Now in its sixth edition, the text continues to empower students with
• a variety of problems for practice. Sections end with Fundamental Problems that require students to focus on basic
applications to develop their problem-solving skills. Other problems of various levels of difficulty engage students in thinking
through real-life situations. Review Problems at the end of each chapter improve learning outcomes by encouraging students
to revisit the most important topics covered.
• the Procedure for Analysis feature. This provides students with a logical and orderly method for applying theory, guiding
students toward mastery of principles and greater confidence.
• an emphasis on free-body diagrams. Specific sections and problems are devoted to drawing free-body diagrams, a skill
essential to solving equilibrium problems.
New to This Edition
• Over 400 new problems have been added to test students’ understanding of basic concepts.
• The answer section has now been expanded with additional information related to the solution of select Fundamental
Problems.
• More photos and photorealistic art throughout the book illustrate how principles apply to real-world situations and how
materials behave under load.
• Videos, developed by the author, that summarize key concepts discussed in the text, demonstrate solving problems, and
model the best way to reach a solution
• Tutorial homework problems that emulate the instructor’s office-hour environment, guiding students through concepts in
multi-step problems
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Hibbeler
• Enhanced feedback, specific to student errors, that make it easier to solve the tutorials and many end-of-section problems,
while also improving the learning experience. Moreover, optional hints, which break problems down into simpler steps,
empower students to make more informed choices while working on the solutions
SIXTH EDITION
IN SI UNITS
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teaching and learning platform that empowers instructors to personalize learning for every student. When combined with
Pearson’s trusted educational content, this optional suite helps deliver the desired learning outcomes. The Mastering course
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Statics and Mechanics
of Materials
Russell Hibbeler’s Statics and Mechanics of Materials combines two of the author’s bestselling texts, Engineering Mechanics: Statics,
and Mechanics of Materials, with a firm focus on concepts that are commonly encountered in engineering practice. This text
features a thorough presentation of the theory and applications of the most fundamental topics of these two important branches
of mechanics.
GLOBAL
EDITION
GLOB AL
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GLOBAL
EDITION
This is a special edition of an established title widely used by colleges
and universities throughout the world. Pearson published this exclusive
edition for the benefit of students outside the United States. If you
purchased this book within the United States, you should be aware that
it has been imported without the approval of the Publisher or Author.
Statics and Mechanics
of Materials
SIXTH EDITION IN SI UNITS
R. C. Hibbeler
03/01/24 5:13 PM
STATICS AND
MECHANICS
OF MATERIALS
SIXTH EDITION IN SI UNITS
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STATICS AND
MECHANICS
OF MATERIALS
SIXTH EDITION IN SI UNITS
R. C. HIBBELER
SI Conversion by
Jun Hwa Lee
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Product Management: Neelakantan K. K., Aurko Mitra, and Deeptesh Sen
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© 2024 by R. C. Hibbeler
The right of R. C. Hibbeler to be identified as the author of this work has been asserted by him in accordance with the Copyright, Designs
and Patents Act 1988.
Authorized adaptation from the United States edition, entitled Statics and Mechanics of Materials, Sixth Edition, ISBN 978-0-13-796489-5,
by Russell C. Hibbeler, published by Pearson Education, Inc. © 2023.
All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted in any form or by any
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ISBN 10 (Print): 1-292-46020-2
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To the Student
With the hope that this work will stimulate
an interest in Engineering Mechanics and
Mechanics of Materials and provide
an acceptable guide to its understanding.
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PREFACE
This book represents a combined abridged version of two of the author’s books,
namely Engineering Mechanics: Statics, Fifteenth Edition in SI Units, and Mechanics
of Materials, Eleventh Edition in SI Units. It provides a clear and thorough
presentation of both the theory and application of the important fundamental
topics of these subjects, that are often used in many engineering disciplines. The
development emphasizes the importance of satisfying equilibrium, compatibility
of deformation, and material behavior requirements. The hallmark of the book,
however, remains the same as the author’s unabridged versions, and that is, strong
emphasis is placed on drawing a free-body diagram, and the importance of selecting
an appropriate coordinate system and an associated sign convention whenever
the equations of mechanics are applied. Throughout the book, many analysis and
design applications are presented, which involve mechanical elements and structural
members often encountered in engineering practice.
New to this Edition
• Expanded Answer Section. The answer section in the back of the
book now includes additional information related to the solution of select
Fundamental Problems in order to offer the student some guidance in solving
the problems.
• Re-writing of Text Material. Some concepts have been clarified further
in this edition, and throughout the book the accuracy has been enhanced, and
important definitions are now in boldface throughout the book to highlight their
importance.
• New Photos. The relevance of knowing the subject matter is reflected
by the real-world applications depicted in the many new or updated photos
placed throughout the book. These photos generally are used to explain how
the relevant principles apply to real-world situations and how materials behave
under load.
7
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8
PREFACE
• New Problems. There are approximately 30% new problems that have
been added to this edition, which involve applications to many different fields
of engineering.
• New Videos. Three types of videos are available that are designed to
enhance the most important material in the book. Lecture videos serve to test
the student’s ability to understand the concepts. Example problem videos are
intended to review these problems, and fundamental problem videos guide the
student in solving these problems that are in the book.
Hallmark Features
Besides the new features just mentioned, other outstanding features that define the
contents of the text include the following.
Organization and Approach. Each chapter is organized into well-defined
sections that contain an explanation of specific topics, illustrative example problems,
and a set of homework problems. The topics within each section are placed into
subgroups defined by boldface titles. The purpose of this is to present a structured
method for introducing each new definition or concept and to make the book
convenient for later reference and review.
Chapter Contents. Each chapter begins with a photo demonstrating a broadrange application of the material within the chapter. A bulleted list of the chapter
contents is provided to give a general overview of the material that will be covered.
Emphasis on Free-Body Diagrams. Drawing a free-body diagram is
particularly important when solving problems, and for this reason, this step is strongly
emphasized throughout the book. In particular, within the statics coverage some
sections are devoted to show how to draw free-body diagrams. Specific homework
problems have also been added to develop this practice.
Procedures for Analysis. A general procedure for analyzing any mechanics
problem is presented at the end of the first chapter. Then this procedure is customized
to relate to specific types of problems that are covered throughout the book. This
unique feature provides the student with a logical and orderly method to follow when
applying the theory. The example problems are solved using this outlined method in
order to clarify its numerical application. Realize, however, that once the relevant
principles have been mastered and enough confidence and judgment have been
obtained, the student can then develop his or her own procedures for solving problems.
Important Points. This feature provides a review or summary of the most
important concepts in a section and highlights the most significant points that should
be realized when applying the theory to solve problems.
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PREFACE
9
Conceptual Understanding. Through the use of photographs placed
throughout the book, the theory is applied in a simplified way in order to illustrate
some of its more important conceptual features and instill the physical meaning
of many of the terms used in the equations. These simplified applications increase
interest in the subject matter and better prepare the student to understand the
examples and solve problems.
Fundamental Problems. These problems may be considered as extended
examples, since the key equations and answers are all listed in the back of the book.
Additionally, when assigned, these problems offer students an excellent means of
preparing for exams, and they can be used at a later time as a review when studying
for the Fundamentals of Engineering Exam.
Homework Problems. Apart from the Fundamental type problems men-
tioned previously, other types of problems contained in the book include the
following:
• General Analysis and Design Problems. The majority of problems in the
book depict realistic situations encountered in engineering practice. Some of
these problems come from actual products used in industry. It is hoped that this
realism will both stimulate the student’s interest in engineering mechanics and
provide a means for developing the skill to reduce any such problem from its
physical description to a model or symbolic representation to which the principles
of mechanics may be applied.
In any set of problems, an attempt has been made to arrange the problems in order
of increasing difficulty, except for the end of chapter review problems, which are presented in random order. Problems that are simply indicated by a problem number
have an answer given in the back of the book. However, an asterisk (*) before every
fourth problem number indicates a problem without an answer.
Accuracy. As with the previous editions, apart from the author, the accuracy of
the text and problem solutions has been thoroughly checked in part by Kai Beng
Yap and Jun Hwa Lee, along with a team of specialists at EPAM, including Georgii
Kolobov, Ekaterina Radchenko, and Artur Akberov.
Contents
The book is divided into two parts, and the material is covered in the traditional
manner.
Statics. The subject of statics is presented in six chapters.The text begins in Chapter 1
with an introduction to mechanics and a discussion of units. The notion of a vector and
the properties of a concurrent force system are introduced in Chapter 2. Chapter 3
contains a general discussion of concentrated force systems and the methods used to
simplify them. The principles of rigid-body equilibrium are developed in Chapter 4
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10
PREFACE
and then applied to specific problems involving the equilibrium of trusses, frames, and
machines in Chapter 5. Finally, topics related to the center of gravity, centroid, and
moment of inertia are treated in Chapter 6.
Mechanics of Materials. This portion of the text is covered in 10 chapters.
Chapter 7 begins with a formal definition of both normal and shear stress, and a
discussion of normal stress in axially loaded members and average shear stress caused
by direct shear; finally, normal and shear strain are defined. In Chapter 8, a discussion
of some of the important mechanical properties of materials is given. Separate
treatments of axial load, torsion, bending, and transverse shear are presented in
Chapters 9, 10, 11, and 12, respectively. Chapter 13 provides a partial review of the
material covered in the previous chapters, in which the state of stress resulting from
combined loadings is discussed. In Chapter 14, the concepts for transforming stress
and strain are presented. Chapter 15 provides a means for a further summary and
review of previous material by covering design of beams based on allowable stress.
In Chapter 16, various methods for computing deflections of beams are presented,
including the method for finding the reactions on these members if they are statically
indeterminate. Lastly, Chapter 17 provides a discussion of column buckling.
Sections of the book that contain more advanced material are indicated by a
star (*). Time permitting, some of these topics may be included in the course.
Furthermore, this material provides a suitable reference for basic principles when it
is covered in other courses, and it can be used as a basis for assigning special projects.
Alternative Method for Coverage of Mechanics of Materials.
It is possible to cover many of the topics in the text in several different sequences.
For example, some instructors prefer to cover stress and strain transformations first,
before discussing specific applications of axial load, torsion, bending, and shear.
One possible method for doing this would be to first cover stress and strain and
its transformations, Chapter 7 and Chapter 14, then Chapters 8 through 13 can be
covered with no loss in continuity.
Acknowledgments
Over the years, this text has been shaped by the suggestions and comments of many
of my colleagues in the teaching profession. Their encouragement and willingness
to provide constructive criticism are very much appreciated and it is hoped that they
will accept this anonymous recognition. A note of thanks is also given to the
reviewers of both my Engineering Mechanics: Statics, and Mechanics of Materials
texts. Their comments have guided the improvement of this book as well.
In particular, I would like to thank: S. Apple, A. Bazar, Fullerton, M. Hughes,
R. Jackson, E. Tezak, H. Zhao,, K. Dennehy, A. Lutz, M. Walter, M. Zhang,
A. Asgharatal, S. Ahmad, J. Aurand, D. Boyajian, J. Callahan, D. Dikin, I. Elishakoff,
R. Hendricks, F. Herrera, J. Hilton, H. Kuhlman, K. Leipold, C. Roche, M. Rosengren,
R. Scott, J. Tashbar, M. Bailey, B. Smith, and J. Nadeau.
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PREFACE
11
During the production process I am thankful for the assistance of Rose Kernan,
my production editor for many years, and to my wife, Conny, for her help with the
proofreading and typing that was needed to prepare the manuscript for publication.
A special note of thanks also goes to Jun Hwa Lee who provided a careful reading
of the manuscript and also checked some of the problems. Through the years,
however, Kai Beng Yap supported me in this regard, but unfortunately his support
has come to an end due to his untimely passing. His contribution to this effort and
his friendship will be deeply missed. Finally, I would also like to thank all my students
who have used the previous edition and have made comments to improve its
contents; including those in the teaching profession who have taken the time to
e-mail me their comments.
I would greatly appreciate hearing from you at any time if you have any comments
or suggestions regarding the contents of this edition.
Russell Charles Hibbeler
hibbeler@bellsouth.net
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12
PREFACE
Acknowledgments for the Global Edition
Pearson would like to thank and acknowledge the following for their work on the
Global Edition.
Contributor
Jun Hwa Lee
Jun has a PhD in Mechanical Engineering from the Korea Advanced Institute of
Science and Technology.
Reviewers
Konstantinos Baxevanakis, Loughborough University
Akbar Afaghi Khatibi, RMIT University
Payam Khazaeinejad, Kingston University
Murat Saribay, Istanbul Bilgi University
We would also like to thank Kai Beng Yap for his contributions to the previous
Global Edition. He was a registered professional engineer working in Malaysia and
had a BS degree in Civil Engineering from the University of Louisiana-Lafayette
and an MS degree from Virginia Polytechnic Institute.
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PREFACE
Mastering Engineering
This online tutorial and assessment program allows you to integrate dynamic homework
and practice problems with automated grading of exercises from the textbook. Tutorials
and many end-of-section problems provide enhanced student feedback and optional
hints. Mastering Engineering™ allows you to easily track the performance of your
entire class on an assignment-by-assignment basis, or the detailed work of an individual
student. For more information visit mlm.pearson.com.
Resources for Instructors
Instructor’s Solutions Manual This supplement provides complete solutions
supported by problem statements and problem figures. The Instructor’s Solutions
Manual is available in the Instructor Resource Center.
PowerPoint Slides A complete set of all the figures and tables from the textbook
are available in PowerPoint format.
Resources for Students
Videos Developed by the author, three different types of videos are now available
to reinforce learning the basic theory and applying the principles. The first set
provides a lecture review and a self-test of the material related to the theory and
concepts presented in the book. The second set provides a self-test of the example
problems and the basic procedures used for their solution. And the third set provides
an engagement for solving the Fundamental Problems throughout the book. The
videos can be accessed in the Pearson eText or from a website available for purchase
separately at www.pearsonglobaleditions.com.
13
your work...
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With the Power of Mastering Engineering for
Statics and Mechanics of Materials
MasteringTM is the teaching and learning platform that empowers every student.
By combining trusted authors’ content with digital tools developed to engage students and
emulate the office hours experience, Mastering personalizes learning and improves results for
each student.
your answer
feedback
0.000844 m 3
Empower Each Learner
Each student learns at a different pace. Personalized learning, including optional hints and
wrong-answer feedback, pinpoints the precise areas where each student needs practice, giving
all students the support they need — when and where they need it — to be successful.
Learn more at www.pearson.com/mastering/engineering ®
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CONTENTS
1
General Principles
31
Chapter Objectives
31
1.1
Mechanics
31
1.2
Fundamental Concepts
1.3
The International System of Units
1.4
Numerical Calculations
1.5
General Procedure for Analysis 40
32
36
38
2
Force Vectors
45
Chapter Objectives
45
2.1
Scalars and Vectors
45
2.2
Vector Operations
46
2.3
Vector Addition of Forces
2.4
Addition of a System of Coplanar
Forces 58
2.5
Cartesian Vectors
2.6
Addition of Cartesian Vectors
2.7
Position Vectors
2.8
Force Vector Directed Along a Line
2.9
Dot Product
48
67
70
77
79
86
17
18
CONTENTS
3
Force System
Resultants 101
Chapter Objectives
101
3.1
Moment of a Force—Scalar
Formulation 101
3.2
Principle of Moments
3.3
Cross Product
3.4
Moment of a Force—Vector
Formulation 114
3.5
Moment of a Force about a
Specified Axis 124
3.6
Moment of a Couple
3.7
Simplification of a Force and Couple
System 142
3.8
Further Simplification of a Force and
Couple System 152
3.9
Reduction of a Simple Distributed
Loading 162
103
111
132
4
Equilibrium of a
Rigid Body 175
Chapter Objectives
175
4.1
Conditions for Rigid-Body
Equilibrium 175
4.2
Free-Body Diagrams
4.3
Equations of Equilibrium
4.4
Two- and Three-Force Members
4.5
Free-Body Diagrams
4.6
Equations of Equilibrium
4.7
Characteristics of Dry Friction
4.8
Problems Involving Dry Friction
177
186
192
201
206
214
219
19
CONTENTS
5
Structural Analysis
237
Chapter Objectives
237
5.1
Simple Trusses
237
5.2
The Method of Joints
240
5.3
Zero-Force Members
246
5.4
The Method of Sections
5.5
Frames and Machines
252
261
6
Center of Gravity, Centroid,
and Moment of Inertia 285
Chapter Objectives
285
6.1
Center of Gravity and the Centroid of a
Body 285
6.2
Composite Bodies
6.3
Moments of Inertia for Areas
6.4
Parallel-Axis Theorem for an Area
6.5
Moments of Inertia for Composite
Areas 314
298
306
307
20
CONTENTS
7
Stress and Strain
325
Chapter Objectives
325
7.1
Introduction
325
7.2
Internal Resultant Loadings
7.3
Stress
7.4
Average Normal Stress in an Axially
Loaded Bar 340
7.5
Average Shear Stress
7.6
Allowable Stress Design
7.7
Deformation
7.8
Strain
326
338
347
358
373
374
8
Mechanical Properties of
Materials 391
Chapter Objectives
391
8.1
The Tension and Compression Test
391
8.2
The Stress–Strain Diagram
8.3
Stress–Strain Behavior of Ductile and Brittle
Materials 397
8.4
Strain Energy
401
8.5
Poisson’s Ratio
410
8.6
The Shear Stress–Strain Diagram
393
412
CONTENTS
21
9
Axial Load
423
Chapter Objectives
423
9.1
Saint-Venant’s Principle
423
9.2
Elastic Displacement of an Axially Loaded
Member 425
9.3
Principle of Superposition
9.4
Statically Indeterminate Axially Loaded
Members 438
9.5
The Force Method of Analysis for Axially
Loaded Members 445
9.6
Thermal Stress
9.7
Stress Concentrations
438
451
458
10
Torsion
467
Chapter Objectives
467
10.1
Torsional Deformation of a Circular
Shaft 467
10.2
The Torsion Formula
470
10.3
Power Transmission
478
10.4
Angle of Twist
10.5
Statically Indeterminate Torque-Loaded
Members 502
*10.6
Solid Noncircular Shafts
10.7
Stress Concentration
488
509
512
22
CONTENTS
11
Bending
523
Chapter Objectives
523
11.1
Internal Loading as a Function of
Position 523
11.2
Graphical Method for Constructing Shear
and Moment Diagrams 530
11.3
Bending Deformation of a Straight
Member 547
11.4
The Flexure Formula
551
11.5
Unsymmetric Bending
564
11.6
Stress Concentrations
572
12
Transverse Shear
Chapter Objectives
581
581
12.1
Shear in Straight Members
581
12.2
The Shear Formula
12.3
Shear Flow in Built-Up Members
582
598
23
CONTENTS
13
Combined Loadings
Chapter Objectives
611
611
13.1
Thin-Walled Pressure Vessels
611
13.2
State of Stress Caused by Combined
Loadings 618
14
Stress and Strain
Transformation 639
Chapter Objectives
639
14.1
Plane-Stress Transformation
639
14.2
General Equations of Plane Stress
Transformation 644
14.3
Principal Stresses and Maximum In-Plane
Shear Stress 647
14.4
Mohr’s Circle—Plane Stress
14.5
Absolute Maximum Shear Stress
14.6
Plane Strain
14.7
General Equations of Plane-Strain
Transformation 676
14.8
Mohr’s Circle—Plane Strain
*14.9
Absolute Maximum Shear Strain
659
670
675
14.10 Strain Rosettes
684
691
693
14.11 Material Property Relationships
697
24
CONTENTS
15
Design of Beams and
Shafts 715
Chapter Objectives
715
15.1
Basis for Beam Design
715
15.2
Prismatic Beam Design
718
16
Deflection of Beams and
Shafts 733
Chapter Objectives
733
16.1
The Elastic Curve
733
16.2
Slope and Displacement by
Integration 737
* 16.3
Discontinuity Functions
16.4
Method of Superposition
16.5
Statically Indeterminate Beams and
Shafts—Method of Superposition 772
753
764
25
CONTENTS
17
Buckling of Columns
Chapter Objectives
789
789
17.1
Critical Load
789
17.2
Ideal Column with Pin Supports
17.3
Columns Having Various Types of
Supports 798
*17.4
The Secant Formula
792
808
Appendices
A.
Mathematical Review and
Expressions 820
B.
Geometric Properties of an Area and
Volume 824
C.
Geometric Properties of Structural
Shapes 826
D.
Slopes and Deflections of Beams
Fundamental Problem
Solutions 830
Answers to Selected
Problems 865
Index
879
828
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CREDITS
Cover: Ivanova Ksenia/Shutterstock
Chapter 1:
30 ALAMY IMAGES: Gaertner/Alamy Stock Photo
34 SHUTTERSTOCK: NikoNomad/Shutterstock
Chapter 2:
44 ALAMY IMAGES: Marcus Hofmann/Alamy Stock Photo
Chapter 3:
100 SHUTTERSTOCK: Zbynek Jirousek/Shutterstock
Chapter 4:
174 ALAMY IMAGES: Stephan Scherhag/Panther Media GmbH/Alamy Stock Photo
Chapter 5:
236 ALAMY IMAGES: Ray Hardinge/Alamy Stock Photo
Chapter 6:
284 ALAMY IMAGES: Olaf Speier/Alamy Stock Photo
Chapter 7:
324 SHUTTERSTOCK: Ilya Akinshin/Shutterstock
Chapter 8:
390 123RF GB LIMITED: John Vlahidis/123RF
Chapter 9:
422 SHUTTERSTOCK: BalLi8Tic/Shutterstock
Chapter 10:
466 SHUTTERSTOCK: Vadim Ratnikov/Shutterstock
Chapter 11:
522 123RF GB LIMITED: Aydo8/123RF
Chapter 12:
580 123RF GB LIMITED: Claudiodivizia/123RF
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28
CREDITS
Chapter 13:
610 123RF GB LIMITED: Ganna Poltoratska/123RF
Chapter 14:
638 SHUTTERSTOCK: ID1974/Shutterstock
Chapter 15:
714 SHUTTERSTOCK: Sasin Paraksa/Shutterstock
Chapter 16:
732 SHUTTERSTOCK: Lev Kropotov/Shutterstock
Chapter 17:
788 SHUTTERSTOCK: Ann Baldwin/Shutterstock
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STATICS AND
MECHANICS
OF MATERIALS
SIXTH EDITION IN SI UNITS
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CHAPTER
1
Cranes such as this one are required to lift extremely large loads. Their design
is based on the basic principles of statics and dynamics, which form the subject
matter of engineering mechanics.
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GENERAL
PRINCIPLES
Lecture Summary and Quiz,
Example, and Problemsolving videos are available
where this icon appears.
CHAPTER OBJECTIVES
■
To provide an introduction to the basic quantities and idealizations
of mechanics.
■
To state Newton’s Laws of Motion.
■
To review the principles for applying the SI system of units.
■
To examine the standard procedures for performing numerical
calculations.
■
To present a general guide for solving problems.
1.1
MECHANICS
Mechanics can be defined as that branch of the physical sciences
concerned with the state of rest or motion of bodies that are subjected to
the action of forces. In this book we will study two important branches of
mechanics, namely, statics and mechanics of materials. These subjects form
a suitable basis for the design and analysis of many types of structural,
mechanical, or electrical devices encountered in engineering.
Statics deals with the equilibrium of bodies, that is, it is used to
determine the forces acting either external to the body or within it that
are necessary to keep the body in equilibrium. Mechanics of materials
studies the relationships between the external loads and the distribution
of internal forces acting within the body. This subject is also concerned
with finding the deformations of the body, and it provides a study of the
body’s stability.
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32
CHAPTER 1
1
GENERAL PRINCIPLES
In this book we will first study the principles of statics, since for the
design and analysis of any structural or mechanical element it is first
necessary to determine the forces acting both on and within its various
members. Once these internal forces are determined, the size of the
members, their deflection, and their stability can then be determined using
the fundamentals of mechanics of materials, which will be covered later.
Historical Development. The subject of statics developed very
early in history because its principles can be formulated simply from
measurements of geometry and force. For example, the writings of
Archimedes (287–212 b.c.) deal with the principle of the lever. Studies of
the pulley and inclined plane are also recorded in ancient writings—at
times when the requirements for engineering were limited primarily to
building construction.
The origin of mechanics of materials dates back to the beginning of the
seventeenth century, when Galileo performed experiments to study the
effects of loads on rods and beams made of various materials. However, at
the beginning of the eighteenth century, experimental methods for testing
materials were vastly improved, and at that time many experimental and
theoretical studies in this subject were undertaken primarily in France, by
such notables as Saint-Venant, Poisson, Lamé, and Navier.
Over the years, after many of the fundamental problems of mechanics
of materials had been solved, it became necessary to use advanced
mathematical and computer techniques to solve more complex problems.
As a result, this subject has expanded into other areas of mechanics, such
as the theory of elasticity and the theory of plasticity. Research in these
fields is ongoing, in order to meet the demands for solving more advanced
problems in engineering.
1.2
FUNDAMENTAL CONCEPTS
Before we begin our study, it is important to understand the definitions
of certain fundamental concepts and principles.
Mass. Mass is a measure of a quantity of matter that is used to compare
the action of one body with that of another. This property provides a
measure of the resistance of matter to a change in velocity.
Force. In general, force is considered as a “push” or “pull” exerted by
one body on another. This interaction can occur when there is direct
contact between the bodies, such as a person pushing on a wall, or it can
occur through a distance when the bodies are physically separated.
Examples of the latter type include gravitational, electrical, and magnetic
forces. In any case, a force is completely characterized by its magnitude,
direction, and point of application.
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1.2
FUNDAMENTAL CONCEPTS
33
Newton’s Three Laws of Motion. Engineering mechanics is
formulated on the basis of Newton’s three laws of motion, the validity
of which is based on experimental observation. These laws apply to the
motion of a particle as measured from a nonaccelerating reference frame.
They may be briefly stated as follows.
1
First Law. A particle originally at rest, or moving in a straight line with
constant velocity, tends to remain in this equilibrium state provided the
particle is not subjected to an unbalanced force, Fig. 1–1a.
F1
F2
v
F3
Equilibrium
(a)
Second Law. A particle acted upon by an unbalanced force F
experiences an acceleration a that has the same direction as the force and
a magnitude that is directly proportional to the force, Fig. 1–1b.* If the
particle has a mass m, this law may be expressed mathematically as
F = ma
(1–1)
a
F
Accelerated motion
(b)
Third Law. The mutual forces of action and reaction between two
particles are equal, opposite, and collinear, Fig. 1–1c.
force of A on B
F
F
A
B
force of B on A
Action–reaction
(c)
Fig. 1–1
*Stated another way, the unbalanced force acting on the particle is proportional to the time
rate of change of the particle’s linear momentum.
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34
CHAPTER 1
GENERAL PRINCIPLES
Newton’s Law of Gravitational Attraction. Shortly after
formulating his three laws of motion, Newton postulated a law
governing the gravitational attraction between any two particles. Stated
mathematically,
1
F =G
m1 m2
r2
(1–2)
where
F = force of gravitation between the two particles
G = universal constant of gravitation; according to experimental
evidence, G = 66.73 ( 10 −12 ) m 3 ( kg s 2 )
m1 , m2 = mass of each of the two particles
r = distance between the two particles
⋅
Weight. According to Eq. 1–2, any two particles or bodies have a
mutual attractive (gravitational) force acting between them. In the case
of a particle located at or near the surface of the earth, however, the
only gravitational force having any sizable magnitude is that between
the earth, because of its very large mass, and the particle. Consequently,
this force, called the weight, will be the only gravitational force we will
consider.
From Eq. 1–2, if the particle has a mass m1 = m, and we assume the
earth is a nonrotating sphere of constant density and having a mass
m2 = M e, then if r is the distance between the earth’s center and the
particle, the weight W of the particle becomes
The astronaut’s weight is diminished since
she is far removed from the gravitational
field of the earth.
W =G
mM e
r2
If we let g = GM e r 2 , we have
W = mg
(1–3)
If we allow the particle to fall downward, then neglecting air resistance,
the only force acting on the particle is its weight, and so Eq. 1–1 becomes
W = ma. Comparing this result with Eq. 1–3, we see that a = g. In other
words, g is the acceleration due to gravity. Since it depends on r, then
the weight of the particle or body is not an absolute quantity. Instead,
its magnitude depends upon the elevation where the measurement was
made. For most engineering calculations, however, g is determined at sea
level and at a latitude of 45°, which is considered the “standard location.”
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1.2
FUNDAMENTAL CONCEPTS
35
Idealizations. Models or idealizations are used in mechanics
in order to simplify application of the theory. Here we will consider three
important idealizations.
1
Particle. A particle has a mass, but a size that can be neglected.
For example, the size of the earth is insignificant compared to the size of its
orbit, and therefore the earth can be modeled as a particle when studying
its orbital motion. When a body is idealized as a particle, the principles
of mechanics reduce to a rather simplified form since the geometry of the
body will not be involved in the analysis of the problem.
Rigid Body. A rigid body can be considered as a combination
of a large number of particles in which all the particles remain at a fixed
distance from one another, both before and after applying a load. This
model is important because the body’s shape does not change when a load
is applied, and so we do not have to consider the type of material from
which the body is made. In most cases the actual deformations occurring
in structures, machines, mechanisms, and the like are relatively small, and
the rigid-body assumption is suitable for analysis.
Three forces act on the ring. Since these
forces all meet at a point, then for any
force analysis, we can assume the ring to
be represented as a particle.
Concentrated Force. A concentrated force represents the effect
of a loading which is assumed to act at a point on a body. We can represent
a load by a concentrated force, provided the area over which the load is
applied is very small compared to the overall size of the body. An example
would be the contact force between a wheel and the ground.
Steel is a common engineering material that does not
deform very much under load. Therefore, we can consider
this railroad wheel to be a rigid body acted upon by the
concentrated force of the rail.
M01_HIBB0208_06_GE_C01.indd 35
Refer to the companion website for
Lecture Summary and Quiz videos.
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36
CHAPTER 1
GENERAL PRINCIPLES
TABLE 1–1
1
SI System of Units
Name
Length
Time
Mass
Force
International
System of Units
SI
meter
second
kilogram
newton*
m
s
kg
N
 kg · m 


s2 
*Derived unit.
1.3
THE INTERNATIONAL SYSTEM
OF UNITS
The four basic quantities—length, time, mass, and force—are not all
independent from one another; in fact, they are related by Newton’s
second law of motion, F = ma. Because of this, the units used to measure
these quantities cannot all be selected arbitrarily. The equality F = ma is
maintained only if three of the four units, called base units, are defined
and the fourth unit is then derived from the equation.
For the International System of Units, abbreviated SI after the
French “Système International d’Unités,” length is in meters (m), time
is in seconds (s), and mass is in kilograms (kg), Table 1–1. The unit of
force, called a newton (N), is derived from F = ma. Thus, 1 newton is
equal to a force required to give 1 kilogram of mass an acceleration of
1 m s 2 ( N = kg · m s 2 ) .
If the weight of a body located at the “standard location” is to be
determined in newtons, then Eq. 1–3 must be applied. Here measurements
give g = 9.806 65 m s 2 ; however, for calculations the value g = 9.81 m s 2
will be used. Thus,
W = mg
1 kg
9.81 N
Fig. 1–1
M01_HIBB0208_06_GE_C01.indd 36
( g = 9.81 m s 2 )
(1–4)
Therefore, a body of mass 1 kg has a weight of 9.81 N, a 2-kg body weighs
19.62 N, and so on, Fig. 1–1. Perhaps it is easier to remember that a small
apple weighs one newton.
Prefixes. When a numerical quantity is either very large or very small,
the units used to define its size may be modified by using a prefix. Some
of the prefixes used in the SI system are shown in Table 1–2. Each represents
a multiple or submultiple of a unit which, if applied successively, moves
12/21/23 3:30 PM
1.3
THE INTERNATIONAL SYSTEM OF UNITS
the decimal point of a numerical quantity to every third place.* For
example, 4 000 000 N = 4 000 kN (kilo-newton) = 4 MN (mega-newton),
or 0.005 m = 5 mm (milli-meter). Notice that the SI system does not
include the multiple deca (10) or the submultiple centi (0.01), which
form part of the metric system. Except for some volume and area
measurements, the use of these prefixes is generally avoided in science
and engineering.
TABLE 1–2
Multiple
1 000 000 000
1 000 000
1 000
Submultiple
0.001
0.000 001
0.000 000 001
37
1
Prefixes
Exponential Form
Prefix
SI Symbol
109
106
103
giga
mega
kilo
G
M
k
10−3
10−6
10−9
milli
micro
nano
m
µ
n
Rules for Use. Here are a few of the important rules that describe
the proper use of the various SI symbols:
• Quantities defined by several units which are multiples of one
another are separated by a dot to avoid confusion with prefix
notation, as indicated by N = kg · m s 2 = kg · m · s −2. Also, m · s
(meter-second), whereas ms (milli-second).
• The exponential power on a unit having a prefix refers to both the
unit and its prefix. For example, µ N 2 = ( µ N) 2 = µ N · µ N. Likewise,
mm2 represents (mm) 2 = mm · mm.
• With the exception of the base unit the kilogram, in general avoid
the use of a prefix in the denominator of composite units. For
example, do not write N mm , but rather kN m ; also, m mg should
be written as Mm kg.
• When performing calculations, represent the numbers in terms of
their base or derived units by converting all prefixes to powers of 10.
The final result should then be expressed using a single prefix. Also,
after calculation, it is best to keep numerical values between 0.1 and
1000; otherwise, a suitable prefix should be chosen. For example,
(50 kN)(60 mm) =  50(10 3 ) N   60(10 −9 )m 
= 3000(10 −6 ) N · m = 3(10 −3 ) N · m = 3 mN · m
*The kilogram is the only base unit that is defined with a prefix.
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38
CHAPTER 1
GENERAL PRINCIPLES
1.4
1
NUMERICAL CALCULATIONS
Numerical work in engineering practice is most often performed by
using handheld calculators and computers. It is important, however, that
the answers to any problem be reported with justifiable accuracy using
appropriate significant figures. In this section we will discuss these topics
together with some other important aspects involved in all engineering
calculations.
Dimensional Homogeneity. The terms of any equation used
to describe a physical process must be dimensionally homogeneous;
that is, each term must be expressed in the same units. Provided this is
the case, all the terms of an equation can then be combined if numerical
values are substituted for the variables. Consider, for example, the
equation s = v t + 12 at 2, where, in SI units, s is the position in meters, m,
t is time in seconds, s, v is velocity in m s, and a is acceleration in m s 2 .
Regardless of how this equation is evaluated, it maintains its dimensional
homogeneity. In the form stated, each of the three terms is expressed
in meters  m, (m s ) s, (m s 2 ) s 2  or solving for a, a = 2 s t 2 − 2v t,
the terms are each expressed in units of m s 2  m s 2, m s 2, (m s) s  .
Keep in mind that problems in mechanics always involve the solution
of dimensionally homogeneous equations, and so this fact can then be
used as a partial check for algebraic manipulations of an equation.
Significant Figures. The number of significant figures contained
in any number determines the accuracy of the number. For instance,
the number 4981 contains four significant figures. However, if zeros
occur at the end of a whole number, it may be unclear as to how many
significant figures the number represents. For example, 23 400 might
have three (234), four (2340), or five (23 400) significant figures. To avoid
these ambiguities, we will use engineering notation to report a result.
This requires that numbers be rounded off to the appropriate number
of significant digits and then expressed in multiples of (10 3 ), such as (10 3 ),
(10 6 ), or (10 −9 ). For instance, if 23 400 has five significant figures, it is
written as 23.400(10 3 ), but if it has only three significant figures, it is
written as 23.4(103).
If zeros occur at the beginning of a number that is less than one,
then the zeros are not significant. For example, 0.008 21 has three
significant figures. Using engineering notation, this number is expressed
as 8.21(10 −3 ). Likewise, 0.000 582 can be expressed as 0.582(10 −3 ) or
582(10 −6 ).
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1.4
NUMERICAL CALCULATIONS
39
Rounding Off Numbers. Rounding off a number is necessary so
that the accuracy of the result will be the same as that of the problem
data. As a general rule, any numerical figure ending in a number greater
than five is rounded up and a number less than five is not rounded up. The
rules for rounding off numbers are best illustrated by example. Suppose
the number 3.5587 is to be rounded off to three significant figures.
Because the fourth digit (8) is greater than 5, the third number is rounded
up to 3.56. Likewise 0.5896 becomes 0.590 and 9.3866 becomes 9.39. If we
round off 1.341 to three significant figures, because the fourth digit (1)
is less than 5, then we get 1.34. Likewise 0.3762 becomes 0.376 and 9.871
becomes 9.87. There is a special case for any number that ends in a 5. As
a general rule, if the digit preceding the 5 is an even number, then this
digit is not rounded up. If the digit preceding the 5 is an odd number, then
it is rounded up. For example, 75.25 rounded off to three significant digits
becomes 75.2, 0.1275 becomes 0.128, and 0.2555 becomes 0.256.
1
Calculations. When a sequence of calculations is performed, it is
best to store the intermediate results in the calculator. In other words, do
not round off calculations until expressing the final result. This procedure
maintains precision throughout the series of steps to the final solution.
In this book we will generally round off the answers to three significant
figures since most of the data in engineering mechanics, such as geometry
and loads, may be reliably measured to this accuracy.
When solving problems, do the work as
neatly as possible. Being neat will
stimulate clear and orderly thinking,
and vice versa.
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40
CHAPTER 1
GENERAL PRINCIPLES
1.5
1
GENERAL PROCEDURE FOR
ANALYSIS
Attending a lecture, reading this book, and studying the example problems
helps, but the most effective way of learning the principles of engineering
mechanics is to solve problems. To be successful at this, it is important to
always present the work in a logical and orderly manner, as suggested by
the following sequence of steps:
• Read the problem carefully and try to correlate the actual physical
situation with the theory studied.
• Tabulate the problem data and draw to a large scale any necessary
diagrams.
• Apply the relevant principles, generally in mathematical form. When
writing any equations, be sure they are dimensionally homogeneous.
• Solve the necessary equations, and report the answer with no more
than three significant figures.
• Study the answer with technical judgment and common sense to
determine whether or not it seems reasonable.
I MPO RTA N T PO I N T S
• Statics is the study of bodies that are at rest or move with
constant velocity.
• A particle has a mass but a size that can be neglected, and a
rigid body does not deform under load.
• A force is considered as a “push” or “pull” of one body on another.
• Concentrated forces are assumed to act at a point on a body.
• Newton’s three laws of motion should be memorized.
• Mass is measure of a quantity of matter that does not change
from one location to another. Weight refers to the gravitational
attraction of the earth on a body or quantity of mass. Its magnitude
depends upon the elevation at which the mass is located.
• In the SI system the unit of force, the newton, is a derived unit.
The meter, second, and kilogram are base units.
• Prefixes G, M, k, m, µ , and n are used to represent large and
small numerical quantities. Their exponential size should be
known, along with the rules for using the SI units.
• Perform numerical calculations with several significant figures,
and then report the final answer to three significant figures.
• Algebraic manipulations of an equation can be checked in part by
verifying that the equation remains dimensionally homogeneous.
• Know the rules for rounding off numbers.
Lecture videos that cover the material in this and most of the other
sections of the book can be obtained at www.pearsonglobaleditions.com.
1.5
EXAMPLE
GENERAL PROCEDURE FOR ANALYSIS
41
1.1
1
Convert 100 km h to m s and 24 m s to km h.
SOLUTION
Since 1 km = 1000 m and 1 h = 3600 s, the factors of conversion are
arranged in the following order, so that a cancellation of the units can
be applied:



100
1000
m
1
h


0.556
0.556
m
m
0.556
1
m
1
ft
ft
0.556
m   1 ft1 ft  
1
h
1000
m
1
h
100
100km
km
km
1000

 
0.556
m ms =s =  
==0.556
100
hh
100
100km
km
km0.556
h===mms s0.556


  
hhh   km
skm
s 3600
0.3048
s sssmm
m m 
km
3600
0.3048
 0.3048
 0.3048
3600
3)31.82
==31.82
ft ft
s s
1.82
100(10
100(10
100(10
)m
)mmftfts=s =1.82
===
===27.8
Ans.
27.8
27.8m
mmsss
3600
3600
3600sss
 


 
0.556
1km
3600
 s1s1ft
m
1km
1 ftm
0.556
m
0.556
0.556
1mft3600
24
mmmm
1km
3600
3600
sft
sm   1 1ftft  
24
24
1km
 0.556
 24
 0.556
0.556
m
smm
mms s= = 
0.556
m ss =0.556
24
m
s0.556
=m
=0.556
24
24
24
s0.556
s == m
s==

  



 sss sss 1000
0.3048
ssm
0.3048
s s  m
0.3048mm 
0.3048
0.3048
1000
mmm
1h
1h
1h
1000
1h
1000
m
m
m 0.3048
3km
= 1.82
ft
sft =s=331.82
ftft ss= =1.82
1.82ftfts s
= 1.82
86.4(10
)1.82
86.4(10
)km
km
86.4(10
86.4(10
)3)km
km
hhhh
====
===86.4
86.4
km
86.4
km
km
=86.4
1000
hhhh
1000
1000
1000
Ans.
NOTE: Remember to round off the final answer to three significant
figures.
EXAMPLE
1.2
Convert the density of steel 7.85 g cm 3 to kg m 3 .
SOLUTION
Using 1 kg = 1000 g and 1 m = 100 cm, and arrange the conversion
factor in such a way that g and cm 3 can be canceled out.
 
  3
 
 
 0.556
1kg
0.556
100
cm
 cm
0.556
mggm
1m
ft1mft
1cm
ft
m
0.556
1m
ft   1 1ftft  
1kg
100
7.85
7.85
1kg
 0.556
 7.85
g 0.556
 100
3
0.556
0.556
m
s 0.556
s30.556
0.556
ms s==   
=s 0.556
=3  m
7.85
gmcm
== mms =
  

3
0.3048
scm
s  1000
s0.3048
0.3048
sgggm
0.3048
1000
1m
cm
1000
1m
s s  m
0.3048
1m
0.3048mm 
 cm
m
 m
   
 
 
   
= 1.82
= 1.82
ft
sft=m
ft
1.82
fts1kg
1.82
fts31scm
3 3 3  
3cm
7.85
gs=1.82
7.85
7.85
gmg1.82
0.556
0.556
100
ft
1kg
1kg
100
0.556
1=sft
m
1=m
ft
1ftcm
ft
m 
0.556
m
1 ft1 ft  
 0.556
 100
 30.556
0.556
0.556
mms s=0.556
0.556
s = 3 3 3   
mms =
=  mms s==0.556
=0.556





3
3
3
  cm
m
 1m
   0.3048
1000
1m
1m
1000
s 1000
0.3048
s sgggm
0.3048
s s mm
mm 
cm
   scm
  0.3048
0.3048
 0.3048
= ==1.82
1.82
ft fts s=3 =)1.82
1.82
ftft3s s = =1.82
1.82
ft fts s
7.85(10
kg
m
Ans.
M01_HIBB0208_06_GE_C01.indd 41
12/21/23 3:30 PM
42
CHAPTER 1
EXAMPLE
GENERAL PRINCIPLES
1.3
1
Evaluate each of the following and express with SI units having an
appropriate prefix: (a) (50 mN)(6 GN), (b) ( 400 mm )( 0.6 MN )2 ,
(c) 45 MN 3 900 Gg.
SOLUTION
First convert each number to base units, perform the indicated
operations, then choose an appropriate prefix.
Part (a)
(50 mN)(6 GN) =  50(10 −3 )N   6(10 9 )N 
= 300(10 6 )N 2
 1 kN   1 kN 
= 300(10 6 ) N 2  3
 10 N   10 3 N 
= 300 kN 2
Ans.
NOTE: Keep in mind the convention kN 2 = ( kN )
2
= 10 6 N 2 .
Part (b)
(400 mm)(0.6 MN) 2 =  400(10 −3 )m   0.6(10 6 )N 
2
=  400(10 −3 )m   0.36(10 12 )N 2 
= 144(10 9 )m · N 2
We can also write
= 144 Gm · N 2
Ans.
2  1 MN   1 MN 
144(10 9 )m · N 2 = 144(10 9 )m · N  6
 10 N   10 6 N 
= 0.144 m · MN 2
Ans.
Part (c)
45 MN 3
45(10 6 N) 3
=
900 Gg
900(10 6 )kg
= 50(10 9 ) N 3 kg
3
 1 kN  1
= 50(10 9 ) N 3  3 
 10 N  kg
= 50 N 3 kg
Ans.
Videos that test your understanding of these and other
example problems throughout the book are available at www
.pearsonglobaleditions.com.
PROBLEMS
43
PROBLEMS
1–1. Round off the following numbers to three significant
figures: (a) 58 342 m, (b) 68.534 s, (c) 2553 N, (d) 7555 kg.
1–13. Using the SI system of units, show that Eq. 1–2 is
a dimensionally homogeneous equation which gives F
in newtons. Determine to three significant figures the
gravitational force acting between two spheres that are
touching each other. The mass of each sphere is 200 kg and
the radius is 300 mm.
1–2. Represent each of the following combinations of
units in the correct SI form using an appropriate prefix:
(a) Mg mm, (b) mN µ s, (c) µ m ⋅ Mg.
1–14. Evaluate each of the following and express with
an appropriate prefix: (a) (430 kg)2, (b) (0.002 mg)2, and
(c) (230 m)3.
1–3. Represent each of the following combinations of
units in the correct SI form using an appropriate prefix:
(a) kN µ s, (b) Mg mN, (c) MN (kg ⋅ ms).
1–15. Evaluate each of the following to three significant
figures and express each answer in SI units using an appropriate
prefix: (a) (200 kN)2, (b) (0.005 mm)2, (c) (400 m)3.
*1–4. Determine the mass of an object that has a weight
of (a) 20 mN, (b) 150 kN, (c) 60 MN. Express the answer to
three significant figures.
*1–16. Evaluate each of the following to three significant
figures and express each answer in SI units using an
appropriate prefix: (a) (212 mN)2, (b) (52 800 ms)2,
The answers to all but every fourth problem (asterisk)
are given in the back of the book.
1–5. Round off the following numbers to three significant
figures: (a) 3.455 55 m, (b) 45.556 s, (c) 5555 N, (d) 4525 kg.
1–6. Represent each of the following as a number between
0.1 and 1000 using an appropriate prefix: (a) 45 320 kN,
(b) 568(10 5 ) mm, (c) 0.005 63 mg.
1–7. Represent each of the following combinations
of units in the correct SI form: (a) Mg ms, (b) N mm,
(c) mN (kg · µ s).
*1–8. Represent each of the following quantities in the
correct SI form using an appropriate prefix: (a) 0.000 431 kg,
(b) 35.3(10 3 ) N, (c) 0.005 32 km.
1–9. Represent each of the following combinations of
units in the correct SI form using an appropriate prefix:
(a) µ MN, (b) N µ m, (c) MN ks 2, (d) kN ms.
1–10. Represent each of the following combinations of
units in the correct SI form using an appropriate prefix:
(a) m ms, (b) µ km, (c) ks mg, (d) km · µ N.
(c) 548(10 6 )
1/ 2
1
ms.
1–17. Evaluate (204 mm)(0.004 57 kg) (34.6 N) to three
significant figures and express the answer in SI units using
an appropriate prefix.
1–18. What is the weight in newtons of an object that has a
mass of (a) 8 kg, (b) 0.04 g, (c) 760 Mg?
1–19. A concrete column has a diameter of 350 mm and
a length of 2 m. If the density (mass volume) of concrete is
2.45 Mg m 3 , determine the weight of the column.
*1–20. Two particles have a mass of 8 kg and 12 kg,
respectively. If they are 800 mm apart, determine the force
of gravity acting between them. Compare this result with
the weight of each particle.
1–21. If a man weighs 690 newtons on earth, specify
(a) his mass in kilograms. If the man is on the moon, where
the acceleration due to gravity is g m = 1.61 m s 2 , determine
(b) his weight in newtons, and (c) his mass in kilograms.
1–11. Represent each of the following combinations of
units in the correct SI form using an appropriate prefix:
(a) GN ⋅ µ m, (b) kg µ m , (c) N ks 2, (d) kN µ s.
*1–12. A rocket has a mass 3.529(10 6 ) kg on earth. Specify
(a) its mass in SI units, and (b) its weight in SI units. If the
rocket is on the moon, where the acceleration due to gravity
is g m = 1.61 m s 2 , determine to three significant figures
(c) its weight in SI units, and (d) its mass in SI units.
M01_HIBB0208_06_GE_C01.indd 43
12/21/23 3:31 PM
CHAPTER
2
This electric transmission tower is stabilized by cables that exert forces on the
tower at their points of connection. In this chapter we will show how to express
these forces as Cartesian vectors, and then determine their resultant.
M02_HIBB0208_06_GE_C02.indd 44
12/21/23 3:29 PM
FORCE
VECTORS
Lecture Summary and Quiz,
Example, and Problemsolving videos are available
where this icon appears.
CHAPTER OBJECTIVES
■
To show how to add forces and resolve them into components
using the parallelogram law.
■
To express force and position in Cartesian vector form and
explain how to determine the vector’s magnitude and direction.
■
To introduce the dot product in order to use it to find the angle
between two vectors or the projection of one vector onto another.
2.1
SCALARS AND VECTORS
Many physical quantities in engineering mechanics are measured using
either scalars or vectors.
Scalar. A scalar is any positive or negative physical quantity that can
be completely specified by its magnitude. Examples of scalar quantities
include length, mass, and time.
Vector. A vector is any physical quantity that requires both
a magnitude and a direction for its complete description. Examples of
vectors encountered in statics are force, position, and moment. A vector
is shown graphically by an arrow, Fig. 2–1. The length of the arrow
represents the magnitude of the vector, and the angle θ between the
vector and a fixed axis defines the direction of its line of action. The head
or tip of the arrow indicates the sense of direction of the vector.
In print, vector quantities are represented by boldface letters such as
A, and the magnitude of a vector is italicized, A. For handwritten work,
it is often convenient
to denote a vector quantity by simply drawing an
→
arrow above it, A.
M02_HIBB0208_06_GE_C02.indd 45
Line of action
A
Magnitude
u Direction
Sense
Fig. 2–1
45
12/21/23 3:29 PM
46
CHAPTER 2
FORCE VECTORS
2.2
VECTOR OPERATIONS
Multiplication and Division of a Vector by a Scalar. If
a vector is multiplied or divided by a positive scalar, its magnitude is
changed by that amount. Multiplying or dividing by a negative scalar
will also change the directional sense of the vector. Graphic examples
of these operations are shown in Fig. 2–2.
2
2A
A
2A
2 0.5 A
Scalar multiplication and division
Fig. 2–2
Vector Addition. When adding two vectors together it is important
to account for both their magnitudes and their directions. To do this
we must use the parallelogram law of addition. To illustrate, the two
component vectors A and B in Fig. 2–3a are added to form a resultant
vector R = A + B using the following procedure:
• First join the tails of the components at a point to make them
concurrent, Fig. 2–3b.
• From the head of B, draw a line parallel to A. Draw another line
from the head of A that is parallel to B. These two lines intersect at
point P to form the adjacent sides of a parallelogram.
• The diagonal of this parallelogram that extends to P forms R, which
then represents the resultant vector R = A + B, Fig. 2–3c.
A
A
A
R
P
B
B
B
R5A1B
Parallelogram law
(a)
(b)
(c)
Fig. 2–3
M02_HIBB0208_06_GE_C02.indd 46
12/21/23 3:30 PM
2.2
VECTOR OPERATIONS
47
We can also add B to A, Fig. 2–4a, using the triangle rule, which is a special
case of the parallelogram law, whereby vector B is added to vector A in
a “head-to-tail” fashion, i.e., by connecting the tail of B to the head of A,
Fig. 2–4b. The resultant R extends from the tail of A to the head of B.
In a similar manner, R can also be obtained by adding A to B, Fig. 2–4c.
By comparison, it is seen that vector addition is commutative; in other
words, the vectors can be added in either order, i.e., R = A + B = B + A .
2
A
B
A
R
R
B
A
B
(a)
R5A1B
R5B1A
Triangle rule
Triangle rule
(b)
(c)
Fig. 2–4
As a special case, if the two vectors A and B are collinear, i.e., both
have the same line of action, the parallelogram law reduces to an
algebraic or scalar addition R = A + B , as shown in Fig. 2–5.
R
A
B
R5A1B
Vector Subtraction. The resultant of the difference between two
Addition of collinear vectors
vectors A and B of the same type may be expressed as
Fig. 2–5
R ′ = A − B = A + ( − B)
This vector sum is shown graphically in Fig. 2–6. Subtraction is therefore
defined as a special case of addition, so the rules of vector addition also
apply to vector subtraction.
2B
A
R9
B
A
or
R9
A
2B
Parallelogram law
Triangle construction
Vector subtraction
Fig. 2–6
M02_HIBB0208_06_GE_C02.indd 47
12/21/23 3:30 PM
48
CHAPTER 2
FORCE VECTORS
2.3
F2
F1
FR
2
VECTOR ADDITION OF FORCES
Experimental evidence has shown that a force is a vector quantity since
it has a specified magnitude, direction, and sense and it adds according
to the parallelogram law. Two common problems in statics involve
either finding the resultant force, knowing its components, or resolving
a known force into two components. We will now describe how each of
these problems is solved using the parallelogram law.
Finding a Resultant Force. The two component forces F1 and
The parallelogram law must be used to
determine the resultant of the two forces
acting on the hook.
F2 acting on the pin in Fig. 2–7a are added together to form the resultant
force FR = F1 + F2 , using the parallelogram law as shown in Fig. 2–7b.
From this construction, or using the triangle rule, Fig. 2–7c, we can apply
the law of cosines or the law of sines to the triangle in order to obtain the
magnitude of the resultant force and its direction.
F1
F1
F1
FR
F2
F2
FR
F2
FR 5 F1 1 F2
y
Fy
F
(a)
u
Fu
(b)
(c)
Fig. 2–7
Finding the Components of a Force. Sometimes it is necessary
Using the parallelogram law the supporting
force F can be resolved into components
acting along the u and v axes.
M02_HIBB0208_06_GE_C02.indd 48
to resolve a force into two components in order to study its pulling or
pushing effect in two specific directions. For example, in Fig. 2–8a,
F is to be resolved into two components along the two members, defined
by the u and v axes. In order to determine the magnitude of each
component, a parallelogram is constructed first, by drawing lines starting
from the tip of F, one line parallel to u, and the other line parallel to v .
These lines intersect the v and u axes, forming a parallelogram. The force
components Fu and Fv are established by simply joining them to the
tail of F, to the intersection points on the u and v axes, Fig. 2–8b. This
parallelogram can be reduced to a triangle, which represents the triangle
rule, Fig. 2–8c. From this, the law of sines can be applied to determine the
unknown magnitudes of the components.
12/21/23 3:30 PM
2.3
v
49
VECTOR ADDITION OF FORCES
v
F
F
Fv
F
Fv
u
u
Fu
(a)
Fu
(b)
(c)
2
Fig. 2–8
Addition of Several Forces. If more than two forces are to be
added, successive applications of the parallelogram law can be carried
out in order to obtain the resultant force. For example, if three forces F1 ,
F2 , F3 act at a point O, Fig. 2–9, the resultant of any two of the forces is
found, say, F1 + F2 , and then this resultant is added to the third force,
yielding the resultant of all three forces; i.e., FR = ( F1 + F2 ) + F3 .
Using the parallelogram law to add more than two forces, as shown here,
generally requires extensive geometric and trigonometric calculation to
determine the magnitude and direction of the resultant. Instead, problems
of this type are easily solved by using the “rectangular-component
method,” which is explained in the next section.
F1 1 F2
FR
F2
F1
F3
O
Fig. 2–9
FR
F1 1 F2
F2
F1
F3
The resultant force FR on the hook requires the addition of F1 + F2,
then this resultant is added to F3.
M02_HIBB0208_06_GE_C02.indd 49
12/21/23 3:30 PM
50
CHAPTER 2
FORCE VECTORS
I MPO RTA N T PO I N T S
• A scalar is a positive or negative number.
• A vector is a quantity that has a magnitude, direction, and sense.
• Multiplication or division of a vector by a scalar will change
the magnitude of the vector. The sense of the vector will
change if the scalar is negative.
2
• Vectors are added or subtracted using the parallelogram law or
the triangle rule.
• As a special case, if the vectors are collinear, the resultant is
F1
formed by an algebraic or scalar addition.
FR
F2
PROCEDURE FOR ANALYSIS
(a)
Problems that involve the addition of two forces can be solved as
follows:
v
F
u
Fv
Fu
(b)
A
c
• If a force F is to be resolved into components along two axes u
B
b
a
C
Cosine law:
C 5 A2 1 B2 2 2AB cos c
Sine law:
A 5 B 5 C
sin a sin b sin c
(c)
Fig. 2–10
Refer to the companion website for
Lecture Summary and Quiz videos.
M02_HIBB0208_06_GE_C02.indd 50
Parallelogram Law.
• Sketch the two “component” forces F1 and F2 added together
according to the parallelogram law, yielding the resultant force
FR that forms the diagonal of the parallelogram, Fig. 2–10a.
and v, then start at the head of force F and construct lines parallel
to the axes, thereby forming the parallelogram, Fig. 2–10b. The
sides of the parallelogram represent the components, Fu and Fv .
• Label all the known and unknown force magnitudes and the angles
on the sketch and identify the two unknowns as the magnitude
and direction of FR , or the magnitudes of its components.
Trigonometry.
• Redraw a half portion of the parallelogram to illustrate the
triangular head-to-tail addition of the components.
• From this triangle, the magnitude of the resultant force can
be determined using the law of cosines, and its direction is
determined from the law of sines. The magnitudes of two
force components are determined from the law of sines. The
formulas are given in Fig. 2–10c.
12/21/23 3:30 PM
2.3
EXAMPLE
VECTOR ADDITION OF FORCES
51
2.1
The screw eye in Fig. 2–11a is subjected to two forces, F1 and F2 .
Determine the magnitude and direction of the resultant force.
108
F2 5 150 N
A
150 N
108
158
2
658
1158
F1 5 100 N
FR
3608 2 2(658)
5 1158
2
100 N
u
158
908 2 258 5 658
(a)
(b)
SOLUTION
Parallelogram Law. The parallelogram is formed by drawing a line
from the head of F1 that is parallel to F2 , and another line from the
head of F2 that is parallel to F1 . The resultant force FR extends to
where these lines intersect at point A, Fig. 2–11b. The two unknowns
are the magnitude of FR and the angle θ (theta).
FR
Trigonometry. From the parallelogram, the vector triangle is
constructed, Fig. 2–11c. Using the law of cosines
FR =
=
150 N
212.6 N
=
sin θ
sin 115°
sin θ =
158
100 N
(c)
10 000 + 22 500 − 30 000( −0.4226) = 212.6 N
= 213 N
Applying the law of sines to determine θ ,
1158
u
f
(100 N) 2 + (150 N) 2 − 2(100 N)(150 N) cos115°
150 N
Ans.
Fig. 2–11
150 N
(sin 115°)
212.6 N
θ = 39.8°
Thus, the direction φ (phi) of FR, measured from the horizontal, is
φ = 39.8° + 15.0° = 54.8°
Ans.
NOTE: The results seem reasonable, since Fig. 2–11b shows FR to have
a magnitude larger than its components and a direction that is between
them.
M02_HIBB0208_06_GE_C02.indd 51
12/21/23 3:30 PM
52
CHAPTER 2
EXAMPLE
FORCE VECTORS
2.2
Resolve the horizontal 600-N force in Fig. 2–12a into components
acting along the u and v axes and determine the magnitudes of these
components.
2
u
u
B
Fu
30
30
30
30
A
600 N
Fv
120
30
600 N
120
Fu
30
120
30
Fv
600 N
C
v
(a)
v
(c)
(b)
Fig. 2–12
SOLUTION
The parallelogram is constructed by extending a line from the head
of the 600-N force parallel to the v axis until it intersects the u axis at
point B, Fig. 2–12b. The arrow from A to B represents Fu . Similarly,
the line extended from the head of the 600-N force drawn parallel to
the u axis intersects the v axis at point C, which gives Fv .
The vector addition using the triangle rule is shown in Fig. 2–12c. The two
unknowns are the magnitudes of Fu and Fv . Applying the law of sines,
600 N
Fu
=
sin 120°
sin 30°
Fu = 1039 N
Ans.
600 N
Fv
=
sin 30°
sin 30°
Fv = 600 N
Ans.
NOTE: The result for Fu shows that sometimes a component can have
a greater magnitude than the resultant.
M02_HIBB0208_06_GE_C02.indd 52
12/21/23 3:30 PM
2.3
EXAMPLE
VECTOR ADDITION OF FORCES
53
2.3
Determine the magnitude of the component force F in Fig. 2–13a and
the magnitude of the resultant force FR if FR is directed along the
positive y axis.
2
y
y
45
F
FR
200 N
45
F
45
30
45
60
75
30
30
(a)
F
45
FR
60
200 N
200 N
(b)
(c)
Fig. 2–13
SOLUTION
The parallelogram law of addition is shown in Fig. 2–13b, and the
triangle rule is shown in Fig. 2–13c. The magnitudes of FR and F are
the two unknowns. They can be determined by applying the law of
sines.
200 N
F
=
sin 60°
sin 45°
F = 245 N
Ans.
200 N
FR
=
sin 45°
sin 75°
FR = 273 N
M02_HIBB0208_06_GE_C02.indd 53
Ans.
Refer to the companion website for
a self quiz of these Example problems.
12/21/23 3:30 PM
54
CHAPTER 2
FORCE VECTORS
F UNDAMEN TAL PRO B L EM S
Partial solutions and answers to all Fundamental Problems are given in the back of the book. In addition, review
videos of these and other fundamental problems are available at www.pearsonglobaleditions.com.
F2–1. Determine the magnitude of the resultant force and
its direction measured clockwise from the positive x axis.
2
F2–4. Resolve the 30-N force into components along the
u and v axes, and determine the magnitude of each of these
components.
v
30 N
15
x
458
608
30
u
2 kN
6 kN
Prob. F2–1
F2–2. Two forces act on the hook. Determine the
magnitude of the resultant force.
Prob. F2–4
F2–5. The force F = 450 N acts on the frame. Resolve this
force into components acting along members AB and AC,
and determine the magnitude of each component.
30
A
C
45
308
450 N
408
200 N
500 N
Prob. F2–2
B
F2–3. Determine the magnitude of the resultant force and its
direction measured counterclockwise from the positive x axis.
y
Prob. F2–5
F2–6. If force F is to have a component along the u axis
of Fu = 6 kN , determine the magnitude of F and the
magnitude of its component Fv along the v axis.
u
800 N
F
458
1058
x
308
v
600 N
Prob. F2–3
Prob. F2–6
55
PROBLEMS
PROBLEMS
2–1. If θ = 60° and F = 450 N, determine the magnitude
of the resultant force and its direction, measured
counterclockwise from the positive x axis.
2–6. If FB = 2 kN and the resultant force acts along the
positive u axis, determine the magnitude of the resultant
force and the angle θ .
2–2. If the magnitude of the resultant force is to be 500 N,
directed along the positive y axis, determine the magnitude
of force F and its direction θ .
2–7. If the resultant force is required to act along the
positive u axis and have a magnitude of 5 kN, determine the
required magnitude of FB and its direction θ .
2
y
y
F
FA
u
u 30
x
15
700 N
3 kN
x
A
B
u
FB
Probs. 2–1/2
Probs. 2–6/7
2–3. Determine the magnitude of the resultant force
FR = F1 + F2 and its direction, measured clockwise from
the positive u axis.
*2–8. Two forces are applied at the end of a screw
eye in order to remove the post. Determine the angle
θ (0° ≤ θ ≤ 90°) and the magnitude of force F so that the
resultant force acting on the post is directed vertically
upward and has a magnitude of 750 N.
*2–4. Resolve the force F1 into components along the u
and v axes and determine the magnitudes of the components.
y
2–5. Resolve the force F2 into components along the u
and v axes and determine the magnitudes of the components.
F
500 N
u
308
x
v
308
758
F1 5 4 kN
308
u
F2 5 6 kN
Probs. 2–3/4/5
M02_HIBB0208_06_GE_C02.indd 55
Prob. 2–8
12/21/23 3:30 PM
56
CHAPTER 2
FORCE VECTORS
2–9. If θ = 60°, determine the magnitude of the resultant
force and its direction measured clockwise from the
horizontal.
2–13. If θ = 30° and T = 6 kN, determine the magnitude of
the resultant force acting on the eyebolt and its direction
measured clockwise from the positive x axis.
2–10. Determine the angle θ for connecting member A to
the plate so that the resultant force of FA and FB is directed
horizontally to the right. Also, what is the magnitude of the
resultant force?
2–14. If θ = 60° and T = 5 kN, determine the magnitude of
the resultant force acting on the eyebolt and its direction
measured clockwise from the positive x axis.
2
2–15. If the magnitude of the resultant force is to be 9 kN
directed along the positive x axis, determine the magnitude
of force T acting on the eyebolt and its angle θ .
FA 5 8 kN
u
A
y
T
u
408
B
x
FB 5 6 kN
45
Probs. 2–9/10
8 kN
Probs. 2–13/14/15
2–11. Determine the magnitude of the resultant force
FR = F1 + F2 and its orientation θ , measured clockwise
from the positive x axis.
*2–12. Determine the magnitude of the resultant
force FR = F1 + F3 and its orientation θ , measured
counterclockwise from the positive x axis.
*2–16. The pelvis P is connected to the femur F at A using
three different muscles, which exert the forces shown on
the femur. Determine the resultant force and specify its
orientation θ , measured counterclockwise from the positive
x axis.
y
F3 5 250 N
y
308
120 N
x
60 N
13 12
5
80 N
308
458
F1 5 400 N
F2 5 360 N
A
308
x
F
Probs. 2–11/12
M02_HIBB0208_06_GE_C02.indd 56
P
Prob. 2–16
12/21/23 3:30 PM
57
PROBLEMS
2–17. Determine the magnitude and direction of the
resultant force, FR measured counterclockwise from
the positive x axis. Solve the problem by first finding the
resultant F ' = F1 + F2 and then forming FR = F ' + F3 .
2–18. Determine the magnitude and direction of the
resultant force, FR measured counterclockwise from
the positive x axis. Solve the problem by first finding the
resultant F ' = F2 + F3 and then forming FR = F ' + F1 .
2–21. Determine the magnitude of the two towing forces
FB and FC if the resultant force has a magnitude FR = 10 kN
and is directed along the positive x axis. Set θ = 15° .
2–22. If the resultant FR of the two forces acting on the jet
aircraft is to be directed along the positive x axis and have
a magnitude of 10 kN, determine the angle θ of the cable
attached to the truck at B so that FB is a minimum. What is
the magnitude of force in each cable when this occurs?
2
y
F1 400 N
C
FC
F2 200 N
90º
208
x
x
150º
u
A
FB
B
F3 300 N
Probs. 2–21/22
Probs. 2–17/18
2–19. Determine the magnitude and direction of the
resultant FR = F1 + F2 + F3 of the three forces by first
finding the resultant F ' = F1 + F2 and then finding
FR = F ' + F3 .
*2–20. Determine the magnitude and direction of the
resultant FR = F1 + F2 + F3 of the three forces by first
finding the resultant F ' = F2 + F3 and then finding
FR = F ' + F1 .
2–23. Two forces act on the screw eye. If F1 = 400 N
and F2 = 600 N, determine the angle θ (0° ≤ θ 180°)
between them, so that the resultant force has a magnitude
of FR = 800 N.
*2–24. Two forces F1 and F2 act on the screw eye. If their
lines of action are at an angle θ apart and the magnitude of
each force is F1 = F2 = F, determine the magnitude of the
resultant force FR and the angle between FR and F11.
F1
y
F1 5 30 N
5
3
F3 5 50 N
4
x
u
208
F2 5 20 N
F2
Probs. 2–19/20
M02_HIBB0208_06_GE_C02.indd 57
Probs. 2–23/24
12/21/23 3:30 PM
58
CHAPTER 2
FORCE VECTORS
2.4
ADDITION OF A SYSTEM
OF COPLANAR FORCES
When a force is resolved into two components along the x and y axes,
the components are then called rectangular components. For analytical
work we can represent these components in one of two ways, using either
scalar or Cartesian vector notation.
Scalar Notation. The rectangular components of force F shown in
Fig. 2–14a are found using the parallelogram law, so that F = Fx + Fy .
Because these components form a right triangle, they can be determined
from
2
y
Fx = F cos θ
Fy = F sin θ
and
F
Fy
u
x
Fx
Instead of using the angle θ , however, the direction of F can also
be defined using a small “slope” triangle, as in the example shown in
Fig. 2–14b. Since this triangle and the larger shaded triangle are similar,
the proportional length of the sides gives
Fx
a
=
F
c
(a)
y
or
Fx
Fy
b
 a
Fx = F  
 c
x
and
c
a
Fy
F
F
(b)
Fig. 2–14
=
b
c
or
 b
Fy = − F  
 c
Here the y component is a negative scalar since Fy is directed along the
negative y axis.
It is important to keep in mind that this positive and negative
scalar notation is to be used only for calculations, not for graphical
representations in figures. Throughout the text, the head of a vector
arrow in any figure indicates the sense of the vector graphically; algebraic
signs are not used for this purpose. Thus, the vectors in Figs. 2–14a and
2–14b are designated by using boldface (vector) notation.* Whenever
italic symbols are written near vector arrows in figures, they indicate the
magnitude of the vector, which is always a positive quantity.
*Negative signs are used only in figures with boldface notation when showing equal but
opposite pairs of vectors, as in Fig. 2–2.
M02_HIBB0208_06_GE_C02.indd 58
12/21/23 3:31 PM
2.4
59
ADDITION OF A SYSTEM OF COPLANAR FORCES
Cartesian Vector Notation. It is also possible to represent
the x and y components of a force in terms of Cartesian unit vectors
i and j. They are called unit vectors because they have a dimensionless
magnitude of 1, and so they can be used to designate the directions of the
x and y axes, respectively, Fig. 2–15.*
Since the magnitude of each component of F is always a positive
quantity, which is represented by the (positive) scalars Fx and Fy , then
we can express F as a Cartesian vector,
y
j
F
Fy
x
Fx
F = Fx i + F y j
i
Coplanar Force Resultants. We can use either of the two
methods just described to determine the resultant of several coplanar
forces. To do this, each force is first resolved into its x and y components,
and then the respective components are added using scalar algebra since
they are collinear. The resultant force is then formed by adding the
resultant components using the parallelogram law. For example, consider
the three concurrent forces in Fig. 2–16a, which have x and y components
shown in Fig. 2–16b. Using Cartesian vector notation, each force is first
represented as a Cartesian vector, i.e.,
2
Fig. 2–15
y
F2
F1
x
F3
(a)
F1 = F1 x i + F1 y j
y
F2 = − F2 x i + F2 y j
F3 = F3 x i − F3 y j
The vector resultant, Fig. 2–16c, is therefore
F2y
F2x
F1x
F3x
FR = F1 + F2 + F3
(b)
= ( F1 x − F2 x + F3 x ) i + ( F1 y + F2 y − F3 y ) j
= ( FR ) x i + ( FR ) y j
If scalar notation is used, then indicating the positive directions of
components along the x and y axes with symbolic arrows, we have
+
(F ) = F − F + F
+↑
R x
1x
2x
x
F3y
= F1 x i + F1 y j − F2 x i + F2 y j + F3 x i − F3 y j
→
F1y
3x
( FR ) y = F1 y + F2 y − F3 y
Fig. 2–16
y
F 4 F3
F2
F1
x
Notice that these are the same results as the i and j components of FR
determined above.
*For handwritten work, unit vectors are usually indicated using a circumflex, e.g., iˆ and ĵ.
Also, realize that Fx and Fy in Fig. 2–15 represent the magnitudes of the components, which
are always positive scalars. The directions are defined by i and j. If instead we used scalar
notation, then Fx and Fy could be positive or negative scalars, since they would account for
both the magnitude and direction of the components.
M02_HIBB0208_06_GE_C02.indd 59
The resultant force of the four cable forces
acting on the post can be determined by
adding algebraically the separate x and y
components of each cable force. This
resultant FR produces the same pulling effect
on the post as all four cables.
12/21/23 3:31 PM
60
CHAPTER 2
FORCE VECTORS
In general then, the components of the resultant force of any number
of coplanar forces can be represented by the algebraic sum of the x and y
components of all the forces, i.e.,
( FR ) x = ΣFx
( FR ) y = ΣFy
2
Once these components are determined, they may be sketched along
the x and y axes with their proper sense of direction, and the resultant
force can be determined from vector addition, Fig. 2–16c. From this
sketch, the magnitude of FR is then found from the Pythagorean theorem;
that is,
y
FR
(FR)y
u
(FR)x
(c)
Fig. 2–16 (cont.)
(2–1)
x
FR =
( FR ) x2 + ( FR ) 2y
Also, the angle θ , which specifies the direction of the resultant force, is
determined from trigonometry:
θ = tan −1
( FR ) y
( FR ) x
The above concepts are illustrated numerically in the examples which
follow.
I MPO RTA N T PO I N T S
• The resultant of several coplanar forces can easily be
determined if an x, y coordinate system is established and the
forces are resolved into components along the axes.
• The direction of each force is specified by the angle its line of
action makes with one of the axes, or by a slope triangle.
• The orientation of the x and y axes is arbitrary, and their
positive direction can be specified by the Cartesian unit vectors
i and j.
• The x and y components of the resultant force are simply the
algebraic addition of the components of all the coplanar forces.
• The magnitude of the resultant force is determined from the
Refer to the companion website for
Lecture Summary and Quiz videos.
M02_HIBB0208_06_GE_C02.indd 60
Pythagorean theorem, and when the resultant components are
sketched on the x and y axes, Fig. 2–16c, the direction θ of the
resultant can be determined from trigonometry.
12/21/23 3:31 PM
2.4
EXAMPLE
2.4
Determine the x and y components of F1 and F2 acting on the boom
shown in Fig. 2–17a. Express each force as a Cartesian vector.
y
F1 5 200 N
SOLUTION
308
Scalar Notation. By the parallelogram law, F1 is resolved into x and y
components, Fig. 2–17b. Since F1x acts in the − x direction, and F1y acts
in the + y direction, we have
F1x = −200 sin 30° N = −100 N = 100 N ←
Ans.
F1y = 200 cos 30° N = 173 N = 173 N ↑
Ans.
The force F2 is resolved into its x and y components, as shown in
Fig. 2–17c. From this “slope triangle” we could obtain the angle θ ,
5
e.g., θ = tan −1 12
, and then proceed to determine the magnitudes
of the components in the same manner as for F1 . The easier method,
however, consists of using proportional parts of similar triangles, i.e.,
2
F2x
12
=
260 N
13
 12 
= 240 N
F2x = 260 N 
 13 
x
u
13
5
12
F2 5 260 N
(a)
( )
y
F1 5 200 N
F1y 5 200 cos 308 N
308
Similarly,
x
F1x 5 200 sin 308 N
 5 
F2y = 260 N 
= 100 N
 13 
(b)
Notice how the magnitude of the horizontal component, F2x, was
obtained by multiplying the force magnitude by the ratio of the
horizontal leg of the slope triangle divided by the hypotenuse;
whereas the magnitude of the vertical component, F2y, was obtained
by multiplying the force magnitude by the ratio of the vertical leg
divided by the hypotenuse. Using scalar notation to represent the
components, we have
F2x = 240 N = 240 N →
Ans.
F2y = −100 N = 100 N ↓
Ans.
Cartesian Vector Notation. Having determined the magnitudes
and directions of the components of each force, we can express each
force as a Cartesian vector.
M02_HIBB0208_06_GE_C02.indd 61
61
ADDITION OF A SYSTEM OF COPLANAR FORCES
F1 = { −100 i + 173 j } N
Ans.
F2 = { 240 i − 100 j } N
Ans.
y
( ( x
12
F2x 5 260 —
— N
13
( (
5
5 N
F2y 5 260 —
—
13
13
12
F2 5 260 N
(c)
Fig. 2–17
12/21/23 3:31 PM
62
CHAPTER 2
EXAMPLE
FORCE VECTORS
2.5
y
F1 5 600 N
F2 5 400 N
458
The link in Fig. 2–18a is subjected to two forces F1 and F2 . Determine
the magnitude and direction of the resultant force.
SOLUTION I
308
x
2
Scalar Notation. First we resolve each force into its x and y
components, Fig. 2–18b, then we sum these components algebraically.
+ ( F ) = ΣF ;
→
R x
x
(a)
( FR ) x = 600 cos 30° N − 400 sin 45° N
= 236.8 N →
y
+ ↑ ( FR ) y = ΣFy ;
F1 5 600 N
F2 5 400 N
458
= 582.8 N ↑
The resultant force, shown in Fig. 2–18c, has a magnitude of
308
x
FR =
(236.8 N) 2 + (582.8 N) 2
= 629 N
(b)
and a direction,
y
FR
582.8 N
( FR ) y = 600 sin 30° N + 400 cos 45° N
 582.8 N 
θ = tan −1 
= 67.9°
 236.8 N 
Ans.
Ans.
SOLUTION II
Cartesian Vector Notation. From Fig. 2–18b, each force is first
expressed as a Cartesian vector.
u
236.8 N
x
F1 = { 600 cos 30°i + 600 sin 30°j } N
F2 = { −400 sin 45°i + 400 cos 45°j } N
(c)
Fig. 2–18
Then,
FR = F1 + F2 = (600 cos 30° N − 400 sin 45° N)i
+ (600 sin 30° N + 400 cos 45° N) j
= { 236.8 i + 582.8 j } N
The magnitude and direction of FR are determined in the same
manner as before.
NOTE: Comparing the two methods of solution, notice that the use of
scalar notation is more efficient since the components can be found
directly, without first having to express each force as a Cartesian
vector before adding the components. Later, however, we will
show that Cartesian vector analysis is very beneficial for solving
three-dimensional problems.
M02_HIBB0208_06_GE_C02.indd 62
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2.4
EXAMPLE
63
ADDITION OF A SYSTEM OF COPLANAR FORCES
2.6
The end of the boom O in Fig. 2–19a is subjected to three concurrent
and coplanar forces. Determine the magnitude and direction of the
resultant force.
y
2
3
F2 5 250 N
458
F3 5 200 N
5
4
F1 5 400 N
O
x
y
250 N
200 N
(a)
458
3
5
4
SOLUTION
Each force is resolved into its x and y components, Fig. 2–19b.
Summing the x components, we have
+ ( F ) = ΣF ;
→
R x
x
( FR ) x = −400 N + 250 sin 45° N − 200
x
400 N
O
(b)
( 45 ) N
= −383.2 N = 383.2 N ←
Summing the y components yields
+ ↑ ( FR ) y = ΣFy ;
( FR ) y = 250 cos 45° N + 200
y
( 53 ) N
FR
296.8 N
= 296.8 N ↑
The resultant force, shown in Fig. 2–19c, has a magnitude of
FR =
u
( −383.2 N) 2 + (296.8 N) 2
= 485 N
(c)
Fig. 2–19
Ans.
NOTE: Application of this method is more convenient, compared to
using two applications of the parallelogram law, first to add F1 and F2,
then adding F3 to this resultant.
M02_HIBB0208_06_GE_C02.indd 63
x
Ans.
From the vector addition in Fig. 2–19c, the direction angle θ is
 296.8 
= 37.8°
θ = tan −1 
 383.2 
O
383.2 N
Refer to the companion website for
a self quiz of these Example problems.
12/21/23 3:31 PM
64
CHAPTER 2
FORCE VECTORS
F U NDAMEN TAL PR O B L EM S
F2–7. Resolve each force acting on the post into its x and y
components.
y
F2–10. If the resultant force acting on the bracket is to
be 750 N directed along the positive x axis, determine the
magnitude of F and its direction θ .
y
F1 5 300 N
F2 5 450 N
5
2
F3 5 600 N
325 N
13
4
12
F
5
3
458
x
u
x
458
Prob. F2–7
600 N
F2–8. Determine the magnitude and direction of the
resultant force.
3
Prob. F2–10
y
250 N
5
F2–11. If the magnitude of the resultant force acting on
the bracket is to be 80 N directed along the u axis, determine
the magnitude of F and its direction θ .
400 N
4
308
y
x
F
300 N
u
5
90 N
Prob. F2–8
F2–9. Determine the magnitude of the resultant
force acting on the corbel and its direction θ measured
counterclockwise from the x axis.
y
x
50 N
45
4
u
3
Prob. F2–11
F2–12. Determine the magnitude of the resultant force
and its direction θ , measured counterclockwise from the
positive x axis.
y
F3 600 N
4
F2 400 N
5
3
F1 700 N
30
F1 5 15 kN
5
3
x
F2 5 20 kN
5
4
3
F3 5 15 kN
4
x
Prob. F2–9
M02_HIBB0208_06_GE_C02.indd 64
Prob. F2–12
12/21/23 3:31 PM
65
PROBLEMS
PROBLEMS
2–25. Determine the magnitude of the resultant force and
its direction, measured clockwise from the positive x axis.
2–27. Resolve each force acting on the gusset plate into its
x and y components, and express each force as a Cartesian
vector.
*2–28. Determine the magnitude of the resultant force
acting on the gusset plate and its direction, measured
counterclockwise from the positive x axis.
y
2
400 N
B
y
30
F3 5 650 N
x
3
F2 5 750 N
5
4
45
458
800 N
x
F1 5 900 N
Probs. 2–27/28
Prob. 2–25
2–26. Express each of the three forces acting on the
support in Cartesian vector form and determine the
magnitude of the resultant force and its direction, measured
clockwise from positive x axis.
2–29. Resolve F1 and F2 into their x and y components.
2–30. Determine the magnitude of the resultant force and
its direction measured counterclockwise from the positive
x axis.
y
y
F1 50 N
60
30
5
4
F1 400 N
4
3
x
F3 30 N
15
45
F2 80 N
F2 250 N
Prob. 2–26
M02_HIBB0208_06_GE_C02.indd 65
x
Probs. 2–29/30
12/21/23 3:31 PM
66
CHAPTER 2
FORCE VECTORS
2–31. Determine the magnitude of the resultant force and its
direction measured counterclockwise from the positive x axis.
y
F3
2–35. Three forces act on the ring. Determine the range of
values for the magnitude of P so that the magnitude of the
resultant force does not exceed 2500 N. Force P is always
directed to the right.
8 kN
y
F2
5 kN
F1
4 kN
1500 N
60
2
600 N
45
608
x
458
P
Prob. 2–31
Prob. 2–35
*2–32. The three forces are applied to the bracket.
Determine the range of values for the magnitude of force P so
that the resultant of the three forces does not exceed 2400 N.
y
*2–36. Express F1, F2 , and F3 as Cartesian vectors.
2–37. Determine the magnitude of the resultant force and its
direction measured counterclockwise from the positive x axis.
y
800 N
3000 N
x
908
F3 5 750 N
458
608
P
x
3
x
5
4
308
F1 5 850 N
F2 5 625 N
Probs. 2–36/37
Prob. 2–32
2–33. Determine the x and y components of F1 and F2.
2–34. Determine the magnitude of the resultant force and its
direction measured counterclockwise from the positive x axis.
2–38. The three concurrent forces acting on the post
produce a zero resultant force FR = 0 . If F2 = 21 F1, and
F1 is to be 90° from F2 as shown, determine the required
magnitude of F3 expressed in terms of F1 and the angle θ .
y
F2
u
y
458
F3
x
F1 5 200 N
F1
308
F2 5 150 N
x
Probs. 2–33/34
M02_HIBB0208_06_GE_C02.indd 66
Prob. 2–38
12/21/23 3:31 PM
2.5
2.5
67
CARTESIAN VECTORS
CARTESIAN VECTORS
z
The operations of vector algebra, when applied to solving problems in
three dimensions, are greatly simplified if the vectors are first represented
in Cartesian vector form. In this section we will present a general method
for doing this; then in the next section we will use this method for finding
the resultant force of a system of concurrent forces.
y
x
Right-Handed Coordinate System. We will use a right-handed
coordinate system to describe the theory of vector algebra that follows.
Specifically, a rectangular coordinate system is said to be right-handed
if the thumb of the right hand points in the direction of the positive z
axis when the right-hand fingers are curled about this axis and directed
from the positive x towards the positive y axis, Fig. 2–20.
2
Fig. 2–20
z
Rectangular Components of a Vector. In general, a vector
A may have one, two, or three rectangular components along the x, y, z
coordinate axes, Fig. 2–21. These components are determined using two
successive applications of the parallelogram law; that is, A = A' + A z
and then A' = A x + A y . Combining these equations to eliminate A',
A is represented by the vector sum of its three rectangular components,
A = Ax + A y + Az
Az
A
(2–2)
Ay
Cartesian Vector Representation. In three dimensions, the
set of Cartesian unit vectors, i, j, k, is used to designate the directions
of the x, y, z axes, respectively, Fig. 2–22. Using these vectors the three
components of A in Fig. 2–23 can be written in Cartesian vector form as
Ax
y
A9
x
Fig. 2–21
z
Az k
z
A
k
k
Ax i
j Ay j
x
Fig. 2–23
M02_HIBB0208_06_GE_C02.indd 67
j
i
x
y
i
y
Fig. 2–22
12/21/23 3:31 PM
68
CHAPTER 2
FORCE VECTORS
z
A = Ax i + Ay j + Az k
A zk
There is a distinct advantage to writing vectors in this manner.
Separating the magnitude and direction of each component vector will
simplify the operations of vector algebra, particularly in three dimensions.
A
Az
A
Ayj
2
A xi
Magnitude of a Cartesian Vector. If A is expressed as
a Cartesian vector, then its magnitude can be determined. As shown in
Fig. 2–24, from the blue right triangle, A = A' 2 + Az2 , and from the gray
right triangle, A' = Ax2 + Ay2 . Combining these equations to eliminate
A' yields
y
Ax
A9
Ay
x
(2–3)
Fig. 2–24
A=
Ax2 + Ay2 + Az2
(2–4)
z
Hence, the magnitude of A is equal to the positive square root of the sum
of the squares of the magnitudes of its components.
A zk
A
Coordinate Direction Angles. We will define the direction of
uA
A by the coordinate direction angles α (alpha), β (beta), and γ (gamma),
measured between the tail of A and the positive x, y, z axes, Fig. 2–25.
Note that regardless of where A is directed, each of these angles will be
between 0° and 180°.
To determine α , β , and γ , consider the projection of A onto the x, y, z
axes, Fig. 2–26. Referring to the shaded right triangles shown in the
figure, we have
g
b
a
Ay j
y
Axi
x
Fig. 2–25
cos α =
z
Az
A
uA =
b
908
Ax
cos β =
Ay
A
cos γ =
Az
A
(2–5)
These numbers are known as the direction cosines of A. Once they
have been obtained, the coordinate direction angles α , β , γ can then be
determined from the inverse cosines.
An easy way of obtaining these direction cosines is to form a unit
vector uA in the direction of A, Fig. 2–25. To do this, divide A by its
magnitude A, so that
g
a
Ax
A
Ay
y
Ay
A
A
A
= x i+
j+ z k
A
A
A
A
(2–6)
By comparison with Eqs. 2–5, it is seen that the i, j, k components of uA
represent the direction cosines of A, i.e.,
x
Fig. 2–26
M02_HIBB0208_06_GE_C02.indd 68
u A = cos α i + cos β j + cos γ k
(2–7)
12/21/23 3:32 PM
2.5
CARTESIAN VECTORS
69
Since the magnitude of u A = cos α i + cos β j + cos γ k is one, then
from this equation an important relation among the direction cosines is
cos 2 α + cos 2 β + cos 2 γ = 1
(2–8)
Therefore, if only two of the coordinate angles are known, the third
angle can be found using this equation.
Finally, if the magnitude and coordinate direction angles of A are
known, then A may be expressed in Cartesian vector form as
A = Au A
= A cos α i + A cos β j + A cos γ k
= Ax i + Ay j + Az k
2
(2–9)
Horizontal and Vertical Angles. Sometimes, the direction of A
can be specified using a horizontal angle θ and a vertical angle φ (phi),
such as shown in Fig. 2–27. The components of A can then be determined
by applying trigonometry first to the light blue right triangle, which yields
Az = A cos φ
A' = A sin φ
and
Now applying trigonometry to the dark blue right triangle,
Ax = A' cos θ = A sin φ cos θ
Ay = A' sin θ = A sin φ sin θ
Therefore, A written in Cartesian vector form becomes
A = A sin φ cos θ i + A sin φ sin θ j + A cos φ k
This equation should not be memorized; rather, it is important to
understand how the components were determined using trigonometry.
z
Az
f
A
Ax O
Ay
u
y
x
A9
Fig. 2–27
M02_HIBB0208_06_GE_C02.indd 69
12/21/23 3:32 PM
70
CHAPTER 2
FORCE VECTORS
2.6
ADDITION OF CARTESIAN
VECTORS
The addition (or subtraction) of two or more vectors is greatly simplified
if the vectors are expressed in terms of their Cartesian components. For
example, if A = Ax i + Ay j + Az k and B = Bx i + B y j + Bz k, Fig. 2–28,
then the resultant vector, R, has components which are the scalar sums of
the i, j, k components of A and B, i.e.,
R = A + B = ( Ax + Bx )i + ( Ay + B y ) j + ( Az + Bz )k
2
If this is generalized and applied to a system of several concurrent
forces, then the force resultant is the vector sum of all the forces in the
system and can be written as
FR = ΣF = ΣFx i + ΣFy j + ΣFz k
(2–10)
Here ΣFx, ΣFy, and ΣFz represent the algebraic sums of the respective x,
y, z or i, j, k components of each force in the system.
z
(Az 1 Bz)k
R
B
(Ay 1 By)j
A
y
(Ax 1 Bx)i
x
Fig. 2–28
I MPO RTA N T PO I N T S
• A Cartesian vector A has i, j, k components along the x, y, z axes.
If A is known, its magnitude is A =
Ax2 + Ay2 + Az2 .
• The direction of a Cartesian vector can be defined by the three
coordinate direction angles α , β , γ , measured from the positive
x, y, z axes to the tail of the vector. To find these angles,
formulate a unit vector in the direction of A, i.e., u A = A A,
and determine the inverse cosines of its components. Only two
of these angles are independent of one another; the third angle
is found from cos 2 α + cos 2 β + cos 2 γ = 1.
Refer to the companion website for
Lecture Summary and Quiz videos.
M02_HIBB0208_06_GE_C02.indd 70
• The direction of a Cartesian vector can also be specified using
a horizontal angle θ and vertical angle φ .
12/21/23 3:32 PM
2.6
EXAMPLE
71
ADDITION OF CARTESIAN VECTORS
2.7
Express the force F shown in Fig. 2–29a as a Cartesian vector.
z
SOLUTION
The angles of 60° and 45° defining the direction of F are not coordinate
direction angles. Two successive applications of the parallelogram law
are needed to resolve F into its x, y, z components. First F = F' + Fz,
then F' = Fx + Fy , Fig. 2–29b. By trigonometry, the magnitudes of the
components are
F 100 N
2
60
y
45
Fz = 100 sin 60 ° N = 86.6 N
F' = 100 cos 60 ° N = 50 N
x
(a)
Fx = F' cos 45° = 50 cos 45° N = 35.4 N
Fy = F' sin 45° = 50 sin 45° N = 35.4 N
z
Fz
Realizing that Fy is in the − j direction, we have
F = { 35.4 i − 35.4 j + 86.6 k } N
Ans.
F 100 N
To show that the magnitude of this vector is indeed 100 N, apply Eq. 2–4,
F =
Fy
Fx2 + Fy2 + Fz2
45
F¿
=
60
(35.4) 2 + (35.4) 2 + (86.6) 2 = 100 N
Fx
x
(b)
If needed, the coordinate direction angles of F can be determined
from the components of the unit vector acting in the direction of F.
Fy
F
F
F
u =
= xi+
j+ zk
F
F
F
F
=
35.4
35.4
86.6
i−
j+
k
100
100
100
y
z
F 100 N
30.0
111
= 0.354 i − 0.354 j + 0.866 k
y
69.3
so that
α = cos −1 (0.354) = 69.3°
β = cos −1 ( −0.354) = 111°
γ = cos −1 (0.866) = 30.0°
x
(c)
Fig. 2–29
These results are shown in Fig. 2–29c.
M02_HIBB0208_06_GE_C02.indd 71
12/21/23 3:32 PM
72
CHAPTER 2
EXAMPLE
FORCE VECTORS
2.8
Two forces act on the hook shown in Fig. 2–30a. Specify the magnitude
of F2 and its coordinate direction angles so that the resultant force FR
acts along the positive y axis and has a magnitude of 800 N.
z
F2
1208
y
608
2
458
F1 5 300 N
x
(a)
SOLUTION
To solve this problem, the resultant force FR and its two components,
F1 and F2 , will each be expressed in Cartesian vector form. Then, as
shown in Fig. 2–30b, it is necessary that FR = F1 + F2 .
Applying Eq. 2–9,
F1 = F1 cos α 1 i + F1 cos β 1 j + F1 cos γ 1 k
= 300 cos 45° i + 300 cos 60° j + 300 cos 120° k
z
= { 212.1i + 150 j − 150 k } N
F2 5 700 N
g2 5 77.68
b2 5 21.88
a2 5 1088
x
F1 5 300 N
(b)
Fig. 2–30
F2 = F2 x i + F2 y j + F2 z k
Since FR has a magnitude of 800 N and acts in the + j direction,
FR = (800 N)( + j) = { 800 j } N
FR 5 800 N
y
We require
FR = F1 + F2
800 j = 212.1i + 150 j − 150 k + F2 x i + F2 y j + F2 z k
800 j = (212.1 + F2 x )i + (150 + F2 y ) j + ( −150 + F2 z )k
To satisfy this equation, the i, j, k components of FR must be equal to
the corresponding i, j, k components of ( F1 + F2 ) . Hence,
0 = 212.1 + F2 x
F2 x = −212.1 N
800 = 150 + F2 y
F2 y = 650 N
0 = −150 + F2 z
F2 z = 150 N
The magnitude of F2 is thus
F2 =
( −212.1 N) 2 + (650 N) 2 + (150 N) 2
= 700 N
Refer to the companion website for
a self quiz of these Example problems.
M02_HIBB0208_06_GE_C02.indd 72
We can use Eq. 2–9 to determine α 2 , β 2 , γ 2 .
−212.1
cos α 2 =
;
α 2 = 108°
700
650
cos β 2 =
;
β 2 = 21.8°
700
150
cos γ 2 =
;
γ 2 = 77.6°
700
Ans.
Ans.
Ans.
Ans.
These results are shown in Fig. 2–30b.
12/21/23 3:32 PM
73
FUNDAMENTAL PROBLEMS
F UN DAMEN TAL PR O B L EM S
F2–13. Determine the coordinate direction angles of the
force.
F2–16. Express the force as a Cartesian vector.
z
F 50 N
z
2
45
5
4
3
x
y
y
45
Prob. F2–16
30
x
F2–17. Express the force as a Cartesian vector.
F 75 N
z
Prob. F2–13
F 5 750 N
F2–14. Express the force as a Cartesian vector.
z
F 5 500 N
458
608
608
608
y
x
Prob. F2–17
x
y
F2–18. Determine the resultant force acting on the hook.
z
Prob. F2–14
F1 500 N
5
F2–15. Express the force as a Cartesian vector.
4
z
x
458
608
F 5 500 N
y
45
y
F2 800 N
x
Prob. F2–15
M02_HIBB0208_06_GE_C02.indd 73
30
3
Prob. F2–18
12/21/23 3:32 PM
74
CHAPTER 2
FORCE VECTORS
P R OBLEMS
2–39. The bolt is subjected to the force F, which has
components acting along the x, y, z axes. If the magnitude
of F is 80 N, and α = 60° and γ = 45°, determine the
magnitudes of its components.
2
z
2–41. The force F acts on the bracket within the octant
shown. If F = 400 N, β = 60°, and γ = 45° , determine the
x, y, z components of F.
2–42. The force F acts on the bracket within the octant
shown. If the magnitudes of the x and z components of F
are Fx = 300 N and Fz = 600 N, respectively, and β = 60°,
determine the magnitude of F and its y component. Also,
find the coordinate direction angles α and γ .
z
Fz
g
F
g
b
F
y
Fy
a
Fx
b
a
x
x
y
Prob. 2–39
Probs. 2–41/42
*2–40. Determine the magnitude and coordinate direction
angles of the force F acting on the support. The component
of F in the x−y plane is 7 kN.
2–43. Determine the magnitude and coordinate direction
angles of the resultant force, and sketch this vector on the
coordinate system.
z
z
F2 125 N
3
F
y
y
30
20
40
7 kN
60
45
60
x
x
Prob. 2–40
M02_HIBB0208_06_GE_C02.indd 74
5
4
F1 400 N
Prob. 2–43
12/21/23 3:32 PM
75
PROBLEMS
*2–44. Determine the magnitude and coordinate direction
angles of F3 so that the resultant of the three forces acts
along the positive y axis and has a magnitude of 600 N.
*2–48. The stock mounted on the lathe is subjected to a
force of 60 N. Determine the coordinate direction angle β
and express the force as a Cartesian vector.
2–45. Determine the magnitude and coordinate direction
angles of F3 so that the resultant of the three forces is zero.
z
60 N
z
F3
2
45
b
60
F1 180 N
y
x
40
y
30
x
Prob. 2–48
2–49. The bracket is subjected to the two forces shown.
Express each force in Cartesian vector form and then
determine the resultant force FR. Find the magnitude and
coordinate direction angles of the resultant force.
z
F2 300 N
Probs. 2–44/45
608
F2 5 400 N
458
1208
2–46. Express each force in Cartesian vector form and
then determine the resultant force. Find the magnitude and
coordinate direction angles of the resultant force.
358
2–47. Determine the coordinate direction angles of F1 .
z
458
458
F2 5 525 N
608
458
608
x
F2 5 500 N
Probs. 2–46/47
M02_HIBB0208_06_GE_C02.indd 75
F1 5 250 N
z
y
x
x
Prob. 2–49
2–50. Determine the magnitude and coordinate direction
angles of the resultant force, and sketch this vector on the
coordinate system.
F1 5 300 N
608
1208
y
258
1208
y
4 5
3
F1 5 450 N
Prob. 2–50
12/21/23 3:32 PM
76
CHAPTER 2
FORCE VECTORS
2–51. Express each force as a Cartesian vector.
*2–52. Determine the magnitude and coordinate direction
angles of the resultant force, and sketch this vector on the
coordinate system.
2–54. Specify the magnitude F3 and directions α 3, β 3, and γ 3
so that the resultant force of the three forces is
FR = { 9 j } kN.
z
2
F2 5 10 kN
F3
z
F3 5 200 N
F2 5 150 N
5
13
g3
b3
12
a3
y
308
3
F1 5 12 kN
608
5
F1 5 90 N
y
4
x
458
Prob. 2–54
x
Probs. 2–51/52
2–55. The pole is subjected to the force F, which has
components acting along the x, y, z axes as shown. If the
magnitude of F is 3 kN, β = 30°, and γ = 75°, determine the
magnitudes of its three components.
2–53. Determine the magnitude and coordinate direction
angles of the resultant force, and sketch this vector on the
coordinate system.
*2–56. The pole is subjected to the force F which has
components Fx = 1.5 kN and Fz = 1.25 kN. If β = 75°,
determine the magnitudes of F and Fy.
z
Fz
z
F1 = 400 N
F
g
60
b
Fy
y
a
135
20
60
60
Fx
x
y
x
F2 500 N
Prob. 2–53
M02_HIBB0208_06_GE_C02.indd 76
Probs. 2–55/56
12/21/23 3:32 PM
2.7
2.7
POSITION VECTORS
77
POSITION VECTORS
z
In this section we will introduce the concept of a position vector. It will
be shown that this vector is of importance in formulating a Cartesian
force vector directed between two points in space.
x, y, z Coordinates. Throughout the book we will use the
convention followed in many technical texts, which requires the positive
z axis to be directed upward (the zenith direction) so that it measures the
height of an object or the altitude of a point. The x, y axes then lie in the
horizontal plane, Fig. 2–31.
Points in space are located relative to the origin of coordinates, O,
by successive measurements along the x, y, z axes. For example, the
coordinates of point A are obtained by starting at O and measuring
x A = +4 m along the x axis, then y A = +2 m along the y axis, and finally
z A = −6 m along the z axis, so that A(4 m, 2 m, − 6 m).
2m
O
y
4m
2
6m
x
A
Fig. 2–31
Position Vector. A position vector r is defined as a fixed vector
which locates a point in space relative to another point. For example, if r
extends from the origin of coordinates, O, to point P(x, y, z), Fig. 2–32a,
then r can be expressed in Cartesian vector form as
r = xi + yj + zk
Note how the head-to-tail vector addition of the three components
yields vector r, Fig. 2–32b. Starting at the origin O, one “travels” x in the
+ i direction, then y in the + j direction, and finally z in the + k direction
to arrive at point P(x, y, z).
z
z
zk
P(x, y, z)
r
r
yj
O
xi
x
y
xi
x
(a)
P(x, y, z)
zk
O
y
yj
(b)
Fig. 2–32
M02_HIBB0208_06_GE_C02.indd 77
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78
CHAPTER 2
FORCE VECTORS
z
B(xB, yB, zB)
In the more general case, the position vector may be directed from
point A to point B in space, Fig. 2–33a. By the head-to-tail vector
addition, using the triangle rule, we require
r
rB
A(xA, yA, zA)
rA + r = rB
rA
y
Solving for r and expressing rA and rB in Cartesian vector form yields
x
(a)
2
r = rB − rA = ( x B i + yB j + zB k ) − ( x A i + y A j + z A k )
or
r = ( x B − x A ) i + ( y B − y A ) j + ( zB − z A ) k
z
B
r
(xB 2 xA)i
Thus, the i, j, k components of r may be formed by taking the coordinates
of the tail of the vector A( x A, y A, z A ) and subtracting them from the
corresponding coordinates of the head B( x B, yB, zB ) . We can also form
these components directly, Fig. 2–33b, by starting at A and moving
through a distance of ( x B − x A ) along the positive x axis ( + i ) , then
( yB − y A ) along the positive y axis ( + j) , and finally ( zB − z A ) along the
positive z axis ( + k ) to get to B.
A
(yB 2 yA)j
x
(zB 2 zA)k
(2–11)
y
(b)
Fig. 2–33
A
u
r
B
M02_HIBB0208_06_GE_C02.indd 78
If an x, y, z coordinate system is
established, then the coordinates
of two points A and B on the cable
can be determined. From this
the position vector r acting along
the cable can be formulated. Its
magnitude represents the distance
from A to B, and its unit vector,
u = r / r , gives the direction defined
by α , β, γ .
12/21/23 3:32 PM
2.8
2.8
79
FORCE VECTOR DIRECTED ALONG A LINE
FORCE VECTOR DIRECTED
ALONG A LINE
z
F
Quite often in three-dimensional statics problems, the direction of a
force is specified by two points through which its line of action passes.
Such a situation is shown in Fig. 2–34, where the force F is directed along
the cord AB. We can formulate F as a Cartesian vector by realizing that
it has the same direction and sense as the position vector r directed from
point A to point B on the cord. This common direction is specified by the
unit vector u = r r , and so once u is determined, then
 ( x − x )i + ( y − y ) j + ( z − z )k 
 r
B
A
B
A
B
A
F = Fu = F  = F

2
2
2
r
(
)
(
)
(
x
−
x
+
y
−
y
+
z
−
z

B
A
B
A
B
A) 
Although we have represented F symbolically in Fig. 2–34, note that it
has units of force, unlike r, which has units of length.
r
B
u
A
y
2
x
Fig. 2–34
F
u
r
The force F acting along the rope can
be represented as a Cartesian vector by
establishing x, y, z axes and first forming a
position vector r along the length of the rope.
Then the corresponding unit vector u = r r
that defines the direction of both the rope
and the force can be determined. Finally, the
magnitude of the force is combined with its
direction, so that F = F u.
IMPORTANT POI N T S
• A position vector locates one point in space relative to another
point.
• The easiest way to formulate the components of a position
vector is to determine the distance and direction that one must
travel along the x, y, z directions—going from the tail to the
head of the vector.
• A force F acting in the direction of a position vector r can
be represented in Cartesian form if the unit vector u of the
position vector is determined and it is multiplied by the
magnitude of the force, i.e., F = Fu = F ( r r ) .
M02_HIBB0208_06_GE_C02.indd 79
Refer to the companion website for
Lecture Summary and Quiz videos.
12/21/23 3:32 PM
80
CHAPTER 2
EXAMPLE
FORCE VECTORS
2.9
z
The man shown in Fig. 2–35a pulls on the cord with a force of 350 N.
Represent this force acting on the support A as a Cartesian vector and
determine its direction.
A
SOLUTION
Force F is shown in Fig. 2–35b. The direction of this vector, u, is
determined from the position vector r, which extends from A to B.
Rather than using the coordinates of the end points of the cord, r can
be determined directly by noting in Fig. 2–35a that one must travel
from A { −6 k } m, then { −2 j } m, and finally { 3i } m to get to B. Thus,
2
7.5 m
2m
1.5 m
B
y
3m
r = { 3i − 2 j − 6 k } m
The magnitude of r, which represents the length of cord AB, is
r =
x
(a)
z¿
g
A
b
F 350 N
a
x¿
u
r
B
(b)
Fig. 2–35
y¿
(3 m) 2 + ( −2 m) 2 + ( −6 m) 2 = 7 m
Forming the unit vector that defines the direction and sense of both
r and F, we have
r
3
2
6
u = = i− j− k
r
7
7
7
Since F has a magnitude of 350 N and a direction specified by u, then


12  3312 8 22 8 2466 24



(350
FFF
=uu=F==u70
=lb70
lb i −ii −i− −j j−j −−j −kk  k 
(350
N)
FFF===
uF
 N)


28 7728 28 77 28 2877 28
 
ii 20
100
N
−−
=150
i−
150
100
jj−−−k300
300
N
==={ {30
i{ −30
j −20j60
Ans.
}60lbkkk}} lb
The coordinate direction angles are measured between the tails of
r (or F) and the positive axes of a localized coordinate system with
origin placed at A, Fig. 2–35b. From the components of the unit
vector:
 
312
33  
1112
−
1
−
1
−
−
αααα====cos
64.6°
cos
64.6°
cos
cos    ====64.6°
64.6°
Ans.
 28
77
728
 
12
−−
22 
 12
−
1
−
1
−
1
αα
64.6°
64.6°
Ans.
ββ === cos
cos
cos   === 107°
107°
 28
28
77 
 
12
−−
66 
 12
−
1
−
−
1
1
αα
64.6°
64.6°
γγ === cos
cos
cos   === 149°
149°
Ans.
28
28
77 
NOTE: These results make sense when compared with the angles
identified in Fig. 2–35b.
M02_HIBB0208_06_GE_C02.indd 80
12/21/23 3:32 PM
2.8
EXAMPLE
81
FORCE VECTOR DIRECTED ALONG A LINE
2.10
The roof is supported by two cables as shown in the photo. If the
cables exert forces FAB = 100 N and FAC = 120 N on the wall hook
at A as shown in Fig. 2–36a, determine the resultant force acting at A.
Express the result as a Cartesian vector.
SOLUTION
The resultant force FR is shown graphically in Fig. 2–36b. We can
express this force as a Cartesian vector by first formulating FAB and
FAC as Cartesian vectors and then adding their components. The
directions of FAB and FAC are specified by forming unit vectors u AB
and u AC along the cables. These unit vectors are obtained from the
associated position vectors rAB and rAC . With reference to Fig. 2–36a,
to go from A to B, we must travel { −4 k } m , and then { 4 i } m . Thus,
2
z
A
FAB 5 100 N
rAB = { 4 i − 4 k } m
rAB =
FAC 5 120 N
4m
(4 m) 2 + ( −4 m) 2 = 5.66 m
y
r 
4
 4

FAB = FAB  AB  = (100 N) 
i−
k
 5.66
5.66 
 rAB 
4m
B
FAB = { 70.7 i − 70.7 k } N
C
2m
To go from A to C, we must travel { −4 k } m, then { 2 j } m, and finally
{ 4 i } . Thus,
x
(a)
z
rAC = { 4 i + 2 j − 4 k } m
rAC =
A
(4 m) 2 + (2 m) 2 + ( −4 m) 2 = 6 m
FAB
r 
2
4 
4
FAC = FAC  AC  = (120 N)  i + j − k 

6
6
6 
 rAC 
FAC
rAC
rAB
= { 80 i + 40 j − 80 k } N
B
C
The resultant force is therefore
FR = FAB + FAC = { 70.7 i − 70.7 k } N + { 80 i + 40 j − 80 k } N
= { 151i + 40 j − 151k } N
y
FR
Ans.
x
(b)
Fig. 2–36
Refer to the companion website for
a self quiz of these Example problems.
M02_HIBB0208_06_GE_C02.indd 81
12/21/23 3:33 PM
82
CHAPTER 2
FORCE VECTORS
F UN DAMEN TAL PR O B L EM S
F2–19. Express the position vector rAB in Cartesian
vector form, then determine its magnitude and coordinate
direction angles.
z
z
B
2
A
F 5 900 N
3m
rAB
3m
3m
F2–22. Express the force as a Cartesian vector.
B
2m
4m
2m
y
y
7m
4m
x
A
2m
Prob. F2–22
x
Prob. F2–19
F2–20. Determine the length of the rod and the position
vector directed from A to B. What is the angle θ ?
F2–23. Determine the magnitude of the resultant force
at A.
z
z
A
2m
FB 5 840 N
B
FC 5 420 N
3m
4m
u
B
O
2m
4m
3m
x
A
x
6m
2m
C
y
y
Prob. F2–23
Prob. F2–20
F2–21. Express the force as a Cartesian vector.
F2–24. Determine the resultant force at A.
z
z
2m
2m
A
A
2m
x
FB 600 N
4m
3m
F
FC 490 N
630 N
4m
6m
y
M02_HIBB0208_06_GE_C02.indd 82
4m
3m
B
x
4m 2m
B
Prob. F2–21
C
4m
y
Prob. F2–24
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83
PROBLEMS
PROBLEMS
2–57. Determine the lengths of wires AD, BD, and CD.
The ring at D is midway between A and B.
2–59. The door is held opened by means of two chains. If
the tension in AB and CD is FA = 300 N and FC = 250 N,
respectively, express each of these forces in Cartesian vector
form.
z
C
z
2
1.5 m
2.5 m
C
FC = 250 N
A
B
D
A
2m
0.5 m
FA = 300 N
0.5 m
y
2m
D
30
1.5 m
0.5 m
1m
x
B
y
x
Prob. 2–57
Prob. 2–59
2–58. Determine the magnitude and coordinate direction
angles of the resultant force acting at A.
z
B
*2–60. Determine the length of the connecting rod AB
by first formulating a position vector from A to B and then
determining its magnitude.
y
C
B
2m
FB
0.5 m
3.5 m
600 N
300 mm
A
FC
450 N
O
1.5 m
1m
1.5 m
y
x
Prob. 2–58
M02_HIBB0208_06_GE_C02.indd 83
308
0.5 m
x
A
150 mm
Prob. 2–60
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84
CHAPTER 2
FORCE VECTORS
2–61. Determine the magnitude and coordinate direction
angles of the resultant force.
z
*2–64. At a given instant, the position of a plane at A and
a train at B are measured relative to a radar antenna at O.
Determine the distance d between A and B at this instant.
To solve the problem, formulate a position vector, directed
from A to B, and then determine its magnitude.
0.75 m
A
z
A
FAB 5 250 N
2
FAC 5 400 N
3m
5 km
y
408
2m
608
358
2m
B
C
O
1m
y
408
258
2 km
x
x
B
Prob. 2–61
2–62. The 8-m-long cable is anchored to the ground at A.
If x = 4 m and y = 2 m, determine the coordinate z to the
highest point of attachment along the column.
2–63. The 8-m-long cable is anchored to the ground at A.
If z = 5 m, determine the location +x, +y of the support
at A. Choose a value such that x = y.
z
Prob. 2–64
2–65. If FB = 560 N and FC = 700 N, determine the
magnitude and coordinate direction angles of the resultant
force acting on the flag pole.
2–66. If FB = 700 N , and FC = 560 N, determine the
magnitude and coordinate direction angles of the resultant
force acting on the flag pole.
z
B
6m
A
FB
FC
2m
z
B
3m
y
x
y
A
x
Probs. 2–62/63
M02_HIBB0208_06_GE_C02.indd 84
x
3m
2m
y
C
Probs. 2–65/66
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85
PROBLEMS
2–67. Determine the magnitude and coordinate direction
angles of the resultant force acting at point A on the post.
2–71. Determine the magnitude and coordinate direction
angles α , β , γ of the resultant force acting on the pole. Set
x = 4 m, y = 2 m.
z
FAC 150 N
C
4
5
2–70. Determine the position (x, y, 0) for fixing cable BA so
that the resultant force exerted on the pole is directed along
its axis, from B toward O. Also, what is the magnitude of the
resultant force?
A
z
FAB 200 N
2
B
3m
F2 5 250 N
3m
3
F1 5 350 N
y
O
B
4m
4m
F3 5 300 N
2m
2m
3m
x
Prob. 2–67
O
3m
y
x
x
A
y
*2–68. Represent each cable force as a Cartesian vector.
Probs. 2–70/71
2–69. Determine the magnitude and coordinate direction
angles of the resultant of the two forces acting at point A.
*2–72. Position vectors along the robotic arm from O to
B and B to A are rOB = { 100 i + 300 j + 400 k } mm and
rBA = { 350 i + 225 j − 640 k } mm, respectively. Determine
the distance from O to the grip at A.
2–73. If rOA = { 0.5 i + 4 j + 0.25 k } m and rOB = {0.3i +
2 j + 2 k} m, express rBA as a Cartesian vector.
z
z
2m
C
2m
E
B
B
FE 5 350 N
3m
FC 5 400 N
FB 5 400 N
x
D
2m
A
O
rBA
A
y
3m
y
x
Probs. 2–68/69
M02_HIBB0208_06_GE_C02.indd 85
Probs. 2–72/73
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86
CHAPTER 2
FORCE VECTORS
2.9
A
u
2
B
Fig. 2–37
DOT PRODUCT
Occasionally in statics one has to find the angle between two lines or
the components of a force parallel and perpendicular to a line. In two
dimensions, these problems can readily be solved by trigonometry since
the geometry is easy to visualize. In three dimensions, however, this is
often difficult, and consequently vector methods should be employed
for the solution. The dot product, which defines a particular method for
“multiplying” two vectors, can be used to solve the above-mentioned
problems.
The dot product of vectors A and B, written A ⋅ B and read “A dot B,”
is defined as the product of the magnitudes of A and B and the cosine
of the angle θ between their tails, Fig. 2–37. Expressed in equation form,
A ⋅ B = AB cos θ
(2–12)
where 0° ≤ θ ≤ 180°. The dot product is often referred to as the scalar
product of vectors since the result is a scalar and not a vector.
Laws of Operation.
The following three laws of operation apply.
1. Commutative law: A ⋅ B = B ⋅ A
2. Multiplication by a scalar: a ( A ⋅ B) = ( aA ) ⋅ B = A ⋅ ( aB)
3. Distributive law: A ⋅ (B + D) = ( A ⋅ B) + ( A ⋅ D)
Cartesian Vector Formulation. If we apply Eq. 2–12, we can
find the dot product for any two Cartesian unit vectors. For example,
i ⋅ i = (1)(1) cos 0° = 1 and i ⋅ j = (1)(1) cos 90° = 0 . If we want to find
the dot product of two general vectors A and B that are expressed
in Cartesian vector form, then we have
A ⋅ B = ( Ax i + Ay j + Az k ) ⋅ ( Bx i + B y j + Bz k )
= Ax Bx ( i ⋅ i ) + Ax B y ( i ⋅ j) + Ax Bz ( i ⋅ k )
+ Ay Bx ( j ⋅ i ) + Ay B y ( j ⋅ j) + Ay Bz ( j ⋅ k )
+ Az Bx ( k ⋅ i ) + Az B y ( k ⋅ j) + Az Bz ( k ⋅ k )
Carrying out the dot-product operations, the final result becomes
A ⋅ B = Ax Bx + Ay B y + Az Bz
(2–13)
Thus, to determine the dot product of two Cartesian vectors, multiply their
corresponding x, y, z components and sum these products algebraically.
The result will be either a positive or negative scalar, or it could be zero.
M02_HIBB0208_06_GE_C02.indd 86
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2.9
Applications. The dot product has two important applications.
• The angle formed between two vectors or intersecting lines. The
A
angle θ between the tails of vectors A and B in Fig. 2–37 can be
determined fromEq. 2–12 and written as
A
θ = cos −1
⋅B 


AB 
87
DOT PRODUCT
u
B
Fig. 2–37 (Repeated)
0° ≤ θ ≤ 180°
2
Here A ⋅ B is found from Eq. 2–13. As a special case, if A ⋅ B = 0,
then θ = cos −1 0 = 90° so that A will be perpendicular to B.
ub
• The components of a vector parallel and perpendicular to a line.
The component of vector A parallel to or collinear with the line aa in
Fig. 2–38 is defined by Aa = A cos θ . This component is sometimes
referred to as the projection of A onto the line, since a right angle
is formed in the construction. If the direction of the line is specified
by the unit vector u a, then since ua = 1 , we can determine the
magnitude of A a directly from the dot product (Eq. 2–12); i.e.,
Aa = A ⋅ u a = A cos θ
u
ur
The angle θ between the rope and the beam
Hence, the scalar projection of A along a line is determined from the can be determined by formulating unit
dot product of A and the unit vector u a which defines the direction of vectors along the beam and rope and then
using the dot product u b ⋅ u r = (1)(1) cos θ .
the line. Notice that if this result is positive, then A a has a directional
sense which is the same as u a , whereas if Aa is a negative scalar,
then A a has the opposite sense of direction to u a .
The component A a represented as a vector is therefore
A a = Aa u a
The perpendicular component of A can also be obtained, Fig. 2–38.
Since A = A a + A ⊥ , then A ⊥ = A − A a . There are two possible
ways of obtaining A⊥ . One way would be to determine θ from the
dot product, θ = cos −1 ( A ⋅ u A A), then A⊥ = A sin θ . Alternatively,
if Aa is known, then by the Pythagorean theorem we can also write
A⊥ = A 2 − Aa 2 .
ub
Fb
F
A'
a
A
u
Aa 5 A cos u ua
Fig. 2–38
M02_HIBB0208_06_GE_C02.indd 87
ua
a
The projection of the cable force F along the
beam can be determined by first finding the
unit vector u b that defines this direction. Then
apply the dot product, Fb = F ⋅ ub.
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88
CHAPTER 2
FORCE VECTORS
I MPO RTA N T PO I N T S
• The dot product is used to determine the angle between two
vectors or the projection of a vector in a specified direction.
• If vectors A and B are expressed in Cartesian vector form, the
dot product is determined by multiplying the respective x, y, z
scalar components and algebraically adding the results, i.e.,
A ⋅ B = Ax Bx + Ay B y + Az Bz .
2
• From the definition of the dot product, the angle formed
between the tails of vectors A and B is θ = cos −1 ( A ⋅ B AB).
• The magnitude of the projection of vector A along a line aa
Refer to the companion website for
Lecture Summary and Quiz videos.
EXAMPLE
whose direction is specified by u a is determined from the dot
product Aa = A ⋅ u a .
2.11
Determine the magnitudes of the projections of the force F in Fig. 2–39a
onto the u and v axes.
v
v
F 5 100 N
F 5 100 N
(Fv )proj
Fv
158
458
u
Fu
(Fu)proj
u
Components of F
(b)
Projections of F
(a)
Fig. 2–39
SOLUTION
Projections of Force. The graphical representation of the projections
is shown in Fig. 2–39a. From this figure, the magnitudes of the projections
of F onto the u and v axes can be obtained by trigonometry:
( Fu ) proj = (100 N)cos 45° = 70.7 N
Ans.
( Fv ) proj = (100 N)cos15° = 96.6 N
Ans.
NOTE: These projections are not equal to the magnitudes of the
components of force F along the u and v axes found from the
parallelogram law, Fig. 2–39b. They would only be equal if the u and v
axes were perpendicular to one another.
M02_HIBB0208_06_GE_C02.indd 88
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2.9
EXAMPLE
DOT PRODUCT
89
2.12
The frame shown in Fig. 2–40a is subjected to a horizontal force
F = { 300 j } N. Determine the magnitudes of the components of this
force parallel and perpendicular to member AB.
z
z
FAB
B
F 5 {300 j} N
3m
A
B
uB
F
A
y
2
F
y
2m
6m
x
x
(b)
(a)
Fig. 2–40
SOLUTION
The magnitude of the projected component of F along AB is equal
to the dot product of F and the unit vector u B , which defines the
direction of AB, Fig. 2–40b. Since
uB =
then
rB
=
rB
2 i + 6 j + 3k
(2) 2 + (6) 2 + (3) 2
= 0.286 i + 0.857 j + 0.429 k
FAB = F cos θ = F ⋅ u B = (300 j) ⋅ (0.286 i + 0.857 j + 0.429 k )
= (0) (0.286) + (300) (0.857) + (0) (0.429)
= 257.1 N
Ans.
Since the result is a positive scalar, FAB has the same sense of direction
as u B , Fig. 2–40b.
Expressing FAB in Cartesian vector form, we have
FAB = FAB u B = (257.1 N)(0.286 i + 0.857 j + 0.429 k )
= { 73.5 i + 220 j + 110 k } N
Ans.
The perpendicular component, Fig. 2–40b, is therefore
F⊥ = F − FAB = 300 j − (73.5 i + 220 j + 110 k )
= { −73.5 i + 79.6 j − 110 k } N
Its magnitude can be determined either from this vector or by using
the Pythagorean theorem, Fig. 2–40b:
F⊥ =
M02_HIBB0208_06_GE_C02.indd 89
2
F 2 − FAB
=
(300 N) 2 − (257.1 N) 2 = 155 N Ans.
12/21/23 3:34 PM