans. 40.6 rpm.
1240. A steel ball is placed in a smooth
parabolic bowl.
The equation of a section
of the bowl is x? = 4y. At what speed is
the bowl turning if the ball climbs to a
point where x = 4 ft.?
Ans. 38.3 rpm.
1241. To the shaft AB, Fig. 562, are
attached
the rods AC
and BCD,
At the
PROBLEMS:
MOTION
IN HORIZONTAL
547
PLANE
tively. The resulting force system is coplanar and d = 3 ft. The fit of the pins is
such that the pins / and D take all the
vertical reaction.
Determine the components of the pin reactions at C, D, H, and F.
Ans. D, = F, = 128.8 lb., Hz = 180 I|b.,
F, = 2310 lb., Cz = 435 lb., Dz = 2560 lb.
Fig. 562.
Problem 1241.
end of rod BCD is a solid sphere D which
weighs 64.4 Ib. and is 1 ft. in diameter.
If the shaft AB rotates at 120 rpm, determine the tension in AC and the horizontal
and vertical components of the reaction at
B. The rods have negligible weight and
the pins are smooth.
Fig. 564.
Problem 1244.
1245. What force acting toward the center will make a body of weight W move
with constant velocity in a 6-ft. diameter
circle with such a velocity as to complete
a revolution in 6 sec.?
Fig. 563.
Problems
1246. An automobile is driven around a
circular curve on a horizontal road at 60
mph.
(a) If the maximum coefficient of
friction is 0.8, find the least radius of
curvature for which the car will not skid.
(b) What would be the maximum and
1242, 1243.
minimum speeds without
vehicle were going around
1242. The framework of Fig. 563 rotates
at a constant speed of 600 rpm about the
The weight of AB is 32.2 lb.
axis EG.
and the weight of the other members is
negligible
considered
this
problem.
for the purposes
of
are
the
If @ = 60°,
what
forces at the pins A and B? What are the
radial reactions at the bearings # and G?
The construction is such that the forces are
essentially coplanar.
Ans. Fa = 13,700 lb., Bz = 5000 1b., By =
11,880 lb., G, = 5880 Ib., H, = 5960 lb.
1243. For
Fig.
563,
the
conditions
determine
the
in 1242
and
at
the
reactions
pins D and C.
1244. In Fig. 564, rotation occurs about
We neglect the
the axis AB at 90 rpm.
effects of the weights of all members except
those of the spheres J and K, which weigh
W, = 96.6 lb. and W2 = 32.2 lb., respec-
radius
found
in (a), when
skidding, if a
a curve of the
ice covered
the
road and f = 0.05?
Ans. (a) 306 ft., (b) 15.1 mph.
1247. The same as the example of § 240
except that the radius of the curve is 125 ft.
1248. (a) If an airplane makes a turn in
a horizontal plane without sideslip at a
constant speed of 400 mph, at what angle
must the plane be banked when the radius
of the path is 500 ft.? (b) If the pilot can
stand an acceleration of 9g for a short time,
would this turn be too sharp?
Ans. (a) 87.3°.
1249, The top speed of a combat plane
is 660 mph.
The pilot is not to be sub-
jected to an acceleration greater than 9g.
(a) What is the corresponding minimum
radius for making a turn in a horizontal
plane, and (b) at what angle must the
plane be banked for zero sideslip?
848
PROBLEMS: MOTION IN HORIZONTAL PLANE [Ch. XVI
1250. A motorcycle is traveling at 90 mph
on a circular track of 300-ft.
radius.
(a)
What should be the inclination of the track
if there is no side frictional force?
(b) If
the weight of rider and cycle is 350 lb. and
if the inclination of the track is actually
80°, what frictional force is necessary to
keep the cycle from sliding? In what direction does sliding tend?
What coefficient
of friction is necessary to prevent sliding?
What is the inclination of the plane of the
motorcycle with the vertical?
Ans. (a) 60.8°; (b) 235 lb., f = 0.346.
1251. The same as 1250 except that the
track radius is 200 ft.
1252. A stunt
rider wants
to build
a
40-ft. diameter cylindrical wall on which to
ride a motorcycle.
He and the machine
together weigh 483 lb., the coefficient of
friction of the tires on the wall is 0.5, and
both tires together have a safe load limit of
1600 lb. Assuming that he travels in a
horizontal circle, calculate (a) the minimum
safe speed and the angle that the machine
and rider make with the horizontal, and
(b) the maximum safe speed and the angle
that the machine and rider make with the
horizontal.
(c) Recommend
to the rider
a good cruising speed, with reasons.
Ans. (a) 24.5 mph, 26.6°; (b) 30.6 mph,
eons
1253. A maximum speed of 70 mph at
impending skidding on a highway curve of
1200-ft. radius is to be allowed.
The width
of the road is 24 ft. For an estimated
minimum f = 0.15, what is the superelevation in inches?
For this superelevation,
what is the most comfortable speed for the
passengers?
Ans. 2.8 ft., 46 mph.
1254. A train is to travel around a 1600ft. curve at 60 mph with zero flange pressure.
(a) If a tie is 9 ft. long, how much higher
should the outer extremity be than the
inner one?
(b) For a car weighing 200,000
Ib. on this track, what is the total flange
pressure at a speed of 50 mph?
At 70 mph?
1255. A railway curve of 1200-ft. radius
is superelevated for zero flange pressure at
a speed of 50 mph.
(a) If the rail gage is
4 ft. 8.5 in., what is the superelevation of
the outer rail? For a car weighing 80 tons,
what is the total flange pressure when the
speed is (b) 30 mph, (c) 70 mph?
Ans. (a) 7.79 in.; (b) 14,100 Ib.; (c) 21,150 Ib.
1256. If the 80-ton car in 1255 is symmetrically loaded, how high could its center
of gravity be above the rails without overturning at 70 mph.
1257. A 30-ft. highway has a 600-ft.
curve at which the superelevation is 3 ft.
(a) If f = 0.18, what is the speed of a car
at impending slipping?
(b) If the tread of
a car is 5 ft. and if its eg is 30 in. above
the ground, at what speed would it be on
the point of turning over?
What is the
minimum coefficient of friction for this
speed?
Ans. (a) 50.6 mph; (b) 105 mph; f = 0.996.
1258. An unbanked highway curve is to
permit a speed of 50 mph without skidding.
(a) If f= 0.6, what should be the minimum
radius of curvature?
(b) If the tread of a
car is 4.7 ft. and if its eg is 24 in. above the
ground, what is the speed of the car when
it is on the point of turning over, for the
radius found in (a)?
1259-1270. These numbers may be used
for other problems.
Chapter
XVII
ROTATION AND
RIGID BODIES
PLANE
MOTION
OF
In this chapter we shall study force systems which
243. Introduction.
produce an angular acceleration of rigid bodies which move in rotation
In subseor in a combination of rotation and translation (plane motion).
quent articles which establish the conditions for these motions, observe
that the equations obtained follow from the application of Newton’s second
law (§ 220). Thus, we shall be dealing with no new principles, but only
with an extension of Newton’s laws of motion.
244. Rotation of a Rigid Body. This analysis is limited to a body whose
center of gravity has plane motion; moreover, the body is symmetric about
the plane in which the center of gravity
Thus, a body B of mass m,
moves.
Fig. 565, is symmetric about a plane
through its center of gravity and perpendicular to the axis of rotation O.
The body has an angular velocity of
and an angular acceleration of a.
Choose at random a particle of mass
dm, which passes through the body
perpendicular to the plane of the
paper with an axis parallel to the axis
of rotation O, and show the; effective
forces (not the reversed effective forces)
acting
;
on
it.
The
normal
and
tan‘
gential components of the acceleration
Rotating Body. The length of
Fig. 565.
element dm, perpendicular to the
the
plane of the paper, is equal to the thick-
ness of the body. The effective force
dma, is the tangential component of the
yesultant on dm; ete. for dmay.
of this random mass dm are a, and a.
dm a,, normal
Then the effective forces, equal to mass tvmes acceleration, are
dm a,, tangent
to the path and directed toward the center of rotation, and
= ra. From
a,
on
accelerati
l
to the path and in the sense of the tangentia
direction,
Newton’s second law, the sum of the forces on the particle in the x
849
ROTATION AND PLANE MOTION OF RIGID BODIES [Ch. XVII
350
for example, is equal to the effective force in that direction; that is, for the
particle dm,
(a)
—dk, = dma,sin 6 — dma, cos 6,
where the negative sign is given to the dR, because we know that the x
component of the resultant force on the particle must act toward the left
in order to cause the particle in its present position to move in the circular
path b-b. Notice particularly that the x axis passes through the center of
gravity and the axis of rotation, a choice that results in the simplest equations.
From Fig. 565, we see that sin 6 = y/r and cos 6 = 2/r. Using these relations and the relations a, = ra and a, = rw? in equation (a), we get
Sahl I
(b)
(dm) (ra) (")— (dm) (re) (2),
—dR, = aydm — wx dm.
A sum of the resultant forces dR, for all the particles of the body is equal
to the z component of the resultant R of the external forces only (D’Alembert’s principle, § 226). Letting this
component be &,, we have, from (b),
-Re
(c)
= afydm — ot [xam,
—R, = aym — wrm,
where we recognize that fy dm = ym
and fedm = %m.
For the particular
axes shown in Fig. 565, 7 = 0 since
the x axis passes through the center
of gravity and consequently aym = 0
Fig. 565. Repeated.
in equation (c). Moreover, = becomes
7, which is the distance between the
center of rotation and the center of gravity, and
wim
where
we
gravity.
(44)
observe
= wim = miw? = maz,
that fw? = d,, the normal
acceleration
of the center of
Thus, using the symbol =F, for R,, we have from (c)
=F, = Mw? = man.
where the subscript
is a reminder that the direction for the sum of forces
is normal to the path of the center of gravity at the instantaneous location
being studied. This result, equation (44), agrees as it should with the
findings of § 226.
Now we make a sum in the y direction, Fig. 565, with the positive direction upward, and proceed as for the sum in the z direction.
Noting that the
§ 244 |ROTATION
OF A RIGID BODY
Bay
y component of the resultant force on the particle dm must act downward
to accord with the sense of a shown we find
—dR,
=
=
(d)
Ry
—dma,cos 6 — dma, sin 0
a
—dm ta — — am re ZeYy=,
afedn + ot [ydm.
As before, if the x axis passes through the center of gravity and the center
of rotation, fydm = 0 and fedm = %m =m,
so that R,, now
written
as 2F,, becomes
LF, = mra = m4,
(45)
where the subscript ¢ is a reminder that equation (45) applies only for a
sum of the forces in a direction tangent to the path of the center of gravity
at some instantaneous location and where fa = 4d, is the tangential acceleration of the center of gravity.
Finally, since this body has an angular acceleration, it is evident that the
Inasmuch
sum of the moments of the external forces about O is not zero.
as the component dm a, is directed through 0, all the moments of the various
forces dm dp, about O are zero, and therefore the moment (d//,) of the effective forces on the particle dm about the axis O is equal to the moment of
dm a, about the axis O.
Thus
dM, = (dma,)r = (dm) (ra) (r) = ar? dm.
By D’Alembert’s principle, the sum of the moments of the effective forces
on all the particles is equal to the sum of the moments of the external forces,
all about the same axis.
Let 2M,
designate the sum of the moments of the
{(dM,) = 2M, about
external forces
torque or turning moment) ;then
the axis of rotation O (that is, the
SM, = afrtdm = Tas,
(46)
where fr? dm is recognized as the moment of inertia I, of the mass m about
indithe axis of rotation O. In making the sums of forces and moments
for all external
cated by equations (44), (45), and (46), be sure to account
at the axis of
reactions
the
and
forces, including the weight of the body
rotation.
From the foregoing analysis, we have three equations
1):
rotating body (x and y axes in Fig. 565 are now n and
rays
—
miu?
—
MGn,
mony, =
mMmra
=
Mat,
applicable to a
>M,
where the axes n and ¢ are chosen specifically as stated above.
=
Lee,
The axis O
respect to
is the fixed center of rotation, and both M, and J, must be with
852
ROTATION AND PLANE MOTION OF RIGID BODIES [Ch. XVII
the same axis O.
about. O.
Recall that 7, = mk,?, where k, is the radius of gyration
245. Rotation about an Axis through the Center of Gravity. If the center
of rotation O coincides with the center of gravity, 7 = 0. Therefore, equations (44) and (45) become
(e)
nl, = 0)
and
ain == (0),
[CENTER OF ROTATION COINCIDES WITH CG]
Moreover, in (46), I, becomes J, the moment of inertia of the body about
a gravity (centroidal) axis perpendicular to the plane of rotation.
Then
(46) becomes
(f)
=M = Ta,
where 2M is the sum of the moments of the external forces about the gravity
axis. This special case of rotation about a gravity axis includes a large
percentage of rotating machine elements, such as gears, pulleys, and flywheels.
In every instance, it is well to choose the center of rotation (when it is fired)
as the center of moments, and to use the corresponding moment of inertia
on the other side of the equation (46).
246. Example—Rotation about the Center of Gravity.
A 4-ft. flywheel, which
weighs 3220 lb. and has a radius of gyration of 15 in., is subjected to a torque of
2400 in-lb. Determine the tangential acceleration of a point on its rim.
SoLutTion.
We first find the angular acceleration from
=M = Ia = mk*a,
2400
(3220\/15\"
pie eae
whence a = 1.28 rad. per sec.2_
2 ft. is
ahealas
The tangential acceleration of a point at a radius of
a, = re = (2)(1.28) = 2.56 fps
247. Example—Rotation about the Center of Gravity. In Fig. 566, a crank A
is attached to a gear B. The gear B meshes with a gear C. The gear C is attached
to a disk D. A constant force F applied to the crank speeds up the disk with
constant acceleration to 1000 rpm in 40 sec. The disk and gear C weigh 400 lb. and
together have a radius of gyration of k, = k = 10in. about the axis of rotation.
The gear B and the arm A weigh 75 lb. and have a radius of gyration of k, = & = 14 in.
about their axis of rotation. Neglecting friction, determine the magnitude of the
force F’,
Sotution.
The turning force between the teeth of the gears is designated by Q,
Fig. 566(b). Without friction, the bearing reaction R, passes through the
center of
the shaft and has no moment about this axis. Thus [see Fig. 566(b)]
1000 rpm =
(1000) (27)
60
= 104.7 rad. per sec.
ABOUT THE CENTER OF GRAVITY
§ 248 |EXAMPLE—ROTATION
353
The angular acceleration of disk D (and C) is
ee
ER
t
40
a
= 2.62 rad. per sec.
For gear C and disk D, we have
ko = Ge
4
10\
= (A5\(5)
= 8.63 slug-ft?
Then, using equation (46), we have
pS
=
4.5Q
= (8.63)(2.62),
5
Toa,
Q = 60.3)b.,
speed
which is the force that must be applied to the teeth of C in order to bring its
instant
any
at
C
and
B
gears
the
of
to 1000 rpm in 40 sec. The angular velocities
Fig. 566.
acceleration of
are inversely proportional to their diameters; therefore, the angular
36
36
;
For crank A and gear B, we have
lop = mk. =
From Fig. 566(c),
2
(Nw)
BONOY
sas = 1(2) a!)
22
Using 2M, = Ia, we find
226
12
whence F = 50.5 lb.
= 3.17 slug-ft.?
NAY
(60.3) (18)
12
18
= (3.17)(0.655),
. A 2 x 4-ft. cast-iron
248. Example—Rotation about the Center of Gravity
a load A of W4 = 25 |b.
s
drum B, Fig. 567, has wound on it a cable which support
ent of friction is 0.015.
coeffici
their
and
r
The supporting bearings are 6 in. in diamete
= k = 8.5in. How
k,
n
gyratio
of
radius
a
has
The drum weighs W, = 5650 lb. and
rest?
from
starts
it
if
sec.
20
in
far does the load A move
854
Sotution.
ROTATION AND PLANE MOTION OF RIGID BODIES [Ch. XVII
The needed free bodies are shown in the end view of Fig. 567, where,
for the purposes of this problem, the forces may be considered as coplanar.
The total
normal force at the bearings is N = N, + N». With little error, it may be assumed
that the frictional force F' is balanced by a reaction Q passing through the axis O.
pa
ee
Fig. 567.
In the free body of load A, choose the downward sense as positive (this is the
sense
of the velocity) and sum forces vertically.
(g)
PSs
Ra
W
eeogo
Observe that the acceleration of A is the same as the tangential
acceleration of a
point on the surface of the drum; hence, a = a/r, or a = @
since r = 1 ft. Thus
equation (g) becomes
DE
(h)
25 = NS —
39.9 ° &,
Considering the free body of the drum, and equating the
vertical sum of the forces
to zero since there is no motion of the drum in this direction,
we have
zr, =N—-W,-T=0,
N=W,+T
= 5650+ Tb.
F = fN = 0.015(5650 + 7) Ib.
Now the sum of the moments of the forces on the
drum about O is equal to Ija.
Hence
;
I, = mk,? = (S82)
2M, = T(1)
(i)
= 88 slug-ft.?
r(4) sie
2M, = T — (Ot) eso se)
SS ox
A simultaneous solution of equations (h)
and (i) may now be made for T and
a;
a = 0.0418 rad. per sec.2
a = 0.0418 fps?,
and
T = 24.97 lb.
The distance moved by A in 20 sec. Is
al
0.0418) (20)?
sai ak
ee
ee
249. Example—Rotation Not about
the Center of Gravity. A steel sector
, with
a central avgle of 60°, a thickness of 6
in., and a radius of 4 ft., is in a position
with its
355
NOT ABOUT THE CENTER OF GRAVITY
§ 249 |EXAMPLE—ROTATION
under the action
center line OB at 6 = 30° with the vertical (Fig. 568), having moved
the sector has
If
vertical.
was
OB
line
of gravity from the position where its center
tangential comrotated from rest about the smooth axis O, what are the normal and
ponents (R, and R,) of the reaction at the axis?
The external forces acting on the body, as shown in Fig. 568, are
So.ution.
the reference
W, R., and R,. In order to use equations (44) and (45), we recall that
through
passes
also
them
of
one
that
and
rotation,
of
axes pass through the center
is neceswork
ry
Prelimina
n.
and
¢
by
axes
these
Designate
gravity.
of
the center
ft.)
sary to determine W, 7, and J,. The weight of the sector is (490 Ib. per cu.
7A? \ f 1
aoo(*)(5) = 2052 |b.
i
we get
Using the equation for = ( = 7) in § 118 (where B = 30° = 7/6rad.),
2r sin 30°
es
S761
= (2) (4)
4)(6)
|
rege
= 2.55 ft.
mr?/2 (§ 160); hence
The moment of inertia of this sector about the O axis is
mr? (2052)(16) _
510 slug-ft.?
I,=
2
(32.2)(2)
Applying 2M, = I.a, we find
=M, = Wr sin 30° = (2052)(2.55)(0.5) = 510a,
(j)
of the velocity and accelerafrom which, a = 5.13 rad. per sec. Letting the sense
in the tangential direction, we
tion determine the positive sense, and summing forces
get (downward direction along t axis positive)
>F, = —R. 4+ W sin 30° = mira
2052
=>=R,-+ (2052)(0.5) = & )\2.55)65.18,
from which, R; = 193 lb., the positive result
indicating that R, is shown in the correct
(Note that if the upward sense along
sense.
the t axis had been taken positive, « would
To make a sum
have been used as —5.13.)
to determine
need
we
on,
directi
in the normal
w (in order to evaluate mfw?).
0/7
To do this, we
a as
must return to equation (j) and express
568,
Fig.
from
Thus,
a function of 6.
3
6’
510a,
=M, = Wrsine = (2052)(2.55) sin @ =
a = 10.28 sin 6,
center
where 6 is the variable angle that the
since a is a variable we use w dw = a dé and
Now,
al.
vertic
line OB makes with the
obtain
30°
o dw =
1028
|
sin 6 dé,
0
it)
30°
oi= 1025]— cos | = 10.28(— 0.866 + 1),
n
0
856
ROTATION AND PLANE MOTION OF RIGID BODIES [Ch.
XVII
from which w? = 2.75. Next, choosing the sense of the normal acceleration of the center
of gravity (Fw) as positive, and summing forces in the n direction, we get
W
=F, = W cos 30° — Rk, = i Sie
2052
(2052) (0.866) — R, = ( \2.55)2.75),
32.2
from which R, = 1330 lb., the positive sign again showing that the correct sense was
chosen. The resultant # and its direction may now be easily found if desired.
Nors. If the upward direction OB had been chosen as positive, then rw? would
have been negative. For the direction OB positive, the sum becomes
=F, = Rk, —
W cos 30 = Tr-2.55)(2.78) |
which is the same as the previous equation.
Decide upon the positive sense, and
then consider the sign of every term with respect to your choice.
250. Line of Action of the Reversed Effective Force, Mra.
If it is desired
to simulate a condition of equilibrium for rotating bodies, by imposing the
reversed effective forces on the free body, it is necessary that these inertia
forces be placed on their proper lines of action.
Since the inertia (centrifugal)
force m7w” passes through the center of rotation, this force is easily located
on the line joining the center of rotation and the center of gravity and it
points away from the center of rotation (opposite to the sense of the normal
acceleration 7w° of the cg). See mfw? in Fig. 569(b).
(b) Simulated Equilibrium.
Fig. 569.
Location of the Reversed Effective Forces.
Suppose that a rotating body is acted upon by the
external forces shown
in Fig. 569(a). We know that the resultant of this
system in the ¢ direction
is so located that it has a turning moment or
torque about O, because the
body has an angular acceleration.
We also know that the magnitude of this
component of the resultant is ma (§ 244) and
that its sense is determined
by the angular acceleration a. Assume that
mFa acts with some moment
arm L about the axis O. Since it is known that
the sum of the moments
2M, of the external force system about O
is equal to J o@, the moment
mfFa about O is
of
§ 250 |LINE OF ACTION OF THE REVERSED EFFECTIVE FORCE, Mra
(k)
mraL
357
= I,a = mk,?a,
because ma is the only component of the resultant which has a moment
about O. From equation (kK), we find
k 2
(1)
Lae,F
that is, the line of action of the effective (and the reversed effective) force
Thus,
mfa must be at a distance L = k,?/F from the center of rotation.
shown,
as
the free body in Fig. 569(b) is in equilibrium when mfa is located
opposite in sense to the t component of the resultant.
If only the external forces are used in the free body, Fig. 569(a), the
equations of motion are
aA ae’ Mier
ZF, = mra,
Lia tw",
If the reversed effective forces are added to the
as previously explained.
free body, as explained for Fig. 569(b), the x and y axes of reference may
be taken in any direction; in which case, the following equations apply:
=F, = 0,
=F, = 0,
where =M represents the sum of the moments
the inertia forces, about any convenient point.
of all the forces, including
(b) Simulated Equilibrium.
(a)
Fig. 570.
=M =0,
and
Transforming mra into a Force and a Couple.
couple [§ 33(c)].
The inertia force mfa may be replaced by a force and a
mfa, each acting
Thus, in Fig. 570(a), apply the equal and opposite forces
There is now the equivalent
through the center of gravity, as shown.
mfa(L — 7), whose
of a force mfa opposed to the sense of a, and a couple
evaluated more easily
sense is opposite to that of a. This couple may be
if it is transformed as follows.
mira(L — ft) = mraL — mira.
on (k); hence
The term mfaL is seen to be equal to I,a, by equati
mfo(L — 7) = (Ip — mP)a = Ta,
about the gravity
since I, — mf? is equal to the moment of inertia J
if to the external
re,
Therefo
(eg) axis (from the transfer formula of § 158).
358
ROTATION AND PLANE MOTION
OF RIGID BODIES [Ch. XVII
force system we add the inertia forces m7w? as before and miro through the center
of gravity, and if we also add a couple whose magnitude is Ja and whose sense
is opposite to that of a, we have placed the body in equilibrium.
The
complete force system is shown in Fig. 570(b), which should be compared
with Fig. 569(b).
To recapitulate: a simulated condition of equilibrium may be obtained in
a rotating body by either of two plans. First plan [Fig. 569(b)]:
1. Add to the free body the reversed effective force mfw? with a sense
opposite to that of d, and with a line of action On passing through the
center of rotation and the center of gravity.
2. Add to the free body the reversed effective force mfa with a line of
action perpendicular to the line On (through the center of rotation and
the center of gravity) and passing through a point a distance of L =
k,?/7 from the center of rotation with a sense opposite to that of 4,,
the tangential acceleration of the eg.
Second plan [Fig. 570(b)]:
1. Add to the free body the reversed effective force mfw2 with a sense
opposite to that of a, and with a line of action On passing through the
center of rotation and the center of gravity (as in the first plan).
2. Add to the free body the reversed effective force mFa passing through
the center of gravity (cg) with a line of action perpendicular to the line
On (through the center of rotation and the center of gravity) and with
a sense opposite to that of 4@,.
3. Add to the free body a couple of magnitude Ja with a sense opposite to
that of a, where J is the moment of inertia of the body about a gravity
axis perpendicular to the plane of rotation.
251. Example.
SotuTion.
Solve the example of § 249, using the reversed effective
forces.
From § 249 we get
8 = 30°, W = 2052 lb., r = 2.55 ft., J, = 510 slug-ft.2
To find k.?, we note that I, = mk,? ; therefore
F, _ (510)(82.2) _
Nm
Then
m
ke
IE,
_—
2
ad
2052
8
2 5b
=
reat.
3.14 ft.,
which locates the line of action of ma (Fig. 571).
The solution may now follow
closely the procedure of § 249 or it may be varied
somewhat.
In the free body of
Fig. 571, choosing the point P as the center of moment
s, because it eliminates from
the resulting equation three forces, one of which
is unknown and one of which contains
the unknown angular acceleration, we have
=Mp = W(0.59 sin 0) — RL = 0,
§ 252 |]COMPOUND
whence
R, _
as before.
859
PENDULUM
(2052) (0.59) (0.5) =y1931be
3.14
Taking moments about point O, we find
=M, = (mra)L — Wi sine = 0,
whence
(m = W/g)
a _ g sin é ae 32.2
sin 9 = 10.28 sin @.
L
3.14
Proceeding as in § 249, we find w? = 2.75. Finally,
=F, = W cos 30° — Ra — mru* =
(2.55) (2.75) = 0,
= (2052) (0.866) — Ra — (5)
32.2
from which R, = 1330 lb.
A compound pendulum consists of any
252. Compound Pendulum.*
axis.
body of finite dimensions oscillating about a frictionless horizontal
displaced
is
it
If
Let the body in Fig. 572 be suspended from the axis O.
@ and then
from its position of equilibrium by some angular displacement
effects
released, it will oscillate indefinitely with some period 1, if frictional
accelerangular
the
examples,
of all kinds are neglected. As in the preceding
ent 6. For
ation a varies as some function of the instantaneous displacem
about O.
the position of the body shown in Fig. 572, let us take moments
In doing so, let counterclockwise displacement be chosen as positive; then clockwise
moments are negative.
=M, =
— W (f sin 6) = I,a = mk,’a,
whence
a=
Fig. 572. Compound Pendulum.
Point P is the center of oscillation.
= sin 6.
If @ is very small, sin @ is very nearly equal to
§, and we may write
(m)
26
(Cl Wie =
gF
boas
Hence the equation
nts.
For a particular pendulum, 7 and k, are consta
replaces L, the
k,2/f
that
(m) is the same as equation (b) of § 235, except
It follows that the results found for the
length of the simple pendulum.
to a compound pendulum if we
simple pendulum (§ 235) may be applied
to solve the problem of the compound
*Christian Huygens (1629-1695) was the first
This was before the invention
um.
pendul
seconds
a
of
length
the
pendulum and compute
in this field is believed to
work
his
and
clock
um
of the calculus. He patented a pendul
a pendulum clock
devised
Galileo
.
pments
be entirely independent of Galileo’s develo
footnote in § 312.
the
also
See
made.
be
could
one
but died before
860
ROTATION AND PLANE MOTION OF RIGID BODIES [Ch. X VII
substitute k,?/7 for L in the equations for a simple pendulum.
For a simple
pendulum 7 = 2r(L/g)"*; therefore, for a compound pendulum,
(n)
;
T=
keg2\ Yl?
22%) ,
when the amplitude @ is small.. We observe that a simple pendulum of
length L = k,’/r, sometimes called the equivalent simple pendulum, has
the same period as the compound pendulum.
If we lay out a distance L
along the line O-cg, a point P is located which is called the center of oscillation in the case of a pendulum.
(See also § 253.) We observe also that the
point P is a point on the line of action of the reversed effective force mra (§ 250).
If the compound pendulum is suspended from the point P instead of from the
point O, the period is unchanged.
From J, = mk,*, we get k,? = I,/m = gI,/W.
Squaring both sides
of (n) and using this value of k,?, we get
L
en
A4r’gI,
gO
TgW
’
whence
er aVWa
I, =
Tp
This expression suggests an experimental method of determining the moment
of inertia of a body about some axis O. If (which may be determin
ed by
balancing the body) is known, the period 7 of a complete oscillatio
n (to
and fro) about the axis can be observed, after which J, can be
computed.
253. Center of Percussion.
The point P, Fig. 573, through which the
reversed effective force mfa acts, is also called the
center of percussion.
The significance of this term is evident if we imagine that
a force F, collinear
with mfa, acts through the point P. In this event,
the addition of the
reversed effective forces to the free body shows that
there will be no horizontal
reaction at the pivot O in Fig. 573(a), since F and
mfa are the only horizontal
forces and since therefore they must be equal and
opposite (=F, = 0 for the
free body of Fig. 573(a) ). Inasmuch as the only
reaction possible at O is
the vertical reaction R,, there is no tendency
for the point O on the body
to move in a horizontal direction.
However, if F is not collinear with mra, the two
forces F and mfa could
not be in equilibrium, so that a horizontal reacti
on at O is then essential
to a simulated equilibrium.
If we suppose that F' acts below the point
P,
Fig. 573(b), the point O on the body tends
to move toward the left. Therefore there is a reaction R, on the body at O
toward the right. If we suppose
that F acts above the point P as in Fig.
573(c), the point O on the body
tends to move toward the right, in which
case, the reaction R, at O is toward
the left.
§ 253 |CENTER OF PERCUSSION
361
Imagine that an object, such as a baseball, is struck by a long slender rod,
such as a baseball bat. If the object is struck by one end of the rod while
the rod is held by the other end, a distinct stinging sensation is produced
in the hands. A heavy steel rod causes more severe stinging than a light
wooden rod. This sting is the physical evidence of the reaction at the point
If the object
of support (the hands) discussed in the previous paragraph.
is struck so that the line of action of the applied force F passes through the
center of percussion, no sting (no reaction in the tangential direction) is
felt by the hands.
In (c), 2F, = F — Rp, — ma = 0.
= F + Ri — mra = 0.
The horizontal reaction at O is zero only when F is
Fig. 573. Center of Percussion.
and mya are assumed to be in a plane of symmetry
F
vectors
The
colinear with mra.
Since the reversed effective forces (dynamic reactions)
parallel to the plane of the paper.
about the center of
always pass through the center of percussion, the sum of the moments
percussion is always zero.
In (a),
2F,
= F — mfa = 0.
In (b), DF.
Observe that the points P and O are related; that is, if the axis O is moved
This relation is
to another position, the center of percussion is changed.
L = iiote Le
have
we
By equation (1) of § 250,
easily determined.
parallel
Fig. 573, 7 is shown as the dimension g. The transfer formula for
axes yields
k2=R+rP
=k?
+ @,
(cg, Fig. 573)
where & is the radius of gyration with respect to the gravity axis
we have
parallel to the axis O. Using these relations and Fig. 573,
ent
fi Bie
k,2
K2
+
2
Oo}
qd
=
qd
’
from which we find
Pp =
ke
—q )
or
qd =
k?
—,
Pp
or
PY
=/ 2
862
ROTATION AND PLANE MOTION OF RIGID BODIES [Ch.
XVII
(a)
Fig. 574.
Reproduced by permission from Mechanics, by F. W. Sears, published
by Addison-Wesley Press.
Multifiash Photograph of a Body Being Struck a Horizontal Blow.
The body
consists of two spheres connected by a rod and suspended from
a cord.
Its center of
mass is indicated by a black band on the rod.
The pictures were obtained by taking a
series of flash photographs.
In (a), the blow is at the center of percussion.
Observe that the swing is smooth with
the axis of the body remaining virtually collinear with the string.
In (b), the blow is at the center of mass.
The body’s initial motion is one of translation
(when the resultant force is through the cg, a body moves in
translation), as may be seen
by comparing the first two positions in (b). Soon after the blow,
the force of the cord
gives the body an angular motion.
In (c), the blow is above the center of percussion.
Observe the initial motion of the
body and visualize the sense of the reaction at the point of support
if the cord were a
rigid part of the body.
In (d), the blow is below the center of percussion.
Again observe the initial angular
motion of the body.
§ 254] EXAMPLE—BEARING
REACTIONS FROM UNBALANCED
INTERTIA FORCES
363
We see then that if a body is suspended at some point O, its center of
percussion P is at a distance L = k,”/7 from O. We also observe that the
center of percussion lies on the axis designated as the center of oscillation in
a compound pendulum and that both “centers” are on the line of action of
the reversed effective force mfa, Fig. 569(b). See also Fig. 574 and its
caption.
254. Example—Bearing Reactions from Unbalanced Inertia Forces. A steel
plate A, Fig. 575, is attached to a shaft B. At a certain instant the angular
What are
velocity of these members is o = 30 rad. per sec., and a = 20 rad. per sec.?,
acceleraangular
the
producing
the reactions at the bearings C and D, if the moment
shaft.
the
of
weight
the
Neglect
reactions?
tion is applied so as not to affect these
Fig. 575.
of the plate A,
The reactions at the bearings are due to the weight W
Sorution.
plate. Little
the
of
n
rotatio
the
from
arising
mfw*
and
mfa
and to the inertia forces
which
through
hole
small
ly
error will result from neglecting the effect of the relative
is
plate
the
of
the shaft passes. Neglecting this hole, the weight
_ (4)(8)(44) 490) = 400 lb.
1728
in. = 1 ft.
From Fig. 575, 7 is seen to be approximately 12
The inertia forces are
4
mio? = (22) neon = 11,180 lb.,
Tr
Mra
=>
400
[SSS
on2 0) =
(5
24
Srolbe
.
°
vertical plane, the total force on the
as shown in the two views of Fig. 575. In the
shaft at A is
miw? + W = 11,580 lb.
summing moments of the forces in the
The vertical reactions are then obtained by
vertical plane as follows:
9650 Ib.
D5
whence
=Me¢ = (11,580)(50) — D, 60 = 0,
1930 lb.
GE
e
whenc
0,
=
60
C,
—
=Mp = (11,580)(10)
In the horizontal plane, we find
=M¢ = (248.5)(50) — Dz 60 = 0,
=Mp = (248.5)(10) — Cz 60 = 0,
whence
whence
DE=,207ilb.
C, = 41.4 lb.
864
ROTATION AND PLANE MOTION OF RIGID BODIES [Ch.
XVII
The total reaction at each bearing may now be found, if desired, by methods already
explained (§ 106). Observe that the major source of the reactions on the bearings
is the centrifugal force mw.
255. Example—Rod Inclined to Axis of Rotation.
A vertical shaft OD (Fig.
576) rotates with a constant angular velocity of w rad. per sec. A uniform rod OB
is attached at the center line of this shaft and is free within limits to swing outward
from the vertical shaft. Determine the line of action of the inertia force mfw*, where
m is the mass of the rod OB.
So.tutTion.
The rod swings away from the axis to
some angle 6, the value of @ being a function of w.
Consider a differential element of the rod of mass dm,
shown in Fig. 576. The center of gravity of this
element is at a distance r from the axis of rotation.
The differential mass is equal to the differential volume
times the density p; dm = pA dz, where A is the
sectional area of the rod. We know that the total
inertia force is m7w?. We also know that the moment
of this force is equal to the sum of the moments of
all the differential inertia forces dm rw?, of which the
one shown in Fig. 576 is typical. Let us assume that
Fig. 576.
a point on the line of action of mw? is on the center
line of the arm, at some distance b from OQ. Thus,
the moment of mfw? about O is mfw2b cos 6, where 6 is unknown at this
time. The
centrifugal force of a differential mass dm is
dm rw” = (pA dz)(x sin 0) (w?),
where
r = xsin6.
The
moment
of all the differential
inertia
forces
about
O is
L
J(oA dx) (x sin 0) (w?) (x cos 0) = pAw sin 6 cos 6 | x? dx
0
pAw*(sin 6 cos 6) =
2
ll (oaly(L sin )wn(?
L cos )
=
fe
= mnor(? L cos )
Equating this expression to the moment mfwb cos @
determined above, we find
2
Miwb
cos @ = m2 L cos ),
from which b = 2/3.
That is, the point through which the resultant
inertia force
mro” acts is at a distance of 2/3 from the pivot
O.
As a check on the foregoing elaborate solutio
n, locate the center of percussion of
a long slender rod suspended at one end and
compare results.
256. Example—Variable Acceleration.
A 3-ft., solid cylindrical drum D, Fig.
577, weighs W = 3220 lb. Wound on it is a
130-{t. cable which weighs w : 10 Ib.
per ft. of length. At a certain instant, 10 ft.
of cable hangs from the drum, when its
angular velocity is 20 rpm. After another 100
ft. have run off the drum, what is its
§ 256 |EXAMPLE—VARIABLE
365
ACCELERATION
angular velocity? Neglect friction and the effect of the size of the cable on the
moment of inertia of the rotating parts.
Using a portion of the cable of length y as a free body, Fig. 577(b), we
Sotution.
get
P= te
=F, = wy —
(0)
where w is the weight per foot of cable and a, is the acceleration of the cable, which is
The moment of
equal to the tangential acceleration of the drum (a; = ra = 1.5z).
drum (mr*/2)
the
of
inertia
of
moment
the
to
equal
is
parts
rotating
inertia of the
SSS
SSS
(c)
(b)
(a)
Fige 377
plus the moment of inertia of the cable on the drum.
Assuming the cable to be at a
approximately,
mean diameter of 3 ft., we find that its moment of inertia is,
eee
g
Fig. 577(c), we find
Now using 2M, = Ive,
gth of cable.
where L is the total len
ly? ecu (L — y)r
=M, = Pr = (F
g
29
SAE
lf,
).
using « = a;/r, and solving for
Substituting in this equation the value of P from (0),
at, we get
At
aps
RS
—
W/2+ wl
Since a, = v dv/dy, we have
y=110
v2
NG dv(etSoa=
o1
ver — ve
59
orles
a
W/2+ wl
gw
W/2+wl
|
J,=10
y dy'Y,
pee = aor
2
“
the other known data, we find
Substituting v1 = Dn. = 1(3)(20)/60 = « fps and
run off the drum), and
have
cable
of
feet
nal
additio
100
v2 = 36.5 fps (when
w2
Les poe = 24.3 rad. per sec.
r
1.5
366
ROTATION AND PLANE MOTION OF RIGID BODIES [Ch. XVII
257. Plane Motion.
A body in plane motion, in the sense of this article,
is not rotating about a fixed point, nor is it in rectilinear translation. Its
movement is a combination of rotation and translation.
It was shown in
§ 226 that for any plane motion of a body
Li _ = Ma,
and
ZF, = may,
where @, and da, represent x and y components of the absolute acceleration
of the center of gravity. However, in addition to these two equations, a
moment equation is generally needed for the solution of problems in plane
motion.
Let the body B, Fig. 578, be in plane motion.
As in previous cases,
Fi
this body is symmetric about the
plane of motion of its center of
gravity.
Choose any point O on
the body as the reference point,
EaVy
whose acceleration
~
B
Fig. 578.
Plane Motion.
is a,.
Let the x
axis be parallel to the direction of the
acceleration a,. (This simplifies
our analysis somewhat.)
Now consider any particle dm. The accel-
eration of this particle is equal to
the acceleration of the point (or
particle) O vectorially plus the accel-
eration of dm relative to O. Since dm and O are two
points on a rigid body,
the acceleration of dm relative to O is composed of:
1. A normal
acceleration
a, = rw*, where
r is the distance
between
dm
and O, and where w is the angular velocity of the
body.
2. A tangential acceleration a, = ra, where
@ is the angular acceleration
of the body.
Thus, the absolute acceleration of dm is compo
sed of the components Qo;
rw, and ra. The corresponding effective forces
are dm Qo, dm rw, and dm ra.
The particle is shown in Fig. 578 with these
effective forces (three components of the resultant force on dm), the
moment of which about O is
(dm a.)y + (dm ra)r.
We have already learned that the sum of
the moments of the effective forces
on all the particles of the body is equal
to the sum of the moments of the
external forces 2M, when taken about the
same axis. Thus
2M, = a. ydm + «fr dm,
(P)
2M, = mya, + Iya,
§ 258 |REVERSED
567
EFFECTIVE FORCES AND COUPLES
where fy dm is recognized as being equal to my and where fr’ dm is recognized as the moment of inertia J, of the body about an axis O. In general
7 is the distance from the line through the reference point O and in the direction of the acceleration of O, to the center of gravity of the body, and it must
be so interpreted if equation (p) is used.
However, particular locations of the reference point O, such that either 7
Thus,
or a, in the first term of (p) is zero, simplify the solution of problems.
with 7 = 0 or a, = 0, equation (p) becomes
(47A)
DM, = Ia.
[SEE LIMITATIONS BELOW. ]
The application of this equation must be limited to the situations:
1. When the reference point is the center of gravity C, in which case
7 = 0 and the moment equation is
=M = I,,a = Ia,
(47B)
where =M is the sum of the moments of the external forces about a gravity
axis C perpendicular to the plane of motion and I is the moment of inertia
about the same axis.
2. When the absolute acceleration of the reference point is directed through
the eg. Such a reference point is 0’, Fig. 578. Again, 7 = 0, because the
x axis for O’ as the reference point is CO’. Then, equation (p) becomes
DINet Oa ae
3. When the reference point has zero absolute acceleration, a, = 0. We
get from equation (p)
YM, = I,a.
From the foregoing discussion, we see that we have three conditions obtained from a force analysis of a body in plane motion:
SF, = Mas,
DF) = ma,
2M = Ia.
reference
With a few exceptions, the center of mass C is the most convenient
point.
Let us suppose that some
258. Reversed Effective Forces and Couples.
of the body B
external force system Fi, F2, Fig. 579, produces plane motion
a free
simulate
to
desired
is
it
If
with a clockwise angular acceleration.
system:
force
body in equilibrium, there must be added to the external
center
1. An inertia force ma, acting in the opposite sense to a; through the
of gravity.
868
ROTATION AND PLANE MOTION OF RIGID BODIES [Ch.
XVII
2. An inertia force md, acting in the opposite sense to a, also through the
center of gravity.
3. A couple Ia with a sense opposite to that of a.
With these dynamic reactions added to the free body, as in Fig. 579, the sum
of the forces in any direction is zero and the sum of the moments about any
point is zero.
Fig. 579.
Simulated Equilibrium—Plane Motion.
259. Example.
A force of Q = 10 lb. acts through the center of a 40-Ib. (= W)
homogeneous sphere which rolls without slipping on a rough horizontal plane. Neglect
rolling resistance and find (a) the acceleration of the sphere’s center, (b) the frictional
force, (c) the minimum value of the coefficient of friction to prevent slipping.
Sotution.
The free body for this example is shown in Fig. 580.
of the frictional foree
F may be determined
The direction
on the basis of the following reasoning.
If the sphere is rolling toward the right, it is turning clockwise and the
angular acceleration, if there is any, will be clockwise.
Thus, the effective moment Ja is clockwise and so the moment of the external forces about the center of gravity
C must
also be clockwise. Thinking of C as the center of moments, we see
that F must act
toward the left to produce a clockwise moment.
If the wheel is rolling and the acceleration of C is a, the tangentia
l acceleration
of a point P relative to C is also @. Thus, the angular acceleration is @ =
a r.
(a) Let the center of gravity be the reference point and apply the
equations of
§ 257. The acceleration in the vertical or y direction is zero.
DL
Te,
Q — F = ma,
(Oia
=F, = May,
N—-W
0,
Ks
=| W = 40)b.
es
4
g
M4 a —
Es) =
ll
| Rg
§ 260 |EXAMPLE
869
where the centroidal moment of inertia of a sphere is (2/5)mr? (§ 161).
this value of F into the equation obtained from =F, = mdz, we get
Substituting
yen (2)(*) ND. 407,
SI\ 9
g
from which @ = 5.75 fps’.
OLE) Oem
_ (b) The frictional force may now be found from
Since its sign is positive, we assumed F in its correct direction.
(c) For the sphere to roll, the frictional force must be 2.86 lb.
minimum coefficient of friction must be
Therefore the
ie ORS
fin = 5p = Go = 0.0715.
The actual f may be larger.
We recall that the point of contact O for a rolling body is
ALTERNATE SoLuTION.
the instantaneous center where v = 0 and that the point of zero velocity is not necesThe aceeleration of point O may be expressed as
sarily the point of zero acceleration.
Ao =
Ac +> Ao/e
=
De 4 Atofe+> Anole-
For rolling, the tangential acceleration of O with respect to C’ (d10/<) 1s equal in magnitude to the absolute acceleration of C (a.) but it is directed in the opposite sense.
Therefore, these two vectors, a- and dio/-, cancel one another and the foregoing
equation gives
Ao
=
Ano/c}
that is, the absolute acceleration of O is equal to its normal acceleration relative
to C. The vector dno/e is directed O to C. This situation is the same as defined in
paragraph number 2 of p. 367; namely, that if the absolute acceleration of a point O
is directed through the cg, then
(47)
2M, = Iua.
From the parallel axis formula, we know that
IL=i+md@?
= =m? + mr = Cnr
7\(40\ in
(42,
sis = 10 = t= (°K
Taking moments about O, we have Q = 10 lb.)
that a
from which a = 5.75 fps? as before. The slight advantage of this approach is
ma;
=
=f,
equations
The
avoided.
was
simultaneous solution of two equations
the
that
Remember
solution.
the
complete
to
used
be
now
and =F, = ma, may
equation (47)
the
applying
in
moments
of
center
a
as
used
be
may
O
contact
of
point
only when the body is rolling.
A 2-ft. cylinder A, Fig. 581, weighing 322 lb., is rolled up a rough
260. Example.
makes a
plane inclined at an angle @ = 30° with the horizontal. A weightless cable
and
guide
smooth
a
over
passes
frictionless connection with the axis of the cylinder,
A
of
n
acceleratio
the
(a)
Find
lb.
193.2
=
Wz
thence downward to a weight of
3870
ROTATION AND PLANE MOTION OF RIGID BODIES [Ch. XVII
and B, (b) the tension in the cable, (c) the frictional force on A, (d) the displacement
of A from rest during 30 sec.
Sotution.
(a) Let the direction of motion be positive and let it be designated as
the y direction. Thus, the positive y direction is upward parallel to the plane and
aia vertically downward after passing over the smooth guide. While it is not
necessary to have the positive directions related
to one another like this, it is probably a safer
procedure for the beginner because then you can
choose your positive direction and stay with it
for F and a.
Since Wz, is greater than Wa, sin 6, the
cylinder will move up the plane for the manner
of connection shown (if rolling resistance is
neglected).
From the free body of A, we get
(upward positive)
=F, =P —Wasineg—F
= va
The rotational direction of motion for the cylinder is clockwise; therefore we take
the clockwise direction as positive for moments.
(We decided to use the direction
of motion as positive.) Thus, « is positive. Taking moments about the center of
gravity of A, we get
2M = F(1) = Ia = (35 )ora,
where a = G,/r = Gy.
(r)
From this equation, we find
F = 54,.
This value of F substituted in (q) gives
(s)
P = 161 + 154,.
In order to find either P or d,, we must consider the free body B. The
downward
direction is taken as positive and the acceleration of B is the same in
magnitude as
that of the center of gravity of A, @, = az. Wem ay write
Wee
q
=F, = 193.2 — P = a
(t)
Oy
P = 198.2 — 64,.
A simultaneous solution of equations (s) and (t) yields a,
= 1.53 fps?.
(b) The value of P is now obtained from either (s)
or (t).
P = 193.2 — (6)(1.53) = 184 lb.
(c) The value of the frictional force is found from
(r) to be
F = 54, = (5)(1.53) = 7.65 lb.
From (t), we get
871
§ 261 | EXAMPLE
(d) Using s = at?/2, we find
1.53)
(30)?
the displacement of A during 30 sec.
In accordance with the discussion in the example of § 259,
Avrernats Sotution.
H as a center of moments, since this cylinder is rolling.
point
contact
the
we may use
+ md?)
Applying the moment equation =M, = Ia, we get (Uo = I
=M, = Pr — Wa(r sin @)
op
(= ap mv?)a=
a\
Co
Fi
\&,
P — 161 = 154,,
of obtaining equation
which is the same as the preceding equation (s). This method
of some advantage.
e
therefor
is
and
used,
method
the
(s) is somewhat shorter than
In a four-link mechanism, Fig. 582, the constant angular velocity
261. Example.
The link AB weighs 50 lb. and is symmetric
of the crank AF iswar = 20rad.persec.
shown, the horizontal reabout axes through its center of gravity. In the position
and A, for the position
A.,
B,,
ne
action at Bis B, = 200 1b. toward the left. Determi
582.
and the dimensions given in Fig.
the
to apply
Sotution.
equations
In order
oF, = Ma:,
=F, = M4a,,
=M = Ia,
determine
the various
first
must
we
accelerations involved.
The linear
speed of point A is
va = rw = (0.5)(20) = 10 fps.
The instantaneous center of the link AB
is found at O (§ 198), from which the
instantaneous radii of points A and B
as points in link AB are measured as
ra (in AB) = 3.59 ft. and rg (in AB)
— 5.69 ft. The velocities of A and B,
as points in AB, are proportional to
these radii. Hence
)
UB
=
YR
—
vs
=
5.69
3 59
10 ———
84
15.84
Ips .
f
Moreover, we have
ee
v
TA
10
=
2,79 ad, per sec.
Fig. 582.
zi
B59)
ed from vz, where B is considered as a
The angular velocity of the link DB is obtain
point in DB.
UB
15.84
l
7.92 rad. per sec.
“a DED
872
ROTATION AND PLANE MOTION OF RIGID BODIES [Ch. XVII
In the equations*
ap = Aatb aB/a,
OnB + Qip = Ana
Gia 1D AnBs aD GB/A,
we know that
OyB = Tone)
ip = Renae
Ata
GngiA
=
=
=
(2)07.92)2 I 125.3 fps’,
(Opieoy
200 fps?,
J
0,
= Y(was)? =
(AB)(waz)?
=
(3.3)(2.79)? =
25.7 ips(
and we know the line of action of the accelerations a:g and a:g;4.
shown a solution of this equation:
aac
In Fig. 582(b) is
125.3 4 diz = 200 4 25.7 4 aizya,
from which a;g/4 = 107 fps? in the sense
shown.
From this value, the angular
acceleration of AB is
QAB
=
QtBiA
_ 107
ABT
38
= 32.4 rad. per sec?
Now the tangential acceleration of the
center of gravity C relative to A is
obtained from the equation
Qicja = Ta = (AC)aap
= (1.65)(32.4) = 53.5 fps®.
Moreover, we have
Qnc/A
OB /A
O
OnB/A
-
(b)
(a)
Fig. 582.
Repeated.
=
To
=
(AC) (waz)?
= (1.65)(2.79)? = 12.84 fps?.
Thus the acceleration of C is defined by
the following components,
Ao = Gat? Anosa > Mesa,
where a4 = dna is directed vertically downward.
If the x and y axes are horizontal
and vertical, we need the horizontal and vertical components of de. The horizontal
component @, is the sum of the two quantities
ducya Sin 21° = (53.5)(0.358) = 19.15 fps? (left),
Anes COS 21° = (12.84)(0.934) = 12 fps? (left),
from which
Gd: =
19.15 + 12 = 31.15 fps? (left).
*Relative accelerations have many subscripts, but it is good
engineering shorthand,
38738
§ 261 | EXAMPLE
The vertical component 4G, is the algebraic sum of
dic) 4C08 21° = (53.5)(0.934) = 50 fps? (up),
Anca sin 21° = (12.84)(0.358) = 4.6 fps? (down),
aa = 200 fps® (down),
or
a, = 200 + 4.6 — 50 = 154.6 fps? (down).
information and the
A free body is now made of link AB, showing the essential
forces (Fig. 583). We obtain
=F, =
eo ee
A, —B,
g
=A, =) 200= (3) 31.15),
whence
Az = 151.6 lb.
is directed toward the left and
The acceleration G, is given a negative sion because it
ring the link AB as a slender
Conside
.
positive
the sense toward the right is taken as
164,
§
rod, we have by
i
zu
mL? _ (50)(8.8)? = 1.41 slug-ft.?
arto, e'(S2.2) (12)
=
Tap,
= (1.41)(82.4)
(200) (0.591) — (B,)(1.54) + (A,)(1.54) + (A,)(0.591)
This equation
taken as positive.
where the sense of « (counterclockwise) has been
reduces to
vertically,
Summing
forces
choosing
the downward
positive, we have
Ay a Wes
By =
=“
and
sense as
Fe
a
—
B
:
y
W=50 lb.
1.54 /
Ae
dy
50
A, + 50+ B, = (Fe),
(v)
= 200 hb.
BB,
SB
|
st
Be
~)
ay
ye
1.54
t
= =154.6 fps
G,=154.6
Fig. 583.
fps?
A,+ B, = 190.
By addition, we find, from (.) and (v),
p=
140.b1b:
Then
‘A, = 190. —B,
= 190 — 14716 = 424 Ib:
AB may be placed in equiSonution By Reversep Errective Forces. The link
of gravity and opposite
center
librium by imposing the forces ma, and ma, through the
sense opposite to that
a
with
Za
couple
the
to the corresponding accelerations, and
874
ROTATION AND PLANE MOTION OF RIGID BODIES [Ch. XVII
of a (§ 258). The free body for this arrangement of forces is shown in Fig. 584(a).
The solution of the problem with this free body is virtually the same as the previous
solution, and we shall not repeat it.
Another free body using reversed effective forces may be obtained by using the
components of the acceleration of C as given by the relative acceleration equation
ae
=
aah
AnC/A +4 tiC/A-
These components of the acceleration a¢ point as shown in Fig. 584(b).
Correspond-
ing to each component is a reversed effective force:
50
MAanc/A
=
(3
)azss =
19.95
lb.,
which acts away from the center A.
50 200 = 311)b.,
maa= (32)
(ea
which acts upward through C.
Maca
jmay
50
=
(sox
s
=
2
Syl (oy
B B,.= B= 200
\e48 Ee
SZ
(a)
Fig. 584.
The line of action of this inertia force is at a distance of k4?/F from
A, since the corresponding tangential acceleration a:c;4 is relative to A (§ 250).
Thus, from § 164,
we find
mL? _ (50)(3.3)
Ia
=
3
=
(32.2)(3)
im
5.64
ha ee fie eeMES)
ka
3.64.
EE eee
oe
).64 slug-ft2
slug-ft
= 5.64
Aee Ee
i
1.65
ah
which defines the line of action of mra = Maic;a, a8
Shown in Fig. 584(b). With
these inertia forces added, the free body of F ig. 584(b)
is now in equilibrium, so that
moments may be taken about any point and forces
may be summed in any direction.
The reader should complete the solution with the free
body of Fig. 584(b) and check
the results already obtained.
It is interesting to note the magnitude of the
md, inertia force (240 lb.)
as compared to the static force W (50 Ib.) and
speculate on the importance
of dynamic loads in the design of the bearin
gs and other parts of moving
machines, even at moderate speeds. The
design of parts where there are
875
§ 263 |EXAMPLE
unbalanced inertia forces ought always to be based on a consideration of
these inertia forces, rather than on consideration of static forces only.
262. Combined Sliding and Rolling.
If a body (wheel, cylinder, sphere)
is partly rolling and partly sliding, the conditions
and
oF, = ma,
LF, = ma,
are applicable as usual, since they hold true for any plane motion.
equation
=M = Ia,
The
also applies. If any sliding occurs, however, the acceleration of the center
of gravity @ is not equal to ra, where r is the radius of the body; that is, there
is no ready means of relating the absolute linear acceleration of the center
of gravity and the angular acceleration of the body.
If some sliding is suspected, the first step is to make very sure of the
If sliding occurs, the frictional force is the maximum value
conditions.
f,N, where f, is the coefficient of kinetic friction. If pure rolling occurs,
the limiting static frictional force must be large enough to prevent slipping.
maximum
If the frictional force required for pure rolling is greater than the
The maximum attainable
attainable value, some sliding necessarily occurs.
t of static friction.
coefficien
the
is
f,
where
value of F is f,N, theoretically,
force can be defrictional
Where there is sliding, the value of the kinetic
a.
termined and then the equation >M = Ia may be used to determine
the horizontal,
263. Example. A 3-ft. solid cylinder.on a plane inclined at 45° with
through a
shown,
direction
the
in
it
to
applied
Ib.
50
=
Q
force
Fig. 585, has a constant
= 100 lb.
W
is
cord that wraps around the cylinder. The weight of the cylinder
= 0.16.
f,
is
friction
kinetic
of
The coefficient of static friction is f, = 0.2, and that
ion of
accelerat
linear
the
(b)
cylinder,
the
of
Determine (a) the angular acceleration
cylinder
the
after
contact
of
point
the
at
speed
rubbing
the
(c)
its center of gravity,
§ 209.)
has moved from rest for 10 sec. (Read again the example of
585, we
Assuming the frictional force to act in the sense shown in Fig.
So.ution.
positive)
is
plane
the
to
have (upward direction parallel
=F, = Q cos 30° + F — W sin 45° = maz
= (50)(0.866) + F — (100)(0.707) = ae aise
F — 27.4 = 3.105 Gz,
some slid- %:
an expression that holds whether pure rolling or
(w)
The moment of inertia of the cylinder about
ing occurs.
its geometric axis is
eens
(100) (1.5)?
qi? eaen22)@)
= 3.459 slug-ft.’
find
Choosing the clockwise sense as positive, we
eM = Q(1.5) — F(1.5) = Ta
(x)
=
(50)(1.5) — 1.5 F = 3495a.
Fig. 585.
876
ROTATION AND PLANE MOTION OF RIGID BODIES [Ch. XVII
Now if pure rolling occurs, dz = ra = 1.5 a.
This value of d, in (w) gives
F — 27.4 = 4.66a.
A simultaneous solution of this equation and equation (x) for F gives
F = + 42.5 lb.
[needed for rolling]
which is the force needed for pure rolling, the positive sign indicating that / is used in
its correct sense. Now summing forces in the y direction, we have
=F, = Q sin
30° — W cos 45° + N = 0,
N = 70.7 — 25 = 45.7 lb.
The maximum available frictional force is
F =f.N = (0.2)(45.8) = 9.16 lb.
Since 9.16 is less than 42.5, some slipping occurs.
force is
f= {N=
(O01 6)045
Sse
[available static].
Therefore, the actual frictional
la.
[actual].
Now that the actual frictional force is known, we may use equations (w) and (x) to
determine @, and a.
(a) From equation (x), we have
75 — (1.5)(7.33) = 3.495,
i
or a = 18.3 rad. per sec.? in a clockwise sense.
(b) From equation (w), we get
7.33 — 27.4 = 3.1054.,
3
Or Gz = —6.47 fps?, where the negative sign indicates
that this acceleration is down the plane, rather than up
,Repeated.
the plane.
45
Fig. 585.
(c) The velocity of the point C after
10 sec. may
be found from the equation a = dv/dt = 6.47, or (downward along the plane now
taken as positive)
10
Ve =}, adt = (6.47)(10) = 64.7 fps.
0
The velocity of the point G, in contact with the ground, relative to the center C is
Yaic = Tw, Where w is found from the equation a = dw/dt = 18.3.
Thus
10
w =i a dt = 183 rad. per sec.
0
Varo = Tw = (1.5)(183) = 274.5 fps.
The rubbing speed at G is the absolute velocity of G (its velocity relative to
the
ground).
Since we know that
Veg
=VceH
Vaile,
and since vg and vg,c are both in the same sense (parallel to the plane,
pointing down),
the rubbing speed is
Vq@ = 64.7 + 274.5 = 339.2 fps
after 10 sec., the body having started from rest.
PROBLEMS:
ROTATION
ABOUT CENTER
877
OF GRAVITY
To summarize the procedure if sliding is suspected:
1. Find the value of the frictional force needed for pure rolling.
2. Find the maximum possible value of the frictional force, using the static coefficient of friction fs.
If the needed
3. If the needed F is less than the maximum F, pure rolling occurs.
Then
occurs.
F is greater than the maximum /’, some sliding
4. For sliding and rolling, the actual F is f,N.
Ta,
5. Compute this actual F and proceed with the use of 2F = ma and =M =
as necessary.
It will be worth while to review this and the two previous
264. Closure.
chapters and notice that the three equations
SF, = maz,
Ne
=M = Ia,
ine
of any
properly interpreted, will give an infallible approach to the solution
(no
type problem discussed in these chapters. If the body is a particle
two
other
the
and
ce,
finite dimensions), J is zero, >M = Ia has no significan
If the body has finite dimensions and is in translation,
equations apply.
body is rotating
rectilinear or curvilinear, a = 0 and again >M = 0. If the
= Ia,* where,
2M
and
0,
=
2h
about. its ce, a, =a, — 0; so 2F,—0,
Thus, if you know
if wis zero, the problem reduces to one of equilibrium.
a complete free
how to apply these equations and if you know how to make
a problem.
body, you should never be at a loss for an immediate attack on
points
various
the
of
location
the
As a refinement, you could master
simpler
find
thereby
and
I,x,
suitable for the center of moments in 2M, =
Also, because of its advantage in some probsolutions to some problems.
rium by adding
lems, you should know how to simulate a condition of equilib
the inertia forces and couples to the free body.
ng equations is treThe variety of possible applications of the foregoi
of these chapters are
mendous, so that the examples and special applications
skill in your use of
suggestive only. The way to achieve confidence and
ms—the more, the better.
these equations is by the solution of many proble
Problems
ROTATION
ABOUT
1271. Solve the example of § 247, considering the arm A and the gear B as
weightless.
What is the percentage of error
caused by-making this assumption?
1272. A 720-lb. solid, cylindrical cast\ron disk, whose radius is 18 in., is keyed
to a shaft.
A constant
force of 20 lb. is
applied tangent to the disk. Neglecting the
friction in the bearings and the weight of
the shaft, determine the angular accelera-
CENTER
OF
GRAVITY
How long would it take for the speed
tion.
of the disk to change from 60 rpm to 180
rpm?
Ans. 1.193 rad. per sec., 10.53 sec.
1273. Figure 586 represents a 6-ft. fly-
wheel Band a 3-ft. crank disk A, each
keyed to a horizontal shaft. The weight
of all rotating parts is 6440
lb. and
the
radius of gyration of the whole is 2.5 ft.
What constant force Q, acting on the disk
at an angle of @ = 15° with a tangent at
PROBLEMS: ROTATION ABOUT CENTER OF GRAVITY [Ch.
378
the point of application and in the plane of
motion, will increase the speed from 30 rpm
to 210 rpm during 600 revolutions of the
shaft?
Ans. 54.2 lb.
1274. Figure 586 represents a 6-ft. flywheel B and a 3-ft. disk A, each keyed to
a horizontal shaft.
The weight of all rotating parts is 6440 lb. and the radius of
gyration of the whole is 2.5 ft. If B is
rotating clockwise at 210 rpm and @ = 15°,
find the braking force Q that will stop
XVII
fluid friction of the steam and air about the
turbine blades, determine the elapsed time
and the total
coming to rest.
number of revolutions in
Ans. 1.33 hr., 71,800 rev.
rotation in 60 sec.
Fig. 587.
(
Fig. 586.
y
Problems 1273-1275.
tangent to a crank disk A, as shown.
The
radius of gyration of all the rotating parts,
including the flywheel B, is 2.5 ft. If the
speed increases from 60 rpm to 240 rpm
during a rotation of 300 revolutions, what
is the weight of the rotating parts?
7
Ans. 14,290 lb.
1276. A 1140-rpm electric motor may be
coupled by a clutch to its load. The rotating elements on the motor side of the clutch
have a moment of inertia of 3.5 slug-ft.?
and a radius of gyration of 6.5 in. The
rotating elements on the load side of the
clutch have a moment of inertia of 20
slug-ft.? and a radius
of gyration of 15 in.
Find the constant torque (produced by an
to the motor)
required
to
bring the motor to speed in 4 sec. (a) if
the clutch is disengaged and (b) if the
clutch is engaged.
Ans. (a) 104.4 ft-lb.; (b) 701 ft-lb.
1277. The same as 1276 except that the
accelerating time is 7 sec,
1278. The rotor in a steam turbine weighs
20,000 Ib. and has a radius of gyration
of
14 in.
When
1279. A force Q = 60 lb. is applied as
f = 1/3)
(12%. Th Fig. 586, a constant force Q =
~ acts at an angle of 6 = 15° with a
current
Problems 1279-1282.
*
n in Fig. 587 to engage the brake shoe
and stop the rotation of the 2576-Ib., 6-ft.
diameter, steel disk from:a counterclockwise
speed of 360 rpm.
For the brake shoe,
tl i
electric
ee
the steam is shut off, the
rotor coasts to a stop from a speed of 1800
rpm.
The average coefficient of friction
in the bearings is 0.005 and the diamet
er
of the bearings is 8 in, Neglecting the
If the brake arm pivot pin is in
position A, find (a) the time required to
stop the disk and (b) the horizontal and
vertical components of the reaction at A.
Ans. (a) 45.2 sec. (b) +100 lb., —300 Ib.
1280. A force Q = 60 lb. is applied as
shown in Fig. 587 to engage the brake
shoe and stop rotation of the 2576-lb., 6-ft.
diameter, steel disk from a counterclockwise
speed of 360 rpm.
For the brake shoe,
f = 1/3. The brake arm pivot pin is in
position A’ and e =9 in. Find (a) the
time required to stop the disk, and (b) the
horizontal and vertical components of the
reaction at A.
1281. The same as 1280 except that the
disk rotates clockwise.
1282. A 2576-lb., 6-ft.
diameter, steel
disk, Fig. 587, is rotating counterclockwise
at'360 rpm.
For the brake shoe, f = 1/3.
If the disk is to be stopped with constant
acceleration in 30 sec., develop an equation
for Q as a function of eccentric distance é.
For what value of e is the brake theoretically self-acting (the brake applies itself
with Q = 0), neglecting the weight of the
brake?
The pivot point is A’.
Anismeu—wA
v5 its
1283. In Fig. 588, W = 32.2 lb. and the
644-lb. steel disk B has a diameter D = 24
in. Neglect friction and the weight of the
cable and the shaft. For W starting from
rest, find (a) the tension in the cable and
(b) the displacement of W in 6 sec.
PROBLEMS:
ROTATION
ABOUT
CENTER
379
OF GRAVITY
1284. The same as 1283 except that the
disk has an initial counterclockwise speed
of 5 rpm.
Ans. (a) 29.3 lb.; (b) 49.6 ft.
1285. In Fig. 588, W = 32.2 lb. and the
30-in. drum B weighs We = 96.6 lb. If
the tension in the cord is 12.2 lb., find the
Neglect
radius of gyration of the drum.
the weight of the cord and friction.
Fig. 589.
Problems 1289, 1290.
rotating parts A and B? (c) what is the
tension in the cord attached to W?
Ans. (a) 98.4 Ib., (b) 196.4 lb.; (c) 3030 Ib.
1290. The same as 1289 except that
W = 100 lb.
Fig. 588.
Problems 1283-1288.
1286. In Fig. 588, W = 32.2 lb. and the
36-in. drum B weighs Wz = 96.6 lb. Neg-
lect the friction and the weight of the cord.
If the radius of gyration of the drum is 1 ft.,
Ans. 18.4 lb.
find the tension in the cord.
1287. In Fig. 588, W = 32.2lb.
The 24-
in. drum B is a homogeneous steel cylinder
weighing 490 Ib. per cu. ft. and having a
length L. If W is released from rest and
found to descend 20 ft. in 3 sec., find the
Neglect friction and
length of the drum.
the weight of the shaft.
1288. In Fig. 588, a weight of W = 1610
Ib. is attached to a cable of negligible
weight. The cable wraps about a D =
of
30-in. drum, whose weight and radius
gyration, including the shaft, are 2415 lb.
The coefficient
and 12 in., respectively.
of friction at the d = 6-in. bearings is f =
The power applied to raise the load
0.02.
W is shut off when W is moving upward at
10 fps. Assuming that there is no friction
except that in the bearings, determine (a)
to
the distance moved by W before coming
Fig. 590.
Problems 1291, 1292.
1291. In Fig. 590, the weight of the
rotating parts B is 6440 |b., their radius of
gyration isk = 3 ft., and the axis is frictionless. Also Wa = 12,880 lb.,fa = 1/4, and
@ = 30°. If the body A moves 100 ft.
with a constant acceleration wp the plane
from rest during 40 sec., (a) what is the
weight of C?
the cord AB?
(b) What is the tension
(c) In the cord BC?
in
1292. The same as 1291 except that the
acceleration of A is down the plane.
Ans. (a) 1760 lb.; (b) 3590 Ib.; (ce) 1767 Ib.
rest, and (b) the tension in the cable.
Ans. (a) 3.02 ft.; (b) 780 lb.
1289. In Fig. 589, a weightless cord supports a weight W = 200 Ib., as shown, and
A
wraps about a sheave B. The members
a
on
unit
a
as
ise
clockw
rotate
and B
of
frictionless axis at O and have a radius
is
gyration of & =2 ft. The sheave C
of
assumed to be weightless. If the speed
A and B increases from 30 rpm to 240
n
rpm during 2 min., (a) what is the tensio
in the cord BC and
(b) the weight of the
Fig. 591.
Problem 1293.
1293. A 128.8-lb. body A, Fig. 591, is on
At a
a @ = 15° incline where f =0.15.
certain instant, the solid cast-iron cylinder
B is rotating at 40 rpm and the block A is
being moved up the incline by virtue of the
cable connection shown. The cable CB
PROBLEMS: ROTATION ABOUT CENTER OF GRAVITY [Ch. XVII
580
wraps around the 2-ft. cylinder B. After
A moves 20 ft. up the incline, it comes to
rest.
What is the weight of the cylinder B?
Neglect the axial friction for B and C and
the mass of the pulley,C.
Brake Shoe
at the fixed peg, fe = 1/7. A force Q =
130 lb. What is the weight of A?
Ans. 796 Ib.
1297. In Fig. 592, the body A weighs
200 Ib. and has a speed at a certain instant
of v4 = 10 fps. The members C are on a
frictionless axis, weigh 500 Ib., and have a
radius of gyration of k = 10 in. At the
brake shoe, f = 0.4. At the fixed peg,
fe = 0.35. A force Q = 100 Ib. Determine (a) the acceleration of A, (b) the
tension in the cord AB, and (c) the tension
in the cord BC.
1298. The
Wa
Fig. 592.
same
as
1297
except
that
=A4Akips.
Problems 1294-1298.
1294. In Fig. 592, the members (C, rotating on an axis for which the friction may
be neglected, weigh 500 Ib. and have a
radius of gyration of k = 10 in. At the
brake shoe, f. = 0.3. A weightless cord
wraps about the 12-in. sheave, passes over
a fixed peg B, and attaches to a weight
Wa = 300 lb.
At B, fg =1/x.
If A is
moving downward with a speed of v4 =
30 fps, what force Q will bring it to rest
with a constant deceleration of 5 fps??
Ans. 93.7 lb.
1295. The same as 1294 except that
va = 30 fps upward.
1296. In Fig. 592, the body A comes to
rest with uniform acceleration in 10 sec.
from an original downward speed of 20 fps.
The members C are on a frictionless axis,
weigh 500 lb., and have a radius of gyration
of k = 10 in.
At the brake shoe, f, = 0.3;
ROTATION
NOT
1299.
Fig. 593.
Problem 1299.
The mass
of all members
in Fig.
593 except that of the spheres C is to be
neglected.
The spheres each weigh 64.4
Ib., and the whole rotates about the gravity
axis AB.
What torque on the axis AB will
increase the speed from 60 to 128 rpm
during 20 sec.?
Ans. 6.41 ft-lb.
ABOUT
CENTER
OF
GRAVITY
1300. A 6-ft. slender rod, Fig. 594, weighSing 70-Tb., falls from rest from the vertical
position and rotates about a smooth axis O
Fig. 594.
Problems 1300-1304.
atoneend.
Determine the angular velocity
of the rod and the horizontal and vertical
reactions at O when 6 = 90°.
Ans. 4.01 rad. per sec., 105 Ib., 17.5 Ib.
1301. The same as 1800 except that
Goals
1302. The same as 1300 except that
a= 1202
1303. The same as 1300 except that in
the initial position the rod has an initial
angular velocity of w; = 10 rad. per sec.
clockwise.
Ans. 10.77 rad. per sec., 758 Ib., 17.5 lb.
1304. The rod described in 1300 oscillates
as a compound pendulum with a displace-
PROBLEMS:
NOT
ROTATION
ABOUT
CENTER
581
OF GRAVITY
ment of 6° on each side of the vertical.
Determine its period, the center of percussion, the center of oscillation, and the length
of the-equivalent simple pendulum.
iat
steel eccentric
shown
in Fig.
, with D = 14 in., rotates about a hor-
izontal axis O for which friction may be
neglected for the purposes of this problem.
At a particular instant, » = 50 rpm and
an applied torque produces an acceleration
of a = 5 rad. per sec.?, both counterclockwise. Steel weighs 490 Ib. per cu. ft. For
9 = 210°, determine (a) the lines of action
of the reversed effective forces and (b) the
horizontal and vertical components of the
Ans. (b) 12.8 lb., 166.5 lb.
reaction at O.
1306. The
@ = 45°.
same
as
1305
except
that
Fig. 596.
Problem 1310.
1310. An eccentric A, Fig. 596, is 12 in.
in diameter, weighs 48.3 lb., and is supported on a smooth axis at O. If this disk
starts from rest at a position @, = 150°,
what is the speed of point B after the disk
has rotated under the action of gravity to a
position 6. = 45°? What are the horizontal
and vertical reactions at O at this instant?
Solve with and without REF.
A
=
Fig. 597.
Fig. 595.
i‘)
Problems 1305-1309.
1307. The eccentric described in 13065 is
mounted midway between bearings which
What are the horare 14 in. on centers.
izontal and vertical bearing reactions when
C' and O are in the same horizontal plane,
other conditions remaining as given in 1305?
does
The torque producing the acceleration
Problem 1311.
1311. A slender rod, 3 ft. long and weighing 30 Ib., is supported in a horizontal
position on a smooth fixed axis at A and a
support at B, Fig. 597. The support B is
At the instant after
suddenly removed.
its removal, what is the reaction on axis A?
ji not affect the bearing reactions.
Ps
Ans. (b) 18.7 lb., 62.1 lb.
in
1308. A 14-in. steel eccentric shown
for
Fig. 595 rotates about a horizontal axis
for the
which friction may be neglected
rcounte
falls
it
If
m.
proble
this
purposes of
the
clockwise from rest at @ = 0 under
the noraction of gravity alone, determine
the reacof
nents
compo
tial
tangen
and
mal
tion at O when @ = 120°.
Ans. 220 lb., 68.6 lb.
disk, D = 2m
tric
eccen
1309. An
shown
in Fig. 595,
weighs
48.3
Ib. and
on a
oscillates as a compound pendulum
of
mass
the
cting
Negle
shaft.
horizontal
ation, the
the shaft, find the center of oscill
d, if the
center of percussion, and the perio
5°. How
maximum displacement is # =
?
ulum
pend
e
simpl
alent
equiv
an
long is
Ans. 8.5 in., 9.33 sec.
Fig. 598.
Problems 1312, 1313.
\
1312. A slender rod, 4 ft. long and weighing 48.3 lb., is attached to a smooth fixed
axis at one end, Fig. 598. Starting from
rest in a horizontal position, it moves under
the action of gravity about the fixed axis.
When 06 = 60°, determine (a) its angular
velocity, (b) its angular acceleration, and
(c) the horizontal and vertical components
of the reaction at the axis O.
Ans. (a) 4.58 rad. per sec.; (b) 6.04 rad. per
sec.2; (c) —47.0 lb., 93.5 Ib.
1313. The same as 1312 except that
ij) = OO.
PROBLEMS: ROTATION NOT ABOUT CENTER OF GRAviTY [Ch.
382
XVII
1314. A 3-ft., homogeneous ring of weight
W is suspended from a fixed point O, Fig.
599, so that it may
oscillate either in the
plane of the ring or perpendicular
to the
ring. If the amplitude does not exceed 5°,
find the period of oscillation and the length
of the equivalent simple pendulum for (a)
oscillation in the plane of the ring and (b)
oscillation perpendicular to the plane of the
ring.
Ans. (a) 1.92 sec., 3 ft.; (b) 1.66 sec., 2.25 ft.
Fig. 600.
Problem 1317.
1318. A 6-ft. drum with a 150-ft. chain
wound around it is mounted on a horizontal
shaft supported by bearings at each end.
Initially, 4 ft. of chain hang vertically from
the drum, the total weight of the rotating
Fig. 599.
Problem 1314.
1315. A 600-lb. steel disk, whose diameter is 6 ft., is in a horizontal plane and
rotates about a smooth vertical axis, which
is 1 ft. from the center of gravity of the
disk.
During 90 sec., the angular velocity
increases
constant
from
rate.
cu.
(a) Determine
ft.
parts, exclusive of the chain, is 966 lb.,
with a radius of gyration of 2 ft. Also, the
initial speed of the drum is 10 rpm and the
speed after an additional 100 ft. of chain
have run off the drum is 120 rpm.
Assuming that the chain has a uniform weight
per unit length, determine the weight of
the chain in pounds per foot.
Neglect
friction but allow for the variation in the
moment of inertia of the rotating parts.
20 rpm to 180 rpm at a
Steel weighs 490 lb. per
the
magnitude
of
the force producing this result if it passes
through the center of percussion and remains perpendicular to the line joining the
center of rotation and the eg. (b) When
n = 180 rpm, what are the components of
the reaction at the axis in the horizontal
plane?
Ans. (a) 3.47 lb.; (b) 6620 Ib. (radial), zero
(tangential).
1316. In 1315 if the vertical axis is supported by a radial bearing A which is 1 ft.
above the disk center and a combination
radial and thrust: bearing B which is 2 ft.
below the disk center, find the rectangular
components of the reactions at A and B
when » = 180 rpm.
1317. A compound pendulum is composed of a slender steel rod, 12 in. long,
1 in. wide, and» 1/2 in. thick, Fig. 600,
attached to a 4-in. steel sphere.
The maximum displacement during its oscillation
is 5°. Determine (a) the period of oscillation, (b) the center of oscillation, (¢) the
center of percussion,
at O at the
displacement.
and
position
(d) the reaction
of the
maximum
Fig. 601.
Problems 1319, 1320.
1319. A 32.2 lb. sector and a 16.1 lb.
slender rod are attached to a horizontal
rotating shaft as indicated in Fig. 601.
The
sector
has a radius
of 1 ft., and
its
geometric axis is the axis of rotation.
At
a given instant, the rod is vertical as shown,
6 = 30°, w» = 20 rad. per sec., and a =5
rad. per sec.*. The shaft is driven by a
couple applied through a clutch. N eglecting the mass of the shaft and the effects
of
friction, deflections, and vibration, calcula
te
(a) the driving couple, (b) the horizontal
and vertical reactions at bearings A and B.
PROBLEMS:
ROTATION
NOT
ABOUT
CENTER
Ans. (a) 3.96 ft-lb.; (b) 107.2 Ib. and — 108.5
Ib. for A, 26.2 lb. and 30.9 lb. for B.
1320. The same as 1319 except that
6 = 60°.
1321. The connecting rod of a locomotive
weighs 800 lb.
883
OF GRAVITY
ing horizontal and vertical reactions at the
pin C? Assume that the construction 1s
such as to yield a coplanar system forces.
Ans. 277 rpm. 3176 lb.. 1080 Lb.
The distance from the cen-
ter of the crank pin to the cg of the rod is
3.5 ft. When this connecting rod is suspended from the crank pin, it is found that
it makes 25 complete oscillations in 90 sec.
Assuming that the effect of friction is negligible, determine the moment of inertia of
the rod about the axis of the crank pin
and about a parallel gravity axis.
Ans. I = 616 slug-ft.?
Fig. 603.
Problem 1324.
1325. The homogeneous slender rod of
weight W shown in Fig. 604 is suspended by
a weightless cord and made to oscillate as
Find (a) the center
a compound pendulum.
(6 ~ 0)
of percussion and (b) the period.
(c) If ABC is to make a complete revolu-
tion about A with ABC remaining in a
straight line, what must be its angular
velocity in the position shown?
Fig. 602.
Problems 1322, 1323.
602,
1322. A 30-in. uniform rod OB, Fig.
al
vertic
a
to
cted
conne
is
It
lb.
75
weighs
is held
shaft at the frictionless pin O and
tless
in position, where @ = 30°, by a weigh
(a)
OB.
rod_AB which is perpendicular to
on AB
force
the
is
ty
veloci
ar
angul
At what
(b) What are the horizontal
equal to zero?
reaction at
and vertical components of the
ant at
O when the angular velocity is const
600 rpm?
Ans.
(a) 4.73 rad. per sec.;
1323. The
same
as
1322
(b) 2170 Ib.,
2140 lb.
that
except
6 = fo.
hed to a
1324. A 30-lb. sphere is attac
The link AB
link at one end B, Fig. 603.
ed as shown.
weighs 50 lb. with its cg locat
the vertical
about
s
occur
ion
rotat
Steady
, what
shown
ion
posit
the
axis CD. For
2000 lb. on
speed (rpm) causes a load of
the cable AD?
Veen
(
What are the correspondPLANE
Fig. 604.
Problems 1325, 1326.
1326. The 16.1-lb. slender rod shown in
Fig. 604 is suspended at rest by a weight
ration
accele
r
angula
the
Find
less cord.
a of the rod if it is acted on by a horizontal
2-lb. force applied so that the rod and cord
be
remain in a straight line. What would
the
reaction
at A at this instant?
See
Fig. 574 (a).
MOTION
e is
1327. A 64.4-lb., 2-ft. diameter spher an
with
ce
surfa
ontal
yolling on a horiz
s to rest
initial speed of 30 fps. If it come
in a distance of 200 ft., find the coefficient
of rolling resistance, other factors being
Ans. 1.17 in.
(See § Ute)
negligible.
584
\
PROBLEMS: PLANE MOTION
[Ch. XVII
1328. A 3-ft. cylinder A, Fig. 605, weighs
1332. A sphere, a cylinder, and a thin-
83 lb.and has wrapped around its mid-
wall cylindrical shell have identical diameters and weights, and they are made of
homogeneous materials.
If all three are
simultaneously released on an inclined plane
section a weightless cord, the other end of
which is fixed. The cylinder is released
from rest in the position shown.
Determine
(a) its angular acceleration, (b) the tension
in the cord, and (c) the speed of the center
of gravity of A after it has moved 30 ft.
Ans. (a) 14.3 rad. per sec.?; (b) 161 lb.;
(c) 35.9 fps.
ay!
Fig. 605.
Problems 1328, 1329.
1329. A cylinder similar to that of Fig.
605 has radius r, initial velocity 5,, and it
unwraps from a weightless, frictionless cord
under the force of gravity.
Develop in
terms of m and r the general expressions for
(a) the angular velocity w at any time, (b)
the linear acceleration d, and the tension
in the cord 7. Solve with and without
the REF.
and permitted
to roll down
50 ft.
Problems 1330, 1331.
Fig. 607.
~ 1330. An 805-lb. homogeneous steel cylinder A,‘18 in. in diameter, rolls down a
rough plane inclined at @ = 30°, Fig. 606.
(a) If the initial
speed
of the
center
of
gravity is 10 fps, what is its speed after
8 sec.?
(b) What is the frictional force
between the plane and the cylinder?
(ce)
What minimum coefficient of friction is
necessary?
Ans. (a) 96 fps; (b) 134. lb.; (ce) 0.193.
1331. The same as 1330 except that the
body A is an 18-in. sphere of the same
weight.
(no
(a) What is the tension in the cable?
(b) If slipping is impending,
coefficient of friction?
Fig. 606.
the plane
sliding), in what order would they reach
the bottom of the plane? Show proof of
your answer.
1333. A sphere and a cylinder of identical
diameters roll without slippage down an
inclined plane with the same acceleration.
If the sphere weighs 32.2 lb., what must
the cylinder weigh?
1334. What is the maximum inclination
of a plane down which a 966-lb. sphere will
roll without slipping when f = 0.2? The
diameter of the sphere is 4 ft. Does this
inclination depend upon the size or upon
the weight of the sphere?
Aqsa Ome
1335. The same as 1334 except that the
rolling member is a solid cylinder.
1336. A 644-lb. sphere whose radius is
2 ft. rolls down a 10° incline.
If the initial
speed of its eg is 5 fps, how far does it roll
to reach a speed of 20 fps? What is the
frictional force iff = 0.5?
Ans. 46.9 ft., 32 lb.
1337. A 644-lb. sphere whose radius is
2 ft. is rolled up a 15° incline by a pull of a
horizontal cable attached to a frictionless
gravity axis. The speed of the eg increases
from 10 fps to 30 fps in a displacement of
what
is the
Problems 1338, 1360.
1338. A 48-in. disk A weighing 322 lb.
rolls down inclined tracks on integral 18-in.
wheels, Fig. 607. Slipping impends and
f = 0.15. Neglecting the effect of the 18-in.
wheels, determine (a) the inclination of the
plane, and (b) the angular velocity after a
displacement of the eg of 20 ft. from rest.
Aus. (a) 10.9°; (b) 9.75 rad. per sec.
1339. A spool A, Fig. 608, weighs 161
Ib. and Kas a radius of gyration & of 9 in.
Aforce Q = 75 lb. is applied to a weightless
cord which wraps about the mid-section of
the spool, as shown.
If the spool rolls
PROBLEMS:
PLANE
Fig. 608.
385
MOTION
Problems 1339, 1340, 1361.
without slipping, what is the acceleration
of its eg? What coefficient of friction is
required.
1340. The same as 1339 except that the
line of action of Q is along BC, Fig. 608.
Ans. 18.06 fps?, 0.095.
Fig. 610.
Problems 1343, 1344.
1345. In Fig. 611, a 6441b. cylinder A
is shown on a rough inclined plane for which
@ = 30°. A weightless cord wraps about
the cylinder and passes over a weightless
and frictionless disk C to a 200-lb. weight
B. If the cylinder rolls, determine (a) the
tension in the cord, (b) the acceleration of
the eg of A, (c) the frictional force between
A and the plane where f = 0.8.
Fig. 609.
Problems 1341, 1342, 1362, 1365,
1366.
1341. A 4-ft.(= Ds») cylinder is supported
by
2-ft.
(= D1)
cylinders
which
run
on
tracks, one on each side, Fig. 609. A constant force Q = 75 lb. is applied to a weightless cord which wraps about the midsection
The weight of
of the cylinder, as shown.
the whole is 200 lb. and the radius of gyraIf noslipping occurs, what
tion & = 10in.
is the acceleration of the cg? What is the
In what direcnecessary frictional force?
tion does this cylinder move?
1342. If the coefficient of static friction
for the cylinder and track described in
1341 is f = 0.5, what may be the maximum
value of Q if no slipping occurs?
Ans. 45.9 lb.
24-in. rolling
=
D,
the
610,
Fig.
In
1343.
wheel A weighs 96.6 lb. and has a radius of
gyration k = 8in. Body B weighs 32.2 lb.
The weightless cord wraps
and f. =0.
around a diameter D, = 12 in. Find (a)
the acceleration @ of A, (b) the acceleration
of B, (ec) the tension in the cord, (d) the
frictional force necessary for rolling and the
corresponding coefficient of friction. (e) If
f = 0.1, would the wheel roll or slide and
roll?
Ans. (a) 3.51 fps?; (b) 1.75 fps?; (©) 30.4 lb.;
(d) 19.9 lb., 0.206; (e) slide and roll,
1344. The same as 1343 except that
D2
==
11) in.
Fig. 611.
Problems 1345-1347, 1363, 1364.
1346. The same
We = 100 lb.
as
1345
except
that
Ans. (a) 2.88 fps?; (b) 117.9 lb.; (c) 146.5 lb.
1347. The same as 1345 except that the
solid disk C weighs 240 lb. The axial
friction of C may be neglected.
Fig. 612.
Problems 1348-1350, 1367.
1348.
In Fig. 612, a 1288-lb. cylinder A,
on a 30°
(= 6) rough incline, has attached
to its gravity axis a cable whose weight
may be neglected. This cable passes about
386
PROBLEMS: PLANE MOTION
[Ch. XVII
weightless, frictionless disks C and D, as
shown.
A weight B is attached to the
smooth axis of the disk D. The coefficient
of friction between A and the inclined
plane is f = 1/3. What weight We will
cause A to be on the point of slipping while
it is rolling up the incline?
What is the
tension in the cable?
How far does B
move in 10 sec.?
Ans. 4940 Ib., 1761 lb., 465 ft.
1349. The same as 1348 except that the
solid disk C weighs 200 lb.
1350. The same as 1348 except that A is
a sphere instead of a cylinder.
Fig. 614.
Problems 1354, 1355.
1354. A slender uniform rod AB, Fig.
614, weighs 80.5 lb. A force Q is such as
to maintain the speed of the point A
constant at v4 = 8 fps. For the instanta-
Fig. 613.
neous position shown, (a) determine the
acceleration of the point B and then the
tangential acceleration a:;4, of B relative
to A; (b) find the angular velocity and the
angular acceleration of AB; (c) compute
the value of Q@ and the reactions at A and
B if the surfaces are smooth.
Ans. (a) 12.5 fps?, 7.5 fps?; (b) 1 rad. per
Problems 1351, 1352.
1351. The connecting rod AB of a fourlink mechanism ABDE, Fig. 613, weighs
120 lb. and has a moment of inertia about
an axis perpendicular to the plane of the
paper and through the cg at point C of 2
slug-ft.2 For the given position, the absolute acceleration of the cg is a. = 139 fps?
at an angle of 14° with the vertical, as
shown.
The horizontal component of the
pin reaction at A is A, = 62 lb. toward the
right. The tangential acceleration of A
relative to C, a:a/c, is 18 fps?. Determine
the other components of the reactions at
the pins A and B.
Ans.Bz = 63.21b., By = 208lb.,Ay = 176]b.
1352. The same as 1351 except that A;
acts toward the left.
1353. The connecting rod in the example
of § 213 may be considered as a long, slender,
ROLLING
1355. The same as 1354 except that the
surfaces are rough, with f = 0.3.
Fig. 615.
;
Problem 1356.
1356. A 12-ft., 32.2-lb. uniform ladder
rests against a wall as shown in Fig. 615
where @ = 60°. The surfaces are very slip-
uniform rod.
Its weight is 30 lb. For the
conditions defined in § 213, determine (a) the
acceleration of the eg, (b) the horizontal and
vertical components of the reaction at the
crankpin and the normal reaction
crosshead.
Neglect friction.
sec. c., 0.75 rad. per sec.2, cc; (c) 22.38 Ib.
64.88 lb., 22.38 lb.
pery (neglect friction) and the ladder slips.
When 6’ = 45°, w = 1.848 rad. per sec.
counterclockwise.
Find a and the reaction
at A’ at this instant.
at the
Ans. 2.79 rad. per sec.? ec, 3.4 lb.
AND
SLIDING
1357. A 3-ft. sphere that weighs 4000 lb.
moves from rest down a rough 30° incline,
where the coefficient of static friction is
fs = 0.12 and the coefficient of kinetic fric-
tion is f, = 0.1.
acceleration,
Determine (a) its angular
(b) the linear acceleration
of
PROBLEMS:
ROLLING
AND
587
SLIDING
its eg, (c) its angular velocity and its
linear speed after it has moved 30 ft. down
the incline.
Ans. (a) 4.65 rad. per sec.?; (b) 13.31 fps?;
(c) 9.85 rad. per sec., 28.2 fps.
1358. The same as 1357 except that the
slope of the incline is 15°.
1359. A block and a sphere of the same
weight and material are released simultaneously on an inclined plane. Which will
reach the bottom of the incline first if the
block slides and the sphere slides and rolls?
Explain your answer.
1360. The disk A of Fig. 607 weighs 322
Ib. and
moves
down
the inclined
tracks,
where 6 = 45°, fe = 0.14, and f, = 0.12.
Determine its linear and angular accelerations and the number of revolutions it
makes during 5 sec.
Ans. 20.07 fps?, 1.023 rad. per sec.’, 2.04 rev.
1361. On the 161-lb. spool of Fig. 608,
a force Q = 150 lb. is acting. Let k =9
in., f, = 0.2, and determine its linear and
angular accelerations.
1332. Let f, = 0.25 for the 200-Ib. cyl-
inder in Fig. 609. The force Q = 75 lb.,
and k = 10 in. Determine the linear and
angular velocities after a period of 10 sec.
if the cylinder starts from rest. Use other
data as needed from 1341.
Ans. 40.25 fps, 232 rad. per sec.
1363. In Fig. 611, Wa = 644]b., 6 = 30°,
fi = 0.25 between the cylinder A and the
plane. Also We = 400 Ib. and We = 0.
Neglecting friction at the axis of the disk
C, determine (a) the frictional force between
A and the plane, (b) @ for A and a for B,
(c) the tension in the cord.
1364. The
same
as
1363
except
that
O = 45%,
1365. In Fig. 609, a homogeneous 32.2-lb.
a diameter
of Dy = 24
cylinder
C with
diameter
D, for C to roll to the left, (b)
in. may roll on weightless wheels of diameter
D, that have a coefficient of friction f= 1/2
with the rails. A force Q is applied to the
Find (a) the largest wheel
horizontal cord.
the smallest
wheel
diameter
roll to the right, and
D, for C to
(c) the diameter
dD,
that would cause C to slide to the right
(d) lf Q = 101b. and D; =
without rolling.
1.5 ft., find the displacement of C from rest
im 3 Sec.
Ans. (a) Di <Ds; (b) Di>Dz;
(c) Di = D2;
(d) 7.94 ft.
1366. In Fig. 609, the homogeneous
32.2-lb. cylinder C of diameter D, = 24 in.
is mounted on a track on weightless wheels
of diameter D, = 6 in. for which f = 1/2.
If force Q on a cord is gradually increased,
find the value of Q that will give C the
maximum acceleration (a) to the left and
(b) to the right. In each case, show whether
the body rolls, slides, or both.
1367. In Fig. 612, a 1288-Ib. cylinder A,
on a 30° (= 6) rough incline, has attached
to its gravity axis a cable whose weight
The cable passes about
may be neglected.
frictionless and weightless disks C and D,
A 2.5-ton weight B is attached
as shown.
to the smooth axis of disk D. The coefhcient of friction between A and the inclined
plane is f = 1/3. Determine the angular
velocity of A after 3 sec., when A and B
Ans. 55.8 rad. per sec.
start from rest.
1368-1380. These numbers may be used
for other problems.
Chapter
XVIII
WORK,
KINETIC
ENERGY,
POWER
265. Introduction.
Where bodies are undergoing an acceleration, there
are three possible approaches to the force analyses: first, the method of force,
mass, and acceleration as suggested by Newton’s second law, =F = ma,
which is presented in the three preceding chapters; second, the principle of
work and kinetic energy, which is discussed in this chapter; and third, the
principle of impulse and momentum, the subject for discussion of the next
chapter. Some problems may be analyzed with virtually equal ease by
either of the three principles or methods.
Some problems are much more
easily solved by one of the three principles than by either of the other
two.
Hence the study of the material of this chapter not only teaches the
concepts
of work, kinetic energy, and power, but also gives us an additional
powerful
tool that will be useful in the analyses of many diverse problems
in engineering,
266. Work.
Work is a form of energy.
In thermodynamics, where the
concept of work is explored more fully, work is spoken of
as energy in transition. The meaning of this phrase may be discussed
from several viewpoints,
but for present purposes, we might interpret it to
mean that, in order for
work to be done, a body must move.
That is, a force must undergo a displacement.
In this sense, it is a transitory form of energy.
No matter how
large a force may be acting on a rigid body,
if the body does not move, no
work is done.
The technical definition of work is in terms of
force and displacement.
If
a force acts on a particle which undergoes a
displacement ds, the work done
by the force is the product of the displacement
and the component of the force in
the direction of the displacement.
In the general case, should the particle
undergo curvilinear motion with a varying
force, we may write the work U as
(48)
C= ih
F ds,
where the component
(/’) in the direction of the displacement
is consi
dered
to be constant for an infinitesimal displaceme
nt ds. If the force F is variable,
we must of course be able to express it in
terms of s in order to integrate (48).
358
§ 267 | POSITIVE AND NEGATIVE
889
WORK
It happens that there are many instances in engineering where the force F is
constant, or so nearly constant that no serious error is involved in assuming
For example,
that it is so. In this event, the integral is readily evaluated.
let a constant force F act on the rigid body of weight W (Fig. 616), and let
the
the body move in a straight line through a distance s. In this illustration,
force acts in the direction of the line
of motion; hence, the work done by
a single constant force F is
|
4
F
7
(a)
Fig. 616.
isa |OSuaalisy
0
the direction of
where s is the distance moved by the point of application in
is equal to
the constant force. For rectilinear motion, the displacement
the distance moved.
617, we must
If the force F does not act in the line of motion, as in Fig.
component
This
ement.
use the component in the direction of the displac
of motion.
line
is F cos 6, where 6 is the angle between the force F and the
If this component force is constant, we find
Ss
(b)
U="F cos ofds = (F cos 6)s.
0
a straight line or
The work of a force acting on a particle which moves in
on a curved path
which moves through an infinitesimal displacement ds
may also be defined as the magnitude of the force times the component of the displacement of its
point of application in the direction of the force. Thus, in Fig.
617, the component of the disa
Ss
~J'b
placement in the direction of the
work.
does
0
cos
F
nent
Compo
Fig. 617.
force is ab = s cos 6, which gives
Component F sin 9 does no work.
U = F(s cos 9),
ent s cos @ of the diswhich agrees with the previous result. This compon
Observe that the fundaplacement s is called the effective displacement.
of application must
mental condition for a force to do work is that its point
ent of the force. In
have a displacement in the direction of some compon
icular to the line of
perpend
Fig. 617, the component F sin 6, which acts
cement of its point
motion, does no work because there is no effective displa
of application in the sense of this component.
quantity, it
267. Positive and Negative Work. Since work is a scalar
positive
either
be
to
red
is not represented by a vector. It may be conside
of a
work
the
ics,
or negative, as may be convenient. Usually in mechan
890
WORK, KINETIC ENERGY, POWER
[Ch.
XVIII
force which has a component in the same sense as the displacement is taken
as positive. Such a force tends to increase the speed of a particle or body.
For this scheme of signs, work done on the body is positive.
If the force has
a component which tends to reduce the speed of the body on which it acts,
its work is taken as negative.
Forces which have the same sense as the velocity and which therefore tend
to accelerate a body or to keep a body in motion (even though it is slowing
down) are motivating forces.
Forces which act in the opposite sense to the
velocity are resisting forces.
If the foregoing convention of signs is adopted,
motivating forces produce positive work and resisting forces produce negative
work. At any rate, it is still advisable to be careful of signs and to make
complete free bodies.
268. Work of a System of Forces Acting on a Rigid Body. The work of
a system of forces acting on a rigid body is equal to the algebraic sum of the
works of the individual forces. For example, the body A in Fig. 618 is moving
up the inclined plane under the action of the constant forces shown.
The
work of the force Q during a displacement S up the plane is
Ug = +(Q cos 6)s;
”\n
Motion
7
the work of the frictional force F is
Up
=>
ss
the work of the force of gravity W is
Uw = —(W sin 6)s;
WwW
rg hoileho tetee and the work of
the normal plane reaction N is
Fig. 618.
zero,
Un= 0!
since its sense is normal to the direction of motion
.
Thus, the resultant or
net work U on the rigid body A is the algebraic
sum of the above work
quantities, or
(c)
U = (Q cos 6)s — Fs — (IW sin 6)s,
where s is the straight-line displacement of A
for which the work is desired.
This equation may be written in the form
(d)
U = (Q cos
0 — F — W gin 0)s,
where the part in parentheses is seen to be the
resultant force R acting on
the body, which is >/,, inasmuch as 2F,=
0,
Thus, we may say that for
a particle or for a rigid body in rectilinear transl
ation (where the force system
is concurrent), the net work done is the work done
by the result
ant force.
The
§ 269 |PRINCIPLE OF WORK
AND KINETIC ENERGY
891
ularly convenient for bodies
point of view expressed by equation (d) is partic
complete free body, as we
in rectilinear motion since we can simply make a
forces in the direction of motion
have already learned to do, and then sum the
it is constant) and the disto get the resultant Rk. The product of R (if
placement s is the net work,
Net work = U = Rs
[RECTILINEAR TRANSLATION]
While
rectilinear translation.
a relation which holds for any rigid body in
are often interested in the
this procedure gives the net work, we nevertheless
n or the work of gravity.
frictio
of
work of some single force, such as the work
Let a particle of mass dm be
269. Principle of Work and Kinetic Energy.
from Newton’s laws, we have
acted upon by a resultant force R. Then
= v dv/ds, we find
R = (dm)a. In this expression, setting a
(e)
pr ds = [Camm dv.
nt, which is the usual case in
In case the mass dm of the particle is consta
ating between any two speeds v1
engineering, we find from (e), by integr
and v2,
? _ (dW)v.2 zt (dW)vi?
(dm)v2? — (dm)vy
ae
ee
Gs
pr
29
29g
(49)
of the particle. The left-hand side
where dm = dW/g, dW being the weight
done on the particle. The exof (49), [R ds, is recognized as the net work
c energy KE of a particle. Therepression (dW)v?/(2g) is called the kineti
Thus,
e of kinetic energy AKE.
fore, the right-hand side of (49) is the chang
y,
namel
ple,
of a very useful princi
equation (49) is a mathematical statement
to the change of kinetic energy
The net work done on a particle is equal
of the particle.
In symbol form,
(f)
Unes = [R ds = AKE = KE, — KE.
s
d as the value at the second point minu
The symbol AKE is always interprete
is a special case of the law of conserthat at the first point. This principle
tempera-
net work is not affected by
vation of energy; it holds true only if the
d change the particle’s internal energy,
ture changes of the particle which woul
Although (f) holds only if all
energy.
by chemical actions, or by electrical
and the change of kinetic energy are
energy changes except the net work
useful to the engineer.
negligible or zero, this principle is very
892
WORK, KINETIC ENERGY, POWER [Ch. XVIII
270. Kinetic Energy of a Rigid Body in Translation.
If a rigid body A,
Fig. 619, is in translation, all particles dm, dm», etc., have the same velocity
v. Therefore the kinetic energy of the body is
Bis dye"
CLE,
Fig. 619.
_ [ (dm) 2 gl 2
:
ea il 2
ey
2
= ee
um
2
2g?
Kinetic Energy
where m = W/g is the mass of the whole body.
The energy Wv?/(2g) is the kinetic energy that a
body possesses at a particular speed v. If the speed is v1, the kinetic energy
is Wv;"/(2g); if the speed is v2, then KH, = Wv»?/(2g); ete.
Since the force system on a body in translation necessarily reduces to
a
single resultant force,* the net work done on the body is JR ds, or Rs,
if R is
constant.
Hence, the principle of work and kinetic energy applied to a
rigid body in translation yields [see equation (f)]
—Translation.
U.a = AKE = KE, — KE,
a Wo
iSeana
Wor
a
[TRANSLATION]
271. Units of Work.
The unit of work is evidently a force unit times a
space unit. For the type of work in which we use the princip
les of this
chapter, the unit is usually a foot-pound (ft-lb.), althoug
h many other units,
such as an inch-pound, a mile-ton, a dyne-centimeter,
could be used.
The
work as obtained in foot-pounds may be converted
to any other energy unit.
The units of some convenient conversion factors are
as follows:
ft-lb.
ce Btu ’
hp-hr.
oc kw-hr.’
28,000
hp-min.
idee hp-hr.’
ee
ft-lb.
lbs
hp-sec.
ft-lb.
1,980,000 ire
550
Btu |
hp-hr. ’
-lb.
dtelb 4
kw-hr.
Btu
3413 aes
2,650,000
1910
_etss
low-see.
where Btu stands for British thermal unit,
hp for horsepower, hr. for hour,
kw for kilowatt.
When a large amount of work is invol
ved, as in the instance of power-generating equipment,
the horsepower-hour (hp-hr.) and
kilowatt-hour (kw-hr.), which are
large units of work, are preferred.
In
dealing with heat and its conversion,
we often prefer the Btu.
Since work and kinetic energy are equiv
alent, the unit of kinetic energy
*R may be zero,
893
§ 273 |POTENTIAL ENERGY
of the
must match that of the corresponding work. Ordinarily, the units
various terms in Wv?/(2g) are
Wer?
(Ib.)(ft./sec.)? _
Aedes
ft./(sec.)?
Zoe
70 tons, starts
272. Example. A freight train consisting of 60 cars, each weighing
pull (i.e., the
drawbar
constant
The
mph.
15
of
speed
up a 1.5% grade with an initial
resistance (intrain
the
and
tons
97
is
train)
the
on
exerts
ve
locomoti
force that the
15 Ib. per ton of weight.
eluding rolling resistance, axial friction, and air resistance) is
(a) How long is the
mph.
30
is
speed
the
At the top of the constant 1.5% erade,
Express the work
(b)
wheels.
the
of
rotation
of
energy
grade? Neglect the kinetic
against gravity?
done
work
of the draw-bar pull in terms of hp-hr. (c) What is the
(a) Let Fig. 620 (where 6 is
Souution.
y\
exaggerated for clearness) represent a
(97)(2000) Ib.
pull
r
draw-ba
the
free body of the train:
D.B.P.
is (97)(2000) Ib., the total resistance is
(60)(70) (15) Ib.
(60)(70)(15) Ib., and the total weight is
W = (60)(70)(2000) lb. Since the body is
in translation only, the net work is Rs =
(60)(70)(2000) Ib.
(=F,)s. Now using the principle Une, =
Fig. 620.
AKE, and the fact that 15 mph = 22 fps,
we have
60) (70) (200
22%),
[97)(2000) — (60)(70)(15) — (60)(70)(2000)sin 4] = oo (442 —
where sin 6 = 0.015.*
Solving for s from this equation, we get
3 = 37,900 ft. = 7.17 mi.
(b) The work of the draw-bar pull is
U = (97)(2000) (37,900) = 7,352,600,000 ft.-Ib.
7,352,600,000 ft-lb.
= 3710 hp-hr.
U
ft-lb. /(hp-hr.)
~ 7,980,000
of W parallel to the plane times
(c) The work done against gravity is (component
distance)
(sin 6) (37,900) ft-lb. _ 2410 hp-hr.
(60) (70) (2000)
1,980,000 ft-lb./hp-hr.
of
energy and the potential energy (§ 278)
There is an increase in both the kinetic
the train.
against gravity in part (c) of the
273. Potential Energy. The work done
The change
e of potential energy.
preceding example is also called the chang
the body is elevated; it is a decrease
of potential energy is an increase when
when the body is lowered.
h above some datum plane, it is
If a particle of weight W has an elevation
ence to the chosen datum, the amount
said to have potential energy with refer
in 100 ft. measured on the level.
*A 1.5% grade means a 1.5 ft. rise
tan 6,
lly equal to
0.015, but for small angles sin @ is virtua
Strictly tan 6 =
894
WORK, KINETIC ENERGY, POWER [Ch. XVIII
being Wh, where h is the vertical distance through which the particle would
travel in passing from its original position to the datum level.
In particular,
this energy is the potential energy of elevation and is due to the configuration
of two bodies, the earth and the particle whose weight is W.
The potential energy of elevation of a finite body with respect to some
datum is work that would be done by the force of gravity (W) on the body
as its center of gravity passes from one elevation to the elevation of the
datum.
To prove this statement, let a body, rigid or otherwise (Fig. 621),
be composed of particles of weights Wi, Wo, etc., at elevations of Y1, Y2, ete.,
above some chosen datum.
The potential
energy PE of the body is the sum of the
A
potential energies of all its particles, that is,
PE
=
Wiyi
+ Woy
+
tai
But recalling (§ 109) that
Wiyi + Wey2 +
ee =
Wy,
we see that the potential energy of a finite
body above a particular datum is
Fig. 621.
Potential Energy.
(g)
PE = Wy.
Observe that the change of potential energy is independent of the
path
taken by a body to reach its new position.
The path may be curved in any
manner, yet the change of potential energy is always WY, where
7 is the
vertical distance moved by the body’s center of gravity.
If the displacement of the body has a component upward, the body gains potential
energy
(as does the train of § 272); if the body’s displacement
has a component
downward, the body loses potential energy. *
The energy that is stored in a body by virtue of the configu
ration of its
particles is sometimes called the potential energy of the body.
For example,
a compressed spring has stored energy because of the
internal stresses induced
in the material of the spring by the compression.
This stored energy is
given up when the spring is released ; in this sense
the energy is potential.
Similarly, the high-pressure steam entering a steam
turbine has stored energy
due to the configuration of its molecules; hence
the energy is potential in
form.** Summarizing, we note that kinetic energy
is that energy which a
body possesses by virtue of its motion, whereas
potential energy is energy
*Torricelli (1608-1647) knew that a system
of bodies would move under the action of
gravity only if the center of gravity of the bodies
descended.
It was Torricelli who invented
the first (liquid) barometer and measured
**However,
molecules.
the pressure of the atmosphere.
the steam also has energy in the form of
the kinetic energy of its moving
395
§ 275 | EXAMPLE
the earth or by virtue of
that a body has by virtue of its position relative to
the configuration of its molecules.
on of the water surface above
274. Example. Ina hydroelectric plant, the elevati
(Fig. 622). The elevation of
level
ce
referen
certain
a
above
the plant is y: = 400 ft.
level is y2 = 100 ft. How much
the water below the plant above this same reference
through the plant if all of the
passing
work may be done by 100,000 cu. ft. of water
efficient?
100%
are
blades
processes are frictionless and if the turbine
y is numerically
Fig. 622. The loss of potential energ
on neglected.
fricti
,
energy
c
kineti
of
gain
equal to the
c energy is
kineti
of
gain
the
In the ideal case assumed,
so the work is equal
converted into work in the turbine;
If one were thinking
to the change of potential energy.
energy, the energy
of
on
rvati
conse
of
law
in terms of the
= 0; that is, up to
relationship would be APE + AKE
leaves the system, so
the turbine, no energy enters or
zero.
be
must
e
chang
that the net
Ground
Fig. 623.
is
all of the change of potential energy
SotuTion. For the conditions specified,
have
converted into work. Hence we
U = Wh = (100,000)(62.4)(400 — 100)
= 1,872,000,000 ft-lb.
ty in lifting the material to
What is the work done against gravi
275. Example.
high, 20 ft. OD,
masonry chimney which is 250 ft.
its place during the building of a
623).
(Fig.
ft.
cu.
s 100 lb. per
and 12 ft. 1D? The masonry weigh
the
me of thickness dy at a height of y from
SotuTion. Choose a differential volu
me weighs
ground, This differential volu
0 dy.
dW =wdV = c00(=)ian— 122) dy = 20,15
ase in potential
this differential weight is its incre
The work done against eravity on
ase in potential
is (20,150 dy) (y) and the total incre
energy. From the ground level, this
energy is
250
9)
PE = 20,150| ydy = es
2
= 628,000,000 ft-lb.
0
in potential
directly by noting that the increase
This result may be obtained more
896
WORK, KINETIC ENERGY, POWER
[Ch. XVIII
energy of the material in the chimney, as measured from ground level, is W7, where
W is its total weight and 7 is the elevation of its center of gravity. Thus we find
h
©
wV 5 = (100) p (202 — 12?)(250)
PE
250
ae
628,000,000 ft-lb.
276. Work of a Couple. The two equal forces F, Fig. 624, form a couple
with a moment arm 2r. If this couple is rotated through a small angle dé,
each force F acts through a displacement ds = rd@ and therefore does the
work F'ds = Frd@.
Therefore, the total work for both forces F is
(h)
v=
[ards = [P2rao = [aras,
where M = F(2r), the moment of the couple.
(i)
If M in (h) is constant, we find
6
v= M fas = m0,
0
the work done by a constant couple, where @ is in radians
and M is generally
in foot-pounds to give U in foot-pounds.*
If M in (h) is not constant, the
integration may be made when M is expressed as a functio
n of 8.
Fig.624.
Work of a Couple.
Fig.625.
Kinetic Energy—Rotation.
277. Kinetic Energy of a Rotating Body.
A body A (Fig. 625) is rotating
about a fixed center O. A particle dm of this
body has kinetic energy whose
value is (dm)v?/2 at some instant, where v
is the speed of the particle (§ 269).
For a rotating body, the speed of a partic
le is v = rw, where w is the same for
all particles of the body and therefore is a
constant that can be taken out of
an integral. Since the sum of the kineti
c energies of all its particles is the
total kinetic energy of a rotating body, we
have
KE Sy
- [oor £ $f hpiae 32
*The reader should observe that the
units of a couple and the units of work
may be
Moreover, each one may be found from
a force times
a distance. Yet the conceptions
of a couple (or any moment) and
of work are entirel
mye
Be certain that each idea is clearly
in mind in order to avoid mistakes
due to
confusion,
the same, foot-pounds, for example.
§ 277 |KINETIC ENERGY
897
OF A ROTATING BODY
or
2
epi a '
(50)
[ROTATION]
of the body about
where we recall that fr? dm = I,, the moment of inertia
Thus, the kinetic energy of a
the axis O, the center of rotation in this case.
angular velocity o.
rotating body is ,w?/2 (foot-pounds in this book) for any
; ete., where the
I,wi2/2
is
If the angular velocity is w:, the kinetic energy
reference point O is the center of rotation.
net work done on a
As in the case of a body in rectilinear translation, the
To prove this, we
rotating body is equal to its change of kinetic energy.
be equal to the
will
body
g
rotatin
a
recall that the resultant couple acting on
of rotation.
center
the
sum of the moments of all the forces on the body about
244) that
(§
have learned
Let this resultant be represented by M,. We
M, = 1,0. Using a = w dw/dé, we find
(j)
[pteaa =
[ted
n, J, is constant, so that
For a particular body with a fixed center of rotatio
on (j) between the limits of
the integration of the right-hand side of the equati
any two angular velocities 1 and w» gives
(Rae
(51)
po dé
=
ao
Rae
xe “a
which accords with equation (f), Une = AKE.
the left-hand integral in (51) becomes
ft-lb.,
Ifthe moment M, is constant,
|M,d0 = M, |do = M8,
designated 2M, the sum of the
where M, is the same as we have previously
t to the center of rotation O.
moments of all the external forces with respec
side as the net work (§ 276)
In equation (51), we recognize the left-hand
It is implicit
c energy (AKE).
and the right-hand side as the change of kineti
than one
more
is
ng body, if there
in (51) that each couple acting on the rotati
49; otherwise, the left-hand side
couple, has the same angular displacement
works of the individual couples,
of (51) must be found from a sum of the
JM, d0. + JM2d6 + °°".
the units of [,w?/2, recall that a
If you care to satisfy yourself concerning
fore has no units (see § 351). It
radian is a ratio of like quantities and there
the units of Iw? = (W/g)k*w? are
follows that the unit for w is 1/sec. Then
found from
Wee» _Ublft.2)(/see.4) _ gy
ib vue
(ye)
pal
898
WORK, KINETIC ENERGY, POWER
[Ch. XVIII
278. Example. A weight A is supported from a cable which is wound about a
4-ft. drum (Fig. 626).
An 8-ft. flywheel turns with the drum.
The total weight of
the rotating parts is 1288 lb. and the radius of gyration is 2.5 ft.
Flywheel.
While A travels 80
ft. vertically downward, the speed of the
rotating parts changes from 10 rpm to 120
rpm. The frictional force in the bearings
acting tangentially to the 6-in. shaft is
I
Ss
et
a
F = 70 lb.
What is the weight W of A?
Sotution.
Two bodies are involved,
one in rotation, one in translation.
The
initial and final angular speeds of the flywheel B are
«©
{.
Oe
2rn
2710
2rn
27120
= er
ey: at eae I=
Fig. 626.
ue
ge
1.047 rad. per sec.,
eee 12.58 rad. per sec.
The corresponding linear velocities of the weight A are
v1 = Tw = (2)(1.047) = 2.09 fps
and
V2 = (2)(12.58) = 25.2 fps.
Considering the free body of A and letting the downward direction be
positive,
we have
CaN
(I
KER
a
Jo
7
29
9).2
29
(W — P)s0 = pgV (25:2 — 2.09%) = 9.7917,
from which
(k)
W = 1.138 P.
While the weight A moves 80 ft., the rotating parts turn through
an angle of
‘
@=-=
i
$0
pie 40 rad.
ee
From the free body of the rotating parts, we get the sum
of the moments about O as
Ms = (P)(2) — (F)(0.25)
= 2) (70) (0:25) ee
TG.
Now using the principle of equation (51) with MW, constan
t and using for J, the value
‘
1288
I, = mk? = ( 5 “N25 = 250 slug-ft?.,
we find
[rw = M,6 = oe (wo? — w,?
250
(2P — 17.5)40 = "(12 58? — 1.0472),
from which
P = 254 lb.
ALTERNATE Sotution.
Using this value of P in (k), we get W = 289 lb.
In setting up the energy relations for a series of
connect
ed
899
§ 279 |BODIES IN PLANE MOTION
bodies, it is not always necessary to take each member of the series as a free body.
For example, in this case, imagine that the cable is whole, so that the force P is solely
an internal force. Now-we have implicitly assumed in this chapter that the internal
forces do no net work, that is, that they are in equilibrium among themselves (§ 226).
For example, the work of one force P downward is cancelled by the work of the other
force P upward.
In applying the principle
Un. = AKE = Kis KE
to a series of connected bodies, the safest procedure is usually to find the work done
by each force (or couple) separately and to consider separately also the changes of
kinetic energy of each body in the system. Thus, neither the normal reaction N at
work
the bearings nor the weight 1288 lb. of the rotating parts does work. The
done by W for s = 80 ft. is
Uw = + Ws = + 80W.
The work done by the frictional force F is
—(70)(0.25)(40) = —700 ft-lb.,
Up = —Moe = — Fro=
where the minus sign indicates that this work tends to retard the bodies.
the net work is
Unee = SOW — 700 ft-lb.
Therefore
The change in kinetic energy of A is
Wor
Woe
= 5 (25.22 — 2.092) = 9.79 W ft-lb.
AKE, =
2g
=
The change in kinetic energy of the rotating parts is
.
250
Tow?
Tw?
(12.582 — 1.0472) = 19,600 ft-lb.
=e
oe
ce a
AKEs
=
connected
Equating the net work to the total change of kinetic energy of all the
we have
SOW — 700 = 9.79W + 19,600 ft-lb.,
from which
W = 289 lb.
If the internal force P in the cable were now
parts,
desired, it
plan of considering a
would be necessary to use a free body of either A or B. This
simplification of a
some
in
results
eroup of connected bodies as a free body often
.
quantities
energy
all
include
to
sure
solution, but it requires care in being
Let the rigid body A (Fig. 627) be in
279. Bodies in Plane Motion.
The path followed by any point is in general a plane
random plane motion.
say the point O.
curved path. Choose any convenient reference point,
at the instant
that
assume
While point O is moving in a curved path, we shall
Any
shown.
as
n,
under consideration, it is moving in a horizontal directio
an absolute velocity
differential mass dm, located at some point B, will have
VB
=
Vo + UB/o
by the principle of relative velocities (§ 204).
The velocity of B relative to
since the body is rigid
Q is necessarily perpendicular to the radius BO =r,
(§ 208).
Hence
va). = 7,
where
w is the angular velocity of A.
The
400
WORK, KINETIC ENERGY, POWER
[Ch. XVIII
algebraic expression for vp is obtained by applying the law of cosines to the
vector triangle shown at B; thus
VB? = Vo? + UB/o2 — 2VB/o COS(180 —
4)
= 0,7 + r2w? + 2v,rw cos 0.
Since the kinetic energy of the particle at B, by definition, is (dm)v,?/2, the
total kinetic energy of the body A is
KE -{% (v2? + rw? + 2v,rw cos 6)
pe
ft
2
ow?
dm + a
dm + vw | (r cos @)dm.
We see that (Fig. 627)
[oecos #)dm =f dm = my,
Fig. 627.
Kinetic Energy — Plane Motion.
Point O is not a center of rotation; it isa random reference point whose velocity is known.
Fig. 628.
and we recognize that fr’dm = I,. From the foregoing expression, we then
get the kinetic energy of a body in any kind of plane motion as
(1)
KE
=
MV,”
9
Tae
a
oer +- VowmMy.
Equation (1) gives the kinetic energy of a body in plane motion when the
reference point O, whose velocity is known, is chosen at random with the
x
axis through O in the direction v,. However, it is expedient in most engineering situations to choose one of the following two reference points; because
they simplify the solutions.
1. The center of gravity of the body, in which case y = 0, the last term drops
out, and equation (1) becomes
(52)
KE = ae + —
[PLANE MOTION]
where @ is the velocity of the center of gravity (or centroid)
and 7 is the
moment of inertia of the body about a gravity axis perpend
icular to the
plane of motion.
401
§ 280 |FRICTIONAL FORCE IN PLANE ROLLING
the first
2. The instantaneous center of the body, in which event v, = 9,
and last terms drop out, and equation (1) becomes
KE = ee
(53)
2
[PLANE MOTION]
the plane of motion
where J, is the moment of inertia about an axis normal to
(Note that O is
through the instantaneous center (point of zero velocity).
no longer a random point.)
is, from (52),
For example (Fig. 628), the kinetic energy of a rolling cylinder
To?
mv?
Ug
mv?
mre”
3m?
i
es
2D iy By aay
slipping and where
where 3 = rw is the velocity of the eg when there is no
I = mr?/2.
To use (53), we find
=
Te
amie
mr
ao ot 08
and then the kinetic energy is (6 = rw)
KE
el
=
5
eel
SiiTaN)
(257)
ponies
=
4
taneous center of a rolling
the same answer as obtained from (52). The instan
628). Equation (52) sugbody is its point of contact with the ground (Fig.
translatory motion and a
a
as
of
gests that plane motion may be thought
is, the total kinetic energy
rotational motion about the center of gravity; that
kinetic energy of rotation
is the kinetic energy of translation mv?/2 plus the
Tw?/2.
= AKE, being a special case
The principle of work and kinetic energy, U,..
bodies in any kind of plane
of the law of the conservation of energy, holds for
motion.
body slides, the frictional
980. Frictional Force in Plane Rolling. When a
does work. When a body rolls
force undergoes displacement and therefore
without slipping, there is
between the rolling body and
no displacement of the point of contact
the ground;
pure rolling does no work.
hence the frictional force which induces
Hesitate
that the frictional force does not
long enough here to “‘see”’ in your own mind
does not slide.
“move through a distance”’ if the body
in
which work must be done
There is a rolling resistance (§ 77) against
d be
In a careful analysis, this resistance shoul
order to maintain motion.
corresponding to this resistance is
estimated; but in many cases, the work
402
WORK, KINETIC ENERGY, POWER
[Ch. XVIII
relatively small and is often neglected, or lumped together with other frictional resistance.
In general, in problems involving a combination of translation and rotation,
it will be safer to find the work of each individual force which does work and
then add algebraically these work quantitres.
281. Example.
A 4-ft. cylinder, which weighs 966 lb., rolls down a 15° incline from
rest. What is its speed after it has rolled 50 ft.?
15°
Fig. 629.
Ws
hy
For pure rolling,
F does no work.
Fig. 630.
Sotution.
The forces acting on the cylinder are W, N, and F, as shown in Fig. 629.
The force N does no work because it acts normal to the direction of motion.
The
force F does no work because the cylinder is rolling. Hence the only work quantity
involves that component of W which is parallel to the incline.
Using (aa
AKGr)
and equation (52), we have
(m)
(W sin 15°)(50) = lies: — 0:7) + <(w2? — w?).
2
29
In this example, v; = 0, w1 = 0, w2 = v2/2, and
=
mr
©
_
(966)(4)
7T2 ea (82.2)0)
Using these values in (m), we find
(966) (0.259) (50)
=
= 60 slug-ft.2
ak
966 _, ‘i(2
Beas
2) 4?
from which v2 = 23.6 fps. Observe that the work done by gravity on the cylinder
is equal to its loss of potential energy.
ALTERNATE SOLUTION.
Using the instantaneous center O (Fig. 629) as the reference
point, and applying equation (53), we find
S
966) (4
96
I, = > + mr = Sails, - oo
= 180 slug-ft.?
Tow,
(WW sin
sin 15°)(50)
15°)(50)p== =
9
(180\ 52
QE =
from which 02 = 23.6 fps, as before.
282. Example.
A 4-ft. cylinder
A, Fig. 630, has a central 2-ft. groove
which is wound a weightless inextensible cord.
about
This cord passes parallel to the 30°
403
§ 282 |EXAMPLE
rd to a body B which weighs
incline and over a smooth post D, thence vertically downwa
its radius of gyration about
and
Ib.
Wes = 300 lb. The cylinder weighs Wa = 500
to cause the cylinder to roll.
nt
sufficie
is
force
al
itsaxisC isk, = & = 1ft. The friction
(b) Determine
motion occur?
(a) If the system is released, in what direction does
ft. from rest.
15
moves
B
r after
the tension in the cord and the velocity of the cylinde
(c) What is the acceleration of the body B?
upon the rela(a) The direction in which the cylinder moves depends
Sotution.
se moment
clockwi
counter
the
of
and
O
about
Q
of
moment
se
tive values of the clockwi
the force in
shown,
not
some force
of W4 about O. If the cylinder is held at rest by
ft-lb. The
900
=
(300)(3)
is
moment
the cord isQ = Wea = 300 lb. and the clockwise
counterclockwise moment is W ar sin 30° =
Since the clockwise
(500)(1) = 500 ft-lb.
, the cylinder will turn clockwise
moment is greater than the counterclockwise moment
are acting.
(or move up the incline) when only the forces shown
C of A. Since rolling occurs,
eravity
of
(b) Let d4 represent the speed of the center
Fig. 630. Inasmuch as the
O,
contact
of
point
the
the instantaneous center is at
distances from the instantaneous
velocities of points in A are proportional to their
is found from
center, the velocity vg of a point on the cord
=
ine
vqg= a
or
z
Ose
a
of a point on the cord; hence vz =
And of course the speed of B is the same as that
(3/2)oa.
te (Une, = AKE)
y
For the free body A, Fig. 630, we ma wri
(n)
(Q)(15) — (Wa sin 30°)(10) =
W
AD We
es Ae
me
e
component Wa
the
il
wh
ce
ft.
15
an
of
where the force Q acts through a dist
acts through 10 ft.; also
sin 30°
pgaD
500
-
=
2g
I = mk? =—,
g
2
s” on A are not the same
force
the
by
ed
Observe particularly that the distances “mov
ately for each force.
separ
found
be
to
have
and that therefore the work quantities
Substituting known values into (n), we have
(0)
For the free body B, we find
(p)
(500)(042) , (500)(6.4
: 2) _ 312.5542
15Q — 2500
(Wz = 300)
(300 — Q)(15) =
4"
)7..4"42 __ 337.50ae
3/2)?0
0)(3/2
(300)(
e? _ (30
Wvpve?
29
29
Adding equations (0) and (p), we get
g
pate
2000 =
aes
[or (p)], we find
Using this value of 64 in equation (0)
from which 64 = 9.95 fps.
2500 + 961
ae
= 230.o1b:
obtained from v dv = ads, we find that
(c) Using the kinematic relation 02 = 2as,
cylinder is
the acceleration of the axis of the
da?
da
=
2s
99.2
(2)(10)
4.96 fps’.
404
WORK, KINETIC ENERGY, POWER
[Ch.
XVIII
Since the acceleration of B is to the acceleration of the axis of A as the ratio of the
instantaneous radii (3/2), we have ag = (3/2)(4.96) = 7.44 fps?.
ALTERNATE SotuTIon.
(b) A direct solution for the velocity of A (or B) may be
made by considering the two bodies together as a free body.
In this event, the net
work is the algebraic sum of the works of the forces W4 and Wz, only (Fig. 631).
(Note that Q is now an internal force.)
kinetic energy of both A and B.
This net work is equal to the total change of
Thus
bee
(Ws)(15)
=
a
ano
— (Wa sin+ 30°)(10)
AKE
loa
yy W -at, +3
Wate
F be
Using vz = (3/2)04 and the values of the other terms as found above, we get i4 =
9.95 fps, as before. The tension in the cord may now be found from a free body of
either A or B, and the other quantities called for in the problem are found as in the
first solution.
No Movement
of These Points
of Application No Work
=9'
Wz =1610 lb.
Wa =1610 Ib.
Fig. 632.
Fig. 631.
283. Example.
A counterweight B of 1610 lb. is to help hold a load A of 1610 Ib.
as shown in Fig. 632. The radius of gyration of B is k = 2 ft. The
pulley C is
frictionless and weightless. After the load A has moved down 20 ft. from rest
under
the action of the forces shown in Fig. 632, what is the speed of B?
SotuTion. The first thing to recognize is that if B moves up a distance
of y, A
moves down a distance 2y; that is, if A moves 20 ft., B moves 10
ft. See Fig. 632.
It follows also that the velocity of A is twice that of B, va = 2vg.
Let the work done
by the weight A be positive; then that by B is negative. Using equation
(52) for
the kinetic energy of B, we get
W ava?
Unet
xi W 30?
—
29
29
a
Hee
Oy?
) , Wada? , mok%(0n/2.5)
20W 4— 10W» = Was
29
9
Ds
5)
(20)(1610) — (10)(1610) = (1610)4es%) , 161005% , (1610)(4)o"
from which dg = 10.7 fps.
29
29
(29) (6.25) ”
405
§ 284] VARIABLE FORCES
may
284. Variable Forces. As previously mentioned, a force doing work
engineer
in
force
variable
a
of
‘The most common example
not be constant.
spring
a
in
as
ment,
ing is that in which the force varies directly as the displace
231), the
(§ 231). If the scale of the spring in Fig. 633 is K Ib. per ft. (§
distance of
magnitude of F after the spring has been gradually compressed a
the force
dy,
ment
displace
imal
y ft. is F = Ky. During a further infinites
stretchin
(or
ing
Ky is virtually constant, so that the work done in compress
total
the
and
ing) a spring an infinitesimal amount is dU = F dy= Ky dy,
is
work of deforming a spring an amount of s ft. from its free length
s
U=
xf y dy =
2
AS feb.
0
of kinetic energy,
When the work done by a variable force involves a change
the principle U,., = AKE may be used.
,
\"
O
‘|
Free
Length
Fig. 633.
mals
Fig. 634.
Spring.
magnitude, the general problem
If the force varies in direction as well as in
In Fig. 634, a
be evaluated.
is to set up for the work an integral that can
The
which moves ina path AB.
force Q acts on the body M at point A,
the tangent mn to the curve at A.
line of action of Q makes an angle @ with
along the curve a distance ds,
If the point of application of the force Q moves
the work done is
Q cos 6 ds,
If Q, 6, and s all vary, the
no work.
inasmuch as the component Q sin 6 does
the three variables are expressed in
integral may be evaluated if any two of
Q are in terms of s; or if all three
terms of the third, for example, if 6 and
er variable, such as
variables can be expressed in terms of anoth
that
this connection, we might recall
ae
an = (aye + ar = [1+ (GE) |
2771/2
@)
«and y.
In
406
WORK, KINETIC ENERGY, POWER
[Ch. XVIII
285. Example. <A 90-lb. body A, Fig. 635, rests upon a 30° incline wheref = 0.2.
It is in contact with a spring which has been compressed 10 in. and whose scale is
K = 30]b. perin.
The lower end of the spring is attached to a fixed wall, and at the
instant the spring reaches its free length, it ceases to act upon the body A.
(a) What
is the speed of the body at the instant the spring reaches its free length?
(b) How
far up the incline does the body go before coming to rest?
(a,b)
N
es
fe
i
90 lb.
7” 30°
x
Fig. 635.
Fig. 636.
Sotution.
(a) A free body of A is shown in Fig. 635. The work of the spring on
A is the same as the work of compressing the spring 10 in. = 10/12 ft. Assuming that
this work is positive, we have
10/12
hy = «|
é
10
2
1
Hoy = aocn( 2)(:)= 125 ft-lb.,
(PEN?
where (30)(12) is K, the scale of the spring, converted to pounds per foot.
normal force is obtained from a sum normal to the plane; we find
The
N = (90)(cos 30) = 77.94 lb.,
from which
F =fN = (0.2)(77.94) = 15.59 lb.
The work of the frictional force through the 10-in. displacement is
U,=
—
10
05507) =
—
13 ft-lb.,
which is taken as negative because it opposes the work of
the spring.
against gravity or change of potential energy is
The work
10
The net work is then the algebraic sum of the individual work
quantities, or
One, = Ur + U2 + Us = 125 — 13 — 87.5
= 74.5 ft-lb.
Using the relation U,., = AKE
= mv*/2, we have
74.5 = 29
2
from which v = 7.3 fps, the speed of A when the spring
regains its free length.
(b) As the body A comes to rest, its kinetic energy
(= 74.5 ft-lb.) is used up by
work against friction and gravity. If the displacement
is s ft. during this action, we
have
(90 sin 30° + 15.59)s = 74.5,
407
§ 286 |EXAMPLE
spring in its free length or
from which s = 1.23 ft. = 14.75 in. from the end of the
.
14.75 + 10 = 24.75 in. (nearly) from its initial position
acts on a body for only part
One of the lessons of this example is that when a force
intervals involving the
those
for
ely
separat
made
be
of its motion, the analysis must
is when a body moves from a
same force systems. Another illustration of this idea
as when a car goes downhill
plane with one slope onto a plane with another slope,
system on the car while it
force
The
and then moves onto a stretch of level road.
the level.
on
is
car
the
while
is
it
as
same
is going downhill is not the
from the
A particle moves along a parabolic curve, 2? = 4cy,
286. Example.
variable
a
by
upon
acted
is
It
origin to a point defined by the coordinates (a,b).
n are
and
m
where
nz?,
=
Q,
and
mx
=
Q,
force Q which is defined by the components
Q?
by
done
What is the work
constants.
the coordinates
Considering the particle in any position, defined by
So.ution.
(zy), Fig. 636, we see that
ve |Q cos 6 ds,
In this expression, the value of Q is
+ nit) = a(m? + n2x2)"?,
@ = 02+ O77)? = (nt
to find ds in terms of «.
We now need to use equation (q) of § 284
the equation of the curve 2? = 4cy, we get
Then,
dy
eS
or
2a dx = 4c dy,
a
21/2
1
1/2
ee )| dx —
=z) dx1 =
(1+ a4
ds — E + (34
2\1/2 didx
Ges 2 + 2?)1?
—
in Fig. 636 that @ = ¢ — ¥To find 6 as a function of x, we observe
Qo)
ann 3 = a
Differentiating
i
Also
and
tan ¢
Substituting these values in
tan @ = tan(¢ — 7) =
eEnd
tan @ — tan y
1 + tan ¢ tan 7’
x(2e+nm)
;
tan ¢ = —
na? — 2cm
of cos @ is
If this expression is tan 8, the value
A
cos 0 =
-
(4c? +
na? — 2cm
:
x?) 2m? +
—————*
neg?) 12
@ into the integral for the work, we find
Putting these values of Q, ds, and cos
ial
Ge ;
y¢ ix)
ey oe
:
Iom)(x da dx) = — : (a’n — 4 23cm),
—_ |, (nx 2 — cm) (x
U = 3e
nces in feet.
in pounds and the dista
where the unit is ft-lb. if Q 1s expressed
408
WORK, KINETIC ENERGY, POWER
287. Graphical
Representation
of Work.
Suppose
we
[Ch. XVIII
plot a curve
in
which the ordinate is the force F and the abscissa is displacement, Fig. 637.
Now consider a differential area between this curve and the z axis. Its
width is dz, its height is #', and its area is PFdx. The total area under the
curve between any two values 2; and zp is
Ze
(r)
Area = / F dz.
However, we recognize {F dx as the expression for the work done by the force F. Hence,
we conclude that the area under a curve whose
je)
5 Ra
a
coordinates are displacement and force repreR
=
ae
sents the work done by the force.
Engineers
Xo
=|
Displacement
find this knowledge very useful in a number
Fig. 637.
of circumstances.
In making tests on various
kinds of reciprocating engines, we often obtain
what is called an ¢ndicator diagram or indicator card (Fig. 638), which is a
record of the variation of the pressure p in the cylinder.
Since the area
A of the piston over which the pressure acts is constant, the indicator
diagram is also a record of the variation of the force F(= pA) on the piston.
F
Force,
se:
Atmospheric
Pressure
i
I,
ri}
s
Fig. 638.
Indicator
Card.
internal combustion
engine.
c to d, doing work.
In this diagram, the area abcda represents work done by
the gases,
This is a typical indicator
card taken from
and the area aefa represents work done on the gases
(usually considered
work in thermodynamics).
a slow-speed
The fuel begins to burn at b, and hot gases expand
from
to be negative
The negative work occurs because the piston must push
out
the exhaust gases and draw in a new charge.
by area abcda minus area aefa.
detail in heat power courses.
The net work of the engine is represented
Indicator cards and their uses are covered
in some
The scale of the area is the product of the scale of the
ordinate and the scale
of the abscissa.
Thus, if the ordinate scale is 1 in. = 300 Ib. and the
abscissa
scale is 1 in. = 1/2 ft., then one square inch of area
represents (300) (0.5) =
150 ft-lb., the scale of the area. After the area under
a curve or the area
enclosed within a force-displacement diagram has
been found, this area in
square inches times the scale of the area is the work done.
409
§ 288 |EXAMPLE
force, the
If the ordinate is pressure in pounds per square inch instead of
Let
area.
piston
of
inch
square
area of the diagram represents the work per
of
scale
the
Then
1 in. = 50 psi (ordinate) and 1 in. = 0.5 ft. (abscissa).
area, which means
the area is (50 psi)(0.5) = 25 ft-lb. per sq. in. of piston
of work for each
that each square inch of diagram area represents 25 ft-lb.
the diagram is 2
of
area
the
that
Thus, suppose
square inch of piston area.
work is then
total
sq. in. and that the area of the piston is 150 sq. in. The
(25)(2)(150) = 7500 ft-lb. as represented by the diagram.
we may someIn finding the area under a curve or within a closed curve,
that we can
such
are
d
involve
times integrate for it when the curve or curves
, the
feasible
not
is
readily express / as a function of x. If this procedure
by
area may be found by using a planimeter or
’s
Simpson
as
such
using some approximate rule,
R
rule.
isc
curve
F-x
the
spring,
a
for
Kx
=
F
Since
os
straight line whose slope is K, the scale of the
iso]
4
the
represents
639,
spring. The curve O-f, Fig.
area
force-deflection curve for a spring, so that any
by the
ie)
under this curve represents work done on or
|
[fsa x,
spring
a
in
stored
energy
the
e,
For exampl
spring.
Xe
in
its free
when it is compressed a distance %q from
639. Work of a Spring.
Fig.
—
Oab.
area
ular
length is represented by the triang
Since the ordinate ab = Kx,, we have
(s)
Kae
yes
U = Area Oab = 5(te)Ka
y
?
Observe in Fig. 639
ation.
which agrees with the result obtained by integr
hed from a deflection of ra to a
that if the spring is compressed or stretc
during this particular deflection is
deflection x, the work done on the spring
to be F,,,(%_ — %a), where F,, is
represented by the area defc, which is seen
2], or
the average force [F.y = (K% + Kzxa)/
Finn
Gin = Sen
(Sanita)
2
Tm
2
ee i Ate.
equation (s), we see that the work
Interpreting this expression in terms of
d to c is represented by
done in compressing the spring from
Area Ocf minus Area Ode = Area defc;
done on the spring (Ocf) minus
that is, this work (d to ¢) is the total work
(Ode).
the work that had already been done
in a steam engine is sometimes assumed to
988. Example. The expansion of steam
of
re in the cylinder times the displacement
follow the law px = C; that is, the pressu
410
WORK, KINETIC ENERGY, POWER
[Ch. XVIII
the piston is a constant.
This law is represented by the curve of Fig. 640, where we
see that the pressure falls from p, at a displacement x, to pa at a displacement of 2».
Let x. = 1 ft., pe = 100 psi, 2, = 3 ft. (= the stroke of the piston), and let the diameter of the piston be 20 in. For the assumption px = C, what is the work done by
the steam on the piston?
SoLution.
The area under the curve cd is fp dz,
where p = C/x. Using this value of p, we find
%
[ow=cf
~
sheds
Fig. 640.
at
z, 2
be
2
is
For the data given, C = P-€a = (100)(144)(1) = 14,400,
where the 144 converts pounds per square inch to
pounds per square foot. Hence the work per square
foot of piston area is
Clog. = = (14,400)log. 3 = 15,800 ft-lb. per sq. ft.
The area of the piston is rD?2/4 = 1(20/12)2/4 sq. ft.; hence the total work performed
by the expansion of the steam from c to } is
af 20\?
U = (2) (15,800) = 34,400 ft-lb.
289. Power.
Power is the time rate of doing work. A gasoline engine or
a steam turbine, for example, is capable of doing repeatedly
a particular
amount of work during each unit of time.
Thus the amount of work per-
formed by a prime mover depends at least in part on how
long it operates.
To say that an engine does, for example, 100,000 ft-lb. of work,
tells nothing
of its size or capabilities, since the work may have been performe
d in a second
or in a week.
Power units therefore invariably involve some time
unit.
The most common unit of power in English-speaking
countries is the horsepower.
If an engine delivers one horsepower, it is, by definition
, doing
work at the rate of 33,000 ft-lb. per min.
For electrical machinery and in countries using the
¢.g.s. system of units,
the power is often expressed in kilowatts (equal
to 1000 watts), where a
kilowatt is the same as 44,250 ft-lb. per min. Since
Power =
—.—.,
Work. = (Power) (Time).
If therefore we multiply a power unit by time,
we obtain a work unit. As
pointed out previously (§ 271), the work ‘units
of horsepower-hour or kilowatt-hour are convenient for large work quanti
ties.
Suppose the work done
is 1000 hp-hr. This work may be done by a 1000-h
p engine running for one
hour, by a 500-hp engine running for two hours,
or by a 3000-hp engine
§ 291 | EQUATIONS
All
FOR HORSEPOWER
running for 20 minutes (1/3 hr.).
equivalent to (see § 271).
At any rate, the 1000 hp-hr. of work is
(1000 hp-hr.)(1,980,000 Ae) = 1,980,000,000 ft-lb.,
or
Btu
(1000 hp-hr.)( 2545 hp-hr. )= 2,545,000 Btu.
290. Conversion Factors.
the following conversions:
:
Some
horsepower
are
shown
in
ft-lb. \ _ ft-lb.
ft-lb. \ _ ft-lb.
cnp)(33,000a)
equivalents
(np)(950 —
Sram
~ sec.’
2
Btu \ _, Btu
2545 a
(np) (
~ Wai
In the generation of electricity, the kilowatt is the common
kw
unit of power.
See also § 271 for other conversion equivalents.
When the need arises, the engineer
291. Equations for Horsepower.
its definition if the work
should, of course, be able to find horsepower from
Hence the reader should
done in a unit of time can be readily calculated.
g to a method of
look upon the discussion in this article as simply pointin
reasoning.
point of application
Suppose that a constant force of F lb. does work, its
If the distance s is
F.
of
moving through a distance s ft. in the direction
application of F is ym = s/1
covered in one minute, the speed of the point of
Similarly, if the speed of the
fpm and the work done is Fv», ft-lb. per min.
at the rate of Fv, ft-lb. per
point of application of F is v, fps, the work is
sec.; hence
ji
(t)
Fo,
hp = 33000 — 550
is represented by 60 ft-lb.,
We recall that work done on a rotating body
ounds and @ is the angle
foot-p
in
where M is the constant torque on the body
d (§ 276). If this angle @ is
in radians through which the torque is applie
, the corresponding angular
turned through in one second or one minute
per min.; hence
velocities are w, rad. per sec. and wm rad.
Ma, is work in foot-pounds per second and
Mw,» is work in foot-pounds per minute and
(u)
Mas
Mom
ED lags (yas 33,000
412
WORK, KINETIC ENERGY, POWER [Ch. XVIII
The work done on a rotating body may also be determined from a tangential
force. Suppose a constant driving force of F lb. always acts tangent to a
circular path in a rotating body at a constant radius
of r ft., Fig. 641. The work done in one revolution
is (Force)(Distance) = F(2zr) ft-lb.
If the body
makes n,, revolutions per minute, the work done in
one minute is F(2 xr)n, ft-lb. per min.; or if it makes
nm; revolutions per second, the work is F(2 zr)n,
ft-lb. per sec.; thus
ane
Fig. 641.
20
= 33,000
2rrn,f
550
M (277m)
33,000 ’
where M = Fr is the torque in foot-pounds.
In a hydroelectric power plant, some of the po-
For a practical
view, consider that
Aand
Rats gears, with gear
A driving gear B, and exae A seeneaa forces
tential energy of water in a reservoir is converted
:
:
.
into kinetic energy by the fall of the water through
pipes to a hydraulic turbine.
Then part of the
kinetic energy of the water as it enters the turbine
is converted into work done on the turbine shaft,
which is finally converted into electrical energy.
The amount of water
which reaches the turbine is generally measured in cubic feet per some unit
of time or in pounds per unit of time. Suppose that V cu. ft. per min. (cfm)
of water arrives at a turbine.
If the water weighs 62.4 lb. per cu. ft., the
weight of water is 62.4V lb. per min. After falling a distance of h ft., the
water has a kinetic energy equal to the loss of potential energy, in the ideal
case where there are no frictional losses.
Thus, with a loss of potential
energy of 62.4Vh ft-lb. per min., the horsepower developed in the
ideal
turbine is
_ 62.4Vh
hp
33,000
We not only use the horsepower and other power units to express
the rate
at which work is done by an engine which generates power,
but we also often
use such units to express the rate at which work is lost ; for
example, in a
brake (see § 294 and see Fig. 642).
292. Efficiency. The mechanical efficiency is a term
used to express the
losses that occur in machines due to friction between
parts which have >
relative motion.
The single word efficiency has a variety of technical
meanings and is generally defined by its context, or by such
phrases as indicated
thermal efficiency.
Detailed knowledge of thermal efficiencies is gained
from
works on thermodynamics.
In this book, we shall sometimes use the simple
term efficiency e, defined by the equation
Output
(w)
ioe or [output and input in energy units],
418
§ 293 |EXAMPLE
kinds
an expression that can be interpreted to apply to most of the various
by
done
work
of
amount
the
hoist,
of efficiencies. For example, in a geared
only
not
hoist)
the
the operator (or electric motor or other engine driving
supportraises the load, but also overcomes the frictional losses in the bearings
That is,
ing the gears and sheaves and the losses in the meshing gear teeth.
the
done,
work
useful
the
than
greater
the input work of the operator is
driving
the
on
operator
the
The work of
difference being the frictional work.
done on the body being moved is
work
the
into
divided
hoist
the
of
shaft
termed the efficiency of the hoist.
the brake horseA typical setup of a prony brake for measuring
Fig. 642. Prony Brake.
in the upper right-hand corner
wheel
ng
clampi
the
e
Observ
engine.
oil
power of a small
and, on the left side of the illustration, observe
which clamps the brake to the brake drum;
the end of the brake arm when the engine
at
the scales which weigh the force exerted
is running.
is defined as the work delivered
The mechanical efficiency of a steam engine
d by the work done by the steam
by the shaft (called the brake work) divide
. The ratio of the brake horsein the cylinder (called the indicated work)
(ihp) gives the same result; thus,
power (bhp) and the indicated horsepower
ency is
for a steam engine, the mechanical effici
é
~
Brake work
Indicated work
_ bhp
— ihp-
hp from its crankshaft.
An automobile engine is delivering 100
293. Example.
wheel bearings, is 80%.
ing
, ete., includ
The efficiency of the transmission, differential
What is the tractive
er.
diamet
in
in.
are 30
The car is going 60 mph and the wheels
road?
force exerted between the wheels and
41h
SoLution.
WORK, KINETIC ENERGY, POWER [Ch.
XVIII
The output (to the wheels) is found from
_ Output _ oe
Input
or the output is 80 hp.
Output
lOO
At 60 mph = 88 fps, the tractive force F does work
U = Fv, = 88F ft-lb. per sec.
The horsepower corresponding to this work per second is 88F /550, which is equal
to 80 hp as found above; that is,
a 2 88F
550’
from which F = 500 lb.
F
es
Fig. 643.
294. Example.
Fig. 644.
A single-block brake (Fig. 643) is used to lower a load at a constant
speed with the 30-in. brake-drum B which turns at 60 rpm. If the foreeQ
= 120 lb.,
a = 4in., b = 20 in., c = 50 in., and the coefficient of kinetic friction
f = 1/3, what
is the frictional horsepower (fhp)?
So.ution.
Observing that N = F'/f = 3F and taking moments
pivot A in the free body of the lever, Fig. 648, we find
=M4 = Fa
—-Nb+Qc
about the fixed
=0
(F)(4) — (3F)(20) + (120)(50) = 0,
from which, F = 107.1 lb.
The peripheral speed of the drum is
30
Um = a7Dn = “(33Joo = 471 fpm.
The frictional loss in horsepower is
Fum
INP = 33 900
}
=
(107.1) (471)
=
33,000
= 1.538 hp.
The brake either must be able to rid itself of heat at this rate
(1.53 X 2545 Btu per hr.)
without overheating, or it must be used intermittently.
Brakes are usually rated in
terms of the braking torque; in this case, the braking
torque exerted is
My; = (107.1)(15) = 1610 in-Ib. = 134 ft-lb.
295. Example. A belt is transmitting 300 hp
at a belt speed of 2400 fpm. If the
coefficient of friction between the belt and the
pulley is f = 0.4 and if the angle of
contact is @ = 180°, what is the value of the tight
tension F,? Consider that the belt
415
§ 296 |CLOSURE
is on the point of slipping and that the speed of the belt is low enough that the centrifugal force of the belt may be neglected with little error.
The net driving force on the pulley is Fi — Fs (Fig. 644), the value of
Sotution.
which is found from the horsepower equation hp = Fv,/33,000. This gives
(x)
F=F,-—F,
__ 33,000 hp _ (33,000) (300)
‘ieee
2400
= 4125 lb.
the
Since the centrifugal force is negligible and slipping is imminent, we may use
relation F,/F2 = ef? (§ 71), and find
Free Fy
Fy
Fy
ef8 —— @0-4)(m)
Bro
belt is
where x radians are equivalent to 180° and where it is understood that the
get
we
(x),
equation
in
F.
of
value
this
Using
pulley.
the
on
in limiting friction
from which the tight tension is F; = 5780 lb.
In applying the principle U,.. = AKE, the beginner must
296. Closure.
and
be particularly careful (1) that he has accounted for the work of each
correctly
has
he
that
(2)
every force in a system in obtaining the net work,
and
determined “the distance through which each force acts” in doing work,
In finding the
(3) that he has included all changes of kinetic energies.
conveniently
may
we
motion,
plane
in
change of kinetic energy of a body
case
which
in
point,
use either the center of gravity as a reference
We
KE = 7
(52)
. Iw?
ee
Or we may use the instantaneous center, in which case
Tw
(53)
:
Ki
=
If a wheel
.
where the subscript o refers to the axis of instantaneous rotation
contact.
of
point
slips on a surface, the instantaneous center is not at the
Iw?/2 represent
It is convenient to memorize that the terms mv?/2 and
We
Then it is a matter of understanding their application.
kinetic energy.
:
might write
vn (8) +0(%)
2
Ik
2
If work
energy.
and apply this to all situations regarding work and kinetic
kinetic
of
the change
is done on a system of bodies, this work is equal to
may be computed
energy of the system and the change of kinetic energy
tion (w = 0),
transla
in
is
body
a
If
separately for each body in the system.
equation (y) reduces to
mv?
Fons (tt
[TRANSLATION]
416
PROBLEMS: TRANSLATION [Ch. XVIII
as found in § 270. If a body is rotating about a fixed center, the term
A(mv?/2) for translation drops out of (y) to give
th 2
OR
=
A
(¥)
[ROTATION]
which agrees with (51) of § 277 when J is taken with respect to the center of
rotation.
Then for general plane motion, the right-hand side of (y) is taken
in accordance with (52). Also, we need to know that U = fF ds and
U = JM dé, from which, if fF and M are constant, we get U = Fs and
U = Me.
Our modern concept of work (and energy) evolved rather slowly. As
typical in the early stages of the development of any science, there were
vocabulary difficulties. Galileo variously called the quantity wv momentum,
impulse, or energy.
Descartes called wv the quantity of motion.
Newton,
with the conception of mass, called mv the quantity of motion (now called
momentum).
Leibnitz called mv? vis viva or living force. Coriolis decided
to call mv?/2 vis viva. Later, to avoid confusion, Belanger proposed the
term living power for mv?/2, which is now called kinetic energy.
Coriolis did
use the name work for F's. Newton used the word force in its modern sense.
The writer has not learned just when the word power took on its present day
meaning, but students (and others) still have difficulty in keeping clear
conceptions in their minds of such terms as force and moment, work and power.
Problems
notE.
The following problems are to be solved using work and energy principles.
TRANSLATION
1381. In Fig. 645,
W = 100 lb., Q =
50 Ib., 6 = 30°, f = 1/4, and the body is
moving with an initial velocity of 28 fps.
For a displacement of 20 ft. to the right,
find the resultant force on the body and
the net work done.
Check your answer by
adding algebraically the work done by each
force.
Ans. 24.5 lb., 491 ft-lb.
1383. In Fig. 646, W = 100 lb., @ = 30°,
Q = 120 lb., 6 = 15°, f = 1/4, and the
body is displaced 8 ft. up the incline. Find
the resultant force on W and the net work
done.
Check your answer by adding algebraically the work done by each force. Does
the velocity increase or decrease?
Ans. 52 lb., 416 ft-lb.; increases.
1382. The same as 1381 except that Q =
10 lb. What does a negative answer for
work mean?
y
q
—le-
Fig. 645.
Problems 1381, 1382.
Fig. 646.
Problems 1383-1387.
PROBLEMS:
iy
TRANSLATION
1384. The same as 1383 except that
ij =O;
1385. The same as 1383 except that
W = 200 lb.
1386. In Fig. 646, Q = 46 lb., 6B = 0,
shipping platform from a second floor which
These cases are expected
is 12 ft. above.
to be given an initial speed of about 2 fps
down the chute and is is desired that they
shall move down the chute, then out on
the level for a distance of about 10 ft., at
which point they are to come to rest under
the action of friction alone. The coefficient
of friction between the chute and cases is
expected to be about f = 0.3. Neglecting
the effect of a short curved path between
the incline and the horizontal, determine
the constant slope of the incline.
Ans. 21.65°
1396. The same as 1395 except that
f = 0.5.
6 = 15°, f = 0.3, and the net work is —200
ft-lb. as the body is displaced 12 ft. to the
right. Find the weight of the body.
Ans. 114 lb.
1387. The same as 1386 except that the
net work done is +200 ft-lb.
1388. A 900-Ilb. body lies on a 30° incline
for which the kinetic coefficient of friction
is f = 0.1. What is the least force that
will keep the body in motion up the plane?
What is the work done by this force in a
distance of 250 ft.? With this force acting,
what is the change in velocity of the body?
Ans. 528 lb., 132,000 ft-lb., no change in v.
1389. The same as 1388 except that
ie Ol25
4
1390. A 3000-ton train has a resistance of
14 Ib. per ton and a drawbar pull of 135,000
Ib. produced by a Virginian steam locoFig. 647.
Problems 1397, 1398.
With what minimum speed must
motive.
the train “hit”? the foot of a 1-mile long,
1.5% grade to clear the top with a speed of
1397)In Fig. 647, f = 1/3 for both planes,
80 fps? Would you suggest a smaller locoh>=-10 ft., and the box comes to rest when
s
motive or the use of two of these locomotive
s = 18 ft. Find the initial velocity.
to attain the objective of 80 fps?
Ans. 5.35 fps.
1391. An-ore car, weighing 18 tons, is
1398. In Fig. 647, W = 64.4 lb.,f = 1/3,
~—_loaded-with 50 tons of ore. The total
the initial velocity is v. = 10 fps, and the
resistance to motion of the car is represented
box comes to rest when s = 18 ft. Find
by a force of 500 lb. parallel to the track.
the initial height h, and the kinetic energy
If the car is on a level track, what constant
at the foot of the incline.
drawbar pull is necessary to change its
of
distance
a
in
fps
15
to
fps
speed from 5
200 ft.2 The kinetic energy of rotation_of
Ans. 2610 lb. )
the wheels is to be neglected.
1392. The same as 1391 except that theof
car is being moved up a 2% grade.
1393. The same as 1391 except that the
car is being moved down a 2% grade.
1394. A, freight car becomes uncoupled
from a train which is moving up a 1.5%
Problems 1399-1401.
Fig. 648.
grade at a speed of 10 fps. It is acted
of
motion
upon by a constant resistance to
(a) What is its
1399. In Fig. 648, Wa = 1000 lb., fa =
10 lb. per ton of weight.
1-mile
point
a
fs = 0.6, and the guide C for the
s
0.15,
reache
it
speed in mph when
The bodies A
it
weightless cable is smooth.
down grade from the point at which
and B are moving leftward with an initial
(b) If the car moves
became uncoupled?
is
speed of 20 fps with the cable taut. After
onto a level track after this 1 mile
level
the
on
go
it
these bodies have each moved 160 ft.,
does
traveled, how far
is
neither one changing its direction, their
before it comes to rest? Its motion
n.
speed is 10 fps. (a) What is the weight
frictio
for
except
ructed
unobst
(b) What are the
Wap of the body B?
Ans. (a) 40 mph; (b) 2.04 mi.
Use
and BC?
AC
cables
the
in
tensions
1395. It is desired to design a steel chute
only energy methods in solving this part.
a
to
cases
ng
packi
r
delive
to be used to
eee
418
PROBLEMS: TRANSLATION
Ans. (a) 1100 lb.; (b) 627 lb.; (¢) —113,000
ft-lb.
1400. The
same
as
1399
except
that
and then slides 50 ft. along the floor.
Cig fp = 011.
Problems 1402, 1403.
1402. In Fig. 649, Wa = 1000 lb., f =
1/3, and the pulleys C and D are to be
considered frictionless and weightless. (a)
If A moves 60 ft. from rest up the incline
14 ft. in diameter,
in 12 sec., what is the weight Wes of the
the change of potential energy of A?
(a) 430
Ib.;
(b) 815
Of B?
lb., 407.5
lb.;
(c) 30,000 ft-lb., —51,500 ft-lb.
ROTATION
ABOUT
1407. The driver of a car turns a 17-in.
steering wheel through 90°. Each hand
exerts a constant tangential force of 10 lb.
at the perimeter of the wheel.
What is the
work done by the hands?
Ans. 267 in-lb.
1408. To speed up a flywheel, a constant
net torque of 1200 in-lb. is applied through
8 revolutions.
What net work is done?
1409. A screw jack with a 1.5-in. square
thread is used to raise a load of 2000 lb.
through a distance of 6 in. The mean
diameter of the screw is 1.35 in., there are
3 single threads per inch, and the coefficient
of friction is f = 0.1. What torque must
be applied to the serew?
How much work
is done?
See § 74.
Ans, 244 in-lb., 2290 ft-lb.
1410. The same as 1409 except that
if = O15,
1411. A 500-lb. flywheel with a radius of
gyration of 18 in. must be speeded up from
148 rpm to 150 rpm in 0.7 of a turn.
What
constant torque must be applied?
below.
The
diameter
of the Jarger base is 30 ft. and the depth
of the tank is 20 ft. The water in the full
cistern is to be pumped to a level 30 ft.
above the bottom of the cistern.
The friction head is equivalent to 8 ft. of lift.
What is the work done in horsepower-hours?
body B?
(b) What are the forces on the
cables attached to A and B?
(c) What is
Ans.
(a)
What happened to the initial potential
energy?
(b) What happened to the initial
kinetic energy?
(c) Find the coefficient of
kinetic friction between the book and the
floor. State the assumptions under which
your solution is valid.
1405. A water tank in the shape of a
hemispherical shell, 96 ft. in diameter, is to
be filled with water from a lake whose
surface is 120 ft. below the top surface of
the tank. The intake is at the bottom of
the tank and the connecting pipe is 12 in.
in diameter.
Assume that there is no loss
due to friction and that the water weighs
62.4 lb. per cu. ft. What is the work done
in filling the tank?
Ans. 745 hp-hr.
1406. A water cistern is in the form of a
frustum of a cone with the smaller base,
1401. The same as 1399 except that the
guide C is not smooth.
The value of f at
Fig. 649.
XVIII
1403. The same as 1402 except that the
displacement of A is 48 ft. down the incline.
1404. A book is thrown horizontally at a
height of 4 ft. above a floor. It travels
20 ft. horizontally before striking the floor,
(c) What is the change of potential energy
of the system?
f, oO
[Ch.
CENTER
OF
GRAVITY
1412) The rotor of a steam turbine weighs
2400-4b. and has a radius of gyration of 15 in.
It is supported in bearings 10 in. in diameter
for which the coefficient of friction is —
0.007.
The steam is shut off while the
turbine ‘is rotating at 1800 rpm.
If there
is no resistance to rotation except the frictional forces in the bearings, how many
turns does the rotor make before coming
to rest? How long does it take?
(Actually, the friction of a connected generator,
the fluid friction of fanning the steam in
the turbine casing, ete., would result in a
much shorter time.)
Ans. 47,100 rev., 52.4 min.
_ 1413. The
k= a
same
as
1412
except
that
€ 1414.) The weight of the rotating drum
assembly B, Fig. 650, is 2576 lb. and its
radius of gyration with respect to the axis
of rotation is 14 in. The weight W is
suspended from a cable which wraps about
the D = 32 in. diameter.
While W moves
downward through a distance of 40 ft., the
e
PROBLEMS:
ROTATION
ABOUT
CENTER
A19
OF GRAVITY
speed. of the drum is increased from 20 rpm
to 40 rpm. If the frictional effects are
negligible, what is the weight Ww?
Ans. 18 lb.
moved
10 ft. from rest, what is the speed
of A and of B? (b) What is the acceleration
of A and of B, and the angular acceleration
of C? (c) What is the change of potential
energy of the system?
Ans. (a) 1.59 fps, 4.77 fps; (b) 0.379 fps’,
1.132 fps?, 0.252
rad. per sec.?; (ec) — 167
ft-lb.
1419. The same
We = 400 lb.
Fig. 650.
as
1418
except
that
Problems 1414-1417.
1415. A 193.2-Ib. body W, Fig. 650, is
suspended from a cable which wraps about
weighs
a D = 32-in. drum B. The drum
on of
gyrati
of
radius
a
has
and
Ib.
2576
(a) After
14 in. with respect to its axis.
is the
W has moved 20 ft. from rest, what
(b) What
angular velocity of the drum?
is the acceleration of W?
bly
1416. In Fig. 650, the rotating assem
t
weigh
the
and
ft.,
2
=
D
lb.,
200
s
weigh
Neglect friction and the mass
W =32.2lb.
If W is released from rest
of the cable.
(a) the
and descends 20 ft. in 4 sec., find
of
radius
the
(b)
and
cable
the
in
tension
bly.
assem
ng
rotati
gyration of the
Ans. (a) 29.7 lb.; (b) 1.38 ft.
1417. The same as 1416 except that the
rotating assembly weighs 161 lb.
|
|
Fig. 651.
Problems 1420-1422.
1420. In Fig. 652, the rotating elements
B, which weigh 1288 lb. and have a radius
120
of gyration of k = 2.5 ft., are turning
While it moves 80 ft. downward, the
rpm.
the
278-Ib. weight A is brought to rest by
constant frictional force at the brake shoe
C, where f = 1/3. The shape of the brake
arm is such that e = 0. What is the value
of the force Q applying the brake?
Ans. 84 lb.
that the
except
1421. The same as 1420
brake
weight A is moving upward when the
d.
is applie
that
1422. The same as 1420 except
Ib.
69.8
Ans.
é = on:
Fig. 653.
Problems 1418, 1419.
A and B
1418. In Fig. 651, the bodies
150 lb., and
weigh Wa = 500 |b., We =
C weighs
D =9 ft. The rotating part
ion of 3 ft.
600 lb. and has a radius of gyrat
with respect to its axis.
Fig. 652.
(a) After B has
Problem 1423.
1423. A 128.8-lb. body A, Fig. 653, is on
a
a 9 = 15° incline where f = 0.15. At
er
certain instant, the solid cast-iron cylind
A is
B is rotating at 40 rpm and the block
the
of
e
virtur
by
incline
the
up
moved
being
The cable CB
cable connection shown.
A
wraps around the 2-ft. cylinder B. After
420
PROBLEMS: ROTATION ABOUT CENTER OF GRavity [Ch.
XVIII
moves 20 ft. up the incline, it comes to rest.
What is the weight of the cylinder B?
Neglect the axial friction for B and C and
the mass of the pulley C. See problem 1293.
Fig. 655.
Fig. 654,
Problems 1426, 1427.
Problems 1424, 1425.
1424. The drum D, the brake wheel C,
and their supporting shaft, Fig. 654, weigh
2576 lb. and have a radius of gyration of
k =3 ft. The 2000-lb. body A is moving
down the incline, where f4 = 1/3, at a
speed of 40 fps. The flexible brake band
subtends 180° on the brake wheel, where the
coefficient of friction is fe = 0.25.
When
the brake is applied by a constant force Q,
the speed of A decreases to 10 fps while it
moves 170 ft. What is the magnitude of
Q when a = 30 in.? The weight of the
cable is negligible.
AMS llialine
1425. The same as 1424 except that the
initial speed of A is 80 fps.
1426. Figure 655 represents diagrammatically a traction drive for an elevator, in
which A is a 6000-Ib. cage and B is a 5000Ib. counterweight.
The cable from the cage
passes over the driving sheave C, thence
around the idler sheave D, back around Or
and thence to the counterweight.
The
ROTATION
NOT
power is delivered to the axis of the driving
sheave C. Each sheave is 30 in. in diameter
and each has a moment of inertia of 5
slug-ft.2 about its axis. If the elevator
attains an upward speed of 10 fps in a
distance of 10 ft. with constant acceleration,
what torque must be applied to the axis
of C? What is the change of potential
energy of the system?
Ans. 3420 ft-lb., +10,000 ft-lb.
1427. The same as 1426 except that the
elevator attains a speed of 5 fps.
1428. Steam leaves a nozzle at a speed
of 2000 fps and enters the blades of a turbine. The rate of discharge is 0.5 lb. per sec.
The rotor carrying the blades rotates at
1800 rpm.
If all of the kinetic energy of
the steam is converted into work on the
shaft of the turbine, what is the torque
exerted on the shaft?
Ans. 165 ft-lb.
1429, The same as 1428 except that the
velocity of the steam leaving the turbine
is 500 fps. Other Josses are to be neglected.
ABOUT
CENTER
OF
GRAVITY
1430. A slender rod, which weighs 16.1
Ih. andis L = 4 ft. long, is pivoted at one
end. It rotates from rest under the action
of gravity only, starting from a vertical
position, Fig. 656. (a) What is the speed of
its center
of gravity
after
it has
turned
through @ = 120°?
(b) What is its kinetic
energy at this instant?
(c) What is its
change of potential energy?
Ans. (a) 12.04 fps; (b) 48.3 ft-lb.; (ce)
48.3 ft-lb.
1431. A homogeneous slender rod of
length L is pivoted at one end in a vertical
position, Fig. 656. In this position, its
angular velocity is w, = 0. Derive an ex-
Fig. 656.
Problems 1430-1432,
ABOUT
CENTER
pression for 52 when @ = 120°,
being forced by gravity alone.
rotation
PROBLEMS:
NOT
ROTATION
1432. The same
@> = 4 rad. per sec.
421
OF GRAVITY
1434. The same as 1433 except that D
weighs 322 lb.
Ans. 32 = 3(2gL)?/4.
as 1431 except that
1433. In Fig. 657, the uniform
bar A
weighs Wa = 48.3 lb., the drum B weighs
128.8 lb. and has a radius of gyration k =
sheave
10 in., the
C and
cable
the
are
considered weightless, the body D weighs
If
644 lb., and all friction is neglected.
bar A is released from rest in the horizontal
position shown, (a) how much kinetic energy does the system have when A strikes
the stop Q? (b) What is the velocity of
Fig. 657.
D when impact occurs?
ROTATION
TRANSLATION
AND
the cylinder? What coefficient of friction
is necessary for rolling?
1435.)A solid homogeneous cylinder, 16
jn-in diameter and weighing 322 Jb., rolls
on
a
rough
horizontal
plane
action of a constant horizontal
under
Problems 1433, 1434.
Ans.
the
(a) 8 fps;
force OQ) =
100 Ib., which acts through the center of
If the initial speed of its cg is 5
gravity.
fps, what is its speed after it has moved
20 ft.2 What is the acceleration of the cg?
Ans. 17.1 fps, 6.67 fps?.
1436. The same as 1435 except that the
rolling body is a solid homogeneous sphere.
1437. A solid homogeneous cylinder, 24
in. in diameter and weighing 483 lb., rolls
(b) 4.26
rad.
per
sec.?;
(c) 5.35 lb., 0.0384.
as 1489 except that
1440. The same
Q = 224.5 |b.
1441. The same as 1439 except that the
rolling body A is a sphere.
Ans. (a) 8.3 fps; (b) 4.6 rad. per sec.’;
(ec) 4.6 lb., 0.033.
down a 30° rough incline under the action
of gravity. If it starts from rest, how far
has it moved when the speed of its center
Ans. 29.1 ft.
of gravity is 25 fps?
1438. The same as 1437 except that the
rolling body is a cylindrical thin shell.
0
Problems 1439-1441.
in a horizontal position.
1439. The body A, Fig. 658, is a 161-lb.
It is being
cylinder, 12 in. in diameter.
rolled up the incline, where 6 = 30°, by a
constant
force
Q = 96.5 lb.
(a) What
Problems 1442, 1443, 1444.
1442. In Fig. 659, the body A is a solid
homogeneous cylinder with a weightless
One
cord wrapped about its midsection.
end of the cord is attached to a fixed surface
at B. If the cylinder is released from rest
in the position shown and moves vertically
downward, what is the speed of its cg after
a displacement of 15 ft.? The axis remains
Q
Fig. 658.
Fig. 659.
is
a
the speed of its center of gravity after
(b) What
displacement of 15 ft. from rest?
(c) What is
is its angular acceleration?
the frictional force between the plane and
What is its change
of potential energy?
1443. The same as 1442 except that A
is a solid homogeneous sphere.
Ans. 26.2 fps, —15W.
1444. The same as 1442 except that A is
a hollow eylinder with a thin shell.
21445.A 322-lb. grooved cylinder,
Fig.
660, is rolled toward the right by a constant
PROBLEMS: ROTATION AND TRANSLATION
Fig. 660.
[Ch. XVIII
Problems 1445, 1446.
force @ until the displacement of its cg is
54 ft. The force Q acts on a weightless
cord along the line AQ which. wraps about
a groove whose diameter is D; = 18 in.
The diameter of the cylinder is Dz = 36 in.
and its moment of inertia is J = 10 slug-ft.?
If the motion starts from rest and if the
final speed is 45 fps, what is the magnitude
of Q? There is no force along the line BE.
Ans. 181 lb.
Fig. 662.
Problems 1448, 1449,
1449. The same as 1448 except that the
sheave C has a moment of inertia J¢ = 0.3
slug-ft.?
;
Ans. (a) 10.5 fps, 11 fps?; (b) 32.9 lb.
1446. The same as 1445 except that the
force Q acts along the line BE and the
cord wraps around the groove in a clockwise direction.
There is no force along
the line AQ.
Fig. 663.
orn
_
ae
Fig. 661.
it
uae
z
ee
——
Problem 1447.
1447, The body shown in Fig. 661 rolls
on the part with a D, = 2-ft. diameter.
A
cord wraps about the D, = 4-ft. diameter,
as shown.
The force Q = 160 Ib., the body
weighs 644 lb., and the moment of inertia
I = 12 slug-ft.2.
What is the speed of C
after C has been displaced 10 ft., the body
having started from rest?
Ans. 10 fps.
1448. A disk A, Fig. 662, has a weightless
cord wrapped about its midsection.
This
cord passes over a frictionless and weightless
sheave C, and thence downward to a 50-lb.
weight B.
Let W4 = 80]b., 0 = 30°. I, =
4 slug-ft.?, and let the displacement of B
be 20 ft. (a) If the system starts from
rest, determine the final speed of the cg of
A and the acceleration of B. (b) What is
the tension in the cord?
Problems 1450, 1451.
1450. In Fig. 663, the grooved cylinder
A weighs 200 lb. and has a moment of
inertia of 14 = 6 slug-ft.2 Let D, = 2 hte,
D, = 3 ft., We = 32.2 lb., and f. = 0 (that
is, the fixed peg C is smooth).
(a) Determine the speed of the eg of A and the acceleration of B after B has moved downward
through 20 ft. (b) What is the tension in
the cord?
(ce) How would the acceleration
of the cg of A vary as the diameter D,
increases, other conditions remaining the
same?
Ans. (a) 11.95 fps, 0.397 fps?; (b) 31.8 lb.;
(c) inversely as Dj.
1451. The same as 1450 except that the
peg C is not smooth and f, = 0.3.
1452. A 2-ft. diameter sphere whose mass
is 5 slugs travels 10 ft. from rest down a
100% grade for which the coefficient of
friction is 0.3. Find the work of friction
and the final kinetic energy. Did the sphere
roll or roll and slide?
Ans. KE, = 797 ft-lb.; rolls and slides.
1453. The same as 1462 except that
the grade is 10%.
PROBLEMS:
SPRINGS—VARIABLE
423
FORCES
SPRINGS—VARIABLE
1454. A force of 100 lb. is required to
compress a spring 4 in. Sketch a forcedisplacement diagram and use it in finding
(a) the scale
(or modulus)
of the spring,
(b) the work and maximum
force required
to compress the spring a total of 6 in., and
(c) the potential energy stored in the spring
when it is compressed 4 in.
Ans. (a) 25 lb. per in.; (b) 450 in-lb., 150 lb.;
(ec) 200 in-lb.
1455. A loaded freight car,weighing
80,000 Ib. and moving on a horizontal track
at 3 mph, strikes a nest of springs on a
bumper post. The car’s brakes are applied
so that the total resistance to motion is
If
constant at 300 lb. per ton of weight.
the springs are compressed 2 in. in bringing
the car to rest, what is the combined scale
How far back does the car
of the springs?
Neglect the kinetic
move on the rebound?
energy of rotation of the car’s wheels and
the effect of the mass of the springs.
1456. A 100-lb. body falls 24 in. from rest
and strikes the free end of a helical spring
whose scale is 30 lb. per in. and whose axis
is vertical.
Find
(a) the maximum
com-
pression of the spring, (b) the maximum
kinetic energy of the falling body, and (c)
the velocity of the body when the spring is
compressed 12 in. Neglect friction and
the mass of the spring. Sketch a velocitydisplacement curve for the motion of the
body from the moment it contacts the
spring until the maximum compression of
the spring is reached.
Ans. (a) 16.43 in.; (b) 214 ft-lb.; (c) 8.8 fps.
1457. A helical spring whose scale is
1000 Ib. per in. is compressed 4 in. with its
axis in a vertical position. A body weighing
20 Ib. is placed on the compressed spring,
Neglecting air
which is then released.
resistance and assuming that the spring acts
on the 20-lb. body only until it regains its
free length, determine the kinetic energy of
the 20-lb. body at the instant the spring
has regained its free length, and determine
the height to which the body rises as
measured from its original position.
Ans. 660 ft-lb., 400 in.
1458. A 40-Ib. body falls freely and strikes
whose scale is 5000 Ib. per a, Abi
spring
a
the spring is compressed 6 in., through
what total vertical distance does the 40-lb.
body move before it is brought to rest?
What is the maximum kinetic energy of the
body during this period?
Ans. 187.5 ft., 7480 ft-lb.
FORCES
1459. The same as 1458 except that the
scale of the spring is 40 Ib. per in.
Fig. 664.
Problems 1460-1462.
1460. In Fig. 664, a coiled spring A,
whose scale is 30 lb. per in., is compressed
10 in. against a fixed surface with a 90-lb.
body B at the free end.
The coefficient of
friction between B and the plane is f = 0.2,
and 6 = 30°. When it is released, the
spring acts on B until the free length is
reached.
How far up the incline from the
point of release does B go?
movement,
what
During this
is B’s maximum
kinetic
energy?
Ans. 1.23 ft., 114.8 ft-lb.
1461. A 100-lb. body B slides from rest
down a 6 = 45° incline, where f = 0.2, for a
distance of 10 ft. At this instant, it strikes
a spring A, Fig. 664, which is compressed
3 in. before B comes to rest. Determine
(a) the scale of spring, (b) the maximum
kinetic energy of the body during this
period, and (c) the distance the block will
For
go up the plane on the rebound.
uniformity, let zero displacement be at the
upper end of the free spring.
Ans. (a) 1547 Ib. per in.; (b) 565.6 ft-lb.,
(c) 6.59 ft.
that
except
1461
as
same
The
1462.
= ao,
1463. A coil spring has a scale of 70 lb.
per in. From an initial deflection of 2 in.,
Integrate
it is compressed 1.5 in. more.
for the addional work done on the spring,
and make a drawing to show an area which
Ans. 288 in-lb.
represents this work.
1464. From an initial deflection of 2 in.,
a coiled tension spring is stretched an additional 3 in. The work necessary for this
action is 60,000 ft-lb.
Determine the scale
of the spring by integration and check by
using a force-deflection diagram.
1465. The work done upon a coiled spring
in deflecting it 0.7 in. from an initial deflection of 0.3 in. is 500 in-Ib. If an additional
1500 in-Ib. of work is done upon the spring,
determine the final total deflection and the
magnitude of the final force.
Ans. 1.93 in., 2120 lb.
1466. A gun weighing 200,000 lb. has an
initial speed of recoil of 10 fps. The recoil
PROBLEMS: SPRINGS—VARIABLE
424
is resisted by a nest of springs whose scale
is 50,000 lb. per in. What is the distance
of recoil? The gun moves in a horizontal
plane.
1467. A 900-lb. body moves on a horizon-
FORCES
[Ch.
XVIII
1474. The same as 1473 except that Q is
variable and Q = 10s + 10, where s is the
displacement in feet of the point of application of Q. A solution by one method only
is required.
Ans. 715 ft-lb.
tal plane, where f = 0.25, under the action
of a horizontal force Q = 3x? + 250 Ib.,
where x is the displacement in feet. If
the body starts from rest, determine the
net work and the speed of the body after a
displacement of 10 ft.
Ans. 1250 ft-lb., 9.45 fps.
1468. A 500-lb. body moves up a 30°
incline, where f = 0.2, under the action of
a force Q = 2x? + 42 + 400 lb. which is
directed upward and parallel to the plane.
The variable z is the displacement in feet
parallel to the plane. If the body starts
from rest, determine the net work and the
speed of the body after a displacement of
20 ft.
Ans. 6712 ft-lb., 29.4 fps.
1469. The same as 1468 except that the
500-lb. body is moving down the incline
and the force Q is directed downward.
1470. A particle moves along the parabolic curve x? = 16y from the origin to
the point where x = 4 ft. It is acted upon
by a resultant force, Q, = 22 lb. (Q, = 0.)
What work is done?
(Solve by the methods
§ 286. Do you see a simpler solution?)
Ans. 16 ft-lb.
1471. The force on a pneumatic device
varies according to Fs°-8§ = C. If a force
of 100 lb. is required when s; = 1 in., find
the force required when the displacement
is Ss) = 4in. How much work is done from
81 tO So?
Fig. 666.
Problems 1475, 1476.
1475. A 6-ft. drum, Fig. 666, has wound
about it a cable whose total length is 100 ft.
The cable weighs 10 lb. per ft. and 10 ft.
of it hang from the drum when rotation
starts. The shaft and drum together weigh
6440 lb. and have a radius of gyration of
2.5 ft. Assume that the bearing friction is
negligible, that all coils of the cable have a
mean diameter of 6 ft., and that the cable
does not slip on the drum.
What is the
angular velocity of the drum after the 90 ft.
of cable have run off?
Ans. 8.03 rad. per sec.
1476. The same as 1475 except that the
cable weighs 4 lb. per ft.
Ans. 30.8 lb., 154 in-lb.
1472.
The same as 1471 except that the
relation between F and s is F’s¥/3 = C,
Fig. 667.
Fig. 665.
1473.
Problems 1473, ‘474.
A constant
force Q = 30 lb. acts
on the cord shown in Fig. 665. The cord
passes over a smooth peg A of negligible
diameter and is attached to the body B.
Determine, by two methods, one of which
uses the calculus, the work done on B by Q
while B is displaced a distance of 15 ft.
Problem 1477.
1477. The 6-ft. sheave B, Fig. 667, weighs
644 lb. and has a radius of gyration of 2.5
ft. A cable A lies across the sheave, overhanging each side, with a = 60 ft. and
b = 40 ft. The cable weighs 5 lb. per ft.
Neglect that part of the cable in contact
with the sheave and neglect the friction in
the bearings.
If the cable does not slip
on the sheave and starts from rest, what
is the angular velocity of the sheave after
the length a has increased from 60 ft. to
100 ft.
Ans. 9.51 rad. per sec.
PROBLEMS:
SPRINGS—VARIABLE
426
FORCES
1478. The force-displacement diagram for
a punch while it is punching a hole in a
metal plate is similar to the solid curve
OABC, Fig. 668.
It is seen that the work
area under the curve is represented roughly
by the triangular area ODC, at least closely
enough for some engineering purposes.
The
maximum force necessary to punch a 3/4-in.
hole in a 1/2-in. thick, soft steel plate is
about Fmax = 50,000 lb.
(a) What
is the
work done?
(b) It is frequently assumed
that the flywheel supplies all this energy.
If the flywheel has a moment of inertia of
34 slug-ft.? and is turning at 150 rpm at the
beginning of the punching, what is its
angular velocity just as the punching is completed? Assume that the frictional losses
and the power supplied by the driving
motor during this short time element are
negligible.
1479. The work of punching a 1/2-in. hole
in a 1/4-in. steel plate may be found as
described in 1478. The maximum force for
this hole is approximately Pmax = 18,000 Jb.
The flywheel of this machine, turning at
Fig. 668.
Problems 1478, 1479.
200 rpm, is to supply the energy for punching the hole. Its speed should not fall below
(a) What is the
170 rpm during punching.
minimum permissible moment of inertia of
Frictional losses are negligithe flywheel?
ble.
(b)
Assuming
that
the
moment
of
inertia of the hub and spokes are negligible,
that the mean diameter of the rim of the
flywheel is 30 in., and that the width of the
rim is 4 in., determine the minimum thickness of the cast-iron rim.
Ans. (a) 3.05 slug-ft.; (b) approx. 5/8 in.
POWER
1480. If a 3000-lb. automobile coasts at
a constant speed of 30 mph down a 0.5%
grade, what is the total resistance to motion
as measured by a force parallel to the grade?
How much is this resistance expressed in
kilowatt units of power?
Ans. 15 |b., 0.895 kw.
1481. How long does it take a 200-Ib.
man to run up a stairway and increase his
elevation 60 ft. if his average net power
(Ina
output during this period is 0.2 kw?
similar manner, a student may determine
his individual power output for a short
period.)
1482. (a) What is the maximum speed
at which a 10-hp motor can lift a 1000-lb.
(b) If friction consumes 25%
elevator?
of the motor’s power, what is the maximum
speed?
Ans. (a) 5.5 fps; (b) 4.125 fps.
1483. An 1800-rpm motor has an output
It is 85% efficient. Find (a) the
of 2hp.
torque
on
its shaft
and
(b) the
cost
to
operate it 30 min. at 2 cents per kw-hr.
1484. What is the greatest head against
which a 20-hp motor can pump 1000 gpm
of water that weighs 62.1 lb. per cu. ft.
The overall efficiency of the operation is
Ans. 63.6 ft.
80%.
1485. A 12-in. gear is transmitting 36 hp
What is the driving force on
at 100 rpm.
the gear teeth?
1486. A 3220-lb. automobile on a level
road utilizes 60 hp to travel at a constant
60 mph and 75 hp to travel at 75 mph.
Find
the
acceleration
(a) at the
instant
when the car is going 60 mph and 75 hp is
applied and (b) at the instant when the
car is going 75 mph and 60 hp is applied.
(Do not assume that 1 hp per mph is a
It may be true for
constant relationship.
two points by coincidence.)
Ans. (a) +0.93 fps?; (b) —0.75 fps’.
1487. An autombile engine develops a
The
maximum of 100 hp at 3600 rpm.
30-in.
diameter
rear
wheels
turn
1/4 as
fast as the engine, and 75% of the engine
For the
power is delivered to the wheels.
maximum power, find (a) the torque on
the rear wheels, (b) the driving force between the wheels and the road, and (¢) the
least weight that may be on the rear wheels
if they are not slipping when f = 0.2.
Ans. (a) 5250 in-lb.; (b) 350 lb.; (¢) 1750 lb.
1488. A 4000-lb. automobile with a fluid
torque-converter transmission might conceivably have its engine so throttled that it
continuously delivers 100 hp to the rear
PROBLEMS:
426
wheels.
Observe that the driving force is
variable and assume that the wheels do not
slip. (a) Find the speed after the car has
traveled 500 ft., measured from the instant
when its speed was 30 mph.
(b) How much
net work is done in this distance?
Ans. (a) 90.8 fps; (b) 391,000 ft-lb.
1489. The same as 1488 except that the
initial speed of the car is 45 mph.
1490. A locomotive exerts a constant
drawbar pull of 50,000 lb. on a freight train
whose gross weight is 2000 tons. The total
train resistance is 15 lb. per ton. This train
starts up a 1% grade with a speed of 60
mph.
(a) If the grade is 2 miles long, what
is the train’s speed in mph when it reaches
the top? What is the maximum horsepower
developed
at
the
drawbar
during
this period? What is the drawbar horsepower at the instant that the train reaches
the top of the grade?
(b) If the grade is
4 miles long, what is the speed at the top?
Ans. (a) 44.9 mph, 8000 hp, 5980 hp; (b)
21 mph.
1491. In problem 1428, what is the horsepower being developed by this ideal turbine?
Ans. 56.5 hp.
1492. What maximum horsepower must
be delivered by an electric motor which operates a mine hoist, when the total load is 22
tons and when the hoist is uniformly accelerated upward from 5 fps to 20 fps in 10 sec.?
Assume that the guides are frictionless.
1493. A hoist with its load weighs 40,000
lb. It attains a speed of 10 fps in a distance
of 10 ft. with constant acceleration.
The
constant friction in the guides is 200 lb.
What is the maximum horsepower delivered
by the driving motor during this period?
What horsepower does the motor develop
when the elevator is on the point of starting?
1494. A motor truck with its load weighs
8.05 tons and is to be accelerated on a level
highway from 10 mph to 45 mph in a
distance of 2000 ft. The resistance to motion is 500 lb. If the acceleration is constant throughout this period, (a) what maximum horsepower must be delivered to the
rear wheels; (b) what is the delivered horsepower when the speed is 10 mph; (c) what
is the tractive force for the 36-in. driving
wheels?
Ans. (a) 122 hp; (b) 27.2 hp; (c) 1018 Ib.
1495. If a hoisting engine is to be able
to lift a 50-ton girder through 400 ft. in
5 min. at constant speed, what horsepower
must it deliver, friction negligible?
1496. The rotating parts of an electric
generator weigh 14 tons and have a radius
POWER
[Ch. XVIII
of gyration of 7ft. These parts are brought
up to a speed of 600 rpm in 30 sec. with
uniform acceleration.
What maximum
horsepower is necessary, if the friction is
neglected?
Ans. 10,190 hp.
1497. The same as 1496 except that a
constant total frictional force of 200 lb.
acts at the surface of the 6-in. journals.
1498. What average horsepower is needed
in filling a water reservoir 100 ft. long,
100 ft. wide, and 8 ft. deep in 4 hr.? The
water is pumped against a constant head
of 30 ft. Neglect the friction.
Ans. 18.9 hp.
1499. At a point where the cross section
of a river is 900 sq. ft. in area, the average
speed of flow is 3 mph.
In this vicinity a
fall of 30 ft. is available.
If the efficiency
of the hydraulic turbines is 80%, what
horsepower could be obtained from the
total flow of the river?
1500. The ideal pumping engine in problem 1405 develops 100 hp. How long will
it take to fill the hemispherical tank?
Ans. 7.45 hr.
1501. The work done to operate an actual
screw jack in raising a 2000-lb. load through
6 in. is 2300 ft-lb.
What.is the efficiency
of the jack? See problem 1409.
1502. The same as 1501 except that the
work done is 3000 ft-lb.
ANS ooo ue
1503. The General Electric Company
rates the 2400-lb.,
J-47 turbo-jet engine
at a static thrust of 5000 lb. for take-off and
at 10,000 hp at 750 mph.
(a) What is the
power output when the plane is on the point
of starting down
the runway?
(b) What
is the thrust at 750 mph?
(c) If the thrust
is constant at all speeds, what is the horsepower developed at 375 mph?
See Fig. 678.
Ans. (a) 0; (b) 5000 Ib.; (c) 5000 hp.
1504. An Air Force F-86 fighter plane
weighs 12,000 Ib. loaded.
If its jet engine
can produce a constant thrust of 5000 lb.
and if this thrust causes the speed to change
from 500 mph to 600 mph, find (a) the net
work done, (b) the final power output in
horsepower, and (e) the altitude that might
have been gained if the pilot had chosen to
increase the potential energy instead of the
kinetic energy of the plane.
(These data
do not reflect the true performance of the
plane.)
Ans. (a) 4.42 x 107 ft-lb.; (b) 8000 hp;
(c) 3680 ft.
1505. An Air Force F-86 fighter plane
weighs 12,000 lb. loaded and climbs at 5700
fpm, maximum.
(a) Find the average “‘lift-
PROBLEMS:
427
POWER
(b) If the “lifting
ing horsepower” needed.
horsepower’? were applied to produce horizontal acceleration when the plane’s speed
is 480 mph, what would be the instantaneous
horizontal unbalanced force on the plane?
(c) What instantaneous acceleration of the
plane would be produced by this unbalanced
force?
1506. A B-36 bomber with a bomb load
weighs 320,000 lb. From a cruising speed
in level flight of 300 mph, the pilot maneuvers the plane to gain 500 ft. of altitude in
a horizontal distance of 2 miles, and then
the plane levels out at a speed of 240 mph.
Find (a) the increase of potential energy,
(c) From a
(b) the loss of kinetic energy.
consideration of (a) and (b) decide whether
the pilot increased or decreased the power
output of the engines during this maneuver.
Ans. (a) 1.6 X 108 ft-lb.; (b) 3.46 X 108 ftIb.; (ce) decrease.
Fig. 670.
Problems 1509, 1510.
(b) If two
represented by this diagram.
such diagrams are completed during each
revolution of a double-acting engine, what
horsepower corresponds to an angular speed
of 120 rpm?
Ans. (a) 17,480 ft-lb.; (b) 127 hp.
1510. The same as 1509 except that Fa =
25,000 Ib. and F, = 5000 lb.
F
BS]
S
3S
So
ba
ma
_
1’=0.75 ft.
Fig. 669.
$
Problems 1507, 1508.
1507. Figure 669 is an indicator diagram
for a two-stroke-cycle diesel engine. The
enclosed area A represents to seale the
work done in a cylinder during each revoluSuppose that the area
tion of the engine.
is
A is 0.55 sq. in., that the force scale
1 in. = 110,000 lb., and that the displacement
seale
is 1 in.e= 0.75 ft.
“(a) What
work is done per revolution in each cylinder?
(b) If the engine has 3 cylinders and turns
at 200 rpm, what horsepower is the engine
developing?
1508. The same as 1507 except that the
area of the diagram A is 0.86 sq. in.
Ans. (a) 71,000 ft-lb.; (b) 1290 hp.
1509. The diagram abcde of Fig. 670
shows what is termed a conventional diaIt is something of
gram for a steam engine.
the
an idealized picture of the variation of
ns.
force on a piston at different piston positio
hat
The actual work of the engine is somew
less than that represented by the enclosed
Suppose that the magnitude of the
area.
force at the point a is Fa = 10,000 Ib. and
that F, = 1600lb.
(a) Determine the work
Fig. 671.
Problem 1511.
1511. In Fig. 671, C is a rim which is
connected
to turn with a shaft, which we
shall call shaft 2. The weights A and B
through a connection by pins in slots are
arranged to turn with a shaft 1, which is
collinear with shaft 2. As shaft 1 speeds
up, centrifugal force causes the weights
A and B to move outward and contact
the rim C. Faster speed finally results in
a frictional foree large enough to turn C
and shaft 2. This figure represents a clutch,
the object of which is to pick up the load
gradually with little shock to shaft 2. The
contact surfaces of the weights are faced
with asbestos for which f = 0.4 and the
weights weigh 32.2 lb. each. The center of
gravity of each of the weights A and B is
9 in. from the axis of the shafts. Ifn = 300
rpm, what horsepower may be transmitted
Ans. 33.8 hp.
through this clutch?
1512. In the brake shown in Fig. 672,
assume that the pressure is uniformly dis-
428
PROBLEMS: POWER
[Ch.
XVIII
resistance to motion is 15 lb. per ton. A
cable from A wraps about a drum whose
diameter is D, = 3 ft. Attached to the
drum is an 8-ft. (= D2) brake wheel, for
which f = 0.3. The car A is to be brought
to rest in a displacement of 50 ft. with
uniform deceleration.
What force Q must
be applied?
What is the maximum rate at
which power is absorbed by the brake?
Express as fhp (frictional horsepower).
Ans. 697 lb., 179 hp.
1515. Figure 674 is a diagrammatic representation of a prony brake, which is a
device to measure the output of engines.
A brake is clamped to the flywheel of an
engine, but it is prevented from rotating
with the flywheel by connecting the brake
arm to a scale which is itself fixed. All the
power of the engine is absorbed by friction
at the brake blocks.
Observing that the
Fig. 672.
moment
Problems 1512, 1513.
tributed on the brake shoe. The force
Q = 100 lb., f = 0.85 for both shoe and
band, and D = 20 in. (a) If the brake
wheel turns counterclockwise, what braking
torque is exerted?
(b) If the speed of
rotation is 200 rpm, what power is dissipated
as heat?
Ans. (a) 389 ft-lb.; (b) 14.8 hp.
1513. The same as 1512 except that the
wheel turns clockwise.
1514. An ore car A, Fig. 673, weighing
50,000 lb. with its load is moving down a
10° incline with a speed of 10 fps. The
of the net scale reading
compute the “brake” horsepower of an
engine.
Suppose that the brake arm is
a = 60 in. long, the scale reading is 40 lb.,
and the flywheel turns 300 rpm.
What is
the brake horsepower?
See Fig. 642, p. 413.
Ans. 11.4 hp.
1516. In Fig. 674, the brake arm is a =
3 ft. long, the scale reading is 250 Ib., and
the angular speed of the flywheel (and
engine) is 275 rpm.
What is the brake
horsepower?
See problem 1515.
LD
Flywheel
Fig. 674,
(the force
at the end of the brake arm) about the
center of the flywheel must be equal to
the moment of the frictional forces at the
brake about the same point, we can easily
Problems 1515, 1516.
PROBLEMS:
4 29
POWER
1517. For checking the horsepower output of a small
engine,
a band
brake
was
rigged up, as shown diagrammatically in
Fig. 675. The brake drum A was attached
to the shaft of the engine.
One end of
the band was connected to a scale and the
other end supported a weight of W = 50
lb. If the scale reading was 200 lb. and
the angular speed of the engine was 300
rpm, determine the brake horsepower.
Ans. 4.28 hp.
1518-1530. These numbers may be used
for other problems.
Fig. 675.
Problem 1517.
Chapter XIX
IMPULSE
AND
MOMENTUM
297. Introduction.
The principles to be explained in this chapter are
additional valuable tools for the engineer.
Not only is it possible and sometimes convenient to solve types of problems already discussed by the principles
of impulse and momentum, but these principles are essential for the solution
of certain other problems involving suddenly applied forces of relatively high
intensity which act for very short periods of time and for the solution of
certain problems involving non-rigid bodies such as fluids.
The first tool of kinetics which we learned about involves force, mass, and
acceleration (/ = ma). The second tool (U = AKE) involves force, displacement, velocity, and mass.
The third tool or general principle, impulse
and momentum, involves, as we shall learn, force, time, velocity, and mass.
Which of these tools is easiest to use in a given instance often depends upon
the known facts at the beginning of a solution, or on the desired results.
For
example, given data which includes force, displacement, and mass suggest
the use of the work-kinetic energy principle; given data which includes force,
time, and mass suggest use of the impulse-momentum principle.
298 Impulse and Momentum.
From Newton’s laws for a body in any
kind of plane motion, we know that =F = md, where @ is the acceleration
of the center of gravity of the body.
The force vector SF (equal to the resultant force R) and the acceleration vector @ have the same sense.
Using
a@ = dv/dt and =F = R in this equation, we find
k=
_ ad
man
or
(54)
[e di =m [av = Mv. — mii,
when the integral fdd is made between the limits of 3, and dp. Equation
(54) introduces another grouping of symbols.
Just as we found it convenient to have names for { ds and mv?/2, so we shall find it advantageous
to name the symbol groups in (54). The integral JR dt is called the linear
impulse of the force R; and the product md is called the linear momentum
430
431
§ 299 | IMPULSE
of the body whose mass is m. Observe that impulse is an action that has
duration, inasmuch as there must be some time interval (dt) during which
the impulse exists. On the other hand, momentum is an instantaneous
property of a body or particle in motion; that is, if a particle of mass m has a
velocity of v at a particular instant, then it also has a linear momentum of
mv. At another instant somewhat later, the velocity may have changed, in
Equation (54) shows that an
which case the momentum also has changed.
impulse is essential to a change in momentum, and vice versa.
Impulse is a vector quantity which has the same sense as
299. Impulse.
the force R in [Rdt. Moreover, the line of action of the impulse coincides
with the line of action of the force R. As a vector, impulse may be resolved
In fact, if a system of forces is involved, each force F in
into components.
the system has an impulse of JF di, and the vector sum of the various imthe
pulses fF dt is equal to the resultant impulse; that is, the vector sum of
individual impulses is equal to the impulse J & dt of the resultant force. These
of a
statements are significant because we often wish to know the impulse
the
of
direction
the
not
is
that
particular force or the impulse in a direction
integrate
may
we
resultant force. To evaluate the impulse of some force /,
be exfF dt when F is constant in magnitude and direction or when /’ can
we have
pressed as a function of the time. For the case of a constant force,
Impulse = .Y = P fa = F At,
where At = te — t; is the time interval during which the impulse of
the constant force F is desired. In
your written work, let the symbol for
Ek
impulse be a script JS in order to
;
a
distinguish it from a lettered J which
Area
Time-Force Diagram.
Fig. 676.
:
:
npaleel
ceeentsti
Pe
Be
stands for moment of inertia.
careful about this distinction.
to plot a time-force
If the force varies in magnitude, it may be possible
ntial area dA of this
diagram as shown in Fig. 676. Considering a differe
,
diagram, we see that it is equal to F' dt, the impulse
dA = F di.
The average
impulse.
Hence the area “under” a time-force curve represents
this average height
height of this area, Fig. 676, is given by A/At. However,
impulse of a force varying in
is also the average force. It follows that the
to the average force Fa,
magnitude (not direction) during time A¢ is equal
multiplied by the time:
Impulse = G = F,, At lb-sec.
IMPULSE AND MOMENTUM
482
[Ch. XIX
If F = f(t), then it is generally a simple matter to integrate J f(é)dt. The
unit of impulse is a compound unit, a force unit times a time unit. Usually,
we find the unit of pound-second most convenient in this work.
300. Example. A 100-lb. body is on a horizontal plane, where the coefficient of
kinetic friction isf= 0.2. A force of Q = 60 lb. acts as shown in Fig. 677.
(a) What
is the impulse of Q in the horizontal direction during 2
sec.? (b) What is the net impulse on the body during the
same time?
Sotution.
(a) The value of Q, is
Q, = (60)(cos 30°) = 52 lb.
Loz = QeAt = (52)(2) = 104 Ib-see.
toward the right, the impulse of Q, during 2 sec.
(b) Since the body has no motion in the vertical direction, >, = 0. Thus the
impulse of the force N acting upward is exactly canceled by the impulses of W and
the vertical component of Q acting downward.
Consequently, the net or resultant
impulse on the body is in a horizontal direction. From =F, = 0, we find N = 130 lb.;
then fF = fN = (0.2)(180) = 26 lb. Now summing horizontally, we get
I-{R dt = (2F,)At = (Q cos 30° — F)At = [(60)(0.866) — 26](2) = 51.9 Ib-sec.
acting toward the right.
301. Principle of Impulse and Momentum.
Linear momentum* also is
a vector quantity.
It has the same sense as v in the product mv.
If a
particle of mass m has an absolute velocity of v, its total momentum is Md =
mv, its momentum in the z direction is _ //, = mv;, and its component in the
y direction is wh = mv,, Where v, and v, are the x and y components, respectively, of the total velocity v. (Again note that the use of the script dM
for momentum in your written work will distinguish it from moment whose
symbol is a lettered J.)
Sometimes it may be necessary to use equation (54) in the differential form**
(a)
a.
Rdt = mdv,
which can be integrated for a body of mass m when R is constant or can be
*When we speak of the momentum and impulse without a modifying adjective, linear
momentum and linear impulse are intended.
Later (§ 305), we shall deal with angular
momentum and angular impulse.
**An impulse may occur, not only because of a change of velocity, but also because of a
change of mass or of mass and velocity. If the equation Rt = mv is differentiated with
the mass considered as variable, we get
R dt
= mdv +vdm,
where, if the mass is constant, dm = 0 and this equation reduces to (a). In general,
those problems in this book which involve a variable mass can be solved by use of the
instantaneous values rather than by integration of this general equation.
§ 301 ] PRINCIPLE OF IMPULSE AND MOMENTUM
433
expressed as a function of ¢ or v. When F& can be taken as constant, the
following brief symbol form will be convenient:
(b)
R At = Ami,
and md are the same, the change of
where, if the directions of the vectors
momentum Ami is always the second value of momentum algebraically minus
Having pointed out the importance of direction in dealing
the first value.
with impulse and momentum, we may now generalize equation (54) in words.
From this equation, we may state that the net linear impulse on a body in a
particular direction is equal to the change of linear momentum of the body in
that direction, where the momentum of a body in a particular direction is the
product of the body’s mass and the component of the velocity of the mass
This statement is known as the principle of linear
center in that direction.
Observe that the change of momentum occurs
impulse and momentum.
In applying this principle, we should
during the interval of the impulse.
be positive and we should remember
to
is
decide early upon the sense which
sign
that all vector quantities pointing in this sense shall be given a positive
a
given
be
and that all vector quantities pointing in the opposite sense shall
be in
negative sign. If an unknown vector quantity is originally assumed to
the
in
points
it
number,
positive
a
be
to
the positive sense and is found
The
sense.
negative
the
in
positive sense; if found to be negative, it points
usual unit of momentum is obtained from
(Ib.) (ft. /sec.)
Wo
ae ae
(ft. /sec.)?
Ae
which is the same as that for impulse.
We may also write
pr dt =
mvz—> m1,
mass is m, at
where v, and v2 are the absolute velocities of the body, whose
momentums is
the instants 1 and 2 and where a vector difference of the
of the
The sense of the resultant impulse fR dt is the same as that
found.
sum
to
ent
conveni
more
ly
— mv:. Nevertheless, it is general
vector mv,
sly
previou
as
n
impulses and momentums in a particular « or y directio
no
is
there
If there is no acceleration (no change of velocity),
suggested.
= 0.
change of momentum, and
the operation
The principle of impulse and momentum is used to explain
In the case of a turbo-jet engine, atmospheric
of a jet engine or a rocket.
Then the burning of fuel in the combustion
air is taken in and compressed.
then pass through the
chamber increases the volume of the gases, which
(see Fig. 678). The
turbine doing enough work to run the compressor
is high. The increase
velocity of the gases as they emerge from the jet nozzle
IMPULSE AND MOMENTUM
484
[Ch. XIX
: doit Rotor
Axial Compressor
—s
Combustion Chambers
bE
Exhaust
Nozzle
i
Courtesy Allison Division, General Motors, Indianapolis.
Fig. 678. Allison Turbo-Jet Engine.
This engine is designed for military aircraft use
and is rated at 5000 lb. static thrust.
A multistage axial compressor is used.
The combustion chambers are spaced around the periphery as shown. A single-stage turbine
develops sufficient power to operate the compressor.
The speed-up of the gases (rate of
change of momentum) produces the thrust.
in gas speed, having occurred because of a pressure drop, results in a change
of momentum m(vz — v1) equal to the impulse,
fe dt = m(ve — 0),
where v2 — v; 1s the increase in speed of the gas, m is the mass of gas, and R
is the thrust developed by the engine.
If FR is constant, as it may be assumed
for a particular operating condition, we get from R At = Amv
from which we see that the magnitude of the thrust produced by the engine
depends on the mass rate of flow of the gas, that is, upon the quantity of gas
involved and the time it takes to speed up from 2; to v2. If the flow of fluid
is in slugs per second, then At = 1 sec. in the foregoing equation.
As the
time interval At becomes shorter, the thrust increases for a particular quantity
m of gas. Recall that if the engine does not move, no work is done by it and
it develops no power.
Thus, if the thrust R remains constant while a plane
with such an engine starts from rest, the power generated is zero until motion
begins, and the power increases with the speed in accordance with equation
(Peed Le
_ Ro,
P = me
[equation (t), p. 411].
The maximum power that such an engine might develop is limited, among
other things, by the time rate at which fuel can be burned (which in turn is
limited by metallurgical considerations), the rate at which the gas can be
made to flow through the engine, and the resistance to motion of the vehicle
in which it is mounted.
435
§ 303 | EXAMPLE
A 200-lb. body is moving toward the left with a velocity 1: = 50 fps
302. Example.
(Fig. 679) at the instant that a force Q = 50 lb. acting toward the right is applied.
There is a constant resistance to motion of F = 40 lb. What is the body’s velocity
after 20 sec.?
Fig. 679.
Sotution.
[Fig. 679(a)]
Let the rightward
sense
be positive.
Using Rh At = Amv, we have
Q + F)at = " (v2 — 11),
(50 + 40)(20) = = [vs — (— 50)],
) sense. From
where 0; is substituted as — 50 because it is in the leftward (negative
positive sign
the
this equation, we find v2 = + 240 fps, an incorrect answer although
toward
moving
starts
body
the
When
shows that the body is moving toward the right.
be in
must
solution
this
Hence
changes.
F
force
the right, the sense of the resisting
the velocity
find
then
rest,
to
come
to
body
the
for
time
the
find
two parts. First,
body of Fig. 679(a)
gained in the remaining time. Thus we have from the free
(50 + 40)(at) = = one - (0 — (= 50)],
time, 20 — 3.45 = 16.55 sec., the
from which At = 3.45 sec. During the remaining
as shown in Fig. 679(b). Using
is
body
body moves toward the right and the free
right is just starting (v1 = 0),
this free body and remembering that the motion to the
we get, from Kh At = Amd,
00
(50 — 40)(16.55) = — "
from which v2 = 26.6 fps.
still a valuable asset.
303. Example.
body is
Observe that the ability to make a correct free
per sec., issues
A jet of steam, flowing at the rate of W = 1.5 lb.
fps (Fig. 680).
from a nozzle with a velocity of vs1 = 500
it is discharged.
and is turned through an angle of 135° before
It enters a fixed blade
What are the horizontal
upon the fixed blade?
and vertical components of the force exerted
nless.
frictio
is
blade
the
across
passage of steam
Assume that the
steam is equal and opposite
The force exerted upon the blade by the
Sonution.
Hence we may
the blade (Newton’s law).
to the force exerted upon the steam by
the jet of
upon
d
exerte
forces
the
680 as
consider the forces Q, and Q, shown in Fig.
of steam
e
passag
the
If
body.
free
the
steam, thereby considering the steam as
ed as shown in
direct
is
v2
but
fps,
500
=
Ys1
=
¥.2
across the blade is frictionless,
Fig. 680.
In problems of this type, any convenient time
unit may be employed.
Let
436
IMPULSE AND MOMENTUM [Ch. XIX
At = 1 sec. Then the mass of steam involved is m = 1.5/32.2 slugs per sec.
expression R,At = m(Vz2 — Vz1),
In the
Vz2 = — Vs2 Sin 45° = — (500)(0.707) = — 353.5 fps;
and
Mies ==
—Q.,
where in each case the negative sign shows that the
sense is toward the left. Thus we have, for At = 1 sec.
(v.21 =
Vs.
=
+6500),
ily
(— Q:)(1) = 39,9 (— 353.5 — 500) Ib-sec.,
from
Fig. 680.
Stationary
ae
Blade.
which
Q, = 39.8
lb.
Considering
the vertical
direction, we find
&
Consider this as a free body
of the steam. There is a
in which
erted on the steam by the
CO™ponent of v1 is zero), and vy: = —v,2 cos 45° =
resultant force Q (or R) ex-
blade whose components are
Q. and Q,.
R, At = m(v,2 — v1),
Ry = — Q,, %y1 = 0 (that is, the vertical
~— 353.5 fps (the negative signs indicating the downward sense). Using these values, we find
1.5
(—Q,)(1) = 355
(— 353.5 — 0) Ib-sec.,
32.2
from which Q, = 16.5 lb., the positive value showing that we
assumed the correct
sense for Q, and that it does act downward on the jet of steam.
The resultant force on the steam is
R=
at an angle of
Q.? + Q,7)¥2 = (89.8? + 16.5°)’2 = 43 Jb.
16.5
Q,
6 —= tan =0. =< tan as
39 8
—
22.9 2
with the horizontal, pointing downward toward the left. The
force of the steam on
the blade is equal and opposite to R.
In practice, the components of the resultant in a situati
on like this may be more
useful than the resultant itself. If, as the designe
r, we wished to hold the blade
stationary, we might very well consider supports
at right angles to one another,
designing one to withstand the Q, load and the other
to withstand the Q, load. If this
blade were a stationary blade in a turbine, it would
be one of many, and it might
very well be fixed at one end with no other support.
ALTERNATE Souution.
For the light it might throw on the princi
ple of impulse
and momentum, we shall explain a graphical solutio
n, using the vector form
R At = mv > m3.
The initial momentum
is
m0, = MV
acting toward the right.
1.5
= (35,)00) = 23.3 lb-sec.,
The final momentum is
15
MV. = M.o = (43
5,600 = 23.3 lb-sec., %
437
S 304 | EXAMPLE
To find
acting 45° below the horizontal toward the left (in the same sense as 0,2).
vector
a
OA,
out
lay
and
681,
Fig.
the vector difference, choose a convenient origin O,
23.3 units long. To subtract mv, from
_A
O- mv,=23.3
this vector mv, lay it out tail to tail
|
addifor
as
point,
to
tail
of
(instead
tion) and get OB = 23.3 units. The
vector AB shows the resultant impulse, and also the resultant force,
since At = 1 sec., in magnitude and
el Hayat
a
ee
@Bos
Firection « scaling Ape weund R=
eae
43 \b., as before. The components
of R, designated Q, and Q,, may be
found in the usual manner.
Fig. 681.
The vector difference of the momen-
tums OA and OB gives the impulse AB.
resultant of Q, and Q, is R.
The
The data of this
304. Example.
blade is moving, and % =
example are the same as those in § 303 except that the
Fig. 682, and also find the
Q,,
and
Q,
reactions
the
Find
right.
200 fps toward the
horsepower being developed in driving this blade.
force on the blade
When both the fluid and the blade are moving, the
So.ution.
relative velocity
the
of
change
the
upon
depends
fluid
or the force of the blade on the
blade depends
the
striking
fluid
of
amount
The
blade.
the
to
of the fluid with respect
by imagining
nt
stateme
this
upon this relative velocity. (You can sense the truth of
should
stream
the
in
car
one
If
n.
a stream of automobiles going in the same directio
velocity
stream
the
occur—
would
impact
an
stream,
go faster or slower than the
vines
Bae
ee
Nozzle
Us1
Us/oizUsir7Vb
remaining
constant.
Vs/b1 =
Vs1 — Vd =
The seriousness of the
wreck would depend in part upon the relative
velocity of impact.)
Since v.; and v are in the same sense, the
initial steam velocity relative to the blade is
500 —
200 =
300 fps.
Now the magnitude of the exit relative velocity
v.92 is also equal to 300 fps when there is no
friction in the blade.
682. Moving Blade. No loss
Fig.
of relative velocity occurs when
there
is no
friction,
Vs = Us/rx
component
velocity is
(The absolute exit velocity of the steam may
The fluid is the free body and forces
and opposite forces act on the blade.
fortis
Equal
relative
sq gos 45° = —(300)(0.707) = —212.1 fps.
be found from
Q, and Q, act on the fluid.
However, the horizontal
of this exit
Vs. =
Vs/b2 +
Ub, if desired.)
The absolute rate of flow (1.5) in equation
n) (Absolute velocity) (Specific weight).
Rate of flow = Av.w = (Area of sectio
that is, the rate at which fluid enters
Similarly, the rate of flow relative to the blade,
the blade, would be
n)(Relative velocity)(Specifie weight);
Fluid striking blade = (Area of sectio
is proportional to the relative velocity:
hence the amount of fluid striking the blade
Relative velocity
W,, lb. striking blade
ute velocity’
Absol
1.5, lb. flow
W, = «.si(5) = 0.9 lb. per sec.
488
IMPULSE AND MOMENTUM
[Ch. XIX
We are now ready to apply the impulse-momentum principle to that quantity (W, =
0.9) of steam striking the blade.
From R,At = Amv,, rightward direction positive,
Fig. 682, we get
Moving
Nozzle
Blade
Ys
Be
aaert
‘
Us/g1=Us1 > U b
yy
ae ya
=
J
-
arene
orQ; = 14.3lb.
Similarly, in the vertical direc-
tion, upward direction positive, we find
‘ig
(= SOU@)0)ye=a5 ( ee ),
we
or Q, = 5.92 lb.
Va
a (— 9191. 300)
If this blade is attached
to
a turbine rotor, Q, = 5.92 lb. is the thrust in
Fig. 682. Repeated.
an axial direction and therefore does no work.
If this thrust is unopposed by other fluid
forces, it must be taken care of in the
design of the bearings.
See Fig. 683 for
a type of turbine wheel used in some small
turbines.
The force Q, = 14.3 lb. is directed circumferentially and therefore does work
when the rotor turns. The rate at which
Q, traverses distance is the speed of the
blade, v = 200 fps. Thus the horsepower is
ae
iH
Qt» _ (14.3)(200)
550
=
550
ili |
= 5.2 hp.
2
_ Leaves
a
~
(NoTE. If the sense of the force Q is
not readily apparent to the reader, he may
assume that its components both act in the
positive sense.
Then
~ Rotor for
“ Exhaust
a negative answer
indicates that the wrong sense was chosen
and that the force acts in the negative
sense. Remember that the principle of
impulse and momentum is being applied to
the stream of fluid and the forces thus
obtained are those acting on the fluid. The
signs for the velocity components must be
correct, unless, of course, a force is known
and a velocity is unknown.
Observe that
while there are no frictional losses, the
efficiency of this conversion is not 100%.
The horsepower represented by the original
kinetic energy of the jet is
We
-—>Buckets
| Steam
_ (1.5)(500)?
29(550) — (2)(g)(550)
= 10.6 hp.
Thus. the efficiency of this process of converting kinetic energy of the jet into work
Courtesy of The Terry Steam
Turbine Co., Hartford, Conn.
Fig. 683. A Reentry Turbine. Observe how
the velocity of the steam is nearly reversed
every time it enters the buckets on the
wheel.
This reversal produces a change
of momentum which is converted into an
impulse.
In the actual construction, there
is a casing covering the wheel which
contains the guides which re-direct the
steam back into the buckets.
While most
turbines are constructed differently from
the one illustrated, they all operate on
the same principle.
§ 305 ] ANGULAR
IMPULSE
AND ANGULAR
439
MOMENTUM
Also, in this
on the ideal turbine shaft is 5.2/10.6 = 49% in this particular case.
the steam
that
is
100%
than
less
efficiency
an
for
reason
particular case, the sole
equivalent
exact,
be
to
energy,
kinetic
residual
ble
considera
a
with
blade
leaves the
be set at
to 10.6 — 5.2 = 5.4 hp. In actual turbines, the axis of the nozzle has to
e even
impossibl
is
efficiency
100%
event,
an angle with the plane of rotation, in which
does
always
It
energy.
kinetic
absolute
no
with
blades
the
if the steam should leave
though.)
energy,
kinetic
some
leave with
We recall that for a
305. Angular Impulse and Angular Momentum.
of rotation
rigid body 2M, = I,«, where the point O is either (1) the center
in plane
or
rotation
in
body
a
of
gravity
of
of a rotating body, (2) the center
tion
accelera
absolute
whose
motion, (3) a point, on a body in plane motion,
plane
in
body
a
on
is directed through the center of gravity, or (4) a point,
purposes,
motion, whose absolute acceleration is zero ($ 257). For present
moments about
let the symbol M/, (instead of =M,) represent the sum of the
have
some point O and use a = dw/dt. Then we
dw
M,=I1 a)dt’
(c)
or
[ota = [Ide.
If J, is a constant, we may integrate fdw from «1 to
w. and get
(55)
ie dt =
I ,we
=
Toone
Fig. 684.
have a moment
On the left-hand side of this equation, we observe that we
y /M,dt is
quantit
This
of a force (or torque) MM, times a time interval At.
If it is recalled that
called the angular impulse with respect to the point O.
to be the moment
a moment is force times a moment arm, the fM, dt is seen
t force / acting
constan
the
684,
For example, in Fig.
of a linear impulse.
magnitude
whose
tangentially to the circle at all times has a linear impulse
O is
is F At. The moment of this impulse about the point
F Atr = Fr At = M, At.
given later (§ 306),
If the reference point O is chosen to conform to rules
ty of a body with an
the expression J,w represents an instantaneous proper
tum or the moment of
angular velocity w which is called the angular momen
r
In accordance with equation (55), the unit of angula
linear momentum.
For the unit
tum.
impulse must be the same as the unit of angular momen
? is
(W/g)k
=
I
of
of angular impulse, we usually use ft-lb-sec. The unit
lla, X<
sec.”
x ft.2 = ft-lb-sec.’
ft
440
IMPULSE AND MOMENTUM
The unit of w is 1/(time unit) = 1/sec.
Therefore,
momentum
[Ch.
XIX
the unit of angular
is
{t-lb-sec.2 < 2s —tt-lb=sec.
sec.
To evaluate equation (c), M,dt = I, dw, M, must be expressed as a function of time or of angular velocity, or it must be considered as constant.
To evaluate {M, dt, either 1, must be expressed as a function of time, or
it must be constant, or the average value of /, for the interval of time must
be obtainable.
If WM, is constant,
Angular impulse = .Yn = M, |
a = M, 4,
where At is the time interval during which the value of the angular impulse
is desired.
In this expression, M, may be the moment of a single force or it
may be the resultant moment of several forces, depending upon what is
wanted.
306. Angular Momentum.
To determine the reference point O for which
the angular momentum J, is equal to the angular impulse M7, At, consider
a particle B, Fig. 685, whose mass is dm and
which is located in a rigid body moving
Choose the reference
with plane motion.
point or axis O at random, but choose the
dmuo
/BN
=
Jo
ws fue|
‘
dmrw
)
| x and y axes so that the direction of the
cD
;
:
:
SS
ee
Se
a
Yx70=TW
velocity of O is along the x axis.
Then
from the principle of relative velocities, the
velocity of B is
y
Fig. 685.
Up
Up ate Umie:
The absolute linear momentum (dm)vz of the particle B may be considere
d
as the vector sum of the momentums (dm)v, and (dm)vp/o, Where vp/> = rw,
Fig. 685. Using the principle that the moment of the resultant is equal
to
the sum of the moments of the resultant’s components, we find the
moment
of the momentum (angular momentum) as
Angular momentum
(d)
=
7, = Z[(dm v,)y + (dm rw)r]
= Uo2ty dm + w=r? dm,
AMn = vomy + Iw.
We see that in general the angular momentum
of a body with reference to a
particular axis O consists of two terms (when v, is
in the x direction), the
vomy term and the J,w term.
The vemy term is zero only when one of the
following locations is chosen for O:
4Al
§ 306 ]ANGULAR MOMENTUM
(1) When O is the instantaneous center, in which case », = 0 and
(e)
M,At = AI,w
[M, constant]:
(2) When O is the center of gravity, in which case 7 = 0 and
(f)
Mat = Alw
[M constant];
(3) When O is a point whose velocity is directed through the center of
gravity. In this case, again, 7 = 0 and
(g)
M At = AI
[M, constant).
With these limitations on the location of the reference point O in mind,
we see that equation (55) expresses the following principle; The resultant
to the
angular impulse on a rigid body im any kind of plane motion is equal
the
to
reference
with
computed
being
both
change of angular momentum,
same axis O.
Remember that angular impulse and angular momentum are names given
linear impulse
to moments of vector quantities—moments, respectively, of
is an angular
time)
times
(force
a
of
moment
Any
and of linear momentum.
produce a
exist)
not
does
ium
Some such moments (when equilibr
impulse.
change of angular momentum in accordance with (55).
be sure to conJust as you would not add the moment of a force to a force,
impulses or
sider separately linear impulses or momentums and angular
(1) the
either:
are
O
points
e
The most convenient referenc
momentums.
gravity.
of
fixed or instantaneous center of rotation, or (2) the center
as positive,
Either a clockwise or a counterclockwise direction may be taken
momentum Jw 1s
as in moments of forces. Then the sign of the angular
governed by the angular sense of w.
He gives himself a certain initial
Imagine a diver leaving a springboard.
the only force acting
angular velocity for a fancy dive. Once he is in the air,
ble), whose line of action
on him is the force of gravity (air friction negligi
no moment with respect
re
passes through the diver’s cg. There is therefo
on (f),
equati
Thus, in
to his center of gravity and fM dt =0.
Alw = 0
or
Iw = a constant.
t of inertia J is a maximum
If the diver stretches to full length, his momen
for that particular dive. If
and his angular velocity w will be a minimum
close to his body, his moment
he should double up, drawing his knees and arms
on decreases and his angular
of inertia J with respect to his axis of rotati
be able to make one or more
velocity increases; he turns faster and may
Alw = 0. See also § 311.
somersaults in a very short time, all because
of gravity is along a parabola,
Observe that the path of the diver’s center
tions are.
as for any projectile, no matter what his contor
IMPULSE AND MOMENTUM
442
[Ch. XIX
307. Example. A 644-Ib. disk, 6 ft. in diameter, is keyed to a 322-lb. shaft which
The shaft, turning at 720 rpm, is supported by two bearings,
is 6 in. in diameter.
where f = 0.02. If there is no torque on the rotating members except that due to the
frictional force in the bearings, how long does it take for the angular speed to be
reduced to 120 rpm? See Fig. 686.
Soxtution.
For the shaft,
=
mr? _
I; —| 5 =
(322)(1/4)?
(32.2)(2)
=
ci 1
i
=i tee
0.313
slug
For the disk,
le
mr?
(644)(3)?
1
De
(PEVO
= 90 slug-ft2
The total moment of inertia of the rotating parts is then J = J, + 7, = 90.3 slug-ft.2
(Observe that /, for the shaft is practically negligible.)
The total load on the bearings
is 644 + 322 = 966 lb., and the frictional force is
1 = iN = (OLAS)
S 1B
IIo,
The moment of F about the axis of rotation is
Wes lg = = «9.0(1) = — 4.83 ftlb.,
where the negative sign is introduced to indicate that this moment is in the opposite
sense to that of the rotation.
Now applying the relation
M At = Alw, we have
120 X 2n — 720 X 20
C— 4°83) (Ai) = 003 (
60
60
:
from which At = 1173 sec. = 19.6 min. Both w; and w were substituted as positive
numbers because the sense of the motion was taken as the positive sense.
Fig. 686. The weight vector passes
through the center of gravity of the
system and F = F, + F,,
308. Example.
In Fig. 687, the weightless cord wraps about a central
groove
12 in. in diameter, and passes over a smooth peg (fe =
0) to a 80-lb. body iy The
central groove is in a 24-in. disk A which weighs W4
= 96.6 Ib. and has a radius of
gyration with respect to its axis of k = 9 in. (a)
If the disk rolls without slipping
what is the speed of B after 6 sec. when the bodies start
from rest? What is the fone
sion in the cord?
(b) What is the frictional force F?
SoLution.
(a) Since the instantaneous center of rotation is
at O, we have
UB
18
HE?
20
448
§ 308 |]EXAMPLE
speed of B, which is
where 6.4 is the speed of the center of gravity of A and vz is the
of A is then
speed
angular
The
cord.
the
on
point
a
the same as the speed of
GA
DA
2QvB
r
(8)()
= Amv,
Considering the free body of B, Fig. 687, we find, from R At
(30 — Q)At = ma(Vpe2 — Vai),
zo,
180 — 6Q = 0.9320
(h)
free body of A, we may
where vg1 = 0, At = 6sec., and Wz = 30 |b. Considering the
O as the center of
center
neous
instanta
the
using
T,wa,
=
At
M,
e
apply the principl
In this expression,
moments.
I, = mak? + mar*;
and
M.= (5) ft-lb.
thus
VN
tee lipeAcy
q(is)(6) _ [ (96.6\/
NS22
|
12e
9 Y +, (286
12\" | 2vne
ant
32.2 /\ 12
N12
9Q = 3.125vz2.
(i)
get
Solving equations (h) and (i) simultaneously, we
C= 20-75 0:
and
Vpo = 59./ Ips
must include all forces which have
(b) In summing forces for linear impulses, we
of the cg of the disk A at position
speed
The
sum.
components in the direction of the
2 is
_ 02)669.7) = 39.8 fps.
12,
EE et
ee
18
is, 041 = 0, the application of RAt = Amv
Since the initial speed of the disk is zero, that
to A yields
(Q c= F)At
=
mad a2 =
da);
96.6
A
(20.73 — F)6 = 32.2 (39.8),
that it is shown in its correct sense in
or F = +0.83 lb., the positive sign indicating
Fig. 687.
by taking moments about the cg of A.
The frictional force might also be found
Since
fae
O)O.15);,
we find from Mat =
TAw
and
reat
(50-+- ir)b=
AS a = 39.8,
Iwo,
(2 i r)s= (3)(0.75)2(39.8),
or F = 0.83 lb. as before.
of this example,
If you check the numerical computations
two
must be rather accurate to make these
as you should, you will find that you
444
IMPULSE AND MOMENTUM
[Ch. XIX
methods of obtaining F produce the same answers, because F is a relatively small
number obtained in each case from the difference of two much larger numbers. In
such a circumstance, a small error in one of the larger numbers results in a large
percentage error in the difference.
309. Conservation of Linear Momentum.
Since {R, dt = Amv,, it follows
that if R, is zero, that is, if there is no component of the resultant force in
the x direction, then there is no change of momentum in the 2x direction.
When two bodies strike each other, the impulse on one body is equal to and
of opposite sense to the impulse on the other body.
This being so, the decrease of momentum of one body is equal to the increase of momentum of
the other body, provided no forces except the force of impact are acting in
the direction of the impulse.
In other words, the total momentum of the
two bodies is not changed by the impact.
This conclusion is expressed in:
The law of the conservation of momentum: If a body or a system of
bodies is not acted upon by a net external force in a particular direction,
the linear momentum of the body or system of bodies in that direction
remains unchanged.
Notice that if two or more bodies are involved, the free body is the system or
group of bodies being considered.
Suppose a body A with a mass W.4/g and a velocity v4; strikes a body B
whose mass is Wz/g and whose velocity is vg1, where both va; and vp, are,
say, in the x direction.
After impact, their velocities are va. and vgo, again
in the x direction. Then the momentum of A and B before impact is equal
to their momentum after impact, both computed in the x direction.
In
equation form, we may write
(j):
Wavar
;
a, Weve:
; =
[SUM TAKEN
Wavas
a , Wavee
-
IN 2 DIRECTION]
In applying this equation, we may drop g, since it disappears if we multiply
both sides by g. We should also be careful that the correct sign for each
known velocity is carried into the equation.
If an unknown velocity, assumed in the positive sense, is computed as a negative number, the
negative
sign means that it is directed in the negative sense.
When two bodies come together with an impact that lasts
for only an
instant, we may frequently assume that existing external
forces have a
negligible effect on the instantaneous action of the bodies.
For example,
if a box resting on a rough plane is struck by a bullet,
it may be assumed
that the initial speed of the box is such that the moment
um of the box and
bullet after impact is the same as the original moment
um of the bullet just
before impact, which is to say that the frictional
force between the box and
the plane does not begin to act until the box
has almost instantaneously
LAB
§ 310 |EXAMPLE
the conservation of
acquired its initial speed in accordance with the law of
force of impact
the
This assumption is safely made because
momentum.
that these other forces
is usually very large as compared to other forces, so
are negligible for the instant of impact.
to use components
In the algebraic solution of problems, it is convenient
on (j). Howequati
in
of momentums in a particular direction, as indicated
of bodies comprising a
ever, the total or absolute momentum of a system
sum; that is,
free body remains constant in accordance with a vector
(mv) a1 +
(mv) B1
=
where (mv) a1 is the initial momentum
of B, ete.
(mv) a2 are (mv) po,
of A, (mv)z: is the final momentum
moving toward
A box of sand 4, Fig. 688(a), weighing 10 Ib., is
310. Example.
where
plane,
the left on a horizontal
f = 0.3, with a velocity of 5 fps when
Pet
v,
N
Uy
Up
a
B weighing . 0.2
ae
it is struck by a bullet
1
Ween B
.
.
SS
right
lb., which is moving toward the
C77
8
73
B
with a velocity of 2000 fps. The
WwW
W,
box.
the
in
itself
bullet embeds
(b)
(a)
(a) What is the resulting velocity of
the box? (b) As measured from the
instant of impact, when does the box
come to rest?
Fig, 688. It is assumed that the impact is
consummated in an instant, so that the effect
during impact is negligible.
How far does it move _ of friction
from the point of impact?
right; then 21 = + 2000 fps
(a) Let the positive sense be toward the
Sonution.
have
of the conservation of momentum, we
and va1 = —5 fps. From the principle
mavar + Mave
=
(Mat Mp)V2,
bullet and box after impact.
where v2 is the common velocity of the
Substituting
find
known values and canceling g, we
)v2,
(10)(— 5) + (0.2)(+ 2000) = (10.2
toward the
ive sign indicating that the motion is
from which v2 = + 34.3 fps, the posit
right.
after impact is shown in Fig. 688(b).
(b) The free body of A and B immediately
h
whic
from
We see that N = W = 10.2 Ib.,
F = (0.3)(10.2) = 3.06 lb.
we get
Using the principle & At = Ami,
10.2
(0 — 34.3 Ns
39.9 (0
( =— 3.06)At = ——
is equal to zero.
where the final velocity, say 1,
find
we
ions,
relat
c
mati
kine
From the
V2
2 eei
eae
34.3
=.
35D
9.00
sec.
From this equation, At = 3.55
1DS
9.66)(3.55)?
2
. = (34.3)(3.55) — Meee ae = 60.9 ft.
IMPULSE AND MOMENTUM
446
[Ch. XIX
311. Conservation of Angular Momentum.
By reasoning analogous to
that of § 309, we see that if the external moment M, about a particular
axis on a body or a system of bodies is zero, there will be no change of angular
momentum of the body or system of bodies with respect to this same axis;
that is, if [M, dt = 0, then
|. di = Alc
= 0,
or
Iw = a constant.
This is a statement of the law of the conservation of angular momentum.
Applied to two bodies A and B with moments of inertia [4 and Jz, with
initial angular velocities of w4; and wz1, and with final angular velocities of
wa. and wg, we have
(k)
T4@a1 + Ipwpi = Tawa. + Ipwpr.
The moments of inertia are about the same axis and the angular velocities
should be given signs in accordance with their senses.
312. Direct Central Impact.
The principle of the conservation of momentum is applicable over any finite time interval, provided the external moments
and forces are zero, as already explained.
But this law is especially convenient in impact problems where there is an unknown impulse and force
between two bodies.*
Waa
ee
UT
(a) Approach
DO
(b) Deformation
and Restitution
Fig. 689.
Direct Central Impact.
the same
sense before and after impact, (2) in the same
anes sbiaps
(c) Separation
The velocities of the bodies A and B may be (1) in
sense before impact and in the
opposite sense after impact, (3) in the opposite sense before impact and in the
same
sense after impact, or (4) in the opposite sense before and after impact.
Direct impact occurs when the velocities of the two colliding bodies are
each directed normal to the surfaces at the point of impact.
Central impact
*Christian Huygens (1629-1695) advanced our formal knowledge
of elastic impact
probably more than any one else.
Born in The Hague, he was a noted astronomer and
mathematician, and he did much original thinking in the fields
of optics and mechanics.
He invented an improved method of grinding lenses, developed
the wave theory of light,
and produced writings on centrifugal force which aided Newton.
With his improved
lenses, he discovered the rings of Saturn, one contribution
from many in astronomy.
He
spent 15 years of his professional life in France as a protégé
of Louis XIV.
His paper
on the laws of collision of elastic bodies was presented
before the Royal Society, London,
in 1669. Others who made substantial contributions to
our knowledge of colliding bodies
include Sir Christopher Wren (architect of the famous
St. Paul’s Cathedral), Edmé Mariotte
(a French priest, who independently discovered the
thermodynamic law we call Boyle’s
law),
and Newton
(who went so far as to determine experimental
values of the coeflicient
of restitution for several materials).
All these men were contemporaries.
Those were
the “good old days” when many of the fundamentals
of mechanics were being discovered
and formulated.
44?
§ 312] DIRECT CENTRAL IMPACT
occurs when the force of impact between two bodies is along the line joining
For example, if two homogeneous
the centers of gravity of the bodies.
spheres A and B of equal size, Fig. 689, are moving in the same line of motion,
they experience a direct central impact when they collide. The velocities
that these bodies have after impact depend not only on the law of the conservation of momentum but also on a factor called the coefficient of restituThis factor is defined as the ratio of the relative velocity of separation
fone.
(that is, after impact) fo the relative
velocity of approach. Thus, if the subscrip 1 denotes the velocities of A and
B before impact and the subscript 2 denotes their velocities after impact, the
coefficient of restitution is defined by
Relative velocity of separation
Relative velocity of approach
2
ipa
Oe
Ue
Vp.’
or
Up 2mm UAL
=
66).6sae
Sy
er
UBL ae UAL
VAa2 —
UB2
VAN
UB
ame
In practice, e is always positive and
For perfectly elastic
less than one.
bodies, the value of eis 1, the maximum
or ideal value; that is, they separate
with the same relative velocity with
For plastic
which they approached.
bodies, the value of e is 0, the minimum possible value, in which case, the
bodies have no velocity of separation
the
and therefore move together after
itself
The bullet embedding
impact.
le
in the box of sand, § 310, is an examp
V2.
=
vzz2
and
0
of a case where e =
The reason that ¢ is always less than
of
unity is that there is some loss
on
kinetic energy of the moving bodies
ated
gener
heat
impact because of the
deforby internal friction during the
of e
mation of the bodies. The value
for
varies with the material. It is high
lower
is
It
glass and hardened steel.
cs by
Reproduced by permission from Mechani
Press.
F.W. Sears, published by Addison-Wesley
Fig. 690. Flash Picture of Rebound.
This is a series of flash photographs of a
golf ball
as
it falls
and
bounces
twice
from a stationary iron plate. If the plate
in
is taken as body B, then vsi and Uz:
The velocity with
equation (56) are zero.
may
which the golf ball A strikes the plate
y
be obtained from v? = 2gh. The velocit
same
of rebound can also be found by the
expression applied to the measured height
of rebound, which
suggests an easy way
to determine the coefficient of restitution.
ts
(Why not compute e from measuremen
Since
made on this picture just for fun?)
s was
the time interval between picture
sive
constant, the spaces between succes
to
positions of the ball are proportional
Recall that the ball
the velocity squared.
lic
is moving as a projectile in a parabo
path.
448
IMPULSE AND MOMENTUM [Ch. XIX
for soft steel and quite low for lead.
This property is used to inspect hard-
ened balls for ball bearings.
The balls are allowed to roll down an incline
onto a flat steel plate from which they bounce into a container.
If the
hardness (and size) is right, they fall into a certain container.
If they are
too hard or too soft, they bounce into other containers further from or closer
to the steel plate and are rejected. The size of the ball also affects its bounce.
See Fig. 690.
Impact may be divided into two processes.
The first process is one of
deformation.
It begins at the instant the bodies touch, at which time the
bodies begin to deform or compress each other at the area of contact.
During this interval, the force between the surfaces is increasing and becomes a
maximum at the instant that
compression ceases.
At this instant the bodies are moving with
the same velocity v. The second
process is one of restitution.
If
the bodies have any elasticity
(e > 0), the deformed surfaces
will return to their original shape
or will tend to do so. During this
interval, the force between the
surfaces is decreasing, and it
becomes zero at the instant of
separation.
Little of the details
Cousieed H. E. Edgerton, author of Flash.
Fig. 691.
Golf Ballin Contact with Club.
This
picture caught the ball practically at the end of
deformation period.
After the ball leaves the
club, it will undergo for a while a sort of oscilla-
tion,
a swelling
aps ee
g
and shrinking
8
along
8
the axis
of
of these
processes
can
be deter-
mined.
If the initial
velocities
e
known,
common
oat
we
can
find
velocity v, which
:
the instant
the
occurs
:
that deformation
ceases, from the principle of conservation of momentum:
(1)
MaVai
+ MsBvsi = (m4 + mag)v.
On the other hand, if the jinal velocities are known,
we can find v from the
equation
(m)
(ma + mz)v = mavao + mpvpo.
Then the impulse during the deformation period
is
(n)
ma(v — v1)
or
ma(v — vB1);
and the impulse during the restitution period
is
(0)
Ma(Vss
=
v)
or
Me(vp2
=
v).
449
§ 313 ] EXAMPLE
The impulse on body A is opposite in sense to the impulse on B. The stateof the
ments in connection with the expressions (n) and (0) are true because
In all of
principle that the impulse is equal to the change of momentum.
should
velocity
particular
a
of
value
the preceding expressions, the numerical
691.
Fig.
be substituted with the proper regard for its sign. See
Assume that two freight cars A and B have a direct central im313. Example.
pact. The car A, moving toward the right
(Fig. 692) with a velocity of 3 mph, weighs
60,000 lb. The car B, moving toward the left
with a velocity of 4 mph, weighs 80,000 lb.
If e = 0.5, determine (a) the velocity of
each car just after impact, (b) the impulse
Fig. 692.
during the period of deformation, and (c) the
loss of kinetic energy during impact.
Since the
(a) In this example, vai = + 3 mph and vg1 = — 4 mph.
SoruTion.
does not
v
of
unit
the
(56),
and
velocity occurs in every term of both equations (k)
we get
um,
moment
of
ation
conserv
of
law
the
need to be converted to fps. Thus, from
Warn
Wat
=
W ava2 + Woave.,
(60,000)(3) + (80,000)(— 4) = 60,000 v.42 + 80,000 va»,
3u42 + 4ua2 =
(p)
From the definition of e, we get
—7.
VB2
pa
VB1
—
—
Va2
—4-—(+3)
Val
Vaz — UBQ =
—
=5
a
3.9.
(q)
ly, we find vg. = + 0.5 mph, which is
Solving the equations (p) and (q) simultaneous
which is directed toward the left.
directed toward the right, and v.42 = —3 mph,
end of the deformation period is obthe
at
cars
(b) The common velocity of the
tained from the equation
W avai + W avn
whence
=
(Wa =e Ws),
(3) + (80,000)(—4) = 140,000 »v;
(60,000)
mation period is
Hence, by (n), the impulse during the defor
» = —1 mph.
ma(v — Vai) =
5280\ _
60,000
39.9 C(a1=- 3)(ean)= — 10,900 lb-sec.,
impulse on car A is toward the left.
where the negative sign indicates that the
of A and B are
(c) Before impact, the kinetic energies
5280/3600)? = 18,100 ft-lb.
(0,000)(3
2g
2
2
— 42,900 ft-lb.
(80,000) (4 ae
he
Total kinetic energy before impact = 61,000 ft-lb.
After impact, the kinetic energies are
(60,000)(3 x 5280/3600)?
2g
KEa =
2
(80,000) > 5280/3600)
KEp =
= 18,100 ft-lb.
=
6,700 ft-lb.
Total kinetic energy after impact = 24,810 ft-lb.
450
IMPULSE AND MOMENTUM [Ch. XIX
Thus the loss of kinetic energy is 61,000 — 24,810 = 36,190 ft-lb., the amount
of
energy dissipated into heat because of internal and external frictional effects.
314. Oblique Impact.
If two bodies collide when they do not move on a
common line of motion, they are said to have an oblique impact.
In this
event, it is advisable to apply the law of the conservation of momentum to
the direction which is normal to the surfaces of contact, and it is essential
that in the defining equation for the coefficient of restitution, the components
of the velocities in the direction normal to the surfaces of contact be used.
The following example shows the application of these principles.
315. Example.
Two smooth spheres A and B (Fig. 693) with velocities of v4; =
8 fps and vs: = 10 fps, collide when these velocities are directed at angles of 45° and
30° with the line of centers, as shown.
The spheres are of equal size but A weighs
Wa=5
lb. and B weighs Wz = 2 lb. The coefficient of restitution is e = 0.8.
(a) What are the absolute velocities of these spheres immediately after impact?
(b) What is the loss of kinetic energy?
U4,
(a)
Fig. 693.
(b)
Imagine that these spheres are on a horizontal surface and that the
above
views are those looking down on the surface.
SoLuTion.
(a) The simplest plan is to choose the line normal to the surfaces of
contact as one of the reference axes. Choosing this line to be the x
axis, Fig. 693(a),
we find that the x and y components of the initial velocities of the
spheres are*
UVgiz = — (10)(cos 30°) = — 8.66 fps,
Vaiz = (8)(cos 45°) = + 5.656 fps,
Vay = (10)(sin 30°) = 5 fps,
Vary = (8)(sin 45°) = 5.656 fps.
Now if the spheres are smooth, the y components of
these velocities will not be
changed by the impact because there is no force to cause
acceleration; that is,
VBiy = Vay = 5 fps
and
Vay = Vary = 5.656 fps,
each acting downward in Fig. 693. Thus, we already
know one component of each
absolute velocity after impact. The other compon
ent of each is obtained from the
equations
and
MaVAiz TF Mad Biz = MaAVAre + MBV Bo»
e
=
UB2r
—
VAre .
VBlc
=
VAlx
*The symbols may seem to be unduly long.
However, they have the distinct advantage
of suggesting precisely what they stand for
in a mnemonic sense.
We could use a single
letter for each velocity, introducing some obstac
les to the thinking process, but vpiz says
so clearly that it stands for the initial velocity
of B in the z direction.
461
§ 316 ] EXAMPLE
we have
Substituting numerical values, but using weights instead of masses,
(5)(5.656) + (2)(— 8.66) = 5vaox + 20522,
ograstes
VB22
—
VA2x
ER66 — (15.650)
5)
= — 1.71 fps and Upo2 =
Solving these two equations simultaneously, we find vax
+ 9.74 fps. The absolute velocity of A is
var = [(Vare)? + (vary)? = 5.9 fps,
and its direction [see Fig. 693(b)] is given by
VA _ tan Si 5.656
71 ze ionlome°
a == tan _, yi
The absolute velocity of B is
vp2 = [(vBex)? + (vaw)*|? = 10.93 fps,
and its direction is given by
UB2
g = tan? —— = tan
VU Box
5
9.74
= Oy OP
:
period are pictured approximately in
The conditions just at the end of the restitution
Fig. 693(b).
s is
(b) The original kinetic energy of the two sphere
ae
Wavar
aT oa
, Woaver _
BO:
(2)(10)? ts
(5)(8)?
a
ae
After impact, the kinetic energy is
0.93)?.__
W vee _ (5)(5.9)? , (2)(1Gl
— Waval
6.42 ft-lb.
=F
OLA
ae Bae
2
KE,
8.08 — 6.42 = 1.66 ft-lb. or 1.66/8.08 =
The loss of kinetic energy due to impact is
of kinetic energy is typical of all actual
20.6% of the original kinetic energy. A loss
kinetic energy always decreases except
ved,
impacts. Although momentum is conser
by any actual materials.
when e = 1, an ideal value never realized
is 3 ft. long and weighs 32.2 lb. While
316. Example. A slender rod A, Fig. 694,
it is struck at the point G at a disrest,
at
it is hung from a fixed, smooth pivot and
B which weighs 2 lb. and has a
object
an
by
tance of r = 30 in. from the pivot point
velocity of 25 fps in a horizontal direction.
The coefficient of restitution is e = 0.7.
its
the mass center of A as measured from
What is the maximum height reached by
initial position?
object B is
The initial linear momentum of the
So.urion.
2)
Mpvpi = (sen) =
50
() Ib-sec.
the
the pivot O is obtained by multiplying
The moment of this momentum about
r:
arm
linear momentum by its moment
Mppir
50
= (Bes
= 3.88 ft-lb-sec.
IMPULSE AND MOMENTUM
452
[Ch. XIX
Immediately after the impact, the velocity of B is vz. and the moment of its linear
momentum
(angular momentum) is then
Mapper = h2)(2.9) VB2
32.2
=
0.15520 go ft-lb-sec.
The moment of inertia of the 3-ft. rod A with respect to O is (see § 164)
I, =
mL?
(32.2)(3)?
3
(32.2)()
= 3slug-ft?
The rod’s initial angular momentum (moment of momentum) is I,w; = 0, because
w, = 0. The rod’s final angular momentum is (using w. = v¢2/7)
Tve2
Iwo =
~~ (3)ve2
M995
=
1.2v¢@2 ft-lb-sec.,
where vq is the linear velocity of the point of contact @
in the rod, Fig. 694. Using the principle that the angular
momentum of A and B with respect to O before impact
is equal to their angular momentum with respect to O
after impact, we have
(r)
3.88
+
0 =
0.1552v
z5 a
1.20 @.
In applying the definition of the coefficient of restitution to this case, use the velocities of B and of the
point of contact G@ on the rod; thus
ay
vB2 —
ad
Yar
as VE
VBi — Ve
eas
25 — 0
0-7,
from which
(s)
Veo — Veo = 17.5
A simultaneous solution of (r) and (s) gives vg. = 4.87 fps.
A just after impact is then
VE2
ee
aie
4.87
Te
The angular velocity of
1.95 rad. per sec.
All the kinetic energy imparted to A by the impact is used to do work against
gravity
before the link A reaches its highest position. The kinetic energy of
A after impact is
(3) (1.95)?
ow
KE =
= 5.7 ft-lb.
2000
2
Let h be the increase in elevation of the center of gravity
of A. Then the work
against gravity is W 4h, which is equal to the loss of kinetic
energy; we have Wah =
AKE = 5.7 ft-lb., or
h
(NorE.
_AKE _ (5.7)(12)
Wat
982.0
= 2.11 ink
If the rod A had had an initial angular speed w1,
it would have been given
a positive sign if in a counterclockwise sense,
a negative sign if in a clockwise sense.
This action would agree with the assumption made,
namely, that the positive sense
is toward the right at the point G.)
PROBLEMS: LINEAR IMPULSE AND MOMENTUM
458
Although the resultant impulse is equal to the vector
317. Closure.
to
change of momentum, it is convenient in the solution of most problems
axes
y
and
x
is,
that
m;
consider collinear changes of impulse and momentu
= Am,,
are chosen and the principle is then applied in the form fR,dt
In any case,
where R, and v; are the x components of FR and v respectively.
against
insuring
in
helpful
most
be
a complete free body diagram will again
the unintentional omission of some force.
While the frictional force between a plane and a rolling body does no work
when
(§ 280), it does have an effect on impulse; hence it must be included
the principles of this chapter are used.
y need to
When the impulse is due to a sharp blow or impact, we generall
free body
e
consider only the impulsive force. In this case, the complet
of velocities and
diagram is not so useful, but a sketch showing the senses
ly.
material
errors
the chosen positive sense will reduce
le of work and
You might recall at this time that we obtained the princip
value from the kinekinetic energy from F = ma by substituting for a the
principle of linear immatic relation v dv = ads; and that we obtained the
the kinematic relation
pulse and momentum by using the value of a from
a = dv/dt in F = ma.
tion, the three
For a body in plane motion, that is, rotation and transla
t or R is a function of ¢)
conditions which apply (when # and m are constan
~.
it
iRy dt = m(G2 — Bx)
(u)
[® dt = m(0,2 — dyi);
(v)
ie dt =
I,(w2 — 1),
y conveniently either the
where the reference point O for M, and T, is usuall
of gravity ($306). Note
instantaneous center of rotation or the center
tion of the coefficient of elasticity,
that these equations, together with the defini
ion (v) does not apply (in the
summarize the theory of this chapter. Equat
translation; equations (t) and
sense that w is zero) for a body in rectilinear
a fixed center of gravity.
(u) do not apply for a body in rotation about
Problems
by the principles explained in this chapter,
Nore. The following problems are to be solved
these principles cannot be applied or where
together with kinematic relations, except where
be used.
may
ples
the statement indicates that other princi
454
PROBLEMS: LINEAR IMPULSE AND MOMENTUM
LINEAR
IMPULSE
1531. A 3000-lb. automobile coasts down
a 2% grade for 3 sec. The constant resistance parallel to the road is 10 lb. What is
the net impulse in the direction of motion?
Ans. 150 Ib-sec.
1532. A force of F = 12t2? +¢-+8 lb.,
where ¢ is in seconds, acts on a body for 0.1
min.
What is the corresponding impulse?
Ans. 930 lb-see.
1533. A force acting on a body is fF =
6? + 2¢ lb., where ¢ is in seconds.
What
is the linear impulse of this force during
10 sec.?
1534. How long would it take a falling
body to speed up from 10 fps to 90 fps, if
the air resistance were negligible?
1535. A block slides from rest down a
plane whose slope is 3/4. The coefficient
of friction is 0.1. What is the impulse in
the direction of motion during the 4th sec.?
What velocity does the block have at the
end of 4 sec.?
Ans. 0.52W Ib-sec., 67 fps.
1536. A constant force Q acts horizontally to move a 500-lb. body up a 30°
incline, where f = 0.25.
If the speed of the
body changes from 5 fps to 15 fps in 45 sec.,
what is the magnitude of Q?
Ans. 488 lb.
1537. A 1000-Ib. body slides down a 30°
incline, where f = 0.2, under the action of
gravity.
How long does it take for its
speed to change from 20 fps to 50 fps?
Ans. 2.85 sec.
1538. A 1000-lb. weight, sliding on a
rough horizontal surface, comes to rest from
an initial speed
of 100 fps in 10 sec.
If
friction is the only retarding force, what is
the coefficient of friction?
Fig. 695.
Problems 1539-1541.
1539. A 2000-lb. body A is connected to
a 3000-Ib. body B by a weightless rigid rod
C, Fig. 695, and they are on a @ = 30° in-
cline.
The coefficients of friction are fa=
0.1 and fg = 0.2. If the bodies start from
rest, how long will it take for them to acquire
a speed of 30 fps? What is the force on the
rod C?
Ans. 2.58 sec., 103 lb. (C).
AND
[Ch. XIX
MOMENTUM
1540. The
same
as
1539
except
that
@ = 45°.
1541. The same as 1539 except that
aloe
1542. A 6000-lb. airplane, moving at 300
mph, crashes into a “rigid” tree. If the
plane is brought to rest in 2 sec., what is
the average force on the tree?
1543. A planing machine table and the
part on it weigh 10 tons and reciprocate in
a horizontal plane.
If the average controlling force is 500 lb., how long does it
take to change the velocity of the table
from 60 fpm toward the right to 180 fpm
toward the left?
Ans. 4.97 sec.
1544.
A 30,000-ton steamship, moving at
0.1 mph, is about to dock.
If it strikes the
dock and comes to rest in 10 sec., what is the
average force on the dock during this time?
From the results obtained, would you think
that the docking of such a large boat
appears to offer difficulties?
Explain.
1545. The initial speed of a body weighing W lb. is 5 fps on a horizontal surface,
where f = 1/3. A force of F = 5¢3/2 Ib.,
where ¢ is in seconds, acts to increase the
speed to 15 fps in 9 sec. What is the
weight W?
Ans. 146.8 lb.
1546. The same as 1545 except that
1
feinles
1547. A 160-lb. man is standing in a
motionless boat.
He walks the 20-ft. length
of the boat with a speed of 3 fps relative
to the boat.
The boat weighs 2000 Ib.
How far does the frictionless boat move
while the man is walking?
Ans. 1.6 ft.
1548. A 20-lb. body initially at rest is
given a sharp horizontal blow whose duration is estimated at 0.5 sec. The body
moves on a rough horizontal plane, coming
to rest with constant acceleration in a
distance of 20 ft. and after 2 sec.
What
was the average value of the force of the
blow?
1549. If a machine gun fires 300 bullets
per minute, each weighing 0.06 Ib, and
having a muzzle velocity of 2000 fps, what
is the average resisting force necessary to
keep the gun in place? The reaction from
the expansion of the gases is to be neglected.
Ans. 18.61 Ib.
1550. A 3-in. fire hose discharges water
at 1500 Ib. per min. through a 1-in, nozzle.
What force is required to hold the nozzle?
1551. A cylindrical jet of water, 3 in. in
diameter and moving with an absolute
PROBLEMS:
LINEAR
IMPULSE
AND
456
MOMENTUM
1558. Water enters a 4-in. pipe at a point
300 ft. above the discharge point. The
discharge is into a hemispherical bowl which
reverses the direction of the velocity of the
The friction in the bowl is to be
water.
neglected, but the friction in the pipe results
If
in a loss of 20% of the available head.
velocity of 30 fps, strikes a flat plate whose
surface is at right angles to the jet, Fig. 696.
What is the force exerted on the plate by
the water, (a) if the plate is fixed (vp = 0),
(b) if the plate moves at vp = 8 fps in the
same sense as the water, (c) if it moves at
vp = 8 fps in the sense opposite to that of
the water?
the bowl is fixed, what is the force exerted
(a) The same
1559.
as 1558 except that
the bowl is used as a bucket on a hydraulic
turbine and moves in the same direction
as the jet with a speed of 40 fps. (b)
Determine the horsepower being developed.
(c) If a valve at the lower end of the pipe
Plate
——
Uw1
Fig. 696.
Ans. 5230 lb.
on it by the water?
Ans. (a) 85.7 lb.; (b) 46.2 lb.; (c) 187 lb.
were closed in 1 sec., what would be the
Would a closaverage force on the valve?
ing in this time seem advisable?
Ans. (a) 2400 lb.; (b) 175 hp;
Problems 1551-1554.
(c) 2600 lb., no.
1552. A jet of water, whose cross-sectional
area is 15.7 sq. in., strikes a flat surface
which is moving at vp = 15 fps in the same
sense
as the water,
If the dis-
Fig. 696.
charge from a stationary nozzle is 600 cfm,
what horsepower is being developed in driving the flat surface?
1553. A jet of water, 4 in. in diameter and
moving at vw = 60 fps, strikes a flat plate
which is at right angles to the jet, Fig. 696.
What is the speed of the plate, if the force
on it due to the jet is 500 lb.? Ans. 5.7 fps.
1554. A vane, constructed as a flat surby
face and moving at vp, = 20 fps, is struck
a jet of water which has a sectional area
the
of 18 sq. in. and a velocity of 120 fps in
and
vane
the
of
y
velocit
the
as
sense
same
(a)
at right angles to the vane, Fig. 696.
the
What horsepower is being developed at
(b) For what speed of the vane
vane?
horsepower be a maximum?
the
would
1555. A 90-mph wind strikes the vertical
ft.
wall of a house whose area is 600 sq.
|b.
The specific weight of the air is 0.07
per cu. ft. and the wind moves horizontally.
r
If the direction of the wind is perpendicula
on
wind
the
of
force
the
is
what
wall,
to the
the wall?
Ans. 22,700 lb.
the
1556. The same as 1555 except that
the
direction of the wind is at 60° with
plane of the wall.
di1557. A sailboat moving at 5 knots
spread
rectly with the wind has its sail
the wind
perpendicular to the direction of
specific
The
mph.
30
is
ity
veloc
whose
ft. and
weight of the air is 0.07 Ib. per cu.
What is
the area of the sail is 250 sq. ft.
the force of the wind on the sail?
(1 knot
=
1 nautical mph = 6080 ft. per hr.)
Ans. 688 lb.
VAS
Fig. 697.
1560.
Problems 1560-1562.
A jet of water issues from a nozzle
with a velocity of vw: = 200 fps and at the
absolute rate of 5 lb. per sec.
fixed
blade
(vg = 0), shaped
It enters a
as shown
in
Fig. 697 where 0 = 60°, and it passes through
the blade with a negligible frictional loss.
What are the horizontal and vertical com-
ponents of the force exerted
on the blade?
Ans. 46.6 lb., 26.9 Ib.
1561. The data are the same as in 1560
except that vs = 80 fps. Determine the
horizontal and vertical components of the
force on the blade and the horsepower being
Define the resultdeveloped at the blade.
ant force on the blade in magnitude and
direction.
1562. In Fig. 697, the fluid approaching
the blade is steam, whose absolute velocity
is 1000 fps and whose specific weight is 0.05
lb. per cu. ft.
The area of the section of the
stream is 1 sq. in.; also,
500
fps.
Determine
the
9 = 60° and vg =
horizontal
and
vertical components of the force on the
blade and the horsepower being developed.
Ans. 4.04 lb., 2.33 lb., 3.68 hp.
456
PROBLEMS: LINEAR IMPULSE AND MOMENTUM
[Ch. XIX
some higher pressure by a so-called diffuser
section.
If the water enters at 20 fps at
right angles to the steam flow and if 10
lb. of water are pumped per pound of steam,
determine the speed of the mixture just
after thorough mixing (without frictional
loss) and compute the loss of kinetic energy.
Ans. 182 fps, 56,500 ft-lb. per lb. of steam.
Fig. 698.
Problems 1563, 1564.
1563. A jet of water with an absolute
velocity of v1 = 100 fps flows at the absolute rate of 50 lb. per sec. and enters a
blade shaped as shown in Fig. 698 where
= 60°. The velocity of the blade is vg =
40 fps. Determine the horizontal and vertical components of the force on the blade
and the horsepower being developed at the
blade.
Ans. 27.95 lb., 48.4 Ib., 2.03 hp.
1564. The
same
as
1563
except
that
@ = is.
y
Fig. 699.
Problem 1565.
1565. A jet of water A, whose velocity
is v4 = 50 fps, strikes a blade B which is
moving at vg = 20 fps in the same direction
as A, Hig. 699.
Six pounds per second of
water strike the blade, half passing without
friction along the upper part, half across
the lower.
What force in the y direction
is necessary to prevent movement of the
blade in that direction?
What is the horizontal thrust on the blade?
Ans. 1.025 lb., 9.4 Ib.
1566. A weight W is projected up a 30°
incline with an initial velocity of 32.2 fps.
Let f, =0.5 and f, = 0.3. After 5 sec.,
(a) what is the body’s velocity; (b) what
is its displacement from the starting point?
1567. The same as 1566 except that
fz = 0.6.
1568. An injector is a pumping device.
Steam is expanded to a low pressure where
it has a high velocity, say, 2000 fps. This
low pressure region is connected to a source
of water, say, which enters the low pressure
region and mixes with the fast-moving
steam.
ip
Fig. 700.
The
mixture
is then brought to
1569. A
1000-lb.
Problem 1569.
body
A,
Fig.
700,
moves with rectilinear motion on a horizontal plane, where f = 0.25. It is driven
by a variable force Q = 360 cos @ lb., where
6 = rt/120, t being in seconds.
When Q is
initially horizontal (@ = 0), the body is at
rest. How long will it be before the body
attains its maximum speed?
Ans. 18 sec.
1570. A 4000-lb. automobile with a fluid
torque-converter transmission might conceivably have its engine so throttled that it
continuously delivers 100 hp to the rear
wheels.
Observe that the driving force is
variable and assume that the rear wheels
do not slip. What is the speed of the car
after 7.12 sec. if the initial speed is 30 mph?
See problem 1488.
Ans. 90.8 fps.
1571. The same as 1570 except that the
initial speed is 30 fps.
1572. A perfectly flexible chain 10 ft.
long weighs 20 lb. per ft. Initially, it hangs
with its lower end 15 ft. above the floor.
If the chain is released and falls freely,
find the maximum force on the floor.
Ans. 1200 lb.
1573. A jet engine on an airplane takes
in air at 600 mph, the speed of the plane,
and the exhaust gases leave at a speed of
1200 mph, both relative to the plane. If
100 lb. of air are taken in per pound of
kerosene burned, find the fuel consumption
in pounds per second and gallons per minute
in order to sustain a 5000-lb. thrust.
The
kerosene weighs 6.5 Ib. per gal.
1574. Scientists estimate that exhaust
gases leave a certain rocket at 52,500 fps.
Its fuel and oxygen are consumed at a rate
of 20 lb. per sec. Find the thrust (a) just
prior to take-off and (b) in flight at 1000
mph.
Ans. (a) 32,600 Ib.; (b) same,
PROBLEMS:
ANGULAR
IMPULSE
ANGULAR
AND
45?
MOMENTUM
IMPULSE
AND
MOMENTUM
1575. In Fig. 701, a force Q = 100 lb. is
applied to the braking lever, as shown.
The brake wheel and other attached rotating
wraps about a solid disk B which weighs
Neglecting
644 lb. and is 2 ft. in diameter.
frictional effects, determine the time taken
for A to attain a speed of 20 fps from an
initial speed of 10 fps, and determine the
De=A2eins\.e = 0 (brake
jn b =6ainga
arm is bent so that the pivot is on a tangent
to the brake wheel), and fz = 0.25 at the
tension in the cord.
parts weigh 193.2 Ib. and have a radius of
gyration of 9 in. Other data are: a = 18
brake shoe. If the speed is 10 rad. per sec.
in a counterclockwise sense at the instant
the brake is applied, how long will it be
before the rotating parts are brought to
rest? What is the angular acceleration?
Ans. 1.862 sec., 53.6 lb.
1579. The same as 1578 except that
Vie —slOniulins
1580. The same as 1578 except that the
peg C is not smooth and fc = O3
Fig. 703.
Problem 1581.
1581. In Fig. 703, Wa = 64.4 lb., fe =
0.2 for A, @ = 30°, We = 966 lb., and D =
4 ft. The cord from A wraps about the
Fig. 701.
Problems 1575-1577.
cylinder B. Find the velocity of A 5 sec.
after it is released from rest.
Ans. 6.18 fps.
1576. In Fig. 701, let a = 24 in., b=6
and
in, D = 18 in., e = 2 in., Q = 50 lb.,
fp = 1/3 at the brake shoe. The suportcoefhiing shaft is 3 in. in diameter and the
0.01.
is
s
bearing
the
in
friction
of
cient
The rotating parts weigh 6440 lb. and have
a radius of gyration of 2 ft. If the speed
is
is 120 rpm clockwise when the brake
come
wheel
the
does
time
what
in
,
applied
Determine the angular accelerato test?
by the
tid and the number of turns made
wiibel during this period.
eve
Ans. 2.6 min., 0.081 rad. per sec.’, 15D
the
1577. The same as 1576 except that
e.
ckwis
erclo
count
is
on
rotati
Fig. 702.
Problems 1578-1580.
702, is
1578. A 64.4 lb. body A, Fig.
that passes
suspended from a weightless cord
then
= 0) and
over a smooth peg at C (fe
Fig. 704.
Problems 1582, 1583.
r
1582. In Fig. 704, the uniform slende
=
W
weight
and
ft.
6
=
L
rod, of length
posi32.2 lb., rotates from rest in a vertical
Neglecting
tion under the force of gravity.
90°
friction, determine the angular velocity
e
after rotation begins. What is the averag
?
angular impulse during this displacement
The principle of work and kinetic energy
may be used if desired.
Ans. 4.01 rad. per see., 48.1 ft-lb-sec.
or
1583. Find the time (approximate
in
exact) required for the 90° of rotation
1582.
1584. A D = 18-in. solid cylinder A
g
weighs 1288 Ib. It rolls without slippin
458
PROBLEMS: ANGULAR IMPULSE AND MOMENTUM
[ Ch. X LX
down a 0 = 30° incline, Fig. 705, its eg
having an initial speed of 10 fps. How long
is it before its eg has a speed of 30 fps?
What is the force of friction between the
cylinder and the plane?
z
Ans. 1.875 sec., 215 lb.
Fig. 707.
Problem 1590.
1590. In Fig. 707, a force Q is exerted
on a cord which wraps about the midsection
Fig. 705.
Problems 1584-1587.
1585. A homogeneous sphere, D = 12 in.,
rolls down a rough 6 = 30° inclined plane,
Fig. 705.
If the initial speed of its center
of gravity is 10 fps down the plane, find
the speed 8 sec. later, and find the minimum
value of the coefficient of friction that will
cause pure rolling.
Ans. 102 fps, 0.165.
of the cylinder.
Cylindrical hubs, with
D, = 2 ft., project from each side of the
D», = 4-ft. cylinder, and rest on tracks, on
which the member rolls. The weight of
the whole is 800 lb. and its radius of gyration
is kK= 1.5 ft.
If, starting from rest, the
cg reaches a speed of 20 fps in 4 sec. with
constant acceleration, determine the magnitude of Q, the frictiona] force at the
tracks, and the acceleration of the eg.
Ans. 404 lb., 280 Ib., 5 fps? to left.
1586. The same as 1585 except that the
body is a 12-in. cylinder.
1587. The same as 1585 except that the
body is a 12-in. cylindrical shell of negligible
thickness.
Fig. 708.
Problems 1591, 1592.
1591. In Fig. 708, the 4-ft. solid cylinder
A weighs 644 lb. and @ = 30°. The weight
of B is 193.2 lb. and the pulley C has negligible weight and friction.
The system
starts from rest.
Determine the velocity
of the cg of A after 10 sec., the tension Q
in the cord, and the frictional force F
between A and the incline.
Fig. 706.
Problems 1588, 1589.
1588. A 3-ft. solid disk A is supported
by a weightless cord which wraps about its
midsection, Fig. 706.
It weighs 128.8 lb.,
and starting from rest, it rolls downward
in a vertical plane.
What is its angular
velocity after 5 sec.? Determine the tension
in the cord and the acceleration of the cg.
Ans. 71.5 rad. per sec., 42.9 lb., 21.44 fps?.
1589. The same as 1588 except that k =
1.25 ft. instead of the body being a solid disk.
Ans. 12 fps, 179 Ib., 167 Ib.
1592. The same as 1591 except that
We = 96.6 lb.
1593. A fireboat has a nozzle on its bow
ejecting water in a starboard direction and
at 30° above the horizontal.
The nozzle
stream has a diameter of 2 in. and a discharge rate of 800 gal. per min. at 8.2
lb.
per gal. Find the downward thrust on
the
bow and the turning moment on the
boat
about its mass center, which is 25 ft.
astern
from the nozzle.
The intake is from
reservoir in the hold of the boat.
a
PROBLEMS:
CONSERVATION
459
OF MOMENTUM
1594. A gun weighing 160,000 lb. fires a
900-lb. projectile whose muzzle velocity is
1400 fps. Neglecting the reaction due to
the expanding gases, determine the maximum speed of recoil.
If the axis of the
gun is horizontal, what is the average value
of the force resisting recoil when the recoil
is 6 ft.?
Ans. 7.86 fps, 25,700 lb.
1595. A 70-ton gun is mounted on a 30ton railway car.
The resistance to motion
of the car and gun is represented by a force
of 15 lb. per ton parallel to the track. The
gun is elevated 30° above the horizontal and
is fired with the car on a level track. The
projectile weighs 1200 Ib. If the car ran
50 ft. in recoil, what was the muzzle velocity
of the shell?
1596. An 8-lb. body A moving with a
velocity of 12 fps is struck by a 16-lb. body
B moving with a velocity of 18 fps in the
same sense as the velocity of A. If the
bodies move together after impact, what
Ans. 16 fps.
is their velocity?
1597. A 10-lb. body A moving at 18 fps
toward the right is struck by a 30-lb. body
B moving at 8 fps toward the left. If the
bodies move together after impact, what is
their velocity?
1598. A 0.75-oz. bullet moving horizontally with a velocity of 1800 fps imbeds
itself in an 8-lb. block of wood which rests
on a horizontal plane, where f = 0.3. How
What is the loss
far does the block move?
of kinetic energy at impact?
Ans. 5.78 ft., 2246 ft-lb.
1599. A 10-lb. projectile moving horizontally imbeds itself in a 966-lb. body which
is at rest on a horizontal plane, where
The body, with the projectile,
f =0.25.
moves 12 ft. along the plane after the
impact.
(a)
What
is the
speed
of
the
(b)
projectile when it strikes the body?
What is the loss of kinetic energy at impact?
Ans. (a) 1360 fps, (b) 283,000 ft-lb.
1600. A 5-lb. body A with a velocity of
20 fps toward the right strikes and attaches
to a body B whose velocity is 10 fps toward
Their common velocity after impact is 2 fps toward the left. What is the
weight of B?
1601. A 100-lb. projectile with an absolute velocity of 2000 fps upward and parallel
to a 30° incline embeds itself in a 1900-Ib.
(a)
body which is at rest on the incline.
after
time
of
length
what
for
0.25,
=
f
If
the impact will the body (and projectile)
the left.
MOMENTUM
OF
CONSERVATION
move?
(b)
What
is the
loss
of kinetic
energy at impact?
Ans. (a) 4.33 sec.; (b) 5,909,000 ft-lb.
1602. The same as 1601 except that the
body struck by the projectile weighs 4900 lb.
1603. A 60-lb. block of material B is
moving with a velocity of 20 fps up a
rough 15° incline when it is struck by a
0.25-Ib. projectile moving parallel to and
upward along the incline with a velocity of
1800 fps. The projectile imbeds itself in
the block.
The coefficient of friction is
f = 0.3. (a) During what Jength of time
does
the
block
move
after
(b) How far does it move?
the
impact?
(c) How much
kinetic energy is lost at impact?
Ans. (a) 1.55 sec.; (b) 21.3 ft.; (c) 12,260 ft-lb.
1604. The same as 1603 except that the
block B is moving down the incline.
1605. The same as 1603 except that the
velocity of the projectile is horizontal, with
a component wp the incline.
1606. If bodies A and B collide with
plastic impact, v2 = vz2, prove that
Loss of KE
1 ma mp (vA — VB1)?
= 9
ae
1607. Two rotating, 6-ft. disks may be
One disk A
suddenly coupled together.
weighs 600 lb. and turns clockwise at 90
rpm. The other disk B weighs 900 lb. and
(a) What
turns counterclockwise at 20rpm.
is their common angular velocity at the
(b) If, at this instant,
instant of coupling?
a braking force of 30 lb. is applied tangentially to the disks, in what time are the
disks brought to rest?
Ans. (a) 2.51 rad. per sec., c; (b) 5.85 sec.
1608. The same as 1607 except that both
disks initially turn clockwise.
1609. Two shafts whose axes are collinear
The shaft A
may be suddenly connected.
with its attached rotating parts turns at
120 rpm, weighs 9660 Ib., and has a radius
of gyration of 4 ft. relative to its axis.
The shaft B with its attached rotating parts
turns at 40 rpm, weighs 12,880 Jb., and has
a radius of gyration of 5 ft. Both shafts
At the instant
turn in the same sense.
after these shafts are coupled, a constant
braking force of 50 lb. acts tangentially on
a brake drum at a radius of 2 ft. What
time is required to bring the shafts to rest?
Through how many revolutions do they turn
during this interval?
Ans. 17.05 min., 565 rev.
1610. The same as 1609 except that the
shafts are initially turning in opposite senses.
460
PROBLEMS: CONSERVATION
1611. A man is standing on a rotating
platform and holds a weight in each hand.
With his hands close to his body, the
moment of inertia of his body and the
weights with respect to the axis of rotation
is 1 slug-ft.2 With the weights extended at
arm’s length, the moment of inertia is
increased to 5 slug-ft.2 If he turns at 40
rpm with his arms down and then suddenly
extends his arms, what is his resulting rpm?
Ans. 8 rpm.
1612. A revolving stage floor is a cylindrical disk 24 ft. in diameter weighing
1288 lb. It rotates on a vertical shaft at
its geometric axis. Friction may be neglected.
It is rotating at such a speed that
a 161-lb. man standing on its rim is traveling
at 24 fps tangential velocity.
No power is
being applied.
The man proceeds to walk
to the center of the rotating stage floor.
What is the angular velocity of the stage
after the man, considered as a particle,
reaches its center?
What net work does
the man do?
1613. A fancy diver has a moment of
OF MOMENTUM [Ch. XIX
the sphere is hanging at rest from a weightless flexible
cord
10 ft. long.