Engineering Mechanics 9782759829026

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Current Natural Sciences

Ping YI, Jun LIU and Feng JIANG

Engineering Mechanics

Printed in France

EDP Sciences – ISBN(print): 978-2-7598-2901-9 – ISBN(ebook): 978-2-7598-2902-6 DOI: 10.1051/978-2-7598-2901-9 All rights relative to translation, adaptation and reproduction by any means whatsoever are reserved, worldwide. In accordance with the terms of paragraphs 2 and 3 of Article 41 of the French Act dated March 11, 1957, “copies or reproductions reserved strictly for private use and not intended for collective use” and, on the other hand, analyses and short quotations for example or illustrative purposes, are allowed. Otherwise, “any representation or reproduction – whether in full or in part – without the consent of the author or of his successors or assigns, is unlawful” (Article 40, paragraph 1). Any representation or reproduction, by any means whatsoever, will therefore be deemed an infringement of copyright punishable under Articles 425 and following of the French Penal Code. The printed edition is not for sale in Chinese mainland. Customers in Chinese mainland please order the print book from Science Press. ISBN of the China edition: Science Press 978-7-03-073762-5 Ó Science Press, EDP Sciences, 2022

Foreword

This book presents and illustrates the concepts, principles and analytical problem-solving methodologies in engineering mechanics. It is divided into ten chapters. Chapter 1 gives an introduction to engineering mechanics and its fundamental concepts. Since the international system of units (SI units) receives worldwide recognition, SI units are also introduced in chapter 1 and used throughout the book. As a convenient tool, vector algebra is then presented in chapter 2. Chapter 3 discusses the simplification of a system of forces and couples or a distributed loading. The conditions and equations of equilibrium for a particle and a rigid body are developed and applied, especially to equilibrium problems of a system of rigid bodies, such as frames, machines and trusses, in chapter 4. Applications to equilibrium problems involving frictional forces are discussed in chapter 5. The kinematics of particles is treated in chapter 6, followed by a discussion of planar kinematics of rigid bodies in chapter 7. Kinetics are discussed in chapter 8 (equations of motion), chapter 9 (work and energy), and chapter 10 (impulse and momentum). In each of these three chapters, applications to only particles’ motion are considered first and then more difficult and practical applications associated with the motion of rigid bodies are addressed. Each chapter begins with the objectives of the chapter. Then it consists of theory sections, example problems and homework problems. The example problems serve double purposes of amplifying the theories and demonstrating the neat work that students should cultivate in their own solutions, and are presented in a logical and orderly manner. Sometimes multiple solutions are provided for a problem, trying to inspire students to broaden their thinking and better understand various principles and methods. For some examples, there is a “RETHINK” section, where in-depth discussions or enlightening questions are presented. Most of the problems are of practical nature and should appeal to engineering students. They are meticulously designed to help students master the principles of mechanics and their applications. DOI: 10.1051/978-2-7598-2901-9.c901 Ó Science Press, EDP Sciences, 2022

Foreword

IV

This book is financially supported by Dalian University of Technology and DUT-BSU Joint Institute, which are gratefully acknowledged. During the preparation of this book, many people have helped in its development. First, the authors are particularly grateful to Dr. Dzianis Marmysh in Belarusian State University and Dr. Ming Li in Dalian University of Technology for their specific help. Also, many thanks to our graduate students who have taken the time to prepare and check the solutions to some of the homework problems of the book. They are: Baopei Cai Xinyue Ma Xiaoqian Wang Na Li

Qingkang Wang Dongchi Xie Yuhan Zhang Zhixuan Li

Linlang Feng Weining Yang Huiqing Liu Juncheng Li

Yongke Li Wenjian Nie Xinshuai Guo Yan Xing

Zuo Zhu Hongzhi Wang Shaopeng Bai Chengcheng Fan

The authors also thank the staff at EDP Sciences and Science Press for their great support and dedication during the preparation of this book. Lastly, the authors wish to thank many of our colleagues and students for their helpful comments and suggestions. Since this list is too long, it is hoped that they will accept this anonymous recognition. If you have any comments, problems, or suggestions related to the book, please feel free to contact us. Your feedback will surely help us complement and improve it. Ping Yi, Jun Liu, Feng Jiang Dalian University of Technology June, 2021

Contents Foreword . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

III

CHAPTER 1 . . . . . . . .

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1 1 2 2 3 3 4 6

Vectors and Vector Operations . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1 Vectors and the Parallelogram Law . . . . . . . . . . . . . . . . . . 2.2 Cartesian Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2.1 Cartesian Vector Representation . . . . . . . . . . . . . . 2.2.2 Addition and Subtraction of Cartesian Vectors . . . . 2.3 Vector Directed Along a Line . . . . . . . . . . . . . . . . . . . . . . 2.4 Dot Product and Cross Product . . . . . . . . . . . . . . . . . . . . 2.4.1 Cartesian Vector Formulation of the Dot Product . 2.4.2 Applications of the Dot Product . . . . . . . . . . . . . . . 2.4.3 Cartesian Vector Formulation of the Cross Product

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37 37 38 43 43 47 52 52 54

Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1 Mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2 Fundamental Concepts and Principles . . . . . . . . . 1.2.1 Fundamental Concepts . . . . . . . . . . . . . . . 1.2.2 Newton’s Laws of Motion . . . . . . . . . . . . . 1.2.3 Newton’s Law of Gravitational Attraction 1.3 Units of Measurement . . . . . . . . . . . . . . . . . . . . . 1.4 Procedure of Problem Solving . . . . . . . . . . . . . . .

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CHAPTER 2

CHAPTER 3 Simplification of Force Systems . . . . . . . . . . . . . . 3.1 Moment of a Force about a Point . . . . . . . 3.1.1 Vector Formulation . . . . . . . . . . . . 3.1.2 Principle of Moments . . . . . . . . . . . 3.1.3 Scalar Formulation . . . . . . . . . . . . . 3.2 Moment of a Force about a Specified Axis 3.3 Moment of a Couple . . . . . . . . . . . . . . . . . 3.3.1 Vector Formulation . . . . . . . . . . . . 3.3.2 Scalar Formulation . . . . . . . . . . . . .

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Contents

VI

3.4

3.5

Simplification of a System of Forces and Couples . . . . . . . . . . . . . . 3.4.1 Equivalent System . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.4.2 Reduction to One Force and One Couple . . . . . . . . . . . . . . . 3.4.3 Further Simplification . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Application of Simplification of Parallel Forces . . . . . . . . . . . . . . . . 3.5.1 Center of Gravity, Center of Mass, and Centroid of a Body . 3.5.2 Center of Gravity, Center of Mass, and Centroid of Composite Bodies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.5.3 Reduction in Distributed Loading . . . . . . . . . . . . . . . . . . . .

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58 58 59 63 70 70

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CHAPTER 4 Equilibrium of Rigid Bodies . . . . . . . . . . . . . . . . . . . . . . . 4.1 Conditions for Rigid-Body Equilibrium . . . . . . . . . . 4.2 Free-Body Diagrams . . . . . . . . . . . . . . . . . . . . . . . . 4.2.1 Support Reactions . . . . . . . . . . . . . . . . . . . . 4.2.2 Procedure of Drawing a Free-Body Diagram . 4.3 Equilibrium Problems of a Rigid Body . . . . . . . . . . 4.4 Two- and Three-Force Members . . . . . . . . . . . . . . . 4.5 Constrains and Statical Determinacy . . . . . . . . . . . 4.6 Equilibrium Problems of Structures . . . . . . . . . . . . . 4.7 Simple Trusses . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.7.1 The Method of Joints . . . . . . . . . . . . . . . . . . 4.7.2 Zero-Force Members . . . . . . . . . . . . . . . . . . . 4.7.3 The Method of Sections . . . . . . . . . . . . . . . .

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105 105 109 109 114 120 127 129 132 142 143 148 149

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171 171 172 174 175 182

Kinematics of Particles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.1 General Curvilinear Motion . . . . . . . . . . . . . . . . . . . . . . . 6.2 Curvilinear Motion: Rectangular Components . . . . . . . . . 6.3 Curvilinear Motion: Tangential and Normal Components 6.3.1 The t-n-b Coordinate System . . . . . . . . . . . . . . . . 6.3.2 Velocity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.3.3 Acceleration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.4 Absolute Dependent Motion Analysis of Particles . . . . . . 6.5 Relative-Motion Analysis Using Translating Axes . . . . . .

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193 194 196 203 204 205 205 212 218

CHAPTER 5 Friction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1 Characteristics of Friction . . . . . . . . . . . . . 5.1.1 Laws of Sliding Friction . . . . . . . . . 5.1.2 Angles of Friction and Self-Locking 5.2 Problems Involving Sliding Friction . . . . . . 5.3 Rolling Resistance . . . . . . . . . . . . . . . . . . .

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CHAPTER 6

Contents

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CHAPTER 7 . . . . . . . . .

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233 235 237 237 239 249 251 253 265

Kinetics: Equations of Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.1 Newton’s Second Law of Motion . . . . . . . . . . . . . . . . . . . . . . 8.2 Equation of Motion for a Particle . . . . . . . . . . . . . . . . . . . . . 8.3 Equation of Motion for a System of Particles . . . . . . . . . . . . 8.4 Mass Moment of Inertia . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.4.1 Parallel-Axis Theorem and Radius of Gyration . . . . . . 8.4.2 Mass Moment of Inertia of Composite Bodies . . . . . . . 8.5 Planar Kinetic Equations of Motion . . . . . . . . . . . . . . . . . . . 8.5.1 Equations of Motion for Translation . . . . . . . . . . . . . . 8.5.2 Equations of Motion for Rotation about a Fixed Axis 8.5.3 Equations of Motion for General Plane Motion . . . . . .

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289 289 290 303 305 309 311 314 319 324 329

Kinetics: Work and Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.1 Work and Power . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.1.1 Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.1.2 Power . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.2 Kinetic Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.2.1 Kinetic Energy of a Particle . . . . . . . . . . . . . . . . . . . . . 9.2.2 Kinetic Energy of a System of Particles . . . . . . . . . . . . 9.2.3 Kinetic Energy of a Rigid Body in Planar Motion . . . . 9.3 Principle of Work and Energy . . . . . . . . . . . . . . . . . . . . . . . . . 9.3.1 Principle of Work and Energy for a Particle . . . . . . . . . 9.3.2 Principle of Work and Energy for a System of Particles 9.4 Conservative Forces and Conservation of Energy . . . . . . . . . . . 9.4.1 Conservative Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.4.2 Potential Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.4.3 Principle of Conservation of Energy . . . . . . . . . . . . . . .

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351 351 351 362 364 364 365 366 370 370 375 385 385 387 388

Planar Kinematics of Rigid Bodies . . . . . . . . . . . . . . . . . . . . . . . . 7.1 Translation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.2 Rotation about a Fixed Axis . . . . . . . . . . . . . . . . . . . . . . . . 7.2.1 Angular Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.2.2 Motion of a Particle on a Rotating Body . . . . . . . . . 7.3 Absolute Dependent Motion Analysis of Bodies . . . . . . . . . . 7.4 General Plane Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.4.1 Relative-Motion Analysis: Velocity and Acceleration . 7.4.2 Instantaneous Center of Rotation . . . . . . . . . . . . . . . CHAPTER 8

CHAPTER 9

Contents

VIII

CHAPTER 10 Kinetics: Impulse and Momentum . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.1 Principle of Linear Impulse and Momentum for a Particle . . . . 10.1.1 Linear Momentum and Linear Impulse . . . . . . . . . . . . 10.1.2 Principle of Linear Impulse and Momentum . . . . . . . . 10.2 Principle of Angular Impulse and Momentum for a Particle . . . 10.2.1 Angular Momentum . . . . . . . . . . . . . . . . . . . . . . . . . . 10.2.2 Principle of Angular Impulse and Momentum . . . . . . . 10.3 Principle of Impulse and Momentum for a System of Particles . 10.3.1 Linear Momentum of a System of Particles . . . . . . . . . 10.3.2 Angular Momentum of a System of Particles . . . . . . . . 10.3.3 Angular Momentum of Rigid Bodies in Planar Motion 10.3.4 Principle of Impulse and Momentum for Rigid Bodies in Planar Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.4 Principle of Conservation of Momentum . . . . . . . . . . . . . . . . .

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Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 455 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 465

Chapter 1 Introduction Objectives
Get an idea of engineering mechanics and the fundamental concepts in mechanics.
Grasp the importance of Newton’s laws of motion and gravitation in engineering mechanics.
Learn the SI system of units and the rules of using a prefix.
Understand the general procedure of problem solving in engineering mechanics.

1.1

Mechanics

Mechanics is a branch of the physical science that describes the state of rest or motion of bodies under the action of forces. It consists of rigid-body mechanics, deformable-body mechanics, and fluid mechanics. This book “Engineering Mechanics” treats only rigid-body mechanics. In engineering, we often need to analyze bridges, cranes, buildings, aerocraft, vehicles or components of mechanical systems to determine whether they are in equilibrium or perform accelerated motion. Of course, these real structures or mechanical components are always deformable under load to a certain extent. However, they can be idealized as rigid bodies because their deformations are generally small and do not appreciably affect their state of rest or motion. So rigid-body mechanics forms a suitable basis for the design and analysis of many engineering structures and mechanical devices. It also provides necessary backgrounds for the study of deformable-body mechanics and fluid mechanics. Engineering mechanics is usually divided into statics, kinematics and kinetics. Statics deals with forces acting on particles or bodies that are at rest or moving with a constant velocity; kinematics deals with only the geometric aspects of the motion without referring to forces as a cause, whereas kinetics relates the accelerated motion and the forces involved. Thus, kinetics is much more difficult and the content in this book is arranged from the easy to the difficult according to cognitive laws. DOI: 10.1051/978-2-7598-2901-9.c001 © Science Press, EDP Sciences, 2022

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Mechanical insight was gradually gained from the application of tools and devices in the living and working experiences. Several cornerstones on statics were laid in “Mohist Canon” (388 B.C.) and the works of Aristotle (384–322 B.C.) and Archimedes (287–212 B.C.). From well-designed experiments, Galileo Galilei (1564–1642) proposed some important principles of dynamics. Then the contributions of Johannes Kepler (1571–1630) and Christian Huygens (1629–1695) led to the formulation of the laws of motion by Isaac Newton (1643–1727). These laws and principles were later modified by d’Alembert (1717–1783), Lagrange (1736–1813), and Hamilton (1805–1865). The validity of their work was also challenged when Einstein (1879–1955) formulated the theory of relativity (1905). Engineering mechanics is based on Newton’s three laws of motion and it is also considered as Newtonian mechanics. Although its limitations have long been recognized, Newtonian mechanics is still the basis of today’s engineering science and continues to enlarge its field of application.

1.2 1.2.1

Fundamental Concepts and Principles Fundamental Concepts

There are four fundamental quantities used throughout mechanics, i.e., length, time, mass and force. Length. Length is a description of the size or measurement of something from one end to the other. Time. Time is measured in seconds, minutes, hours, etc., and it is used to describe a succession of events. Mass. Mass is a property of matter that does not change with the location. It provides a quantitative measure of the matter’s resistance to a change in velocity. Force. A force is the mechanical action of one body on another. It can be exerted by direct contact, like a push or pull a person acts on a cart, or at a distance, like a gravitational or magnetic force. A force is characterized by its magnitude, direction, and point of application and it is represented by a vector (§ 2.1). In engineering mechanics, we will study the conditions of rest or motion of particles and rigid bodies in terms of the four above-mentioned basic quantities. Particles and rigid bodies are idealized models in engineering mechanics in order to simplify the application of its theories to practice. Particle. A particle has a mass but negligible size. When the size and shape of an object has insignificant influence on its state of rest or motion, this object can be idealized as a particle.

Introduction

3

Rigid body. A rigid body consists of a large number of particles occupying fixed positions with respect to one another before and after applying a force. It means a rigid body does not deform under load.

1.2.2

Newton’s Laws of Motion

Engineering mechanics is formulated on the basis of Newton’s three laws of motion. They apply to the motion of an object whose velocity, measured from a nonaccelerating reference frame, is far less than the speed of light. First Law. A particle will remain at rest (if originally at rest) or move with constant velocity in a straight line (if originally in motion) when the resultant force acting on the particle is zero. Second Law. When a particle is acted upon by a nonzero force F, it experiences an acceleration proportional to the magnitude of the force and in the same direction as the force. This law can be formulated as F ¼ ma

ð1:1Þ

where m and a represent the mass and acceleration of the particle, respectively. Third Law. The forces of action and reaction between two objects have the same magnitude, opposite sense, and same line of action, as shown in figure 1.1.

FIG. 1.1 – Action and reaction forces between two objects.

1.2.3

Newton’s Law of Gravitational Attraction

This law states that two particles of mass M and m are mutually attracted by equal and opposite forces F and −F, whose magnitude is given by F ¼f

Mm r2

ð1:2Þ

where r is the distance between the two particles and f is the universal constant of gravitation. From experiments, f = 66.73 × 10−12 m3/(kgs2) [1, 2]. According to Newton’s law of gravitational attraction, any two objects have mutual attractive forces acting between them, as shown in figure 1.2. This law

Engineering Mechanics

4

introduces the idea of an interaction exerted at a distance and it is in accordance with Newton’s third law: the action and reaction forces are equal, opposite, and collinear.

FIG. 1.2 – Illustration of Newton’s law of gravitational attraction.

For an object located on or near the surface of the Earth, the only gravitational force having considerable magnitude is that between the Earth and the object. This force exerted by the Earth on the object is called the weight, W, of the object. Setting M equal to the mass of the Earth, and r equal to the distance between the center of the Earth and the object, and introducing the term g¼f

M r2

ð1:3Þ

we can express the weight of an object of mass m as W ¼ mg

ð1:4Þ

Comparing equations (1.1) and (1.4), g is called the acceleration of gravity. Since g depends on r, the weight of an object is not an absolute quantity and its magnitude varies with the position of the object. However, for most engineering problems where the object remains located on or very near the Earth’s surface, it is sufficiently accurate to assume that g = 9.81 m/s2 [3].

1.3

Units of Measurement

The four fundamental quantities—length, time, mass and force—in mechanics are related by equation (1.1). Therefore, the units used to measure them cannot be chosen independently. Three of the units can be selected arbitrarily and are called base units. The fourth unit, however, is derived from equation (1.1) and referred to as a derived unit. The units must be chosen in this way to form a consistent system of units. The international system of units (SI Units) is such a consistent system of units receiving worldwide recognition. This system of units will be used throughout the book. In the SI system, the base units are the units of length, time and mass. It specifies length in meters (m), time in seconds (s), and mass in kilograms (kg). The unit of force, called newton (N), is a derived unit. From F = ma, a force of 1 N gives an acceleration of 1 m/s2 to a mass of 1 kg (1 N = 1 kgm/s2).

Introduction

5

When a numerical quantity is very large or very small, a prefix shown in table 1.1 is generally used along with the fundamental SI units. Each prefix represents a multiple or submultiple of a unit which successively moves the decimal point of the numerical quantity three places to the right or to the left. For example, 8.4 Mm = 8 400 km = 8 400 000 m and 2 ms = 0.002 s. It should be noted that the kilogram is the only fundamental unit that is defined with a prefix [1]. TAB. 1.1 – Prefixes. Multiplication factor 1012 109 106 103 10−3 10−6 10−9 10−12

Prefix tera giga mega kilo milli micro nano pico

Symbol T G M k m μ n p

Besides the submultiples listed in table 1.1, deci (d, 10–1) and centi (c, 10–2) are also often used for the meter, namely, decimeter (dm) and centimeter (cm). From definition, we have 1 dm ¼ 101 m;

1 cm ¼ 102 m

Accordingly, the submultiples of the unit of area are 1 dm2 ¼ ð1 dmÞ2 ¼ ð101 mÞ2 ¼ 102 m2 ;

1 cm2 ¼ ð1 cmÞ2 ¼ 102 m


2

¼ 104 m2

The corresponding submultiples of the unit of volume are
3
3 1 dm3 ¼ ð1 dmÞ3 ¼ 101 m ¼ 103 m3 ; 1 cm3 ¼ ð1 cmÞ3 ¼ 102 m ¼ 106 m3 Although the SI unit of time is second (s), the minute (min), and hour (h) are also permitted for use. It is known that 1 min = 60 s and 1 h = 60 min = 3600 s. The SI unit of angle is radian (rad). However, the degree (°) and revolution (rev) are also used and we have 1 rev = 2π rad = 360°. It is worth noting that for a composite unit defined by multiplication of several fundamental units, a “” should be used to separate the fundamental units to avoid confusion with prefix notation. For example, ms is meter-second, whereas ms represents milli-second. At the same time, for any combination of fundamental units, a prefix should generally be used for the first fundamental unit, not the following one. For example, do not use Nmm and kg/mm, but rather mNm and Mg/m. The commonly used SI units in engineering mechanics are summarized in table 1.2.

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6

TAB. 1.2 – SI units commonly used in engineering mechanics. Quantity Length Time Mass Force Moment of a force Area Volume Density Velocity Acceleration Angle Angular velocity Angular acceleration Work Energy Power Pressure Frequency Linear impulse Linear momentum Angular impulse Angular momentum

1.4

Unit meter second kilogram Newton Newton-meter square meter cubic meter kilogram per cubic meter meter per second meter per second squared radian radian per second radian per second squared Joule Joule Watt Pascal Hertz Newton-second kilogram-meter per second Newton-meter-second kilogram-meter squared per second

Symbol m s kg N

Formula

kgm/s2 Nm m2 m3 kg/m3 m/s m/s2

rad

J J W Pa Hz

rad/s rad/s2 Nm Nm J/s N/m2 s–1 Ns = kgm/s kgm/s = Ns Nms = kgm2/s kgm2/s = Nms

Procedure of Problem Solving

When solving a problem in engineering mechanics, it is important to first understand and correlate the problem with the relevant principle or principles. A neat diagram showing all quantities involved would be very helpful to formulate the problem. For example, always draw a free-body diagram to account for all the forces acting on the object studied before solving an equilibrium problem. When writing equations for a problem, you should pay attention to dimensional homogeneity and use a consistent set of units. An equation can be partially checked by verifying that the equation remains dimensionally homogeneous. For example, if an equation is expressed as A ¼ B þ C  D, each term (A, B, C and D) must have the same units to maintain dimensional homogeneity. For numerical calculations, the accuracy of the solution of a problem cannot be better than the accuracy of the problem data. The accuracy of a numerical quantity is specified by the number of significant figures it contains. Since the data for lengths, masses, and other measurements in engineering problems are often reported with an accuracy not higher than 0.2% [3], a calculated result should be “rounded off” to an appropriate number of significant figures. In this book, we generally

Introduction

7

perform intermediate calculations to four significant figures and then report the final answer to three significant figures. When rounding off a calculated result to n significant figures, the following rules apply [1]:
If the n + 1 digit is smaller than 5, round down. For example, 3.5284 rounded off to n = 4 significant figures should be 3.528.
If the n + 1 digit is greater than 5, round up. For example, 3.5284 rounded off to n = 3 significant figures becomes 3.53.
If the n + 1 digit is equal to 5 with any nonzero digit following it, round up. For example, 3.52503 rounded off to n = 3 significant figures should be 3.53.
If the n + 1 digit is equal to 5 without any nonzero digit following it, round off the nth digit to an even number. For example, 3.525 and 3.53500 rounded off to n = 3 significant figures should be 3.52 and 3.54, respectively. After the solution, do not forget to study carefully the answer with technical judgment and common sense. Example 1.1. An astronaut has a mass of 70 kg on the Earth. Find (1) his weight on the Earth, (2) his mass on the Moon, and (3) his weight on the Moon, where the acceleration of gravity is gm ¼ 1:63 m=s2 . Express the results to three significant figures. Solution: (1) W ¼ mg ¼ 70  9:81 ¼ 686:7 ¼ 687 N (2) mm ¼ m ¼ 70 kg (3) Wm ¼ mm gm ¼ 70  1:63 ¼ 114:1 ¼ 114 N Example 1.2. Two spheres located near the Earth’s surface are touching each other. Each sphere has a mass of 10 kg and a radius of 10 cm. Determine the force of gravitational attraction acting between them and compare this result with their weight. Solution: The force of gravitational attraction between the two spheres is F ¼f

Mm 10  10 ¼ 66:73  1012 ¼ 1:668 25  109 ¼ 1:67 nN r2 ð2  10  102 Þ2

The weight of each sphere is W ¼ mg ¼ 10  9:81 ¼ 98:1 N It can be seen that for these two spheres on the Earth, the gravitational force between them is very small, much smaller than that between the Earth and the object. This is the reason that in engineering mechanics, the only gravitational force we should consider is that due to the Earth, i.e., the weight.

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Example 1.3. The radius of a steel ball is 8 mm and steel has a density of 7.85 g/cm3. What is the weight of the steel ball? Remember to use at least 4 significant figures for intermediate calculations and 3 for the final answer. Solution: First, represent the unit of the density in the correct SI form.     g  1 kg 100 cm 3 3 q ¼ 7:85 g/cm ¼ 7:85 3 ¼ 7:85  103 kg/m3 ¼ 7:85 Mg/m3 cm 1000 g 1m This is because for a composite unit defined by dividing a fundamental unit by another fundamental unit, a prefix should only be used in the numerator, not in the denominator. Then determine the volume, the mass, and finally the weight of the steel ball.
3 4 4 V ¼ pr 3 ¼  3.1416  8  103 ¼ 2144:665 6  109 ¼ 2:145  106 m3 3 3 m ¼ qV ¼ 7:85  103  2:145  106 ¼ 16:83825  103 ¼ 16:84  103 kg W ¼ mg ¼ 16:84  103  9:81 ¼ 165:2004  103 ¼ 0:165 N Example 1.4. Represent the following composite units in the correct SI form using appropriate prefixes: (1) MN/ms, (2) g/mm3, and (3) mN/(Mgμs). Solution: MN 106 N ¼ ¼ 109 N/s = GN/s ms 103 s g 103 kg ¼ ¼106 kg/m3 = Gg/m3 (2) mm3 ð103 mÞ3 mN 103 N ¼ 3 = N/ðkgsÞ (3) Mgls 10  106 ðkgsÞ (1)

Example 1.5. A car is travelling at 110 km/h. Determine its speed in m/s.    110 km 6 1000 m 16h 110 000 m Solution: 110 km/h ¼ ¼ 30:6 m/s ¼ 6h 1 km 6 3600 s 3600 s Example 1.6. The speed of rotation can be expressed by n (revolution per minute, rev/min) or angular velocity ω (radians per second, rad/s). If n = 20 rev/min, determine ω.     rev 1 min 2p rad Solution: 20 rev/min ¼ 20 ¼ 2:09 rad/s min 60 s 1 rev

Introduction

9

PROBLEMS 1.1 Express the following numbers to three significant figures: (1) 36.87, (2) 255.38, (3) 44.6503, (4) 126.5, (5) 127.5. 1.2 Represent the following quantities using an appropriate prefix and round off the answers to three significant figures: (1) 0.000 839 2 kg, (2) 78 670 N, (3) 0.035 67 km. 1.3 Determine the mass of a body on the Earth that has a weight of (1) 60 mN, (2) 25 N, (3) 120 kN, (4) 700 MN. Report the result to three significant figures and use an appropriate prefix if necessary. 1.4 Determine the weight of a body on the Earth that has a mass of (1) 0.85 g, (2) 710 kg, and (c) 366 Mg. Report the result to three significant figures and use an appropriate prefix if necessary. 1.5 A train is traveling at 340 km/h. Determine its speed in meters per second. Report the result to three significant figures. 1.6 A train is traveling at 15 m/s. Determine its speed in kilometers per hour. Report the result to three significant figures. 1.7 A rocket has a mass of 4.2 Gg on the Earth. Determine (1) its weight on the Earth, (2) its mass on the Moon, and (3) its weight on the Moon, where the acceleration of gravity is gm ¼ 1:63 m=s2 . Report the result to three significant figures and use an appropriate prefix if necessary. 1.8 A concrete column has a radius of 180 mm and a length of 3 m. If the density of concrete is 2.45 g/cm3, determine the weight of the column. Remember to use at least 4 significant figures for intermediate calculations and 3 for the final answer. 1.9 Represent the following combinations of units in the correct SI form using appropriate prefixes: (1) Nms, (2) g/cm3, (3) N/mm2 and (4) kNμm. 1.10 If a spring stretches 35 mm under a load of 150 N, express the spring constant k in the correct SI form. 1.11 If a body is rotating with a speed of rotation n = 15 rev/min, determine its angular velocity ω (rad/s). 1.12 If a body is rotating with an angular velocity ω = 6 rad/s, determine its speed of rotation n (rev/min).

Chapter 2 Vectors and Vector Operations Objectives
Add vectors and resolve them into components using the parallelogram law.
Express a vector in Cartesian vector form and perform the addition and subtraction of Cartesian vectors.
Formulate a Cartesian force vector directed along a line using the position vector.
Comprehend the dot product and its application.
Understand the cross product.

2.1

Vectors and the Parallelogram Law

Physical quantities in mechanics are generally expressed by means of scalars and vectors. Scalar. A scalar, such as mass and length, only has a magnitude and is represented by a plain number. Scalars are indicated by letters in italic type, for example, a mass m and a length l. The addition of scalars is just ordinary arithmetic or algebra. Vector. A vector, such as force and velocity, has a magnitude and direction. In this book a vector will be symbolized by an italic boldface letter, such as a force F. In handwriting, a vector should be represented by an italic letter with an arrow over it, ~ The magnitude of the force is designated by jFj or simply F or such as the force F. ~ jFj in handwriting. A vector is represented graphically by an arrow, as shown in figure 2.1. The magnitude of the vector is indicated by the length of the arrow and the direction is defined by the line of action and the sense. Point A is called the tail of the vector and point B the head of the vector. Then the sense is from the tail to the head, or indicated by the arrowhead.

DOI: 10.1051/978-2-7598-2901-9.c002 © Science Press, EDP Sciences, 2022

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FIG. 2.1 – Graphical representation of a vector.

Vector addition. Vector addition is performed according to the parallelogram law. To add vector Q and vector S shown in figure 2.2a, two vectors are joined at their tails. Then from the head of Q, draw a line parallel to vector S, and from the head of S, draw a line parallel to vector Q. Two parallel lines intersect at a common point P and form the adjacent sides of a parallelogram. The diagonal of the parallelogram is therefore the resultant R, as shown in figure 2.2b. Vector addition can also be performed using the triangle method. To add S to Q, first draw vector Q. From the head of Q, then draw vector S. Thus, the resultant R extends from the tail of Q to the head of S, as shown in figure 2.2c. It can be seen that this triangle is actually the upper half of the parallelogram. In a similar manner, we can also add Q to S by connecting the head of S to the tail of Q to get R, as shown in figure 2.2d. This triangle is the bottom half of the parallelogram. Here we always follow a “head to tail” fashion, i.e., the head of Q to the tail of S shown in figure 2.2c, or the head of S to the tail of Q shown in figure 2.2d. Apparently, vector addition is commutative, which means R = Q + S = S + Q.

FIG. 2.2 – Vector addition.

Vector subtraction. The subtraction of a vector can be treated as the addition of the corresponding negative vector. Thus, vector subtraction is a special case of vector addition and we have R0 ¼ Q  S ¼ Q þ ðSÞ

ð2:1Þ

The parallelogram law or the triangle method is then applied to vector subtraction, which is shown graphically in figure 2.3.

Vectors and Vector Operations

13

FIG. 2.3 – Vector subtraction.

Vector resolution. A vector can be resolved into two “components” along designated axes using the parallelogram law. For example, F shown in figure 2.4a is to be resolved into components along axes u and v, which are coplanar with F. Starting from the head of F, construct a line parallel to u until it intersects v and a line parallel to v until it intersects u, figure 2.4b. Therefore, a parallelogram is formed and the sides of the parallelogram represent the two components Fu and Fv, as shown in figure 2.4c.

FIG. 2.4 – Resolution of a vector along two arbitrary axes.

If the angles between F and the two axes u and v are α and β, respectively, we can get the magnitudes of the two components Fu and Fv from the trigonometry. From figure 2.5, we have F Fu Fv ¼ ¼ sin½180  ða þ bÞ sin b sin a

ð2:2Þ

Thus, the magnitudes of the two components are Fu ¼

F sin b ; sinða þ bÞ

Fv ¼

F sin a sinða þ bÞ

ð2:3Þ

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FIG. 2.5 – Components of a vector along two arbitrary axes. More often, we will resolve vector F along two mutually perpendicular axes, for example, x and y axes shown in figure 2.6, in which α + β = 90°. For this case, we actually get a rectangle when using the parallelogram law and it is easy to determine the magnitudes of these two rectangular components. They are Fx ¼ F cos a;

Fy ¼ F sin a

From the two rectangular components, we can also get qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Fx F ¼ Fx2 þ Fy2 ; cos a ¼ F

ð2:4Þ

ð2:5Þ

FIG. 2.6 – Rectangular components of a vector.

2.2

Cartesian Vectors

As shown in figure 2.7a, there are three forces F1, F2, F3 acting at point O in three dimensions. To obtain the resultant force FR of these three forces, we can first determine the resultant force of F1 and F2 using the parallelogram law. Then using the parallelogram law again, this resultant F12 is added to F3 to obtain the resultant force of all three forces, as shown in figure 2.7b. If four, five or even more concurrent forces are to be added, the parallelogram law can be applied successively in this way. However, the solution to this type of problems can be greatly simplified if the forces are first resolved into rectangular components shown in figure 2.7c and then represented in Cartesian vector form. So, in this section we will first learn to represent a vector in Cartesian vector form, and then add or subtract vectors using Cartesian vector notation.

Vectors and Vector Operations

15

FIG. 2.7 – Determination of the resultant force of concurrent forces.

To represent a vector in Cartesian vector form and develop the theory of vector algebra, a right-hand coordinate system will be used. An orthogonal coordinate system is said to be a right-hand coordinate system if the right-hand thumb points in the direction of the positive z axis when the right-hand fingers are curled from the positive x to positive y, as shown in figure 2.8a. Cartesian unit vectors, i, j and k, shown in figure 2.8b are used to designate the directions of the positive x, y and z axes, respectively. For a two-dimensional problem, we usually set x horizontal to the right and y vertically upward, as shown in figure 2.9. Then for a right-hand coordinate system, the z axis would be directed outward, perpendicular to the x–y plane.

FIG. 2.8 – Right-hand coordinate system.

FIG. 2.9 – Two-dimensional coordinate system.

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2.2.1

Cartesian Vector Representation

From two successive applications of the parallelogram law, a vector can be resolved into three components along the right-hand coordinate axes, i.e., Q = Qxy + Qz and then Qxy = Qx + Qy, as shown in figure 2.10a. Combining these two equations, Q is written as Q ¼ Qx þ Qy þ Qz

ð2:6Þ

Actually, these three rectangular components, Qx, Qy, Qz, form the three adjacent sides of a cuboid and vector Q is the diagonal of the cuboid, as shown in figure 2.10b. The magnitude of vector Q, i.e., the length of the diagonal of the cuboid, is qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð2:7Þ Q ¼ Qx2 þ Qy2 þ Qz2

FIG. 2.10 – Three rectangular components of a vector.

If a unit vector uQ having a magnitude of 1 and the same direction as Q, shown in figure 2.11a, is utilized, vector Q can be expressed in terms of its magnitude and direction separately as follows: Q ¼ QuQ

ð2:8Þ

When the magnitude Q ≠ 0, the unit vector can be represented by uQ ¼

Q Q

ð2:9Þ

Vectors and Vector Operations

17

FIG. 2.11 – Unit vector of a vector and its coordinate direction angles. It should be noted that vector Q is of a certain type of physical quantity and a proper set of units should be used for both vector Q and its magnitude Q. For example, a force vector Q has the unit newton (N) and its magnitude may be Q = 10 N. However, the unit vector uQ is dimensionless since the units of vector Q and scalar Q will cancel out from equation (2.9). Since the three rectangular components of Q have the magnitudes Qx, Qy, and Qz, and point in the directions of i, j and k respectively, we can express each component in terms of its magnitude and direction separately and equation (2.6) becomes Q ¼ Qx i þ Qy j þ Qz k

ð2:10Þ

This is the Cartesian vector form of vector Q. Besides the unit vector uQ, the direction of Q can also be defined by the coordinate direction angles α, β and γ, that Q forms with the x, y, z axes, as shown in figure 2.11b. Each of these angles is between 0° and 180°. Referring to the colored right triangle DOA1 A2 shown in figure 2.11b, we have cosγ = Qz/Q. Similarly, cos a ¼

Qx ; Q

cos b ¼

Qy ; Q

cos c ¼

Qz Q

ð2:11Þ

They are called the direction cosines of Q. From equations (2.9) and (2.10), we have uQ ¼

Q Qx Qy Qz ¼ iþ jþ k ¼ cos ai þ cos bj þ cos ck Q Q Q Q

ð2:12Þ

Since uQ has a magnitude of 1, then cos2 a þ cos2 b þ cos2 c ¼ 1

ð2:13Þ

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It means if two coordinate direction angles are given, the third one can be determined from equation (2.13) when vector Q lies in a known octant (see example 2.2). From equations (2.10) and (2.11), we have Q ¼ Qx i þ Qy j þ Qz k ¼ Q cos ai þ Q cos bj þ Q cos ck

ð2:14Þ

Example 2.1. A force is shown in figure 2.12a. The magnitudes of the three components are Fx = 3 N, Fy = 4 N, Fz = 5 N. Write the force in Cartesian vector form and determine its magnitude and coordinate direction angles.

FIG. 2.12 – Writing a force in Cartesian vector form. Solution: Since the three components Fx, Fy, and Fz of the force F point in the directions of +i, +j and −k, we can write it in Cartesian vector form as F ¼ f3i þ 4j  5kg N According to equations (2.7) and (2.11), the magnitude and coordinate direction angles of the force are qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi F ¼ Fx2 þ Fy2 þ Fz2 ¼ 5 2 N cos a ¼

Fx 3 ¼ pffiffiffi ¼ 0:4243 F 5 2

)

a ¼ 64:9

cos b ¼

Fy 4 ¼ pffiffiffi ¼ 0:5657 F 5 2

)

b ¼ 55:6

cos c ¼

Fz 5 ¼ pffiffiffi ¼ 0:7071 F 5 2

)

c ¼ 135

Vectors and Vector Operations

19

The angles are shown in figure 2.12b. It should be noted that when calculating cos c, “5” should be used in the numerator since Fz is in the −k direction.

2.2.2

Addition and Subtraction of Cartesian Vectors

If two vectors have been written in Cartesian vector form, for example, A ¼ Ax i þ Ay j þ Az k and B ¼ Bx i þ By j þ Bz k, we can easily obtain   A þ B ¼ ðAx þ Bx Þi þ Ay þ By j þ ðAz þ Bz Þk   A  B ¼ ðAx  Bx Þi þ Ay  By j þ ðAz  Bz Þk Following this concept of vector addition, for a system of concurrent forces shown in figure 2.7, we can first write each force in Cartesian vector form, i.e., Fi ¼ Fxi i þ Fyi j þ Fzi k ði ¼ 1; 2; 3Þ. Therefore, the resultant force can be written as X X X X FR ¼ Fi ¼ Fxi i þ Fyi j þ Fzi k ð2:15Þ P P P Here Fxi , Fyi and Fzi represent the algebraic sums of the respective i, j, and k components of all the forces in the system. In this way, vector addition and subtraction of several vectors are greatly simplified. Example 2.2. Three forces shown in figure 2.13a act at point O. If F1 = 400 N and F2 = 500 N, determine the third force F3 so that the resultant force has a magnitude of 200 N and acts in the positive z direction.

FIG. 2.13 – Addition of concurrent forces in Cartesian vector form.

Solution: For the force F1, only two coordinate direction angles are specified and the third angle γ should be determined from equation (2.13). cos2 45 þ cos2 60 þ cos2 c¼ 1

)

cos c¼  0:5

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Hence, two possibilities exist, namely, c ¼ 60 or c ¼ 120 By inspection of figure 2.13a, we have c ¼ 120 . Using equation (2.14), F1 is expressed in Cartesian vector form as F1 ¼ f400 cos 45 i þ 400 cos 60 j þ 400 cos 120 kg N ¼ f282:8i þ 200j  200kg N For the force F2, the angles of 30° and 45° are not coordinate direction angles. The parallelogram law (or the triangle method) are utilized twice to resolve F2 into its x, y, z components, figure 2.13b. By trigonometry, the magnitudes of the components are F2z ¼ 500 cos 45 ¼ 353:6 N F2x ¼ F2xy sin 30 ¼ 500 sin 45  sin 30 ¼ 176:8 N F2y ¼ F2xy cos 30 ¼ 500 sin 45  cos 30 ¼ 306:2 N Realizing that F2x has a direction defined by −i, we have F2 ¼f176:8i þ 306:2 j þ 353:6kg N The Cartesian vector form of F3 can be written as F3 ¼ F3x i þ F3y j þ F3z k Since the resultant force has a magnitude of 200 N and acts in the positive z direction, its Cartesian vector form is FR ¼ f200kg N. From equation (2.15), we have   200k ¼ ð282:8  176:8 þ F3x Þi þ 200 þ 306:2 þ F3y j þ ð200 þ 353:6 þ F3z Þk   ¼ ð106 þ F3x Þi þ 506:2 þ F3y j þ ð153:4 þ F3z Þk To satisfy this equation, the corresponding i, j, and k components on the left side and on the right side must be equal. Hence, 0 ¼ 106 þ F3x

)

F3x ¼ 106 N

0 ¼ 506:2 þ F3y

)

F3y ¼ 506:2 N

200 ¼ 153:4 þ F3z

)

F3z ¼ 46:6 N

Vectors and Vector Operations

21

F3 expressed in Cartesian vector form is F3 ¼f106i  506:2j þ 46:6kg N Its magnitude and coordinate direction angles can also be determined. Readers can finish the calculations by themselves.

2.3

Vector Directed Along a Line

In the previous section, we could write a vector extending from the origin of the coordinate system in Cartesian vector form. However, the force vector in figure 2.14 is extending from an arbitrary point A, not the origin O. To write this vector in Cartesian vector form, the concept of position vector should be introduced first. A position vector is defined as a fixed vector which locates one point relative to another. Consider two points, A(xA, yA, zA) and B(xB, yB, zB), in figure 2.15. The position vector directed from A to B, is designated by the symbol rAB (This position vector is also denoted as rB=A in §6.5 and §7.4). Position vectors rOA and rOB, which extend from the origin O, are also designated by rA and rB, respectively. rA and rB can be expressed in Cartesian vector form as rA ¼ Ax i þ Ay j þ Az k; rB ¼ Bx i þ By j þ Bz k From the triangle method, we can get the relationship of the three vectors in figure 2.15 as follows: rA þ rAB ¼ rB Solving for rAB yields

  rAB ¼ rB  rA ¼ ðBx  Ax Þi þ By  Ay j þ ðBz  Az Þk

FIG. 2.14 – Vector directed along a line.

ð2:16Þ

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22

FIG. 2.15 – Position vector.

Equation (2.16) means the i, j, and k components of the position vector rAB are formed by subtracting the corresponding coordinates of the tail A from that of the head B. The unit vector having the same direction as rAB is represented by uAB ¼

rAB rAB

Therefore, the force vector shown in figure 2.14 can be expressed in terms of its magnitude and direction separately as F ¼ FuAB ¼ F

rAB rAB

ð2:17Þ

Note that the force vector F has the unit of force, rAB the unit of length, and the unit vector uAB is dimensionless. From figure 2.14, the coordinates of the two points are A(0, 3 m, 0) and B(2 m, 2.5 m, −2 m). Then the position vector from A to B is rAB ¼ ð2  0Þi þ ð2:5  3Þj þ ð2  0Þk ¼ f2i  0:5j  2kgm The magnitude of rAB is qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi rAB ¼ 22 þ ð0:5Þ2 þ ð2Þ2 ¼ 2:872 m The unit vector that defines the direction of both rAB and F is uAB ¼

rAB ¼ 0:6963i  0:1741j  0:6963k rAB

Then, the Cartesian vector form of the force F having a magnitude of 200 N is F ¼ FuAB ¼ 200ð0:6963i  0:1741j  0:6963kÞ ¼ f139i  34:8j  139kg N Here the position vector is used to formulate a Cartesian force vector directed along a line passing through two arbitrary points in space. It will be shown later that the position vector is also of importance in finding the moment of a force in chapter 3 and the angular momentum in chapter 10.

Vectors and Vector Operations

23

Example 2.3. Two forces are acting at the top of a pile along two cords AB and AC, figure 2.16a. If F1 = 150 N and F2 = 80 N, determine the magnitude and coordinate direction angles of the resultant force.

FIG. 2.16 – Adding two concurrent vectors directed along lines.

Solution: First formulate unit vectors uAB and uAC from the associated position vectors rAB and rAC. From figure 2.16a, the coordinates of the three points are A(0, 0, 5 m), B(4 m, −7 m, 0) and C(−1.928 m, 2.298 m, 0). Therefore rAB ¼ f4i  7j  5kgm, rAC ¼ f1:928i þ 2:298j  5kgm rAB ¼

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 42 þ ð7Þ2 þ ð5Þ2 ¼ 9.487 m, rAC ¼ ð1:928Þ2 þ 2:2982 þ ð5Þ2 ¼ 5:831 m

uAB ¼

uAC ¼

rAB ¼ 0:4216i  0:7379j  0:5270k rAB

rAC ¼ 0:3306i þ 0:3941j  0:8575k rAC

Then write F1 and F2 as Cartesian vectors: F1 ¼ F1 uAB ¼ 150ð0:4216i  0:7379j  0:5270kÞ ¼ f63:24i  110:7j  79:05kgN F2 ¼ F2 uAC ¼ 80ð0:3306i þ 0:3941j  0:8575kÞ ¼ f26:46i þ 31:53j  68:60kgN The resultant force, which also acts at A as shown in figure 2.16b, is FR ¼ F1 þ F2 ¼ f36.78i  79.17j  147:6kgN

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The magnitude and coordinate direction angles of the resultant force are qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 þ F 2 þ F 2 ¼ 171:5 N FR ¼ FRx Ry Rz

2.4

cos a ¼

FRx ¼ 0:2145 FR

cos b ¼

FRy ¼ 0:4617 FR

) b ¼ 117

cos c ¼

FRz ¼ 0:8607 FR

) c ¼ 149

) a ¼ 77.6

Dot Product and Cross Product

There are two types of vector multiplication: dot product and cross product. The dot product of two vectors Q and S, written as Q  S, and read “Q dot S ”, yields a scalar. It is expressed as Q  S ¼ QS cos h

ð2:18Þ

where θ (0 h 180 ) is the angle formed between the two vectors, as shown in figure 2.17. However, the cross product of two vectors Q and S, written as Q  S, and read “Q cross S ”, yields another vector R. It is expressed as R¼QS The magnitude of R is R ¼ QS sin h and the direction of R is perpendicular to both Q and S and specified by the right-hand rule, i.e., curling the right-hand fingers from Q to S, the thumb then points in the direction of R, figure 2.18. Using a unit vector uR to define the direction of R, the cross product can be written as R ¼ Q  S ¼ ðQS sin hÞuR

ð2:19Þ

The dot product and cross product are often referred to as the scalar product and vector product, respectively, since their results are a scalar and a vector, respectively.

FIG. 2.17 – Dot product.

Vectors and Vector Operations

25

FIG. 2.18 – Cross product.

2.4.1

Cartesian Vector Formulation of the Dot Product

The dot product obeys the following two laws: (1) Commutative law: QS¼SQ (2) Distributive law: P  ðQ þ SÞ ¼ ðP  QÞ þ ðP  SÞ From equation (2.18), it is easy to prove these laws and readers are recommended to finish the proof by themselves. For the Cartesian unit vectors shown in figure 2.8b, we get i  i ¼ 1  1  cos 0 ¼ 1 and i  j ¼ 1  1  cos 90 ¼ 0 from equation (2.18). In a similar manner, we can get the following nine expressions: i  i ¼ 1; i  j ¼ 0; j  i ¼ 0; j  j ¼ 1; k  i ¼ 0; k  j ¼ 0;

ik¼0 jk¼0 kk¼1

For two general vectors Q ¼ Qx i þ Qy j þ Qz k and S ¼ Sx i þ Sy j þ Sz k, their dot product is     Q  S ¼ Qx i þ Qy j þ Qz k  Sx i þ Sy j þ Sz k ¼ Qx Sx ði  iÞ þ Qx Sy ði  jÞ þ Qx Sz ði  kÞ þ Qy Sx ðj  iÞ þ Qy Sy ðj  jÞ þ Qy Sz ðj  kÞ þ Qz Sx ðk  iÞ þ Qz Sy ðk  jÞ þ Qz Sz ðk  kÞ Substituting the nine expressions of the dot product of the unit vectors, the final result becomes Q  S ¼ Qx Sx þ Qy Sy þ Qz Sz

ð2:20Þ

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2.4.2

Applications of the Dot Product

The dot product has the following two important applications. (1) The angle between two vectors Q and S can be determined from equation (2.18) as follows:   1 Q  S h = cos 0 h 180 QS Note that if Q  S ¼ 0, we have h = cos1 0 ¼ 90 , i.e., Q is perpendicular to S. (2) The components of a vector parallel and perpendicular to an axis can be determined using the dot product. As shown in figure 2.19, vector Q can be resolved into two components Qk and Q? , which are parallel and perpendicular to the axis a, respectively. Using a unit vector ua to specify the direction of the positive axis, we have Q  ua ¼ Q  1  cos h ¼ Q cos h ¼ Qk

ð2:21Þ

Qk is a scalar and often referred to as the projection of Q onto the axis. It may be positive when the vector Qk is in the +ua direction (0 h\90 ) or negative when the vector Qk is in the −ua direction (90 \h 180 ). Therefore, the vector Qk can be expressed in terms of its magnitude and direction separately as Qk ¼ Qk ua ¼ ðQ cos hÞua ¼ ðQ  ua Þua The perpendicular component Q? can then be obtained from the vector subtraction, i.e., Q? ¼ Q  Qk . From the Pythagorean theorem, the magnitude of Q? is qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Q? ¼ Q 2  Qk2 ð2:22Þ Of course, if the angle θ between vector Q and the axis is known, we also have Q? ¼ Q sin h

FIG. 2.19 – Projection of a vector onto an axis.

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27

Example 2.4. Two vectors r1 and r2 are shown in figure 2.20. The magnitude of r2 is r2 = 4 m. Determine the angle θ between the two vectors, the projection of r2 along r1, and the magnitude of the projected component of r1 along r2.

FIG. 2.20 – The angle between two vectors. Solution: The two vectors can be written in Cartesian vector form as r1 ¼ f1:5i  2:598j þ 5kgm 

 12 5 jþ k ¼f3:692j þ 1:538kgm r2 ¼ 4 13 13 The magnitude of r1 is r1 ¼

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1:52 þ ð2:598Þ2 þ 52 ¼ 5:831 m

Then we can determine the angle θ between the two vectors.

  1 r1  r2 1 1:5  0 þ ð2:598Þ  3:692 þ 5  1:538 h ¼ cos ¼ cos 5:831  4 r1 r2 ¼ cos1 ð0:0816Þ ¼ 94:68 The unit vectors along r1 and r2 are u1 ¼

r1 1:5i  2:598j þ 5k ¼ 0:2572i  0:4456j þ 0:8575k ¼ 5:831 r1 u2 ¼

r2 3:692j þ 1:538k ¼ 0.9231j þ 0:3845k ¼ 4 r2

Therefore, the projection of r2 along r1 is r2k ¼ r2  u1 ¼ 3:692  ð0:4456Þ þ 1:538  0:8575 ¼ 0:326 m The magnitude of the projected component of r1 along r2 is   r1k  ¼ jr1  u2 j ¼ jð2:598Þ  0:9231 þ 5  0:3845j ¼ j0:476j ¼ 0:476 m

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Note that the magnitude of the projected component is always positive, whereas the projection can be positive or negative. Furthermore, since the angle θ between these two vectors is known now, the projection of r2 along r1 can also be obtained as follows: r2k ¼ r2 cos h ¼ 4  cos 94:68 ¼ 0:326 m Example 2.5. The frame OAB is subjected to a force F from cable BC, figure 2.21a. If F = 180 N, determine the angle θ between F and the segment BA, and the magnitudes of the two components of F, which are parallel and perpendicular to BA, respectively.

FIG. 2.21 – Parallel and perpendicular components of a force to a bar.

Solution: First formulate unit vectors uBA and uBC from the associated position vectors rBA and rBC. From figure 2.21a, the coordinates of the three points are A(0, 0, 3 m), B(−1.8 m, 2.5 m, 5 m) and C(4 m, 3 m, 0). Therefore rBA ¼ f1:8i  2:5j  2kgm, rBC ¼ f5:8i þ 0:5j  5kgm rBA ¼

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1:82 þ ð2:5Þ2 þ ð2Þ2 ¼ 3.673 m, rBC ¼ 5:82 þ 0:52 þ ð5Þ2 ¼ 7:674 m

uBA ¼

rBA ¼ 0:4901i  0:6807j  0:5445k rBA

uBC ¼

rBC ¼ 0:7558i þ 0:0652j  0:6516k rBC

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The force vector F in Cartesian vector form is F ¼ FuBC ¼ 180ð0:7558i þ 0:0652j  0:6516kÞ ¼ f136i þ 11:73j  117:3kgN Then we can determine the angle θ between F and the segment BA.  

1 F  rBA 1 136  1:8 þ 11:73  ð2:5Þ þ ð117:3Þ  ð2Þ ¼ cos h ¼ cos ¼ 47:1 FrBA 180  3:673 Actually, this angle is also the angle between the two unit vectors uBC and uBA. Readers are recommended to determine the angle from these two unit vectors by themselves. The force F is resolved into two components, as shown in figure 2.21b. The magnitude of the parallel component can be determined from the dot product. FBA ¼ F  uBA ¼ 136  0:4901 þ 11:73  ð0:6807Þ þ ð117:3Þ  ð0:5445Þ ¼ 123 N

The magnitude of the perpendicular component is obtained from the Pythagorean theorem, qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 ¼ F? ¼ F 2  FBA 1802  1232 ¼ 132 N Since θ has been obtained, the magnitudes of the two components can also be determined directly from trigonometry. FBA ¼ F cos h ¼ 180  cos 47:1 ¼ 123 N F? ¼ F sin h ¼ 180  sin 47:1 ¼ 132 N

2.4.3

Cartesian Vector Formulation of the Cross Product

The cross product obeys the distributive law: P  ðQ þ SÞ ¼ ðP  QÞ þ ðP  SÞ However, unlike the dot product, the cross product does not obey the commutative law, i.e., Q  S 6¼ S  Q. Actually, we have Q  S¼  ðS  QÞ This is because Q  S and S  Q have the same magnitude of QS sin h and opposite directions, as shown in figure 2.22. For the Cartesian unit vectors shown in figure 2.23, we get i  j ¼ ð1  1  sin 90 Þk ¼ k and i  i ¼ 0 (since ji  ij ¼ 1  1  sin 0 ¼ 0) from equation (2.19). In a similar manner, we can get the following nine expressions:

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30 i  i ¼ 0; j  i ¼ k;

i  j ¼ k; j  j ¼ 0;

i  k ¼ j jk¼i

k  i ¼ j;

k  j ¼ i;

kk¼0

FIG. 2.22 – Non-commutativity of cross product.

FIG. 2.23 – Cross product of Cartesian unit vectors.

For two general vectors Q ¼ Qx i þ Qy j þ Qz k and S ¼ Sx i þ Sy j þ Sz k, their cross product is     Q  S ¼ Qx i þ Qy j þ Qz k  Sx i þ Sy j þ Sz k ¼ Qx Sx ði  iÞ þ Qx Sy ði  jÞ þ Qx Sz ði  kÞ þ Qy Sx ðj  iÞ þ Qy Sy ðj  jÞ þ Qy Sz ðj  kÞ þ Qz Sx ðk  iÞ þ Qz Sy ðk  jÞ þ Qz Sz ðk  kÞ

Substituting the nine expressions of the cross product of the unit vectors and combining terms yields Q  S ¼ ðQy Sz  Qz Sy Þi þ ðQz Sx  Qx Sz Þj þ ðQx Sy  Qy Sx Þk

ð2:23Þ

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31

The cross product can also be determined from the expansion of a determinant form as    i j k   ð2:24Þ Q  S ¼  Qx Qy Qz   Sx Sy Sz  PROBLEMS 2.1 Determine the x and y components of the resultant force of the four coplanar forces.

Prob. 2.1

2.2 The force F has components of Fx = 1.5 kN and Fy = 2.5 kN. If c ¼ 30 , determine Fz and F.

Prob. 2.2

2.3 A force F is acting inside a cylinder as shown. If F = 100 kN and h ¼ 30 , express the force as a Cartesian vector.

Prob. 2.3

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2.4 Determine the magnitude and coordinate direction angles of the resultant force of the two forces.

Prob. 2.4

2.5 The magnitude of the force F as shown is 2 kN. If h ¼ 30 and u ¼ 40 , express F as a Cartesian vector and determine its coordinate direction angles.

Prob. 2.5

2.6 Three forces are acting on a cuboid as shown. If F1 ¼ 500 N, F2 ¼ 700 N and F3 ¼ 400 N, express the three forces in Cartesian vector form.

Prob. 2.6

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33

2.7 Find the angle θ between the pillar AO and the cable AB.

Prob. 2.7

2.8 Determine the magnitude and coordinate direction angles of the resultant force of the two forces.

Prob. 2.8

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2.9 Express the position vector r in Cartesian vector form and specify its magnitude and coordinate direction angles.

Prob. 2.9

2.10 Two position vectors r1 and r2 are shown in the figure. The magnitude of r2 is r2 = 5 m. Determine the angle θ between the two vectors, the projection of r1 along r2, and the projection of r2 along r1.

Prob. 2.10

2.11 Determine the angle θ between the two position vectors r1 and r2, the projection of r1 along r2, and the projection of r2 along r1.

Prob. 2.11

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35

2.12 Determine the Cartesian vector form of the resultant force of the two forces acting at A and the unit vector along this resultant force.

Prob. 2.12

2.13 Determine the angles cable OA makes with the X, Y, Z axes.

Prob. 2.13

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2.14 The bar AO is subjected to a force F along line AB as shown. If F = 100 N, find the angle θ between F and AO, and the magnitudes of the two components of F, which are parallel and perpendicular to AO, respectively.

Prob. 2.14

2.15 The right-angled frame OAB is subjected to a force F from cable BC as shown. Find the angle θ between F and BA, and the magnitudes of the two components of F, which are parallel and perpendicular to BA, respectively.

Prob. 2.15

Chapter 3 Simplification of Force Systems Objectives
Determine the moment of a force using vector formulation and scalar formulation.
Find the moment of a force about a specified axis.
Define a couple and determine the couple moment.
Find an equivalent force-couple system for a system of forces and couples and discuss the further simplification.
Understand the concepts of center of gravity, center of mass, and centroid and specify their locations.
Determine the equivalent resultant force for a simple distributed loading.

3.1

Moment of a Force about a Point

A force F lying on the x–y plane is acting on the handle OA, figure 3.1a. From experience, we know that this force tends to rotate the handle about point O. This rotation effect is called the moment of a force or a torque, denoted as MO(F). From point O, draw a line perpendicular to the line of action of the force and get the perpendicular or shortest distance from point O to the force. This distance is called the moment arm. It can be seen that the moment arm d = lsinθ in figure 3.1a, however, d = l and d = 0 in figure 3.1b and c, respectively. Apparently, the larger the force F or the moment arm d is, the greater the turning effect is. Then the magnitude of the moment is represented as jM O ðFÞj ¼ Fd

ð3:1Þ

Therefore, the unit of the moment should be the unit of force times the unit of distance, i.e., Nm in SI unit system. The direction of the moment is defined by its moment axis, which passes through point O and is perpendicular to the plane containing both the force F and O. There is a “right-hand rule” for the moment axis.

DOI: 10.1051/978-2-7598-2901-9.c003 © Science Press, EDP Sciences, 2022

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FIG. 3.1 – Moment of a force about a point. With the right-hand fingers curled such that they follow the sense of rotation, as shown in figure 3.1, the thumb then points along the moment axis (here the z axis).

3.1.1

Vector Formulation

For an arbitrary force in three dimensions shown in figure 3.2, it is not easy to identify the moment arm and the moment axis. Therefore, to represent the moment of a force in three dimensions, the following vector cross product formulation will be used. M O ðF Þ ¼ r  F

ð3:2Þ

where r is a position vector from point O to a point on the line of action of the force F. The magnitude of the cross product is jM O ðFÞj ¼ jr  Fj ¼ rF sin h ¼ F r sin h ¼ Fd where θ is the angle between r and F, figure 3.2. This magnitude agrees with equation (3.1). From the concept of the cross product, the direction of r  F is also specified by the right-hand rule, i.e., curling the right-hand fingers from vector r to vector F, the thumb then points in the direction of r  F. This direction is perpendicular to the shaded plane containing both r and F and just indicates the moment axis. Then we can see that indeed the vector formulation r  F represents the magnitude and direction of the moment M O ðFÞ. The moment vector is usually illustrated by an arrow with a curl around it to distinguish it from a force vector, as shown in figure 3.2. This curl represents the sense of rotation around the moment axis. Since the cross product is not commutative, it is very important that the proper order of r and F be maintained in equation (3.2). From Cartesian vector formulation of the cross product given in equation (2.24), the moment M O ðFÞ can also be determined from the expansion of a determinant as


i j k


M O ðFÞ ¼ r  F ¼

rx ry rz

¼ ðry Fz  rz Fy Þi þ ðrz Fx  rx Fz Þj þ ðrx Fy  ry Fx Þk
Fx Fy Fz


Simplification of Force Systems

39

where rx, ry, and rz represent the x, y, and z components of the position vector r, and Fx, Fy, and Fz are the x, y, and z components of the force vector F, respectively. If a body is acted upon by a system of forces, figure 3.3, the resultant moment of all the forces about point O, representing the resultant turning effect about O, is determined by vector summation of the moments of all the forces as follows: X X M RO ¼ M O ðFi Þ ¼ ðri  F i Þ

FIG. 3.2 – Vector Formulation.

FIG. 3.3 – Resultant moment. Example 3.1. A force F = 120 N acts at the end of rod OA through cable AB, figure 3.4. Determine the moment of F about the base point O.

FIG. 3.4 – Determining the moment of a force about a point using vector formulation.

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Solution: The coordinates of the points are A(3.5 m, 3.5 m, 0) and B(4.5 m, 0, −1.5 m). Therefore rAB ¼ fi  3:5j  1:5kg m rAB ¼

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 12 þ ð3:5Þ2 þ ð1:5Þ2 ¼ 3.937 m

Write F in Cartesian vector form: F¼F

rAB i  3:5j  1:5k ¼ f30.48i  106.7j  45.72kgN ¼ 120 3.937 rAB

The moment of F about point O are determined by the cross product rA × F.

i j
3:5 M O ðFÞ ¼ rA  F ¼

3:5
30:48 106:7


k

0

¼ f160i þ 160j  480kgNm 45:72


RETHINK: If we pull on cable AB with the force F at point B, the force should produce the same turning effect about the base point O. Then, the moment of F about point O can also be determined by the cross product rB × F.

i
M O ðFÞ ¼ rB  F ¼

4:5
30:48

j 0 106:7


k

1:5

¼ f160i þ 160j  480kgNm 45:72


Two expressions obtain the same moment about point O. Actually, we can pull on cable AB with the force F at any point lying on the cable and the turning effect (moment) of this force about O should always be the same. This means a force is a sliding vector and can therefore act at any point on its line of action and still produce the same turning effect (also the same translating effect, see §§ 3.4.1), which is referred to as the principle of transmissibility. From this principle, the position vector r in M O ðFÞ ¼ r  F can be directed from point O to any point on the line of action of F. It should be pointed out that the moment of a force indicates the tendency of rotation, but does not always cause a rotation. In this example, the force F tends to rotate the rod about the base point O, but the fixed support at O prevents the rotation. Example 3.2. A force F = 750 N is acting along line AB, figure 3.5a. Determine the moments of this force about point O and about point P, respectively.

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41

FIG. 3.5 – Determining the moments of a force about different points. Solution: First, write the force F along line AB in Cartesian vector form using position vector rAB. From figure 3.5a, the coordinates of the points are A(−3 m, 4 m, 0) and B(4 m, 6 m, −6 m). qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi rAB ¼ f7i þ 2j  6kg m; rAB ¼ 72 þ 22 þ ð6Þ2 ¼ 9.434 m F¼F

rAB 7i þ 2j  6k ¼ f556:5i þ 159:0j  477.0kgN ¼ 750 9.434 rAB

When calculating the moment of the force about point O, the position vector rA (rOA) shown in figure 3.5b can be used.


i j k


4 0

¼ f1908i  1431j  2703kgNm M O ðFÞ ¼ rA  F ¼

3
556:5 159:0 477:0
To calculate the moment of the force about point P, the position vector rPA can be used.


i j k


2 0

¼ f954i  3339jgNm M P ðFÞ ¼ rPA  F ¼

7
556:5 159:0 477:0
RETHINK: From the principle of transmissibility, the position vector rB (rOB) can also be used to calculate the moment of the force about point O, i.e., M O ðFÞ ¼ rB  F. The position vector rPB can also be used to calculate the moment of the force about point P, i.e., M P ðFÞ ¼ rPB  F. Readers are recommended to finish this by themselves. Example 3.3. Determine the resultant moment of the two forces shown in figure 3.6 about point O.

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FIG. 3.6 – The resultant moment of two forces about a point in 3-D. Solution: The coordinates of the points are A(0, 1.3 m, 0), B(−1 m, 0, 2.4 m), C(1 m, 0, − 2 m) and D(0, 2.3 m, 0). Therefore rAC ¼ fi  1:3j  2kg m; rBD ¼ fi þ 2:3j  2:4kg m qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi rAC ¼ 12 þ ð1:3Þ2 þ ð2Þ2 ¼ 2.587 m qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi rBD ¼ 12 þ 2:32 þ ð2:4Þ2 ¼ 3:471 m Then write F1 and F2 as Cartesian vectors: F 1 ¼ F1

F2 ¼ F2

rAC i  1:3j  2k ¼ f7:732i  10:05j  15:46kgkN ¼ 20 2:587 rAC

rBD i þ 2:3j  2:4k ¼ f4:321i þ 9:939j  10:37kgkN ¼ 15 3:471 rBD

The resultant moment of the two forces about O is therefore M RO ¼

X

M O ðF i Þ ¼

X



i j

¼
0 1:3

7:732 10:05

ðri  Fi Þ ¼ rA  F1 þ rD  F2





i j



2:3 0
þ
0

15:46

4:321 9:939 k

¼ f43:96i  19:99kgkNm





0

10:37
k

Simplification of Force Systems

3.1.2

43

Principle of Moments

There is an important principle about the moment of a force and it is extensively used in the solution of problems and proofs of theorems afterwards. It states that the moment of a force about a point is equal to the sum of the moments of the force’s components about the same point, which is referred to as the principle of moments. As shown in figure 3.7a, three concurrent forces F1, F2, and F3 can be simplified to a resultant force FR, i.e., FR = F1 + F2 + F3. Consider the moment of the resultant force FR about a fixed point O shown in figure 3.7b, we have M O ðFR Þ ¼ r  FR ¼ r  ðF1 þ F2 þ F3 Þ X M O ðFi Þ ¼ r  F1 þ r  F2 þ r  F3 ¼ Thus, the principle of moments is verified.

FIG. 3.7 – Verification of the principle of moments.

3.1.3

Scalar Formulation

Many problems in mechanics involve coplanar force systems and it is more convenient to view these problems in two dimensions. For example, the moments of forces in the x–y plane at the top of figure 3.8 are given at the bottom in two-dimensional view. In figure 3.8a, the moment vector MO(F) points to +z using the right-hand rule in three-dimensional view, which can be simply represented by the counterclockwise curl in two-dimensional view. However, in figure 3.8b, the moment vector MO(F) points to –z, which is simply represented by the clockwise curl in two-dimensional view. Therefore, for a system of forces lying in the x–y plane, considering positive moments as counterclockwise (in the +z direction), the moment of a force can be represented by scalar formulation as follows: MO ðFÞ ¼ Fd

ð3:3Þ

Therefore, the resultant moment about point O of the coplanar force system shown in figure 3.9 can be determined by simply adding the moments of all the forces algebraically as X X MRO ¼ MO ðFi Þ ¼ Fi di ¼ F1 d1 þ F2 d2  F3 d3

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FIG. 3.8 – Scalar formulation.

FIG. 3.9 – Resultant moment in 2-D. Example 3.4. Determine the moment of the force F ¼ 110 N shown in figure 3.10a about point O.

FIG. 3.10 – Application of the principle of moments.

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45

Solution: The moment arm d of the force F about point O is shown in figure 3.10b. It is not very easy to determine. However, if the force is resolved into its two rectangular components Fx and Fy , the moment arms of Fx and Fy are easy to identify. They are 1.5 m and 3.5 m, respectively. Then considering positive moments as counterclockwise, we can determine the moment of the force F from the principle of moments as follows: MO ðFÞ ¼ MO ðFx Þ þ MO ðFy Þ ¼ Fx  1:5  Fy  3:5 ¼ 110  cos 30  1:5  110  sin 30  3:5 ¼ 335 Nm We will find that the principle of moments is widely used, since it is often easier to determine the moment arms of a force’s components rather than the moment arm of the force itself. Example 3.5. Determine the resultant moment of two forces shown in figure 3.11 about point P.

FIG. 3.11 – The resultant moment of two forces about a point in 2-D. Solution: Assume that positive moments act in the counterclockwise direction. Resolve each force into its x and y components and calculate the resultant moment using the principle of moments. 3 4 MRP ¼ MP ðF1 Þ þ MP ðF2 Þ ¼ 40   1 þ 40   1  30 cos 30  3  30 sin 30  1 5 5 ¼ 36:9 Nm

The negative sign means the resultant moment is clockwise. Example 3.6. Determine the angle θ (0° ≤ θ ≤ 180°) of the force F shown in figure 3.12a so that the magnitude of the moment of the force about point O is (1) the minimum and (2) the maximum. If F = 100 N, compute the moments in each case.

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FIG. 3.12 – The moment of a force with variable direction. Solution: (1) When the force F is along line OB, figure 3.12b, the moment arm is zero and the force F produces zero moment (the minimum moment) about point O. For this case, h ¼ u ¼ arctan

3:5 ¼ arctan 0:4375 ¼ 23:6 8 Mmin ¼ 0

(2) When the force F is perpendicular to line OB, figure 3.12c, the moment arm is the biggest and the force produces the maximum moment about point O. For this case, h ¼ 90 þ u ¼ 90 þ 23:6 ¼ 113:6 Mmax ¼ F  OB ¼ 100 

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 3:52 þ 82 ¼ 873 Nm

Example 3.7. Determine the moment of the force F = 600 N shown in figure 3.13a about point O.

FIG. 3.13 – Multiple solutions of the moment of a fouce about a point. Solution 1: Scalar formulation The force can be resolved into its x and y components, as shown in figure 3.13b. Taking positive moments as counterclockwise, i.e., in the +z (+k) direction, clockwise moments will be negative. According to the principle of moments, we have

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47

MO ðFÞ ¼ MO ðFx Þ þ MO ðFy Þ ¼ Fx  0:8  Fy  0:6 ¼ 600 cos 50  0:8  600 sin 50  0:6 ¼ 584 Nm If the moment arm d shown in figure 3.13c is found by trigonometry, we can also use MO ðFÞ ¼ Fd to determine the moment of the force. Solution 2: Vector formulation Using a Cartesian vector approach, the force and position vectors shown in figure 3.13c can be represented as r ¼ f0:6i þ 0:8jg m F ¼ fF cos 50 i  F sin 50 jg ¼ f385.7i  459.6jg N The moment is therefore



i
M O ðFÞ ¼ r  F ¼

0:6
385:7

j 0:8 459:6


k

0

¼ f584kgNm 0


In solution 1, scalar formulation of the moment is used and we get a negative number, which means the turning effect is clockwise. In solution 2, vector formulation of the moment is used and we get a vector with only –k component, which means the moment vector points to –z. These two formulations are in accordance with each other. RETHINK: By comparison, scalar formulation is more convenient than vector formulation for this case, since the moment axis (indicating the direction of the moment) and the moment arm (related to the magnitude of the moment) are easy to determine for coplanar force system. Therefore, vector formulation is generally recommended for only three-dimensional problems, where the moment axis and moment arm are often difficult to establish.

3.2

Moment of a Force about a Specified Axis

In the previous section, the moment of a force is computed about a point. This moment measures the turning effect or tendency of rotation about the point, or actually about the moment axis passing through this point. However, sometimes we want to find the moment of a force about a specified axis, which measures the turning effect about this specific axis. For example, consider the bent rod shown in figure 3.14a, which lies in the horizontal (x–y) plane and is subjected to a vertical force F at point A. In practice, we may want to determine the moment of the force about the a axis, since the rod is only able to rotate about this axis. From figure 3.14b, this moment, denoted as Ma ðFÞ, has a magnitude of

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48

jMa ðFÞj ¼ Fda ¼ 250  0:4 sin h ¼ 250  0:4 

3 ¼ 60 Nm 5

Here da is the perpendicular or shortest distance from the line of action of the force to the axis. The sense of direction, which is also shown in figure 3.14b, is indicated by the right-hand thumb when the right-hand fingers are curled in accordance with the sense of rotation. A unit vector ua is used to define the direction of the a axis. If the sense of direction of Ma ðFÞ is the same as that of ua , Ma ðFÞ is positive, whereas it is negative. Ma ðFÞ in figure 3.14b should be negative since its sense of direction is opposite to ua . Thus, Ma ðFÞ ¼ 60 Nm.

FIG. 3.14 – The moment of a force about a specified axis. Generally, when the perpendicular distance da is determined, the moment of a force F about a specified axis can be represented by the following scalar formulation: Ma ðFÞ ¼ Fda

ð3:4Þ

It should be noted that a force will not create a moment about a specified axis if the force passes through the axis (da ¼ 0) or is parallel to the axis. However, the perpendicular distance da is often not easy to determine. For example, for another force F 0 at point A shown in figure 3.14b, the perpendicular distance from its line of action to the a axis is difficult to establish. For this case, a vector analysis, which is shown afterwards, will be used. The vector formulation of the moment of the force F about point O is


i j k


M O ðFÞ ¼ rA  F ¼

0 0:3 0

¼ f75igNm
0 0 250
The projection of this moment vector along the a axis can be determined from the dot product. Mk ¼ ua M O ðFÞ ¼ f0:8i  0:6jg f75ig ¼ 60 Nm

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49

Then we can see that this projected component is just the moment of the force about the a axis. This two-step solution of first finding the moment of the force about a point on the axis and then finding the projected component of the moment along the axis can also be written as a triple scalar product.


uax uay uaz


ry rz

ð3:5Þ Ma ðFÞ ¼ ua M O ðFÞ ¼ ua  ðrA  FÞ ¼

rx
Fx Fy Fz
where uax, uay, and uaz

specify the x, y, and z components of the unit vector along the a axis; rx, ry, and rz are the x, y, and z components of the position vector r drawn from any point on the a axis to any point on the line of action of the force; Fx, Fy, and Fz are the x, y, and z components of the force vector F. Using this triple scalar product, the moment of the force F 0 in figure 3.14b about the a axis is


0:8 0:6 0


Ma ðF 0 Þ ¼ ua M O ðF 0 Þ ¼ ua ðrA  F 0 Þ ¼

0 0:3 0

¼ 22:8 Nm
80 80 95
The negative sign means the sense of direction of Ma ðF 0 Þ is opposite to ua.

Example 3.8. The handle ABCDE lies in the horizontal (x–y) plane, figure 3.15a. The AB segment is perpendicular to the BC segment and the BC segment is perpendicular to the CDE segment. A force F, which is perpendicular to the y axis and makes an angle a with the plumb line, is acting at point D. If AB ¼ BC ¼ l and CD ¼ a, determine the moments of the force F about the x, y, and z axes, respectively.

FIG. 3.15 – The moments of a force about x, y and z axes. Solution 1: Scalar analysis Since the force is perpendicular to the y axis, it can be resolved into its x and z components and Fx ¼ F sin a, Fz ¼ F cos a, figure 3.15b. The perpendicular

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50

distances of these component forces to the x, y, and z axes are easy to determine. Then, from the principle of moments, we have Mx ðFÞ ¼ Mx ðFx Þ þ Mx ðFz Þ ¼ Mx ðFz Þ ¼ ðl þ aÞF cos a My ðFÞ ¼ My ðFx Þ þ My ðFz Þ ¼ My ðFz Þ ¼ lF cos a Mz ðFÞ ¼ Mz ðFx Þ þ Mz ðFz Þ ¼ Mz ðFx Þ ¼ ðl þ aÞF sin a

You should pay attention to the negative signs. In the above calculation, Mx ðFx Þ ¼ Mz ðFz Þ ¼ 0 and My ðFx Þ ¼ 0, since a force will not create a moment about an axis when the force is parallel to or passes through the axis. Solution 2: Vector analysis First, determine the moment vector of the force F about point A.



i j k



M A ðFÞ ¼ rAD  F ¼
l l þa 0

F sin a 0 F cos a
¼ ðl þ aÞF cos ai  lF cos aj  ðl þ aÞF sin ak Then the i, j and k components of this moment vector are just the three moments about the x, y, and z axes, namely, Mx ðFÞ ¼ ½M A ðFÞx ¼ ðl þ aÞF cos a My ðFÞ ¼ ½M A ðFÞy ¼ lF cos a Mz ðFÞ ¼ ½M A ðFÞz ¼ ðl þ aÞF sin a Of course, we can also use the triple scalar product.



1 0 0


¼ ðl þ aÞF cos a l þa 0 Mx ðFÞ ¼ iM A ðFÞ ¼ i ðrAD  FÞ ¼

l

F sin a 0 F cos a
Readers are recommended to finish the moments about the y and z axes using the triple scalar product by themselves. Example 3.9. A force F is acting on the rod at point B, figure 3.16a. Determine the moment of the force F about the OA axis. Solution: A vector analysis using MOA ðFÞ ¼ uOA  ðr  FÞ will be performed because the perpendicular distance from the line of action of F to the OA axis is difficult to determine. In the triple scalar product, the position vector r can be drawn from any point on the OA axis to any point on the line of action of the force (Why?). Thus, the position vector can be rOC ðrC Þ, rOB ðrB Þ, rAC or rAB , as shown in figure 3.16b.

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51

FIG. 3.16 – Illustration of the triple scalar product Here the solutions using rOC and rAC are illustrated. Readers are suggested to finish the solutions using the other two position vectors by themselves. From figure 3.16a, the coordinates of the points are A(0.4 m, −0.3 m, 0), B(0.3 m, 0.2 m, 0.7 m) and C(0, 0.5 m, 0). Therefore rOA ¼ f0:4i  0:3jg m; rOC ¼ f0:5jg m; rAC ¼ f0:4i þ 0:8jg m; rBC ¼ f0:3i þ 0:3j  0:7kg m rOA ¼

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 0:42 þ ð0:3Þ2 ¼ 0.5 m; rBC ¼ ð0:3Þ2 þ 0:32 þ ð0:7Þ2 ¼ 0.8185 m uOA ¼

F¼F

rOA 0:4i  0:3j ¼ 0:8i  0:6j ¼ 0:5 rOA

rBC 0:3i þ 0:3j  0:7k ¼ f65:97i þ 65:97j  153.9kgN ¼ 180 0.8185 rBC

Then, using the triple scalar product yields

0:8 0:6
MOA ðFÞ ¼ uOA  ðrOC  FÞ ¼

0 0:5
65:97 65:97

MOA ðFÞ ¼ uOA  ðrAC



0:8
 FÞ ¼

0:4
65:97

0:6 0:8 65:97


0

0

¼ 61:6 Nm 153:9

0

0

¼ 61:6 Nm 153:9


The negative sign indicates that the sense of direction of the moment is opposite to that of uOA.

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RETHINK: How about the moment of the force F about the AO axis? Please analyze this by yourself.

3.3 3.3.1

Moment of a Couple Vector Formulation

To turn the wheel when driving, we usually exert two noncollinear parallel forces that have the same magnitude but opposite directions, figure 3.17a. Since the resultant force (vector summation) of these two forces is zero, the only effect these two forces produce is a rotation or tendency of rotation. Look at the general case shown in figure 3.17b, two equal, opposite, and parallel forces F and F 0 that are separated by a perpendicular distance d, only create turning effect and are termed as a couple. The turning effect of the couple is defined by the couple moment, denoted as MðF; F 0 Þ. The magnitude of the couple moment should be Fd (also twice the area of the shaded triangle as shown). The direction of the couple moment is perpendicular to the plane containing these two forces and defined by the right-hand rule, figure 3.17b.

FIG. 3.17 – Couple and couple moment. The couple moment can be determined by finding the sum of the moments of the two couple forces about any arbitrary point. For example, draw position vectors rA and rB from an arbitrary fixed point O to points A and B, which lie on the lines of action of F and F 0 , respectively, figure 3.18a. The sum moment about point O is M O ðF; F 0 Þ ¼ M O ðFÞ þ M O ðF 0 Þ ¼ rA  F þ rB  F 0 ¼ ðrA  rB Þ  F ¼ rBA  F It is easy to prove that rBA  F has the magnitude of Fd and the same direction as that of the couple moment MðF; F 0 Þ. Therefore, the vector formulation of the couple moment is MðF; F 0 Þ ¼ rBA  F

ð3:6Þ

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53

where rBA is a position vector from any point on the line of action of F 0 to any point on the line of action of F due to the principle of transmissibility. Since rBA , not rA and rB directed from the fixed point O, is used to determine MðF; F 0 Þ, the couple moment can act at any point and it is a free vector. However, the moment of a force is a fixed vector, which should be fixed at the point about which the moment is measured. For example, M O ðFÞ should be fixed at point O. Moreover, it has been mentioned in example 3.1 that the force F is a sliding vector.

FIG. 3.18 – Vector formulation of the couple moment. From figure 3.18b, it can be seen that MðF; F 0 Þ ¼ rBA  F ¼ rAB  F 0

ð3:7Þ

It means the couple moment is equal to the position vector directed from any point on one force to any point on the other force cross the other force. Thus, it is easy to get the couple moment M 1 ðF1 ; F 01 Þ in figure 3.19a and M 2 ðF2 ; F 02 Þ in figure 3.19b.

FIG. 3.19 – Determining the couple moment using vector formulation.

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Since couple moments are free vectors, two couples are said to be equivalent if their couple moments have the same magnitude and the same direction. For example, the couple (20 N, 20 N) on plane I in figure 3.20a is equivalent to the couple (40 N, 40 N) on plane Ⅱ in figure 3.20b. They have the same magnitude of 40 N m and the same direction, which is perpendicular to plane I/Ⅱ and points to –y. Also because couple moments are free vectors, several couples acting on a body shown in figure 3.21 can be moved to any arbitrary point and added together vectorially to get the resultant couple moment, namely, X M ¼ M1 þ M2 þ    þ Mn ¼ Mi ð3:8Þ

FIG. 3.20 – Equivalence of two couples in 3-D.

FIG. 3.21 – Couple system in 3-D.

3.3.2

Scalar Formulation

Many problems in mechanics involve coplanar force systems. As shown in figure 3.22, three couples ðF1 ; F 01 Þ, ðF2 ; F 02 Þ and ðF3 ; F 03 Þ lie in the x–y plane. All of the three couple moments act perpendicular to the x–y plane and point to +z (counterclockwise in two-dimensional view) or z (clockwise in two-dimensional view). Therefore, considering positive couple moments as counterclockwise, the couple moment in a coplanar force system can be represented by scalar formulation as follows:

Simplification of Force Systems

55 M ðF; F 0 Þ ¼ Fd

ð3:9Þ

where d is the perpendicular distance or moment arm between the two couple forces. Consequently, the resultant couple moment for the coplanar couple system shown in figure 3.22 can be determined by simply adding the couple moments algebraically as X   X MR ¼ M Fi ; F 0i ¼ Fi di ¼ F1 d1  F2 d2 þ F3 d3

FIG. 3.22 – Couple system in 2-D. Example 3.10. A couple acts on the wheel, figure 3.23a. The radius of the wheel is r = 0.2 m. Replace it by an equivalent couple shown in figure 3.23b and determine the magnitude of the two forces.

FIG. 3.23 – Equivalence of two couples in 2-D. Solution: In figure 3.23a, the couple has a magnitude of Fd ¼ 10  2r ¼ 4 Nm and a direction that is perpendicular to the wheel and outward since the forces tend to rotate the wheel counterclockwise. In figure 3.23b, the couple forces also create a counterclockwise turning effect. To get an equivalent couple, the magnitude of the couple should also be 4 Nm. Therefore, M ðF; F 0 Þ ¼ Fd ¼ F  0:2 ¼ 4 Nm F ¼ 20 N

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Since two couples in figure 3.23a and b are equivalent, afterwards we may use an arc line with an arrowhead to represent a couple, as shown in figure 3.23c. The arc line lies in the plane containing the couple forces and the arrowhead indicates the sense of rotation. Example 3.11. Each force shown in figure 3.24a has a magnitude of F = 10 kN. Determine the couple moment.

FIG. 3.24 – Determination of the couple moment in 2-D. Solution: Here it is somewhat difficult to determine the perpendicular distance between the two parallel forces and compute the couple moment as M = Fd. Instead, we can resolve each force into its horizontal and vertical components, as shown in figure 3.24b. The magnitudes of the component forces are Fx ¼ 3F=5 ¼ 6 kN;

Fy ¼ 4F=5 ¼ 8 kN

This means the couple ðF; F 0 Þ in figure  3.24ais replaced by two component   0 couples in figure 3.24b, i.e., Fx ; F x and Fy ; F 0y . Considering positive moments as counterclockwise, we have     M ðF; F 0 Þ ¼ M Fx ; F 0x þ M Fy ; F 0y ¼ 6  2:5 þ 8  3 ¼ 9 kNm RETHINK: The couple moment can be determined about any point, for example, about point O, or A, or B. Readers are suggested to finish the calculations by themselves and understand which solution is easier. Example 3.12. Each force shown in figure 3.25 has a magnitude of F = 800 N. Determine the couple moment and express the result as a Cartesian vector.

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57

FIG. 3.25 – Determination of the couple moment in 3-D. Solution 1: The coordinates of the points are A(−2 m, 3.3 m, 2 m), B(1 m, −1 m, 2 m) and C(2 m, 2.5 m, 0). Then, we can write the position vectors and the force vector. rAC ¼ f4i  0:8j  2kg m; rBA ¼ f3i þ 4:3jgm F¼F

rAC 4i  0:8j  2k ¼ 800 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ f704:4i  140:9j  352.2kgN rAC 42 þ ð0:8Þ2 þ ð2Þ2

Applying the vector cross product yields


i j k


4:3 0

¼ f1514i  1057j  2606kgNm M ¼ rBA  F ¼

3
704:4 140:9 352:2
Solution 2: The couple moment can be determined by adding the moments of the two couple forces about any point. If point O is chosen, we have M ¼ M O ðFÞ þ M O ðFÞ ¼ rA  F þ rB  ðFÞ



i i j j k







1 ¼
2 3:3 2
þ
1



704:4 140:9 352:2

704:4 140:9





2

352:2
k

¼ f880i þ 704j  2042kg þ f634i  1761j  564kg ¼ f1514i  1057j  2606kgNm Apparently, solution 1 is much easier than solution 2.

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3.4 3.4.1

Simplification of a System of Forces and Couples Equivalent System

When a number of forces and couples are applied on a body, it is difficult to identify their overall effect on the body. If a complicated force system shown in figure 3.26a can be simplified into one force plus one couple moment in figure 3.26b having the same “external” effects, their overall effect is easier to understand and these two systems are called equivalent systems. Here external effects are opposite to internal effects. Internal effects include deformation, heating etc., whereas, external effects mean the effects of translating and rotating a body. In engineering mechanics, we only consider external effects. In this subsection, we will first observe two simple cases of equivalent system having the same external effects. One case is to move a force along its line of action, as shown in figure 3.27. In example 3.1, it has been illustrated that a force is a sliding vector and can therefore act at any point on the body along its line of action and still creates the same effect of rotation. Apparently, it also creates the same translational effect. So, the equivalency has been maintained when moving the force F along its line of action, from point A, figure 3.27a, to any other point B on the body, figure 3.27b. This has been referred to as the principle of transmissibility.

FIG. 3.26 – Equivalent force system.

FIG. 3.27 – Moving a force along its line of action. Another case is to move a force off its line of action, as shown in figure 3.28. A force F is acting at point A, and another point B on the body is located at a distance d from the force line of action, figure 3.28a. In order to move the force from

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59

point A to point B without changing the external effects on the body, we can add equal but opposite forces F and –F at B first, figure 3.28b. Then the two forces with slashes across them form a couple. Using a position vector rBA, the couple moment is determined by M ¼ rBA  F ¼ M B ðFÞ It should be pointed out that although the couple moment M ¼ M B ðFÞ, M is essentially a free vector and it may be applied at any point on the body, not just point B. Finally, F acts at point B as required, but with an additional couple moment M, figure 3.28c. The equivalency has been maintained during this procedure, as indicated by the equal signs between the diagrams.

FIG. 3.28 – Moving a force off its line of action.

3.4.2

Reduction to One Force and One Couple

Using the abovementioned procedure of moving a force off its line of action, we can simplify a complicated system of forces and couples shown in figure 3.29a. To do this, choose an arbitrary point, for example, point O, as the simplified point. Then move the forces F1 and F2 off their lines of action to point O and the corresponding couple moments M 1 ¼ M O ðF1 Þ and M 2 ¼ M O ðF2 Þ have to be applied to the body to produce an equivalent effect, figure 3.29b. For the couple moments Mc1 and Mc2, they can just be moved to point O since they are free vectors. Moreover, all the forces can be reduced to a resultant force vector and all the couples can be reduced to a resultant couple vector as follows: F 0R ¼ F1 þ F2 M RO ¼ M O ðF1 Þ þ M O ðF2 Þ þ M c1 þ M c2 So, the previous complicated force system shown in figure 3.29a is equivalent to one force at point O plus one couple shown in figure 3.29c. Note that M RO is a free vector and can act at any point on the body, for example, at point P shown in figure 3.29d.

60

Engineering Mechanics

If another point O′ is chosen as the simplified point, move the forces to point O′ and add the compensating couple moments M 01 ¼ M O0 ðF1 Þ and M 02 ¼ M O0 ðF2 Þ. 0 Finally, we will get the same P resultantPforce F R ¼ F1 þ F2 , but different resultant couple moment M RO0 ¼ M O 0 ðFi Þ þ M cj , as shown in figure 3.29e. So, it is concluded that the resultant force is independent of the simplified point; however, the resultant couple moment depends upon this point.

FIG. 3.29 – Simplifying a 3-D force system to a resultant force and a resultant couple. The above method of simplifying a system of several forces and couples to a resultant force and a resultant couple moment can be mathematically represented by the following equations. P F 0R ¼ P Fi P ð3:10Þ M RO ¼ M O ðFi Þ þ M cj Equation (3.10) states that the resultant force is the vector summation of all the forces; and the resultant couple moment is the sum of the moments of all the forces about the simplified point plus all the couple moments. For a coplanar force system, similar simplifying procedure can be utilized, as shown in figure 3.30. However, for this case, the couple moments Mcj and the compensating couple moments MO ðFi Þ when moving the forces to point O can both be represented by scalars, since they are always along the z axis. Therefore, for a coplanar force system, equation (3.10) can also be rewritten as

Simplification of Force Systems 0 FRx ¼ 0 FRy ¼

MRO ¼

61 X X X

Fxi Fyi M O ðF i Þ þ

X

ð3:11Þ Mcj

FIG. 3.30 – Simplifying a 2-D force system to a resultant force and a resultant couple. Example 3.13. A beam is subjected to three forces: F1 ¼ 2 kN, F2 ¼ 1 kN and F3 ¼ 5 kN, figure 3.31. If l = 3 m, a = 15 cm, and b = 20 cm, replace this force system by an equivalent force-couple system at point O.

FIG. 3.31 – Replacing a 3-D force system by an equivalent force-couple system at a point.

Solution: There is no couple in this system. Expressing the forces as Cartesian vectors, we have F1 ¼ f2kg kN; F2 ¼ f1ig kN

Engineering Mechanics

62  F3 ¼

 20 15 5 i 5 k ¼ f4i  3kg kN 25 25

Then, the resultant force of the system is the vector summation of all the forces. X Fi ¼ F1 þ F2 þ F3 ¼ f3i  5kg kN F 0R ¼ The resultant couple moment of the system is the sum of the moments of all the forces about point O. X M RO ¼ M O ðFi Þ ¼ rA  F1 þ rB  F2 þ rB  F3






i k

j k



i j j k



i






¼
0 1:5 0
þ
0 3 0
þ
0 3 0







0 0 2

1 0 0

4 0 3
¼ f12i  9kgkNm Example 3.14. Replace the force system shown in figure 3.32a by an equivalent force-couple system at point O.

FIG. 3.32 – Replacing a 2-D force system by an equivalent force-couple system at a point.

Solution: The x and y components of the resultant force are X 5 0 FRx ¼ Fxi ¼ 20   28 sin 60 ¼ 16:6 kN 13 0 ¼ FRy

X

Fyi ¼ 20 

12  28 cos 60 ¼ 5:96 kN 13

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63

Then, the magnitude and coordinate direction angle of the resultant force are qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi FR0 ¼ ð16:6Þ2 þ 5:962 ¼ 17:6 kN b ¼ arccos

0 FRy 5:96 ¼ 70:2 ¼ arccos 17:6 FR0

To calculate the resultant couple moment, the principle of moments will be applied to the two forces and the moments of their rectangular components will be considered. Taking moments as positive when counterclockwise, i.e., in the +z direction, we have X X MRO ¼ M O ðF i Þ þ Mcj ¼ 20 

12  3 þ 28 sin 60  3:5 þ 28 cos 60  1  8 ¼ 146 kNm 13

Therefore, the force system in figure 3.32a is equivalent to a force at point O and a counterclockwise couple moment in figure 3.32b.

3.4.3

Further Simplification

In the previous subsection, a complicated system of forces and couples have been simplified to a resultant force plus a resultant couple. Here we will discuss possible further simplification. Case 1 F 0R ?M RO If the resultant force F 0R is perpendicular to the resultant couple moment M RO , figure 3.33a, we can further simplify by moving F 0R off its line of action and get a single resultant force FR ¼ F 0R at another point P, figure 3.33b. Point P is located on the axis perpendicular to both F 0R and M RO , and at a distance d from point O. To maintain the equivalence between the two diagrams, the distance d should be d¼

MRO FR0

ð3:12Þ

With FR so located, the moment of FR about point O has the same magnitude and the same direction as M RO . For a coplanar force system, the resultant couple moment MRO is always along the z axis and then perpendicular to the resultant force F 0R , figure 3.34a. So, a coplanar force system can always be further reduced to a single resultant force by positioning FR a distance d from point O so as to create the same moment MRO about O, figure 3.34b. The distance d is also calculated by equation (3.12).

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Besides coplanar force system, both concurrent and parallel force systems can always be reduced to a single resultant force. The concurrent force system has been discussed in chapter 2. The reduction of the parallel force system will be illustrated in example 3.16.

FIG. 3.33 – Further simplification to a signle force.

FIG. 3.34 – Further simplification to a signle force in 2-D. F 0R

Case 2 k M RO If the resultant force F 0R is parallel to the resultant couple moment M RO , figure 3.35a, this cannot be further simplified. This combination of a collinear force and couple moment is referred to as a wrench. The line of action of the force is also called the axis of the wrench. The wrench tends to cause both translation along the axis and rotation about the axis. When you use a screwdriver to put a screw into the wood, figure 3.35b, you are exerting a wrench, i.e., a force to push in the screw and at the same time a couple to rotate the screw.

FIG. 3.35 – A wrench.

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65

Case 3 F 0R making an arbitrary angle θ with M RO In a general case, the resultant force F 0R would make an arbitrary angle θ with the resultant couple moment M RO , figure 3.36a. For this case, resolve M RO to two components: one component M ? that is perpendicular to F 0R , and the other component M k that is parallel to F 0R , figure 3.36b. As mentioned earlier, the perpendicular component M ? can be eliminated by moving F 0R to point P, figure 3.36c. To maintain the equivalency, the distance d from P to O should be d¼

M? FR0

ð3:13Þ

Since M k is a free vector, it can just move to point P, and a combination of collinear force and couple, or a wrench, is obtained, figure 3.36c. The axis of the wrench passes through P. Therefore, a general system of forces and couples can usually be reduced to a wrench.

FIG. 3.36 – Further simplification to a wrench. Example 3.15. The frame is subjected to a coplanar force system, figure 3.37a. Replace this loading system by a single resultant force and specify its line of action.

FIG. 3.37 – Replacing a 2-D force system by a single resultant force.

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Solution: First, replace this force system equivalently by a resultant force acting at point A and a resultant couple. Two rectangular components of the resultant force are X 0 FRx ¼ Fxi ¼ 50 sin 30 ¼ 25 N 0 ¼ FRy

X

Fyi ¼ 10  20  50 cos 30 ¼ 73:3 N

Then, the magnitude and direction angle of the resultant force are qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi FR0 ¼ ð25Þ2 þ ð73:3Þ2 ¼ 77:4 N


F 0
25
Rx
¼ 18:8 h ¼ arctan
0
¼ arctan
FRy
73:3 Taking moments as positive when counterclockwise, the resultant couple moment is X X MRA ¼ MA ðFi Þ þ Mcj ¼ 20  2 þ 50 sin 30  5  50 cos 30  5 þ 200 ¼ 68:5 Nm Therefore, the force system in figure 3.37a is equivalent to a force at point A and a counterclockwise couple in figure 3.37b. This force and this couple can be further reduced to a single resultant force FR, as shown in figure 3.37c. The line of action of the single resultant force can be represented by the distance d away from point A and the distance is d¼

MRA 68:5 ¼ 0:885 m ¼ 77:4 FR0

We can also assume the single resultant force passes through an arbitrary point P (x, y). From the equivalency between the diagrams, the moment about point A of the resultant force at P in figure 3.37c should be equal to MRA in figure 3.37b. Therefore, we have MRA ¼ MA ðFR Þ ¼ 25y  73:3x ¼ 68:5 This is the expression of the line of action of the single force FR. From the principle of transmissibility, the position of the line of action can also be represented by two intersecting distances, xR and yR, on the two axes, as shown in figure 3.37c. Setting x = 0 yields 25yR  73:3  0 ¼ 68:5

)

yR ¼ 2:74 m

Simplification of Force Systems

67

Setting y = 0 yields 25  0  73:3xR ¼ 68:5

)

xR ¼ 0:935 m

Example 3.16. Prove that the parallel force system acting on the plate in figure 3.38a can be reduced to a single resultant force. Determine the resultant force and locate its point of application on the plate.

FIG. 3.38 – Replacing a parallel force system by a single resultant force.

Solution 1: Vector analysis First, replace this parallel force system by an equivalent force-couple system at point O. X F 0R ¼ Fi ¼ 350k þ 500k  550k ¼ f400kgN M RO ¼

X



i

¼
4

0

M O ðFi Þ ¼ rA  FA þ rB  FB þ rC  FC




k

k



i j j k



i j




0
0
þ
4 3 0 0
þ
0 3




0 350

0 0 500

0 0 550


¼ f150i þ 3600jgNm From their Cartesian vector forms, the resultant force F 0R is perpendicular to the resultant couple moment M RO . Actually, when each force, which has only k component, is moved off its line of action to any arbitrary point, the compensating couple moment will have only i and j components. Therefore, the resultant force having only k component will always be perpendicular to the resultant couple moment and the parallel force system can always be reduced to a single resultant

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force. Assume the single force FR acts at point P (x, y) on the plate, as shown in figure 3.38b. From the principle of moments, we have X M O ðFi Þ ¼ M RO M O ðFR Þ ¼

i

x

0

j y 0


k

0

¼ 400yi þ 400xj ¼ 150i þ 3600j 400


Equating the corresponding i and j components yields 400y ¼ 150 400x ¼ 3600 Solving these equations yields the coordinates of point P. x ¼ 9 m; y ¼ 0:375 m Solution 2: Scalar analysis A parallel force system can be reduced to a single resultant force FR . This force is also parallel to every force, i.e., parallel to the z axis in this problem. Its magnitude and sense of direction is determined as follows: X þ " FR ¼ Fi ; FR ¼ 350  550 þ 500¼  400 Nð#Þ Assume FR acts at point P (x, y) on the plate, figure 3.38b. From the principle of moments, the moment of the single resultant force about the x axis should be equal to the sum of the moments of all the forces shown in figure 3.38a about the same axis. Using the right-hand rule, where positive moments act in the +i direction, we have X M x ðF R Þ ¼ Mx ðFi Þ;  400y ¼ 550  3 þ 500  3 ) y ¼ 0:375 m Similarly, taking positive moments as in the +j direction, a moment equation can be written about the y axis. X My ðFR Þ ¼ My ðFi Þ; 400x ¼ 550  4 þ 350  4 ) x ¼ 9 m RETHINK: The coordinates of point P can also be determined by realizing that the moment of the single resultant force about point P is equal to the sum of the moments of all the forces shown in figure 3.38a about P. Readers are recommended to finish this by themselves. Example 3.17. Prove that the force system acting on the thin plate in figure 3.39a will be reduced to a wrench. Specify the force and the couple moment of the wrench and the point P (y, z) where the wrench’s line of action intersects the plate.

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69

FIG. 3.39 – Replacing a 3-D force system by a wrench.

Solution: First, replace this force system by an equivalent force-couple system at point O. X F 0R ¼ Fi ¼ FA þ FB ¼ f80i þ 100jgN M RO ¼

P

M O ðFi Þ þ M c ¼ rA  FA þ rB  FB þ M c




i j k

j k



i




¼
0 0 8
þ
0 10 4
þ f10kg ¼ f400i  640j  10kgNm;




80 0 0

0 100 0


The angle between F 0R and M RO is

  0 F R M RO 1 ð80Þ  ð400Þ þ 100  ð640Þ þ 0  ð10Þ h ¼ cos1 ¼ cos 128  755 FR0 MRO  ¼ 109 So, the force system will be reduced to a wrench. Assume the wrench’s line of action intersects the plate at point P (y, z), figure 3.39b. Because M RP should have the same direction as F 0R , we assume M RP ¼ kM ð80i þ 100jÞ ¼ f80kM i þ 100kM jgNm Since the wrench in figure 3.39b is equivalent to the force system in figure 3.39a, we have


i j k


M RO ¼ rP  F 0R þ M RP ¼

0 y z

þ ð80kM i þ 100kM jÞ
80 100 0
¼ ð100z þ 80kM Þi  ð80z  100kM Þj þ 80yk

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Substituting equation (1) and equating the corresponding i, j and k components yield  ð100z þ 80kM Þ ¼ 400  ð80z  100kM Þ ¼ 640 80y ¼ 10 Solving these equations, we get y ¼ 0:125 m, z ¼ 5:56 m, kM ¼ 1:95 Thus, the couple moment of the wrench is M RP ¼ f156i  195jgNm

3.5 3.5.1

Application of Simplification of Parallel Forces Center of Gravity, Center of Mass, and Centroid of a Body

If a body is divided into a number of particles, the weights of the particles comprise a system of parallel forces, as shown in figure 3.40a.1 From the previous section, we know that a parallel force system can always be reduced to a single resultant force. The magnitude of the resultant force, or the weight of the whole body, is the sum of the weights of all the particles. X P¼ Pi Assume the resultant weight acts at point C ðxC ; yC ; zC Þ, and this point is referred to as the center of gravity. From the principle of moments, the moment about an axis (or a point) of the resultant weight should be equal to the sum of the moments about the same axis (or the same point) of all the particle weights. Thus, summing moments about the x and y axes, respectively, yields P P X X Pi yi Pi yi Mx ðPÞ ¼ ¼ P Mx ðPi Þ; PyC ¼  Pi yi ) yC ¼ P Pi

1

It should be noted that in the exact sense, the weights are not parallel to each other, but concurrent at the Earth’s center due to Newton’s law of gravitational attraction. At the same time, the acceleration of gravity g is not constant and depends on the distance from the Earth’s center to the particle. However, since the Earth is far larger than the rigid body, which is located near the surface of the Earth in most engineering problems, these effects are generally neglected.

Simplification of Force Systems

My ðPÞ ¼

X

My ðPi Þ;

PxC ¼

71 X

P P i xi

)

xC ¼

P P i xi P i xi ¼ P P Pi

Since all the weights make zero moment about the z axis, we cannot get zC directly. Imaging that the coordinate system along with the body fixed in it is rotated 90º about the x axis, we have figure 3.40b. For this case, summing moments about the x axis yields P P X X Pi z i Pi zi ¼ P Mx ðPÞ ¼ Mx ðPi Þ; PzC ¼ Pi zi ) zC ¼ P Pi So, the coordinates of the center of gravity C are P P P P P P Pi x i Pi x i Pi yi Pi y i Pi zi P i zi ¼ P ¼ P ¼ P ; yC ¼ ; zC ¼ ð3:14Þ xC ¼ P P P Pi Pi Pi

FIG. 3.40 – Determining the center of gravity from the principle of moments Generally, a rigid body is divided into an infinite number of particles, as shown in figure 3.41. Then, integration rather than a discrete summation should be utilized and equation (3.14) becomes R R R R R R ~z dP ~z dP x~dP x~dP y~dP y~dP xC ¼ P ; yC ¼ P ; zC ¼ P ð3:15Þ ¼ RP ¼ RP ¼ RP P P P dP dP P P P dP where x~; y~; ~z xC, yC, and zC

are the x, y, and z coordinates of the center of gravity of the differential element; are the x, y, and z coordinates of the center of gravity of the rigid body.

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FIG. 3.41 – Center of gravity. Substituting dP ¼ gdm into equation (3.15) yields R R R ~gdm ~dm x~dm mx mx R R ¼ ¼ m xC ¼ m gdm dm Rm Rm R ~ y gdm y~dm y~dm m m yC ¼ R ¼ R ¼ m m gdm dm Rm Rm R ~z gdm ~z dm ~z dm zC ¼ Rm ¼ Rm ¼ m m gdm dm m m

9 > > > > > > > > = > > > > > > > > ;

ð3:16Þ

Here the acceleration of gravity g is considered constant throughout the body, taken out of the integration, and then cancelled from both the numerators and denominators. The point determined by equation (3.16) is the body’s center of mass, which is a very important concept in dynamics. By comparison, the center of gravity coincides with the center of mass. However, note that the center of gravity only exists when under the influence of a gravitational attraction, whereas the center of mass is independent of gravity. If γ represents the specific weight of the body, measured as a weight per unit volume, substituting dP = γdV into equation (3.15) yields R R R ~z cdV x~cdV y~cdV V V xC ¼ R ; yC ¼ R ; zC ¼ RV ð3:17Þ V cdV V cdV V cdV If the material composing the body is homogeneous or uniform, γ will be constant throughout the body. Therefore, γ can be taken out of the integrals and cancelled from equation (3.17) and we get R R R R R R ~dV ~dV ~dV ~dV ~z dV z dV V x V x V y V y V ~ R R R ; yC ¼ ; zC ¼ xC ¼ ¼ ¼ ¼ V V V V dV dV dV V V V ð3:18Þ Equation (3.18) defines the centroid or geometric center of a body or volume. By comparison, the centroid coincides with the center of gravity/mass when the material is homogeneous.

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73

An area can be treated as a plate with unit thickness. Then substituting dV = 1dA into equation (3.18) yields the coordinates of the centroid of an area: R R R R R R ~z dA ~z dA x~dA x~dA y~dA y~dA xC ¼ RA ; yC ¼ RA ; zC ¼ RA ¼ A ¼ A ¼ A A A A dA dA dA A A A ð3:19Þ A line can be treated as a thin rod or wire with unit cross-section area. Then substituting dV = 1dL into equation (3.18) yields the coordinates of the centroid of a line: R R R R R R ~dL ~dL ~dL ~dL ~z dL z dL Lx Lx Ly Ly L~ R R R xC ¼ ; yC ¼ ; zC ¼ ð3:20Þ ¼ ¼ ¼ L L L L dL dL dL L L L The procedure of determining the center of gravity/mass of an body or centroid of a shape is summarized as follows: (1) Specify an appropriate coordinate system. For instance, polar coordinates are generally appropriate for shapes having circular boundaries (see examples 3.18 and 3.20). (2) Choose an appropriate differential element for integration. For an area, the differential element dA is often a rectangle having a finite length and a differential width (see examples 3.19 and 3.21); For a revolution body, the differential element dV can be a circular disk having a finite radius and a differential thickness, or a cylindrical shell having a finite length and radius and a differential thickness (see examples 3.22 and 3.23). (3) Express the weight dP, or mass dm, or volume dV, or area dA, or length dL of the differential element and determine the coordinates ðx~; y~; ~z Þ of the center of gravity/mass or centroid of the element. (4) Perform the integration on the entire region using appropriate equation (equations (3.15) through (3.20)). It should be noted that in some cases the centroid is not located on the object, as shown in figure 3.42. If an object has a symmetric axis, then the centroid lies on this axis, as shown in figure 3.43.

FIG. 3.42 – Centroid for a curved line.

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FIG. 3.43 – Centroid for a symmetric area. Example 3.18. Find the location of the centroid of the circular arc shown in figure 3.44.

FIG. 3.44 – Centroid of a circular arc. Solution: Since the y axis is a symmetrical axis, we have xC ¼ 0 and only yC has to be determined by integration. Utilize the polar coordinate system and choose a differential arc element as shown. Its length is dL ¼ Rdh and its centroid is located at y~ ¼ R cos h. Then, from equation (3.20), we have R Ra Ra ~dL R2 a cos hdh 2R2 sin a R sin a Ly a R cos hRdh yC ¼ ¼ ¼ ¼ ¼ 2Ra a R2a 2Ra L When a ¼ p2, i.e., the arc is a semicircle arc, we obtain yC ¼ 2R p. Example 3.19. Determine yC , the y coordinate of the center of gravity of the uniform triangular thin plate shown in figure 3.45.

FIG. 3.45 – Center of gravity of a uniform plate.

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75

Solution: For a uniform thin plate, the center of gravity coincides with the centroid of the area. A differential element parallel to the x axis is used. This differential element can be considered as a rectangle. The centroid of the element is located at y~ ¼ y. Its height is dy and its length b0 can be determined by the proportional relationship. Then the area of the differential element is dA ¼ b0 dy ¼ b

hy dy h

Then, from equation (3.19), we have R h ðhyÞ R yb dy h ~dA Ay yC ¼ ¼ 0 1 h ¼ A 3 2 bh Example 3.20. Determine the centroid of the circular section shown in figure 3.46a.

FIG. 3.46 – Centroid of a circular section. Solution 1: Since the y axis is a symmetrical axis, we have xC ¼ 0 and only yC has to be determined by integration. Utilize the polar coordinate system and choose a differential circular sector element as shown in figure 3.46a. Neglecting higher order differentials, this differential element can be considered as a triangle. Its centroid is located at y~ ¼ 2R cos h=3 from example 3.19 and its area is 1 1 1 dA ¼ RdL ¼ RRdh ¼ R2 dh 2 2 2 Then, from equation (3.19), we have Ra 2 R R cos h 1 R2 dh 23 R3 sin a 2R y~dA A sin a ¼ a 3R a 1 2 2 ¼ yC ¼ R ¼ 3a R2 a A dA a 2 R dh When a ¼ p2, i.e., the area is a half-circle, we get yC ¼ 4R 3p .

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Solution 2: The differential element may be chosen as a circular arc having a thickness dr, figure 3.46b. According to example 3.18, the centroid of this differential element is located at y~ ¼ r sin a=a and its area is dA ¼ 2ardr Then, from equation (3.19), we have R R r sin a R 2ardr 23 sin aR3 2R ~dA Ay R sin a yC ¼ ¼ ¼ ¼ 0 R Ra 3a R2 a 2ardr A dA 0

Example 3.21. Determine the centroid of the shaded area bounded by two curves, figure 3.47a.

FIG. 3.47 – Centroid of an area bounded by two curves. Solution 1: A vertical rectangular element of differential thickness dx shown in figure 3.47b is considered. The element intersects the curves at points (x, y1) and (x, y2), and it has a height (y1–y2). The area of the element is dA = (y1–y2)dx, and its centroid is located at x~ ¼ x; y~ ¼

y1 þ y2 2

Then, from equation (3.19), we have R 4:5 pffiffiffiffiffi 2  R 4:5 R x 2x  x dx 4:05 x ðy1  y2 Þdx x~dA 0 A ¼ 1:8 m xC ¼ R ¼ R0 4:5 pffiffiffiffiffi 3  ¼ ¼ R 4:5 2:25 2x  23 x dx A dA 0 ðy1  y2 Þdx 0  R 4:5 y1 þ y2 R 4:5  R ðy1  y2 Þdx 12 0 2x  49 x 2 dx 3:375 y~dA yC ¼ RA ¼ 1:5 m ¼ 0 R 4:52 ¼ R 4:5 pffiffiffiffiffi  ¼ 2:25 2x  2 x dx ðy1  y2 Þdx A dA 0

0

3

Solution 2: A horizontal rectangular element having a differential thickness dy shown in figure 3.47c is considered. The element intersects the curves at the points (x1, y) and (x2, y) and it has a length (x2–x1). The area of the element is dA = (x2–x1)dy, and its centroid is located at

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77

x~ ¼

x1 þ x2 ; y~ ¼ y 2

Then, from equation (3.19), we have

 R  R 3 x1 þ x2 R y4 1 3 9 2 y  ðx2  x1 Þdy 2 0 4 x~dA 4 dy 4:05  ¼ 1:8 m ¼ xC ¼ RA ¼ R  ¼ 0 R 32 2 3 3 y 2:25 dA ðx2  x1 Þdy A y  dy 0

0

2

2

 R 3 3 R3 R y2 y y  y ðx2  x1 Þdy y~dA 0 2 dy 2 3:375  yC ¼ RA ¼ 1:5 m ¼ R  ¼ ¼ R0 3 2 3 3 y 2:25 dA ðx2  x1 Þdy A y  dy 0

0

2

2

Example 3.22. Locate the centroid of the revolution body, which is generated by revolving the shaded area shown in figure 3.48a about the z axis.

FIG. 3.48 – Centroid of a revolution body.

Solution 1: Due to symmetry, we have xC ¼ yC ¼ 0 and only zC has to be determined by integration. A differential disk element having a thickness dz and a radius r = y is chosen, figure 3.48b. The volume of the disk element is dV = (πy2)dz = πzdz, and its centroid is located at ~z ¼ z. From equation (3.18), we have R R1 R1 2 ~z dV z pzdz z dz V 0 ¼ R1 zC ¼ R ¼ R01 ¼ 0:667 m V dV 0 pzdz 0 zdz Solution 2: As shown in figure 3.48c, a cylindrical shell element having a radius r = y and a thickness dy is chosen. The element intersects the generating curve at point (0, y, z). Thus, the height of the cylindrical shell is (1–z). The volume of the shell element is dV = 2πydy (1–z) = 2πy(1–y2)dy, and its centroid is located at ~z ¼ z þ

1  z 1 þ z 1 þ y2 ¼ ¼ 2 2 2

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Then, applying equation (3.18) yields  R R1 R1  2 2 4 z dV V ~ 0 ð1 þ y Þ2py ð1  y Þdy 0 y 1  y dy R zC ¼ ¼ R1 ¼ 0:667 m ¼ R1 2 0 y ð1  y 2 Þdy 2 0 2py ð1  y 2 Þdy V dV Example 3.23. The revolution body shown in figure 3.49a is made from a material having a density that varies with its height, i.e., ρ = 10(3.5 −z). Locate the center of mass of the body if R = 1.8 m and h = 3 m.

FIG. 3.49 – Center of mass of a revolution body with variable density. Solution: Due to symmetry, we have xC ¼ yC ¼ 0 and only zC has to be determined by integration. A differential disk element having a thickness dz and a radius r = y is chosen, figure 3.49b. The centroid of the disk element is located at ~z ¼ z and its mass is   2pð3:5  zÞz 4 dm ¼ qdV ¼10ð3:5  zÞ py 2 dz ¼ dz 5 Then, applying equation (3.16) yields R 3 2pð3:5zÞz 4 R3 R dz ð3:5  zÞz 5 dz z dm 0 z 5 m~ R zC ¼ ¼ R03 ¼ 2:32 m ¼ R 3 2pð3:5zÞz 4 ð3:5  zÞz 4 dz dz m dm 0

5

0

RETHINK: Can we use a cylindrical shell element (see example 3.22) to solve this problem? Why?

3.5.2

Center of Gravity, Center of Mass, and Centroid of Composite Bodies

Some engineering bodies can be sectioned into several simple-shaped bodies, such as semispheres, cylinders and cuboids. Such bodies are called composite bodies. For a body made from homogeneous material, the center of gravity/mass is just the centroid of the body. Since the centroid and the volume/area/length of simple shapes are usually known (such information can be found in table 3.1), we can then

Shape

Geometric property L ¼ 2ar xC ¼

r sin a a

p 2r If a ¼ (semicircular arc), xC ¼ 2 p

1 A ¼ bh 2 1 yC ¼ h 3

Shape

Geometric property 1 A ¼ ab 3 3 xC ¼ a 4 3 yC ¼ b 10

Simplification of Force Systems

TAB. 3.1 – Geometric properties of simple shapes.

4 V ¼ pr 3 3

2 V ¼ pr 3 3 A ¼ ab

3 zC ¼ r 8

79

Shape

Geometric property

80

TAB. 3.1 – (continued). Shape

Geometric property

A ¼ ar 2 xC ¼

2 r sin a 3 a

V ¼ pr 2 h

p 4r If a ¼ (semicircular area), xC ¼ 2 3p

2 A ¼ ab 3 3 xC ¼ a 5 3 yC ¼ b 8

1 V ¼ pr 2 h 3 1 zC ¼ h 4

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81

locate the centroid of a composite body by just treating each simple-shaped body as a particle and following the procedure mentioned in the previous subsection as follows: P P P P P P Vi x i Vi x i Vi yi Vi y i Vi zi V i zi ¼ P ¼ P ¼ P xC ¼ ; yC ¼ ; zC ¼ ð3:21Þ V V V Vi Vi Vi where xi ; y i ; z i xC, yC, and zC

are the x, y, and z coordinates of the centroid of the i-th simple-shaped body; are the x, y, and z coordinates of the centroid of the composite body.

In equation (3.21), the V’s can be replaced by L’s, A’s, P’s or m’s to locate the centroid or the center of gravity/mass of composite lines, areas, or bodies. Example 3.24. Determine the centroid of the area shown in figure 3.50a.

FIG. 3.50 – Centroid of a composite area.

Solution 1: The area can be divided into two rectangular segments as shown in figure 3.50a. The location of the centroid and area of each rectangular segment are easy to obtain. For the sake of brevity, the calculations are tabulated as follows: Segment 1 2 P

Thus,

Ai (cm2) 6 8 14

xi (cm) 0.5 5

yi (cm) 3 0.5

Aixi (cm3) 3 40 43

P Ai xi 43 ¼ 3:07 cm ¼ xC ¼ P 14 Ai

Aiyi (cm3) 18 4 22

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P Ai yi 22 ¼ 1:57 cm yC ¼ P ¼ 14 Ai Solution 2: The area can also be treated as a big rectangle ABDE minus a small rectangle A0 B 0 D0 E , as shown in figure 3.50b. Here the area of thePsmall P rectangle is considered “negative”, and the negative area will be used in A , Ai xi and i P Ai yi . The calculations are tabulated as follows: Segment 1 2 P

Ai (cm2) 54 –40 14

xi (cm) 4.5 5

yi (cm) 3 3.5

Aixi (cm3) 243 –200 43

Aiyi (cm3) 162 –140 22

The following calculation is the same as above and omitted. In practice, we may encounter irregular shaped bodies or even heterogeneous bodies. For this kind of bodies, the center of gravity/mass may be determined through experiments, for example, the suspension method shown in figure 3.51.

FIG. 3.51 – Determining the center of gravity through the suspension method.

3.5.3

Reduction in Distributed Loading

The surface area of a body is often subjected to a distributed loading such as that caused by wind, snow, or the weight of objects over the body’s surface, figure 3.52. In this subsection we will consider a common case of a distributed loading on a flat surface, as shown in figure 3.53a. The intensity of the loading at point (x, y) on the surface is defined as the pressure p = p(x, y), which represents the force per unit area and is measured in pascals (Pa), where 1 Pa = 1 N/m2. On a differential area dA, located at point (x, y), there is a differential force. dF ¼ pðx; yÞdA

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83

FIG. 3.52 – Distributed loadings in practical.

FIG. 3.53 – Distributed loading on a flat surface. All the differential forces on the entire surface area constitute a system of parallel forces infinite in number. This parallel force system can always be simplified to a single resultant force FR, as shown in figure 3.53b. The magnitude of FR is therefore expressed as an integral: Z Z Z FR ¼ dF ¼ pðx; y ÞdA ¼ dV ð3:22Þ A

V

where pðx; y ÞdA ¼ dV is the volume of the colored differential element. Therefore, the magnitude of FR is equal to the total volume under the distributed-loading diagram p = p(x, y). The location of FR is determined by using the principle of moments. Equating the moment of FR to the sum of the moments of all the differential forces about the y or x axis, respectively, we will obtain R R R R xdV ydV xpðx; y ÞdA ypðx; y ÞdA V A A ¼ R ; y¼ R ¼ RV ð3:23Þ x¼ R A pðx; y ÞdA V dV A pðx; y ÞdA V dV

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This means FR passes through the centroid of the volume under the distributed-loading diagram. Sometimes, the distributed loading is uniform along one axis of a flat rectangular surface. An example of such a loading is the water pressure on a flat rectangular sluice gate, as shown in figure 3.54. Here the loading function is a function of only x, i.e., p = p(x), since the pressure is uniform along the y axis, as shown in figure 3.54a2. Multiplying p(x) by the width b of the rectangular area along the y axis, we get q(x) = p(x)b. This q(x) should have the units of N/m, which represents the force per unit length along the x axis. Consequently, the distributed-loading diagram for q(x) can be represented by a coplanar parallel force system, figure 3.54b. This parallel force system can always be simplified to a single resultant force FR. Following the similar procedure mentioned above, the magnitude and location of FR are determined by the following formulae: Z Z FR ¼ q ðx Þdx ¼ dA ¼ A ð3:24Þ L

A

R R R xdA L xq ðx Þdx A xdA R R x¼ ¼ ¼ A A q ð x Þdx dA L A

ð3:25Þ

FIG. 3.54 – Simple distributed loading with uniform intensity along an axis. Therefore, the magnitude of FR is equal to the total area under the distributed-loading diagram q = q(x), and its line of action passes through the centroid of the area. However, it should be noted that the resultant force FR is actually acting at ðx; 0Þ on the surface by symmetry, figure 3.54a. For the distributed loading in a simple shape, such as rectangle or triangle, the information of the area and the centroid can be found in table 3.1. Thus, integration is not necessary and the results are shown in figure 3.55a and b. For a trapezoidal distributed loading shown in figure 3.55c, the trapezoid can be divided into a rectangle and a triangle. Then the distributed loading is simplified to two parallel forces For water pressure shown in figure 3.54, the intensity p0 ¼qgh, where ρ is the density of the water, g is the acceleration of gravity, and h is the depth of the point from the fluid surface. 2

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85

FR1 and FR2. Of course, these two parallel forces can be further reduced to a single resultant force. However, usually, we just use these two parallel forces to represent the trapezoidal distributed loading to participate in the subsequent calculations.

FIG. 3.55 – Simplification of distributed loading in simple shape.

Example 3.25. A distributed loading acts on the beam, figure 3.56a. Determine the magnitude and location of the equivalent concentrated force.

FIG. 3.56 – Replacing a distributed loading by an equivalent concentrated force.

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Solution: Considering the colored differential element shown in figure 3.56a and applying equations (3.24) and (3.25), we obtain Z Z 4 pffiffiffi 80 x dx ¼ 427 N FR ¼ q ðx Þdx ¼ L

0

R4 R pffiffiffi x 80 x dx 1024 xq ðx Þdx 0 L ¼ 2:4 m ¼ ¼ x¼ R 427 427 L q ðx Þdx The resultant force is shown in figure 3.56b. RETHINK: The results can also be directly determined by using table 3.1. From the table, for a semi-parabolic area of length a = 4 m, height b = 160 N/m, the area and the location of the centroid are 2 2  4  160 A ¼ ab ¼ ¼ 427 N ¼ FR 3 3 3 3 xC ¼ a ¼  4 ¼ 2:4 m ¼ x 5 5 Example 3.26. The cross-section shape of the gravity dam is shown in figure 3.57a. For convenience, the unit length of dam is considered. The specific weights of the dam and water are γd = 23.5 kN/m3 and γw = 9.81 kN/m3, respectively. Replace the force system of weight and water pressure by a single resultant force.

FIG. 3.57 – Simplifying a force system of dam weight and water pressure. Solution: First, divide the dam into two segments: one rectangle and one triangle, as shown in figure 3.57a. The weights of the two segments are P1 ¼ cd Vd1 ¼ 23:5  ð8  50  1Þ ¼ 9400 kN

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87

P2 ¼ cd Vd2 ¼ 23:5 

1 2

  36  50  1 ¼ 21 150 kN

The locations of the weights are defined by xC1 ¼ 4 m; xC2 ¼ 20 m The water pressure forms a triangular distributed loading, figure 3.57a. The intensity at the bottom is q0 ¼ cw hb ¼ 9:81  45  1 ¼ 441:45 kN/m The equivalent resultant force of the water pressure shown in figure 3.57a is 1 1 Fw ¼ q0 h ¼  441:45  45 ¼ 9932 kN 2 2 Its location is defined by yC3 ¼ 15 m Then, replace this force system by an equivalent force-couple system at point O, figure 3.57b. The rectangular components of the resultant force are X 0 FRx ¼ Fxi ¼ Fw ¼ 9932 kN 0 FRy ¼

X

Fyi ¼ P1  P2 ¼ 9400  21 150 ¼ 30 550 kN

Then, the magnitude and direction angle of the resultant force are qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi FR0 ¼ 99322 þ ð30 550Þ2 ¼ 32 124 kN h ¼ arccos

0 FRx 9932 ¼ 72 ¼ arccos 32 124 FR0

Taking moments as positive when counterclockwise, the couple moment is X MRO ¼ MO ðFi Þ ¼ P1 xC1  P2 xC2  Fw yC3 ¼ 9400  4  21 150  20  9932  15 ¼ 609 580 kNm Therefore, the force system in figure 3.57a is equivalent to a force and a clockwise couple moment at point O in figure 3.57b. This force and couple can be further reduced to a single resultant force FR, figure 3.57c. The line of action of this single resultant force can be represented by the distance d away from point O and the distance is


MRO
609 580 d ¼

0

¼ ¼ 19:0 m 32 124 FR

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88

RETHINK: The line of action of the single resultant force can also be defined by two intersecting distances on the two axes, i.e., xR and yR, as shown in figure 3.57c. Readers are suggested to finish the calculation by themselves. PROBLEMS 3.1 A force F = 80 N acts along the diagonal of the cuboid. Use MO = rA × F and MO = rB × F, respectively, to determine the moment of F about point O and verify that a force is a sliding vector and can therefore act at any point along its line of action.

Prob. 3.1

3.2 A force F = 80 N acts along the diagonal of the cuboid. Use MC = rCA × F and MC = rCB × F, respectively, to determine the moment of F about point C.

Prob. 3.2

Simplification of Force Systems

89

3.3 Determine the moment of the force F about point A.

Prob. 3.3

3.4 Two forces FA = 50 N and FB = 45 N are exerted on the rod. Find the moment of each force about joint O. Which way will the rod rotate?

Prob. 3.4

3.5 Determine the moments of the force F about point A and point B, respectively.

Prob. 3.5

Engineering Mechanics

90

3.6 Find the moment of the force F about joint O in the shown two cases.

Prob. 3.6

3.7 If the force F = 20 N makes an angle 30° with line AC, and OA = OC = 5 m, determine the moments of the force about the x, y, and z axes, respectively.

Prob. 3.7

3.8 Determine the moments of the force F about the x, y, and z axes, respectively if F = 5 kN and OA = OC = OD = 1 m.

Prob. 3.8

Simplification of Force Systems

91

3.9 Determine the moments of the force F about the a axis and the y axis, respectively.

Prob. 3.9

3.10 Determine the moment of the force F about the z axis if F = 1 kN.

Prob. 3.10

Engineering Mechanics

92

3.11 Three couples act on the triangular prism as shown. If F1 ¼ 100 kN, F2 ¼ 200 kN, F3 ¼ 300 kN, a ¼ 1 m and b ¼ 0:8 m, determine the resultant couple moment.

Prob. 3.11

3.12 Three bars OA, OB and OC are fixed together and lie in the x–y plane. Three disks A, B and C, which have the radii of 15 cm, 10 cm and 5 cm, respectively, are attached to the three bars and perpendicular to them, respectively. Three couples are acting on the disks as shown. If the resultant couple moment is zero, specify the magnitude of F and the angle a.

Prob. 3.12

Simplification of Force Systems

93

3.13 Five couples act on the member as shown. Each couple has a couple moment of 800 Nm. If tan a ¼ 0:75, find the magnitude and coordinate direction angles of the resultant couple moment.

Prob. 3.13

3.14 Four forces act at point O as show. If F1 ¼ 500 N; F2 ¼ 300 N; F3 ¼ 600 N and F4 ¼ 1000 N, determine the resultant force.

Prob. 3.14

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94

3.15 Replace the force system by a single resultant force. The side length of each square in the picture is 1 m.

Prob. 3.15

3.16 The square plate ABCD has a side length of a. Three forces and a couple are acting on the plate. Replace the force system by a single resultant force.

Prob. 3.16

Simplification of Force Systems

95

3.17 The gravity dam of a unit length is considered. The weight of the dam comprises P1 ¼ 300 kN and P2 ¼ 140 kN. The water pressures of the upstream and downstream are F1 ¼ 250 kN and F2 ¼ 70 kN, respectively. Replace the force system of weight and water pressure by a single resultant force.

Prob. 3.17

3.18 Four forces F1 ¼ F2 ¼ F3 ¼ F4 ¼ F are acting on the square plate and AB ¼ BC ¼ CD ¼ DE ¼ a. Replace this force system by an equivalent force-couple system at point O. How about the result if point B is chosen as the simplified point?

Prob. 3.18

Engineering Mechanics

96

3.19 Three forces F1 ¼ F2 ¼ F3 ¼ F are acting on the regular triangle plate with side length a. Determine the simplified result when point A is chosen as the simplified point. How about the result if point B or point C is chosen as the simplified point?

Prob. 3.19

3.20 Determine the resultant force equivalent to the parallel force system and locate its point of application on the slab.

Prob. 3.20

3.21 The parallel force system is as shown. The side length of each square is 1 cm. Determine the single resultant force and locate its line of action.

Prob. 3.21

Simplification of Force Systems

97

3.22 Three forces act on a cuboid as shown. If F1 ¼ 100 N; F2 ¼ 300 N and F3 ¼ 200 N, replace this force system by an equivalent force-couple system at point O.

Prob. 3.22

3.23 Four forces act on a cuboid as shown. If F1 ¼ 20 kN; F2 ¼ 30 kN; F3 ¼ 60 kN and F4 ¼ 80 kN, replace this force system by an equivalent force-couple system at point O.

Prob. 3.23

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98

3.24 Four forces act on a cube with side length a. If each force has the same magnitude F, replace this force system by an equivalent force-couple system at point O.

Prob. 3.24

3.25 Prove that the force system acting on the plate can be reduced to a single resultant force. Determine this force and the point P (y, z) where its line of action intersects the plate.

Prob. 3.25

3.26 Prove that the force system acting on the plate can be reduced to a wrench. Specify the force and the couple moment for the wrench and the point P (y, z) where the wrench’s line of action intersects the plate.

Prob. 3.26

Simplification of Force Systems

99

3.27 Prove that the force system acting on the plate can be reduced to a wrench. Specify the force and the couple moment for the wrench and the point P (x, y) where the wrench’s line of action intersects the plate.

Prob. 3.27

3.28 Determine the centroid of the wire AOB.

Prob. 3.28

Engineering Mechanics

100 3.29 Locate the centroid of each shaded area.

Prob. 3.29

3.30 The revolution body has a radius R = 1.2 m, and a height h = 2 m. Determine the centroid of the body.

Prob. 3.30

Simplification of Force Systems

101

3.31 A semi-ellipsoid is formed by rotating about the x axis. Locate the centroid of the body.

Prob. 3.31

3.32 The solid cylinder has a radius R and a height h. It is made from a material having a density that varies with the height, i.e., q ¼ k þ az 2 , where k and a are constant parameters. Locate the center of mass of the cylinder.

Prob. 3.32

3.33 Locate the centroid of the composite area.

Prob. 3.33

Engineering Mechanics

102 3.34 Locate the centroid of the shaded area.

Prob. 3.34

3.35 The truss includes nine rods. The weight per unit length of each rod is 2 N/m. Determine the weight of the whole truss and locate its center of gravity.

Prob. 3.35

3.36 Find the centroid of the composite body.

Prob. 3.36

Simplification of Force Systems

103

3.37 The homogeneous assembly is made from a hemisphere and a cone, whose radii are both r. If the center of gravity of the assembly is at the center of the interface between the hemisphere and the cone, determine the height of the cone.

Prob. 3.37

3.38 Replace the distributed loading by an equivalent force-couple system at point A if q = 10 kN/m.

Prob. 3.38

3.39 The distributed loading on the beam consists of a uniform segment along AO and a variable segment along OB. Replace this loading by a single resultant force and dertermine its location measure from point O.

Prob. 3.39

Chapter 4 Equilibrium of Rigid Bodies Objectives
Study equations of equilibrium for two- or three-dimensional force system.
Examine the constraints and draw a proper free-body diagram for a rigid body or a system of rigid bodies.
Solve equilibrium problems of a rigid body or a system of rigid bodies.
Understand the concepts of redundant constraints, improper constraints and statically determinacy or indeterminacy.
Analyze the forces in members of a simple truss.

4.1

Conditions for Rigid-Body Equilibrium

From chapter 3, a general force system acting on a body can be simplified to a resultant force F0R plus a resultant couple moment M RO at an arbitrarily chosen simplified point O. When the resultant force F0R ¼ 0 and the resultant couple moment M RO ¼ 0, the force system is equivalent to zero and imparts neither translational nor rotational effect to the body. Then the rigid body is in equilibrium. From equation (3.10), the necessary and sufficient conditions for equilibrium of a rigid body are (X Fi ¼ 0 X ð4:1Þ MO ¼ 0 P It should be noted that M O is the vector sum of all the couple moments and the moments of all the forces about point O. Resolving forces and moments into their rectangular components, the two vector equilibrium equations can be replaced by the following six scalar equations:

DOI: 10.1051/978-2-7598-2901-9.c004 © Science Press, EDP Sciences, 2022

Engineering Mechanics

106 8X Fx ¼ 0 > >

> :X Fz ¼ 0

X X X

Mx ¼ 0 My ¼ 0

ð4:2Þ

Mz ¼ 0

These six independent scalar equations can be used to solve for at most six unknowns. If it is a three-dimensional (3-D) concurrent force system shown in figure 4.1, three moment equations in equation (4.2) are always satisfied and become useless. Then there are only three force equilibrium equations left: 8X Fx ¼ 0 > >

> :X Fz ¼ 0 P parallel force system shown in figure 4.2, Fx ¼ 0, P P If it is a three-dimensional Mz ¼ 0 in equation (4.2) are always satisfied and become useless. Fy ¼ 0 and Then there are only three independent equilibrium equations left: 8X Fz ¼ 0 > >

> :X My ¼ 0 If it is a three-dimensional couple system shown in figure 4.3, only three moment equilibrium equations remain useful: 8X Mx ¼ 0 > >

> :X Mz ¼ 0

FIG. 4.1 – 3-D concurrent force system.

Equilibrium of Rigid Bodies

107

FIG. 4.2 – 3-D parallel force system.

FIG. 4.3 – 3-D couple system.

(2-D) force system shown in figure 4.4, P P If it is aPcoplanar or two-dimensional Mx ¼ 0 and My ¼ 0 in equation (4.2) are always satisfied and Fz ¼ 0, useless. Then there are only three independent equilibrium equations left: 8X Fx ¼ 0 > >

> :X MO ¼ 0 P Here MO is the algebraic sum of all the couple moments and the moments of all the forces about an axis perpendicular to the x–y plane and passing through an arbitrarily chosen point O on the plane. These equilibrium equations can also be derived from equation (3.11). If it is a two-dimensional concurrent force system shown in figure 4.5, there are only two independent equilibrium equations left: (X Fx ¼ 0 X ð4:7Þ Fy ¼ 0

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108

If it is a two-dimensional parallel force system shown in figure 4.6, there are also only two independent equilibrium equations: (X Fx ¼ 0 X ð4:8Þ MO ¼ 0 If it is a two-dimensional couple system shown in figure 4.7, there is only one independent equilibrium equation: X Mi ¼ 0 ð4:9Þ The number of independent equations of equilibrium for every type of force system should be kept in mind. Only as many unknowns as the number of independent equilibrium equations can be solved.

FIG. 4.4 – Coplanar (2-D) force system.

FIG. 4.5 – 2-D concurrent force system.

FIG. 4.6 – 2-D parallel force system.

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109

FIG. 4.7 – 2-D couple system.

4.2

Free-Body Diagrams

In order to apply the equations of equilibrium for a rigid body or a structure formed by several connected bodies to determine the unknowns, we must first identify all the forces and couples acting on the object under consideration. The best way to account for these forces and couples is to draw the object’s free-body diagram. It is a sketch of the contour of the isolated object, on which all the forces and couples applied to the object and the reactions exerted on the object by its supports or connections are shown. In this section, we will first learn reactions that occur at various types of supports or connections, and then present a detailed procedure of drawing a free-body diagram.

4.2.1

Support Reactions

Each type of support provides specific reactions. As a general rule, if a support prevents translation of the connected body in a given direction, a force is developed on the body in the opposite direction. Similarly, if rotation about an axis is prevented, a couple about this axis is exerted on the body in the opposite direction [1]. The reactions of common types of supports or connections are listed in table 4.1. In this table, concentrated reactive forces and couple moments are given. However, they actually represent the resultants of the distributed loadings that exist at the contacting surface between the support and the connected body. It is important to identify each support and to understand how it produces the reactive forces or couples. For example, the support of ball and socket in table 4.1 prevents the translation of the attached body in any direction. Therefore, a force in any direction may act on the body. This force can be represented by an unknown magnitude F and two unknown coordinate direction angles (the third one can be determined from cos2 a þ cos2 b þ cos2 c ¼ 1). However, it is more generally represented by three unknown rectangular components, Fx, Fy, and Fz. Moreover, since the attached body can rotate freely about any axis, the ball-and-socket joint produces no reactive couple.

Type of support or connection

Reaction

110

TAB. 4.1 – Common types of supports or connections. Characteristics

(1) Cable

One unknown. It is a tensile force which pulls the member along the cable

(2) Weightless link One unknown. It can be a tensile or compressive force along the link

One unknown. It is a compressive force which pushes the member along the common normal line at the contacting point

Engineering Mechanics

(3) Smooth contacting surface

TAB. 4.1 – (continued).

One unknown. It is a compressive force and essentially the reaction of the smooth contacting surface

Equilibrium of Rigid Bodies

(4) Roller or rocker

(5) Ball and socket Three unknowns, including three rectangular force components, or the magnitude and two coordinate direction angles of the reactive force

3-D: n 3-D force system: five unknowns, including three force components and two couple-moment components

(6) Smooth pin

In 2-D force system: two unknowns, including two force components

111

2-D:

TAB. 4.1 – (continued). Reaction

Characteristics

112

Type of support or connection (7) Hinge

Five unknowns: three force components and two couple-moment components

(8) Journal bearing

3-D:

2-D:

In 3-D force system: four unknowns, including two force components and two couple-moment components

In 2-D force system: two unknowns, including one force and one couple moment

In 3-D force system: five unknowns, including three force components and two couple-moment components

(9) Thrust bearing

2-D:

In 2-D force system: three unknowns, including two force components and one couple moment

Engineering Mechanics

3-D:

(10) Journal bearing with square shaft Five unknowns: two force components and three couple-moment components

3-D: In 3-D force system: six unknowns, including three force components and three couple-moment components

(11) Fixed support

2-D:

Equilibrium of Rigid Bodies

TAB. 4.1 – (continued).

In 2-D force system: three unknowns, including two force components and one couple moment

*

It should be noted that the couple moments are generally not considered for smooth pin (6), hinge (7), journal bearing (8) and thrust bearing (9) if several supports are used and properly aligned. This is illustrated in examples 4.4 and 4.5.

113

114

Engineering Mechanics

Due to symmetry, many engineering problems can be solved by projecting all the active loadings and reactive forces acting on an object onto its symmetric plane. For example, the flat rectangular sluice gate with symmetric configuration involves symmetric loadings (its own weight and the water pressure), as shown in figure 4.8a. Therefore, the loadings can be projected onto the symmetric plane, figure 4.8b. The support reactions can also be projected onto this plane and we get a coplanar or twodimensional force system shown in figure 4.8c. Here support A is actually a smooth pin listed in table 4.1. It can provide three force components and two couple-moment components in three-dimensional force system. However, for the two-dimensional force system in figure 4.8c, only two force components FAx and FAy need to be considered. Different reactions should be considered in two- and three-dimensional force systems for some types of supports, which have been pointed out in table 4.1. Constraints in engineering may be complex and the selection of the type of supports becomes a challenging task for an actual constraint in practice. Sometimes the constraint cannot be idealized as any common type listed in table 4.1. Then you should decide the reactions it can provide. The general rule is always that a force is developed when the support prevents translation of the attached body, whereas a couple moment is exerted when rotation of the body is restricted.

FIG. 4.8 – 2-D idealized model due to symmetry.

4.2.2

Procedure of Drawing a Free-Body Diagram

To analyze a practical engineering problem, a corresponding idealized model that approximates as closely as possible the actual situation should be developed. What simplifying assumptions can be made to make the analysis feasible? What loadings should be considered and what can be omitted? Which type of support or connection does an actual constraint belong to? Careful choices have to be made about such things. This process requires both skills and experiences. It will be illustrated in examples 4.1 through 4.4.

Equilibrium of Rigid Bodies

115

Once the idealized model is developed, we can draw its free-body diagram. The steps you must follow in drawing a correct free-body diagram are as follows: (1) Draw the isolated object under consideration. Isolate the object from its supports and connections and draw its contour. (2) Show all the forces and couple moments. Identify all the forces and couple moments that act on the object, including (1) active forces, such as applied loadings and the weight of the object, (2) reactions exerted by the supports or connections. (3) Mark the knowns and unknowns of the forces, couple moments and dimensions. The known forces and couple moments should be labeled with their magnitudes and directions. Unknown forces and couple moments are represented by their rectangular components denoted by letters, such as, FAx, FAy, MBz, etc. Specify the dimensions necessary for calculating the moments of forces. Example 4.1. Draw the free-body diagram of the crane girder shown in figure 4.9a. The mass of the girder is 2 t. The mass of the electric hoist with the suspended cargo is 5 t and the mass of the cab with the operator is 0.3 t.

FIG. 4.9 – Idealized model and free-body diagram of a crane girder. Solution: This problem can be idealized as a problem of coplanar force system due to symmetry. It is reasonable to assume that the girder is a uniform beam and its center of gravity is at the midpoint. The three weights are PGirder ¼ 2  103  9:81 ¼ 19:62  103 N ¼ 19:62 kN PEH ¼ 5  103  9:81 ¼ 49:05  103 N ¼ 49:05 kN PCab ¼ 0:3  103  9:81 ¼ 2:943  103 N ¼ 2:943 kN The supports at the two ends of the girder can be considered as a smooth pin at A and a roller at B. This allows slight horizontal movement, so the girder can expand or contract slightly when the temperature changes. Average dimensions l, a and b are then used to specify the location of the loads and the supports. The idealized model of the crane girder is shown in figure 4.9b. Finally, the free-body diagram of the girder is drawn, figure 4.9c. The pin support at A exerts two force components Ax and Ay and the roller at B applies a vertical force FB on the girder. The magnitudes of these reactive forces are unknown, and their senses are assumed as shown.

116

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Example 4.2. Draw the free-body diagram of the chimney shown in figure 4.10a. Consider its weight and the wind pressure.

FIG. 4.10 – Idealized model and free-body diagram of a chimney.

Solution: This problem can be idealized as a coplanar equilibrium problem due to symmetry. The chimney is fixed to the ground since the support prevents both translation and rotation at the root of the chimney. Building codes should be used to specify the loadings of wind pressure and the chimney weight. Then the idealized model of the chimney is shown in figure 4.10b and its free-body diagram is given in figure 4.10c. Of course, the dimensions should be specified according to the practical situation. In the following examples, sometimes the dimensions are not shown, but always assumed to be known. Example 4.3. Draw the free-body diagram of the traffic sign shown in figure 4.11a. Consider the weight of the sign plate and the wind pressure on it.

FIG. 4.11 – Idealized model and free-body diagram of a traffic sign.

Equilibrium of Rigid Bodies

117

Solution: The sign plate can be idealized as a uniform plate and its center of gravity coincides with its centroid. The weight of the pole is neglected since it is relatively small. The sign plate is fixed to the pole and they are considered together as a “single” body. This body is fixed to the ground since the support prevents both translation and rotation at the root of the pole. The wind pressure is modelled as a uniform distributed loading on the whole surface area of the plate, which is perpendicular to the plate. Therefore, this problem should be idealized as a threedimensional equilibrium problem. The idealized model of the traffic sign is shown in figure 4.11b and its free-body diagram is given in figure 4.11c. The magnitudes of the loadings and the dimensions should be specified according to the practical situation or the design codes. Example 4.4. Draw the free-body diagram of the windlass shown in figure 4.12a.

FIG. 4.12 – Idealized model and free-body diagram of a windlass.

Solution: By inspection, it is reasonable to consider that the windlass is supported by a journal bearing at A and a thrust bearing at B. A force F is applied to the handle to keep the windlass in equilibrium, which holds a bucket weighing P. The idealized model of the windlass is shown in figure 4.12b. The weight of the windlass is not shown in this figure, since the weight is small compared to the loading and neglected. Afterwards, if the weight of a member is neither mentioned nor shown in a problem, it means the weight is neglected. In table 4.1, journal bearing (8) and thrust bearing (9) can provide both force components and couple-moment components. However, two bearings are used here and their force reactions alone are adequate to prevent the rotation about both the y and z axes. In other words, the effect of the couple moments is negligible and can be neglected. So, only force reactions are shown in the free-body diagram of the windlass in figure 4.12c. Actually, the couple moments are generally not considered if several bearings, pins or hinges are used and properly aligned.

118

Engineering Mechanics

Example 4.5. Draw the free-body diagram of the frame shown in figure 4.13a.

FIG. 4.13 – Free-body diagram of a frame.

Solution: By inspection, the frame is supported by a thrust bearing at A and a journal bearing at B. This is a two-dimensional force system. Since the bearings at A and B are aligned correctly, only force reactions are to be considered at these supports. Then the free-body diagram of the frame is shown in figure 4.13b. Example 4.6. Draw the free-body diagram of the uniform smooth cylinder shown in figure 4.14a.

FIG. 4.14 – Free-body diagram of a smooth cylinder.

Solution: The free-body diagram of the uniform cylinder is shown in figure 4.14b. The weight of 10 N is an active force. Since the cylinder is smooth, the reactive forces FA and FB act perpendicular to the contacting surface. Then both FA and FB pass through the centroid of the cylinder and this is a coplanar concurrent force system.

Equilibrium of Rigid Bodies

119

Example 4.7. The three-pinned frame is shown in figure 4.15a. Draw the free-body diagrams for each member and the system.

FIG. 4.15 – Free-body diagrams of each member and the frame system.

Solution: The free-body diagram of each member, including member AC, member BC and the pin C, is shown in figure 4.15b. Notice carefully how Newton’s third law is applied between pin C and the connected members. Pin C is subjected to two force components Cx1 and Cy1 from member AC and two force components Cx2 and Cy2 from member BC. Since the frame including pin C is in equilibrium, we have Cx1 ¼ Cx2 and Cy1 ¼ Cy2 . For this simple case in which a pin connects only two members and no force is acting on the pin, we often only draw the free-body diagrams of the two members, as shown in figure 4.15c. Here Cx and Cy on member AC are equal and opposite to that on member BC. However, you should keep in mind that this is not the direct effect of Newton’s third law; instead, this is from the equilibrium condition of the pin. Combining the free-body diagrams of all the members shown in figure 4.15b or c together yields the free-body diagram of the system, figure 4.15d. Note that the force components at pin C are not shown in this diagram since they form equal but opposite colinear pairs of internal forces and cancel out. Furthermore, the unknown force components at A and B should be exactly the same as that shown in figure 4.15b and c, i.e., acting in the same direction and sense and represented by the same letter.

120

Engineering Mechanics

Example 4.8. Draw the free-body diagram for each member of the frame shown in figure 4.16a.

FIG. 4.16 – Free-body diagram of each member of a frame. Solution: The free-body diagrams of member AB, pin B and member BC are shown in figure 4.16b–d, respectively. Notice carefully how Newton’s third law is applied between pin B and the connected members. Here pin B connects only two members. However, there is a force F acting on the pin. For this case, it is much clearer to draw a free-body diagram of pin B separately. Also note that couple is a free vector and the couple moment M can be moved anywhere on member AB, but not to member BC. RETHINK: How about the free-body diagram of the system? Remember that internal forces should never be shown on the free-body diagram of the system since they always occur in equal but opposite collinear pairs and then cancel out. Readers are recommended to finish this diagram by themselves. It should be kept in mind that before analyzing any equilibrium problem, it is essential to draw a free-body diagram of the object under consideration first to account for all the forces and couples acting on it. The importance of drawing free-body diagrams will be illustrated again and again throughout the study of mechanics.

4.3

Equilibrium Problems of a Rigid Body

Once the free-body diagram is finished and all the known and unknown forces and couples acting on the body are accounted for, we can use the equations of equilibrium described in §4.1 to solve for the unknowns. Example 4.9. The 240-N cargo is suspended by spring AB and cable AC, figure 4.17a. If the spring has an unstretched length of 0.2 m and a stiffness k = 400 N/m, determine the deformed length of spring AB and the force in cable AC.

Equilibrium of Rigid Bodies

121

FIG. 4.17 – Equilibrium of a 2-D concurrent force system. Solution: The free-body diagram of ring A is shown in figure 4.17b. Applying the equilibrium equations for this coplanar concurrent force system, we have X Fy ¼ 0; FAC sin 45  240 ¼ 0 ) FAC ¼ 339:4 N X

Fx ¼ 0;

FAC cos 45  Fk ¼ 0

)

Fk ¼ 240 N

Assuming that the deformed length of the spring is l, the spring force is Fk ¼ k ðl  0:2Þ ¼ 400ðl  0:2Þ ¼ 240 N Then the deformed length of the spring is l ¼ 0:8 m. Example 4.10. The beam is subjected to a force and a distributed loading, figure 4.18a. If P = 200 N and q = 100 N/m, determine the reactions at the supports.

FIG. 4.18 – Equilibrium of a general 2-D force system. Solution: The free-body diagram of the beam is shown in figure 4.18b. The distributed loading is replaced by the resultant force in figure 4.18b. The magnitude of the resultant force is Fq ¼ 100  3=2 ¼ 150 N Summing forces in the x direction yields X Fx ¼ 0; FAx ¼ 0

Engineering Mechanics

122

Summing forces in the y direction will yield an equation with two P unknowns, and we cannot get a direct solution. Applying the moment equation MA ¼ 0 , we will obtain FB directly since the other unknown FAy passes through point A and creates zero moment about this point. Assuming counterclockwise moment to be positive, we have X MA ¼ 0; P  3 þ Fq  1  FB  2 ¼ 0 ) FB ¼ 375 N Using this result and summing forces in the y direction gives X Fy ¼ 0; FAy þ FB  P  Fq ¼ 0 ) FAy ¼ 25 N RETHINK: For this coplanar force system, there are only three independent equations of equilibrium. Three equations have been written to solve for three unknowns. We can write other equilibrium equations, for example, X MB ¼ 0; P  1 þ Fq  1  FAy  2 ¼ 0 ) FAy ¼ 25 N X

MD ¼ 0;

FB  1 þ FAy  3  Fq  2 ¼ 0

)

FAy ¼ 25 N

It can be seen that although more equations than the number of independent equilibrium equations can be written, these equations are not independent and can only be used to check the calculations. Example 4.11. Three weightless links are used to support a beam, figure 4.19a. Two forces, F1 ¼ 20 kN and F2 ¼ 10 kN, are acting on the beam. Determine the reactions at the supports. Neglect the weight of the beam in the calculations.

FIG. 4.19 – A problem showing the advantage of moment equilibrium equation in 2-D. Solution: The free-body diagram of the beam is shown in figure 4.19b. For this problem, if we write a force equilibrium equation first, there are always two or even three unknowns in the equation no matter in which direction we sum forces. Thus, the moment equation of equilibrium is used first. Summing moments about point O1, which is the intersecting point of the lines of action of FA and FB, yields X MO1 ¼ 0; 3  FC  2  F2 ¼ 0 ) FC ¼ 6:67 kN

Equilibrium of Rigid Bodies

123

Proceed to write moment equations and sum moments about point O2 where the lines of action of FB and FC intersect and point O3 where the lines of action of FA and FC intersect, respectively. X pffiffiffi MO2 ¼ 0; 3 2  FA þ 3  F1 þ 1  F2 ¼ 0 ) FA ¼ 16:5 kN X

MO3 ¼ 0;

pffiffiffi 3  F1  3 2  FB þ1  F2 ¼ 0

)

FB ¼ 16:5 kN

Here, three moment equilibrium equations are utilized. This example clearly shows the advantage of the moment equation. Sum moments about a point where as many unknown forces as possible intersect, and therefore these unknowns will not be included in the moment equation. This is likely to get a direct solution and avoid simultaneous calculation. The above computations can be verified in part by a fourth equilibrium equation. X Fx ¼ 0; FA cos45  FB cos 45 ¼ 0 ) FB ¼ 16:5 kN Example 4.12. A tower crane shown in figure 4.20a weighs P = 700 kN. Its carrying capacity is P1 = 220 kN. Determine (1) the balanced load P2 to ensure that the crane will not overturn when it is fully loaded or unloaded; (2) the pressures that the rails A and B act on the crane, when the balanced load P2 = 180 kN and the crane is fully loaded.

FIG. 4.20 – Equilibrium of a 2-D parallel force system. Solution: The supports at A and B can be considered as rollers. Then the force system on the crane is a coplanar parallel force system as shown in figure 4.20b. (1) The balanced load P2 should be in an appropriate range, because if it is too small, the crane may overturn about point B when fully loaded; and if P2 is too large, the crane may overturn about point A when unloaded. First, consider the

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critical overturning state about point B when fully loaded (P1 = 220 kN). For this state, we have FA ¼ 0 and the balanced load is P2 min . Writing the moment equation of equilibrium yields X MB ¼ 0; P2 min ð6 þ 2Þ þ 2P  P1 ð12  2Þ ¼ 0 ) P2 min ¼ 100 kN For the critical overturning state about point A when unloaded (P1 = 0), we have FB ¼ 0 and the balanced load is P2 max . Writing the moment equation of equilibrium yields X MA ¼ 0; P2 max ð6  2Þ  2P ¼ 0 ) P2 max ¼ 350 kN So, to ensure that the crane will not overturn when it is fully loaded or unloaded, the balanced load should be 100 kN\P2 \350 kN (2)

The balanced load P2 ¼ 180 kN is in the appropriate range. Then the crane should be in equilibrium under the action of P, P1 , P2 , FA and FB . X MA ¼ 0; 4P2  2P  14P1 þ 4FB ¼ 0 ) FB ¼ 940 kN X

Fy ¼ 0;

FA þ FB  P  P2  P1 ¼ 0

)

FA ¼ 160 kN

We can check the calculation by summing moments about point B as follows: X MB ¼ 0; 8P2 þ 2P  10P1  4FA ¼ 0 ) FA ¼ 160 kN Example 4.13. The cargo is placed on a weightless plate and hoisted by three cables, figure 4.21. The cable forces are TA = 280 N, TB = 200 N and TD = 180 N. Determine the weight P of the cargo and the coordinates xC and yC of its center of gravity.

FIG. 4.21 – Equilibrium of a 3-D parallel force system.

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125

Solution: This is a three-dimensional parallel force system and we can write three independent equations of equilibrium to solve for three unknowns. A force summation along the z axis yields X Fz ¼ 0; TA þ TB þ TD  P ¼ 0 ) P ¼ 660 N Summing moments of the forces about the x and y axes, respectively, we have X Mx ¼ 0 ; TB  0:65 þ TD  1:15  P  yC ¼ 0 ) yC ¼ 0:511 m X

My ¼ 0 ;

P  xC  TB  0:8  TD  0:5 ¼ 0

)

xC ¼ 0:379 m

RETHINK: How to determine the coordinate zC? This is left for readers to think about. Example 4.14. The bent rod is subjected to a force and a couple, figure 4.22a. Segments OA, AB and BC are in the directions of y, x and z, respectively. Determine the reactions at the fixed support O. The weight of the rod is neglected.

FIG. 4.22 – Equilibrium of a general 3-D force system.

Solution: The free-body diagram of the rod is shown in figure 4.22b. This is a three-dimensional general force system and we can write six independent scalar equations of equilibrium to solve for the six unknown reactions at the fixed support O. However, for this problem, it is better to use the vector analysis since the couple M and the force F are given in Cartesian vector form and the moment of F about point O is easy to obtain from the vector cross product formulation. The vector summation of the forces yields

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126 X

F ¼ 0;



 FOx i þ FOy j þ FOz k þ ð80i  80j þ 95kÞ ¼ 0

  ðFOx þ 80Þi þ FOy  80 j þ ðFOz þ 95Þk ¼ 0 Equating the respective i, j, and k components on the left side to zero yields FOx þ 80 ¼ 0 FOy  80 ¼ 0 FOz þ 95 ¼ 0

) ) )

FOx ¼ 80 N FOy ¼ 80 N FOz ¼ 95 N

The vector summation of the moments about point O yields X   M O ¼ 0; MOx i þ MOy j þ MOz k þ M c þ rC  F ¼ 0   i    MOx i þ MOy j þ MOz k þ 55k þ  1:8  80

 j k  1:6 1:1  ¼ 0 80 95 

Evaluating the cross product and combining terms yields   ðMOx þ 64Þi þ MOy  259 j þ ðMOz  272Þk ¼ 0 Then we have MOx þ 64 ¼ 0

)

MOx ¼ 64 Nm

MOy  259 ¼ 0

)

MOy ¼ 259 Nm

MOz  272 ¼ 0

)

MOz ¼ 272 Nm

RETHINK: The negative signs indicate that FOx, FOz and MOx have an opposite sense to that shown on the free-body diagram, figure 4.22b. Therefore, when drawing the free-body diagram, the sense of a force or couple moment having an unknown magnitude but known line of action can just be assumed. If the solution of the equilibrium equations yields a negative number, it indicates the sense is opposite to that assumed on the free-body diagram. Example 4.15. Six weightless links are used to support a plate in the horizontal position, figure 4.23a. The geometric dimensions are a = b = h = 1 m. A force of 80 N and a couple moment of 40 Nm are acting on the plate. Determine the reactions at the weightless links.

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127

FIG. 4.23 – A problem showing the advantage of moment equilibrium equation in 3-D.

Solution: The free-body diagram of the plate is shown in figure 4.23b. Since the moment equilibrium equation has advantage, six moment equations will be used here. For this three-dimensional force system, we should sum moments about an axis that as many unknown forces as possible are passing through or parallel to. X MAB ¼ 0; F6  1  80  0:5 ¼ 0 ) F6 ¼ 40 N X

X

MAC ¼ 0;

X X X

F5  cos 45  1  40 ¼ 0

MAE ¼ 0;

MHG ¼ 0;

pffiffiffi pffiffiffi 2 2 þ F4  ¼0 F5  cos 45  2 2

MBF ¼ 0;

MFG ¼ 0;

)



F1  cos 45  1  40 ¼ 0

)

F1  cos 45  1 þ 80  0:5 þ F2  1 ¼ 0

F5 ¼ 56:6 N

)

F4 ¼ 40 N

F1 ¼ 56:6 N )

F3  cos 45  1 þ F4  1 þ F2  1 þ 80  0:5 ¼ 0

F2 ¼ 80 N )

F3 ¼ 0

Of course, you can use other equations, such as the force equilibrium equations, to check the answer. Readers are recommended to finish this by themselves. RETHINK: Pay attention to the order when we write the equations of equilibrium. Always try to write one equation including only one unknown and avoid simultaneous solution.

4.4

Two- and Three-Force Members

A member that is in equilibrium and subjected to forces at only two points is called a two-force member. For example, if the three-pinned frame shown in figure 4.24a has negligible weight and only an active force F is acting on the frame, member BC is only subjected to two forces at pins B and C, and it is a two-force member. To keep

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member BC in equilibrium, the two forces FBC and FCB at pins B and C cannot be like that shown in figure 4.24b even if they have the same magnitude, or that shown in figure 4.24c even if they have the same magnitude and opposite direction. To keep member BC in equilibrium, these two forces must be equal, opposite and colinear, like that shown in figure 4.24d. As a result, the direction (line of action) of both forces is known since it should pass through the two points B and C. Hence, only the force magnitude is unknown.

FIG. 4.24 – A three-pinned frame with a two-force member.

There is also a concept of three-force member. If a member in equilibrium is subjected to only three nonparallel forces, these three forces must be coplanar and concurrent at a point and this member is a three-force member. To show the concurrency requirement, consider member AC in the three-pinned frame in figure 4.24a. Since pin C connect only member AC and member BC, the force acting on member AC at pin C should be equal but opposite to that acting on member BC at pin C (see example 4.7). This force FCB acting on member AC intersects with the applied load P F at point O, as shown in figure 4.24e. To satisfy moment equilibrium equation MO ¼ 0, the third force FA at support A must also pass through O. Therefore, the force system is concurrent at O. However, since forces have to be resolved into their x and y components when writing equilibrium equations, we generally do not emphasize the concept of three-force member and still use Ax and Ay for the reaction at support A, as shown in figure 4.24f. Combining the free-body diagrams of member AC in figure 4.24f and member BC in figure 4.24d yields the free-body diagram of the whole frame, figure 4.24g. It should be noted that the pair of FCB cancel each other out and they are not shown in the free-body diagram of the whole frame as internal forces.

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Many mechanical elements can be idealized as two-force members. Typical examples include the links of a tripod in figure 4.25 and the spokes of a bicycle in figure 4.26. Actually, weightless link listed in table 4.1 is just a two-force member. For equilibrium problems of a system of rigid bodies, identify two-force members first can simplify the analysis. This will be illustrated in example 4.17.

FIG. 4.25 – Two-force member: link of a tripod.

FIG. 4.26 – Two-force member: spoke of a bicycle.

4.5

Constrains and Statical Determinacy

To sustain loadings without collapse and ensure equilibrium, a rigid body or a system of rigid bodies must be properly constrained. Some bodies may have more supports than that are necessary for equilibrium; others may not have enough supports or the supports may not be properly arranged that would cause instability [1]. These cases are to be discussed in this section. In practical engineering, a body or a system of bodies may have more supports than that are necessary to hold it in equilibrium. For example, the fixed support at A for the beam AB shown in figure 4.27a can prevent rotation and translation in two directions by providing one couple MA and two reactive forces FAx and FAy,

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figure 4.27b. It is sufficient to hold the beam in equilibrium. Three independent equations of equilibrium can just be used to solve for these three unknown reactions. However, in practice, we may add another support (figure 4.27c) or even more supports (figure 4.27d) to decrease the deformation and ensure a sense of security. These supports are called redundant constraints and they would develop FB and FC, figure 4.27e. Then there are five unknown reactions for this problem and we cannot solve all these five unknowns using just three independent equilibrium equations. To summarize, a problem having redundant constraints will have more unknowns than independent equilibrium equations available for their solution and it is statically indeterminate. Statically indeterminate problems cannot be solved in engineering mechanics. It will be studied in mechanics of materials by considering the deformation.

FIG. 4.27 – Illustration of redundant constraints.

Statical indeterminacy is opposite to statical determinacy, which means a problem having no redundant constraints and the unknown reactions can all be solved by the equations of equilibrium. For example, the structure shown in figure 4.28a is composed of two members, member AC and member CBD. The free-body diagram of each member is shown in figure 4.28b. There are FAx, FAy and MA at the fixed support A, FCx, FCy at the pin connection C and FB at the weightless link B, altogether six unknown reactions. For each member we can write three independent equations of equilibrium. Then the number of independent equilibrium equations is 3 × 2 = 6. The six unknown reactions can all be solved by these six independent equilibrium equations. The problem is statically determinate. If you forget to draw the little circle at joint C, as shown in figure 4.28c, ACBD becomes one single member and the number of independent equilibrium equations becomes 3 × 1 = 3. However, there are four unknown reactions for this member, figure 4.28d, and this problem becomes statically indeterminate. Therefore, the little circle at joint C is very important and it means a pin connection. Do not forget this kind of little circles when drawing the diagram.

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FIG. 4.28 – Illustration of statical indeterminacy. In some cases, there are as many unknown reactions on the body as the available independent equilibrium equations; however, instability still develops because of improper constraining. One case of improper constraining is that the support reactions all intersect a common axis. For example, the door is supported by two hinges and all the hinge reactions pass through the y axis, figure 4.29. When a force F perpendicular to the door as shown is exerted, the door cannot remain in equilibrium and will rotate about the y axis. In 2-D problems of such case, the common axis is perpendicular to the plane of the force system and appears as a point, figure 4.30. Thus, when all the reactive forces are concurrent at a common point for a coplanar force system, the body is improperly constrained and may rotate about this point. Another case of improper constraining is that all the reactive forces are parallel to each other. 3-D or 2-D examples of such case are shown in figures 4.31 and 4.32, respectively. For this case, a nonparallel active force will cause translation. To sum up, proper constraining requires that the reactive forces do not intersect a common axis (or a common point in 2-D problems) and are not parallel to one another.

FIG. 4.29 – Improper constraining: all reactive forces intersect a common axis.

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FIG. 4.30 – Improper constraining: all reactive forces intersect a common point.

FIG. 4.31 – Improper constraining: all reactive forces are parallel to each other. Sometimes a body may not have enough supports to ensure stability and it is only partially constrained. For example, the beam shown in figure 4.32a is only pin-supported at its end A. Two unknown reactions FAx and FAy are developed, figure 4.32b; whereas there are three independent equations of equilibrium for this coplanar force system. To keep the beam in equilibrium, P the applied loadings F and M should satisfy the moment equilibrium equation MA ¼ 0, i.e., Fl  M ¼ 0. Otherwise, the beam will rotate about pin A. So, the beam is only partially constrained. Mechanical devices are all partially constrained.

FIG. 4.32 – A beam partially constrained.

4.6

Equilibrium Problems of Structures

In practice, rigid bodies are often connected to form a structure, such as a frame, machine or truss. The truss is a special type of structure with only two-force members and it will be discussed in the next section. This section considers frames and machines, which have multi-force members. Frames are fully constrained structures, which are stationary and used to support loads. Here only statically determinate frames are considered and statically indeterminate frames with redundant constraints are out of the range of this book. Machines are partially constrained structures, which contain moving parts and are used to transmit the effect of forces. To design a structure, the forces acting at the supports and connections should be determined. When determining such reaction forces, usually the structure should

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133

be disassembled, appropriate analysis object be chosen, and its free-body diagram be drawn. The analysis object could be one member, a portion of the structure including several members, or the entire structure. The choice needs skills and we usually try to find the most direct solution of the problem. This will be explained in the following examples. The general procedure of drawing a free-body diagram has been given in §§4.2.2. Here some important points should be noted:
Identify two-force members in the structure first. For a two-force member, the two forces are equal, opposite and passing through the two points. Then only one unknown, i.e., the force magnitude, needs to be determined for the member.
Forces common to two contacting members are equal, opposite, and shown on the respective free-body diagrams of the members. These action and reaction forces should be denoted by the same letters to indicate their relationship.
When the analysis object is a “system” including several members connected, the forces between the connected members within the system are internal forces and not shown on the free-body diagram. Example 4.16. Due to terrain constraints, the two pin-supports A and B of a threepinned frame are located at different elevations, figure 4.33a. Determine the reactions at pins A and B when q = 3 kN/m.

FIG. 4.33 – Equilibrium of a three-pinned frame with two pin supports located at different elevations.

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Solution: The free-body diagram of the whole frame is shown in figure 4.33b. For this coplanar force system, three independent equilibrium equations can be written: X X

MA ¼ 0;

4FBy þ 4FBx  2  4q ¼ 0

ð1Þ

MB ¼ 0;

2  4q þ 4FAx  4FAy ¼ 0

ð2Þ

X

Fx ¼ 0;

FAx  FBx ¼ 0

ð3Þ

However, there are four unknowns in these three equations and none of them can be obtained. So, it is necessary to dismember the frame. Here we analyze member BC, figure 4.33c. There are also four unknowns in this diagram. Summing moments about C can eliminate the two uninterested reactions at C. X

MC ¼ 0;

2FBy  4FBx  1  2q ¼ 0

ð4Þ

Simultaneous solution of equations (1) and (4) yields FBx ¼ 1 kN;

FBy ¼ 5 kN

Substituting these results into equations (2) and (3) respectively yields FAx ¼ 1 kN;

FAy ¼ 7 kN

RETHINK: If we just want to avoid simultaneous solution, the orthogonal components of the reactions at pin-supports A and B along line AB and perpendicular to AB, as shown in figure 4.33d and e, can be used. Then summing moments about A in figure 4.33d would yield a direct solution for FBη and summing moments about C in figure 4.33e would obtain FBξ directly. Then summing forces along the ξ and η axes in figure 4.33d, respectively, FAξ and FAη can also be obtained directly. We can continue to determine the magnitudes and directions of FA and FB. The results should be the same as those from the x and y components of FA and FB. Readers are recommended to check this. Example 4.17. For the frame shown in figure 4.34a, q0 = 2 kN/m, M = 10 kNm and F = 2 kN. Neglecting the weight of each member, determine the horizontal and vertical force components that pin C exerts on member ABC.

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135

FIG. 4.34 – Equilibrium of a frame with two-force member.

Solution 1: The free-body diagram of the whole frame is shown in figure 4.34b. Summing moments about D can get a direct solution for FA. X 1 MD ¼ 0; 6  FA þ q0  3  4  M þ F  1 ¼ 0 ) FA ¼ 0:667 kN 2 Analyze member ABC and its free-body diagram is shown in figure 4.34c. Note that BD is a two-force member since it is pin-connected at B and D and no other forces act on it. Therefore, the line of action of the force FBD passes through B and D. From the trigonometry, we have sin h ¼ 0:8;

cos h ¼ 0:6

Then applying the three independent equilibrium equations for member ABC yields X

MC ¼ 0;

1 6  FA þ 4  q0  3  M  3  FBD sin h ¼ 0 2

X X

Fx ¼ 0;

Fy ¼ 0;

FCx  FBD cos h ¼ 0

)

FBD ¼ 0:833 kN

) FCx ¼ 0:5 kN

1 FA  q0  3 þ FBD sin h þ FCy ¼ 0 2

)

FCy ¼ 3 kN

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Solution 2: If failing to recognize that BD is a two-force member, we can still solve the problem, but generally with more effort. Just like solution 1, FA is first determined by analyzing the whole frame. Then the free-body diagram of member ABC is shown in figure 4.34d. There are four unknown reactions in this diagram, which cannot all be obtained from three independent equilibrium equations. However, summing moments about B would yield a direct solution for FCy. X 1 MB ¼ 0; 3  FA þ 1  q0  3  M þ 3  FCy ¼ 0 ) FCy ¼ 3 kN 2 From member ABC, we cannot obtain FCx. So, proceed to analyze member CD, figure 4.34e. Summing moments about D can obtain FCx directly. X MD ¼ 0; 1  F  4  FCx ¼ 0 ) FCx ¼ 0:5 kN

RETHINK: It should be noted that FDx1 and FDy1 in figure 4.34e are the horizontal and vertical force components that the pin at D exerts on member CD, whereas FDx and FDy in figure 4.34b are the horizontal and vertical force components that the support at D exerts on the pin. They are different. Readers are recommended to draw the free-body diagram of the pin and observe carefully the forces between pin D and its contacting members or support. Example 4.18. For the structure shown in figure 4.35a, AD = 20 cm, BD = 40 cm and a ¼ 45 . The cord wrapping over a pully with a radius r = 10 cm is used to suspend a cargo weighing P = 1800 N. Neglecting the weight of other members, determine the horizontal and vertical force components at pin A and the force in member BC.

FIG. 4.35 – Equilibrium of a structure with pulley and cord.

Solution: From the free-body diagram of the cargo, figure 4.35b, it is clear that the tension in the vertical segment of the cord is T1 = P for equilibrium. The free-body diagram of the pulley along with the contacting portion of the cord is shown in figure 4.35c. Summing moments about D yields

Equilibrium of Rigid Bodies X

MD ¼ 0;

137 T2  r  T1  r ¼ 0

) T2 ¼ T1 ¼ P

This means the cord force remains unchanged as the cord passes over a pulley in equilibrium. Afterwards, consider beam AB, the pulley along with the cord portion and the cargo together as a “system”, figure 4.35d. Three equilibrium equations are written to solve for the unknowns. X MA ¼ 0; FBC sin a  AB þ P  r  P ðAD þ r Þ ¼ 0 ) FBC ¼ 848 N X X

Fx ¼ 0;

FAx  P  FBC cos a ¼ 0

) FAx ¼ 2400 N

Fy ¼ 0;

FAy  P þ FBC sin a ¼ 0

) FAy ¼ 1200 N

Example 4.19. For the structure shown in figure 4.36a, determine the horizontal and vertical components of forces that the pin at C exerts on member ABC and on member CDE, respectively.

FIG. 4.36 – Forces between a pin and its connected members.

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Solution: Considering member EGH, member GD and member CDE together as a “system”, the free-body diagram is shown in figure 4.36b. Note that member DB is a two-force member. pffiffiffi X pffiffiffi 2 FDB  1 ¼ 0 ) FDB ¼ 12 2 kN MC ¼ 0; 6  2  2 X

X

Fx ¼ 0;

Fy ¼ 0;

FCEx

pffiffiffi 2 FDB ¼ 0  2

FCEy  6 

)

pffiffiffi 2 FDB ¼ 0 2

FCEx ¼ 12 kN

)

FCEy ¼ 6 kN

FCEx and FCEy are the horizontal and vertical force components that pin C exerts on member CDE. For the entire structure, figure 4.36c, X MA ¼ 0; 6  1  FC  3 ¼ 0 ) FC ¼ 2 kN X

Fy ¼ 0;

FA  6 þ FC ¼ 0

)

For pin C along with the wheel, figure 4.36d, X Fx ¼ 0; FCAx  FCEx ¼ 0 ) X

Fy ¼ 0;

FC þ FCAy  FCEy ¼ 0

FA ¼ 4 kN

FCAx ¼ 12 kN )

FCAy ¼ 8 kN

FCAx and FCAy are the horizontal and vertical force components between pin C and member ABC. RETHINK: The solution can be checked by applying equilibrium equations to member ABC, figure 4.36e, pffiffiffi X 2 FDB  FCAx ¼ 0 ) FCAx ¼ 12 kN Fx ¼ 0; 2 X

MA ¼ 0;



pffiffiffi 2 FDB  2 þ FCAy  3 ¼ 0 2

)

FCAy ¼ 8 kN

By the way, it should be noted that this is a structure partially constrained or improperly constrained. If a horizontal force is applied on this structure, it cannot remain in equilibrium. Example 4.20. For the structure shown in figure 4.37a, M ¼ 5 kNm and q ¼ 1 kN=m. Neglecting the weight of each member, determine the reactions at the fixed support C.

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139

FIG. 4.37 – An equilibrium problem with multiple solutions.

Solution 1: By inspection, members ED, EB and EG are all two-force members since they are pin-connected at their end points and no other forces act on them. For member AB, figure 4.37b, pffiffiffi X 2 1 4 4  2 ¼ 0 ) FGE ¼  4:61 kN MB ¼ 0; M þ  q  4  þFGE  2 2 7 3 For pin E, figure 4.37c, X

Fx ¼ 0;

FGE 

pffiffiffi 2 þFED ¼ 0 2

For the entire structure, figure 4.37d, X Fx ¼ 0; FCx þ FED ¼ 0

)

)

FED ¼ 3:26 kN

FCx ¼ 3:26 kN

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140

X

X

Fy ¼ 0;

FCy 

1 q7¼0 2

)

FCy ¼ 3:5 kN

1 7 M þMC þ  q  7  þFED  2 ¼ 0 2 3

MC ¼ 0;

)

MC ¼ 6:65 kNm

Solution 2: For member AB, figure 4.37b, X

MB ¼ 0;

pffiffiffi 2 1 4 4 2¼0 M þ  q  4  þFGE  2 2 7 3

)

FGE ¼  4:61 kN

For the pin at E, figure 4.37c, X

Fy ¼ 0;

FGE

pffiffiffi 2 þFEB ¼ 0  2

)

FEB ¼ 3:26 kN

For member AB and member BC connected as a “system”, figure 4.37e, pffiffiffi X 2 þ FCx ¼ 0 ) FCx ¼ 3:26 kN Fx ¼ 0; FGE  2 X

X

Fy ¼ 0;

MC ¼ 0;

FGE

pffiffiffi 2 1  FCy  FEB   q  7 ¼ 0  2 2

M þ MC þ FGE 

pffiffiffi 2 7 1  5 þ FEB  3 þ   7 ¼ 0 2 3 2

Solution 3: For the entire structure, figure 4.37d, X 1 Fy ¼ 0; FCy   q  7 ¼ 0 2

)

For member AB, figure 4.37b, X 1 4 4 ME ¼ 0; M þ   4  þFBx  2 ¼ 0 2 7 3 For member BC with the pin B, figure 4.37f, X Fx ¼ 0; FCx þ FBx ¼ 0 )

)

FCy ¼ 3:5 kN

)

MC ¼ 6:65 kNm

FCy ¼ 3:5 kN

)

FBx ¼ 3:26 kN

FCx ¼ 3:26 kN

Equilibrium of Rigid Bodies

X

MB ¼ 0;

141

1 3 2 4 3 MC  FCy  3   q  3   3  q  3  ¼ 0 2 7 3 7 2

Solution 4: For the entire structure, figure 4.37d, X 1 Fy ¼ 0; FCy   q  7 ¼ 0 2

)

)

MC ¼ 6:65 kNm

FCy ¼ 3:5 kN

For member BC with the pin B, figure 4.37f, X

MB ¼ 0;

1 3 2 4 3 MC  FCy  3   q  3   3  q  3  ¼ 0 2 7 3 7 2

Analyze the entire structure again, figure 4.37d, X 7 1 MD ¼ 0; M þ MC þ   7  FCx  2 ¼ 0 2 3 Solution 5: For the entire structure, figure 4.37d, X 1 Fy ¼ 0; FCy   q  7 ¼ 0 2 X

MD ¼ 0;

)

)

)

MC ¼ 6:65 kNm

FCx ¼ 3:26 kN

FCy ¼ 3:5 kN

1 7 M þ MC þ  q  7   FCx  2 ¼ 0 2 3

For member AB and member BC connected as a “system”, figure 4.37e, X 1 2 ME ¼ 0; M þ MC   q  7   FCx  2  FCy  3 ¼ 0 2 3

ð1Þ

ð2Þ

Sustituting FCy ¼ 3:5 kN into equation (2) and combining terms yields equation (1). This means equations (1) and (2) are not independent and they cannot be used to obtain MC and FCx . When writing several equilibrium equations by analyzing different objects and trying to get a simultaneous solution, it is not easy to judge whether these equations are independent or not. The best way is to write one equation including only one unknown and get a direct solution. RETHINK: Several solutions are given for this problem. Solutions 1 and 2 use five equations to obtain the three wanted reactions. Solution 3 uses four equations and solution 4 uses only three equations. Solution 4 is the most direct solution using the least equations. It is strongly recommended to come up with several solutions to an equilibrium problem of structures and find the most direct one. This can improve your skills quickly.

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4.7

Simple Trusses

Trusses are fully constrained structures, which consist exclusively of slender members joined together at their end points. They are widely used to support roofs, bridges and towers, figure 4.38.

FIG. 4.38 – Truss structures. If a truss along with its loadings lies in a single plane, as shown in figure 4.39, it is called a planar truss. To prevent collapse under loadings, a truss must be rigid or stable. Since a triangle is the simplest form that is rigid, a simple planar truss begins with a triangular element and is then expanded by adding two members and a joint each time, as shown in figure 4.40.

FIG. 4.39 – A planar truss.

FIG. 4.40 – Simple planar truss.

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To design a truss, the force developed in each member under the expected loadings should be determined. For the analysis, two important assumptions are made: (1) All loadings are applied at the joints. The weight of each member is often neglected since the weight is usually small compared to the applied loadings. If a member’s weight cannot be neglected, it is usually represented by two vertical forces with the magnitude of half weight applied at the member’s two endpoints (see example 4.22). (2) The members are joined together by smooth pins, figure 4.41a. Actually, the members are often connected by welding or bolting their ends to a gusset plate in practice, as shown in figure 4.41b and c. However, this assumption can make the analysis feasible and the error due to it is usually acceptable in practice.

FIG. 4.41 – Joint of a truss. Based on these two assumptions, each truss member is a two-force member, and the forces at the ends of the member must be equal, opposite and directed along the axis of the member. If the forces pull on the member and tend to elongate it, the member is in tension (T), figure 4.42a; whereas if the forces push on the member and tend to shorten it, the member is in compression (C), figure 4.42b. When designing a truss, compressive members generally need to be made thicker than tensile members due to the buckling effect that only occurs in a compressive member. Therefore, it is very important to identify whether a member is in tension or compression.

FIG. 4.42 – Truss members in tension or compression.

4.7.1

The Method of Joints

It has been mentioned that the member forces should be determined before designing a truss. Consider a truss shown in figure 4.43a. If we analyze the entire truss, the forces in the members are internal forces and not shown in the free-body

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diagram of the truss, figure 4.43b. Thus, the member forces cannot be obtained from analyzing the entire truss. Only the support reactions can be determined. X Fx ¼ 0; FAx ¼ 0 X

MB ¼ 0;

1  0:25 þ 1  0:5 þ 1  0:75 þ 0:5  1  FAy  1 ¼ 0

X

Fy ¼ 0;

FAy þ FB  4 ¼ 0

)

)

FAy ¼ 2 N

FB ¼ 2 N

If we analyze the pin at a joint of the truss, such as joint A, the forces in members connected to joint A become external forces on the joint’s free-body diagram, figure 4.43c. Note that for the two unknown member forces, we assume they are “pulling” on the pin, which means we assume the members are in tension. This will be easily understood if the joint with small segments of the members connected to the pin is isolated, figure 4.43d. Applying the equations of equilibrium, we can determine the member forces. X Fy ¼ 0; 2  0:5 þ FAF sin 30 ¼ 0 ) FAF ¼ 3 N X

Fx ¼ 0;

FAC þ FAF cos 30 ¼ FAC þ ð3Þ cos 30 ¼ 0

)

FAC ¼ 2:6 N

The negative sign for FAF means that member AF is actually in compression. It can be seen that for a simple planar truss, the force system acting on each joint is a coplanar concurrent force system and two independent equilibrium equations can be written and used to solve for at most two unknowns. Therefore, we can successively analyze a joint with at most two unknown forces. After joint A, we can then analyze joint F, figure 4.43e.

FIG. 4.43 – The method of joints.

Equilibrium of Rigid Bodies X

Fy0 ¼ 0; X

145

1  cos 30  FFC cos 30 ¼ 0

Fx 0 ¼ 0;

)

FFC ¼ 1 N

FFD  FAF þ FFC sin 30  1  sin 30 ¼ 0

Substituting FFC ¼ 1 N and FAF ¼ 3 N into the latter equation yields FFD ¼ 2 N And then we can analyze joint D, figure 4.43f. X Fx ¼ 0; FDE cos 30  FFD cos 30 ¼ 0 X

Fy ¼ 0;

)

1  FDC  ðFDE þ FFD Þ sin 30 ¼ 0

FDE ¼ 2 N )

FDC ¼ 1 N

This is a symmetrical structure under symmetrical loads. From the symmetry, we have FBE ¼ FAF ¼ 3 N;

FEC ¼ FFC ¼ 1 N;

FBC ¼ FAC ¼ 2:6 N

For the sake of clarity, the results are often illustrated in the form of figure 4.43g, which not only gives the magnitude of each member force, but also indicates whether it is tensile or compressive. In this process, we successively analyze a joint with at most two unknown forces to solve for the member forces. This is called the method of joints. It should be noted that besides the member forces, other forces acting at a joint include applied loads and support reactions. For the member forces, always assume they are tensile, i.e., “pulling” on the pin. Then the solution of equilibrium equations will yield positive scalars for members in tension and negative scalars for members in compression. Example 4.21. For the truss shown in figure 4.44a, the maximum tensile force and the maximum compressive force any member can sustain are 10 kN and 5 kN, respectively. Determine the maximum force P the truss can support at joint C.

FIG. 4.44 – Determining the accepted active force.

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Solution: By inspection, we can start with joint C without determining the support reactions, since there are only two unknown member forces at joint C besides the applied force P, figure 4.44b. The two member forces FCA and FCB are assumed tensile. Applying two independent equilibrium equations, we have X pffiffiffi Fy ¼ 0; FCB sin 45  P ¼ 0 ) FCB ¼ 2PðTÞ X

Fx ¼ 0;

FCB cos 45  FCA ¼ 0

)

FCA ¼ PðCÞ

The negative sign means member CA is actually in compression. Then joint B can be analyzed, since there are only two unknown forces on joint B, i.e., the support reaction of the weightless link and the force in member AB, figure 4.44c. Applying the equilibrium equations yields X Fy ¼ 0; FCB sin 45  FAB ¼ 0 ) FAB ¼ PðCÞ pffiffiffi Set the tensile member force is 10 kN, i.e., 2P ¼ 10 kN, we get P1 ¼ 7:07 kN; set the compressive member force is 5 kN, i.e., P ¼ 5 kN, we get P2 ¼ 5 kN. Then the maximum force P the truss can support at joint C is P ¼ minfP1 ; P2 g ¼ 5 kN Example 4.22. Determine the force in each member of the truss shown in figure 4.45a when the weight of the truss is neglected. If each member of the truss is made of steel with a mass per length of 3.6 kg/m and the weight of each member is represented by two vertical forces with the magnitude of half weight applied at two endpoints of the member respectively, determine the force in each member for this case.

FIG. 4.45 – Comparison of the member forces when neglecting or considering the member weight.

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Solution: The free-body diagram of the entire truss is shown in figure 4.45b. Applying the equations of equilibrium yields X ME ¼ 0 ; FAy  4  6  4 ¼ 0 ) FAy ¼ 6 kN X

Fx ¼ 0;

)

FAx ¼ 0

We can then analyze joint D, since there are only two unknown member forces at joint D, figure 4.45c. Applying the equations of equilibrium gives X Fy ¼ 0; FCD sin 60  6 ¼ 0 ) FCD ¼ 6:93 kNðTÞ X

Fx ¼ 0;

FCD cos 60  FED ¼ 0

)

FED ¼ 3:46 kNðCÞ

Then proceed to analyze joint C, figure 4.45d. Apply the equations of equilibrium, X Fy ¼ 0; FCD sin 60  FCE sin 60 ¼ 0 ) FCE ¼ 6:93 kNðCÞ X

Fx ¼ 0;

FCD cos 60  FCE cos 60  FBC ¼ 0

)

FBC ¼ 6:93 kNðTÞ

By inspection, the structure is a symmetrical structure with symmetrical loading. From symmetry, we have FAB ¼ FCD ¼ 6:93 kNðTÞ;

FBE ¼ FCE ¼ 6:93 kNðCÞ;

FAE ¼ FED ¼ 3:46 kNðCÞ

If each member of the truss is made of steel with a mass per length of 3.6 kg/m, the weight of each member is 4  3:6  9:81 ¼ 141 N. This weight is represented by two vertical forces with the magnitude of half weight applied at two endpoints of the member. Therefore, the free-body diagram of the entire truss is shown in figure 4.45e. Apply the equations of equilibrium, X   ME ¼ 0 ;  FAy  0:141  4  ð6 þ 0:141Þ  4 þ 0:212  2  0:212  2 ¼ 0 ) FAy ¼ 6 kN X

Fx ¼ 0;

)

FAx ¼ 0

Analyze joint D first, figure 4.45f. Apply the equations of equilibrium, X Fy ¼ 0; FCD sin 60  6  0:141 ¼ 0 ) FCD ¼ 7:09 kNðTÞ X

Fx ¼ 0;

FCD cos 60  FED ¼ 0

)

FED ¼ 3:55 kNðCÞ

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Then analyze joint C, figure 4.45g. Applying the equations of equilibrium gives X Fy ¼ 0; FCD sin 60  FCE sin 60  0:212 ¼ 0 ) FCE ¼ 7:34 kNðCÞ X

Fx ¼ 0;

FCD cos 60  FCE cos 60  FBC ¼ 0

)

FBC ¼ 7:22 kNðTÞ

The structure is still a symmetrical structure with symmetrical loading. From symmetry, we have FAB ¼ FCD ¼ 7:09 kNðTÞ;

FBE ¼ FCE ¼ 7:34 kNðCÞ;

FAE ¼ FED ¼ 3:55 kNðCÞ

RETHINK: For this example, the weight of the truss member (141 N) is relatively small compared to the applied loading (6 kN). Comparing the answers for two cases, considering the weight or not makes relatively small difference on the member forces. This is the general case for most practical trusses and that is why the weight of the truss is often neglected.

4.7.2

Zero-Force Members

For a given loading case, some member forces of a truss may be zero and these members are called zero-force members. Zero-force members can be inspected from the following rules: (1) If a joint has only two non-collinear members connected to it and no applied forces act at this joint, figure 4.46a, the two members are P both zero-force members. It is easy to P prove this. From figure 4.46a, applying Fy ¼ 0 yields F2 ¼ 0; and applying Fx ¼ 0 yields F1 ¼ 0. (2) If a joint has only two non-collinear members connected to it and one active force acts at this joint along the axis of one member, figure 4.46b, the other member is zero-force member, i.e., F2 = 0. (3) If a joint has three members connected to it, two of the members are collinear and no active force acts at this joint, figure 4.46c, the third non-collinear member is a zero-force member, i.e., F2 = 0. Readers are recommended to prove the last two rules by themselves. Truss analysis will be greatly simplified if zero-force members are first identified.

FIG. 4.46 – Rules to inspect zero-force members.

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149

Example 4.23. Identify all the zero-force members in the truss shown in figure 4.47a.

FIG. 4.47 – Zero-force members for different loading cases.

Solution: From joint A, we can see that members AB and AC are zero-force members based on the first rule. From joint D, member CD is a zero-force member based on the third rule. Realize that we cannot identify CD as a zero-force member by analyzing joint C, where five members are connected. RETHINK: It should be noted that zero-force members exist for a specific loading case. For example, if the applied force P is moved from joint B to joint A, as shown in figure 4.47b, member AB is no longer a zero-force member. Therefore, zero-force members are not useless. They are used to keep stability and rigidity of the truss under any loading condition.

4.7.3

The Method of Sections

A truss shown in figure 4.48a is subjected to two forces P1 = 10 kN and P2 = 8 kN. Each member has a length of 1 m. We want to determine the forces in members 1, 2 and 3. By inspection, there are more than two unknown forces (including member forces or support reactions) at each joint. The support reactions have to be determined first. From the free-body diagram of the entire truss, figure 4.48b, we have X X

MA ¼ 0 ;

X

Fy ¼ 0;

Fx ¼ 0;

FAx ¼ 0

FB  4  P1  1  P2  2 ¼ 0 FAy þ FB  F1  F2 ¼ 0

)

)

FB ¼ 6:5 kN

FAy ¼ 11:5 kN

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FIG. 4.48 – The method of sections.

Then to determine forces in members 1, 2 and 3 using the method of joints, we have to analyze joints A, C, E, and D in sequence. This process is cumbersome and actually we have another easier method—the method of sections. Use an imaginary section, for example, section mm shown in figure 4.48b, to “cut” the truss into two parts. Then draw the free-body diagrams of the isolated two parts respectively, as shown in figure 4.48c and d. As in the method of joints, we also assume that the member forces are tensile, i.e., “pulling” on the members at the cut section. Since the truss is in equilibrium, any part of the truss is also in equilibrium. If the left part in figure 4.48c is considered, applying the equilibrium equations yields pffiffiffi X 3 MG ¼ 0 ; P1  1  FAy  2  F1  ¼ 0 ) F1 ¼ 15 kNðCÞ 2 X

MD ¼ 0 ; X

pffiffiffi 3 F3  þ P1  0:5  FAy  1:5 ¼ 0 2

Fy ¼ 0;

FAy  P1  F2 sin 60 ¼ 0

)

)

F3 ¼ 14:1 kNðTÞ

F2 ¼ 1:73 kNðTÞ

You can also obtain the same results for these three member forces by considering the right part in figure 4.48d. Readers are recommended to finish this by themselves. From the abovementioned procedure, we can see that if only several member forces are to be determined, the method of sections has advantage over the method of joints. A key step in this method is to decide how to “cut” the truss. Usually, we cut through the members where forces are to be determined to expose these member forces. Since only three independent equilibrium equations can be applied to the isolated part of the truss, the section generally passes through at most three

Equilibrium of Rigid Bodies

151

members whose forces are unknown (an exception can be seen in example 4.24). After the section, you can analyze either part of the sectioned truss and would get the same results for the member forces at the cut section. Example 4.24. Find the force in member CE of the cantilever truss shown in figure 4.49a.

FIG. 4.49 – Advantage of the method of sections. Solution: For this cantilever truss, we do not need to determine the reactions at supports A and B, since we can analyze the right part after the section. By inspection of figure 4.49a, any imaginary section that cuts through CE, such as section mm or nn, will also cut through three other members. However, member forces of EK, FK and DF would intersect at point F if section nn is used, as shown in figure 4.49b. Then summing moments about F would eliminate these three unknowns and yield a direct solution for FCE as follows: X MF ¼ 0 ; 4FCE  6P ¼ 0 ) FCE ¼ 1:5P ðTÞ RETHINK: If the method of joints is used, it is necessary to analyze joints I, G, H, L and E in sequence. However, for the method of sections, we can just cut member CE and determine the wanted member force directly. This is the advantage of the method of sections. Example 4.25. Determine the forces in members 1, 2, 3, 4 and 5 of the truss shown in figure 4.50a.

FIG. 4.50 – Combination of the method of sections and the method of joints.

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Solution: Cut the truss using section mm and draw the free-body diagram of the right part, figure 4.50b. Writing three independent equilibrium equations gives X MF ¼ 0 ; 30  4  F1  3 ¼ 0 ) F1 ¼ 40 kNðTÞ X X

MC ¼ 0 ; Fy ¼ 0;

F3  3 þ 30  6 ¼ 0

)

F3 ¼ 60 kNðCÞ

3 30  F2 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ 0 22 þ 32

)

F2 ¼ 36:1 kNðTÞ

To determine the forces in members 4 and 5, section nn can be used. The freebody diagram of the small upper part of the sectioned truss is shown in figure 4.50c. Apparently this free-body diagram is the same as that for the pin at F from the method of joints. For this concurrent force system, there are only two independent equations of equilibrium. X 2 Fx ¼ 0; F4  F2 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi  F3 ¼ 0 ) F4 ¼ 40 kNðCÞ 2 2 þ 32 X

Fy ¼ 0;

3 F2 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi  F5 ¼ 0 2 2 þ 32

)

F5 ¼ 30 kNðCÞ

Example 4.26. Find the force in member CD of the truss shown in figure 4.51a. ΔABC is an equilateral triangle with a side length of 2a and D, E and G are midpoints of each side.

FIG. 4.51 – Importance of first identifying zero-force member.

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153

Solution: By inspection of figure 4.51a, member DE is a zero-force member from joint E. Then use section mm to cut the truss and draw the free-body diagram of the lower right part of the truss, figure 4.51b. Summing moments about point B will yield FCD directly. pffiffiffi X 3 a ¼ 0 ) FCD ¼ 0:866F ðCÞ MB ¼ 0 ; FCD  a  F  2 RETHINK: If failing to recognize that DE is a zero-force member, could we still solve this problem? In this section, only simple planar truss is discussed. For simple space truss, the method of joints and the method of sections can also be applied. Only the threedimensional geometry is usually more complicated. PROBLEMS 4.1 Draw the free-body diagram of member AB in equilibrium. The weight of the member is neglected unless it is shown. Assume all the contacting surfaces are smooth and the dimensions are known.

Prob. 4.1

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4.2 Draw the free-body diagram of each member. The weight is neglected unless it is shown. Assume all the contacting surfaces are smooth and the dimensions are known.

Prob. 4.2

4.3 A 10-kN cargo is suspended by two weightless cables as shown. If h ¼ 60 and a ¼ 45 , determine the cable forces.

Equilibrium of Rigid Bodies

155

Prob. 4.3

4.4 The 1-kN block is supported by a system of five cords. Determine the force in each cord.

Prob. 4.4

4.5 Non-extensible cord AB is 20 cm long and spring BC has a stiffness k = 50 N/cm. A vertical force of 100 N is acting at ring B and the angle h becomes 60 for equilibrium. Determine the unstretched length of spring BC.

Prob. 4.5

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4.6 The uniform bar AB has a length l and a weight P. It is held in equilibrium by a cargo weighing P1 through the pulley-cable system. If AC = AB and the size of the pulley is neglected, determine the angle a.

Prob. 4.6

4.7 A couple is applied to a beam as shown in the two cases. Determine the support reactions for each case.

Prob. 4.7

4.8 If q ¼ 3 kN=m, F ¼ 6 kN and M ¼ 5 kNm, determine the reactions at the fixed end A.

Prob. 4.8

Equilibrium of Rigid Bodies

157

4.9 Find the reactions at the supports A and B.

Prob. 4.9

4.10 Find the reactions at the supports A and B when F1 ¼ 20 kN and F2 ¼ 50 kN.

Prob. 4.10

4.11 Determine the reactions at the supports A and B when F = 2 kN and M ¼ 1:5 kNm.

Prob. 4.11

4.12 Find the reactions at the supports A and B.

Prob. 4.12

Engineering Mechanics

158 4.13 Find the reactions at the supports A and B.

Prob. 4.13

4.14 Determine the reactions at the fixed support A.

Prob. 4.14

4.15 The car stops on a 20-m long bridge. The load due to the front wheels is 10 kN and the rear-wheel load is 20 kN. The distance between the front and rear axles is 2.5 m. If the reactions at the supports A and B are equal, specify the distance x from the rear wheel to support A.

Prob. 4.15

Equilibrium of Rigid Bodies

159

4.16 The ununiform plate is held in the horizontal position by three rollers A, B and D. If the reactive forces are FA = 320 N, FB = 300 N and FD = 240 N, determine the weight P of the plate and the coordinates xC and yC of its center of gravity C.

Prob. 4.16

4.17 The 2-kN uniform slab is held in the horizontal position by three vertical cables. Determine the tension in each cable.

Prob. 4.17

4.18 The shaft is supported by a thrust bearing at A and a journal bearing at B, which are properly aligned and only force reactions need to be considered. Two identical wheels C and D are fixed to the shaft. On wheel C, there is a vertical tangent force of 5 kN and a radial force of 2 kN. On wheel D, there is a horizontal tangent force of 5 kN and a radial force of 3 kN. Determine the reactions at the bearings.

Prob. 4.18

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4.19 Determine the reactions at the fixed support A when l = 2 m and a = 1 m.

Prob. 4.19

4.20 A force F1 = 500 N along the z axis is acting on the cylinder beam at the bottom point of section A. Another force F2 = 400 N along the x axis is acting at the top point of section B. The dimensions are l1 = l2 = 1.2 m and d = 0.6 m. The reaction at the fixed support is actually a distributed loading on the contacting surface. It can be simplified to one force plus one couple at point O. Determine the x, y and z components of the force and the couple.

Prob. 4.20

4.21 The linkage mechanism as shown is in equilibrium. The weightless members are pin-connected at their ends. If AB = BC = CD = l and M1 ¼ M2 ¼ M , find the reactions at pins A and D.

Prob. 4.21

Equilibrium of Rigid Bodies

161

4.22 The weightless members are pin-connected at their ends. If OA = 40 cm, O1 B = 60 cm and M1 = 1 Nm, specify the couple moment M2 for equilibrium. Also find the force in member AB.

Prob. 4.22

4.23 Two identical concrete cylinders are placed between two slopes AB and BC. Assume all the contacting surfaces are smooth. Determine the reactions at the contacting points D, E and F if P1 ¼ P2 ¼ 40 kN.

Prob. 4.23

4.24 The concrete cylinder is supported by a two-member frame. Neglect the weight of the frame and the friction. Determine the reactions at pins B and C when the weight P = 5 kN and AD = DB.

Prob. 4.24

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4.25 The three-pinned arch is subjected to a vertical force F as shown. Neglecting the weight of the members, find the reactions at supports A and B.

Prob. 4.25

4.26 Determine the support reactions on the compound beam when Q = 10 kN/m and M = 40 kNm.

Prob. 4.26

4.27 Determine the reactions on the compound beam at the fixed support A and weightless link C.

Prob. 4.27

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163

4.28 Determine the reactions on the compound beam at the supports A, B, C and D.

Prob. 4.28

4.29 A crane is placed on a compound beam. Its weight P1 = 50 kN and the cargo weight P2 = 10 kN are applied to the beam through two rollers E and F. Neglecting the weight of the beam, find the reactions at the supports A, B and D.

Prob. 4.29

4.30 The idealized model of a folding ladder is composed of members AB and AC, which are connected by pin A and cable DE. The ladder is placed on the horizontal smooth ground and a vertical force F is acting on the ladder as shown. Neglect the weight of the ladder. If AB ¼ AC ¼ l, determine the cable force.

Prob. 4.30

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4.31 A bottomless hollow cylinder with a radius R is placed on the horizontal smooth ground. Two identical balls with a weight P and a radius r are in the cylinder. Assuming that all the contacting surfaces are smooth, find the minimum weight of the cylinder P1 min to hold the system in equilibrium.

Prob. 4.31

4.32 Which of the shown structures is/are statically determinate or statically indeterminate? Why?

Prob. 4.32

4.33 The uniform cube A with a size length 2a rests on the horizontal smooth ground and next to the stop block C. Cylinder B with a radius R and a weight P1 is placed on the smooth inclined plane. Neglecting the size of the stop block, determine the minimum weight Pmin of cube A to hold the cylinder in equilibrium.

Prob. 4.33

Equilibrium of Rigid Bodies

165

4.34 The four-member frame is subjected to a vertical force F = 12 kN. Determine the reactions at pins A and B.

Prob. 4.34

4.35 The two-member frame is pin-connected at B. Find the horizontal and vertical components of the reaction at support C.

Prob. 4.35

4.36 Determine the horizontal and vertical force components that the pin at B exerts on member BED.

Prob. 4.36

Engineering Mechanics

166 4.37 Determine the forces in members 1, 2 and 3.

Prob. 4.37

4.38 If F = 10 kN, Q = 6 kN/m and M = 188 kNm, determine the reactions at the fixed support C.

Prob. 4.38

4.39 A vertical force F = 300 kN is acting on the composite structure. Determine the reactions at the fixed support A and the force in member DH.

Prob. 4.39

Equilibrium of Rigid Bodies

167

4.40 For the frame as shown, q0 = 2 kN/m, M = 10 kNm and F = 2 kN. Find the horizontal and vertical force components that the pin at D exerts on member CD.

Prob. 4.40

4.41 Determine the force in each member of the truss.

Prob. 4.41

4.42 Determine the force in each member of the truss.

Prob. 4.42

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4.43 Find the forces in members JD, JE and JI of the truss.

Prob. 4.43

4.44 Determine the forces in members IL, JI, JL and CL of the truss.

Prob. 4.44

4.45 Determine the forces in members 1, 2 and 3 of the truss.

Prob. 4.45

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169

4.46 Determine the force in member 1 of the truss.

Prob. 4.46

4.47 A three-member space truss is used to suspend a block weighing P = 1 kN. Members OB and OC are in the horizontal position and OB = OC. Find the force in each member.

Prob. 4.47

4.48 For the space truss, AB = BC = CA = A0 B 0 = B 0 C 0 = C 0 A0 . A force F, which is parallel to member BC and has a magnitude of 50 kN, is acting at joint A. Determine the force in each member.

Prob. 4.48

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4.49 Six weightless links are used to support plate ABCD in the horizontal position. Two forces act on the plate at points A and B. Find the force in each link.

Prob. 4.49

4.50 Six weightless links are used to support plate ABCD in the horizontal position. A vertical force acts on the plate at point D. Determine the force in each link.

Prob. 4.50

Chapter 5 Friction Objectives
Examine characteristics of friction.
Solve equilibrium problems involving friction force.
Understand the concept of rolling resistance.

5.1

Characteristics of Friction

Prior to this chapter, we always assume that the contacting surface is smooth and the friction is neglected. This is because for many engineering problems, the effect of friction is often very small and can be neglected. However, sometimes frictional forces play an important role. For example, a car brake just utilizes the friction force to stop the car; for a gravity dam or a retaining wall, the friction force is an important factor to hold the dam or the slope in equilibrium, figure 5.1. For these cases, the friction cannot be neglected.

FIG. 5.1 – Engineering cases in which friction plays an important role.

DOI: 10.1051/978-2-7598-2901-9.c005 © Science Press, EDP Sciences, 2022

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Friction can be defined broadly as a resistance that occurs between two surfaces in contact when they tend to move relative to each other. Based on the relative movement between the two surfaces, there are two kinds of friction. When the two contacting bodies tend to slide relative to each other, figure 5.2a, there is sliding friction; whereas when they tend to roll relative to each other, figure 5.2b, rolling resistance occurs. This chapter will mainly discuss sliding friction and give a brief introduction to rolling resistance.

FIG. 5.2 – Sliding friction and rolling resistance. Friction can also be classified as fluid friction and dry friction. If the contacting surfaces are separated by a thin layer of liquid or gas, it is fluid friction; whereas dry friction occurs when there is no lubricating fluid between the contacting surfaces. In this book only the effects of dry friction will be considered.

5.1.1

Laws of Sliding Friction

A simple experiment is used to discuss sliding friction. The block having weight P rests on the rough horizontal ground and horizontal force F is exerted on the block, as shown in figure 5.3a. Static equilibrium. When force F is very small, the block would be still in static equilibrium. Two surfaces tend to, but do not, move relatively. To hold the block in equilibrium, there should be a normal reactive force N = P and a tangent reactive force Fs = F from the ground, figure 5.3b. This tangent reactive force, which prevents or retards sliding of the block relative to the ground, is the friction force or static friction force. It should be noted that normal force N is located at distance d to the right of the line of action of P. This distance can be determined by writing the moment equilibrium equation about point O, i.e., Pd – Fh = 0 or d = Fh/P.

FIG. 5.3 – Three states of the block and the corresponding friction forces.

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173

To fully understand these two reactive forces, it is necessary to consider that the contacting surfaces are nonrigid and have irregularities. Then the actual reactions include nonuniform distributed normal reaction qN and nonuniform distributed frictional reaction qFs along the contacting surface, as shown in figure 5.3c. N and Fs are actually the resultants of the distributed normal and frictional forces. Impending Motion. If the magnitude of F is increased gradually, fictional force Fs increases correspondingly, continuing to oppose F, until it approaches a certain maximum value Fsmax and the block is on the verge of sliding, figure 5.3d. If F is further increased, the friction force could not balance it anymore and the block would start sliding. Besides this impending sliding, there is another type of impending motion: impending tipping. As F is increased, distance d (Fh/P) correspondingly increases. If d increases to a/2, i.e., N acts at the right bottom corner of the block, the block will be on the verge of tipping, or impending tipping. To study the characteristics of the sliding friction, we assume that tipping does not occur in this section. This is often the truth for a wide base block, i.e., “a” is large and “h” is small. Motion. When the magnitude of F is increased to be greater than Fsmax, the block begins to slide. After the block is set in motion, the friction force, denoted by Fk and called the kinetic friction force, generally decreases a little first and then remains approximately constant. This is because there is less interpenetration between the irregularities of the contacting surfaces when the surfaces move relative to each other. C. A. Coulomb (1736–1806) extensively studied characteristics of sliding friction and proposed laws of sliding friction. Through experiments, Coulomb concluded that the maximum static friction force Fsmax is proportional to the magnitude of the normal force N, i.e., Fsmax ¼ ls N ð5:1Þ where μs is a constant, called the coefficient of static friction. Similarly, the kinetic friction force Fk is also proportional to the magnitude of the normal force N, i.e., Fk ¼ lk N

ð5:2Þ

where μk is a constant, called the coefficient of kinetic friction. The values of μs and μk depend mainly on the conditions of the surfaces in contact, such as the roughness, cleanliness and materials of the surfaces. Approximate values for the two coefficients found in many engineering handbooks are given in table 5.1. When accurate values are required, the coefficients should be determined directly by experiment. TAB. 5.1 – Coefficients of sliding friction [4]. Material 1–Material 2 Coefficient of static friction (μs) Coefficient of kinetic friction (μk) Steel–Steel 0.15 0.15 Steel–Cast iron 0.3 0.18 Steel–Bronze 0.15 0.15 Steel–Mild steel – 0.2 Cast iron–Cast iron – 0.15 Leather–Cast iron 0.3  0.5 0.6 Mild steel–Wood 0.6 0.4  0.6 Wood–Wood 0.4  0.6 0.2  0.5

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From the experiment discussed above, the friction force acts tangent to the contacting surface in a sense opposite to the motion tendency or relative motion of the body over its contacting surface. There are three cases for the block’s state and the friction force on the block. (1) When applied force F is not large enough, the block is still in static equilibrium. For this case, the friction force is static friction force Fs and its magnitude is determined from equilibrium equations. (2) When applied force F is increased to a critical value, the block is on the verge of sliding. For this case, the friction force is the maximum static friction force Fsmax and Fsmax ¼ ls N . (3) When applied force F is greater than Fsmax, sliding occurs. The friction force is the kinetic friction force and Fk ¼ lk N . These three cases are illustrated in figure 5.4. From this figure, it can be seen that the static friction force is in a range: 0  Fs  Fsmax .

FIG. 5.4 – Changing process of the friction force.

5.1.2

Angles of Friction and Self-Locking

From figure 5.3b, normal force N and static friction force Fs can be combined to get a resultant reactive force FR. FR makes an angle φ with N. As applied force F is increased until sliding becomes impending, angle φ also grows and reaches a maximum value φs, figure 5.3d. This angle φs, which is the resultant reactive force made with the normal force when the block is on the verge of sliding, is called the angle of static friction. From the geometry, we have tan us ¼

Fsmax ls N ¼ ls ¼ N N

or us = arc tan ls

ð5:3Þ

Just as the static friction force is in a range: 0  Fs  Fsmax , angle φ is also in a range: 0  u  us . When sliding occurs, the friction force is kinetic friction force Fk; similarly, the angle the resultant reactive force made with the normal force is the angle of kinetic friction φk, figure 5.3e. From the geometry, we have tan uk ¼

Fk lk N ¼ lk ¼ N N

or

uk = arc tan lk

ð5:4Þ

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Just like coefficients of friction μs and μk, the angles of friction φs and φk depend on the conditions of the contacting surfaces. From table 5.1, it is seen that generally μs ≥ μk, and then φs ≥ φk. A simple experiment can be performed to measure the angle of static friction φs between two materials in contact, as shown in figure 5.5. The board is made of one material and the block is of another material. The block is placed on the board and subjected to no other active force than its weight P. Let the board be tipped up gradually and slowly.

FIG. 5.5 – A method to measure the angle of static friction. (1) When the inclination angle θ is very small, the block rests on the board and is in static equilibrium. This means resultant reactive force FR balances active force P. They are equal, opposite and colinear (θ = φ), figure 5.5a. (2) Keep increasing the angle of inclination until the block is on the verge of sliding. At this time, the angle between resultant reactive force FR and the normal reaches its maximum value φs. Since the block is in unstable equilibrium, resultant reactive force FR continues to balance P (θ = φs), figure 5.5b. (3) With further increase in inclination angle θ, sliding occurs and the angle between the resultant reactive force FR and the normal is the angle of kinetic friction φk, figure 5.5c. FR is not vertical now and cannot balance P anymore. From this experiment, it can be concluded that the inclination angle at the time of impending sliding is just the angle of static friction φs and then the coefficient of static friction is determined from ls ¼ tan us . Furthermore, when angle θ < φs, the block will rest on the inclined plane no matter how large the resultant active force (just its weight P for this experiment) is. This phenomenon is called self-locking. Self-locking is an important concept in friction analysis. In engineering, sometimes we have to avoid self-locking and sometimes we want to utilize self-locking.

5.2

Problems Involving Sliding Friction

For an equilibrium problem involving sliding friction, the force system must satisfy not only the equations of equilibrium but also the necessary frictional equation. Generally, problems involving friction fall into one of the following two types.

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For the first type, all applied forces and coefficients of friction are known; whereas we have no idea whether the body is at rest or moving. For this type of problem, we can assume that the body is at rest and with a tendency of motion. Then, the friction force should be opposite to the motion tendency in the free-body diagram (see example 5.1). Write the equilibrium equations and solve for the normal force and friction force. If the obtained friction force is a negative scalar, it means the actual sense is opposite to that shown in the free-body diagram. Moreover, we have to compare the magnitude of Fs with maximum static friction force Fsmax ¼ ls N . If jFs j  Fsmax , the body really remains at rest and the obtained friction force is just the actual one. If jFs j [ Fsmax , equilibrium cannot be maintained, and motion takes place. Therefore, the actual friction force is the kinetic friction force, i.e., Fk ¼ lk N . For the second type, the motion is known to be impending in a given direction and the applied forces or the required coefficient of friction are to be determined. For this type of problem, the friction force must be drawn in the free-body diagram with its sense opposite to the impending motion. If there is only one frictional contact in the problem and the impending motion occurs at this contact (see examples 5.2 and 5.3), just write the frictional equation for this contact, i.e., Fs ¼ Fsmax ¼ ls N . If there are two or even more frictional contacts, two cases exist. One case is that impending sliding occurs at all contacts, and the frictional equation Fs = Fsmax = μsN can be written for all contacts (see examples 5.4 and 5.5). Another case is that several possibilities of impending motion exist and the actual situation has to be decided. This case will be illustrated in example 5.6. Example 5.1. The coefficients of friction between the inclined surface and the block are μs = 0.2 and μk = 0.18 respectively. The block weighs P = 980 N and force F = 700 N, which is parallel to the inclined surface, is exerted on the block, figure 5.6a. Determine the friction force.

FIG. 5.6 – Determining the friction force when having no idea of the block’s state. Solution: For this problem, two active forces and the coefficients of friction are known; whereas we have no idea whether the block is at rest or sliding. We can assume that the block is at rest and has a tendency of sliding downward along the inclined surface. Then draw the free-body diagram of the block, in which the friction force is upward along the inclined surface, figure 5.6b. Writing two force equilibrium equations yields

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177 X

Fx ¼ 0 ;

X

Fy ¼ 0 ;

F  P sin a þ Fs ¼ 0 N  P cos 30 ¼ 0

) )

Fs ¼ 210 N N ¼ 848:7 N

The maximum static friction force is Fsmax ¼ ls N ¼ 0:2  848:7 ¼ 169:7 N Since jFs j = 210 N [ Fsmax , the assumption that the block is at rest does not hold and the block actually slides on the inclined surface. The negative sign means that the block is actually sliding upward along the surface. Therefore, the friction force is the kinetic friction force, whose direction is downward along the surface and the magnitude is Fk ¼ lk N ¼ 0:18  848:7 ¼ 153 N RETHINK: If the coefficient of static friction is changed from 0.2 to 0.25, the maximum static friction force becomes Fsmax ¼ ls N ¼ 0:25  848:7 ¼ 212.2 N. Then jFs j ¼ 210 N\Fsmax and the block really remains at rest. The static friction force has a magnitude of 210 N, and the negative sign means its actual sense is opposite to that shown on the free-body diagram, i.e., downward along the surface. Example 5.2. The block weighing P is placed on a slope with an inclination angle a, figure 5.7a. The coefficient of static friction is μs and a [ arctan ls . Specify the magnitude of horizontal force F to hold the block in equilibrium.

FIG. 5.7 – The appropriate range of the active force to hold the block in equilibrium. Solution: Since a [ arctan ls , the block will definitely slide downward if no other active force except weight P is applied to the block. However, if applied horizontal force F is too small, the block will still slide downward along the slope; whereas if F is too large, the block will slide upward. Therefore, to keep the block in equilibrium, the magnitude of F should be in an appropriate range. For the lower limit Fmin , the block is on the verge of sliding downward and the friction force is the maximum static friction force Fsmax1 , which is upward along the surface. From the free-body diagram, figure 5.7b, two equilibrium equations can be written:

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178 X

Fx ¼ 0 ;

X

Fy ¼ 0 ;

Fmin cos a  P sin a þ Fsmax1 ¼ 0

ð1Þ

N1  Fmin sin a  P cos a ¼ 0

ð2Þ

Substituting the frictional equation Fsmax1 ¼ ls N1 and solving gives Fmin ¼

tan a  ls P 1 þ ls tan a

For upper limit Fmax , the block is on the verge of sliding upward and the friction force is maximum static friction force Fsmax2 , which is downward along the surface. From the free-body diagram, figure 5.7c, write two equilibrium equations: X Fx ¼ 0 ; Fmax cos a  P sin a  Fsmax2 ¼ 0 ð3Þ X

Fy ¼ 0 ;

N2  Fmax sin a  P cos a ¼ 0

ð4Þ

Substituting frictional equation Fsmax2 ¼ ls N2 and solving equations (3) and (4), we obtain Fmax ¼

tan a þ ls P 1  ls tan a

Therefore, to keep the block in equilibrium, the magnitude of active force F should be in an appropriate range as follows: tan a  ls tan a þ ls P F  P 1 þ ls tan a 1  ls tan a Example 5.3. The brake system is shown in figure 5.8a. The coefficient of static friction between the brake pad and the drum is μs. Determine the required magnitude of vertical force F to stop the drum. Neglect the size of the brake pad.

FIG. 5.8 – The required active force to stop the drum.

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Solution: Assume that the drum is in unstable equilibrium due to vertical applied force F on the level. From the free-body diagram of the drum, figure 5.8b, we get X Pr MO1 ¼ 0; Pr  Fsmax R ¼ 0 ) Fsmax ¼ R Since Fsmax ¼ ls N , the normal force is N ¼

Pr ls R

From the free-body diagram of the lever OAB, figure 5.8c, write the moment equilibrium equation. X MO ¼ 0; Fa  Nb  Fsmax c ¼ 0 Solving, we obtain F¼

  Pr b þc aR ls

This is the minimum magnitude of vertical applied force F to stop the drum. Example 5.4. Neglecting the mass of the wedge, find smallest horizontal force F needed to lift the 100-kg cargo shown in figure 5.9a. The coefficient of static friction at all contacting surfaces is μs = 0.25.

FIG. 5.9 – The needed active force to lift the cargo. Solution: When smallest horizontal force F required to lift the cargo is exerted, all the contacting surfaces are on the verge of sliding. The wedge and the cargo are in unstable equilibrium. Their free-body diagrams are shown in figure 5.9b and c, respectively. Keep in mind that the direction of each friction force should be opposite to the corresponding impending sliding. The dimensions of the cargo and the wedge

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are not important since neither of them will “tip” and the moment equilibrium equation will not be used. Due to impending sliding, the frictional equations can be written. Fs1 ¼ ls N1 ; Fs2 ¼ ls N2 ; Fs3 ¼ ls N3 For the cargo, figure 5.9c, there are two force equilibrium equations. X Fy ¼ 0 ; N2  Fs3  981 ¼ 0 X

Fx ¼ 0 ;

N3  Fs2 ¼ 0

Solving simultaneously yields N2 ¼ 1046 N; Fs2 ¼ 261:6 N; N3 ¼ 261:6 N; Fs3 ¼ 65:4 N For the wedge, figure 5.9b, writing two equilibrium equations and solving yields X Fy ¼ 0 ; N1 cos 20  N2  Fs1 sin 20 ¼ 0 ) N1 ¼ 1225 N X

Fx ¼ 0 ;

F  N1 sin 20  Fs1 cos 20  Fs2 ¼ 0

)

F ¼ 968 N

Example 5.5. Beam AB weighing 200 N is held in the horizontal position by a wedge at B, figure 5.10a. The coefficient of static friction between the wedge and the contacting surfaces is μs = 0.3. Neglecting the weight of the wedge, find horizontal force F required to remove the wedge from under the beam.

FIG. 5.10 – The required active force to remove the wedge. Solution: When minimum force F required to remove the wedge is exerted, the two contacting surfaces are on the verge of sliding. The beam and the wedge are in unstable equilibrium. Their free-body diagrams are shown in figure 5.10b and c, respectively. Due to impending sliding, the frictional equations can be written. FsB ¼ ls NB ; FsB1 ¼ ls NB1 For the beam, figure 5.10b, summing moments about point A yields X MA ¼ 0 ; NB  2  200  1 ¼ 0 ) NB ¼ 100 N

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For the wedge, figure 5.10c, writing two equilibrium equations, substituting the frictional equations and solving yield X Fy ¼ 0 ; NB1 cos 15 þ FsB1 sin 15  NB ¼ 0 ) NB1 ¼ 95:82 N X

Fx ¼ 0 ;

FsB þ FsB1 cos 15  F  NB1 sin 15 ¼ 0

)

F ¼ 32:97 N

Example 5.6. Two identical bricks are stacked on the ground, figure 5.11a. The coefficient of static friction between the two bricks is lsA ¼ 0:3 and that between the brick and the ground is lsB ¼ 0:1. If P ¼ 40 N, a = 240 mm and b = 100 mm, determine the limit magnitude of horizontal force F that can hold the system in equilibrium.

FIG. 5.11 – A problem with several types of impending motion.

Solution: When the applied force F is increased gradually, four types of impending motion may occur: brick A slides relative to brick B; brick A tips over; brick B slides relative to the ground; brick B tips over. Calculate the four magnitudes of the force F corresponding to the four types of impending motion and the smallest one is the limit magnitude of the force F that can still hold the system in equilibrium. The free-body diagram of brick A is shown in figure 5.11b. For the impending sliding of brick A, the friction force Fs1 ¼ Fsmax1 ¼ lsA N1 can be written. Then write the force equilibrium equations: X Fy ¼ 0 ; N1  P ¼ 0 ) N1 ¼ 40 N X

Fx ¼ 0 ;

F1  Fs1 ¼ 0

)

F1 ¼ Fs1 ¼ lsA N1 ¼ 12 N

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For the impending tipping of brick A, d1 ¼ b=2. Summing moments about point O1 yields X MO1 ¼ 0 ; Pd1  F2 a ¼ 0 ) F2 ¼ 8:33 N The free-body diagram of the two bricks together as a “system” is shown in figure 5.11c. For the impending sliding of brick B on the ground, friction force Fs2 ¼ Fsmax2 ¼ lsB N2 can be written. Write the force equilibrium equations: X Fy ¼ 0 ; N2  2P ¼ 0 ) N2 ¼ 80 N X

Fx ¼ 0 ;

F3  Fs2 ¼ 0

)

F3 ¼ Fs2 ¼ lsB N2 ¼ 8 N

For the impending tipping of brick B, d2 ¼ a=2. Summing moments about point O2 yields X MO2 ¼ 0 ; 2Pd2  F4 ða þ bÞ ¼ 0 ) F4 ¼ 28:2 N Therefore, the limit magnitude of horizontal force F that can hold the system in equilibrium is minðF1 ; F2 ; F3 ; F4 Þ ¼ 8 N

5.3

Rolling Resistance

In this section, we will give a brief introduction to rolling resistance. A simple experiment is used to discuss the rolling resistance. Horizontal force F is exerted on a cylinder that is placed on the rough ground, figure 5.12a. The cylinder has a weight P and a radius r. When F is very small and the ground is rough enough, the cylinder will remain at rest. To keep the cylinder in static equilibrium, there should be a normal reactive force N = P and a friction force Fs = F P at the contacting point, figure 5.12b. To satisfy the moment equilibrium equation, MA ¼ 0 , there should be a counterclockwise couple moment Mr = Fr from the ground to balance the clockwise turning effect caused by F. This couple Mr is rolling resistance couple. Just like the friction force, whose value is within a range of 0  Fs  Fsmax , rolling resistance couple moment Mr is also within a range of 0  Mr  Mrmax . When applied force F is increased gradually, rolling resistance couple moment Mr = Fr correspondingly increases. When it attains maximum rolling resistance couple moment Mrmax, the cylinder is on the verge of rolling. If F is further increased, the rolling resistance couple moment cannot balance clockwise moment Fr anymore, and the cylinder starts rolling.

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FIG. 5.12 – Illustration of rolling resistance.

From experiment, it is concluded that maximum rolling resistance couple moment Mrmax is proportional to the magnitude of normal force N, namely Mrmax ¼ dN

ð5:5Þ

where δ is a constant, called the coefficient of rolling resistance. It should be noted that a constant coefficient is generally dimensionless. The coefficients of friction, μs and μk, are both dimensionless. However, this coefficient of rolling resistance has the dimension of length. To understand this, it is necessary to consider that the contacting surfaces are nonrigid or deformable. Then the actual reactions from the ground are nonuniform distributed forces, figure 5.12c. This complicated distributed force system can be simplified to resultant force FRA at point A and resultant couple Mr, figure 5.12d. This resultant couple is just the rolling resistant couple. Resultant force FRA is then divided into normal force N and friction force Fs, figure 5.12b. For normal force N and couple Mr, we can further simplify them by moving N off its line of action and get single force N 0 , figure 5.12e. Distance d is d¼

Mr N

ð5:6Þ

When the cylinder is on the verge of rolling, the rolling resistance couple moment is Mrmax and the distance is also the maximum, i.e., dmax ¼

Mrmax ¼d N

ð5:7Þ

This means δ equals the maximum distance normal force N can be moved from the contacting point due to the rolling resistance. Therefore, coefficient δ has the dimension of length. Just like the coefficients of friction, μs and μk, the coefficient of rolling resistance also depends mainly on the conditions of the surfaces in contact, such as the roughness and materials of the surfaces. Table 5.2 lists some coefficients of rolling resistance commonly encountered in engineering handbooks. Pay attention to the unit: millimeter. So, the coefficient of rolling resistance for cast iron on cast iron is 0.5 mm. The smallest coefficient in this table is 0.05 mm and the biggest one is 10 mm. Therefore, the coefficients of rolling resistance are commonly very small and then maximum rolling resistant moment Mrmax = δ N is also very small. This is the reason why the rolling resistant is often neglected.

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TAB. 5.2 – Coefficients of rolling resistance [5] (Unit: mm). Material 1–Material 2 Cast iron–Cast iron Steel wheel–Steel rail Wood–Steel Wood–Wood

d 0.5 0.5 0.3  0.4 0.5  0.8

Material 1–Material 2 Mild steel–Steel Cork–Cork Steel wheel–Wood Tire–Pavement

d 0.05 1.5 1.5  2.5 2  10

Actual experiences tell us that rolling a cylinder is much easier than sliding it and people in ancient times invented carts for transportation for this reason, figure 5.13. Why is rolling a cylinder much easier than sliding it? The simple experiment shown in figure 5.12 is used again for the analysis.

FIG. 5.13 – An imitation of ancient Chinese cart. When the cylinder approaches impending rolling, the rolling resistant couple moment is the maximum rolling resistant couple moment, i.e., Mr = Mrmax = δN = δP. For this unstable equilibrium state, the moment equilibrium equation should also be satisfied, i.e., Mr = F1r. Then we get d F1 ¼ P r This is the required magnitude of F to roll the cylinder. Similarly, when the cylinder approaches impending sliding, the friction force is the maximum static friction force, i.e., Fs = Fsmax = μsN = μsP. For this unstable equilibrium state, the force equilibrium equation in the horizontal direction should also be satisfied, i.e., Fs = F2. Then we get F2 ¼ ls P This is the required magnitude of F to slide the cylinder. Generally, d=r is much smaller than μs and then F1 is much smaller than F2. For example, the weight of the cylinder is P = 10 kN and the radius is r = 0.1 m. The coefficient of static friction is μs = 0.5, which is the average coefficient of wood on wood. The coefficient of rolling resistance is δ = 0.65 mm, which is the average coefficient of rolling resistance of wood on wood. From the two foregoing expressions,

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185

we have F1 = 65 N and F2 = 5000 N. Therefore, minimum force F to move, actually to roll, the cylinder is 65 N. If we want to slide the cylinder, the required force is 5000 N. Thus, the force needed to roll the cylinder is much smaller than that needed to slide the cylinder. This is the reason why rolling a cylinder is much easier than sliding it. As applied force F is increased gradually, there are three stages for the cylinder as follows: (1) If F < 65 N, the cylinder will remain at rest. The static friction force and the rolling resistance couple moment can be determined from the equilibrium equations. (2) If 65 N < F < 5000 N, the cylinder will be rolling without sliding, which is one important concept in dynamics. When the cylinder is rolling without sliding, the friction force is still the static friction force and this static friction force can be determined from the equations of motion, which will be discussed in chapter 8. At the same time, there is the kinetic rolling resistant couple moment, which can be considered equal to the maximum rolling resistant couple moment Mrmax. However, this couple moment is often neglected since it is very small. (3) If F > 5000 N, the cylinder will be rolling and sliding. There are kinetic friction force and kinetic rolling resistant couple moment at the contacting surface. Still this couple moment is often neglected. PROBLEMS 5.1 Coefficients of friction between the inclined surface and the block are μs = 0.3 and μk = 0.2, respectively. Horizontal force F = 100 N acts on the block as shown. Determine the friction force acting on the block.

Prob. 5.1

5.2 Coefficients of friction between the inclined surface and the block are μs = 0.2 and μk = 0.18, respectively. Force F = 10 kN, which is parallel to the surface, acts on the block. Determine the friction force acting on the block.

Prob. 5.2

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186

5.3 A simple device is used to transport the concrete bucket weighing 25 kN. The coefficient of friction between the bucket and the slideway is 0.3. Find the tension of the rope when the bucket rises or falls at a constant speed, respectively.

Prob. 5.3

5.4 The uniform bar weighing 15 N is resting on the rough ground at A and against the smooth wall at B. The coefficient of static friction between the ground and the bar is μs = 0.78. Specify horizontal force F that must be exerted to move the bar. (Hint: The motion can be rotation about A or sliding at A.)

Prob. 5.4

5.5 In order to rotate a uniform cylinder, which is placed in a V-shaped groove, couple moment M = 1500 Ncm is needed. If the cylinder weighs P = 400 N and its diameter is d = 25 cm, specify the coefficient of friction between the cylinder and the groove.

Prob. 5.5

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187

5.6 The 50-kg uniform cylinder rests against the wall at A and on the ground at B. The coefficients of friction at the contacting points are μsA = 0.6 and μsB = 0.5. Find couple moment M needed to rotate the cylinder.

Prob. 5.6

5.7 The 100-N uniform ladder is placed on the rough ground and against the smooth wall. The coefficient of static friction between the ground and the ladder is 0.3. Determine appropriate angle α to assure that a man with a weight of 700 N can climb to the top of the ladder safely. Neglect the size of the man.

Prob. 5.7

5.8 A block weighing P = 12 N is pressed against the vertical wall by force F = 20 N as shown. The coefficients of friction between the block and the wall are μs = 0.5 and μk = 0.2, respectively. Determine the friction force acting on the block.

Prob. 5.8

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188

5.9 The 500-N drum is placed on the rough ground and against the smooth wall. The coefficient of static friction between the drum and the ground is μs = 0.25. A block is suspended to a cable, which is wrapped around the inner core of the drum. If R = 20 cm, r = 10 cm, find the maximum weight of the block that can keep the drum in equilibrium.

Prob. 5.9

5.10 A uniform bar with weight P and length 2b is placed on the ground and against the semicircular bulge as shown. The coefficient of static friction at the contacting surfaces is μs. Specify the maximum value of angle φ that can hold the bar in equilibrium.

Prob. 5.10

5.11 Vertical force F is acting on block B, which is placed on triangular prism A. The coefficient of static friction between A and B is μs. Neglect the friction at other contacting surfaces and the weights of A and B. Determine (a) the minimum magnitude of force F1 required to prevent block B from falling down; (b) the maximum magnitude of force F1 permitted to prevent block B from moving upward.

Prob. 5.11

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189

5.12 The wheel is subjected to a couple moment of 5 kNm. Determine smallest force F that should be applied to the block brake to stop the wheel from rotating. Neglect the weight of the members and the coefficient of friction between the wheel and the block brake is 0.25.

Prob. 5.12

5.13 Iron plate A weighs 2000 N. A 5000-N block B is resting on it. Block B is attached to the wall with a rope as shown in the two cases. The coefficient of static friction between iron plate A and the ground is lsA ¼ 0:2 and that between iron plate A and block B is lsB ¼ 0:25. Determine the minimum magnitude of horizontal force F to pull the iron plate out for the two cases.

Prob. 5.13

5.14 Two identical uniform bars AB and BC are connected at B by a smooth pin. When ΔABC is an equilateral triangle, the two bars are in unstable equilibrium in the vertical plane on the rough ground. Find the coefficient of static friction between the bars and the ground.

Prob. 5.14

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190

5.15 Block A weighing 100 N and block B weighing 800 N are connected by a rope and placed on a ramp as shown. The coefficient of friction between the ramp and block A is 0.2, and that between the ramp and block B is 0.4. If the angle θ = 20°, determine the friction force acting on block B. 5.16 Two blocks A and B are connected by a rope and placed on a ramp as shown. The coefficient of friction between the ramp and block A is 0.2, and that between the ramp and block B is 0.4. If the angle θ = 20° and the weight of block A is 100 N, determine the minimum weight of block B to hold it in equilibrium. 5.17 Two blocks A and B are connected by a rope and placed on a ramp as shown. The coefficient of friction between the ramp and block A is 0.2, and that between the ramp and block B is 0.6. If the angle θ = 30° and the weight of block B is 200 N, determine the maximum weight of block A to hold the system in equilibrium.

Prob. 5.15/16/17

5.18 The 800-N refrigerator rests on the floor and the coefficient of friction is μs = 0.25. Determine the smallest magnitude of F that will cause impending motion (tipping or sliding) of the refrigerator.

Prob. 5.18

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191

5.19 Horizontal force F is exerted on a cylinder as shown to move it on the rough ground. The cylinder has weight P and radius R. The coefficient of rolling resistance is δ. What condition should the coefficient of static friction meet to ensure that the cylinder is rolling without sliding?

Prob. 5.19

5.20 Member AC is fixed-supported at A and pin-connected with CB at C. A weightless cable passing over a smooth pulley O is attached to block E at one end and to member CB at point D at the other end. Block E weighing P = 6 kN is placed on an inclined surface. The coefficient of static friction between block E and the surface is 0.3. Determine (a) the appropriate range x of the distributed loading to ensure that the system is in equilibrium; (b) the reactions at the fixed support A when x = 2 m.

Prob. 5.20

5.21 Block A weighs 500 N and drum B weighs 1000 N. A weightless rope around the inner hub of drum B is tied horizontally to block A. Another rope around the outer rim of drum B passes over smooth pulley D and is attached to block C. The coefficients of static friction are 0.5 between block A and the plane and 0.2 between drum B and the ground. If r = 5 cm, R = 10 cm, determine the maximum weight of block C to ensure that the system is in equilibrium. Neglect the rolling resistance.

Prob. 5.21

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192

5.22 A weightless rope around the inner hub of drum B passes over smooth pulley A and is attached to a block weighing P1. The system is in equilibrium. The weight of drum B is P. The rope segment AB makes an angle α with the vertical line. Determine the friction force, normal force and rolling resistance couple moment at the contacting point C.

Prob. 5.22

5.23 A 2-kN cylinder is placed on the ramp and against the handle OA, which is in the vertical position as shown and subjected to a horizontal force F at its end. The radius of the cylinder is 10 cm. If the coefficient of friction at the contact points B and D is 0.1, specify the maximum magnitude of the horizontal force to keep the system in equilibrium.

Prob. 5.23

Chapter 6 Kinematics of Particles Objectives
Understand the kinematic concepts of position, displacement, velocity, and acceleration.
Investigate a particle’s curvilinear motion using different coordinate systems.
Study absolute dependent motion of two particles.
Analyze relative motion of two particles using a translating coordinate system. Statics deals with forces that lead to equilibrium of particles or bodies. However, kinematics cares about only the geometrical aspects of the motion, such as position, displacement, velocity and acceleration, without involving forces. The analysis object in kinematics can also be considered as a particle or a rigid body. If the motion analyzed is characterized by the motion of the object’s mass center and the dimensions (size and shape) have no or little influence on the motion, this object can be considered as a particle. For example, when we analyze a plane’s trajectory or its position as a function of time, figure 6.1, it can be considered as a particle although it is not small at all to our naked eyes. But if we analyze the fighter plane’s rotation and flipping, figure 6.2, it has to be considered as a body since the dimensions influence the motion greatly. So, we can see that whether an object can be considered as a particle or not is decided by its motion analysis, not by its physical dimensions. Following the cognitive process of learning from the simple to the complex, this chapter considers the motion of a particle, and the motion of a rigid body will be discussed in the next chapter.

DOI: 10.1051/978-2-7598-2901-9.c006 © Science Press, EDP Sciences, 2022

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FIG. 6.1 – Plane’s trajectory.

FIG. 6.2 – Plane’s flight attitude.

6.1

General Curvilinear Motion

When a particle moves along a straight line, it undergoes rectilinear motion; whereas a particle moving along a curved path undergoes curvilinear motion. Rectilinear motion has been considered extensively in physics and it will be treated as a special case of curvilinear motion. The kinematics of a particle includes specifying the particle’s position, displacement, velocity, and acceleration. Position. A particle is moving along a space curve, figure 6.3. The position of the particle at any instant can be defined by position vector r measured from fixed point O. Apparently, both the magnitude and direction of the vector r will change as the particle moves along its curved path. Then the position vector is a single-valued function of time as follows: r ¼ rðtÞ

ð6:1Þ

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FIG. 6.3 – Path, position, displacement, average velocity and instantaneous velocity. Displacement. The displacement of a particle represents the change in its position. Suppose that at a given instant t, the particle is at point P on the curve and position vector is r, as shown in figure 6.3. After a small time interval Δt, the particle moves to new position P 0 and the position vector becomes r0 . The displacement is then determined by Dr ¼ r0  r. Velocity. The velocity of a particle is defined as the time rate of change in its position. During the time interval Δt, the average velocity of the particle is Dr Dt

v avg ¼

Then the instantaneous velocity is determined by letting Dt ! 0, i.e., Dr dr ¼ Dt!0 Dt dt

v ¼ lim v avg ¼ lim Dt!0

ð6:2Þ

Since the direction of Dr gradually approaches the tangent to the curved path as Dt ! 0, the direction of dr, also the direction of v, is tangent to the path, as shown in figure 6.3. The magnitude of v, also called the speed, is usually expressed in units of m/s or km/h (kilometer per hour). Acceleration. The acceleration of a particle is the time rate of change in its velocity. Suppose that the particle has velocity v at given instant t and velocity v 0 after a small time interval Δt, figure 6.4a. The change in the particle’s velocity during the time interval is Dv ¼ v 0  v and the average acceleration is aavg ¼

Dv Dt

Then the instantaneous acceleration is determined by letting Dt ! 0, i.e., Dv dv d2 r ¼ ¼ Dt!0 Dt dt dt 2

a ¼ lim aavg ¼ lim Dt!0

ð6:3Þ

The magnitude of a is usually expressed in units of m/s2 or km/h2. To analyze the direction of a, the velocity vectors in figure 6.4a are plotted from a fixed point O 0 and their arrowheads touch points on a curve, which is called a hodograph,

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figure 6.4b. It can be seen that the direction of Dv gradually approaches the tangent to the hodograph as Dt ! 0. Hence, the direction of dv, also the direction of a, is tangent to the hodograph. In summary, the acceleration a is tangent to the hodograph, while the velocity v is tangent to the path.

FIG. 6.4 – Hodograph, average acceleration and instantaneous acceleration

6.2

Curvilinear Motion: Rectangular Components

The motion of a particle along a space curve can be described using a rectangular or Cartesian coordinate system Oxyz, with its origin at a fixed point O, figure 6.5. Position. For a particle moving along a curve in Cartesian space, figure 6.5, its position at any instant may be described by rectangular coordinates (x, y, z). Its position vector can be written as r ¼ xi þ yj þ zk

ð6:4Þ

where i; j and k are the unit vectors along x, y and z axes, respectively. Since r ¼ rðtÞ, we have x ¼ xðtÞ;

y ¼ yðtÞ;

z ¼ zðtÞ

ð6:5Þ

It means the rectangular components are all single-valued functions of time. From the rectangular components, we can get the magnitude and direction cosines of r as follows: pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x y z r ¼ x 2 þ y 2 þ z 2 ; cosðr; iÞ ¼ ; cosðr; jÞ ¼ ; cosðr; kÞ ¼ r r r Velocity. From equation (6.2), velocity of the particle is determined by the first time derivative of its position vector r, i.e., v¼

dr d dx dy dz di dj dk ¼ ðxi þ yj þ zkÞ ¼ iþ jþ kþx þy þz dt dt dt dt dt dt dt dt

ð6:6Þ

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FIG. 6.5 – Describing a particle’s curvilinear motion in a rectangular coordinate system. Since the Oxyz reference frame is fixed, both the direction and magnitude of the unit vectors i; j and k do not change with time. Thus, the last three terms in equation (6.6) are zero and we have v¼

dr dx dy dz ¼ iþ jþ k dt dt dt dt

ð6:7Þ

The rectangular components of v are vx ¼

dx _ ¼ x; dt

vy ¼

dy _ ¼ y; dt

vz ¼

dz ¼ z_ dt

ð6:8Þ

_ represents the first time derivative [2, 5]. The The “dot” notation, such as x, magnitude and direction cosines of v can be determined from its rectangular components. qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi vx vy vz v ¼ vx2 þ vy2 þ vz2 ; cosðv; iÞ ¼ ; cosðv; jÞ ¼ ; cosðv; kÞ ¼ ð6:9Þ v v v In addition, the direction of v should be tangent to the path curve, as shown in figure 6.5. Acceleration. From equation (6.3), acceleration of the particle is determined by the first time derivative of v or the second time derivative of r, i.e., a¼

dv dvx dvy dvz d2 x d2 y d2 z ¼ iþ jþ k ¼ 2 iþ 2 jþ 2 k dt dt dt dt dt dt dt

ð6:10Þ

In the above derivation, the fact that the time derivatives of i; j and k are zero is utilized again. The rectangular components of a are ax ¼

dvx d2 x ¼ 2 ¼ x€; dt dt

ay ¼

dvy d2 y ¼ 2 ¼ y€; dt dt

az ¼

dvz d2 z ¼ 2 ¼ €z dt dt

ð6:11Þ

The “double dot” notation, such as x€, represents the second time derivative [2, 5]. The magnitude and direction cosines of a can be determined from its rectangular components.

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a¼

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ax2 þ ay2 þ az2 ;

cosða; iÞ ¼

ax ; a

cosða; jÞ ¼

ay ; a

cosða; kÞ ¼

az a

ð6:12Þ

The direction of a is tangent to the hodograph, not to the path curve, as shown in figure 6.5. It can be seen from the above equations that a curvilinear motion can be divided into three rectilinear motions along x, y and z axes, respectively. Considering these three rectangular component motions, the changes in both the magnitude and direction of the particle’s position, velocity and acceleration are automatically taken into account. Example 6.1. A particle starting from rest moves in a straight line with an acceleration a ¼ ð0:2 þ t 2 Þm/s2 , where t is the time in seconds. Find the velocity and the distance traveled in 5 s. Solution: Since acceleration a is a function of time t and a ¼ dv=dt, the particle’s Rv Rt velocity can be determined from v0 dv ¼ 0 adt. Noting that v = 0 when t = 0, we have Z v Z t   t3 dv ¼ 0:2 þ t 2 dt ) v ¼ 0:2t þ 3 0 0 Velocity v is also a function of time t. From v ¼ dx=dt, the particle’s position can Rx Rt be determined by the integration x0 dx ¼ 0 vdt. Since x = 0 when t = 0, we get  Z x Z t t3 t4 dx ¼ 0:2t þ dt ) x ¼ 0:1t 2 þ 3 12 0 0 Therefore, when t = 5 s, the velocity and the distance traveled are as follows: v ¼ 0:2  5 þ

53 ¼ 42:7 m/s 3

x ¼ 0:1  52 þ

54 ¼ 54:6 m 12

Example 6.2. A particle travels along a straight line with an acceleration a ¼ ð2 þ 0:1x Þm/s2 , where x is in meters. When t = 0, v = 0 and x = 0. Find the particle’s velocity when x = 50 m and the time needed to travel this distance. Solution: Acceleration a is a function of the position coordinate x. From a ¼ dv=dt and v ¼ dx=dt, eliminating dt yields vdv ¼ adx This equation can be used to determine the particle’s velocity when x = 50 m, since it relates a, v, and x. Noting that v = 0 when x = 0, we get

Kinematics of Particles

199 Z

v

Z

x

vdv ¼

0

v 2 =2 ¼ 2x þ 0:05x 2

ð2 þ 0:1x Þdx

0

)

v¼

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 4x þ 0:1x 2

The positive root is chosen since the particle is moving in the +x direction. When x = 50 m, the velocity is pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi v ¼ 4  50 þ 0:1  502 ¼ 21:2 m/s The time needed to travel the distance of 50 m can be obtained using v ¼ dx=dt. pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi dx dx ) dt ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi v ¼ 4x þ 0:1x 2 ¼ dt 4x þ 0:1x 2 Noting that x = 0 when t = 0, we can determine the time needed to travel the distance of 50 m as follows: Z t Z 50 dx pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi dt ¼ 4x þ 0:1x 2 0 0 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi  50 pffiffiffiffiffiffiffi 1 4 t ¼ pffiffiffiffiffiffiffi ln 4x þ 0:1x 2 þ x 0:1 þ pffiffiffiffiffiffiffi ¼ 6:09 s 0:1 2 0:1 0 Example 6.3. A car travels along a straight road with a constant acceleration a = 5000 km/h2, figure 6.6. When t = 0, v ¼ v0 ¼ 10 km/h. Determine the time needed to reach a speed of 100 km/h and the distance traveled during this time interval.

FIG. 6.6 – A problem of rectilinear motion. Solution: Express the acceleration and two velocities in SI units.    km 1000 m 1h 2 a ¼ 5000 km/h2 ¼ 5000 2 ¼ 0.3858 m/s2 1 km 3600 s h    km 1000 m 1h v0 ¼ 10 km/h ¼ 10 ¼ 2.778 m/s h 1 km 3600 s

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v ¼ 100 km/h ¼ 100

   km 1000 m 1h ¼ 27.78 m/s h 1 km 3600 s

For the rectilinear motion with a constant acceleration, we have v ¼ v0 þ at; x ¼ x0 þ v0 t þ

1 2 at 2

So, when v ¼ 100 km/h ¼ 27.78 m/s, the elapsed time is t¼

v  v0 27:78  2:778 ¼ 64:8 s ¼ 0:3858 a

The distance traveled during this time interval is x ¼ 0 þ v0 t þ

1 2 at ¼ 990 m 2

Example 6.4. From the origin O, a boy walks east to point A in 4 s, then north to point B in 6 s, and then west to point C in 10 s, figure 6.7a. Determine the magnitude of the average velocity and average speed. (The average speed is defined as the total distance traveled divided by the total time elapsed.)

FIG. 6.7 – Determining the average velocity and average speed. Solution: The elapsed time is Dt ¼ 4 þ 6 þ 10 ¼ 20 s The displacement during this time interval shown in figure 6.7b is Dr ¼ f3i þ 4jg m Then the magnitude of average velocity is qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð3Þ2 þ 42 jDrj 5 v avg ¼ ¼ ¼ 0:25 m/s ¼ Dt 20 20

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201

The total distance traveled during this time interval is sT ¼ 3 þ 4 þ 6 ¼ 13 m Then the average speed is vavg sp ¼

sT 13 ¼ 0:65 m/s ¼ Dt 20

Example 6.5. The ball at O is kicked with a velocity v0 at an angle θ0 = 30°. It then strikes the slope at P, figure 6.8. Neglect the air resistance and the size of the ball. Determine the initial speed v0 and the speed with which it strikes the slope.

FIG. 6.8 – Free-flight motion.

Solution: After the kick, the motion of the ball is the free-flight motion of a projectile. The rectangular components of a projectile’s acceleration are ax ¼ 0;

ay ¼ g

where g is the acceleration of gravity. In engineering problems, g is a constant, generally taken as 9:81 m/s2 . Then, the ball’s motion is divided into two rectilinear motions: the horizontal motion with zero acceleration and the vertical motion with constant acceleration. The rectangular components of the ball’s velocity and position are vx ¼ v0x ¼ v0 cos h0 ¼ v0 cos 30 ¼ 0:866v0 vy ¼ v0y  gt ¼ v0 sin 30  9:81t ¼ 0:5v0  9:81t x ¼ x0 þ v0x t ¼ 0:866v0 t 1 y ¼ y0 þ v0y t  gt 2 ¼ 0:5v0 t  4:905t 2 2 in which v0x ¼ v0 cos h0 , v0y ¼ v0 sin h0 , x0 ¼ 0 and y0 ¼ 0. At position P, we have x = 20 m and y ¼ 3 m. Then

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202 20 ¼ 0:866v0 t 3 ¼ 0:5v0 t  4:905t 2

solving the equations simultaneously and taking positive root of t yields t ¼ 1:72 s;

v0 ¼ 13:4 m/s

Example 6.6. A wheel of radius r rolls without slipping on the horizontal ground. Its center C is undergoing a rectilinear motion with a constant speed vC. Determine the velocity of a particle on the rim of the wheel when the particle is at the highest position and the lowest position (contacting the ground) respectively.

FIG. 6.9 – A wheel rolling without slipping. Solution: For convenience, consider particle M that is coincident with the origin O of the coordinate system when t = 0. At any arbitrary instant, the location of particle M is shown in figure 6.9. Because the wheel rolls without slipping, we have d ¼ ru, in which u is the angle between the radial line CM and the vertical OA ¼ AM line CA. Since C is undergoing a rectilinear motion with a constant speed vC, we have OA ¼ vC t ¼ ru. Therefore, u¼

vC t r

From the geometry, the coordinates of particle M are

9 vC t > = r vC t > ; y ¼ CA  CB ¼ r  r cos u ¼ r  r cos r

x ¼ OA  MB ¼ ru  r sin u ¼ vC t  r sin

ð1Þ

Taking time differentiation of equation (1) and applying the chain rule of calculus, the components of the velocity are 9 vC t > vx ¼ vC  vC cos = r ð2Þ vC t > ; vy ¼ vC sin r

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The components of the acceleration are determined from time differentiation of equation (2) and application of the chain rule. 9 v2 vC t > > = ax ¼ C sin r r ð3Þ vC2 vC t > > ; ay ¼ cos r r When particle M is at the highest position M1 , figure 6.9, the angle u ¼ vC t=r ¼ p. Substituting this into equations (2) and (3), the components of the velocity and acceleration are v1x ¼ vC  vC cos p ¼ 2vC ; a1x ¼ 0;

a1y ¼ 

v1y ¼ 0

vC2 r

It means that at the highest position M1, the velocity of the particle v1 is in the +x direction and its magnitude is 2vC ; the acceleration of the particle a1 is in the y direction and its magnitude is vC2 =r, figure 6.9. When particle M is at the lowest position M2 , figure 6.9, the angle u ¼ vC t=r ¼ 2p. Substituting this into equation (2) and (3), the components of the velocity and acceleration are v2x ¼ 0;

v2y ¼ 0

a2x ¼ 0;

a2y ¼

vC2 r

It means that at the lowest position M2, i.e., particle M is contacting the ground, the velocity of particle v2 is zero; the acceleration of particle a2 is in the +y direction and its magnitude is vC2 =r, figure 6.9. From the above discussion, it can be seen that for a wheel rolling without slipping, the point of contact with the ground has zero velocity. This point is called the instantaneous center of rotation, which will be discussed in detail in §§ 7.4.2. It should be noted that the acceleration of the instantaneous center of rotation is generally not zero.

6.3

Curvilinear Motion: Tangential and Normal Components

Sometimes the path along which a particle moves is known. For example, a train can only move along the railway and its path is known in advance. Apparently, the particle’s position can be defined by distance, s, along the known curved path from fixed reference point O, as shown in figure 6.10, and s is a single-valued function of time as follows: s ¼ sðtÞ

ð6:13Þ

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For this case, it is generally more convenient to describe the particle’s velocity and acceleration using tangential (t) and normal (n) components than rectangular components. The tangential and normal coordinate system will be introduced first.

FIG. 6.10 – A particle’s motion along a known curved path.

6.3.1

The t-n-b Coordinate System

A particle is moving along a fixed known curve, figure 6.11a. Consider a coordinate system that has its origin located on the moving particle P, i.e., the origin moves with particle P. The tangential axis is tangent to the curve at point P and pointing in the direction of increasing s. Unit vector t is used to designate its positive direction. Geometrically, the curve can be considered to be constructed from a series of differential circular arc segments ds, which has a radius of curvature ρ and the center of curvature O 0 , figure 6.11b. For a space curve, an infinite number of lines can be drawn normal to the tangential axis at P. A principal normal axis is chosen as directed from point P to the center of curvature O 0 . Since O 0 is always on the concave side of the curve, the positive direction of the principal normal axis is always on the concave side of the curve, which is designated by unit vector n. A binormal axis that is perpendicular to both t and n is then constructed, figure 6.11c. A unit vector b is used to designate its positive direction and b ¼ t  n, which means the t-n-b coordinate system is a right-hand coordinate system. Note that since the origin of this t-n-b coordinate system moves with the particle, the three unit vectors t, n and b generally always change their directions with the particle’s motion along the space curve. The plane containing the tangential and principal normal axes is referred to as the osculating plane, figure 6.11c. For a space curve, the orientation of the osculating plane always changes with the particle’s motion. However, if the particle moves along a plane curve, the osculating plane is fixed in the plane of motion.

FIG. 6.11 – The t-n-b coordinate system.

Kinematics of Particles

6.3.2

205

Velocity

Suppose that at a given instant, the particle is at point P on the fixed known curve. After a small time interval Δt, the particle moves a distance Ds along the path curve to a new position P 0. The displacement is Dr, as shown in figure 6.12. From equation (6.2), we have Dr dr ¼ Dt!0 Dt dt

v ¼ lim

Since gradually the direction of Dr approaches the tangent to the path curve and the magnitude of Dr becomes equal to the distance Ds when Dt ! 0, we have lim

Dr

Dt!0 Ds

¼

dr ¼t ds

Therefore, v¼

dr dr ds ds ¼ ¼ t ¼ vt dt ds dt dt

ð6:14Þ

FIG. 6.12 – Velocity vector and unit tangent vector. Note that here v ¼ s_ is an algebraic quantity. s_ [ 0 indicates s is increasing with time t and the particle is moving along the path curve in the +s direction. It means velocity v is in the direction of +t. Likewise, s_ \0 indicates s is decreasing with time t and velocity v is in the direction of t.

6.3.3

Acceleration

The acceleration of the particle is the time rate of change of its velocity. Thus, a¼

dv d dv dt d2 s dt ¼ ðvtÞ ¼ tþv ¼ 2 tþv dt dt dt dt dt dt

ð6:15Þ

The time derivative dt=dt should be determined. When the particle moves from P to P 0 during an infinitesimal time interval dt, the tangential unit vector is changed from t to t0 , as shown in figure 6.13a. t0 and t have the same magnitude of unity; however, their directions are different and t0 makes an angle of dh with t. Construct a

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triangle using the two unit vectors t, t0 and their difference dt ¼ t0  t, figure 6.13b. It can be seen that jdtj ¼ 2  jtj  sin Therefore,

dh dh dh ¼ 2  1  sin  2 ¼ dh 2 2 2

dt jdtj dh dh ds 1 v ¼ ¼ ¼ ¼ v¼ dt dt dt ds dt q q

For the derivation, the expression ds ¼ qdh is applied. Moreover, vector dt is perpendicular to t, or in the direction of n, since angle dh is infinitesimal and the triangle is in the osculating plane. Thus, we have dt v ¼ n dt q

ð6:16Þ

Substituting equation (6.16) into equation (6.15) yields a¼

dv dv dt dv v2 ¼ tþv ¼ t þ n ¼ a t t þ a n n ¼ at þ an dt dt dt dt q

ð6:17Þ

It means acceleration vector a is the sum of two mutually perpendicular components, at and an , which are shown in figure 6.14.

FIG. 6.13 – Unit tangent vectors at adjacent positions.

FIG. 6.14 – Tangential and normal acceleration components.

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207

The tangential acceleration component at is tangent to the path curve. at ¼ v_ ¼ €s is an algebraic quantity. at [ 0 means at is in the +t direction, whereas at \0 means at is in the t direction. It has been mentioned that v ¼ s_ is also an algebraic quantity. When v and at have the same sign, i.e., v and at are in the same direction, the particle is speeding up; and when v and at have opposite signs, i.e., v and at are in opposite directions, the particle is slowing down, figure 6.15. Since an ¼ v 2 =q and it is always positive, the normal acceleration component an is always directed to the +n direction, i.e., to the center of curvature of the path. Then the vector summation of at and an , namely the acceleration a, is always on the concave side of the path curve, as shown in figure 6.15. The magnitude of a is qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a ¼ at2 þ an2 ð6:18Þ From equation (6.17), acceleration a has only tangential (t) and normal (n) components in the t-n-b coordinate system and the binormal (b) component is always zero, i.e., ab  0. From equation (6.14), the velocity v has only a tangential (t) component in the t-n-b coordinate system and the other two components are always zero. However, there are three rectangular components for both v and a in the x-y-z coordinate system. That is why using tangential and normal components is more convenient to describe a particle’s motion along a known path. The comparison of these two methods of describing curvilinear motion is listed in table 6.1.

TAB. 6.1 – Comparison of two methods of describing curvilinear motion. Rectangular components Position r ¼ xi þ yj þ zk x ¼ xðtÞ y ¼ yðtÞ z ¼ zðtÞ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi r ¼ x 2 þ y2 þ z 2

Velocity v ¼ vx i þ vy j þ vz k vx ¼ x_ vy ¼ y_ vz ¼ z_ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi v ¼ vx2 þ vy2 þ vz2

Accelaration a ¼ ax i þ ay j þ az k ax ¼ v_ x ¼ x€ ay ¼ v_ y ¼ y€ az ¼ v_ z ¼ €z qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a ¼ ax2 þ ay2 þ az2

Tangential-normal components Position

Velocity

Accelaration

s = s(t)

v ¼ vt ¼ s_ t v ¼ s_

a ¼ at t þ an n at ¼ v_ ¼ €s 2 an ¼ vq

s

v ¼ s_

FIG. 6.15 – Accelerated or decelerated motion.

a¼

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi at2 þ an2

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To better understand normal and tangential components of acceleration, some special cases are discussed. (1) Accelerated rectilinear motion If the particle moves along a straight line, the radius of curvature q is infinite. _ figure 6.15. For this case, the direction of Then, an ¼ v 2 =q  0 and a ¼ at ¼ v, the velocity does not change and only the magnitude of the velocity changes. Therefore, the tangential acceleration component represents the time rate of change in the magnitude of velocity. (2) Curvilinear motion with a constant speed If the particle moves along a curved path with a constant speed, we have at ¼ v_  0 and a ¼ an ¼ v 2 =q. For this case, the magnitude of the velocity (the speed) does not change. Only the direction of the velocity changes, and therefore, the normal acceleration component represents the time rate of change in the direction of velocity. (3) Accelerated curvilinear motion From at ¼ dv=dt and v ¼ ds=dt, integrations are performed to obtain the speed and the position of the particle on the fixed curve.  Z t Z t Z t v ¼ v0 þ at dt ; s ¼ s0 þ v0 t þ at dt dt ð6:19Þ 0

0

0

where v0 and s0 are, respectively, the initial speed and initial position of the particle at t = 0. Eliminating dt from at ¼ dv=dt and v ¼ ds=dt, yields vdv ¼ at ds Integrating this equation, we have

Z

v 2  v02 ¼ 2

s

s0

at ds

ð6:20Þ

ð6:21Þ

If tangential acceleration component at is constant, equations (6.19) and (6.21) can be rewritten as v ¼ v0 þ at t;

s ¼ s0 þ v0 t þ

1 2 at t ; v 2  v02 ¼ 2at ðs  s0 Þ 2

ð6:22Þ

The differential relations expressed by v ¼ ds=dt, at ¼ dv=dt and vdv ¼ at ds for curvilinear motion of a particle are similar to those developed for rectilinear motion, namely, v ¼ dx=dt; a ¼ dv=dt; vdv ¼ adx If acceleration a of the rectilinear motion is constant, integrations are performed to obtain the following expressions, which are similar to those in equation (6.22).

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209

v ¼ v0 þ at; x ¼ x0 þ v0 t þ

1 2 at ; v 2  v02 ¼ 2aðx  x0 Þ 2

ð6:23Þ

Example 6.7. A cannonball is fired with velocity v 0 at angle h0 , figure 6.16. Determine the radius of curvature of the cannonball’s path at any instant. Neglect the air resistance and the size of the cannonball.

FIG. 6.16 – Cannonball’s motion. Solution: After the firing, the motion of the cannonball is the free-flight motion of a projectile. The rectangular components of a projectile’s acceleration are ax ¼ 0;

ay ¼ g

where g, the acceleration of gravity, is generally considered as a constant and taken as 9:81 m/s2 . The rectangular components of the ball’s velocity are vx ¼ v0x ¼ v0 cos h0 vy ¼ v0y  gt ¼ v0 sin h0  gt The magnitudes of the acceleration and velocity are qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a ¼ ax2 þ ay2 ¼ g v¼

qffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi vx2 þ vy2 ¼ v02 cos2 h0 þ ðv0 sin h0  gtÞ2

The first time derivative of v yields the tangential component of acceleration. at ¼

dv g ¼  ðv0 sin h0  gtÞ dt v

Thus, the normal component of acceleration at any instant is qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi g an ¼ a 2  at2 ¼ v0 cos h0 v Therefore, the radius of curvature of the path at any instant is h i3 2 2 2 2 2 3 v cos h þ ðv sin h  gtÞ 0 0 0 0 v v q¼ ¼ ¼ an gv0 cos h0 gv0 cos h0

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210

Example 6.8. A car passes a bridge with a constant speed of 36 km/h. The bridge curve is defined by y ¼ 4hxðl  xÞ=l 2, where x and y are in meters, h = 1 m and l = 32 m, figure 6.17. Determine the acceleration of the car when it is at the highest point A of the bridge deck. Neglect the size of the car.

FIG. 6.17 – A car passing a bridge.

Solution: Express the speed in SI units.    km 1000 m 1h v ¼ 36 km/h ¼ 36 ¼ 10 m/s h 1 km 3600 s Since the speed is constant, we have at ¼ dv=dt ¼ 0 and a ¼ an ¼ v 2 =q. From calculus, the radius of curvature can be determined by "  2 #32 , 2 d y dy q ¼ 1þ dx 2 dx Here the first and second derivatives of the equation of path are dy 4h 16  x ¼ 2 ðl  2xÞ ¼ ; dx l 128

d2 y 1 ¼ dx 2 128

At the highest point A, x = 16 m and dy=dx ¼ 0. Thus,

2 d y q ¼ 1 2 ¼ 128 m dx The magnitude of the acceleration at point A is a ¼ an ¼

v 2 102 ¼ ¼ 0:78 m/s2 q 128

The direction of acceleration a is the same as that of an, i.e., vertically downward. Example 6.9. The motion of a particle is defined by position vector r ¼ f5 cosð3t Þi þ 5 sinð3t Þj þ 4:5tkgm, where t is in seconds and the arguments for sine and cosine are in radians. Determine the velocity and acceleration at t = 2 s. Also analyze the trajectory of the particle and the radius of curvature.

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211

Solution: (1) The velocity is obtained by the first time derivative of r. v¼

dr d ¼ f5 cosð3t Þi þ 5 sinð3t Þj þ 4:5tkg ¼ f15 sinð3t Þi þ 15 cosð3t Þj þ 4:5kgm/s dt dt

The magnitude of the velocity is qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi v ¼ vx2 þ vy2 þ vz2 ¼ 15:66 m/s This means the magnitude of the velocity is constant during the motion. The direction of the velocity is v 15 sinð3t Þi þ 15 cosð3t Þj þ 4:5k ¼ v 15:66 ¼ 0:958 sinð3t Þi þ 0:958 cosð3t Þj þ 0:287k

uv ¼ When t = 2 s,

uv jt¼2 s ¼ 0:1i þ 0:953j þ 0:287k Note that uv should be just the tangent to the path curve. (2) The acceleration is obtained by the first time derivative of v. a¼

dv d ¼ f15 sinð3t Þi þ 15 cosð3t Þj þ 4:5kg ¼ f45 cosð3t Þi  45 sinð3t Þjgm/s2 dt dt

The magnitude of the acceleration is qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a ¼ ax2 þ ay2 þ az2 ¼ 45 m/s2 Then the magnitude of the acceleration is also constant during the motion. The direction of the acceleration is ua ¼

a 45 cosð3t Þi  45 sinð3t Þj ¼ ¼  cosð3t Þi  sinð3t Þj a 45

When t = 2 s, ua jt¼2 s ¼ 0:995i  0:105j (3) From position vector r, the x, y and z coordinates of the particle are x ¼ 5 cosð3t Þ;

y ¼ 5 sinð3t Þ;

z ¼ 4:5t

Eliminating t from the first two expressions, we have x 2 þ y 2 ¼ 25

ð1Þ

This represents a cylindrical surface with a radius of 5 m, which is parallel to the z axis. Eliminating t from the last two expressions yields

Engineering Mechanics

212   2z y ¼ 5 sin 3

ð2Þ

This represents a curved surface parallel to the x axis, which intersects the y–z plane to form a sine curve. Combining equations (1) and (2), the trajectory of the particle is a spiral line, as shown in figure 6.18. (4) From the above calculation, v ¼ 15:66 m/s and a ¼ 45 m/s2 for any instant. Then we have at ¼ dv=dt ¼ 0 and an ¼ a ¼ 45m/s2 . Thus, the radius of curvature at any instant is q¼

v 2 v 2 15:662 ¼ 5:45 m ¼ ¼ an a 45

FIG. 6.18 – A particle moving along a spiral path.

6.4

Absolute Dependent Motion Analysis of Particles

If two or more particles are connected by one or more inextensible cords that are wrapped around pulleys, the motion of one particle will depend on another particle’s motion. For example, in the pulley-cord system shown in figure 6.19, the downward movement of block A will cause a corresponding upward movement of block B. To analyze this dependency, first specify the location of the blocks using position coordinates sA and sB. Note that each position coordinate axis should be measured from a fixed datum line and in the direction of motion of each block. Then the position coordinates can be related using the total cord length lT. sA þ lCD þ sB ¼ lT where lCD is the length of the cord passing over arc CD. For an inextensible cord, lT and lCD are both constant. Thus, taking the time derivative of the position-coordinate equation yields the velocity equation. dsA dsB þ ¼0 dt dt

or

vB ¼ vA

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213

The negative sign means that when block A has a downward velocity, i.e., moves in the direction of +sA, block B correspondingly has an upward velocity, i.e., moves in the direction of sB . Similarly, time derivative of the velocity equation or second time derivative of the position-coordinate equation yields the acceleration equation. d2 sA d2 sB þ ¼0 dt 2 dt 2

or

aB ¼ aA

Consider a more complicated pulley-cord system shown in figure 6.20. In this case, the position of the center of pulley D from which block A is suspended is denoted as sA, since the center point of pulley D has the same motion (velocity and acceleration) as block A. The position of block B is specified by sB. Note that the origins of sA and sB are different. Each particle’s coordinate axis can have the origin located at an appropriate fixed datum and is directed along its path of motion.

FIG. 6.19 – Pulley-cord system.

FIG. 6.20 – Absolute dependent motion analysis of two particles.

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214

From figure 6.20, the two position coordinates are related by 2sA þ sB ¼ l0 where l0 is the length of the whole cord minus the arc segments wrapped over pulleys C and D. Since these segments do not change length during the motion, l0 also remains constant during the motion. Thus, taking the first and second time derivatives of the position-coordinate equation yields 2

2

dsA dsB þ ¼0 dt dt

d2 sA d2 sB þ ¼0 dt 2 dt 2

or

or

vB ¼ 2vA

aB ¼ 2aA

Example 6.10. Block A of the pulley-cord system shown in figure 6.21 is moving downward with a velocity of 2.8 m/s, which is increasing at 0.5 m/s2. At the same time, block B is moving downward with a velocity of 1.2 m/s, which is decreasing at 0.2 m/s2. Find the velocity and acceleration of block C.

FIG. 6.21 – Absolute dependent motion analysis of three particles. Solution: There are one cord and three blocks in this system. The positions of blocks A, B and C can be defined by coordinates sA, sB and sC, respectively, which are all measured from a horizontal fixed datum as shown. sA, sB and sC can be related as follows: sA þ 2sB þ 2sC ¼ l0 where l0 is the length of the whole cord minus the segments that do not change length during the motion. Taking the first and second time derivatives of the above position-coordinate equation yields

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215

dsA dsB dsC þ2 þ2 ¼0 dt dt dt d2 sA d2 sB d2 sC þ 2 þ 2 ¼0 dt 2 dt 2 dt 2

or

or

1 vC ¼  ðvA þ 2vB Þ 2 1 aC ¼  ðaA þ 2aB Þ 2

Then, when vA ¼ 2:8 m/s (downward) and vB ¼ 1:2 m/s(downward), we have 1 vC ¼  ð2:8 þ 2  1:2Þ ¼ 2:6 m/s 2 The negative sign means v C is actually upward, in the direction of sC . When aA ¼ 0:5 m/s2 (downward) and aB ¼ 0:2 m/s2 (upward), we have 1 aC ¼  ð0:5  2  0:2Þ ¼ 0:05 m/s2 2 The negative sign means block C has an upward acceleration at this instant. However, since its velocity is also upward, block C is actually speeding up at this instant. Example 6.11. For the pulley-cord system shown in figure 6.22a, the motor is drawing the cord at 1.4 m/s, and slowing at 0.5 m/s2. Find the velocity and acceleration of block A at this instant.

FIG. 6.22 – A pulley-cord system with two cords.

Solution 1: There are two cords in this pulley-cord system. Three position coordinates sA, sD and sM shown in figure 6.22b can be used to express the fixed lengths of the two cords as follows: 2sA þ sD ¼ l1 ; ðh  sD Þ þ ðsM  sD Þ ¼ l2

Engineering Mechanics

216

Here l1 is the length of one cord minus two arc segments wrapped over pulleys B and C; l2 is the length of the other cord minus one arc segment wrapped over pulley D; h is the vertical distance between the center of pulley C and the ground. l1 , l2 and h are all constant during the motion. Eliminating sD from the two position-coordinate equations yields 4sA þ sM ¼ 2l1 þ l2  h Taking the first and second time derivatives of this equation yields 4

4

dsA dsM þ ¼0 dt dt

d2 sA d2 sM þ ¼0 dt 2 dt 2

or

or

vA ¼ 

vM 4

aA ¼ 

aM 4

Then, when vM ¼ 1:4 m/s (downward) and aM ¼ 0:5 m/s2 (upward), we have vA ¼ 

1:4 0:5 ¼ 0:35 m/sð"Þ; aA ¼  ¼ 0:125 m/s2 ð#Þ 4 4

It means block A is moving upward at 0:35 m/s and is slowing at 0:125 m/s2 . Solution 2: The three position coordinates sA, sD and sM can also be set as shown in figure 6.22c, since the setting meets the requirements, i.e., each coordinate axis is in the direction of motion of the block and measured from a fixed datum. In this case, the two expressions of the lengths of the two cords are 2sA þ ðh  sD Þ ¼ l1 ; sD þ ðsD  sM Þ ¼ l2 Eliminating sD from the two position-coordinate equations yields 4sA  sM ¼ 2l1 þ l2  2h Taking the first and second time derivatives of this equation yields vM aM vA ¼ ; aA ¼ 4 4 Then, when vM ¼ 1:4 m/s (downward) and aM ¼ 0:5 m/s2 (upward), we have vA ¼

1:4 0:5 ¼ 0:35 m/sð"Þ; aA ¼ ¼ 0:125 m/s2 ð#Þ 4 4

It should be noted that since the sM axis is set as positive upward in this solution, the downward velocity is negative (vM ¼ 1:4 m/s) and the deceleration is upward and positive (aM ¼ 1 m/s2 ). The signs of the velocity and acceleration should be consistent with the sign convention of the position coordinate. RETHINK: From the above solutions, different settings of position coordinates obtain the same results for the velocity and acceleration of block A. Furthermore,

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217

there are two inextensible cords in this example and then we can write two position-coordinate equations from the fixed lengths of the two cords. Actually, as many position-coordinate equations as the cords in the pulley-cord system can be written and the positions (motions) of the particles are then related by these equations. Example 6.12. A collar fits loosely on the vertical fixed shaft. It is attached to a cord which passes over a pulley. The other end of the cord is pulled outward with a constant speed u = 0.2 m/s, figure 6.23. If x = 0 when t = 0, find the position, velocity and acceleration of the collar at t = 10 s.

FIG. 6.23 – A collar moving along a fixed shaft pulled by a pulley-cord system. Solution: Cord segment AB changes both direction and magnitude in this problem. However, the collar is still undergoing rectilinear motion and its position can be defined by the coordinate x. From figure 6.23, we have pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi OB ¼ OC 2 þ CB 2 ¼ 122 þ 52 ¼ 13 m AB ¼ OB  ut ¼ 13  0:2t Then, the position of the collar at any arbitrary instant t is qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 2 x ¼ OC  AC ¼ OC  AB  BC ¼ 12  ð13  0:2tÞ2  52 The first and second time derivatives of x yield the velocity and acceleration of the collar. v¼

dx 1 2  ð13  0:2tÞð0:2Þ 0:2ð13  0:2tÞ ¼   qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi dt 2 2 ð13  0:2tÞ  52 ð13  0:2tÞ2  52

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218

dv d2 x ¼ 2 dt dt qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 ffi  2  ð13  0:2tÞ  ð0:2Þ ð0:2Þ  ð13  0:2tÞ2  52  ð13  0:2tÞ  12  pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð130:2tÞ2 52 ¼ 0:2  qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2

a¼

ð13  0:2tÞ2  52

1 ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi3 ð13  0:2tÞ2  52

When t = 10 s, x ¼ 12 

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð13  0:2  10Þ2  52 ¼ 2:2 m

0:2ð13  0:2  10Þ v ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ 0:225 m/s ð13  0:2  10Þ2  52 1 1 a ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi3 ¼ pffiffiffiffiffi3 ¼ 0:001 m/s2 96 ð13  0:2  10Þ2  52

6.5

Relative-Motion Analysis Using Translating Axes

In the previous sections of this chapter, we observed and described the motion of a particle in a single fixed reference frame. For example, standing still on the ground, we observe the motion of a bus or the motion of a car, as shown in figure 6.24. This type of motion is called absolute motion. However, sometimes, one may move with another reference frame and observe a particle’s motion and this type of motion is relative motion. For instance, the driver of the car observes the motion of the bus. Apparently, the relative motion is related to the absolute motions of the two objects. Determination of this relationship is relative-motion analysis. In this section, we only consider relative-motion analysis of two particles using translating reference frame. Relative-motion analysis using rotating reference frame can be referred to references [2] and [5]. Position. As shown in figure 6.25a, the motion of particle B can be observed from either a fixed reference frame Oxyz or a translating reference frame Ax 0 y 0 z 0 . rA and rB , measured from the origin O of the fixed reference frame, are referred to as absolute positions. The relative position of B with respect to A is denoted by the relative-position vector rB=A , figure 6.25a. The three position vectors can be related: rB ¼ rA þ rB=A

ð6:24Þ

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219

FIG. 6.24 – Fixed reference frame and translating reference frame.

FIG. 6.25 – Relative-motion analysis using translating axes.

Velocity. Taking the first time derivative of equation (6.24), we have drB drA drB=A ¼ þ dt dt dt

ð6:25Þ

From equation (6.2), we have drB =dt ¼ v B and drA =dt ¼ v A , which are called absolute velocities observed from the fixed reference frame. To determine drB=A =dt, rB=A is first expressed in Cartesian vector form as rB=A ¼ x 0 i0 þ y 0 j0 þ z 0 k0 , where x 0 ; y 0 and z 0 are coordinates of particle B in the translating reference frame Ax 0 y 0 z 0 ; i0 ; j0 and k0 are unit vectors of the translating coordinate axes. Then, drB=A d dx 0 0 dy 0 0 dz 0 0 di0 dj0 dk0 ¼ ðx 0 i0 þ y 0 j0 þ z 0 k0 Þ ¼ i þ j þ k þ x0 þ y0 þ z0 dt dt dt dt dt dt dt dt

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220

Since the Ax 0 y 0 z 0 reference frame is a translating frame, both the direction and magnitude of the unit vectors i0 ; j0 and k0 do not change with time. Thus, the last three terms in the above expression are zero and we obtain the relative velocity observed from the translating reference frame. drB=A dx 0 0 dy 0 0 dz 0 0 ¼ i þ j þ k ¼ vB=A dt dt dt dt Substituting this into equation (6.25) yields v B ¼ v A þ v B=A

ð6:26Þ

Equation (6.26) states that the absolute velocity of particle B is equal to the vector sum of the absolute velocity of particle A and the relative velocity of B with respect to A measured from the translating reference frame Ax 0 y 0 z 0 . Using the parallelogram law, equation (6.26) is graphically illustrated in figure 6.25b. Acceleration. Taking the time derivative of equation (6.26) yields the relation between the accelerations as aB ¼ aA þ aB=A

ð6:27Þ

where aB ¼ dv B =dt and aA ¼ dv A =dt are absolute accelerations of particle B and particle A, respectively. aB=A ¼ dv B=A =dt is the relative acceleration of B with respect to A measured from the translating reference frame Ax 0 y 0 z 0 . Note that while deriving aB=A by time differentiation, the fact that i0 ; j0 and k0 do not change with time is utilized again. Equation (6.27) states that the absolute acceleration of particle B is equal to the vector sum of the absolute acceleration of particle A and the relative acceleration of B with respect to A. Using the parallelogram law, equation (6.27) is graphically illustrated in figure 6.25c. Equations (6.26) and (6.27) are both vector equations. Since three vectors in each equation form a parallelogram and must be in the same plane, each of these vector equations can be resolved to obtain two scalar equations. Therefore, at most two unknowns, represented by the magnitudes and/or directions of the vectors, can be solved for from each equation. Example 6.13. At the instant shown in figure 6.26a, car A is traveling on a straight path with a constant speed of 50 km/h. Car B is moving along a curve having a radius of curvature of 0.6 km at 20 km/h, and is slowing at 1000 km/h2. Find the relative velocity and relative acceleration of B with respect to A.

FIG. 6.26 – Relative-motion analysis of two cars.

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221

Solution: The fixed x, y axes are established at an arbitrary point on the ground. Afterwards in this book, the fixed reference frame is always established on the ground, unless otherwise stated. The translating x 0 , y 0 axes are attached to car A, figure 6.26b. Applying the relative velocity equation yields v B=A ¼ vB  v A ¼ ð20 cos 30 i  20 sin 30 jÞ  ð50iÞ ¼ f32:7i  10jg km/h The trigonometric solution is shown in figure 6.26c. The magnitude and direction of the relative velocity can be determined by the law of sines or cosines as follows: qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi vB=A ¼ vA2 þ vB2  2vA vB cos 30 ¼ 34:2 km/h sin h sin 30 ¼ vB vB=A

)

h ¼ 17:0

The magnitude and direction of v B=A can also be determined from its vector form. Readers are recommended to finish this. The tangential component of car B’s acceleration is aBt ¼ 1000 km/h2 and the normal acceleration component is aBn ¼

vB2 202 ¼ q 0:6

)

aBn ¼ 667 km/h2

Applying the relative acceleration equation, we have   aB=A ¼ aB  aA ¼ atB þ a nB  0 ¼ atB þ a nB aB=A ¼ ð1000 cos 30 i þ 1000 sin 30 jÞ þ ð667 sin 30 i þ 667 cos 30 jÞ ¼ f533i þ 1078jg km/h2 The trigonometric solution of aB=A is shown in figure 6.26d. Its magnitude and direction can be determined as follows: pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi aB=A ¼ 10002 þ 6672 ¼ 1202 km/h2 h0 ¼ arctan

667 ¼ 33:7 1000

Readers are recommended to determine the magnitude and direction from the vector form of aB=A . RETHINK: It should be noted that the relative velocity and relative acceleration of B with respect to A can be obtained in this example, since car A is moving along a straight line and we can attach a translating reference frame to it. However, observation of the motion of car A with respect to car B should be obtained using a rotating reference frame attached to car B, which can be referred to in references [2] and [5].

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222

Example 6.14. The river flows with a speed of 1.5 m/s and a man can row his kayak in still water at 3 m/s. Starting at position A, he tries to cross the river to position B, figure 6.27a. Specify the angle θ at which he should direct his kayak and the time needed to make the crossing.

FIG. 6.27 – Relative-motion analysis of the kayak and water.

Solution: The translating x 0 , y 0 axes are attached to the flowing water. The directions of the absolute and relative velocities are shown in figure 6.27b. Applying the relative velocity equation yields vk ¼ vw þ vk=w ¼ ð1:5jÞ þ ð3 cos hi þ 3 sin hjÞ ¼ f3 cos hi þ ð1:5 þ 3 sin hÞjg m/s Because the kayak is from A to B, the direction of v k should be from A to B. Then, 1:5 þ 3 sin h 70 ¼ 3 cos h 100 15 þ 30 sin h ¼ 21 cos h

)

h ¼ 10:81

Therefore, the velocity of the boat is v k ¼ f2:947i þ 2:063jg m/s The magnitude of v k is vk ¼

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2:9472 þ 2:0632 ¼ 3:597 m/s

The time needed to make the crossing is pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi lAB 1002 þ 702 ¼ 33:9 s ¼ t¼ 3:597 vk

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223

Example 6.15. The center player throws a football B with a velocity of v0 = 18 m/s at an angle θ0 = 40°, figure 6.28. Player A is 10 m away from the center player when the center player throws the ball. Find the constant speed at which player A must run so that he would catch the ball at the same elevation at which it was thrown. Also determine the time needed to make the catch.

FIG. 6.28 – Relative-motion analysis of a ball and a player. Solution: The translating x 0 , y 0 axes are attached to player A. The absolute accelerations of player A and the ball at any instant are, respectively, aA ¼ 0;

aB ¼ f9:81jg m/s2

Thus, the absolute velocities of player A and the ball at any instant are respectively v A ¼ fvA0 ig m/s v B ¼ 18 cos 40 i þ ð18 sin 40  9:81t Þj ¼ f13:79i þ ð11:57  9:81t Þjg m/s Applying the relative velocity equation yields the relative velocity of the ball with respect to player A at any instant. vB=A ¼ v B  vA ¼ fð13:79  vA0 Þi þ ð11:57  9:81t Þjg m/s Then the relative position vector is

   rB=A ¼ x00 þ ð13:79  vA0 Þt i þ y00 þ 11:57t  4:905t 2 j m From the problem description, x00 ¼ 10 m; y00 ¼ 0; and at the final instant, x ¼ 0; y 0 ¼ 0. Thus 0

10 þ ð13:79  vA0 Þt ¼ 0 11:57t  4:905t 2 ¼ 0 Solving these equations, we have t ¼ 2:36 s;

vA0 ¼ 9:55 m/s

Engineering Mechanics

224 PROBLEMS

6.1 A particle starting from rest moves along a straight line with an acceleration a ¼ ðt 2 =6Þm/s2 , Where t is in seconds. Determine the distance traveled in 6 s. 6.2 The velocity of a particle moving along a straight line is v ¼ ð0:1s þ 10Þm/s, where s is the distance in meters. If s = 0 when t = 0, determine the time needed to travel the 100-m distance. 6.3 When two jeeps A and B are side by side, they are traveling in the same direction at speeds of 100 km/h and 120 km/h, respectively. Afterwards, A maintains its speed and B begins to decelerate at 1000 km/h2. How long will jeep B stop? Also determine distance d between the two jeeps at this instant.

Prob. 6.3

6.4 A particle is traveling along a straight line at a speed of 3 m/s. Afterwards, it begins to accelerate at a ¼ ð5=v 2 Þm/s2 , where v is the speed in m/s. Determine its velocity and displacement in 2 s. 6.5 A particle moves on a straight path with an acceleration a ¼ cebt , where both c and b are constants. The initial position is x0 and the initial velocity is v0 . Determine the position and velocity at any instant. pffiffiffi  pffiffiffi  6.6 The position of a particle is defined by x ¼ 2t 2 ; y ¼ cos 3 t ; z ¼ sin 3 t , where x, y and z are in meters; t is in seconds and the arguments for sine and cosine are in radians. Find the radius of curvature of the path at t = 1 s. 6.7 A particle travels along a plane curve with an acceleration a ¼ f16 cosð2t Þi 20 sinð2t Þjgcm/s2 , where t is in seconds and the arguments for sine and cosine are in radians. When t = 0, x0 ¼ 4 cm, y0 ¼ 5 cm and v0x ¼ 0, v0y ¼ 10 cm=s. Determine the equation of path. 6.8 The particle moves with a constant speed v = 12 m/s along a plane curve y = ex, where x and y are measured in meters. Determine the x and y components of acceleration at y = 1 m. Also find the tangential and normal components of acceleration at this position. 6.9 The rectangular coordinates of a particle are x ¼ 4t; y ¼ 4t þ t 2 and z ¼ sinðptÞ, where x, y, z are in meters; t is the time in seconds and the argument for sine is in radians. Find the rectangular components of the velocity and acceleration when t = 3 s.

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225

6.10 A jeep travels along a circular path with a radius of 2000 m. Its position is defined by s ¼ ð40t  t 2 Þm, where t is in seconds. Find the magnitudes of velocity and acceleration when s = 400 m. 6.11 A car travels along a circular path with a radius of 1000 m. The initial speed is 54 km/h and then the car increases its speed at a constant rate. In 30 s, the car travels 600 m. Find the magnitudes of velocity and acceleration at this instant. 6.12 The position of a particle moving along a plane curved path is defined by x ¼ f1 ðtÞ, y ¼ f2 ðtÞ. Prove that at any instant the tangential and normal components _ y y€ _ xj jx€ _ xffiffiffiffiffiffiffiffiffiffi _y ffi þ y€ ffiffiffiffiffiffiffiffiffiffi ffi, respectively, and that the radius of and an ¼ p of acceleration are at ¼ px€ 2 2 2 2 x_ þ y_

curvature of the path is q ¼

x_ þ y_

ðx_ 2 þ y_ 2 Þ3=2 _ y y€ _ xj . jx€

6.13 The horizontal coordinate of the particle is specified by x ¼ ðt 2 Þm, where t is in seconds. If the equation of path is y ¼ x 3 =240, determine the acceleration of the particle at t = 2 s. 6.14 The position of a particle is defined by x ¼ 75 cosð4t 2 Þ and y ¼ 75 sinð4t 2 Þ, where x, y are in centimeters; t is the time in seconds and the arguments for sine and cosine are in radians. Find the velocity and the tangential and normal components of acceleration at any instant. 6.15 Starting from rest, a particle moves along a circular path of radius R and increases its speed at a constant rate at. How long does it take for the normal acceleration component equal to the tangential acceleration component? 6.16 The motion of a particle is defined by the position vector r ¼ f3 cosð3t Þi þ 4 sinð3t Þjgm, where t is in seconds and the arguments for the sine and cosine are in radians. Find the velocity and acceleration of the particle at t = 2 s. Also determine the equation of path. 6.17 A breakstone is shot from position A at an angle θ with the horizontal due to directional blasting and lands on position B as shown. Neglect the air resistance and the size of the stone. Express initial velocity v0 in terms of θ, L, H and the acceleration of gravity g.

Prob. 6.17

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226

6.18 A ball is kicked out with an initial velocity v0 ¼ 9:4 m/s at an angle θ0 = 45° from the horizontal as shown. Determine the highest altitude the ball can reach and the required time to reach this altitude. 6.19 A cannonball is fired with a speed v0 ¼ 800 m/s at an angle θ0 = 30° with the horizontal. Find the tangential and normal components of the cannonball’s velocity and acceleration at t = 10 s.

Prob. 6.18/19

6.20 Starting from the same point O, two particles A and B travel in opposite directions along the circular path at speeds of vA ¼ ð3 þ 2t Þm/s and vB ¼ ð5 þ 0:5t Þm/s, respectively, where t is in seconds. How long will they collide? Also find the magnitude of the acceleration of each particle at this instant.

Prob. 6.20

6.21 Block A of the pulley-cord system is moving downward with a velocity of 2 m/s, which is increasing at 1.5 m/s2. Determine the velocity and acceleration of block B.

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227

Prob. 6.21

6.22 Motor M is used to lift crate A. If point B of the cable has an instantaneous velocity of 4 m/s and an acceleration of 1 m/s2, which are both directed to the right, determine the velocity and acceleration of crate A at this instant.

Prob. 6.22

6.23 At a given instant, the cord at B is pulled down at 20 cm/s, and is slowing at 5 cm/s2. Specify the velocity and acceleration of block a at this instant.

Prob. 6.23

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6.24 Collar A is hoisted by motor M as shown. When sA = 1 m, the collar is moving upward at 0.8 m/s and slowing down at 0.2 m/s2. Find the velocity and acceleration of the cable at which it is drawn into motor M at this instant.

Prob. 6.24

6.25 A man walks with a constant velocity v0 ¼ 1 m=s to the right. He pulls end A of the cord to lift box B. When t ¼ 0, the box is on the ground and the AC segment of the cord is at the vertical position A0C. Determine the time needed for box B to reach pulley C.

Prob. 6.25

6.26 At the instant shown, runner A is running at 8.2 m/s along the straight road of the race track and accelerating at 0.6 m/s2. Runner B is running at 7.5 m/s around the curved track while accelerating at 0.5 m/s2. Find the relative velocity and relative acceleration of B with respect to A at this instant.

Prob. 6.26

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6.27 At the instant shown, car A is traveling on a straight path with a constant speed of 100 km/h and car B is moving along a curved path with a radius of curvature of 0.8 km at 40 km/h while decelerating at 800 km/h2. Find the relative velocity and relative acceleration of B with respect to A at this instant.

Prob. 6.27

6.28 A boat can travel at a speed vb in still water. Starting from position A, the boat travels directed perpendicular to the river bank. After 10 min, the boat arrives at position C on the other side, which is 120 m downstream of position A as shown. If the boat wants to arrive at position B, which is just across the river from position A, the boat should be directed with an appropriate angle θ. For this case, the boat arrives at position B after 12.5 min. Determine vb, the speed vw at which the river flows and the width of the river.

Prob. 6.28

6.29 Ore sand is falling from conveyor belt A to conveyor belt B with an absolute velocity of v1 = 4 m/s in the direction shown. Conveyor belt B makes an angle of 15° with the horizontal. If the relative velocity of the ore sand with respect to conveyor belt B is perpendicular to belt B, determine the speed v2 of belt B. Also find the relative velocity of the ore sand with respect to belt B if v2 = 2 m/s.

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Prob. 6.29

6.30 Starting from rest, triangular prism A begins to move in the direction of +x with an acceleration aA ¼ 10 cm/s2 and block B begins to move downward along pffiffiffi the slope of the triangular prism with a relative acceleration ar ¼ 10 2 cm/s2 . The initial absolute position of block B is defined by xB0 ¼ 0; yB0 ¼ h. Determine the absolute position of block B at any instant.

Prob. 6.30

6.31 Ring M fits loosely on both the moving verticle shaft AB and the fixed semicircular loop OD, whose radius is r = 12 cm. When OB = BC = 6 cm, shaft AB is moving to the right with a velocity vA ¼ 3 cm/s and an acceleration aA ¼ 3 cm/s2 . Determine the relative velocity and relative acceleration of ring M with respect to shaft at this instant.

Prob. 6.31

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6.32 Cross slider K fits loosely on both the fixed vertical shaft AB and the moving horizontal shaft CD. Shaft CD is pin-connected to block D, which can move along the inclined slot. The position of block D is defined by AD ¼ s ¼ ½40 þ 32 sinð0:5t Þcm, where t is in seconds. If the inclination angle is h ¼ 60 , determine the absolute accelation of slider K and its relative accelation with respect to shaft CD at t ¼ ðp=3Þs.

Prob. 6.32

Chapter 7 Planar Kinematics of Rigid Bodies Objectives
Classify three types of rigid-body planar motion: translation, rotation about a fixed axis and general plane motion.
Investigate rotation about a fixed axis and velocity and acceleration of particles on a rotating body.
Study absolute dependent motion of bodies undergoing planar motion.
Study velocity and acceleration of particles on a body undergoing general plane motion through a relative-motion analysis using a translating reference frame.
Examine the instantaneous center (IC) of rotation and determine velocity of particles on a body undergoing general plane motion by using IC. In machinery, the gears, cams, and links cannot be treated as particles. Their size and shape must be considered and rotation is an important aspect in the analysis of their motions. These objects are treated as rigid bodies. This chapter discusses only planar motion of rigid bodies. Rigid-body planar motion occurs when all the particles of the body move along paths that are equidistant from a fixed plane. For example, when brushing the blackboard, as long as the brush (neglecting its deformation) does not leave the blackboard, the distance from any particle on the brush to the blackboard remains unchanged during the motion. Then, any particle of the brush moves along a path that is equidistant from the fixed blackboard plane and the brush is undergoing planar motion. Many mechanical parts undergo planar motion, such as gears, cams and linkages shown in figure 7.1.

DOI: 10.1051/978-2-7598-2901-9.c007 © Science Press, EDP Sciences, 2022

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FIG. 7.1 – Rigid bodies undergoing planar motion.

There are three types of rigid-body planar motion and they are defined as follows: Translation. Translation occurs if any line segment inside the body remains parallel to its original orientation during the motion. When particles on the translating body move along straight lines, the body’s motion is called rectilinear translation, figure 7.2a; whereas when the paths of particles are curved lines, the body’s motion is called curvilinear translation, figure 7.2b. The griddle illustrated in figure 7.2b is used to sift mineral. Although any particle on the griddle moves along a circular path, the griddle is always in upside position. The sides are always vertical and the bottom is always horizontal. Thus, the griddle is undergoing curvilinear translation.

FIG. 7.2 – Rectilinear translation and curvilinear translation.

Rotation about a fixed axis. When a rigid body rotates about a fixed axis, the particles of the body move along circles centered on the fixed axis, called the axis of rotation, as shown in figure 7.3.

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235

FIG. 7.3 – Rotation about a fixed axis. General plane motion. General plane motion is a combination of translation and rotation, as shown in figure 7.4. In the mechanism shown in figure 7.5, crankshaft OA is rotating about a fixed axis that is passing through O and perpendicular to the shown plane; slider B is undergoing rectilinear translation; and the connecting link AB is undergoing general plane motion.

FIG. 7.4 – General plane motion.

FIG. 7.5 – A mechanism with bodies undergoing planar motion.

7.1

Translation

For a rigid body undergoing translation, figure 7.6, an arbitrary line segment AB on the body maintains parallel to its original direction during the motion. By inspection, the path curves of A and B have the same shape and are equidistant from each

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other. If you move the path curve of B shown in figure 7.6 upward and to the right a little, it will overlap with that of A. Position. The positions of particles A and B can be defined by position vectors rA and rB, measured from a fixed point O, figure 7.6. Another position vector, the relative-position vector rB/A, can be used to define the position of B with respect to A. These three vectors can be related by rB ¼ rA þ rB=A

ð7:1Þ

FIG. 7.6 – Illustration of translation.

Velocity. Taking the first time derivative of equation (7.1), we have drB drA drB=A ¼ þ dt dt dt

ð7:2Þ

Here, drB =dt ¼ vB , drA =dt ¼ v A , and they are absolute velocities measured from the fixed reference point O. The term drB=A =dt ¼ 0, since the magnitude of rB=A is constant by definition of rigid body, and the direction of rB=A is also unchanged by definition of translation. Thus, equation (7.2) becomes vB ¼ vA

ð7:3Þ

Acceleration. Taking the time derivative of equation (7.3) yields aB ¼ aA

ð7:4Þ

Equations (7.3) and (7.4) indicate that all particles on a translating rigid body move with the same velocity and the same acceleration at any instant, as shown in figure 7.6. As a result, a translating rigid body can be treated as a particle. Any particle’s motion can represent the whole body’s motion.

Planar Kinematics of Rigid Bodies

7.2

237

Rotation about a Fixed Axis

The motion of a fixed pulley, a motor rotor or a door has a common feature. When the body moves, a straight line on the body or on its extension always remains motionless, as shown in figure 7.7a. This fixed line is called the axis of rotation and the body is rotating about this fixed axis. Obviously, all particles of the body, except those on the axis of rotation, moves along circular paths centered on the axis, figure 7.7a.

FIG. 7.7 – Illustration of rotation about a fixed axis.

7.2.1

Angular Motion

The rotation of a body can be specified by angular position, angular velocity, and angular acceleration. Angular Position. Consider an arbitrary particle P on the rotating body. It is moving along a circular path within the shaded plane, which is perpendicular to the axis of rotation and center C of the circular path is on this axis, figure 7.7a. At given instant t, the radial line CP makes an angle φ with its initial position CP0. Apparently, φ is a single-valued function of time as follows: u ¼ uðtÞ

ð7:5Þ

This angle can be used to define the angular position of the rotating body. φ may be measured in radians, or occasionally, in degrees and revolutions, where 2p rad ¼ 360 ¼ 1 rev. Suppose that after time interval Δt, the radial line moves to a new position CP 0 and the angle between the radial line with its initial position CP0 becomes u þ Du. The angular displacement is Δφ, as shown in figure 7.7a. From the top view of the shaded plane, the angular position φ and angular displacement Δφ shown in

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figure 7.7b can be represented by algebraic scalars. Counterclockwise φ and Δφ are generally considered positive. Angular Velocity. The angular velocity of the rotating body, denoted as ω (omega), is the time rate of change in its angular position, i.e., Du du ¼ Dt!0 Dt dt

ð7:6Þ

x ¼ lim

ω is often measured in rad/s. The speed of rotation can also be defined by the speed of revolution n (rev/min, or rpm). The conversion between ω and n is x¼

n  2p np ¼ 60 30

ð7:7Þ

Angular Acceleration. The time rate of change in the angular velocity of the rotating body is the angular acceleration, denoted as α (alpha), i.e., a¼

dx d2 u ¼ 2 dt dt

ð7:8Þ

α is measured in rad/s2. Consistent with that specifying the positive and negative senses of φ and Δφ, ω and α are also considered positive counterclockwise. From a ¼ dx=dt and x ¼ du=dt, integrations are performed to obtain the angular velocity and angular position of the body.  Z t Z t Z t x ¼ x0 þ adt ; u ¼ u0 þ x0 t þ adt dt ð7:9Þ 0

0

0

where x0 and u0 are the initial values of the body’s angular velocity and angular position, respectively. After the solution, the senses of φ, ω and α are indicated by the algebraic signs of their numerical quantities. When ω and α have the same algebraic sign (sense of rotation), the body is rotating with increasing angular speed and when ω and α have opposite signs, the body is rotating with decreasing angular speed. Eliminating dt from a ¼ dx=dt and x ¼ du=dt yields xdx ¼ adu Integrating this equation, we have x2  x20 ¼ 2

Z

u

u0

ð7:10Þ

adu

ð7:11Þ

The differential relations expressed by equations (7.6), (7.8) and (7.10) for angular motion of a body are similar to those for rectilinear motion of a particle (v ¼ dx=dt, a ¼ dv=dt, vdv ¼ adx) and those for curvilinear motion of a particle (v ¼ ds=dt, at ¼ dv=dt, vdv ¼ at ds). If the rotating body has a constant angular acceleration α, equations (7.9) and (7.11) can be rewritten as the following expressions, which are similar to those in equation (6.22) or in equation (6.23).

Planar Kinematics of Rigid Bodies

x ¼ x0 þ at ;

7.2.2

239

u ¼ u0 þ x0 t þ

1 2 at ; 2

x2  x20 ¼ 2aðu  u0 Þ

ð7:12Þ

Motion of a Particle on a Rotating Body

Once angular motion of the rotating body is specified, the velocity and acceleration of any particle on the body can be determined. Since any particle of the body, except those on the rotating axis, moves along a known circular path, it is convenient to describe the particle’s velocity and acceleration using tangential (t) and normal (n) coordinates. Velocity. Consider arbitrary particle P on the rotating body, figure 7.8. Its circular path has radius r, with center C on the rotating axis. The particle’s position can be defined by the distance, s, along the circular path from fixed point P0. Using geometry, we have s ¼ ru From equation (6.14), the velocity is v¼

ds dðruÞ t¼ t¼ dt dt

  du r t ¼ ðrxÞt dt

ð7:13Þ

ð7:14Þ

Then, v ¼ rx and the direction of v is tangent to the circular path, or perpendicular to radial line CP, as shown in figure 7.8. This means the magnitude of any particle’s velocity on the rotating body is proportional to the radius r and the direction is perpendicular to the radial line. These characteristics are illustrated in figure 7.9. Note that the sense of direction of the velocity should be consistent with the sense of rotation of the angular velocity.

FIG. 7.8 – Motion of a particle on a rotating body.

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FIG. 7.9 – Characteristics of velocity of a particle on a rotating body.

Acceleration. From equation (6.17), the acceleration is a¼

  dv v2 dðrxÞ ðrxÞ2 tþ n ¼ tþ n ¼ ðraÞt þ rx2 n ¼ at þ an dt dt q r

ð7:15Þ

The tangential acceleration component has a magnitude of at ¼ ra, which is proportional to radius r. Its direction is tangent to the circular path and the sense of direction is consistent with the sense of rotation of the angular acceleration. The normal acceleration component an is always directed to the +n direction, i.e., toward the center of the circular path and its magnitude is an ¼ rx2 , which is also proportional to the radius r. The two components of acceleration are shown in figure 7.10.

FIG. 7.10 – The t and n acceleration components.

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The magnitude and direction angle of acceleration a are as follows: qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a ¼ at2 þ an2 ¼ ðraÞ2 þ ðrx2 Þ2 ¼ r a2 þ x4 cos w ¼

an rx2 x2 ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 4 2 a r a þx a þ x4

ð7:16Þ

Equation (7.16) means that the magnitude of any particle’s acceleration on the rotating body is proportional to the radius r; and the acceleration is on the concave side of the circular path and makes the same angle w with the radial line. These characteristics are illustrated in figure 7.11.

FIG. 7.11 – Characteristics of acceleration of a particle on a rotating body.

Using right-hand rule, the angular velocity and angular acceleration can be expressed by vectors. The right-hand fingers are curled according to the sense of rotation, then the thumb points toward the vector x or a, figure 7.12a. Set the rotating axis z positive upward and unit vector k is used to designate the direction of the z axis. Then x ¼ xk a¼

dx dðxkÞ dx dk dx ¼ ¼ kþx ¼ k ¼ ak dt dt dt dt dt

ð7:17Þ ð7:18Þ

Thus, when scalars x and a are counterclockwise in the shaded plane (the x–y plane) from the top view, figure 7.12b, vectors x and a are in the +k direction.

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FIG. 7.12 – Representing velocity and acceleration of a particle on a rotating body by cross products.

Using vectors x and a, the velocity and acceleration of any particle on a rotating body can be represented by cross products. For the velocity of particle P, v ¼ x  rP

ð7:19Þ

where rP is a position vector directed from any point on the axis of rotation to particle P, figure 7.12c. The magnitude of this cross product is jx  rP j ¼ xrP sin h ¼ xðrP sin hÞ ¼ xr ¼ v, which agrees with equation (7.14). By the right-hand rule, the right-hand fingers are curled from x toward rP , then the thumb indicates the direction of x  rP , which is tangent to the path, in the direction of motion and just the direction of v. It should be noted that the order of x and rP must be maintained to get the correct sense of direction for v, since the cross product is not commutative. As a special case, position vector r directed from center C to point P can be chosen for rP and the velocity of point P is v ¼xr

ð7:20Þ

Taking the time derivative of equation (7.19), we have a¼

dv dx drP ¼  rP þ x  ¼ a  rP þ x  v dt dt dt

Substituting equation (7.19) into the above equation yields a ¼ a  rP þ x  v ¼ a  rP þ x  ðx  rP Þ

ð7:21Þ

The magnitude of a  rP is ja  rP j ¼ arP sin h ¼ aðrP sin hÞ ¼ ar ¼ at , and the direction of a  rP is just the direction of at by the right-hand rule, figure 7.12c. Similarly, the term x  v has a magnitude of jx  v j ¼ xv sin 90 ¼ xv ¼ x  xr ¼ x2 r ¼ an , and it is in the direction of an, figure 7.12c.

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243

Choosing the position vector r for rP and noting that an is in the direction of r, the acceleration in equation (7.21) can be rewritten by its two components as follows: a ¼ at þ a n ¼ a  r  x 2 r

ð7:22Þ

Example 7.1. Two rods O1 A and O2 B shown in figure 7.13a have the same length l = l m. Initially they are in the vertical position. Then they begin to rotate and the angular position is defined by u ¼ ð0:1t 2 Þrad, where t is in seconds. Determine the velocity and acceleration of particle M on link AB when t = 2 s.

FIG. 7.13 – Determining the velocity and acceleration of a particle on a translating body. Solution: Rods O1 A and O2 B are rotating about fixed axis and link AB is undergoing curvilinear translation. Then we have v A ¼ v M and aA ¼ aM . Since u ¼ ð0:1t 2 Þrad, the angular velocity and angular acceleration of the rods are as follows: x¼

du dx ¼ ð0:2t Þ rad/s, a ¼ ¼ 0:2 rad/s2 dt dt

When t = 2 s, we have u ¼ 0:1  22 ¼ 0:4 rad;

x ¼ 0:2  2 ¼ 0:4 rad/s,

a ¼ 0:2 rad/s2

The position of the mechanism is shown in figure 7.13b. Since A is a particle on the rotating rod O1 A, its velocity and two acceleration components (also the velocity and acceleration components of particle M) are vA ¼ vM ¼ lx ¼ 1  0:4 ¼ 0:4 m=s t ¼ la ¼ 1  0:2 ¼ 0:2 m=s2 aAt ¼ aM n aAn ¼ aM ¼ lx2 ¼ 1  0:42 ¼ 0:16 m/s2

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Their directions are shown in figure 7.13b. The magnitude and direction angle of the acceleration are qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi   n 2 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffi t 2 ¼ 0:22 þ 0:162 ¼ 0:0656 ¼ 0:256 m/s2 aM þ aM aM ¼ h ¼ arc cos

n aM 0:16 ¼ 51:3 ¼ arc cos 0:256 aM

Example 7.2. Pulley O with a radius r = 0.2 m is rotating to lift cargo A, figure 7.14. The angular position of the pulley is defined by u ¼ ð3t  t 2 Þrad, where t is in seconds. Determine the velocity and acceleration of particle P on pulley O when t = 1 s. Also determine the velocity and acceleration of cargo A at this instant.

FIG. 7.14 – The velocity and acceleration of a particle on the pulley and the lifted cargo. Solution: Since u ¼ ð3t  t 2 Þrad, the angular velocity and angular acceleration of pulley O are as follows: x¼

du dx ¼ ð3  2t Þ rad=s; a ¼ ¼ 2 rad=s2 dt dt

When t = 1 s, we have u ¼ 3  1  12 ¼ 2 rad;

x ¼ 3  2  1 ¼ 1 rad=s; a ¼ 2 rad=s2

Planar Kinematics of Rigid Bodies

245

Here ω and α have opposite signs (senses of rotation), the pulley is rotating with decreasing angular speed at this instant. (1) Particle P on rotating pulley O has the following velocity and two acceleration components at t = 1 s: vP ¼ rx ¼ 0:2  1 ¼ 0:2 m=s aPt ¼ ra ¼ 0:2  ð2Þ ¼ 0:4 m=s2 aPn ¼ rx2 ¼ 0:2  12 ¼ 0:2 m=s2

Their directions are shown in figure 7.14. The magnitude and direction angle of the acceleration are qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi  t 2  n 2ffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 2 aP þ aP ¼ ð0:4Þ þ ð0:2Þ ¼ 0:447 m/s2 aP ¼ h ¼ arc cos

aPn 0:2 ¼ 63:4 ¼ arc cos 0:447 aP

(2) Since the inextensible cable is wrapped around the pulley and moves tangent to it, the vertical distance of cargo A is equal to the distance of particle P along its circular path, i.e., xA ¼ sP

Taking successive time derivatives of the above equation yields vA ¼ vP ;

aA ¼ aPt

Thus, when t = 1 s, we have vA ¼ vP ¼ 0:2 m/s;

aA ¼ aPt ¼ 0:4 m/s2

Their directions are shown in figure 7.14. Here v A and aA have opposite signs (senses of direction), cargo A is moving upward and slowing down at this instant. Example 7.3. Wheel O1 shown in figure 7.15a begins to rotate from rest with an angular acceleration a1 ¼ ð0:6t Þ rad/s2 , where t is in seconds. The inextensible transmission belt does not slip on the wheels. The radii of the two wheels are r1 ¼ 30 cm and r2 ¼ 50 cm. Determine the angular acceleration and angular velocity of wheel O2 after wheel O1 has turned ten revolutions.

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FIG. 7.15 – Belt-pulley system. Solution: (1) Angular motion of wheel O1. Since a1 ¼ ð0:6t Þ rad/s2 and the initial values of wheel O1’s angular velocity and angular position are x0 ¼ 0 and u0 ¼ 0, we have Z t Z t   x1 ¼ x0 þ a1 dt ¼ 0:6tdt ¼ 0:3t 2 rad/s 0

u1 ¼ u0 þ x0 t þ

Z t Z 0

0 t

  Z t Z t   a1 dt dt ¼ 0:6tdt dt ¼ 0:1t 3 rad

0

0

0

Convert the ten revolutions to radians.   2p rad u1 ¼ 10 rev ¼ 62:83 rad 1 rev Therefore, when wheel O1 has turned ten revolutions, i.e., 62.83 rad, the time elapsed is rffiffiffiffiffiffiffiffiffiffiffi 3 62:83 t¼ ¼ 8:565 s 0:1 At this instant, the angular acceleration and angular velocity of wheel O1 are a1 ¼ 0:6  8:565 ¼ 5:139 rad/s2 ; x1 ¼ 0:3  8:5652 ¼ 22:01 rad/s (2) Angular motion of wheel O2. Since the inextensible belt does not slip, an equivalent belt length s is unraveled from both wheels at all times. Thus, s ¼ u1 r1 ¼ u2 r2 Taking the first and second derivative yields v ¼ x1 r1 ¼ x2 r2 ;

at ¼ a1 r1 ¼ a2 r2

Then we get the transmission ratio of the two wheels rotating about fixed axis as follows:

Planar Kinematics of Rigid Bodies

i12 ¼

u1 x1 a1 n1 r2 ¼ ¼ ¼ ¼ u2 x2 a2 n2 r1

247

ð7:23Þ

where n is the speed of revolution. The negative sign should be used when the belt is arranged as shown in figure 7.15b, since the two wheels have opposite senses of rotation. From the transmission ratio, the angular acceleration and angular velocity of wheel O2 at t ¼ 8:565 s are a2 ¼

a1 r1 5:139  30 x1 r1 22:01  30 ¼ 3:08 rad/s2 ; x2 ¼ ¼ 13:2 rad/s ¼ ¼ 50 50 r2 r2

Example 7.4. Gear O1 shown in figure 7.16 rotates with an angular acceleration   a1 ¼ 0:3u21 rad/s2 , where φ1 is in radians. The radii of the two gears are r1 ¼ 24 cm and r2 ¼ 16 cm. If gear O1 is initially turning at x0 ¼ 10 rad/s, determine the angular velocity and angular acceleration of gear O2 after gear O1 turns two revolutions.

FIG. 7.16 – Gear system. Solution: (1) Angular motion of gear O1. Since there are 2π rad in one revolution, then u1 ¼ 2  2p ¼ 12:57 rad At this instant, the angular acceleration is a1 ¼ 0:3  12:572 ¼ 47:40 rad/s2 Because the angular acceleration is a function of φ1, we can get the angular velocity of gear O1 from the integration of xdx ¼ adu. Z x1 Z u1 Z 12:57 x1 dx1 ¼ a1 du1 ¼ 0:3u21 du1 ¼ 198:6 ) x1 ¼ 22:30 rad/s 10

0

0

(2) Angular motion of gear O2. When the gear system works properly, the contacting points A and B on gear O1 and on gear O2, respectively, should have the same speed, i.e., vA ¼ vB . Thus, vA ¼ x1 r1 ¼ vB ¼ x2 r2

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Taking the time derivative of the above expression yields aAt ¼ a1 r1 ¼ aBt ¼ a2 r2 Then we get the transmission ratio of the two gears rotating about fixed axis. i12 ¼

x 1 a1 n 1 u 1 r2 Z2 ¼ ¼ ¼ ¼ ¼ x 2 a2 n 2 u 2 r1 Z1

ð7:24Þ

where Z is the number of teeth of the gear. For external gearing shown in figure 7.16, the negative sign should be used, since the two gears have opposite senses of rotation. For inner gearing, the positive sign should be used. From the transmission ratio, the angular acceleration and angular velocity of gear O2 after gear O1 turns two revolutions are a1 r 1 47:40  24 ¼ 71:1 rad/s2 ; ¼ 16 r2 x1 r1 22:30  24 ¼ 33:5 rad/s x2 ¼  ¼ 16 r2

a2 ¼ 

The negative signs mean the angular acceleration and angular velocity of gear O2 are clockwise. Example 7.5. A simple gearbox consists of four gears, as shown in figure 7.17. Gears I and II, III and IV are in mesh. Gears II and III are rigidly connected to the same shaft and turns with it. The number of teeth of each gear is Z1 ¼ 36; Z2 ¼ 112; Z3 ¼ 32 and Z4 ¼ 128, respectively. If the angular velocity of gear I is x1 ¼ 1 50 rad/s, determine the angular velocity of gear IV.

FIG. 7.17 – A simple gearbox. Solution: The transmission ratio between gear I and gear IV is     x1 x1 x2 x3 Z2 Z4 Z2 Z4 112  128 ¼ 12:44 i14 ¼ ¼ ¼  ¼ 1  ¼ 36  32 x4 x2 x3 x4 Z1 Z3 Z1 Z3 Thus, the angular velocity of gear IV is x4 ¼

x1 150 ¼ 12:1 rad/s ¼ i14 12:44

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249

x4 is positive and this means the sense of rotation of gear IV is the same as that of gear I, i.e., counterclockwise.

7.3

Absolute Dependent Motion Analysis of Bodies

In a mechanism, the motion of one body may depend on other body’s motion because of restrictions from connections and supports. For example, an eccentric circular cam rotates about a fixed axis passing through O, figure 7.18. A flat face follower AB maintains contact with the cam due to restriction of the vertical track and the weight D. Then the motion of follower AB depends on the cam’s rotation.

FIG. 7.18 – Dependent motion between an eccentric circular cam and a follower. To describe this dependency, the following procedure should be used.
First, position coordinates should be set from fixed datum lines (or points) to describe the bodies’ motion. In figure 7.18, angular position φ measured from a vertical fixed datum line passing through O is used to define the cam’s rotation. Position coordinate y in the direction of the rectilinear translation of follower AB and measured from a horizontal fixed datum line passing through O is used to define follower AB’s motion.
Second, the coordinates are related to each other using geometry. If the cam in figure 7.18 has radius r and eccentricity OC = e, angular position φ can be related to the position coordinate y as y ¼ r þ e sin u

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Third, take successive time derivatives of the position coordinate equation to get relationship between (angular) velocities and relationship between (angular) accelerations. Taking the first time derivative of the above position coordinate equation yields dy du ¼ e cos u dt dt

)

v ¼ e cos ux

)

a ¼ ea cos u  ex2 sin u

Take time derivative again, dv dx du ¼ e cos u  e sin u x dt dt dt

Example 7.6. Ring M fits loosely on both rocker OA and fixed rod BD, figure 7.19. Starting from the vertical position, rocker OA rotates about O with constant angular velocity ω and pushes ring M to move along rod BD. Find the velocity and acceleration of ring M when it is in any arbitrary position x.

FIG. 7.19 – Dependent motion between a rocker and a ring. Solution: Angular position φ and position coordinate x are set as shown. They can be related as x ¼ h tan u ¼ h tan xt Take successive time derivatives of this position coordinate equation to get the velocity and acceleration of ring M.     dx x2 v¼ ¼ hx sec2 xt ¼ hx 1 þ tan2 xt ¼ hx 1 þ 2 dt h     dv x2 2 2 2 2 2 2 2 a¼ ¼ 2hx sec xt  tan xt ¼ 2xx sec xt ¼ 2xx 1 þ tan xt ¼ 2xx 1 þ 2 h dt

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251

Example 7.7. Crank CB with length r = 0.1 m rotates about C and drives rocker OA to rotate about O through sleeve B, figure 7.20a. The angular position of crank CB is defined by u ¼ ð0:1t Þrad, where t is in seconds. If h = 0.3 m, specify the angular position of rocker OA at any instant.

FIG. 7.20 – Dependent Motion between a rocker and a crank. Solution: Angular position θ measured from a fixed datum shown in figure 7.20a is used to define rocker OA’s rotation. Draw line BD perpendicular to line OC. From geometry, we have tan h ¼

BD r sin u 0:1 sinð0:1t Þ sinð0:1t Þ ¼ ¼ ¼ DO h  r cos u 0:3  0:1  cosð0:1t Þ 3  cosð0:1t Þ

Thus, the angular position of rocker OA at any instant is   sinð0:1t Þ h ¼ arctan 3  cosð0:1t Þ RETHINK: Taking successive time derivatives of the expression of the angular position, we should get the angular velocity and angular acceleration, but the differentials are difficult. For this case, we may use a numerical procedure. Here, using the MATLAB program, we construct the xt and at graphs as shown in figure 7.20b and c respectively.

7.4

General Plane Motion

As shown in figure 7.21a, a rigid body is undergoing general planar motion and any particle on the body moves along a path that is equidistant from a fixed plane— plane Ⅰ. The rigid body can be considered to be composed of an infinite number of straight lines A1A2 that are perpendicular to the fixed plane I. Using plane II, which

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is parallel to plane I, to cut the rigid body yields a section S and to cut line A1A2 yields a point A. By inspection, section S as well as point A will move within plane II and line A1A2 will translate. Then, the motion of A can represent the motion of line A1A2, and likewise the motion of section S can represent the motion of the whole body. From the top view of plane II, section S moves within the x–y plane, figure 7.21b. Apparently, an arbitrary line segment AB can define the position of section S within the x–y plane. To locate line segment AB, three parameters (coordinates) should be used. 9 xA ¼ xA ðtÞ = yA ¼ yA ðtÞ ð7:25Þ ; u ¼ uðtÞ If xA  xA0 and yA  yA0 , it means the body is rotating about a fixed axis perpendicular to the x–y plane and passing through A; if u  u0 , it means the body is translating. These are two special cases of general plane motion. Commonly, general plane motion is a combination of translation and rotation.

FIG. 7.21 – Illustration of general plane motion. To view these “component” motions separately, a relative-motion analysis involving two reference frames will be used. Besides the fixed reference frame Oxy, a translating reference frame Ax 0 y 0 with its origin attached to A, which is called the “base point”, is also utilized. As a translating reference frame, its coordinate axes do not rotate with the body; rather they only translate with respect to the fixed frame. In other words, the x 0 axis is always parallel to the x axis and the y 0 axis is always parallel to the y axis. It should be noted that the base point A is an arbitrarily chosen point. Any point on section S can be chosen as the base point. However, a point having a known motion is generally chosen as the base point.

Planar Kinematics of Rigid Bodies

7.4.1

253

Relative-Motion Analysis: Velocity and Acceleration

Position. Suppose that at given instant t, the body’s position is represented by line segment AB, as shown in figure 7.22. Choose A as the base point. Point B’s motion will be observed from both fixed reference frame Oxy and translating reference frame Ax 0 y 0 . Absolute-position vectors rA and rB , measured in the fixed reference frame, and the relative-position vector rB=A , locating point B with respect to A, can be related as follows: rB ¼ rA þ rB=A

ð7:26Þ

FIG. 7.22 – Translation with the base point A and rotation about A. Displacement. After a differential time interval dt, points A and B undergo displacements drA and drB , respectively, and the body comes to the position of A0 B 0 , figure 7.22. This general plane motion can be divided into two components: translation with the base point A by an amount drA to A0 B 00 and rotation about A by an amount du to A0 B 0 . Due to the rotation about the base point A, drB=A ¼ rB=A du. Then the displacement of B is drB ¼ drA þ drB=A

ð7:27Þ

This vector summation is also shown in figure 7.22. Velocity. Taking the time derivative of equation (7.26), or dividing equation (7.27) by dt, we have drB drA drB=A ¼ þ dt dt dt

or vB ¼ vA þ vB=A

ð7:28Þ

Here drB =dt ¼ v B and drA =dt ¼ vA are absolute velocities measured from the fixed reference frame Oxy. drB=A =dt ¼ v B=A is the relative velocity of B with respect to A observed from the translating reference frame Ax 0 y 0 . The observer only sees point B move along a circular path that has center at A and radius rB=A . In other words, the body appears to be “pinned” at A and rotating about A in the observer’s

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eyes. Therefore, the direction of relative velocity v B=A is tangent to the circular path or perpendicular to rB=A . Its sense of direction is in accordance with the sense of rotation of angular velocity x of the body. The magnitude of v B=A is vB=A ¼

drB=A du ¼ rB=A x ¼ rB=A dt dt

ð7:29Þ

From equation (7.20), the relative velocity can be accounted for by the cross product vB=A ¼ x  rB=A . Hence, v B ¼ v A þ v B=A ¼ v A þ x  rB=A

ð7:30Þ

Equation (7.30) is graphically illustrated in the kinematic diagram of velocity in figure 7.23. Since the three vectors v B ; v A and v B=A are in the same plane—the x–y plane, the vector equation can be resolved to obtain two scalar equations. Therefore, at most two unknowns, represented by the magnitudes and/or directions of the velocity vectors, can be solved for.

FIG. 7.23 – Relative-motion analysis: velocity.

Acceleration. Taking the time derivative of equation (7.28) yields dv B dv A dv B=A ¼ þ dt dt dt

or aB ¼ aA þ aB=A

ð7:31Þ

Here dv B =dt ¼ aB and dv A =dt ¼ aA are absolute accelerations. dv B=A =dt ¼ aB=A is the relative acceleration of B with respect to A observed from the translating reference frame Ax 0 y 0 . As mentioned above, to this observer the body appears to be rotating about A and point B is moving along a circular path that has the center at A and a radius rB=A . Therefore, aB=A can be divided into the tangential and normal components, aB=A ¼ atB=A þ anB=A

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255

where atB=A and anB=A are the relative tangential and normal components of acceleration of point B with respect to the base point A. Their magnitudes are t n ¼ arB=A and aB=A ¼ x2 rB=A . The direction of atB=A is tangent to the circular aB=A path or perpendicular to rB=A . Its sense of direction is in accordance with the sense of rotation of the angular acceleration a of the body. The direction of anB=A is always from point B to the center of the circular path, i.e., to the base point A. Therefore, the relative acceleration equation can be written as aB ¼ aA þ aB=A ¼ aA þ atB=A þ anB=A

ð7:32Þ

From equation (7.22), the relative acceleration can also be accounted for by the cross product aB=A ¼ atB=A þ anB=A ¼ a  rB=A  x2 rB=A . Hence, aB ¼ aA þ atB=A þ anB=A ¼ aA þ a  rB=A  x2 rB=A

ð7:33Þ

Equation (7.33) is graphically illustrated in the kinematic diagram of acceleration in figure 7.24. The four vectors aB , aA , atB=A and anB=A are in the same plane—the x–y plane, since section S moves within the x–y plane. Thus, the vector equation can be resolved to obtain two scalar equations, and at most two unknowns, represented by the magnitudes and/or directions of the acceleration vectors, can be solved for.

FIG. 7.24 – Relative-motion analysis: acceleration.

Example 7.8. Slider A can only move in the horizontal slot and slider B can only move in the vertical slot, figure 7.25a. At the instant shown, slider A moves to the left with a speed vA = 20 cm/s. If the length of link AB is l = 20 cm, find the velocity of slider B and the angular velocity of link AB.

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FIG. 7.25 – A mechanism including two sliders and a link. Solution 1: Scalar analysis Slider A and B are undergoing rectilinear translations and can be treated as particles. Link AB is undergoing general plane motion. From the relative velocity equation, we have Magnitude ?p Direction v B

p

¼ v A p þ v B=A ?p

Here, two question marks “?” mean the magnitudes of v B and v B=A are unknown. The other four magnitudes and/or directions of the vector quantities are known. The direction of v B should be vertical and v B=A should be perpendicular to rB=A . Draw the velocity parallelogram in the kinematic diagram, as shown in figure 7.25b. From trigonometry, we have vB ¼ vA cot 30 ¼ 20 cot 30 ¼ 34:6 cm/s vB=A ¼

vA 20 ¼ ¼ 40 cm/s sin 30 sin 30

Since vB=A ¼ xAB rB=A ¼ xAB l, the angular velocity xAB of link AB is xAB ¼

vB=A 40 ¼ 2 rad/s ¼ 20 l

The sense of rotation of xAB is clockwise since it should be in accordance with the sense of direction of v B=A shown in figure 7.25b. Solution 2: Vector analysis Link AB is undergoing general plane motion. Assume its angular velocity xAB is clockwise and v B is vertically upward, figure 7.25b. From the relative velocity equation using cross product, we have v B ¼ vA þ xAB  rB=A

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257

The kinematic diagram of velocity is shown in figure 7.25b. Express every vector in Cartesian vector form and evaluate the cross product. vB j ¼ ð20iÞ þ ðxAB kÞ  ð20 cos 30 i þ 20 sin 30 jÞ vB j ¼ ð20iÞ þ ð20xAB cos 30 j þ 20xAB sin 30 iÞ Equating the i and j components on both sides yields i: j:

0 ¼ 20 þ 20xAB sin 30 vB ¼ 20xAB cos 30

Solving, we obtain xAB ¼ 2 rad/s, vB ¼ 34:6 cm/s The calculated results of (angular) velocities are both positive. This means the assumption for their sense of direction is true. Example 7.9. In the mechanism shown in figure 7.26a, crankshaft OA is rotating with a constant angular velocity x. If OA = r, AB = l, determine (1) the velocity of slider B and the angular velocity of link AB; (2) the acceleration of slider B and the angular acceleration of link AB in the position shown (crankshaft OA is perpendicular to link AB).

FIG. 7.26 – A mechanism including a crank, a link and a slider.

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258

Solution 1: Scalar analysis Since crankshaft OA is rotating with a constant angular velocity x, the velocity and acceleration of point A can be determined. Their directions are shown in figure 7.26b and c, respectively. Their magnitudes are vA ¼ rx;

aA ¼ aAn ¼ rx2

Link AB is undergoing general plane motion. From the relative velocity equation, we have Magnitude ?p Direction v B

p

¼ v A p þ v B=A ?p

The direction of v B should be horizontal and vB=A should be perpendicular to rB=A . There are only two unknowns in the vector equation, namely, the magnitudes of v B and v B=A . Draw the velocity parallelogram in the kinematic diagram, figure 7.26b. From trigonometry, we have pffiffiffiffiffiffiffiffiffiffiffiffiffiffi vA r2 þ l2 vB ¼ ¼ rx l cos h r r2 vB=A ¼ vA tan h ¼ rx  ¼ x l l Since vB=A ¼ xAB rB=A ¼ xAB l, the angular velocity xAB of link AB is xAB ¼

vB=A r2 ¼x 2 l l

The sense of rotation of xAB is clockwise since it should be in accordance with the sense of direction of v B=A shown in figure 7.26b. From the relative acceleration equation, we have Magnitude ?p Direction aB

p

?

p

¼ aA p þ atB=A p þ anB=A p

n where aB=A ¼ x2AB rB=A ¼ x2 r 4 =l 3 . The direction of every vector is shown in the kinematic diagram, figure 7.26c. The magnitudes of aB and atB=A are unknown, and the sense of direction of these two vectors have been assumed. Since there are only two unknowns in the above vector equation, these two unknowns should be able to be determined. Equating the x and y components on both sides yields

x: y:

t n aB ¼ aA sin h þ aB=A sin h  aB=A cos h

ð1Þ

t n 0 ¼ aA cos h þ aB=A cos h þ aB=A sin h

ð2Þ

pffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffi Substituting sin h ¼ r= r 2 þ l 2 ; cos h ¼ l= r 2 þ l 2 and solving simultaneously yields

Planar Kinematics of Rigid Bodies pffiffiffiffiffiffiffiffiffiffiffiffiffiffi r4 r2 þ l2 aB ¼ x ; l4 2

259

t aB=A



r4 ¼x r 1 4 l



2

t [ 0. This means the assumption for the Apparently, aB [ 0 and when r\l, aB=A t senses of the two accelerations is true. Since aB=A ¼ aAB rB=A ¼ aAB l, the angular acceleration aAB of link AB is   t aB=A r r4 1 4 ¼ x2 aAB ¼ l l l

The sense of rotation of aAB is counterclockwise since it should be in accordance with the sense of direction of atB=A shown in figure 7.26c. Solution 2: Vector analysis Since crankshaft OA is rotating with a constant angular velocity x, the velocity and acceleration of point A are v A ¼ x  rA ¼ ðxkÞ  ðr sin hi þ r cos hjÞ ¼ xr sin hj  xr cos hi aA ¼ anA ¼ x2 rA ¼ x2 ðr sin hi þ r cos hjÞ Link AB is undergoing general plane motion. Assume its angular velocity is clockwise and v B is horizontal to the left, figure 7.26b. From the relative velocity equation using cross product, we have v B ¼ vA þ xAB  rB=A The kinematic diagram of velocity is shown in figure 7.26b. Express every vector in Cartesian vector form and evaluate the cross product. vB i ¼ ðxr sin hj  xr cos hiÞ þ ðxAB kÞ  ðl cos hi  l sin hjÞ vB i ¼ ðxr sin hj  xr cos hiÞ  xAB l cos hj  xAB l sin hi vB i ¼ ðxr cos h þ xAB l sin hÞi þ ðxr sin h  xAB l cos hÞj Equating the i and j components on both sides yields i:

vB ¼ ðxr cos h þ xAB l sin hÞ j:

Solving, we obtain xAB

0 ¼ xr sin h  xAB l cos h

pffiffiffiffiffiffiffiffiffiffiffiffiffiffi xr r2 r r2 þ l2 tan h ¼ x 2 ; vB ¼ x ¼ l l l

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260

From the relative acceleration equation using cross product, we have aB ¼ aA þ aAB  rB=A  x2AB rB=A Draw the kinematic diagram of acceleration, as shown in figure 7.26c. Here assume aAB is counterclockwise and aB is horizontal to the left. Express every vector in Cartesian vector form and evaluate the cross product. aB i ¼ x2 ðr sin hi þ r cos hjÞ þ ðaAB kÞ  ðl cos hi  l sin hjÞ  x2AB ðl cos hi  l sin hjÞ

aB i ¼ x2 ðr sin hi þ r cos hjÞ þ ðaAB l cos hj þ aAB l sin hiÞ  x2AB ðl cos hi  l sin hjÞ     aB i ¼ x2 r sin h þ aAB l sin h  x2AB l cos h i þ x2 r cos h þ aAB l cos h þ x2AB l sin h j

Equating the i and j components on both sides yields i: j:

aB ¼ x2 r sin h þ aAB l sin h  x2AB l cos h

ð3Þ

0 ¼ x2 r cos h þ aAB l cos h þ x2AB l sin h

ð4Þ

Actually, equations (3) and (4) are just equations (1) and (2) in scalar analysis. Substituting xAB and solving, we get pffiffiffiffiffiffiffiffiffiffiffiffiffiffi   r r4 r4 r2 þ l2 1  4 ; aB ¼ x 2 aAB ¼ x2 l l4 l

RETHINK: For this problem, the absolute dependent motion analysis can also be performed. Set the position coordinates u; h and xB as shown in figure 7.26d. The position coordinate equations can be written as qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi xB ¼ r cos u þ l 2  ðr sin uÞ2   r sin u h ¼ arcsin l Taking successive time derivatives of these position coordinate equations and using the chain rule of calculus, we can obtain the velocity and acceleration of slider B and the angular velocity and angular acceleration of link AB at any instant (position), although sometimes the differentials are difficult. However, the relative velocity equation and relative acceleration equation can only be used to determine the velocity and acceleration at a specific instant (position). For this example, the velocities and accelerations are calculated in the position of

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261

crankshaft OA being perpendicular to link AB. For another position, the similar time-consuming procedure has to be performed again. Example 7.10. The disk rotates with a constant angular velocity x0 = 10 rad/s. If OA = 20 cm and AB = 100 cm, find the velocity and acceleration of slider B at the instant shown in figure 7.27a.

FIG. 7.27 – A mechanism including a wheel, a link and a slider. Solution: Since disk OA is rotating with a constant angular velocity, the velocity and acceleration of point A can be determined. Their directions are shown in figure 7.27b and c, respectively. Their magnitudes are vA ¼ x0  OA ¼ 10  20 ¼ 200 cm/s; aA ¼ aAn ¼ x20  OA ¼ 102  20 ¼ 2000 cm/s2 Link AB is undergoing general plane motion. From the relative velocity equation, we have Magnitude ?p Direction v B

p

¼ v A p þ v B=A ?p

The direction of v B should be vertical and v B=A should be perpendicular to link AB. Only the magnitudes of v B and vB=A are unknown. Draw the velocity parallelogram in the kinematic diagram, figure 7.27b. From trigonometry, we have pffiffiffi vA vB ¼ ¼ 200 2 ¼ 283 cm/s  cos 45 vB=A ¼ vA tan 45 ¼ 200 cm/s

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Since vB=A ¼ xAB  AB, the angular velocity xAB of link AB is xAB ¼

vB=A 200 ¼ 2 rad/s ¼ 100 AB

The sense of rotation of xAB is clockwise since it should be in accordance with the sense of direction of v B=A shown in figure 7.27b. From the relative acceleration equation, we have Magnitude ?p Direction a B

p

?

p

¼ a nA p þ a tB=A p þ a nB=A p

ð1Þ

n where aB=A ¼ x2AB  AB ¼ 22  100 ¼ 400 cm/s2 . The direction of every vector is shown in the kinematic diagram, figure 7.27c. Only the magnitudes of aB and atB=A are unknown. The senses of direction of these two vectors have been assumed. Equating the ξ component on both sides of equation (1) yields

n:

n aB cos 45 ¼ aB=A ¼ 400

)

aB ¼ 566 cm/s2

The negative sign indicates the sense of direction of aB at this instant is downward, opposite to that shown on the kinematic diagram. RETHINK: Equating the η component on both sides of equation (1), we should get atB=A and then the angular acceleration of link AB. Readers are recommended to finish this by themselves. Example 7.11. A wheel of radius r = 0.5 m is moving on the rough surface. At the instant shown, it has an angular velocity x ¼ 5 rad/s and angular deceleration a ¼ 3 rad/s2 , and its center has a velocity vC ¼ 8 m/s and deceleration aC ¼ 1 m/s2 , figure 7.28a. Determine the velocity and acceleration of point A, which contacts the ground at this instant.

FIG. 7.28 – A wheel moving on the rough surface. Solution: The wheel is undergoing general plane motion. From the relative velocity/acceleration equation using cross product and the kinematic diagram shown in figure 7.28b, we have

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263

v A ¼ v C þ x  rA=C ¼ ð8iÞ þ ð5kÞ  ð0:5jÞ ¼ f5:5igm/s aA ¼ aC þ a  rA=C  x2 rA=C ¼ ð1iÞ þ ð3kÞ  ð0:5jÞ  25ð0:5jÞ ¼ f0:5i þ 12:5jgm/s2 Since the velocity of the contacting point A is not zero, the wheel is not rolling without slipping. Example 7.12. In the mechanism shown in figure 7.29a, wheel O has radius r = 40 cm and the lengths of link AB and crank BC are 120 cm and 30 cm, respectively. At the instant shown (OA is in the vertical position), wheel O has a clockwise angular velocity xO ¼ 3 rad/s and a counterclockwise angular acceleration aO ¼ 5 rad/s2 . Find the angular velocity and angular acceleration of crank BC at this instant.

FIG. 7.29 – A mechanism including a wheel, a link and a crank. Solution: Since wheel O is rotating with xO ¼ 3 rad/s and aO ¼ 5 rad/s2 , the velocity and acceleration of point A are v A ¼ xO  rA ¼ ðxO kÞ  ð40jÞ ¼ 40xO i ¼ f120igcm/s aA ¼ atA þ anA ¼ aO  rA  x2O rA ¼ ðaO kÞ  ð40jÞ  x2O ð40jÞ ¼ 40aO i  40x2O j ¼ f200i  360jgcm/s2 Link AB is undergoing general plane motion. From the relative velocity/ acceleration equation using cross product, we have

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264 v B ¼ vA þ xAB  rB=A

ð1Þ

aB ¼ aA þ aAB  rB=A  x2AB rB=A

ð2Þ

Since B is also a point on the rotating crank BC, we have v B ¼ xBC  rB=C

ð3Þ

aB ¼ atB þ anB ¼ aBC  rB=C  x2BC rB=C

ð4Þ

Substituting equation (3) into equation (1) yields xBC  rB=C ¼ v A þ xAB  rB=A The kinematic diagram is shown in figure 7.29b. Express every vector in Cartesian vector form and evaluate the cross product. ðxBC kÞ  ð30 cos 45 i þ 30 sin 45 jÞ ¼ ð120iÞ þ ðxAB kÞ  ð120 cos 60 i þ 120 sin 60 jÞ

pffiffiffi pffiffiffi pffiffiffi 15 2xBC j  15 2xBC i ¼ 120  60 3xAB i  60xAB j Equating i and j components on both sides yields pffiffiffi pffiffiffi i : 15 2xBC ¼ 120  60 3xAB j:

pffiffiffi 15 2xBC ¼ 60xAB

Solving simultaneously, we get xBC ¼ 2:07 rad/s, xAB ¼ 0:732 rad/s The negative sign of xBC indicates that the sense of rotation of the angular velocity of crank BC at this instant is clockwise, opposite to that shown on the kinematic diagram. Substituting equation (4) into equation (2) yields aBC  rB=C  x2BC rB=C ¼ aA þ aAB  rB=A  x2AB rB=A Express every vector in Cartesian vector form and evaluate the cross product. ðaBC kÞ  ð30 cos 45 i þ 30 sin 45 jÞ  x2BC ð30 cos 45 i þ 30 sin 45 jÞ ¼ ð200i  360jÞ þ ðaAB kÞ  ð120 cos 60 i þ 120 sin 60 jÞ  x2AB ð120 cos 60 i þ 120 sin 60 jÞ

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265

pffiffiffi pffiffiffi   15 2 aBC þ x2BC i þ 15 2 aBC  x 2BC j

pffiffiffi pffiffiffi ¼ 200  60 3aAB þ 60x2AB i  360 þ 60aAB þ 60 3x2AB j Equating i and j components on both sides yields pffiffiffi pffiffiffi  i : 15 2 aBC þ x2BC ¼ 200  60 3aAB þ 60x2AB j:



pffiffiffi pffiffiffi  15 2 aBC  x2BC ¼  360 þ 60aAB þ 60 3x2AB

Substituting the angular velocities and solving simultaneously, we get aBC ¼ 8:38 rad/s2 ; aAB ¼ 2:45 rad/s2 The negative signs indicate that the senses of rotation of aBC and aAB at this instant are both clockwise, opposite to that shown on the kinematic diagram.

7.4.2

Instantaneous Center of Rotation

The velocity of any point B on a rigid body undergoing general plane motion can be obtained by the relative velocity equation v B ¼ vA þ vB=A ¼ v A þ x  rB=A . If base point A we choose has zero velocity at the instant considered, the relative velocity equation becomes v B ¼ v B=A ¼ x  rB=A . For other arbitrary points, such as C and D, their velocities at this instant are v C ¼ v C =A ¼ x  rC =A and v D ¼ v D=A ¼ x  rD=A , respectively, as shown in figure 7.30. It can be seen that the body appears to be momentarily “pinned” at A and rotating about A. This point A with zero velocity is called the instantaneous center of rotation or instantaneous center of zero velocity (IC), which has been mentioned in example 6.6. It has been derived in that example that the point of contact on the wheel has zero velocity when the wheel rolls without slipping on the ground. Thus, this contacting point is IC and the velocities of other points on the wheel are shown in figure 7.31. Apparently, the point of contact on the wheel is different at different instant. This means the IC for the body always change its position with the time and can only be used to determine other points’ velocities for an instant of time.

FIG. 7.30 – Instantaneous center of zero velocity.

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FIG. 7.31 – IC of a wheel rolling without slipping. If the IC of the body for an instant considered is found, the velocity of any other point on the body at this instant can be obtained easily, but how to find the IC? The fact that the velocity of a point on the body is always perpendicular to the line connecting the point and the IC can be used to locate the IC. There are several cases of locating the IC: (1) If velocity v A of point A and angular velocity x are known, figure 7.32a, the IC must be located on the line drawn perpendicular to vA at A and the distance from A to the IC should be rIC =A ¼ vA =x, since vIC ¼ vA  xrIC =A ¼ 0. At the same time, the IC lies above point A in figure 7.32a such that v A is in accordance with the counterclockwise angular velocity x about the IC. (2) If directions of two nonparallel velocities v A and v B are known, figure 7.32b, the intersection point of two lines that are drawn perpendicular to v A and v B at points A and B, respectively, is just the IC of the body at this instant. (3) If the magnitudes and directions of two parallel velocities v A and vB are known and the velocities are perpendicular to line AB, figure 7.32c and d, the location of the IC is determined by proportional triangles as shown. (4) If directions of two parallel velocities v A and v B are known and the velocities are not perpendicular to line AB, figure 7.32e, the two lines drawn perpendicular to v A and v B at points A and B, respectively, are parallel to each other and intersect at infinity, i.e., rIC =A ¼ vA =x ¼ 1 and then x ¼ 0. Therefore, from v B ¼ v A þ x  rB=A , we obtain v B ¼ v A . Actually, for any arbitrary point M on the body, we have v M ¼ v A ¼ v B . This case is called momentary translation. Momentary translation should be distinguished from translation. For translation, all particles on the body have the same velocity and the same acceleration at any instant. The angular velocity and angular acceleration of the body are always zero. However, for momentary translation, only the velocities of all particles on the body are the same and the angular velocity of the body is zero at the instant considered. The accelerations of particles on the body are generally different and the angular acceleration of the body is not zero. For example, link AB in figure 7.32f are undergoing momentary translation. At the instant shown, v B ¼ v A and xAB ¼ 0, but apparently, aB 6¼ aA and aAB 6¼ 0.

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267

FIG. 7.32 – Cases of identifying IC.

Example 7.13. At the instant shown, the gear racks A and B have the velocities vA = 10 m/s and vB = 2 m/s, figure 7.33a. The radius of the gear is r = 0.8 m. Specify the velocity of the center point C and the angular velocity of the gear at this instant.

FIG. 7.33 – Gear and gear racks.

Solution 1: Method of IC The contact points on the gear should have the same velocities with the gear racks A and B, respectively. Therefore, the kinematic diagram of the gear is shown in

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268

figure 7.33b. For the gear undergoing general plane motion, the magnitudes and directions of two parallel velocities v A and v B are known and the velocities are perpendicular to line AB. Thus, the IC can be determined by proportional triangles as shown in figure 7.33b. rB=IC rB=IC vB 2 ¼ ¼ ¼ vA 10 rA=IC rB=IC þ 1:6

)

rB=IC ¼ 0:4 m

Then the angular velocity of the gear is x¼

vB rB=IC

¼

2 ¼ 5 rad/s 0:4

The velocity of point C is

  vC ¼ xrC =IC ¼ x 0:8 þ rB=IC ¼ 5ð0:8 þ 0:4Þ ¼ 6 m/s

Its direction is perpendicular to rC =IC , as shown in figure 7.33b. Solution 2: Vector analysis From the relative velocity equation using cross product, we have v B ¼ v A þ x  rB=A From figure 7.33c, the above vector equation can be written as 2i ¼ 10if þ ðxkÞ  ð1:6jÞ ¼ ð10  1:6xÞi 2 ¼ 10  1:6x

)

x ¼ 5 rad/s

Then the velocity of point C can also be determined from the relative velocity equation using cross product. vC ¼ v A þ x  rC =A v C ¼ 10if þ ð5kÞ  ð0:8jÞ ¼ f6igm/s Example 7.14. In the stone crusher shown in figure 7.34a, crank OE is rotating about O with a constant speed of revolution n = 100 rpm. If OE = 10 cm, BC = CD = 40 cm and AB = 60 cm, determine the angular velocity of crankshaft AB at the instant shown.

Planar Kinematics of Rigid Bodies

269

FIG. 7.34 – A stone crusher. Solution: Links CE and BC are undergoing general plane motion and the other members are rotating about a fixed axis. Since crank OE is rotating about O with a constant speed of revolution n = 100 rpm, the velocity of point E is vE ¼ xOE  OE ¼

n  2p 100p  10 ¼ cm/s 60 3

The direction of v E is shown in figure 7.34b. Since link CD is rotating about D, we have v C ?CD. Then for link CE, the directions of two nonparallel velocities v C and v E are known. Construct at points C and E lines that are perpendicular to v C and v E , respectively. The intersection point of these two lines is the IC of link CE at the instant shown. From trigonometry, IC1 C ¼

OC 80 þ 20 ¼ 200 cm ¼ sin30 0:5

IC1 E ¼ IC1 O  OE ¼ IC1 C cos 30  OE ¼ 200  cos 30  10 ¼ 163:2 cm Therefore, xCE ¼

vE 100p ¼ 0:641 rad/s ¼ IC1 E 3  163:2

vC ¼ IC1 C  xCE ¼ 200  0:641 ¼ 128:2 cm/s The sense of rotation of xCE and the direction of vC are shown in figure 7.34b. Since crankshaft AB is rotating about A, we have vB ?AB. Then for link BC, the directions of two nonparallel velocities v B and v C are known. Two lines drawn perpendicular to vB and v C intersect at IC2, which is the instantaneous center of rotation of link BC. From trigonometry, vB ¼ IC2 B  xBC ¼ IC2 B 

vC IC2 B ¼ vC ¼ vC cos 30 ¼ 111 cm/s IC2 C IC2 C

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The direction of v B is shown in figure 7.34b. Thus, the angular velocity of crankshaft AB at the instant shown is xAB ¼

vB 111 ¼ 1:85 rad/s ¼ 60 AB

The sense of rotation of xAB is clockwise in accordance with the sense of direction of v B . Example 7.15. A disk of radius r rolls without slipping on the flat ground. At a given instant, it has an angular velocity x and angular acceleration a, figure 7.35. Find the acceleration of the point of contact at this instant.

FIG. 7.35 – Rolling without slipping on the flat ground. Solution: Since the disk rolls without slipping, the point of contact is the IC. Then vC ¼ xr. This expression is always satisfied and we can take the time derivative of it and get aC ¼

dvC dx r ¼ ar ¼ dt dt

ð1Þ

From the relative acceleration equation, we have Magnitude ? Direction aIC ?

p

p

p

¼ aC p þ atIC =C p þ anIC =C p

t n 2 2 where aIC =C ¼ a  rIC =C ¼ ar and aIC =C ¼ x  rIC =C ¼ x r. The direction of every vector on the right side is shown in the kinematic diagram, figure 7.35. Equating x and y components on both sides yields

x: y:

t n aICx ¼ aCx þ aIC =Cx þ aIC =Cx ¼ ar  ar þ 0 ¼ 0 t n 2 2 aICy ¼ aCy þ aIC =Cy þ aIC =Cy ¼ 0 þ 0 þ x r ¼ x r

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271

Therefore, aIC ¼ aICy ¼ x2 r and the direction is from the contacting point toward center C. This means the IC has only zero velocity, and its acceleration is not zero, which has been mentioned in example 6.6. Example 7.16. A disk of radius r rolls without slipping on a circular path of radius R. At a given instant, it has an angular velocity x and angular acceleration a, figure 7.36. Find the acceleration of the point of contact at this instant.

FIG. 7.36 – Rolling without slipping on a circular path. Solution: Since the disk rolls without slipping, the point of contact is the IC. Then vC ¼ xr. This expression is always satisfied and we can take the time derivative of it and get aCt ¼

dvC dx r ¼ ar ¼ dt dt

Compare this equation with equation (1) in example 7.15 and the reason for the difference is that point C is moving along a straight line in example 7.15, but a curved path in this example. The normal acceleration component of point C in this example is aCn ¼

vC2 x2 r 2 ¼ q Rþr

From the relative acceleration equation, we have Magnitude ? Direction aIC ?

p

p

p

p

¼ atC p þ anC p þ atIC =C p þ anIC =C p

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t n 2 2 where aIC =C ¼ a  rIC =C ¼ ar and aIC =C ¼ x  rIC =C ¼ x r. The direction of every vector on the right side is shown in the kinematic diagram, figure 7.36. Equating x and y components on both sides yields

x:

y:

t n t n aICx ¼ aCx þ aCx þ aIC =Cx þ aIC =Cx ¼ ar þ 0  ar þ 0 ¼ 0

t n t n aICy ¼ aCy þ aCy þ aIC =Cy þ aIC =Cy ¼ 0 

x2 r 2 Rr þ 0 þ x2 r ¼ x2 Rþr Rþr

Therefore, aIC ¼ aICy ¼ x2 Rr=ðR þ r Þ and the direction is from the point of contact toward the center C. RETHINK: When R ! 1, the circular path becomes a straight line and aIC ¼ x2 Rr=ðR þ r Þ ! x2 r, which is consistent with the result in example 7.15. Example 7.17. The spool unravels from the cable with an angular velocity x ¼ 0:5 rad/s and angular acceleration a ¼ 2 rad/s2 at the instant shown in figure 7.37a. If R ¼ 0:6 m and r ¼ 0:4 m, find the acceleration of point A at this instant.

FIG. 7.37 – Spool unravelling from the cable. Solution: The spool looks like rolling downward without slipping and the leftmost point on the inner hub is the IC. From example 7.15, the acceleration of center C is aC ¼ ar ¼ 0:8 m/s and its direction is vertically downward. From the relative acceleration equation using cross product, we have aA ¼ aC þ a  rA=C  x2 rA=C The kinematic diagram is given in figure 7.37b. Express the vectors in Cartesian vector form and evaluate the cross product.

Planar Kinematics of Rigid Bodies

273

aA ¼ ð0:8jÞ þ ð2kÞ  ð0:6jÞ  0:25ð0:6jÞ ¼ f1:2i þ 0:95jgm/s2 Example 7.18. Crank OA rotates about O and drives the sector-ring-shaped slider B to move along the circular slot through link AB. At the instant shown, crank OA has an angular velocity xO and angular acceleration aO , figure 7.38a. O1 is the pffiffiffi center of the circular path. If OA ¼ r; AB ¼ 2 3r and O1 B ¼ 2r, Determine the tangential and normal components of acceleration of slider B at this instant.

FIG. 7.38 – A mechanism with a slider moving along a circular slot. Solution: Since crank OA rotates about O with an angular velocity xO, v A is perpendicular to rA=O and its magnitude is vA ¼ xO r. The direction of v B should be tangent to the circular path, as shown in figure 7.38b. Then for link AB undergoing general plane motion, the directions of two nonparallel velocities v A and v B are known. Two lines perpendicular to v A and vB intersect at IC, which is the instantaneous center of rotation of link AB at the instant shown. From trigonometry, rA=IC ¼ AB tan 30 ¼ 2r; rB=IC ¼ rA=IC =sin 30 ¼ 4r Then the angular velocity of link AB is xAB ¼

vA rA=IC

¼

xO r xO ¼ 2r 2

The velocity and normal component of acceleration of slider B are vB ¼ xAB rB=IC ¼ 2xO r; aBn ¼

vB2 ð2xO r Þ2 ¼ ¼ 2x2O r O1 B 2r

Their directions are shown in figure 7.38b. From the relative acceleration equation using cross product, we have aB ¼ aA þ aAB  rB=A  x2AB rB=A

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Since aB ¼ atB þ anB and aA ¼ atA þ anA ¼ aO  rA=O  x2O rA=O , the above equation is rewritten as follows: atB þ anB ¼ aO  rA=O  x2O rA=O þ aAB  rB=A  x2AB rB=A The kinematic diagram and the Oxy coordinate system are shown in figure 7.38b. Express every vector in Cartesian vector form and evaluate the cross product.  t    aB sin 30 i þ aBt cos 30 j þ aBn cos 30 i þ aBn sin 30 j pffiffiffi

pffiffiffi

¼ ðaO kÞ  ðrjÞ  x2O ðrjÞ þ ðaAB kÞ  2 3ri  x2AB 2 3ri 

   aBt sin 30 þ aBn cos 30 i þ aBt cos 30 þ aBn sin 30 j pffiffiffi pffiffiffi ¼ aO ri  x2O rj  2 3aAB rj þ 2 3x2AB ri

ð1Þ

Equating the i component on both sides yields i:

pffiffiffi aBt sin 30 þ aBn cos 30 ¼ aO r þ 2 3x2AB r

Substituting aBn ¼ 2x2O r and xAB ¼ xO =2 and solving, we get pffiffiffi aBt ¼ 3x2O r  2aO r RETHINK: Equating the j component on both sides of equation (1), we should get aAB . Readers are recommended to finish this by themselves.

PROBLEMS 7.1 Crank O1 A is rotating with an angular velocity ω and angular acceleration α at the instant shown. If O1 A ¼ O2 B ¼ CM ¼ R, find the velocity and acceleration of point M.

Prob. 7.1

Planar Kinematics of Rigid Bodies

275

7.2 For the mechanism shown, O1 A ¼ O2 B ¼ AM ¼ 0:2 m and O1 O2 ¼ AB. If the angular position of wheel O1 A is defined by u ¼ ð15pt Þrad, where t is in seconds, determine the velocity and acceleration of point M on link AB when t ¼ 0:5 s.

Prob. 7.2

7.3 A wheel starts to rotate with a constant angular acceleration from rest and its speed of revolution reaches 120 rev/min in 10 min. After rotating at this speed for a certain period of time, the wheel starts to rotate with a constant angular deceleration and stops after 6 min. Totally the wheel has turned 3600 revolutions. Determine the total time of rotation. 7.4 Point A on the rim of the rotating disk has a speed of 50 cm/s and another point B has a speed of 10 cm/s. The distance between point A and the center of rotation O is 20 cm larger than that between point B and O. Find the angular velocity and radius of the disk.

Prob. 7.4

7.5 Gear O1 is in mesh with gear O2 as shown. The radii of the gears are r1 ¼ 10 cm and r2 ¼ 15 cm. If gear O2 starts to rotate with a constant angular acceleration a2 ¼ 2 rad/s2 from rest, find the time required for gear O1 to attain a speed of revolution n1 = 432 rev/min.

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Prob. 7.5

7.6 A simple gearbox as shown consists of four gears. Gears I and II, and gears III and IV are in mesh, respectively. Gears II and III are rigidly connected to the same shaft and turns with it. The numbers of teeth of the gears are Z1 ¼ 8; Z2 ¼ 60 and Z4 ¼ 64, respectively. If the transmission ratio between gear I and gear IV is i14 = 60, determine the number of teeth of gear III.

Prob. 7.6

7.7 In the mechanism as shown, gears I and II are in mesh. Gears II and III are rigidly connected to the same shaft and turns with it. If the speed of revolution of gear I is n1 (rev/min), determine the velocity of the trolley.

Planar Kinematics of Rigid Bodies

277

Prob. 7.7

7.8 The wheel is originally rotating at ω0 = 8 rad/s. It is then subjected to a constant angular acceleration α = 6 rad/s2. Find the velocities and accelerations of points A and B just after the wheel undergoes 2 revolutions.

Prob. 7.8

7.9 Starting from rest when yA = 0, wheel I is given an angular acceleration a ¼ ð2:5 þ 3hÞrad/s2 , where θ is in radians. Wheels II and III are rigidly connected to the same shaft and turns with it. The radii of the three wheels are r1 ¼ 5 cm; r2 ¼ 10 cm and r3 ¼ 6 cm, respectively. Determine the speed of block A when yA = 3 m.

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Prob. 7.9

7.10 Starting from rest, pulley O of radius r = 0.2 m begins to rotate with an angular acceleration a ¼ ð3t þ 0:1t 2 Þrad/s2 , where t is in seconds. Determine the magnitudes of velocity and acceleration of any particle on the rim of the pulley when t = 1 s. Also find the velocity and acceleration of cargo A at this instant.

Prob. 7.10

7.11 Gear A of radius rA = 20 mm is in mesh with gear B of radius rB = 50 mm. Starting from rest, gear A rotates with an angular acceleration aA ¼ ð10  0:1xA Þrad/s2 , where ωA is in rad/s. Find the time needed for gear B to attain an angular velocity xB ¼ 10 rad/s.

Planar Kinematics of Rigid Bodies

279

Prob. 7.11

7.12 Crank O1 A rotates with a constant angular velocity x ¼ 2 rad=s. Sleeve C, which is hinged to rod CD, fits loosely on link AB. If O1 A ¼ O2 B ¼ 10 cm and O1 O2 ¼ AB, find the velocity and acceleration of rod CD when h ¼ 30 .

Prob. 7.12

7.13 Crank OA has a length of 40 cm and is rotating with a constant angular velocity x ¼ 0:5 rad=s. A flat face follower BC maintains contact with the end of crank OA and moves along the vertical track. Find the velocity and acceleration of BC when h ¼ 30 .

Prob. 7.13

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7.14 The right-angled rod OAB rotates with a constant angular velocity x. Sleeve C, which is hinged to rod CD, fits loosely on segment AB. If OA = r, find the velocity and acceleration of rod CD when u ¼ 30 .

Prob. 7.14

7.15 Ring M fits loosely on both the right-angled rod OBC and the fixed rod OA. Rod OBC rotates with a constant angular velocity x ¼ 0:5 rad=s. If OB ¼ 10 cm, find the velocity and acceleration of ring M when h ¼ 60 .

Prob. 7.15

7.16 The right-angled rod BC moves to the left with a constant velocity v0 and pushes crank OA to rotate about pin O. Find the angular velocity and angular acceleration of crank OA at any position θ.

Planar Kinematics of Rigid Bodies

281

Prob. 7.16

7.17 The end A of the bar AB is moving to the left with a constant velocity vA = 0.5 m/s. Find the angular velocity and angular acceleration of the bar at the instant shown.

Prob. 7.17

7.18 Ring M fits loosely on both rocker OA and the fixed circular ring of radius R ¼ 10 cm. Starting from the horizontal position, rocker OA rotates about O with a constant angular velocity x ¼ 0:1p rad/s and pushes ring M to move along the fixed ring. Find the velocity and acceleration of ring M at any instant.

Prob. 7.18

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282

7.19 The cam rotates about O with a constant angular velocity ω. Write the position equation of the rod AB. The radius of the cam is r and OC = e.

Prob. 7.19

7.20 Crank OA with a length r = 10 cm is rotating about O with an angular velocity x ¼ 1 rad/s and angular acceleration a ¼ 1 rad/s2 when h ¼ 30 . Determine the acceleration of the slotted guide BC at this instant. 7.21 The length of crank OA is r = 0.3 m. Slotted guide BC is moving upward with a velocity v = 2 m/s and an acceleration a = 3 m/s2 when h ¼ 60 . Determine the angular acceleration of crank OA at this instant.

Prob. 7.20/21

7.22 A disk of radius r = 0.5 m is cast on the rough surface such that points A and B on the disk have the speeds vA = 3 m/s and vB = 2 m/s, respectively, as shown. Find the velocity of the center point C and the angular velocity of the disk at this instant.

Planar Kinematics of Rigid Bodies

283

Prob. 7.22

7.23 The disk of radius r = 0.3 m is moving on a horizontal ground. At a given instant, it has an angular velocity x ¼ 8 rad/s and angular deceleration a ¼ 5 rad/s2 . Its center C has a velocity vC ¼ 10 m/s and deceleration aC ¼ 1 m/s2 . Find the velocities and accelerations of the highest point A and lowest point B at this instant.

Prob. 7.23

7.24 At a given instant, gear racks A and B have the velocities and accelerations shown. The radius of gear C is r = 0.25 m. Find the angular velocity and angular acceleration of gear C.

Prob. 7.24

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7.25 Crank OA with a length of 20 cm is rotating about O with an angular velocity x ¼ 2 rad/s and angular acceleration a ¼ 1 rad/s2 at the instant shown. For this instant, crank OA is at the vertical position and rod AB with a length of 40 cm is at an angle h ¼ 30 from the horizontal ground. Determine the acceleration of point B and the angular acceleration of rod AB at this instant.

Prob. 7.25

7.26 The two ends of link AB of length l are hinged to two identical wheels of radius R. Wheel O is rolling without slipping on the horizontal ground with a constant angular velocity xO. Wheel A is also rolling without slipping. Determine the angular velocity and angular acceleration of wheel A at the position shown.

Prob. 7.26

7.27 A disk is rolling without slipping on the horizontal ground. Its center C has a constant velocity vC ¼ 10 cm/s. The radius of the disk is R = 20 cm and the length pffiffiffiffiffi of link AB is 20 26 cm. Find the velocity and acceleration of slider B at the position shown.

Planar Kinematics of Rigid Bodies

285

Prob. 7.27

7.28 Crank OA of length r is rotating about O with a constant angular velocity x0. pffiffiffi If AB ¼ 6r and BC ¼ 3 3 r, find the velocity and acceleration of slider C at the given position.

Prob. 7.28

7.29 Crank OA with a length of 30 cm is rotating with a constant speed of revolution nOA ¼ 40 rev/ min. Locate the instantaneous center of rotation of link AB and determine the velocity of member CDB at the position shown, where crank OA is perpendicular to link AB.

Prob. 7.29

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7.30 Crank OA with a length of 40 cm is rotating with a constant speed of revolution n ¼ 180 rev/ min. The length of link AB is l m. Locate the instantaneous center of rotation of link AB and determine the angular velocity of link AB and the velocity of the mid-point M at the given instant.

Prob. 7.30

7.31 Crank OA with a length of 15 cm is rotating with a constant speed of revolution n = 60 rev/min. Disk B with a radius of R = 15 cm is rolling without slipping on the horizontal ground. Locate the instantaneous centers of rotation for link AB and disk B, respectively, at the given instant. Also determine the angular velocity of disk B at this instant.

Prob. 7.31

7.32 In the mechanism, OA ¼ O1 B ¼ AB=2 and crank OA is rotating with an angular velocity x ¼ 3 rad/s. At the instant shown, O, O1 and B are on the same straight line, which is perpendicular to crank OA. Locate the instantaneous center of rotation of rod AB and determine the angular velocities of rods AB and O1 B, respectively, at this instant.

Planar Kinematics of Rigid Bodies

287

Prob. 7.32

7.33 Crank AB rotates with a constant angular velocity x ¼ 1 rad/s. Locate the instantaneous center of rotation of link BC and determine the angular velocity of rod CD at the given instant.

Prob. 7.33

Chapter 8 Kinetics: Equations of Motion Objectives
Relate force and acceleration by equations of motion from Newton’s second law.
Apply equations of motion to solve kinetic problems of a particle using different coordinate systems.
Determine the mass moment of inertia of a rigid body.
Derive planar kinetic equations of motion for a symmetric body undergoing planar motion.
Draw free-body diagram and kinetic diagram for kinetic problems involving a rigid body or a system of connected rigid bodies.
Apply planar kinetic equations of motion and appropriate kinematic relationships to solve kinetic problems for a rigid body or a system of connected rigid bodies. Statics deals with forces acting on particles or bodies that lead to a state of equilibrium. Then kinematics treats only the geometric aspects of the motion, without involving forces. Now we come to kinetics that relates forces acting on particles or bodies to their motions. In this chapter, based on Newton’s second law of motion, the equations of motion on different coordinate systems are first formulated and applied to solve particles’ kinetic problems. Then planar kinetic equations of motion are derived and applied to symmetric bodies undergoing planar motion. The principle of work and energy and the principle of impulse and momentum can both be derived from equations of motion, which will be discussed in the following two chapters, respectively.

8.1

Newton’s Second Law of Motion

As previously stated, engineering mechanics is based on Newton’s three laws of motion. The first and third laws have been used extensively in statics. Although they are still applied in kinetics, Newton’s second law of motion forms the basis of kinetics. It states that when an unbalanced force acts on a particle, the particle has DOI: 10.1051/978-2-7598-2901-9.c008 © Science Press, EDP Sciences, 2022

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290

an acceleration proportional to the magnitude of the force and in the direction of the force, which is expressed mathematically as F ¼ ma ¼ m

d2 r dt 2

ð8:1Þ

where m is the mass of the particle and it is a positive scalar. When F is constant, the bigger m is, the smaller the magnitude of a is. Therefore, m provides a quantitative measure of the resistance of the particle to a change in its velocity, i.e., a quantitative measure of the particle’s inertia. Equation (8.1) is also called the equation of motion, which is one of the most important formulations in mechanics. However, Albert Einstein developed the theory of relativity in 1905 and placed limitations on the use of Newton’s second law. It was proven that time is not an absolute quantity and the equation of motion cannot predict the exact behavior of a particle when the particle’s speed approaches the speed of light. Furthermore, the theory of quantum mechanics developed in the 1920s indicates that Newton’s second law is also invalid when particles are the size of an atom. Fortunately, these situations regarding particle’s speed and size are seldom encountered in engineering problems, and Newton’s laws of motion still remain the basis of today’s engineering mechanics. It should be noted that acceleration a in equation (8.1) should be measured with respect to an inertial reference frame, which is either fixed or translating with a constant velocity. For problems concerned with motions on or near the surface of the earth, the inertial frame is generally chosen to be fixed to the earth, as shown in figure 8.1. However, when studying the motions of space vehicles, the inertial reference frame should be fixed to the stars.

FIG. 8.1 – Describing a particle’s motion by using an inertial reference frame.

8.2

Equation of Motion for a Particle

When several forces act on a particle, as shown in figure 8.2, equation (8.1) can be rewritten as

Kinetics: Equations of Motion

FR ¼

291 X

Fi ¼ ma ¼ m

d2 r dt 2

ð8:2Þ

FIG. 8.2 – Describing a particle’s motion by using a rectangular reference frame. In figure 8.2, an inertial rectangular reference frame Oxyz is used to describe the particle’s motion. Then acceleration a and every force Fi acting on the particle can be expressed in terms of their rectangular components, i.e., i, j, k components. Equation (8.2) can then be rewritten as X    Fxi i þ Fyi j þ Fzi k ¼ m ax i þ ay j þ az k For this vector equation to be satisfied, the respective i, j, k components on the left side must be equal to the corresponding components on the right side. Therefore, the following three scalar equations are obtained. 8 X > d2 x > > ¼ ma ¼ m F xi x > > dt 2 > >

dt 2 > > > > X > d2 z > : Fzi ¼ maz ¼ m 2 dt They are called the equations of motion on rectangular coordinates. When a particle moves along a curved path that is known, it is more convenient to use tangential and normal coordinates to formulate the acceleration components, figure 8.3. Resolving the acceleration and the forces into components in the tangential, normal, and binormal directions, the equation of motion for the particle becomes   X dv v2 tþ n ðFti t þ Fni n þ Fbi bÞ ¼ m ðat t þ an nÞ ¼ m dt q

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This equation is satisfied provided 8P v d2 s > ¼ m < Fti ¼ mat ¼ m d dt2 dt 2 P v ¼ ma ¼ m F n > q : P ni Fbi ¼ 0

ð8:4Þ

They are called the equations of motion on tangential and normal coordinates.

FIG. 8.3 – Describing a particle’s motion by using a t-n-b coordinate system. Procedure for analysis The equations of motion are used to determine the relationship between the forces acting on a particle and its accelerated motion they cause. The procedure for analysis is as follows: (1) Choose an appropriate coordinate system. For a problem involving a particle moving along a known curved path, tangential and normal coordinates is generally considered for the analysis. Otherwise, the rectangular coordinate system is used. P (2) Draw the particle’s free-body diagram to account for all the forces ( Fi ) acting on the particle. (3) Establish the particle’s acceleration a. If the rectangular coordinate system is used, a ¼ ax þ ay þ az and if tangential and normal coordinates are used, a ¼ at þ an . For an unknown component, always assume the sense of this component in the same direction as its positive coordinate axis for mathematical convenience. (4) Apply either equation (8.3) or equation (8.4) to solve for the unknowns. (5) If the velocity or position of the particle is to be found, apply the appropriate kinematic equations once the particle’s acceleration a is determined from the equations of motion.

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293

Example 8.1. A cannonball of mass m is fired from the ground with an initial velocity v0 at angle θ, as shown in figure 8.4a. Neglect the size of the cannonball and the air resistance. Determine the position equation of the cannonball.

FIG. 8.4 – Cannonball’s motion under just its weight. Solution: Since the air resistance is neglected, the weight is the only force acting on the cannonball during the flight. Establishing the fixed x, y coordinates shown in figure 8.4a, we can write the equations of motion X ð þ !Þ Fxi ¼ max ; 0 ¼ max ð þ "Þ

X

Fyi ¼ may ;

mg ¼ may

where g, the acceleration of gravity, is generally considered as a constant and taken as 9:81 m/s2 . Eliminating m, we can rewrite the two equations from kinematics. d2 x dvx ¼0 ¼ dt 2 dt

ð1Þ

d2 y dvy ¼ g ¼ dt 2 dt

ð2Þ

ax ¼

ay ¼

The initial conditions are x0 ¼ 0, y0 ¼ 0, v0x ¼ v0 cos h and v0y ¼ v0 sin h when t = 0. Therefore, separating the variables and integrating for equation (1) yields Z vx dx ¼ v0x ¼ v0 cos h dvx ¼ 0 ) vx ¼ dt v0x Integrating again we obtain Z 0

x

Z dx ¼

t

v0 cos hdt

0

x ¼ v0 t cos h Separating the variables and integrating for equation (2) yields Z vy Z t dy ¼ v0 sin h  gt dvy ¼ gdt ) vy ¼ dt 0 v0y

ð3Þ

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Integrating again we obtain Z y Z t dy ¼ ðv0 sin h  gtÞdt 0

0

1 y ¼ v0 t sin h  gt 2 2

ð4Þ

Equations (3) and (4) are the position equations of the cannonball. Eliminating t yields the trajectory, g y ¼ x tan h  2 x2 2v0 cos2 h It is a parabola. RETHINK: (1) How to determine the maximum height, the furthest distance and the speed at which the cannonball hits the ground? This is left to the readers to think about. (2) If the cannonball is fired from height h with a horizontal initial velocity v 0, as shown in figure 8.4b, the same equations of motion can be written. In other words, we get the same equations (1) and (2) for this case. However, the initial conditions become x0 ¼ 0, y0 ¼ h, v0x ¼ v0 and v0y ¼ 0 when t = 0. Then from integration we get the position equations of the cannonball for this case: 1 x ¼ v0 t; y ¼ h  gt 2 2 They are apparently different from equations (3) and (4). Example 8.2. In the pulley-cord system shown in figure 8.5a, two blocks have masses of mA = 50 kg and mB = 10 kg. The coefficient of kinetic friction between block B and the horizontal ground is 0.3. Neglect the mass of the pulleys and cords. If block A is moving downward with a speed of 3 m/s at a given instant, determine its speed 10 s later.

FIG. 8.5 – Pulley-cord system.

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295

Solution: Coordinates sA and sB shown in figure 8.5a are used to define the positions of blocks A and B. Each coordinate is measured from a fixed datum and in the direction of motion of a block. We can express the length of the cable in terms of these two position coordinates. 2sA þ sB ¼ l Two successive time derivatives of the above position equation give 2aA þ aB ¼ 0

)

aB ¼ 2aA

ð1Þ

For block B, whose free-body diagram is shown in figure 8.5b, write the equations of motion, X ð þ "Þ Fyi ¼ may ; N  mB g ¼ 0 ) N ¼ 98:1 N FB ¼ lk N ¼ 0:3  98:1 ¼ 29:43 N ð þ !Þ

X

Fxi ¼ max ;

T  FB ¼ mB aB

ð2Þ

For block A, whose free-body diagram is shown in figure 8.5c, write the equation of motion, X ð þ "Þ Fyi ¼ may ; 2T  mA g ¼ mA aA ð3Þ Solving equations (1)–(3) simultaneously, we have aA ¼ 4:796 m/s2 Since aA is constant, the velocity of block A in 10 s is vA2 ¼ vA1 þ aA t ¼ 3 þ 4:796  10 ¼ 50:96 m/s Example 8.3. The 10-kg box is moving on a horizontal smooth surface under the action of a horizontal force having a variable magnitude F = 100(1 − t), where t is measured in seconds and F in newtons. The initial velocity is v0 = 20 cm/s, which has the same direction with that of the force at that time, as shown in figure 8.6. Determine the time needed to stop the box. How far has the box traveled?

FIG. 8.6 – Rectilinear motion under a force with variable magnitude.

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Solution: Apparently, the box will move along a straight-line path and can be defined using a single coordinate axis x. Its equation of motion is X ð þ !Þ Fxi ¼ max ; F ¼ 100ð1  tÞ ¼ 10a a¼

dv ¼ 10ð1  t Þ dt

Since v0 = 20 cm/s = 0.2 m/s when t = 0, utilizing definite integral yields Z v Z t dv ¼ 10ð1  t Þdt 0:2

v¼

0

dx ¼ 0:2 þ 10t  5t 2 dt

ð1Þ

Since x0 = 0 when t = 0, we have Z x Z t   dx ¼ 0:2 þ 10t  5t 2 dt 0

0

5 x ¼ 0:2t þ 5t 2  t 3 3

ð2Þ

Substituting v = 0 into equation (1) and solving yield pffiffiffiffiffiffiffiffi 10  104 s t¼ 10 We should use the positive sign in the above expression by its physical meaning and therefore, the time needed to stop the box is pffiffiffiffiffiffiffiffi 10 þ 104 t¼ ¼ 2:02 s 10 Substituting this value into equation (2), we get the distance the box has traveled before it stops. 5 x ¼ 0:2  2:02 þ 5  2:022   2:023 ¼ 7:07 m 3 Example 8.4. A particle of mass m is fired vertically from the earth’s surface with an upward velocity v 0, figure 8.7a. R is the radius of the earth. Neglecting the atmospheric resistance, determine the minimum initial velocity required to escape the earth’s gravitational field.

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FIG. 8.7 – The minimum firing velocity required to escape the earth’s gravitational field. Solution: The universal gravitational attraction is the only force acting on the particle, figure 8.7b. From Newton’s law of gravitational attraction, we have mme F ¼f 2 x where x is the distance from the particle to the earth’s center; f is the universal constant of gravitation; me is the mass of the earth. Since F = mg when x = R, we have mme mg ¼ f 2 ) fme ¼ R2 g R Therefore F¼

m mR2 g fme ¼ 2 x x2

The equation of motion of the particle is as follows: ð þ "Þ

X

Fxi ¼ max ;

a¼

F ¼ 

mR2 g ¼ ma x2

dv R2 g ¼ 2 dt x

ð1Þ

dv ¼ dv dx ¼ v dv Since dt dx dt dx Equation (1) can be rewritten as v

dv R2 g ¼ 2 dx x

)

vdv ¼ 

R2 g dx x2

Now we have finished separating the variables and can perform integration. Since v ¼ v0 and x ¼ R when t ¼ 0, we have

Engineering Mechanics

298 Z

v

v0

Z vdv ¼

x

R

R2 g  2 dx x

)

sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi  ffi 1 1  v ¼ v02 þ 2gR2 x R

If the particle wants to escape the gravitational field of the earth and does not fall back to the earth, it requires that v  0 as x ! 1. Using this requirement along with R = 6371 km and g ¼ 9:81 m/s2 , we have pffiffiffiffiffiffiffiffiffi v0 min ¼ 2gR ¼ 11:2 km=s This is the minimum initial velocity at which a projectile should be shot vertically from the earth’s surface so that it can escape the earth’s gravitational field. It is also called the escape velocity. Example 8.5. A cannonball of mass m is fired with initial velocity v0 at angle u, as shown in figure 8.8. The atmospheric drag resistance is F = mkv, where k is a constant, and v is the speed of the cannonball. Determine the position equation of the cannonball.

FIG. 8.8 – Cannonball’s motion under its weight and the atmospheric drag resistance.

Solution: There are two forces acting on the cannonball, namely, its weight and the atmospheric resistance F. F should always be opposite to v, as shown in figure 8.8. Therefore, the atmospheric resistance force in vector form is F ¼ mkv. Using the initial position of the cannonball as the origin and establishing the fixed x, y coordinates, we can write the equations of motion: X ð þ !Þ Fxi ¼ max ; Fx ¼ mkvx ¼ max ð þ "Þ

X

Fyi ¼ may ;

mg þ Fy ¼ mg  mkvy ¼ may

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Eliminating m yields dvx ¼ kvx dt

ð1Þ

dvy ¼ g  kvy dt

ð2Þ

ax ¼

ay ¼

The initial conditions are x0 ¼ 0, y0 ¼ 0, v0x ¼ v0 cos u and v0y ¼ v0 sin u when t = 0. Therefore, separating the variables and integrating for equations (1) and (2), respectively, yield Z vx Z t dvx ¼ kdt ) vx ¼ ðv0 cos uÞekt ð3Þ v x 0 v0 cos u Z

vy

dvy ¼ g þ kvy v0 sin u

Z

t

)

dt 0

Integrating again we obtain Z x

Z

0

Z

y

0

dy ¼

t

dx ¼

 g g vy ¼ v0 sin u þ ekt  k k

ð4Þ

ðv0 cos uÞekt dt

0

Z t h

v0 sin u þ

0

x¼

g  kt g i e  dt k k

v0 cos u ð1  ekt Þ k

  v0 sin u g g þ 2 ð1  ekt Þ  t y¼ k k k

ð5Þ

ð6Þ

Equations (5) and (6) are the position equations of the cannonball. Eliminating t yields the trajectory,     g g k y ¼ tan u þ x x þ 2 ln 1  kv0 cos u k v0 cos u The trajectory is shown in figure 8.8. u RETHINK: From equation (5), we can see that x ! v0 cos when t ! 1. It means k v0 cos u the trajectory will finally approach an asymptote, x ¼ k . From equations (3) and (4), we can see that vx ! 0 and vy !  gk when t ! 1. It means the cannonball will finally tend to fall down vertically with a constant velocity vlimit ¼  gk. This vlimit is called the terminal velocity of a projectile in damping medium.

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Example 8.6. The small ball having a mass of 1 kg is attached to a cable of length l = 30 cm. The small ball is undergoing a circular motion in the horizontal plane with a constant speed and the cable makes an angle h ¼ 60 with the vertical line, figure 8.9. Determine the speed and the cable force.

FIG. 8.9 – A ball’s motion under its weight and the cable force. Solution: The free-body diagram of the small ball is shown in figure 8.9, where P = 1 × 9.81 = 9.81 N. Since the small ball moves along a circular path with a constant speed, its tangential and normal acceleration components are an ¼

v2 v2 ¼ q l sin h

Fni ¼ man ;

F sin h ¼

at ¼ 0; Write the equations of motion: ð þ -Þ

X

ð þ "Þ

X

Fbi ¼ 0;

P v2 g l sin h

F cos h  P ¼ 0

From equation (1), we get F¼

P 9:81 ¼ ¼ 19:6 N cos h cos 60

Substituting this result into equation (2) yields sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Fgl sin2 h ¼ 2:1 m/s v¼ P

ð1Þ ð2Þ

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Example 8.7. Conveyor is used to deliver packages to a smooth circular ramp with a horizontal velocity v0, figure 8.10a. The radius of the ramp is r. Neglecting the size and shape of the package, specify the angle at which the package begins to leave the ramp.

FIG. 8.10 – Packages moving down a smooth circular ramp. Solution: The free-body diagram of the package at an arbitrary angle θ is shown in figure 8.10b. Write the equations of motion in the tangential and normal directions, X ð þ &Þ Fti ¼ mat ; mg sin h ¼ mat ð1Þ

ð þ .Þ

X

Fni ¼ man ;

FN þ mg cos h ¼ m

v2 r

ð2Þ

From kinematics, we have at ds ¼ vdv. Noting that ds ¼ rdh and at ¼ g sin h from equation (1), we have at ¼

vdv vdv ¼ ¼ g sin h ds rdh

Separating the variables gives vdv ¼ gr sin hdh Since v ¼ v0 at h ¼ 0, we have Z v Z vdv ¼ v0

h

gr sin hdh

0

v 2 ¼ v02 þ grð2  2 cos hÞ When the package leaves the surface of the ramp, FN ¼ 0. Substituting FN ¼ 0 and the above expression into equation (2) and solving yields h ¼ arc cos

2gr þ v02 3gr

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Example 8.8. A cable is used to suspend the 100-kN cargo and moves with the trolley along the fixed bridge girder with a velocity v = 1 m/s, as shown in figure 8.11a. Neglect the mass of the cable and the size of the cargo. The length of the cable is l = 5 m. If the trolley suddenly jams (stops), determine the maximum cable tension during the motion and the tension in the cable when the cargo reaches the highest position.

FIG. 8.11 – The suspended cargo’s motion when the trolley suddenly jams.

Solution: When the trolley suddenly jams, the cargo begins to move along a circular path due to inertia. The free-body diagram of the cargo at a general position φ is shown in figure 8.11b. Write the equations of motion on normal and tangential coordinates: ð þ -Þ

X

Fni ¼ man ;

ð þ %Þ

T  P cos u ¼ X

Fti ¼ mat ;

P P v2 an ¼ g g l P sin u ¼

)

T ¼ P cos u þ

P P dv at ¼ g g dt

Pv 2 ð1Þ gl ð2Þ

dv ¼ dv du ¼ dv x ¼ dv v dt du dt du du l Equation (2) can be rewritten as v dv ¼  sin udu gl Since

Since v = 1 m/s and φ = 0 when t = 0, integration of the above expression yields Z v Z u v dv ¼  sin udu 1 gl 0 v 2 ¼ 2gl ðcos u  1Þ þ 1

ð3Þ

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Substituting this into equation (1) yields

  P 1 P T ¼ P cos u þ ½2gl ðcos u  1Þ þ 1 ¼ 3P cos u  2  gl gl

From the above expression, we can see that T has the maximum value when cos u ¼ 1 or u = 0. This maximum cable tension is T ¼Pþ

P 100 ¼ 100 þ ¼ 102 kN gl 9:81  5

When the cargo reaches the highest position, v = 0. Substituting this into equation (3) yields cos u ¼ 1 

1 ¼ 0:9898 2gl

Therefore, the tension in the cable at this instant is   1 T ¼ 3  100  0:9898  2   100 ¼ 98:0 kN 9:81  5

8.3

Equation of Motion for a System of Particles

Consider a system of n particles within an enclosed region in space, as shown in figure 8.12. Assume the arbitrary i-th particle is subjected to internal force f i and external force Fi . Internal force f i is actually the resultant of all the forces exerted by the other particles in this system on the i-th particle. External force Fi represents the resultant of all the forces exerted by adjacent bodies or particles not included in the system on the i-th particle. For the i-th particle, we can write the equation of motion, f i þ Fi ¼ mi ai Consider each of the other particles of the system and we can write similar equations. Then for the whole system, adding all the n equations vectorially yields X X X Fi ¼ m i ai fi þ P

f i , the sum of all the internal forces, would always be equal to zero because internal forces always occur in equal P but opposite collinear pairs. Therefore, only the sum of all the external forces, Fi , remain and the equation of motion for the system of particles becomes X X Fi ¼ m i ai ð8:5Þ

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From the definition of the center of mass, we have P X mi ri or mrC ¼ mi ri ð8:6Þ rC ¼ m P where m ¼ mi is the total mass of the system and position vector rC locates the center of mass of the system, as shown in figure 8.12. For a system with constant mass, differentiating equation (8.6) twice with respect to time, we get P X m i ai or maC ¼ aC ¼ m i ai ð8:7Þ m Conjoining equations (8.5) and (8.7) yields X Fi ¼ maC

ð8:8Þ

It states that the sum of all the external forces acting on the system of particles is equal to the total mass of the system times the acceleration of its mass center C. Since no restriction is imposed in the way the particles are connected, equation (8.8) can be used to analyze the motion of any kind of system of particles, such as a solid, liquid, or gas system. In this book, we mainly apply it to a rigid body, or a system of rigid bodies. The forces acting on a rigid body generally form a nonconcurrent force system, figure 8.13. This nonconcurrent force system would cause the body to both translate and rotate. The translational aspect of the motion is just governed by equation (8.8) and the rotational aspect of the motion will be discussed in the next section. Before the study of the next section, an important concept, the mass moment of inertia of a body, has to be discussed first.

FIG. 8.12 – A system of particles.

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FIG. 8.13 – Nonconcurrent force system on a body.

8.4

Mass Moment of Inertia

For a rigid body shown in figure 8.14, the mass moment of inertia about the z axis is defined as Z Iz ¼ r 2 dm ð8:9Þ

FIG. 8.14 – Mass moment of inertia.

306

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where r, the perpendicular distance from the arbitrary element dm to the z axis, is referred to as the moment arm. The perpendicular distance from the element dm to different axis is generally different. For example, for the z 0 axis in figure 8.14, the perpendicular distance or the moment arm is r 0 . Substituting r 0 into the integration, R 02 Iz 0 ¼ r dm will be obtained. It means the value of I of the same body for different axis will be different. Just like the mass, the mass moment of inertia is also a positive quantity. The unit of the mass moment of inertia is commonly kg m2. It should be pointed out that choosing appropriate differential element dm requires both skill and experience. Some types of differential elements will be discussed in the following examples. Example 8.9. Determine the mass moment of inertia Iz of the homogeneous rod shown in figure 8.15a. The rod has a mass m and a length l.

FIG. 8.15 – Mass moment of inertia of a homogeneous rod.

Solution: Choose the differential rod element dx as shown in figure 8.15b and then m dm ¼ ql dx ¼ dx l where ql is the mass per unit length of the homogeneous rod. From single integration, we can obtain the mass moment of inertia. Z Z l m ml 2 2 x 2 dx ¼ Iz ¼ x dm ¼ l 3 0 Example 8.10. Determine the mass moment of inertia of the homogeneous disk about the z axis, which is perpendicular to the disk and passes through its center C, figure 8.16a. The total mass of the disk is m and its radius is R.

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FIG. 8.16 – Mass moment of inertia of a homogeneous disk.

Solution: Choose the differential circular arc element having a thickness dr as shown in figure 8.16a. The mass of the differential element is m dm ¼ qA dA ¼ 2prdr pR2 where qA is the mass per unit area of the homogeneous disk. From integration, we can obtain the mass moment of inertia. Z Z R m mR2 2 r 2 2 2prdr ¼ Iz ¼ r dm ¼ pR 2 0 RETHINK: If the mass moment of inertia of the disk about the x or y axis is to be determined, the differential element dm as shown in figure 8.16b can be used. From the definition of the mass moment of inertia, we have Z Z Ix ¼ y 2 dm; Iy ¼ x 2 dm R R R R Then Iz ¼ r 2 dm ¼ ðx 2 þ y 2 Þdm ¼ x 2 dm þ y 2 dm ¼ Iy þ Ix Due to the homogeneous disk’s polar symmetry about its center, we have Ix ¼ Iy . Therefore, Ix ¼ Iy ¼

Iz mR2 ¼ 2 4

Actually, for any thin plate with z axis perpendicular to the plate, the relationship Iz ¼ Iy þ Ix is satisfied. Example 8.11. A semi-ellipsoid is formed by revolving about the x axis, as shown in figure 8.17a. The material has a constant density q ¼ 10 kg=m3. Determine the total mass of this revolution body and its mass moment of inertia about the x axis.

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FIG. 8.17 – Mass moment of inertia of a semi-ellipsoid.

Solution: Choose a differential disk element having a radius y and a thickness dx, figure 8.17b. The differential element has a mass   9 dm ¼ qdV ¼ qpy 2 dx ¼ qp 9  x 2 dx 25 From example 8.10, the mass moment of inertia of this differential disk element about the x axis is 1 dIx ¼ y 2 dm 2 Integrating over the entire semi-ellipsoid yields the mass moment of inertia of the body. 2 Z  Z 1 2 qp 5 9 Ix ¼ y dm ¼ 9  x 2 dx ¼ 108qp ¼ 3393 kg m2 2 2 0 25 The mass of the body is Z Z m ¼ dm ¼ qp

5 0



 9 2 9  x dx ¼ 30qp ¼ 942 kg 25

Example 8.12. The solid cylinder of radius R = 0.2 m and height h = 0.5 m is made from a material that has a density varying from its center as q ¼ ð2 þ r 2 Þkg/m3 , figure 8.18a. Determine the mass of the cylinder and its mass moment of inertia about the z axis.

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FIG. 8.18 – Mass moment of inertia of a cylinder with variable density. Solution: Choose a differential shell element that has height h, radius r and thickness dr, figure 8.18b. The volume of the differential shell element is dV ¼ ð2pr Þhdr ¼ 2prhdr Since the density varies from the center as q ¼ ð2 þ r 2 Þkg/m3 , the mass of the differential shell element is   dm ¼ qdV ¼ 2 2 þ r 2 prhdr Substituting h = 0.5 m and integrating over the entire body yield Z Z 0:2   m ¼ dm ¼ 2 þ r 2 prdr ¼ 0:127 kg 0

Due to its thinness, the mass moment of inertia of the differential shell element about the z axis is     dIz ¼ r 2 dm ¼ 2r 2 2 þ r 2 prhdr ¼ 2pr 3 h 2 þ r 2 dr Substituting h = 0.5 m and integrating over the entire solid cylinder yield Z Z 0:2   Iz ¼ dIz ¼ pr 3 2 þ r 2 dr ¼ 2:55  103 kg m2 0

8.4.1

Parallel-Axis Theorem and Radius of Gyration

Consider a body shown in figure 8.19, where axis zC passes through the body’s mass center C and a parallel z axis lies at distance d away. Select a differential element of

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310

mass dm, which is located at ðxC ; yC ; zC Þ in the CxC yC zC coordinate system and ðx; y; z Þ in the Oxyz coordinate system. From figure 8.19, it can be seen that x ¼ xC and y ¼ yC þ d. From the definition of the mass moment of inertia of a body, we have Z Z Z h i  2  Iz ¼ r 2 dm ¼ x þ y 2 dm ¼ x 2 þ ðyC þ d Þ2 dm Z Z Z Z  2   2  ¼ xC þ yC2 þ 2yC d þ d 2 dm ¼ xC þ yC2 dm þ 2d yC dm þ d 2 dm The first integral is just IzC , the mass moment of inertia of the body about the zC R axis. The second integral equals zero, since yC dm ¼ my C and y C ¼ 0. The third R integral dm ¼ m is the total mass of the body. Therefore, the mass moment of inertia of the body about the z axis can be written as Iz ¼ IzC þ md 2

ð8:10Þ

Note that d is the perpendicular distance between the two parallel axes and one of the axes should be the axis passing through the mass center C. Equation (8.10) is referred to as the parallel-axis theorem. It means if the mass moment of inertia of a body about an axis passing through its mass center C is known, the mass moment of inertia about any other parallel axis can be determined by the parallel-axis theorem. Since md 2  0, it can be seen that among a group of parallel axes, the mass moment of inertia of a body about the axis passing through its mass center is the smallest. The mass moment of inertia of a homogeneous rod about the z axis shown in figure 8.20 has been computed in example 8.9 and is Iz ¼ ml 2 =3. Then from the parallel-axis theorem, the mass moment of inertia about the zC axis is  2 l ml 2 IzC ¼ Iz  m ¼ 2 12

FIG. 8.19 – Parallel-axis theorem.

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FIG. 8.20 – Mass moment of inertia of a rod about two parallel axes.

In engineering, the mass moment of inertia of a body about a specified axis is often reported using the radius of gyration, k. The relationship between k and the body’s mass moment of inertia is rffiffiffiffiffi I k¼ ð8:11Þ or I ¼ mk 2 m Apparently, k is a geometrical property which has unit of length. Just as I, the value of k is different for different axis about which it is reported.

8.4.2

Mass Moment of Inertia of Composite Bodies

The mass moment of inertia of some common shapes, such as disks, rods, and spheres, are tabulated in table 8.1. For a composite body consisting of several simple shapes, we can obtain the mass moment of inertia of the composite body about an axis by first computing the mass moments of inertia of its component parts about the axis and then adding them together algebraically. For a “hole” shown in figure 8.21, a negative quantity should be used and therefore its mass moment of inertia is subtracted from that of a solid plate. The parallel-axis theorem should be applied when the mass center of each component part does not lie on the desired axis. The equation for the calculation is as follows: Iz ¼

X

IzCi þ mi di2



ð8:12Þ

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TAB. 8.1 – Mass moment of inertia and radius of gyration of some simple homogeneous bodies. Shape Rod

Diagram

Mass moment of inertia Iz ¼

1 ml 2 12

1 ma 2 12 1 Iy ¼ mb2 12 1 Iz ¼ mða2 þ b2 Þ 12 Ix ¼

Thin plate

Ring

1 Ix ¼ Iy ¼ mR2 2 Iz ¼ mR2

Disk

1 Ix ¼ Iy ¼ mR2 4 1 Iz ¼ mR2 2

Ix ¼ Iy ¼ m Cylinder

 2  R l2 þ 4 12

1 Iz ¼ mR2 2

Sphere

2 Ix ¼ Iy ¼ Iz ¼ mR2 5

Ix ¼ Iy ¼

3 mð4R2 þ h 2 Þ 80

Cone Iz ¼

3 mR2 10

Ix ¼ Iy ¼ Hemisphere

83 mR2 320

2 Iz ¼ mR2 5

Radius of gyration 1 pffiffiffiffiffi l 12 1 pffiffiffiffiffi a 12 1 pffiffiffiffiffi b 12 1 pffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffi a 2 þ b2 12 1 pffiffiffi R 2 R 1 R 2 1 pffiffiffi R 2 rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 3R2 þ l 2 12 1 pffiffiffi R 2 rffiffiffi 2 R 5

rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 3ð4R2 þ h 2 Þ 80 rffiffiffiffiffi 3 R 10 rffiffiffiffiffiffiffiffi 83 R 320 rffiffiffi 2 R 5

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FIG. 8.21 – Mass moment of inertia of a composite body. Example 8.13. Determine the mass moment of inertia of the assembly shown in figure 8.22 about an axis perpendicular to the assembly and passing through O. The material has a density of ρ = 7.9 × 103 kg/m3.

FIG. 8.22 – Mass moment of inertia of the assembly. Solution 1: The assembly can be treated as a big disk with a radius of 2.5 m minus a small disk with a radius of 1.2 m. The mass of the big disk is m1 ¼ p  2:52  0:3  7:9  103 ¼ 46:53  103 kg The mass moment of inertia of the big disk about the axis passing through O can be obtained from the parallel-axis theorem as follows: 1 IO1 ¼ IC 1 þ m1 d12 ¼ m1  2:52 þ m1  2:52 ¼ 436:3  103 kg m2 2

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314 The mass of the small disk is

m2 ¼ p  1:22  0:3  7:9  103 ¼ 10:72  103 kg The mass moment of inertia of the small disk about the axis passing through O is also obtained from the parallel-axis theorem. 1 IO2 ¼ IC 2 þ m2 d22 ¼ m2  1:22 þ m2  2:52 ¼ 74:72  103 kg m2 2 Since the small disk is a “hole”, its mass moment of inertia should be subtracted and the mass moment of inertia of the assembly is IO ¼ IO1  IO2 ¼ 436:3  103  74:72  103 ¼ 362 Mg m2 Solution 2: By inspection, mass center C of this composite body is easily obtained. So here the mass moment of inertia of the assembly about the axis passing through C can be first calculated and then the parallel-axis theorem is used. The mass moment of inertia of the assembly about the axis passing through the mass center C is   1 1 p  2:52  0:3  7:9  103  2:52  p  1:22  0:3  7:9  103  1:22 2 2 ¼ 145:4  103  7:72  103 ¼ 137:7  103 kg m2

IC ¼

The total mass of the assembly is   m ¼ p 2:52  1:22  0:3  7:9  103 ¼ 35:81  103 kg Then from the parallel-axis theorem, we have IO ¼ IC þ md 2 ¼ 137:7  103 þ 35:81  103  2:52 ¼ 362 Mg m2

8.5

Planar Kinetic Equations of Motion

From chapter 7, we know that for a rigid body undergoing planar motion, all the particles of the body move along paths that are equidistant from a fixed plane. In this section, the analysis will be limited to planar kinetics of rigid bodies which, along with their loadings, are symmetrical with respect to a reference plane and this reference plane is parallel to the aforementioned fixed plane. Then all the forces and couple moments acting on the body can be projected onto this symmetrical reference plane and the mass center C of the body should move within this plane, as shown in figure 8.23a.

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FIG. 8.23 – Derivation of the planar kinetic equations of motion. From §7.4, we know that a rigid body subjected to general plane motion undergoes a combination of translation and rotation. The translational part of motion actually has been considered in §8.3, and equation (8.8) as follows is used to describe it. X Fi ¼ maC This is called the translational equation of motion for a rigid body. It means the sum of all the external forces acting on the body is equal to the body’s mass times the acceleration of its mass center C. Setting an inertial coordinate system Oxy in the symmetrical reference plane, the translational equation of motion may be resolved into two independent scalar equations, namely, (X Fxi ¼ maCx X Fyi ¼ maCy The rotational part of motion will be discussed in detail here. P is an arbitrary point on the symmetrical reference plane of the body and a translating reference frame Px 0 y 0 is set, as shown in figure 8.23a. The fixed reference frame Oxy is also shown in figure 8.23a, but hereafter it is omitted for brevity. The i-th particle of the

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316

body, having mass mi, is subjected to internal force f i and external force Fi , figure 8.23b. The internal force f i represents the resultant of all the forces exerted on the i-th particle by other particles of this body. External force Fi is the resultant of all the forces exerted on the i-th particle by particles not included in this body or other adjacent bodies. Assume the particle’s acceleration is ai and we can write the equation of motion for the i-th particle, f i þ Fi ¼ mi ai Then from figure 8.23b, we have r0i  Fi þ r0i  f i ¼ r0i  mi ai

ð8:13Þ

If the body has an angular velocity x ¼ xk and angular acceleration a ¼ ak, as shown in figure 8.23b, we have ai ¼ aP þ a  r0i  x2 r0i from equation (7.33). Substituting this into equation (8.13) yields   MP ð f i Þk þ MP ðFi Þk ¼ mi r0i  aP þ a  r0i  x2 r0i ¼ mi ½r0i  aP þ r0i  ða  r0i Þ  x2 ðr0i  r0i Þ The term x2 ðr0i  r0i Þ is zero. Carrying out the cross product for the other two terms yields

MP ð f i Þk þ MP ðFi Þk ¼ mi ðxi0 i þ yi0 jÞ  ðaPx i þ aPy jÞ þ ðxi0 i þ yi0 jÞ  ½ak  ðxi0 i þ yi0 jÞ    ¼ mi yi0 aPx þ xi0 aPy þ a xi02 þ yi02 k   ¼ mi yi0 aPx þ mi xi0 aPy þ mi ri02 a k This can be rewritten as follows: MP ð f i Þ þ MP ðFi Þ ¼ mi yi0 aPx þ mi xi0 aPy þ mi ri02 a Letting mi ! dm and integrating on the entire body, we get Z  Z  Z  X X MP ð f i Þ þ MP ðFi Þ ¼  y 0 dm aPx þ x 0 dm aPy þ r 02 dm a P The term MP ðf i Þ must be zero, since for the entire body the internal forces always occur in equal and opposite collinear pairs and thus the moment of each pair of these forces about P cancels. For the first and second integrals on the right side, R R we have y 0 dm ¼ my 0 and x 0 dm ¼ mx 0 , which locate the body’s mass center R C with respect to P, as shown in figure 8.23c. The last integral, r 02 dm ¼ IP , represents the body’s mass moment of inertia about an axis perpendicular to the symmetrical reference plane and passing through point P. Then the above equation becomes X MP ðFi Þ ¼ y 0 maPx þ x 0 maPy þ IP a ð8:14Þ

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Using equation (7.33), we also have aC ¼ aP þ a  r0  x2 r0 Expressing the vectors in Cartesian vector form and carrying out the cross product yield aCx i þ aCy j ¼ aPx i þ aPy j þ ak  ðx 0 i þ y 0 jÞ  x2 ðx 0 i þ y 0 jÞ     ¼ aPx  y 0 a  x 0 x2 i þ aPy þ x 0 a  y 0 x2 j Equating the i and j components yields aCx ¼ aPx  y 0 a  x 0 x2 ;

aCy ¼ aPy þ x 0 a  y 0 x2

Meanwhile, from the parallel-axis theorem, we have 2

2

2

IP ¼ IC þ mr 0 ¼ IC þ mðx 0 þ y 0 Þ Substituting these results into equation (8.14) and simplifying gives X MP ðFi Þ ¼ y 0 maCx þ x 0 maCy þ IC a

ð8:15Þ

P

Here MP ðFi Þ is the sum of the moments of all the external forces acting on the body about point P. A free-body diagram accounting for all the external forces and couple moments, as shown in figure 8.23d, can help for the calculation. The first two terms on the right side are referred to as kinetic moments of the components of maC about P. The third term is the kinetic moment of IC a. maC and IC a are also called inertial terms. A kinetic diagram, as shown in figure 8.23e, can be drawn to visualize the inertial terms. Then to calculate the kinetic moments, you can treat maCx and maCy as sliding vectors, just like “forces”, and treat IC a as a free vector, just like a “couple moment”. This will make the calculation convenient. However, you should keep in mind that maCx , maCy and IC a are not a real force or a real couple moment. They are just the effects of external forces and couple moments acting on the body. Also note that aC is the absolute acceleration of the mass center C of the body with respect to the fixed reference frame Oxy. The sum of the P kinetic moments of the inertial terms maCx , maCy and IC a is also represented by ðMk ÞP, and therefore equation (8.15) can be rewritten in a more general form as follows: X X ðMk ÞP ð8:16Þ MP ðFi Þ ¼ If point P coincides with mass center C of the body, equation (8.15) is simplified to the following form: X MC ðFi Þ ¼ IC a ð8:17Þ Equations (8.15) or (8.16) and (8.17) are all called the rotational equation of motion. Equation (8.17) states that the sum of the moments of all the external

318

Engineering Mechanics

forces about the body’s mass center C is equal to the mass moment of inertia of the body about C P times the body’s angular acceleration. From equation (8.17), we can see that when MC ðFi Þ is constant, the bigger the mass moment of inertia IC is, the smaller the angular acceleration α is. Therefore, the mass moment of inertia is a measure of the resistance of a body to angular acceleration, just like that mass is a measure of a particle’s resistance to acceleration [2, 6, 7]. From the above analysis, two scalar translational equations of motion and a rotational equation of motion are respectively used to describe the translational part and rotational part of the general plane motion of a symmetrical rigid body. In other words, three independent scalar equations can be written for the rigid-body planar motion. 8P < PFxi ¼ maCx Fyi ¼ maCy ð8:18Þ P P :P MP ¼ ðMk ÞP MC ¼ IC a or When mass center C of the body moves along a known curved path, it is much more convenient to describe the acceleration using t and n components rather than x and y components. For example, mass center C of the wheel shown in figure 8.24 moves along a circular path and the inertial terms are then maCt , maCn and IC a. The three scalar equations can be written as 8P < P Fti ¼ maCt ð8:19Þ F ¼ maCn P P : P ni MP ¼ ðMk ÞP MC ¼ IC a or If point A shown in figure 8.24 is chosen to sum moments, we get X X MA ¼ ðMk ÞA ¼ rmaCt þ IC a

FIG. 8.24 – Kinetic diagram of a wheel moving along a known curved path.

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319

In the following subsections, we will discuss the application of equation (8.18) or (8.19) for three types of rigid-body planar motion.

8.5.1

Equations of Motion for Translation

In the case of a symmetrical body undergoing planar translation, the angular acceleration of the body is always zero and its inertial terms reduce to only one vector maC attached at C (figure 8.25). Hence, the equations of motion in this case become 8P < P Fxi ¼ maCx ð8:20Þ F ¼ maCy : P yi MC ¼ 0 However, other points rather than the mass center C can also be chosen to sum moments. For example, if point A or point O on the y axis shown in figure 8.25 is chosen, the following rotational equation of motion applies: X X X X MA ¼ ðMk ÞA ¼ bmaC or MO ¼ ðMk ÞO ¼ bmaC Here we consider positive moments as counterclockwise since counterclockwise moment vectors are directed along the positive z axis (out of the page). Therefore, negative sign appears for the kinetic moment about point O in the above rotational equation of motion.

FIG. 8.25 – Kinetic diagram of a body undergoing rectilinear translation. When the rigid body is subjected to curvilinear translation and the curved path is known, as shown in figure 8.26, the tangential and normal coordinates will be used and the three scalar equations become

Engineering Mechanics

320 8X Fti ¼ maCt > >

X > : MC ¼ 0

ð8:21Þ

FIG. 8.26 – Kinetic diagram of a body undergoing curvilinear translation. Procedure for analysis The procedure of solving kinetic problems of a symmetrical body undergoing planar translation is as follows: (1) Choose an appropriate coordinate system. If the body is subjected to curvilinear translation and the curved path is known, tangential and normal coordinates are generally considered for the analysis. Otherwise, the rectangular coordinate system is used. (2) Draw the free-body diagram to show all the external forces and couple moments acting on the body. (3) Draw the kinetic diagram to visualize the inertial terms maCx and maCy , or matC and manC , or just maC . (4) Apply equation (8.20) or (8.21). To simplify analysis, sometimes the more P the P general rotational equation of motion MP ¼ ðMk ÞP will be used, where point P should be chosen as the intersection of the lines of action Pof as many unknown forces as possible. When calculating the kinetic moment ðMk ÞP , the moments of the inertial terms maCx , maCy or matC , manC , or just maC about point P should be considered. (5) If the velocity or position of the body is to be found, appropriate kinematic equations should be applied. Example 8.14. The 50-kg crate shown in figure 8.27a rests on a flatbed trailer and the coefficient of static friction is 0.25. Determine the largest acceleration the trailer can have without causing the crate to slip or tip.

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FIG. 8.27 – A crate resting on an accelerated trailer. Solution: The free-body diagram and kinetic diagram of the crate are shown in figure 8.27b and c. Assuming that the crate will slip first. When it is on the verge of sliding, Fs ¼ Fsmax ¼ ls N ¼ 0:25N The rotational equation of motion is X ð þ xÞ MC ¼ 0; Fs  1:5  Nb ¼ 0

)

b ¼ 0:375 m

Here b = 0.375 m > 0.3 m, it means the assumption of slipping first is wrong and the crate will tip first. When the crate is on the verge of tipping, we have b = 0.3 m, i.e., the normal force N is acting at the right corner point A. The equations of motion and solutions are X ð þ "Þ Fyi ¼ maCy ; N  mg ¼ 0 ) N ¼ 50  9:81 ¼ 490:5 N ð þ xÞ

X

ð þ !Þ

MC ¼ 0; X

Fs  1:5  N  0:3 ¼ 0

Fxi ¼ maCx ;

Fs ¼ ma

)

)

Fs ¼ 98:1 N

a ¼ 1:96 m/s2

Therefore, the largest acceleration the trailer can have without causing the crate to slip or tip is 1:96 m/s2 . RETHINK: When the crate is on the verge of tipping, applying the rotational equation of motion about the right corner point A can eliminate the unknown N and Fs and get a direct solution of the acceleration. X X ðMk ÞA ; mg  0:3 ¼ ma  1:5 ) a ¼ 1:96 m/s2 ð þ yÞ MA ¼

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Example 8.15. A 2-t truck carries a 3-t cargo. The mass center of truck Ct and the mass center of cargo Cc are shown in figure 8.28a. Neglect the mass of the wheels. The coefficient of kinetic friction on the pavement is 0.3. The truck is traveling at v0 = 60 km/h when the brakes are applied and cause all the wheels to lock. Determine the distance it would skid before stopping and the normal forces on the wheels at A and B.

FIG. 8.28 – A truck with cargo.

Solution: In this problem, we have the mass of the truck and the mass of the cargo. This can be considered as a system of rigid bodies. It is possible to determine the location of the mass center of the system first. However, considering the truck and cargo separately, as shown in the free-body and kinetic diagrams, figure 8.28b and c, would be more convenient. Both parts translate with the same acceleration a. The equations of translational motion and friction are as follows: X ð þ "Þ Fyi ¼ maCy ; NA þ NB  mt g  mc g ¼ 0 ) NA þ NB ¼ ðmt þ mc Þg ð1Þ FA þ FB ¼ lk ðNA þ NB Þ ¼ lk g ðmt þ mc Þ ðþ

Þ

X

Fxi ¼ maCx ;

ðFA þ FB Þ ¼ ðmt þ mc Þa

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323

Solving simultaneously get the acceleration of the truck: a¼

FA þ FB ¼ lk g ¼ 0:3  9:81 ¼ 2:943 m/s2 mt þ mc

The acceleration is constant. From kinematics, we have v 2  v02 ¼ 2a ðx  x0 Þ. 3 Here x0 ¼ 0 and v0 ¼ 60 km/h ¼ 6010 3600 m/s ¼ 16:67 m/s. Thus, the distance the truck skids before stopping is x¼

v 2  v02 0  16:672 ¼ ¼ 47:2 m 2a 2  ð2:943Þ

The rotational equation of motion is written by summing moments about point A. ð þ xÞ

X

MA ¼

X

ðM k ÞA ;

NB  3:6  mt g  1:2  mc g  2:4 ¼ mt a  1:3 þ mc a  2 Substituting the masses and a ¼ 2:943 m/s2 and solving yield NB ¼ 19.1 kN Substituting this value into equation (1) gives NA ¼ 29.9 kN Example 8.16. The 10-kg thin plate is held in the position by two identical weightless links O1A and O2B and a cord at D, figure 8.29a. Determine the acceleration of the plate and the force in each link immediately after the cord was cut.

FIG. 8.29 – A plate which undergoes a curvilinear translation after cutting the cord.

Solution: After the cut, the plate will undergo a curvilinear translation and the particles on the plate move along parallel circles, each with a radius of 1 m. Immediately after the cord is cut, the velocity of the plate is zero and then the normal component of acceleration of any particle is zero. Thus, the acceleration of

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324

any particle, including mass center C of the plate, has only the tangent component. The free-body diagram and kinetic diagram of the plate are shown in figure 8.29b and c. X ð þ .Þ Fti ¼ maCt ; mg cos 30 ¼ maC ) aC ¼ g cos 30 ¼ 8:496 m/s2 ð þ -Þ

X

Fni ¼ maCn ; FA þ FB  mg sin 30 ¼ 0 ð þ xÞ

P

MC ¼ 0;

FA ðcos 30  0:5  sin 30  1Þ  FB ðsin 30  1 þ cos 30  0:5Þ ¼ 0

ð1Þ

ð2Þ

Solving equations (1) and (2) simultaneously yields FA ¼ 52:8 N,

FB ¼ 3:75 N

RETHINK: Applying the rotational equation of motion about point B can eliminate the unknown FB and get a direct solution of FA. P P ð þ xÞ M B ¼ ðM k ÞB ; FA cos 30  1  mg  1 ¼ maC sin 30  0:5  maC cos 30  1 Substituting the value of aC , we get FA ¼ 52:8 N. Substituting this into equation (1) yields FB ¼ 3:75 N.

8.5.2

Equations of Motion for Rotation about a Fixed Axis

Consider a rigid body along with its loads is symmetrical with respect to a reference plane and rotates about a fixed axis perpendicular to this reference plane, figure 8.30a. Because the body’s mass center C moves along a circular path, its acceleration can be represented by the tangential and normal components as follows: aCt ¼ rC a;

aCn ¼ rC x2

Their directions are shown in figure 8.30a, The free-body diagram and kinetic diagram of the body are then shown in figure 8.30b and c. Therefore, the equations of motion for the rotating body can be written as 8X Fti ¼ maCt ¼ mrC a > >

X > : M C ¼ IC a

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325

FIG. 8.30 – Free-body diagram and kinetic diagram for a rotating body.

The rotational equation of motion can be written about the pin at O to eliminate the unknown reactions FOx and FOy . From the kinetic diagram, this rotational equation of motion is X X X   MO ¼ ðM k ÞO ; MO ¼ rC maCt þ IC a ¼ mrC2 a þ IC a ¼ mrC2 þ IC a From the parallel-axis theorem, we have mrC2 þ IC ¼ IO . Thus, the above rotaP tional equation of motion can be rewritten as MO ¼ IO a. Consequently, the three equations of motion for the rotating body become 8X Fti ¼ maCt ¼ mrC a > >

X > : MO ¼ I O a Procedure for analysis The procedure of solving kinetic problems of a symmetrical body undergoing rotation about a fixed axis is as follows: (1) Draw the kinetic diagram to visualize the inertial terms matC , manC and IC a. Note that aCt ¼ rC a and aCn ¼ rC x2 . (2) Draw the free-body diagram to show all the external forces and couple moments acting on the body. (3) Usually, we choose the tangential and normal coordinates for the analysis and apply equation (8.22) or (8.23). But sometimes it is also convenient to use the rectangular coordinates (please see example 8.17). Meanwhile, any point P other than C and P O can also Pbe used to sum moments and get the rotational equation of motion M P ¼ ðM k ÞP . (4) If the angular velocity or angular position of the body is to be found, appropriate kinematic equations should be applied.

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Example 8.17. Uniform rod AB of weight P and length l is held in horizontal position by two weightless links at A and a string at B, figure 8.31a. Then the string is cut and the rod begins to rotate. Determine the forces in the links when the rod turns an arbitrary angle φ.

FIG. 8.31 – The forces in the two weightless links caused by the rotating rod. Solution: The free-body and kinetic diagrams of the rod at a general position φ are shown in figure 8.31b and c, respectively. Here assume that the rod has a clockwise € . Therefore, maCt ¼ Pal=2g angular velocity x ¼ u_ and angular acceleration a ¼ u n 2 and maC ¼ Px l=2g. Write the equations of motion for the rod: X ð1Þ ð þ "Þ Fyi ¼ maCy ; F2 sin 30  P ¼ maCt cos u þ maCn sin u ð þ !Þ

X

Fxi ¼ maCx ;

X

F1  F2 cos 30 ¼ maCt sin u  maCn cos u

ð2Þ

X

l l Pl 2 a ð3Þ ðMk ÞA ; P  cos u ¼ maCt  þ IC a ¼ 2 2 3g P For the rotational equation of motion, MA ¼ IA a can also be applied. For this case, the parallel-axis theorem for the mass moment of inertia should be utilized.  2 l Pl 2 P l 2 Pl 2 IA ¼ IC þ m þ  ¼ ¼ 2 g 4 12g 3g ð þ yÞ

MA ¼

For a given angle φ there are four unknowns in equations (1)–(3), namely F1, F2, ω and α. Therefore, the fourth equation should be obtained. The supplemental equation is usually obtained from kinematics. Solving equation (3) yields a ¼ 3g cos u=2l. Then the following kinematic equation should be used. xdx ¼ adu ¼

3g cos udu 2l

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327

Since x ¼ 0 at u ¼ 0 , we have Z x Z u 3g cos udu xdx ¼ 0 0 2l

)

x2 ¼

3g sin u l

Substituting maCt ¼ Pal=2g ¼ 3P cos u=4 and maCn ¼ Px2 l=2g ¼ 3P sin u=2 into equations (1) and (2) yields F2 ¼ 5P 

F1 ¼ 

9P cos2 u 2

pffiffiffi pffiffiffi 5 3 9 3 9 Pþ P cos2 u þ P sin u cos u 4 2 4

At the instant immediately after the string has been cut, u ¼ 0 ; F1 ¼  pffiffi and F2 ¼ P2 . When u ¼ 90 , F1 ¼  5 2 3 P; and F2 ¼ 5P.

pffiffi 3 4 P;

Example 8.18. The drum has mass mD = 50 kg and radius r = 0.5 m. Its radius of gyration about the pin support (also the mass center) O is kO ¼ 0:3 m. Two blocks A and B having weights of 150 N and 50 N, respectively, are attached to a cord around the drum, figure 8.32a. The system is released from rest. If there is no slipping between the cord and the drum, determine the drum’s angular acceleration.

FIG. 8.32 – The drum’s angular acceleration due to the suspended blocks.

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328

Solution 1: Assuming the drum has a counterclockwise angular acceleration α, block A will have a downward acceleration αr and block B will have an upward acceleration αr. Consider the drum and two blocks separately and their free-body and kinetic diagrams are shown in figure 8.32b–d, respectively. Applying the translational equation of motion to the two blocks, respectively, yields X 150 ð þ "Þ Fyi ¼ maCy ; T1  150 ¼ mA ar ¼  ar ð1Þ 9:81 ð þ "Þ

X

Fyi ¼ maCy ;

T2  50 ¼ mB ar ¼

50 ar 9:81

ð2Þ

Writing the rotational equation of motion for the drum about pin support O to eliminate the reaction forces, we get X ð þ xÞ MO ¼ IO a; ðT1  T2 Þr ¼ IO a ¼ 50  0:32  a ð3Þ Substituting the value of r and solving the above equations simultaneously yield T1 ¼ 110:2 N;

T2 ¼ 63:3 N;

a ¼ 5.21 rad/s2

Solution 2: The cord tensions T1 and T2 can be avoided by considering the drum and two blocks together as a single system. The free-body and kinetic diagrams of the system are shown in figure 8.32e. X X ð þ xÞ MO ¼ ðMk ÞO ; ð150  50Þr ¼ IO a þ mA ar  r þ mB ar  r Substituting and solving this equation also yield a ¼ 5.21 rad/s2 . RETHINK: (1) From equation (3), it is clear that if IO 6¼ 0 and a 6¼ 0, the cord tensions at two sides of the drum are different, i.e., T1 6¼ T2 . Only when IO ¼ 0 (for example, the mass of drum is neglected) or a ¼ 0 (The drum is at rest or rotates with a constant angular velocity), we have T1 ¼ T2 . (2) If the cord is wrapped around the periphery of the drum and attached to a block having a weight of 100 N, as shown in figure 8.32f, you can determine that a ¼ 7:09 rad/s2 . If a force of 100 N, not a block, is applied to the cord, as shown in figure 8.32g, you can determine that a ¼ 11.1 rad/s2 . For the three cases shown in figure 8.32g, f and a, the values of α are becoming smaller and smaller, since blocks have inertia, or resistance to acceleration.

Kinetics: Equations of Motion

8.5.3

329

Equations of Motion for General Plane Motion

The equations of motion applying to a rigid body under general plane motion are just equation (8.18) or (8.19). For this kind of relatively complicated problems, drawing free-body diagrams and kinetic diagrams will be of great help when writing the equations of motion. Procedure for analysis The procedure of solving kinetic problems of a symmetrical body undergoing general plane motion is as follows: (1) Choose an appropriate coordinate system. If the mass center of the rigid body moves along a known curved path, tangential and normal coordinates should be considered for the analysis. Otherwise, the rectangular coordinate system is used. (2) Draw the free-body diagram to show all the external forces and couple moments acting on the body. (3) Draw the kinetic diagram to visualize inertial terms maCx , maCy or matC , manC and IC a. P P (4) Apply equation (8.18) or (8.19). When using MP ¼ ðMk ÞP , P is usually chosen as the intersecting point of the lines of action of as many unknown forces as possible. (5) If the velocity or position of the body is to be found, appropriate kinematic equations should be applied. Example 8.19. The homogeneous disk with mass m and radius r is released from rest on the slope, figure 8.33a. The coefficient of friction between the disk and the slope is μ. Determine the angular acceleration of the disk in the case of (a) rolling without slipping, and (b) rolling and slipping.

FIG. 8.33 – A disk rolling down a slope.

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330

Solution: The free-body diagram and kinetic diagram of the disk are shown in figure 8.33b and c, respectively. (a) No slipping. In this case, we have aC ¼ ar The equations of motion are X ð þ &Þ Fxi ¼ maCx ; ð þ %Þ

X

ð þ yÞ

mg sin h  Fs ¼ maC

ð2Þ

N  mg cos h ¼ 0

ð3Þ

1 Fs r ¼ mr 2 a 2

ð4Þ

Fyi ¼ maCy ;

X

ð1Þ

MC ¼ IC a;

Solving equations (1)–(4) simultaneously yields a¼

2g sin h; 3r

N ¼ mg cos h; Fs ¼

mg sin h 3

ð5Þ

A more direct solution to this problem is to sum moments about point A to eliminate N and Fs. X X 3 ð þ yÞ MA ¼ ðMk ÞA ; mg sin h  r ¼ IC a þ maC  r ¼ mr 2 a 2 From this equation, α can be obtained directly. (b) Slipping. In this case, equation (1) does not apply. However, the friction force is the kinetic friction force and we have Fk ¼ lN The equations of motion become X ð þ &Þ Fxi ¼ maCx ; ð þ %Þ

X

ð þ yÞ

mg sin h  Fk ¼ maC

ð7Þ

N  mg cos h ¼ 0

ð8Þ

1 Fk r ¼ mr 2 a 2

ð9Þ

Fyi ¼ maCy ;

X

ð6Þ

MC ¼ IC a;

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331

Solving equations (6)–(9) simultaneously yields aC ¼ ðsin h  l cos hÞg;

a¼

2lg cos h ; r

N ¼ mg cos h

RETHINK: How about the condition under which the disk would roll without slipping? Because the inequality Fs \lN should be satisfied in the case of rolling without slipping, we can get the condition under which the disk would roll without slipping from equation (5). mg sin h\lN ¼ lmg cos h ) tan h\3l Fs ¼ 3 It means only when the inclination angle θ is small and the coefficient of friction μ is big enough (the slope is rough enough), the disk would roll without slipping. Example 8.20. A cord is wrapped around each of the two identical homogeneous disks with a weight P and a radius r, as shown in figure 8.34a. If they are released from rest, determine the acceleration of the center of disk B. Neglect the mass of the cord.

FIG. 8.34 – Two disks wrapped around by a cord.

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332

Solution 1: Consider disk A and disk B separately and their free-body and kinetic diagrams are shown in figure 8.34b and c, respectively. Using kinematics, aB will be related to aA and aB . As shown in figure 8.34d, we have aB ¼ aCx þ aCy þ atB=C þ anB=C The projection of the above vector equation along the y axis is t aB ¼ aCy þ aB=C ¼ raA þ raB

ð1Þ

Applying the rotational equation of motion about point A for disk A, we have ð þ yÞ

X

MA ¼ IA aA ;

FT r ¼

Pr 2 aA 2g

Applying the equations of motion for disk B yields X P ð þ #Þ Fyi ¼ maCy ; P  FT ¼ aB g ð þ yÞ

X

M B ¼ IB aB ;

FT r ¼

Pr 2 aB 2g

ð2Þ

ð3Þ

ð4Þ

Solving equations (1)–(4) simultaneously, we obtain 4 aB ¼ g 5 Solution 2: Cord tension FT can be avoided by considering two disks together as a single system. The free-body and kinetic diagrams of the system are shown in figure 8.34e. Applying the rotational equation of motion about point A to eliminate the unknown FA, we have ð þ yÞ

X

MA ¼

X

ðMk ÞA ;

P  2r ¼

Pr 2 Pr 2 P aA þ aB þ aB  2r g 2g 2g

ð5Þ

Solving equations (1) and (5) simultaneously, we also obtain 4 aB ¼ g 5 Example 8.21. The uniform rod of mass m and length l rests against the wall at an angle u0 , figure 8.35a. Neglect the friction. Determine its angular velocity and angular acceleration at any arbitrary position after it is released from rest and slides down. Also find the specific angle when the rod begins to leave the wall.

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333

FIG. 8.35 – A rod sliding down the smooth wall and smooth ground. Solution: The free-body diagram and kinetic diagram of the rod are shown in figure 8.35b and c, respectively. Note that we establish the angular position u measured from the fixed ground and clockwise as positive. Therefore, to seek a simultaneous solution, the angular velocity and angular acceleration should also be € , respectively. assumed clockwise at any arbitrary position and they are u_ and u Establishing the fixed x, y coordinate axes, we have l xC ¼ cos u; 2

l yC ¼ sin u 2

Two successive time derivatives of xC and yC yield aCx ¼ x€C ¼ 

 l € sin u þ u_ 2 cos u ; u 2

aCy ¼ y€C ¼

 l € cos u  u_ 2 sin u u 2

Applying the equations of motion, we have X  ml  € sin u þ u_ 2 cos u u ð þ !Þ Fxi ¼ maCx ; FNA ¼  2 ð þ "Þ

ð þ yÞ

X

X

Fyi ¼ maCy ;

MC ¼ IC a;

FNB  mg ¼

 ml  € cos u  u_ 2 sin u u 2

l l ml 2 € FNA  sin u  FNB  cos u ¼ u 2 2 12

ð1Þ

ð2Þ

ð3Þ

Solving equations (1)–(3) simultaneously, we obtain €¼ u

3g cos u 2l

The negative sign indicates that the angular acceleration is actually counterclockwise. From kinematics, we can write

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334

€¼ u

du_ du_ du du_ 3g ¼ ¼ u_ ¼  cos u dt du dt du 2l

Since u_ ¼ 0 at u ¼ u0 , we have Z u_ Z u 3g _ u_ ¼  cos udu ud 2l u0 0

)

rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 3g u_ ¼  ðsin u0  sin uÞ l

By inspection, the actual angular velocity should be counterclockwise. Therefore, _ i.e., the negative sign should be used for the expression of u, rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 3g u_ ¼  ðsin u0  sin uÞ l € When the rod begins to leave the wall, there should be FNA ¼  ml 2 ðu sin u þ € and u_ and solving yield u cos uÞ ¼ 0. Substituting the expressions of u   2 2 sin u ¼ sin u0 ) u ¼ arcsin sin u0 3 3 _2

PROBLEMS 8.1 Crane is used to lift a crate of weight G = 5 kN. The crate accelerates uniformly from rest to reach a velocity of 0.4 m/s in 0.5 s. Then the crate ascends at a constant speed for 3 s. Afterwards it decelerates uniformly until it stops in 0.2 s. Determine the tension in the cable during these three periods. 8.2 The maximum allowable tension in the crane cable is 35 kN. It is used to lift a crate of weight G = 25 kN. If we want to accelerate the crate uniformly from rest to reach a velocity of 0.6 m/s in 0.25 s, is the lifting safe?

Prob. 8.1/2

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335

8.3 A conveyor belt at angle θ with the horizontal is used to transfer gravel. If the acceleration of the belt is a, determine the minimum required coefficient of static friction between gravel and the belt for gravel without slipping on the surface of the belt.

Prob. 8.3

8.4 Starting from rest from origin O, a block on the smooth ground move along a straight line under the action of a force F = (12 − 2t) N, where t is expressed in seconds. Determine the time required for the block to return to origin O.

Prob. 8.4

8.5 A 650-N paratrooper open his parachute from an at-rest position at a very high altitude. The atmospheric drag resistance is F ¼ C qSv 2 , where C is a constant, S is the maximum cross-sectional area on the horizontal plane. q is the air density and generally taken as 1:25 kg/m3 . For a fully open spherical parachute, C = 0.48 and S = 50 m2. Determine the paratrooper’s velocity when he lands on the ground, which can be referred to as the maximum (terminal) velocity and by letting the time of fall t ! 1. If the parachute does not open due to a problem, C = 0.312 and S = 0.4 m2. How about the terminal velocity at this unfortunate case? 8.6 Starting from x = 0 with an initial velocity v0 = 5 m/s, a 10-kg particle is moving along a horizontal straight line in liquid under the action of a drag resistance force F ¼ 20v 2 =ð3 þ x Þ, where F, x and v are expressed in N, m and m/s, respectively. Determine the position equation of the particle.

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336

Prob. 8.6

8.7 Blocks A and B with the same mass of 10 kg are connected by a pulley-cord system as shown. The blocks are on the horizontal ground and the coefficient of friction between the blocks and the ground is 0.2. A horizontal force F = 50 N is acting on block A. Neglecting the mass of the pulleys and cords, determine the accelerations of the two blocks.

Prob. 8.7

8.8 Blocks A and B have masses of 100 kg and 300 kg, respectively. Neglect the mass of the pulleys and cords and the horizontal plane is frictionless. Determine the acceleration of each block when the system is released.

Prob. 8.8

8.9 Starting from point A with an initial velocity v0 = 5 m/s, a small ball begins to move on a smooth inclined plane. The direction of the initial velocity is parallel to base line CD of the slope as shown. If h ¼ 30 , determine (1) the time required for the small ball to reach point B; (2) distance d between points B and D.

Kinetics: Equations of Motion

337

Prob. 8.9

8.10 A cannonball is fired from height h above the ground with horizontal initial velocity v0. The atmospheric drag resistance is F = kmv, where k is a constant, m and v are the mass and speed of the cannonball, respectively. Determine the position equation of the cannonball.

Prob. 8.10

8.11 The 1-kg particle A is initially at rest at the position shown. It is then acted upon by its weight and three forces F1 ¼ f2i þ 6j  2kgN, F2 ¼ f3i  2kgN and F3 ¼ 3i  t 2 j  2k N, where t is in seconds. Find the distance of the particle from the origin in 5 s.

Prob. 8.11

Engineering Mechanics

338

8.12 A small ball of weight P is pin supported by two weightless links AM and BM, which are pin-connected to the rotating vertical rod. AM = BM = l and AB = 2a. If the rod is rotating with a constant angular velocity ω, find the forces in the two links. Neglect the size of the ball.

Prob. 8.12

8.13 A small ball having weight P is suspended by two cables AB and AC in a vertical plane. Then cable AB is cut and the ball begins to swing. Neglecting the size of the ball, determine the force in cable AC, (1) just after cable AB is cut; (2) when cable AC moves to the vertical position.

Prob. 8.13

Kinetics: Equations of Motion

339

8.14 The 30-kg boy on the swing is momentarily at rest when θ = 60°. Find his speed and the force in each of the two cords of the swing when θ = 90°. Neglect the boy’s size and the mass of the seat and cords.

Prob. 8.14

8.15 The 8-kN car is running with a constant speed v0 = 18 km/h on a bumpy track. The radius of curvature of both the convex and concave ground is ρ = 25 m. Find the normal reaction of the ground when the car is moving (1) on the level part of the road; (2) at the highest point of the convex part of the road; (3) at the lowest point of the concave part of the road. Neglect the size of the car.

Prob. 8.15

8.16 The small ball having a mass of 1 kg is attached to two cables. It is undergoing a circular motion in the horizontal plane with a constant speed v ¼ 2:5 m/s. If r ¼ 0:5 m, determine the two cable forces.

Engineering Mechanics

340

Prob. 8.16

8.17 Find the mass moment of inertia of the homogeneous annulus about an axis which is perpendicular to the annulus and passes through O. The inner and outer radii of the annulus are R1 and R2 , respectively. The total mass of the annulus is m. Neglect the mass of the spokes.

Prob. 8.17

8.18 The right circular cone as shown is formed by revolving around the x axis. If the cone has constant density ρ = 7860 kg/m3, find its mass moment of inertia about the x axis.

Kinetics: Equations of Motion

341

Prob. 8.18

8.19 Determine the mass moment of inertia of a uniform triangular sheet of mass m about the z axis.

Prob. 8.19

8.20 Determine the mass moments of inertia of two uniform sheets about the z axis. Their sizes are as shown. The mass of one portion of sheet with area ab is m.

Prob. 8.20

Engineering Mechanics

342

8.21 Find the mass moment of inertia of the frustum about the z axis. Its sizes is as shown and its total mass is m.

Prob. 8.21

8.22 The assembly shown consists of a uniform rod AB and two small balls. Each ball has a weight P and the rod has a weight 2P1 . Neglect the size of the small balls. The assembly is attached to the rotating axis z. Rod AB makes angle h with the z axis. Determine the mass moment of inertia of the assembly about the z axis.

Prob. 8.22

Kinetics: Equations of Motion

343

8.23 A wheel consists of one ring, one disk and eight rods. The large ring, small disk and each of the rods weigh M, m, and mr, respectively. The radii of the ring and the disk are R and r. Find the mass moment of inertia of the wheel about an axis perpendicular to the wheel and passing through its center C.

Prob. 8.23

8.24 A block of weight G is placed on the rough surface of triangular prism ABC. The coefficient of static friction between the block and the surface is fs and fs \ tan h. The triangular prism is moving to the left with constant acceleration a. Determine the smallest acceleration to keep the block relatively resting on the surface. 8.25 A block of weight G is placed on the rough surface of triangular prism ABC. The coefficient of static friction between the block and the surface is fs and fs [ tan h. The triangular prism is moving to the left with constant acceleration a. Determine the biggest acceleration without causing the block to move relative to the surface.

Prob. 8.24/25

8.26 The 3-t van is moving forward with a speed of v0 = 10 m/s when the brakes are applied, causing all four wheels to stop rotating. Then the van skids 7.5 m before it stops. Determine the normal reactions and the friction forces at A and B, respectively, during the skidding.

Engineering Mechanics

344

Prob. 8.26

8.27 An eccentric circular cam has radius R and eccentricity OC = e. The cam is rotating with a constant angular velocity x. A flat face follower, which can move along a vertical track, maintains contact with the cam. Block A of mass m is placed on the follower. Determine the largest value of x without causing the block to leave the surface of the follower.

Prob. 8.27

8.28 Identical cranks O1 A and O2 B can rotate and move the shelf AB to transfer the container M . O1 A ¼ O2 B ¼ l ¼ 1:5 m, O1 O2 ¼ AB. From the position shown at which h ¼ 45 , the cranks begin to rotate from rest with an angular acceleration a ¼ 5 rad/s2 . Determine the smallest coefficient of static friction to keep the container relatively resting on the shelf.

Kinetics: Equations of Motion

345

Prob. 8.28

8.29 The 40-kg square homogeneous plate is held in the vertical plane by three cords. The side length of the plate is 10 cm. Determine the acceleration of the plate center C and tensions in the cords AD and BE immediately after the cord FG is cut.

Prob. 8.29

8.30 The homogeneous disk has a mass of 100 kg and radius r = 2 m. It is pin-connected at A. Find the horizontal and vertical components of reaction at A immediately after the disk is released from rest from the position shown.

Prob. 8.30

Engineering Mechanics

346

8.31 The circular flywheel has been balanced so that its mass center and its geometric center coincide. The radius is R ¼ 50 cm. An inextensible cord is wrapped around the flywheel and attached to an 80-N block. The block is released from rest at a height h ¼ 2 m. It takes t1 ¼ 16 s for the block to reach the ground. To eliminate the effect of the frictional torque of the bearing, another 40-N block is used to repeat the experiment. It takes t2 ¼ 25 s for this block to reach the ground from the same height. Assume that the frictional torque is independent of the weight of the block. Determine the mass moment of inertia of the flywheel and the frictional torque.

Prob. 8.31

8.32 The device as shown is used to measure the mass moment of inertia of an irregular object A about its rotating axis. An inextensible cord is wrapped around drum D, over pulley C and finally attached to block B of mass m. The radius of drum D is r. Neglect the friction and the mass of the drum, pulley and cord. If block B starts from rest and moves down a distance h over a time interval s, determine the mass moment of inertia of object A about the z axis.

Kinetics: Equations of Motion

347

Prob. 8.32

8.33 The 10-kg homogeneous disk has radius r = 0.3 m and is given an initial angular velocity x0 ¼ 6 rad/s when it is placed on the surface. If the coefficient of kinetic friction is 0.3, determine the time it takes for the disk to stop slipping.

Prob. 8.33

8.34 The 100-kg spool has a radius of gyration kC = 0.3 m. A horizontal force P = 800 N is acting on the weightless cord that is wrapped around the spool’s inner hub. If the coefficients of static and kinetic friction between the spool and the ground are μs = 0.2 and lk = 0.15, respectively, determine the angular acceleration of the spool.

Engineering Mechanics

348

Prob. 8.34

8.35 A uniform circular wheel is fixed to a uniform circular shaft. Two centers coincide and the diameter of the shaft is d = 5 cm. The shaft is placed on the inclined plane and begins to roll without slipping from rest. The center travels s = 3 m in 5 min. If h ¼ 20 , determine the radius of gyration of the assembly about an axis that is perpendicular to the page and passing through the center.

Prob. 8.35

8.36 The 80-kg spool has a radius of gyration kC = 0.2 m about its mass center C. A vertical force P = 150 N is applied to the weightless cable that is wrapped around the spool. If the coefficients of static and kinetic friction between the rail and the hub are μs = 0.3 and μk = 0.25, respectively, find the angular acceleration of the spool and the acceleration of C. 8.37 The 50-kg spool has a radius of gyration kC = 0.2 m about its mass center C. A vertical force P = 300 N is applied to the weightless cable that is wrapped around the spool. If the coefficients of static and kinetic friction between the rail and the hub are μs = 0.2 and μk = 0.15, respectively, find the angular acceleration of the spool and the acceleration of C.

Kinetics: Equations of Motion

349

Prob. 8.36/37

8.38 The ends of a 30-kg uniform rod AB are pin-connected to two weightless collars that can slide along smooth fixed rods. If the rod is released from rest at θ = 15°, determine the angular acceleration of the rod and the reactions at A and B immediately after release.

Prob. 8.38

8.39 Uniform bar AB has a mass of 4 kg and is held in the horizontal position by two cords at its ends. Determine the tension in the other cord immediately after one cord has been cut.

Engineering Mechanics

350

Prob. 8.39

8.40 Uniform bar AB has mass m and length l. It is held in the horizontal position by two cords OA and OB. If θ = 45°, determine the acceleration of the middle point C of the bar immediately after cord OB has been cut.

Prob. 8.40

Chapter 9 Kinetics: Work and Energy Objectives
Calculate the work done by a force or a couple.
Understand the concepts of power and efficiency of a mechanical system.
Calculate the kinetic energy of a particle, a system of particles, a rigid body or a system of rigid bodies undergoing planar motion.
Derive and apply the principle of work and energy to solve kinetic problems that involve force, displacement and velocity.
Understand the concept of conservative force and calculate the gravitational and elastic potential energy of a mechanical system.
Solve kinetic problems using conservation of energy. Chapter 8 relates force and acceleration through Newton’s second law, i.e., the equation of motion F = ma. Once F is given, we can solve for the acceleration a from this equation; then from kinematics, the velocity and position of the particle at any time can be obtained from integration of a. However, using the equation of motion together with kinematics allows us to derive the principle of work and energy, which directly relates force, displacement and velocity and make the determination of the acceleration unnecessary. We will discuss the principle of work and energy in this chapter. The concepts of conservative force and potential energy and the principle of conservation of mechanical energy will also be examined.

9.1 9.1.1

Work and Power Work

In mechanics, work has a very specific definition that involves force and displacement. Force F does work on a particle when the particle acted upon by the force undergoes a displacement in the direction of the force. For example, constant force

DOI: 10.1051/978-2-7598-2901-9.c009 © Science Press, EDP Sciences, 2022

Engineering Mechanics

352

F is exerted on a particle that moves along a straight line and has a displacement s, figure 9.1. Then its work equals the magnitude of the force times the component of displacement in the direction of the force, i.e., W ¼ Fs cos h. If position vector r is drawn from the initial position M1 to the final position M2, we can also use the dot product to obtain the work as follows: W1!2 ¼ Fr ¼ Fs cos h

ð9:1Þ

where h is the angle between vectors F and r. Work is a scalar quantity, a positive or negative number. If 0  h\90 , the work is positive and if 90 \h  180 , the work is negative. If h ¼ 90 , i.e., the displacement is perpendicular to the force, the work would be zero. The unit of work in SI unit system is joule (J) and 1 J ¼ 1 N  1 m ¼ 1 Nm.1 1. Work of a variable force Consider a particle, which is acted upon by variable force F, undergoes a finite displacement along a curved path from M1 to M2, figure 9.2. The finite displacement has to be decomposed into infinitesimal displacements and then integration is used. For example, when the particle moves along the path from position M to a very near position M 0 , an infinitesimal displacement dr is obtained. Because this curved segment is very small, we have ds ¼ jdrj and force F can be viewed as constant during this infinitesimal displacement. Then from equation (9.1), the elementary work done by F on this infinitesimal displacement can be written as dW ¼ Fdr ¼ F cos h ds

ð9:2Þ

where h is the angle between the tails of F and dr. Therefore, the work of F during a finite displacement from M1 to M2 is obtained by integrating equation (9.2) along the curved path. Z M2 Z s2 Fdr ¼ F cos hds ð9:3Þ W1!2 ¼ M1

s1

where the variable of integration s is the distance traveled by the particle along the path. Substituting F ¼ Fx i þ Fy j þ Fz k and dr ¼ dxi þ dy j þ dzk into equation (9.3) yields Z M2 Z M2 W1!2 ¼ Fdr ¼ ðFx dx þ Fy dy þ Fz dzÞ ð9:4Þ M1

M1

Equation (9.4) indicates that the work of a force is obtained by a curvilinear integral, which is generally dependent on the curved path of the particle.

The joule (J) is the SI unit of work and energy. Even though J ¼ Nm, the moment of a force must be expressed in Nm, not in joules, since the moment of a force is not a form of energy [2]. 1

Kinetics: Work and Energy

353

FIG. 9.1 – Work of a constant force.

FIG. 9.2 – Work of a variable force. 2. Work of a weight Consider a body of weight P, whose center of gravity moves along the path from M1 to M2, as shown in figure 9.3. Because P ¼ Pk, from equation (9.4) we have Z z2 W1!2 ¼ Pdz ¼ Pðz2  z1 Þ ¼ Pðz1  z2 Þ ¼ PDh ð9:5Þ z1

where Dh is the altitude difference of the center of gravity of the body from the initial position to the final position. Equation (9.5) means that the work is independent of the path and is equal to the body’s weight times the altitude difference of the center of gravity of the body. The work is positive when the body moves downward (Dh [ 0). When the body moves upward (Dh\0), the work is negative.

FIG. 9.3 – Work of a weight.

Engineering Mechanics

354 3. Work of a spring force

When a body attached to fixed point O by an elastic spring moves, as shown in figure 9.4, spring force Fe is a variable force. For a linear elastic spring, the magnitude of Fe exerted by the spring on the body is proportional to the deflection of the spring. Noting that spring force Fe is directed to point O, it can be mathematically expressed as Fe ¼ kðr  l0 Þ

r r

where k is the spring constant or stiffness measured in N/m or kN/m in SI unit system; l0 is the unstretched length, and r is the position vector from point O to the body. Then r  l0 is the deflection of the spring. The rectangular components of the spring force are x y z Fex ¼ kðr  l0 Þ ; Fey ¼ kðr  l0 Þ ; Fez ¼ kðr  l0 Þ r r r

FIG. 9.4 – Work of a spring force. From equation (9.4) we have, Z M2 Z M2   W1!2 ¼ Fe dr ¼ Fex dx þ Fey dy þ Fez dz M1 ZMM1 2 xdx þ ydy þ zdz ¼ kðr  l0 Þ r M1 Because x 2 þ y 2 þ z 2 ¼ r 2 , we have xdx þ ydy þ zdz ¼ rdr. Substituting this into the above expression yields Z r2 i kh W1!2 ¼ kðr  l0 Þdr ¼  ðr2  l0 Þ2  ðr1  l0 Þ2 2 r1

Kinetics: Work and Energy

355

Set d1 ¼ r1  l0 and d2 ¼ r2  l0 , which are the deflections of the spring at position M1 and position M2 , respectively. Then we get the work of a spring force as follows: k W1!2 ¼ ðd21  d22 Þ 2

ð9:6Þ

It can be seen that the work of a spring force is also independent of the path. It just depends upon the initial and final deflections of the spring. Example 9.1. The spring has an unstretched length l0 ¼ R. Its two ends are respectively attached to a fixed point O and a small ring, which is moving along a fixed circular path, figure 9.5. The spring constant is k. Determine the work of the spring force when the ring moves from position A to position B, and the work when the ring moves from position B to position D.

FIG. 9.5 – A spring with its one end moving along a circular path. Solution: When the ring moves from position A to position B, the initial and final deflections of the spring are pffiffiffi dA ¼ OA  l0 ¼ 2R  R; dB ¼ OB  l0 ¼ R Therefore, i k k h pffiffiffi WA!B ¼ ðd2A  d2B Þ ¼ ð 2  1Þ2 R2  R2 ¼ 0:414kR2 2 2 When the ring moves from position B to position D, dB ¼ OB  l0 ¼ R;

pffiffiffi dD ¼ OD  l0 ¼ ð 2  1ÞR

pffiffiffi k k WB!D ¼ ðd2B  d2D Þ ¼ ½R2  ð 2  1Þ2 R2  ¼ 0:414kR2 2 2

Engineering Mechanics

356 4. Work of a gravitational force

Consider a particle of mass m moves along a curved path around another particle of mass ms fixed at position O, as shown in figure 9.6. From Newton’s law of gravitational attraction, the moving particle is subjected to a force of gravitational attraction exerted by the particle at O and the magnitude is ms m F ¼f 2 r where f is the universal constant of gravitation. Since F is directed toward O and the position vector r is directed away from O, the vector form of the force is F ¼ f

ms m r r2 r

The elementary work of the force F on an infinitesimal displacement is dW ¼ Fdr ¼ f

ms m r ms m dr ¼ f 2 dr 2 r r r

ð9:7Þ

In the above derivation, we use rr dr ¼ 2r1 dðrrÞ ¼ 2r1 dðr 2 Þ ¼ dr. From equation (9.7), the work of the gravitational force F on a finite displacement from M1 to M2 (figure 9.6) is   Z r2 ms m 1 1 f 2 dr ¼ fms m  W1!2 ¼ ð9:8Þ r r2 r1 r1 We can use equation (9.8) to determine the work of the force exerted by the earth on a body at a distance r from the earth’s center when r is much larger than the radius of the earth. Equation (9.8) indicates that the work of a gravitational force is also independent of the path. It depends only upon the initial and final distances between the two particles.

FIG. 9.6 – Work of a gravitational force.

Kinetics: Work and Energy

357

5. Work of a force on a rotating body Consider a body rotating about a fixed axis z. The body is subjected to a force F, which is directed relative to the body as shown in figure 9.7. The force can be resolved into three components: Ft , Fn and Fb . When the body has an infinitesimal angular displacement du, the application point of the force has an infinitesimal displacement dr ¼ ds ¼ rdu. Because both Fn and Fb are perpendicular to the infinitesimal displacement and do no work, the work done by the component force Ft is just the work of the force F. Therefore dW ¼ Ft ds ¼ Ft rdu

ð9:9Þ

It should be noted that Ft can be positive or negative here. If Ft is in the direction of +t, Ft is positive. For this case, Ft and dr are in the same direction and dW is positive accordingly. Otherwise, both Ft and dW are negative. Since Ft r is also the moment of the force F about the rotating axis z, i.e., Mz ðFÞ ¼ Ft r, equation (9.9) can be rewritten as dW ¼ Mz ðFÞdu. Therefore, when the body rotates from u1 to u2 , the work of the force is Z u2 Z u2 W1!2 ¼ Ft rdu ¼ Mz ðFÞdu ð9:10Þ u1

u1

FIG. 9.7 – Work of a force on a rotating body.

Engineering Mechanics

358 6. Work of a couple on a rotating body

Consider a body rotating about the fixed axis z is subjected to a couple M, which is parallel to the z axis, figure 9.8. When the body has an infinitesimal angular displacement du, the elementary work of the couple is dW ¼ M du. Therefore, when the body rotates from u1 to u2 , the work of the couple is Z u2 W1!2 ¼ M du ð9:11Þ u1

If couple moment M is constant, the above expression becomes W1!2 ¼ M ðu2  u1 Þ

ð9:12Þ

FIG. 9.8 – Work of a couple on a rotating body. 7. Work of a force on a body undergoing general plane motion Consider a body undergoing general plane motion. A force F is acting on the body at point A, figure 9.9a.2 For a differential displacement, the elementary work of the force is dW ¼ FdrA . If the displacement of point A is not easy to determine, the force can be moved to another point whose displacement is easy to determine. For example, move the force from point A to point B, provided couple moment M is

The line of action of the force should be in the same plane on which point A moves during the motion. If it is not, the projection of the force on this plane should be used.

2

Kinetics: Work and Energy

359

added to maintain an equivalent system, as shown in figure 9.9b. This couple moment equals the moment of F about B, i.e., M ¼ MB ðFÞ. Then the elementary work becomes dW ¼ FdrB þ MB ðFÞdu For a finite displacement of the body, the work of force F is Z B2 Z u2 W1!2 ¼ FdrB þ MB ðFÞdu B1

u1

ð9:13Þ

This strategy of determining the work of a force on a body undergoing general plane motion will be shown in example 9.2.

FIG. 9.9 – Work of a force on a body undergoing general plane motion.

8. Work of support reactions As shown in figures 9.10a–c, the reaction forces at a smooth surface, a roller and an inextensible cord do no work since the reaction forces act in a direction perpendicular to the displacement (cos h ¼ 0). As shown in figure 9.10d and e, the reaction forces (or couple) at a smooth hinge and a fixed support do no work because the displacement is zero (ds = 0, dφ = 0).

FIG. 9.10 – Work of support reactions.

Engineering Mechanics

360

It can be seen that many reaction forces frequently encountered do no work. But as one kind of special reaction force, the friction force needs careful consideration. Kinetic friction forces generally do negative work. Static friction forces may do negative work (figure 9.11a), positive work (figure 9.11b) or no work (figure 9.11c). When a rigid body rolls without sliding on a fixed surface, the friction force at the contacting point is a static friction force and does no work. This is because its application point, i.e., the contacting point, is the instantaneous center of zero velocity and so the elementary work of the friction force is always zero (dr = 0).

FIG. 9.11 – Work of a friction force. Example 9.2. As shown in figure 9.12a, constant force F is acting on the weightless cord that is wrapped around the spool’s inner hub. Force F always makes angle θ with the horizontal. If the spool of weight P rolls without slipping on the ground, determine the total work done by all the forces acting on the spool when its center C moves a distance s.

FIG. 9.12 – Total work done by all the forces on the spool. Solution: All forces acting on the spool are shown in figure 9.12b. The weight P of the spool does no work because its center of gravity moves horizontally. Friction force Fs does no work, since the spool rolls without slipping and the application point of Fs is the instantaneous center of zero velocity (IC) and dr = 0. The normal force FN also does no work because its application point is IC.

Kinetics: Work and Energy

361

Because the displacement of the application point of force F is not easy to determine, the force has been moved to the center C and a clockwise couple moment M = Fr is added to maintain an equivalent system, as shown in figure 9.12b. Since the spool does not slip as it rolls, when the center C moves a distance s, the spool rotates u ¼ s=R. Then the work of force F acting on the cord is the sum of the work of the force at C and the work of the couple. The total work is WF ¼ Fs cos h þ Fru ¼ Fs cos h þ

Frs R

9. Work of internal forces Internal forces always occur in equal but opposite collinear pairs. It can be proven that the total work of the internal forces holding the particles of a rigid body together is zero [2, 6, 7]. As shown in figure 9.13, two particles A and B on a rigid body exert on each other the equal and opposite forces F and F 0 . Although their displacements drA and drB are generally different, the components of their displacements along AB, i.e., dr 0 A and dr 0 B , must be equal; otherwise, the particles cannot remain at the same distance from each other, which is contrary to the concept of rigid body. Therefore, the work of F and the work of F 0 must be equal in magnitude and opposite in sign, making their sum equal to zero. From the above explanation, we can also see that for a nonrigid body or an arbitrary system of particles, internal forces may do work.

FIG. 9.13 – Work of internal forces inside a rigid body. For two rigid bodies connected by a weightless link or an inextensible cord, as shown in figure 9.14a, the work of internal forces FA and FB cancels out due to the above-mentioned reason. For two rigid bodies connected by a smooth pin, as shown in figure 9.14b, the work of FO and the work of FO0 must be equal in magnitude and opposite in sign and the sum work is zero, since FO and FO0 are equal, opposite and applied at the same point. This also applies to the pairs of internal forces for a system of rigid bodies that are in mesh with each other, figure 9.14c. Thus, for all these cases, when the system of connected rigid bodies is considered, the work of internal forces would cancel out.

Engineering Mechanics

362

FIG. 9.14 – Work of internal forces in a system of rigid bodies.

9.1.2

Power

Power is the time rate at which work is done, i.e., the amount of work done per unit of time. It is an important criterion to select a motor or engine that is required to do a certain amount of work in a given time. If the work done during a time interval Dt is DW , the average power during this time interval is P ¼

DW Dt

Letting Dt ! 0, we get the instantaneous power. DW dW Fdr dr ¼ ¼ ¼ F ¼ Fv Dt!0 Dt dt dt dt

P ¼ lim

ð9:14Þ

where F is the force exerted on the particle and v is the velocity of the particle. For a rigid body rotating about a fixed axis and acted upon by a couple moment M, which is parallel to the rotating axis, figure 9.8, we have dW ¼ M du. Therefore, the power is P¼

dW du ¼M ¼ Mx dt dt

ð9:15Þ

where ω is the angular velocity of the body. From the concept of power, its unit is the unit of work divided by the unit of time. In SI unit system, power is measured in J/s, also called watt (W). 1 W ¼ 1 J=s ¼ 1 Nm=s Mechanical efficiency of a machine is defined as the ratio of the output work to the input work: e¼

Work output Work input

ð9:16Þ

If work is done at a constant rate, the efficiency may also be expressed as e¼

Power output Power input

ð9:17Þ

Kinetics: Work and Energy

363

Practically friction forces always exist within a machine and extra work is needed to compensate for the energy losses due to friction. Therefore, efficiency of a machine is always less than 1. Example 9.3. Motor M with an efficiency e ¼ 0:8 is used to lift the 100-kg crate A, figure 9.15a. Determine the needed input power of the motor when point B of the cable (1) is moving to the right at a constant speed of 3 m/s; (2) has an instantaneous velocity of 3 m/s and an acceleration of 1 m/s2, where both are directed to the right. Neglect the mass of the pulleys and cables.

FIG. 9.15 – Output power and the needed input power. Solution: From figure 9.15a, we have 2sA þ sB ¼ l Taking the second time derivative of this equation yields aB aA ¼  2 The negative sign indicates that when point B has an acceleration to the right, i.e., in the direction of þ sB , it causes a corresponding upward acceleration of crate A, i.e., in the direction of sA . From the free-body diagram of crate A, figure 9.15b, we can write the equation of motion, X ð þ #Þ Fyi ¼ may ; mg  2T ¼ maA

(1) When point B of the cable is moving to the right at a constant speed, we have aA ¼ aB =2 ¼ 0. Then from the equation of motion, we have 100  9:81  2T ¼ 0

)

T ¼ 490:5 N

Engineering Mechanics

364

The output power of the motor to draw the cable in at a constant rate of 3 m/s is P ¼ Tv ¼ 490:5  3 ¼ 1472 W Therefore, the needed input power of the motor is Power input ¼

Power output 1472 ¼ ¼ 1840 W e 0:8

(2) When point B has an instantaneous velocity of 3 m/s and an acceleration of 1 m/s2, we have aA ¼ aB =2 ¼ 1=2 ¼ 0:5 m/s2 . Then from the equation of motion, we get 100  9:81  2T ¼ 100  ð0:5Þ

)

T ¼ 515:5 N

The output power of the motor to draw the cable in at an instantaneous rate of 3 m/s is P ¼ Tv ¼ 515:5  3 ¼ 1546:5 W Therefore, the needed input power of the motor is Power input ¼

Power output 1546:5 ¼ ¼ 1933 W e 0:8

RETHINK: It can be seen that the motor needs to deliver more power to produce accelerated motion than to produce motion at a constant velocity.

9.2 9.2.1

Kinetic Energy Kinetic Energy of a Particle

Kinetic energy of a particle of mass m moving with a velocity v is defined as 1 T ¼ mv 2 2

ð9:18Þ

Like work, kinetic energy is a scalar. However, work can be either positive 2 or negative,  but the kinetic energy is always positive. Since 1 kgðm/sÞ ¼ 1 kgm/s2 m ¼ 1 Nm ¼ 1 J, the kinetic energy has the same unit as work, i.e., joules (J) in SI unit system.

Kinetics: Work and Energy

9.2.2

365

Kinetic Energy of a System of Particles

The kinetic energy of a system of particles is the sum of the kinetic energies of all the particles within the system, namely X1 mi vi2 T¼ ð9:19Þ 2 where mi and vi are, respectively, the mass and velocity of an arbitrary particle Mi of the system. As shown in figure 9.16, we establish a fixed reference frame Oxyz and a translating reference frame Cx 0 y 0 z 0 whose origin is attached to and moves with the mass center C of the system of particles. v C is the mass center’s velocity relative to the fixed frame Oxyz, and v i=C is the relative velocity of particle Mi with respect to the translating frame Cx 0 y 0 z 0 . From the relative-motion analysis, we have vi ¼ v C þ v i=C

FIG. 9.16 – Kinetic energy of a system of particles.

Therefore 2 vi2 ¼ vi v i ¼ ðvC þ v i=C Þðv C þ vi=C Þ ¼ vC2 þ 2vC v i=C þ vi=C

Substituting the above expression into equation (9.19), we can express the kinetic energy of the system as X1 X1 2 mi vi2 ¼ mi ðvC2 þ 2v C vi=C þ vi=C T¼ Þ 2 2   X X1 1 X 2 mi vC2 þ v C  mi vi=C mi v i=C þ ¼ 2 2

366

Engineering Mechanics

P In this equation, mi ¼ m is the total mass of the system. For the mass center P C, we have mi ri ¼ P mrC in the fixed frame Oxyz. Similarly, in the translating frame Cx 0 y 0 z 0 , we have mi ri=C ¼ mrC =C ¼ 0. Taking the time derivative yields P mi v i=C ¼ mv C =C ¼ 0. Then the second term on the right side is zero. The third term represents the kinetic energy of the system due to its relative motion with respect to the translating frame Cx 0 y 0 z 0 and can be denoted as Tr . Therefore X1 1 1 2 mi vi=C T ¼ mvC2 þ ¼ mvC2 þ Tr ð9:20Þ 2 2 2 It means the kinetic energy of a system of particles is the sum of the translational kinetic energy due to the translation with the mass center C, and the relative kinetic energy due to the relative motion with respect to the translating frame Cx 0 y 0 z 0 .

9.2.3

Kinetic Energy of a Rigid Body in Planar Motion

For a rigid body undergoing general plane motion shown in figure 9.17, v i=C is the tangential velocity due to the rotation about mass center C, i.e., vi=C ¼ ri=C x. Substituting this into equation (9.20) yields  X1  2 1 1 1 X 2 T ¼ mvC2 þ mi ri=C x ¼ mvC2 þ mi ri=C x2 2 2 2 2

FIG. 9.17 – Kinetic energy of a rigid body in planar motion. P 2 The sum mi ri=C ¼ IC is the mass moment of inertia of the body about the axis passing through C and perpendicular to the plane of motion. Therefore, we have

Kinetics: Work and Energy

367 1 1 T ¼ mvC2 þ IC x2 2 2

ð9:21Þ

It means the kinetic energy of a rigid body in general plane motion is the sum of the translational kinetic energy due to the translation with the mass center C, and the body’s rotational kinetic energy due to the relative rotation about C. The application of equation (9.21) for three types of rigid-body planar motion is discussed below. 1. Translation. When a rigid body undergoes translation, the rotational kinetic energy is zero because x ¼ 0. The kinetic energy of a translating body is therefore 1 T ¼ mvC2 2

ð9:22Þ

2. Rotation about a fixed axis. For a rotating body shown in figure 9.18, we have vC ¼ rC x. Substituting this into equation (9.21) and considering the parallel-axis theorem yield  1 1 1 1 T ¼ m ðrC xÞ2 þ IC x2 ¼ mrC2 þ IC x2 ¼ IO x2 2 2 2 2

ð9:23Þ

where IO is the mass moment of inertia of the body about the fixed axis at O. Both equations (9.21) and (9.23) can be used to calculate the kinetic energy of a rigid body undergoing rotation about a fixed axis.

FIG. 9.18 – The relationship between vC and x for a rotating body. 3. General plane motion. For a rigid body undergoing general plane motion, if its instantaneous center of rotation (IC) for a specific instant is located, figure 9.19, we have vC ¼ rC =IC x. Substituting this into equation (9.21) and considering the parallel-axis theorem yield

368

Engineering Mechanics  2 1 1  1 2 1 mrC =IC þ IC x2 ¼ IIC x2 T ¼ m rC =IC x þ IC x2 ¼ 2 2 2 2

ð9:24Þ

where IIC is the mass moment of inertia of the body about an axis perpendicular to the plane of motion and passing through IC for the instant considered. Both equations (9.21) and (9.24) can be used to calculate the kinetic energy of a rigid body undergoing general plane motion.

FIG. 9.19 – The relationship between vC and x for a body undergoing general plane motion. Example 9.4. The system shown in figure 9.20a consists of a fixed pulley A of weight PA , a movable pulley B of weight PB and a block C of weight PC . Two pulleys are treated as two uniform disks. Knowing that r ¼ R=2 and xA is the angular velocity of pulley A, determine the kinetic energy of the system. Neglect the mass of the cables. Solution: This is a system of connected rigid bodies for which the total kinetic energy is just the sum of the kinetic energies of all its moving components. Depending on the type of planar motion, the kinetic energy of each body is determined by applying equations (9.22)–(9.24), or just (9.21). The kinematic analysis of this system is shown in figure 9.20b. Pulley A is subjected to rotation about a fixed axis and vD ¼ xA R ¼ vE . Pully B is undergoing general plane motion and its IC is the leftmost point. From kinematics, we have vB ¼ vE =2 ¼ xA R=2 and xB ¼ vE =R ¼ xA . Block C is translating and we have vC ¼ vB ¼ xA R=2. The kinetic energy of rotating pulley A is PA x2A R2 1 1 PA R2  x2A ¼ TA ¼ IA x2A ¼  2 2 2g 4g

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FIG. 9.20 – Kinetic energy of a system of bodies.

The kinetic energy of pulley B undergoing general plane motion is  2  2 3PB x2A R2 1 PB 2 1 1 PB 1 1 PB R2 2 xA R þ  vB þ IB xB ¼    x2A ¼ TB ¼ 2 g 2 2 2 2 g 2g 16g The kinetic energy of translating block C is  2 1 PC 2 1 PC 1 PC x2A R2 TC ¼ xA R ¼ vC ¼   2 g 2 2 g 8g Therefore, the total kinetic energy of the system is T ¼ TA þ TB þ TC ¼ ¼

PA x2A R2 3PB x2A R2 PC x2A R2 þ þ 4g 16g 8g

x2A R2 ð4PA þ 3PB þ 2PC Þ 16g

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9.3 9.3.1

Principle of Work and Energy Principle of Work and Energy for a Particle

P Consider a particle of mass m that is acted upon by forces Fi and moving along a curved path, figure 9.21. P From § 8.2, expressing Newton’s second law in the tangential direction yields Fti ¼ mat . From kinematics, we have at ds ¼ vdv or at ¼ vdv=ds. Substituting this into the tangential equation of motion gives   X 1 2 Fti ds ¼ mvdv ¼ d mv 2 Noting that Ft ds ¼ F cos hds ¼ dW and integrating from position M1, where s ¼ s1 and v ¼ v1 , to M2, where s ¼ s2 and v ¼ v2 , we have  Z v2  X Z s2 1 2 1 1 Fti ds ¼ W1!2 ¼ d mv ¼ mv22  mv12 ¼ T2  T1 2 2 2 s1 v1 It is often re-expressed as T1 þ W1!2 ¼ T2

ð9:25Þ

This is the principle of work and energy for a particle, which means the particle’s initial kinetic energy plus the work done by all the forces acting on the particle as it moves from its initial position to the final position equals the particle’s final kinetic energy. It is obvious that equation (9.25) is dimensionally homogeneous since the kinetic energy has the same unit, joules (J), as work.

FIG. 9.21 – Derivation of the principle of work and energy. Note that the speed v used to determine the kinetic energy should be measured with respect to a Newtonian reference frame, since the principle of work and energy is derived from Newton’s second law, which only applies with respect to a Newtonian reference frame.

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Procedure for analysis From the derivation, we know that the principle of work and energy is an inteP grated form of Fti ¼ mat combined with the kinematic equation at ds ¼ vdv. It then gives a direct relationship between force, displacement and velocity. For problems involving such three terms, it is generally more convenient to use the principle of work and energy, since it can avoid the determination of the acceleration and the integration. The procedure for analysis is as follows: 1. Establish an inertial coordinate system, analyze the velocities of the particle at its initial and final positions, and calculate T1 and T2. 2. Draw the particle’s free-body diagram to show all the forces that do work during the particle’s motion. The forces that act on the particle but do no work do not need to be drawn. 3. Calculate the total work W1!2 by algebraically adding the work of all the forces. 4. Apply T1 þ W1!2 ¼ T2 to solve for the unknown. Example 9.5. A 2000-kg jeep is driven down an incline at a speed of 90 km/h, figure 9.22a. Then the brakes are applied and cause a constant total braking force of 7 kN. Find the distance the jeep travels before stopping.

FIG. 9.22 – Determining the distance the jeep travels after the brakes are applied.

Solution 1: The free-body diagram of the jeep is shown in figure 9.22b. The velocity and kinetic energy at the initial position are as follows: v1 ¼ 90 km/h ¼ 25 m/s 1 1 T1 ¼ mv12 ¼  2000  252 ¼ 625  103 J 2 2 The velocity and kinetic energy at the final position are v2 ¼ 0;

T2 ¼ 0

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During the skidding, the normal force FN does no work since it does not move along its line of action. The work done by the other two forces is W1!2 ¼ 7000x þ 2000  9:81  sin 5  x ¼ 5290x Thus, from the principle of work and energy, T1 þ W1!2 ¼ T2 , we have 625  103  5290x ¼ 0

)

x ¼ 118 m

Solution 2: This problem can also be solved by using the equation of motion, where two steps are involved. First, from the free-body diagram, figure 9.22b, applying the equation of motion along the incline yields X ð þ &Þ Fxi ¼ max ; 2000  9:81  sin 5  7000 ¼ 2000a a ¼ 2:645 m/s2 Since a is constant, we have v 2 ¼ v02 þ 2a ðx  x0 Þ. Substituting the corresponding values yields 02 ¼ 252 þ 2  ð2:645Þ  ðx  0Þ

)

x ¼ 118 m

RETHINK: Unlike two steps in solution 2, the principle of work and energy in solution 1 directly relates force, displacement and velocity and make the determination of the acceleration unnecessary. Therefore, solution 1 is easier here than solution 2 using the equation of motion. Example 9.6. Starting from rest at point 1, a 0.1-kg toy car moves along the smooth track, figure 9.23a. Determine (1) the force exerted by the track on the toy car at point 2, where the radius of curvature of the track is 30 cm; (2) the minimum value of the radius of curvature at point 3 without causing the car to leave the track.

FIG. 9.23 – A toy car moving along a smooth track. Solution: (1) Apply the principle of work and energy when the toy car moves from point 1 to point 2,

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373

T1 þ W1!2 ¼ T2 ;

1 0 þ mg  0:5 ¼ mv22 2

)

v22 ¼ g

The free-body diagram and kinetic diagram of the toy car at point 2 are shown in figure 9.23b. Writing the equation of motion in the normal direction yields ð þ "Þ

X

Fni ¼ man ;

N2  mg ¼ man2 ¼ m

v22 q2

)

N2 ¼ 4:25 N

(2) Applying the principle of work and energy between point l and point 3, we obtain T1 þ W1!3 ¼ T3 ;

1 0 þ mg  ð0:5  0:2Þ ¼ mv32 2

)

v32 ¼ 0:6g

The free-body diagram and kinetic diagram of the toy car at point 3 are shown in figure 9.23c. The minimum value of q3 without causing the car to leave the track can be obtained when N3 ¼ 0. Writing the equation of motion in the normal direction, we get ð þ #Þ

X

Fni ¼ man ;

mg ¼ man3 ¼ m

v32 q3

)

q3 ¼ 0:6 m

RETHINK: One advantage of the principle of work and energy is that forces that do no work are eliminated from the solution of the problem. However, this also means this principle cannot be used to directly determine these kinds of forces, such as N2 in this example. For this case, the principle of work and energy are combined with the equation of motion to solve for the force. The principle of work and energy is used to determine the speed of the car, and then the equation of motion is used to determine the normal force. Example 9.7. A small ball is released from rest and begins to fall down from the peak of the smooth semicylinder, as shown in figure 9.24a. Specify the angle φ0 at which the ball starts to leave the surface of the semicylinder.

FIG. 9.24 – A ball falling down a smooth semicylinder.

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Solution 1: The free-body diagram of the small ball at an arbitrary angle φ is shown in figure 9.24b. The normal force FN does no work and only the weight does work. Apply the principle of work and energy, 1 0 þ mgRð1  cos uÞ ¼ mv 2 2

T1 þ W1!2 ¼ T2 ;

v 2 ¼ 2gRð1  cos uÞ

ð1Þ

Write the equation of motion in the normal direction, ð þ .Þ

X

Fni ¼ man ;

FN þ mg cos u ¼ m

v2 R

When the small ball leaves the surface of the semicylinder, FN ¼ 0; hence, the above equation becomes v 2 ¼ gR cos u Substituting this into equation (1) and solving gives the angle at which the ball starts to leave the surface. cos u0 ¼

2 3

u0 ¼ 48:2

)

Solution 2: From the free-body diagram of the small ball shown in figure 9.24b, write the equations of motion on normal and tangential coordinates: ð þ .Þ

X

ð þ &Þ

Fni ¼ man ;

X

Fti ¼ mat ;

FN þ mg cos u ¼ m

v2 R

ð2Þ

mg sin u ¼ mat ¼ m

dv dt

ð3Þ

From kinematics, we have at ds ¼ vdv. Noting that ds ¼ Rdu and at ¼ g sin u from equation (3), we have at ¼

vdv vdv ¼ ¼ g sin u ds Rdu

Separating the variables gives gR sin udu ¼ vdv Since v ¼ 0 at u ¼ 0, we have Z u Z gR sin udu ¼ 0

0

v

vdv

)

v 2 ¼ 2gRð1  cos uÞ

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375

Substituting this into equation (2) yields FN ¼ mg cos u  m  2g ð1  cos uÞ ¼ 2mg  3mg cos u When the ball starts to leave the surface, FN ¼ 0. This gives cos u0 ¼

2 3

)

u0 ¼ 48:2

RETHINK: Comparing the two solutions, it is apparent that the principle of work P and energy provide a convenient substitution for Fti ¼ mat and at ds ¼ vdv.

9.3.2

Principle of Work and Energy for a System of Particles

Consider a system of n particles within an enclosed region in space, figure 9.25. Apply the principle of work and energy to the i-th particle, which is subjected to internal force f i and external force Fi , we obtain ðT1 Þi þ ðW1!2 Þi ¼ ðT2 Þi Applying this principle to each of the other particles of the system, similar equations would be written. Since work and energy are both scalar quantities, for the whole system, all the n equations can be added together algebraically. T1 þ W1!2 ¼ T2

ð9:26Þ

This expression is just like equation (9.25). However, in this case, T1 and T2 are the initial and final kinetic energies of the whole system and W1!2 is the work done by all the external and internal forces and couple moments acting on the system. This is the principle of work and energy for a system of particles.

FIG. 9.25 – Derivation of the principle of work and energy for a system of particles.

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Since no restriction is imposed in the way the particles are connected, equation (9.26) can be used to analyze the motion of any kind of system of particles, such as a solid, liquid, or gas system. However, it should be noted that the expression W1!2 in equation (9.26) represents the work of all the forces and couples acting on the system, whether they are external or internal. In § 9.1, it has been explained that for an arbitrary system of particles, such as a liquid system or a nonrigid body, internal forces may do work; whereas for a rigid body or a system of rigid bodies connected by smooth pins, weightless links or inextensible cables, or in mesh with one another, internal forces do no work. Therefore, if a rigid body or a system of connected rigid bodies is considered, the internal forces can be eliminated from the analysis and only the work of all the external forces and couples need to be considered. Procedure for analysis As mentioned earlier, the principle of work and energy is suitable for solving kinetic problems that involve force, displacement and velocity. Although this principle can apply to any kind of system of particles, it is more convenient to apply it to a rigid body, or a system of connected rigid bodies, since the internal forces do no work for this case and only the work of the external forces and couples need to be considered. In engineering mechanics, we mainly apply it a rigid body, or a system of rigid bodies. The procedure for analysis is as follows: 1. Establish an inertial coordinate system and calculate the initial and final kinetic energies of the system. For every body’s kinetic energy, T ¼ 12 mvC2 þ Tr or the alternative equations should be utilized according to the type of rigid-body planar motion. 2. Draw the free-body diagram of the system to account for all the forces and couples that do work. When several rigid bodies are connected by smooth pins, weightless links or inextensible cables, or in mesh with one another, the entire system rather than separate bodies should be considered to avoid the work of internal forces at the connections. 3. Calculate the total work W1!2 by adding algebraically the work of all the forces and couples. 4. Apply T1 þ W1!2 ¼ T2 to solve for the unknown. Example 9.8. Block A with a mass of 100 kg and block B with a mass of 200 kg are joined by an inextensible cable, figure 9.26a. The coefficient of kinetic friction between block A and the plane is 0.2. If the system is released from rest, determine the velocity of block A after it has moved to the right 1 m. Neglect the mass of the cable and the pulley.

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377

FIG. 9.26 – A pulley-cable system. Solution: Since the cable is inextensible, two blocks should have the same speed at any instant. Analyze the two blocks separately. The free-body diagram of block A is shown in figure 9.26b, in which FA ¼ lk NA ¼ lk mA g ¼ 0:2  100  9:81 ¼ 196:2 N Using the principle of work and energy for block A, we have T1 þ W1!2 ¼ T2 ;

1 0 þ FT  1  FA  1 ¼ mA v 2 2

ð1Þ

The free-body diagram of block B is shown in figure 9.26c. Using the principle of work and energy for block B, we have T1 þ W1!2 ¼ T2 ;

1 0 þ mB g  1  F T  1 ¼ mB v 2 2

ð2Þ

Adding equations (1) and (2), we can see that the work of the forces exerted by the cable on A (FT  1) and on B (FT  1) cancels out. This is why when solving problems using the principle of work and energy, it is generally better to analyze the system to include all the moving bodies, since you do not need to worry about the work of internal forces. Therefore, combining equations (1) and (2) or applying the principle of work and energy to the system of both blocks and the cable altogether, we get 1 1 0  F A  1 þ mB g  1 ¼ mA v 2 þ mB v 2 2 2

)

v ¼ 3:43 m/s

RETHINK: As discussed above, when using the principle of work and energy, it is generally much more convenient to analyze the system including everything that moves. However, when an internal force is to be determined, you need to isolate part of a system. For instance, if the force in the cable is needed in this problem, we have to dismember and analyze the blocks separately.

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Example 9.9. A 10-kg homogeneous disk shown in figure 9.27a is pin supported at its center and acted upon by a couple moment M. Determine the angular displacement it makes when its angular velocity reaches 10 rad/s starting from rest if (1) M ≡ 2 Nm; (2) M = (2θ + 0.5) Nm, where θ is in radians.

FIG. 9.27 – Determining the angular displacement due to a constant or variable couple moment.

Solution 1: Since the disk is initially at rest and then rotates about a fixed axis, T1 ¼ 0

1 1 1 2 2  10  0:4  102 ¼ 40 J T2 ¼ IO x2 ¼ 2 2 2 The free-body diagram of the disk is shown in figure 9.27b. Apparently, the pin reactions and the weight do no work, since they are not displaced. Only the couple moment M does work. (1) If M ≡ 2 Nm, the work done by the constant couple moment, as the disk rotates from rest through an angle h, is Z h Z h W1!2 ¼ M dh ¼ 2dh ¼ 2h 0

0

Apply the principle of work and energy, T1 þ W1!2 ¼ T2 ; T1 þ 2h ¼ T2

)

h ¼ 20 rad

(2) If M = (2θ + 0.5) Nm, where θ is in radians, the work done by the variable couple moment is

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379 Z

W1!2 ¼

h

Z M dh ¼

0

h

ð2h þ 0:5Þdh ¼ h2 þ 0:5h

0

From the principle of work and energy, T1 þ W1!2 ¼ T2 , we have T1 þ h2 þ 0:5h ¼ T2

)

h ¼ 6.08 rad

Note that a negative value of θ is abandoned due to its physical meaning. Solution 2: From the free-body diagram of the disk shown in figure 9.27b, write the rotational equation of motion,   X 1 2  10  0:4 a ¼ 0:8a ð þ yÞ MO ¼ IO a; M ¼ IO a ¼ 2

(1) If M ≡ 2 Nm, the angular acceleration is as follows: a¼

M 2 ¼ ¼ 2:5 rad/s2 0:8 0:8

Since a is constant, from kinematics, we have x2 ¼ x20 þ 2aðh  h0 Þ. Substituting the corresponding values yields 102 ¼ 0 þ 2  2:5  ðh  0Þ

)

h ¼ 20 rad

(2) If M = (2θ + 0.5) Nm, where θ is in radians, the angular acceleration is as follows: a¼

M 2h þ 0:5 ¼ ¼ ð2:5h þ 0:625Þrad/s2 0:8 0:8

a is variable and from kinematics, we have adh ¼ xdx. Since x ¼ 0 at h ¼ 0, integrating and solving yields Z h Z 10 ð2:5h þ 0:625Þdh ¼ xdx ) h ¼ 6.08 rad 0

0

Example 9.10. The 10-kg gear A has radius rA = 24 cm and radius of gyration kA = 15 cm; the 5-kg gear B has radius rB = 16 cm and radius of gyration kB = 10 cm. The system is at rest when a constant couple M = 2 Nm is applied to gear B, figure 9.28a. Determine (1) the number of revolutions gear B must make before its angular velocity reaches 200 rpm; (2) the average tangential force acting on gear A from gear B.

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FIG. 9.28 – The gear system. Solution: (1) From transmission ratio of gear mesh (figure 9.28b), we have hA xA rB 16 2 ¼ ¼ ¼ ¼ hB xB rA 24 3 xB ¼ 200 rpm ¼ 200

   rev 2p rad 1 minute ¼ 20.94 rad/s minute 1 rev 60 s

2 xA ¼ xB ¼ 13.96 rad/s 3 The mass moments of inertia of the two gears about their respective fixed axes are IA ¼ mA kA2 ¼ 10  0:152 ¼ 0:225 kgm2 IB ¼ mB kB2 ¼ 5  0:12 ¼ 0:05 kgm2 Then we can calculate the kinetic energy. The system is initially at rest, so T1 ¼ 0. For the final kinetic energy, 1 1 1 1 T2 ¼ IA x2A þ IB x2B ¼  0:225  13.962 þ  0:05  20.942 ¼ 32.89 J 2 2 2 2 During the motion, only the couple moment M does work. The reactions at pin A and pin B and the two weights of the two gears do no work, since they are not displaced. For the sake of brevity, they are not shown in figure 9.28b. When the angular displacement of gear B is hB , the work done by the constant couple moment M is Z hB Z hB W1!2 ¼ M dh ¼ 2dh ¼ 2hB 0

0

Apply the principle of work and energy,  T1 þ W1!2 ¼ T2 ;

0 þ 2hB ¼ 32.89

)

hB ¼ 16.45 rad = 16:45 rad

1 rev 2p rad

 ¼ 2.62 rev

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381

Here, the system of two gears is chosen to apply the principle of work and energy and then the work of internal forces between two gears cancels out and are not considered. However, when the tangential force acting on gear A from gear B (internal force) is needed, we have to isolate the system and analyze gear A separately. (2) The free-body diagram of gear A is shown in figure 9.28c. Only the tangential force Ft does work. W1!2 ¼ Ft s ¼ Ft ðhA rA Þ ¼ Ft ðhB rB Þ ¼ Ft  16:45  0:16 ¼ 2:632Ft Initially, gear A is at rest, so T1 ¼ 0. The final kinetic energy of gear A is 1 1 T2 ¼ IA x2A ¼  0:225  13.962 ¼ 21.92 J 2 2 Substituting these values into the principle of work and energy gives T1 þ W1!2 ¼ T2 ;

0 þ 2:632Ft ¼ 21:92

)

Ft ¼ 8:33 N

Example 9.11. The uniform disk has weight G and radius r. A spring of stiffness k is attached to it using a segment of rope around its periphery, as shown in figure 9.29a. The spring is initially unstretched. Thereafter the disk rolls from rest without slipping under the action of a constant couple moment M. Find velocity of its center C after C moves s. Neglect the mass of the spring and the rope.

FIG. 9.29 – A disk rolling without slipping under the action of a constant couple moment. Solution: The kinematic diagram of the disk is shown in figure 9.29b. Because the disk rolls without slipping, the point of contact with the ground is the instantaneous center of rotation (IC). Then vC ¼ xr, and when the center C moves s, the spring is stretched 2 s. The disk’s free-body diagram is shown in figure 9.29c. During the motion, weight G and normal force FN do no work since there is no displacement along their lines of action. Friction force Fs also does no work since at any instant its point of application is IC. Therefore, only couple moment M and spring force F do work. The total work is

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382

W1!2 ¼ M u þ

 k 2 s k Ms d1  d22 ¼ M  ð2s Þ2 ¼  2ks2 2 r 2 r

The disk is initially at rest, then T1 ¼ 0. After C moves s, the kinetic energy of the disk is 1 1 1G 2 1 Gr 2 vC 2 3 G 2 vC þ v T2 ¼ mvC2 þ IC x2 ¼ ¼ 2 2 2g 2 2g r 4g C Substituting to the principle of work and energy, T1 þ W1!2 ¼ T2 , yields 0þ

Ms 3G 2  2ks2 ¼ v r 4g C

sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi  ffi 4g Ms  2ks2 vC ¼ 3G r Example 9.12. The 100-kg drum shown in figure 9.30a has radius of gyration kO = 0.25 m. The 20-kg block is moving downward at 5 m/s when a force F = 500 N is applied to the brake arm. Neglect the weight of the brake arm and the size of the brake pad C. If the coefficient of kinetic friction at the brake pad C is 0.3, find the distance of the block before it stops from the instant the brake is applied.

FIG. 9.30 – The rotating drum and the brake.

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383

Solution 1: The free-body diagram of the brake arm is shown in figure 9.30b. X MO1 ¼ 0; F  ð0:8 þ 0:6Þ  FN  0:6 ¼ 0 ) FN ¼ 1167 N Fk ¼ lk FN ¼ 0:3  1167 ¼ 350 N The free-body and kinematic diagram of the drum along with the block is shown in figure 9.30c. x¼

vB 5 ¼ 25 rad=s; ¼ 0:2 0:2

IO ¼ 100  0:252

1 1 1 1 T1 ¼ mB vB2 þ IO x2 ¼  20  52 þ  100  0:252  252 ¼ 2203 J 2 2 2 2 Assuming that the block descends h before it stops, the total work during this motion is   h 3  0:3 ¼ 20  9:81  350  h ¼ 328:8h W1!2 ¼ mB gh  Fk  0:2 2 Utilize the principle of work and energy, T1 þ W1!2 ¼ T2 ;

2203  328:8h ¼ 0

)

h ¼ 6:7 m

Solution 2: As above, Fk ¼ 350 N is determined first. The free-body and kinematic diagram of the drum along with the block is shown in figure 9.30c. X X ð þ xÞ MO ¼ ðMk ÞO ; 0:2mB g  0:3Fk ¼ IO a þ 0:2mB aB Substituting the kinematic equation aB ¼ 0:2a and solving yields a ¼ 9:328 rad=s2 ;

aB ¼ 0:2  ð9:328Þ ¼ 1:866 m=s2

For block B with constant acceleration aB , we have 0  vB2 ¼ 2aB h

)

h ¼ 6:7 m

Solution 3: As above, Fk ¼ 350 N is determined first. The free-body and kinematic diagram of the drum is shown in figure 9.30d. X ð þ xÞ MO ¼ IO a; 0:2T  0:3Fk ¼ IO a The free-body and kinematic diagram of the block is shown in figure 9.30e. X ð þ "Þ Fyi ¼ may ; T  mB g ¼ mB aB ¼ mB ð0:2aÞ

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Eliminating T from the above two equations and solving yield a ¼ 9:328 rad=s2 ;

aB ¼ 0:2  ð9:328Þ ¼ 1:866 m=s2

For block B with constant acceleration aB , we have 0  vB2 ¼ 2aB h

)

h ¼ 6:7 m

RETHINK: The principle of work and energy in solution 1 directly relates force, displacement and velocity and make the determination of the acceleration unnecessary. Therefore solution 1 is much easier here than the other two solutions using the equations of motion. Compared to solution 3, solution 2 can eliminate the cable tension T from the analysis by considering the drum along with block as a single system. Example 9.13. Block B of weight P is attached to an inextensible cord, which is wrapped around disk A of weight P1 and radius r, figure 9.31a. The coefficient of kinetic friction between block B and the ground is μk. Initially the system is at rest and thereafter the disk is acted upon by a couple moment M ¼ M0 þ au þ bu2 . Here a and b are constant parameters and M0 is the initial couple moment that is big enough to overcome the friction at block B and cause the disk to rotate. Neglecting the mass of the cord, determine the angular velocity of disk A after it has turned two revolutions.

FIG. 9.31 – A couple moment varying with the angular displacement. Solution: The free-body diagram of the system is shown in figure 9.31b. The reactions at pin A and weight P1 do no work, since they are not displaced. Normal force FN and weight P do no work since they act in a direction perpendicular to the displacement. When the disk has turned two revolutions, the total work of the couple moment and the kinetic friction force is as follows: Z 4p ðM 0 + au þ bu2 ) du  Fk ðr  4pÞ W1!2 ¼ WM þ WFk ¼ 0

4p ð3M0 þ 6ap þ 16bp2 Þ  4rlk Pp ¼ 3 4p ð3M0 þ 6ap þ 16bp2  3rlk PÞ ¼ 3

Kinetics: Work and Energy

385

The system is initially at rest, so T1 ¼ 0. For the final kinetic energy,   1 1 1 P1 r 2 2 1 P P1 þ 2P 2 2 r x T2 ¼ IA x2 þ mB vB2 ¼ x þ  ðxr Þ2 ¼ 2 2 2 2g 2 g 4g From the principle of work and energy, T1 þ W1!2 ¼ T2 , we get 0þ

4p P1 þ 2P 2 2 ð3M0 þ 6ap þ 16bp2  3rlk PÞ ¼ r x 3 4g sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 16pgð3M0 þ 6ap þ 16bp2  3rlk PÞ x¼ 3ðP1 þ 2P Þr 2

9.4 9.4.1

Conservative Forces and Conservation of Energy Conservative Forces

In § 9.1, it has been derived that the work of a variable force is obtained by a curvilinear integral, Z M2 Z M2 Fdr ¼ ðFx dx þ Fy dy þ Fz dzÞ W1!2 ¼ M1

M1

This curvilinear integral is generally dependent on the path of the force’s point of application, as shown in figure 9.32a. However, there are some forces whose work only depends upon the initial and final positions, and is independent of the path. This kind of force is called conservative force or path-independent force. As stated in § 9.1, the work of a weight is independent of the path and depends only on the altitude difference of the center of gravity of the object at two positions; the work of a spring force is just dependent on the initial and final deflections of the spring; the work of a gravitational force depends only upon the initial and final distances between the two particles. They are examples of conservative forces. In contrast, the work done by a kinetic friction force depends on the path—the longer the path is, the greater the work is. So friction forces are nonconservative. Generally, some of the work done by them would dissipate in the form of heat. Because the work of a conservative force only depends on the initial and final positions, we can write the work of a conservative force as W1!2 ¼ V1 ðx1 ; y1 ; z1 Þ  V2 ðx2 ; y2 ; z2 Þ or for short, W1!2 ¼ V1  V2

ð9:27Þ

Engineering Mechanics

386

Note that if position M2 coincides with M1 , in other words, the particle moves a closed path, as shown in figure 9.32b, we have V1 ¼ V2 and the work is zero. This may be expressed mathematically as aFdr ¼ 0

ð9:28Þ

FIG. 9.32 – Path-dependent or path-independent force. From equation (9.27), it is clear that V1 and V2 have the same unit with the work and energy, i.e., joules (J) in SI unit system. V ðx; y; zÞ is then referred to as the potential energy or potential function. Applying equation (9.27) between two neighboring points M ðx; y; zÞ and M 0 ðx þ dx; y þ dy; z þ dzÞ, the elementary work of a conservative force corresponding to the infinitesimal displacement dr is dW ¼ V ðx; y; zÞ  V ðx þ dx; y þ dy; z þ dzÞ ¼ dV

ð9:29Þ

Thus, the elementary work of a conservative force is the exact differential of its potential function. Recalling that dW ¼ Fx dx þ Fy dy þ Fz dz and the definition of the exact differential, we have   @V @V @V dx þ dy þ dz Fx dx þ Fy dy þ Fz dz ¼  @x @y @z This equation is satisfied provided Fx ¼  Therefore,

@V ; @x

Fy ¼ 

@V ; @y

Fz ¼ 

@V @z

  @V @V @V iþ jþ k F ¼ Fx i þ Fy j þ Fz k ¼  @x @y @z

ð9:30Þ

ð9:31Þ

Kinetics: Work and Energy

387

Equation (9.31) relates a conservative force to its potential function. It can also serve as a mathematical criterion to judge whether a force is conservative or not.

9.4.2

Potential Energy

1. Gravitational potential energy From equation (9.5), the work done by weight P is W1!2 ¼ PDh ¼ Pðz1  z2 Þ ¼ Pz1  Pz2 Compared with equation (9.27), we know that the potential energy with respect to the weight, which is a special case of gravitational force, is Vg ¼ Pz

ð9:32Þ

where z is measured from a selected horizontal datum at which the potential energy is considered as zero. If a particle (or the center of gravity of a body) is located a distance above the datum (+z), as shown in figure 9.33, it has positive gravitational potential energy Vg, since its weight P has the capacity of doing positive work when the particle moves from its position to the datum. Likewise, if the particle is located below the datum (−z), Vg is negative. It should be noted that equation (9.32) for the gravitational potential energy is valid only when the weight P can be assumed constant, i.e., only when the particle is near the surface of the earth. For a space vehicle, however, the general gravitational force considering the variation of the force magnitude with the distance r from the center of the earth (figure 9.6) should be used. From equation (9.8), the work of a general gravitational force is       1 1 fms m fms m W1!2 ¼ fms m  ¼    r2 r1 r1 r2 Compared with equation (9.27), we know that the potential energy Vg with respect to a general gravitational force is Vg ¼ 

fms m r

2. Elastic potential energy From equation (9.6), the work of a linear elastic spring force is k k k W1!2 ¼ ðd21  d22 Þ ¼ d21  d22 2 2 2

ð9:33Þ

Engineering Mechanics

388

Compared with equation (9.27), we know that the elastic potential energy is k Ve ¼ d2 2

ð9:34Þ

Since d is squared, no matter the spring is elongated or compressed, Ve is always positive, as shown in figure 9.34. It means at the deformed position, the spring force always has the “potential” to do positive work on the particle when the spring is moved back to its unstretched position.

FIG. 9.33 – Gravitational potential energy.

FIG. 9.34 – Elastic potential energy.

9.4.3

Principle of Conservation of Energy

As mentioned above, the work of a conservative force, such as the weight or the spring force, can be expressed as a change in potential energy, i.e., W1!2 ¼ V1  V2 . If only conservative forces do work when a particle or body moves, the principle of work and energy, T1 þ W1!2 ¼ T2 , can be expressed in a modified form as T1 þ V1  V2 ¼ T2 or T1 þ V1 ¼ T2 þ V2

ð9:35Þ

Kinetics: Work and Energy

389

This is the principle of conservation of (mechanical) energy. It states that if only conservative forces do work when a particle or body moves, the sum of its kinetic and potential energies remains constant. It can also apply to a system of rigid bodies connected by smooth pins, inextensible cords or weightless links, or in mesh with one another. In all these cases the internal forces at the connections do no work and do not need to be considered when analyzing the whole system. The sum T þ V is referred to as the mechanical energy. We have discussed two types of potential energy: gravitational potential energy, Vg , and elastic potential energy, Ve . So, equation (9.35) can also be rewritten as T1 þ Vg1 þ Ve1 ¼ T2 þ Vg2 þ Ve2 Procedure for analysis Like the principle of work and energy from which it is derived, the principle of conservation of energy is convenient to use for solving kinetic problems involving force, displacement and velocity. However, it applies to problems in which only conservative forces do work. The following procedure for analysis is suggested. 1. Establish an inertial coordinate system and calculate the kinetic energy T for both initial and final positions. 2. Draw the free-body diagram to make sure that only conservative forces do work. 3. To calculate the gravitational potential energy Vg ¼ Pz, be sure to establish a fixed horizontal datum. The elastic potential energy, Ve ¼ k2 d2 , is always positive no matter the spring is elongated or compressed. The potential energy is then determined from V ¼ Vg þ Ve for both initial and final positions. 4. Apply T1 þ V1 ¼ T2 þ V2 to solve for the unknown. Example 9.14. A 1-kg block is attached to a spring with stiffness k = 10 N/m and can move in a vertical smooth slot, figure 9.35a. If the block is released from rest at position A, where the spring is unstretched, determine the speed of the block when y = 1 m.

FIG. 9.35 – A block attached to a spring.

Engineering Mechanics

390

Solution 1: The block’s free-body diagram at any arbitrary position is shown in figure 9.35b. Normal force N does no work and only the weight and the spring force does work. Therefore, the conservation of energy can be used to solve this problem. Set the datum at position A for the gravitational potential energy, then VgA ¼ 0 and VgB ¼ mgy ¼ 9:81 J. Since at position A, the block is at rest and the spring is unstretched, we have TA ¼ 0 and VeA ¼ 0. For position B, pffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 1 1 1 12 þ 12  1 ¼ 0:857 J TB ¼ mvB2 ; VeB ¼ kd2B ¼  10  2 2 2 From the conservation of energy, TA þ VA ¼ TB þ VB , we have 1 0 þ ð0 þ 0Þ ¼ mvB2 þ ð0:857  9:81Þ 2

)

vB ¼ 4:23 m/s

Solution 2: From the block’s free-body diagram, figure 9.35b, the weight and the spring force do work when the block moves from position A to position B. The total work is    1  1 W1!2 ¼ mgy þ k d2A  d2B ¼ 1  9:81  1 þ  10  0  0:4142 ¼ 8:953 J 2 2 Since the block is at rest at position A, we have TA ¼ 0. For position B, 1 TB ¼ mvB2 2 Apply the principle of work and energy, T1 þ W1!2 ¼ T2 ;

1 0 þ 8:953 ¼ mvB2 2

)

vB ¼ 4:23 m/s

Solution 3: From the block’s free-body diagram, figure 9.35b, write the equation of motion in the y direction, X ð þ #Þ Fyi ¼ may ; mg  F sin h ¼ ma From the trigonometry, we have pffiffiffiffiffiffiffiffiffiffiffiffiffi  F ¼ kd ¼ 10 y2 þ 1  1 ;

y sin h ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffi y2 þ 1

Substituting these expressions into the equation of motion and simplifying gives 10y a ¼ 9:81  10y þ pffiffiffiffiffiffiffiffiffiffiffiffiffi y2 þ 1 Since v ¼ 0 at y ¼ 0 and ady ¼ vdv, integrating both sides we have ! Z 1 Z vB 10y 9:81  10y þ pffiffiffiffiffiffiffiffiffiffiffiffiffi dy ¼ vdv ) vB ¼ 4:23 m/s y2 þ 1 0 0

Kinetics: Work and Energy

391

RETHINK: It can be seen that when using the equation of motion, the variation in both the magnitude and direction of the spring force during the motion must be considered. However, when using the work-energy methods, the calculations depend only on data calculated at the initial and final positions, i.e., VeB ¼ 12 kd2B and W1!2 ¼ k2 ðd2A  d2B Þ. So, the work-energy methods are apparently convenient to use here. Example 9.15. The uniform rod AB of weight P and length l is held in horizontal position by two weightless links at A and a string at B, figure 9.36a. Then the string is cut and the rod begins to rotate. Determine the forces in the links when the rod turns an arbitrary angle φ.

FIG. 9.36 – The forces in the two links caused by the rotating rod.

Solution: The free-body diagram of the rod at an arbitrary angle φ is shown in figure 9.36b. The reactions F1 and F2 do no work, since they are not displaced. Only weight P does work. Therefore, the conservation of energy can be used to solve this problem. Set the datum through O1 A for the gravitational potential energy. Then V1 ¼ 0;

l V2 ¼ P sin u 2

For the arbitrary position at φ, assume the rod has an angular velocity x ¼ u_ and € . The initial and final kinetic energies of the rod are an angular acceleration a ¼ u T1 ¼ 0;

1 1 Pl 2 P  x2 ¼ l 2 x2 T2 ¼ IA x2 ¼  2 2 3g 6g

From the conservation of energy, T1 þ V1 ¼ T2 þ V2 , we have 0þ0 ¼

P 2 2 l l x  P sin u 6g 2

)

x2 ¼

3g sin u l

Engineering Mechanics

392

Taking the first time derivative yields 2xa ¼

3g cos ux l

)

a¼

3g cos u 2l

The kinetic diagram of the rod at an arbitrary angle φ is shown in figure 9.36c, in which     P al 3P cos u P x2 l 3P sin u maCt ¼ ; maCn ¼ ¼ ¼ g 2 4 g 2 2 Write the translational equations of motion for the rod: X ð þ "Þ Fyi ¼ maCy ; F2 sin 30  P ¼ maCt cos u þ maCn sin u ð þ !Þ

X

Fxi ¼ maCx ;

F1  F2 cos 30 ¼ maCt sin u  maCn cos u

ð1Þ ð2Þ

Solving these two equations gives F2 ¼ 5P 

F1 ¼ 

9P cos2 u 2

pffiffiffi pffiffiffi 5 3 9 3 9 Pþ P cos2 u þ P sin u cos u 4 2 4

RETHINK: This problem hasPbeen solved in example 8.17, where the rotational P equation of motion, MA ¼ ðMk ÞA , is used to obtain the angular acceleration a at an arbitrary angle u and then the kinematic equation adu ¼ xdx is integrated to get the angular velocity x. Here the conservation of energy is utilized to obtain x at an arbitrary angle u and then taking the time derivative to get a. However, to calculate the reactions F1 and F2 , the translational equations of motion have to be used in both examples. PROBLEMS 9.1 A 20-kg crate is moving on the ground under the action of constant force F. It makes angle h ¼ 30 with the horizontal and its magnitude is F ¼ 98 N. The coefficient of kinetic friction is lk ¼ 0:2. Determine the work done by each force acting on the crate when the crate has a displacement of s ¼ 6 m. 9.2 A crate is moving on a smooth horizontal plane under the action of force F. This force always makes angle h ¼ 30 with the horizontal. Its magnitude is F ¼ ð0:1 þ 4s ÞN, where s is the displacement the crate travels on the plane, measured in meters. Determine the work done by force F when the crate has a displacement s ¼ 6 m.

Kinetics: Work and Energy

393

Prob. 9.1/2

9.3 Blocks A and B are connected by a cord which passes over pulley O of radius r ¼ 0:5 m. Neglect the mass of the cord and the pulley. There is no slipping between the cord and the pulley. Starting from rest, the pulley begins to rotate under the action of a couple moment M ¼ ð4uÞNm, where u is the angular displacement measured in radians. Determine the total work after pulley O has one full turn.

Prob. 9.3

9.4 A block of weight P is released from rest from position M on the incline. It moves down the inclined plane and then along the horizontal ground and finally stops at position M 0 as shown. The coefficient of kinetic friction at all contacting surfaces is μk. Express the distance s in terms of h, μk and θ.

Engineering Mechanics

394

Prob. 9.4

9.5 The spring has stiffness k and undeformed length l0 ¼ a. Determine the work of the spring force when the particle attached to the spring moves from position A to position B, and the work when the particle moves from position B to position C.

Prob. 9.5

9.6 A block of weight G is attached to a spring and rests on the smooth incline. The spring constant is k. Due to an instantaneous disturbance, the block moves down the plane a distance l from the equilibrium position. Determine the total work done by all the forces acting on the block during the motion.

Kinetics: Work and Energy

395

Prob. 9.6

9.7 The bulldozer engine is generating a constant power of 55 kW while its efficiency is e ¼ 0:8. Determine the average resistance acting on the bulldozer when it is moving forward with a constant velocity of 0.5 m/s. 9.8 Composite body OAB consists of two uniform rod segments, each has weight P and length l. The composite body is rotating about vertical axis z with angular velocity x. Determine its kinetic energy.

Prob. 9.8

Engineering Mechanics

396

9.9 Uniform bar AB has weight G and length l. A small ball of weight G is fixed to the bar at A. At a given instant, the bar rotates about O with an angular velocity ω and angular acceleration α as shown. Determine the system’s kinetic energy T.

Prob. 9.9

9.10 Uniform bar OA with weight P and length l is pin-connected to a vertical weightless rod, which is rotating about axis z with an angular velocity x. If the bar makes angle θ with the rotating axis, determine the kinetic energy of the bar.

Prob. 9.10

9.11 Load A of weight P is suspended from an inextensible cord, which passes over pulley O and is attached to the center of disk B. A spring with stiffness k is also attached to the center of disk B. Disk B has weight P1 and radius R and pulley O has

Kinetics: Work and Energy

397

weight P2 and radius r. They are treated as uniform disks. When load A moves downward with velocity v, pulley O rotates and disk B rolls without slipping. Determine the kinetic energy of the system.

Prob. 9.11

9.12 An inextensible cord of negligible mass is wrapped around two uniform disks. Each of them has weight P and radius R. At a given instant, the angular velocities of two disks are xA and xB , respectively. Determine the kinetic energy of the system for this instant.

Prob. 9.12

9.13 For the spring device shown, the spring constant is 2 N/cm and the undeformed length is 20 cm. If the spring is pushed back 10 cm and a 0.3-N ball is shot from rest, determine the velocity when the ball is leaving the ejection chute. Neglect the mass of the spring.

Engineering Mechanics

398

Prob. 9.13

9.14 A jeep is traveling at a speed of 20 km/h when the driver applies the brakes. Thereafter the jeep skids 5 m before stopping. If the jeep is traveling at 70 km/h when the brakes are applied, determine the distance the jeep would skid before stopping. 9.15 At a given instant, a 20-kg bin is moving to the right with a speed of 8 m/s. Thereafter a constant force F = 100 N as shown acts on it. Determine its speed when the bin has moved to the right a distance s = 10 m from this instant. The coefficient of kinetic friction between the bin and the ground is μk = 0.25.

Prob. 9.15

9.16 A nonlinear spring serving as a buffer is installed on the 4500-kg train car. The force of the spring on the train car is F ¼ ðks2 Þ N, where s is the spring deflection in meters. Determine constant k so that the maximum deflection of the spring is less than 0.18 m when the train car, traveling at 3 m/s, strikes the rigid stop. Treat the train car as a particle.

Prob. 9.16

Kinetics: Work and Energy

399

9.17 In the pulley-cord system, two blocks A and B have masses mA = 100 kg and mB = 15 kg, respectively. The coefficient of kinetic friction between the incline and block A is 0.25. The system is released from rest. Find the speed of A after it moves 5 m down the incline. Neglect the mass of pulleys and cords.

Prob. 9.17

9.18 Motor M is used to pull upward a 600-kg cargo starting from rest with a constant acceleration a0 = 1 m/s2 as shown. Neglecting the mass of pulleys and cord, determine the output power of motor M when t = 4.5 s.

Prob. 9.18

Engineering Mechanics

400

9.19 Starting from rest, a 2000-kg car moves along a horizontal straight road with a constant power of 200 kW. Find the distance it travels before reaching a speed of 25 m/s. 9.20 Starting from rest at O, the collar with a mass of 5 kg is lifted along the smooth vertical rod by applying a constant force F = 100 N to the weightless cord. Determine the power developed by the force when the collar moves to position A. Neglect the weight of the pulley.

Prob. 9.20

9.21 The homogeneous bar AB with weight G1 ¼ 60 N and length l ¼ 24 cm is pin-connected at B with a homogeneous disk. The disk has weight G2 ¼ 150 N and radius r = 8 cm. If the system is released from rest from the horizontal position, determine the velocity of point B at its lowest position.

Prob. 9.21

Kinetics: Work and Energy

401

9.22 A small ball of weight P is suspended by an inextensible cord of length l. The ball is released from rest when the cord makes angle h with the vertical line. Then the cord touches fixed peg O1 . OO1 ¼ h and OO1 makes angle b with the vertical line. Determine the smallest value of h for which the ball will describe a circle about the peg. Neglect the size of the ball and the mass of the cord. 9.23 A small ball of weight P is suspended by an inextensible cord of length l. The ball is released from rest when the cord makes an angle h with the vertical line. Then the cord touches the fixed peg O1 . OO1 ¼ h and OO1 makes an angle b with the vertical line. Determine the cord force just before and after the cord contacts the peg O1 . Neglect the size of the ball and the mass of the cord.

Prob. 9.22/23

9.24 The 40-kg square homogeneous plate is held in the vertical plane by three cords at θ = 60°. Then the cord FG is cut and the plate begins to move. Determine the acceleration of the plate center C and tensions in the cords AD and BE when θ = 90°. The side length of the plate is 10 cm.

Prob. 9.24

Engineering Mechanics

402

9.25 The uniform bar OA of length l = 3.27 m can rotate about an axis perpendicular to the page and passing through O. Determine initial angular velocity x0 to make sure that the bar can rotate from the vertical position to the horizontal position.

Prob. 9.25

9.26 A block having mass m2 is placed on an incline and attached to an inextensible cord, which is wrapped around a drum of mass m1 . The drum can be treated as a uniform disk and its radius is r. The coefficient of kinetic friction between the block and the incline is μk. Initially the system is at rest. Thereafter the drum is acted upon by a constant couple moment M and begins to rotate. Determine the angular velocity and angular acceleration of the drum after it has turned an angle u.

Prob. 9.26

9.27 The spool has a mass of 30 kg, radius of gyration kO = 0.4 m, and radius r = 0.3 m for the inner hub. If the system is released from rest, determine the distance the 10-kg block A must fall such that the spool attains an angular velocity x = 10 rad/s. Also find the cord force during the motion.

Kinetics: Work and Energy

403

Prob. 9.27

9.28 A uniform disk of mass m and radius r is released from rest. If the disk rolls without slipping, determine its angular velocity after 2 full turns.

Prob. 9.28

9.29 The uniform disk has a mass of 20 kg and radius r = 0.2 m. It is originally held in equilibrium by a spring of stiffness k = 100 N/m. A constant couple moment M = 18 N·m is then applied to the disk to make it roll without slipping. How far does the center of the disk travel along the incline before it stops?

Engineering Mechanics

404

Prob. 9.29

9.30 Constant couple moment M is acted on drum O to pull spool C up the incline. Both drum O and spool C can be treated as uniform disk. If spool C is rolling without slipping, determine the angular acceleration of the drum O and the cord force.

Prob. 9.30

9.31 In the gear system as shown, a constant couple moment M is acted on gear I to hoist a load D of weight P. Gear II and drum III are rigidly connected to the same shaft and turns with it. Gear I, gear II and drum III are all treated as uniform disk. Their radii are r1 , r2 and r3 , respectively, and r1 ¼ r3 ¼ r2 =2. Their weights are P1 , P2 and P3 , respectively. Determine the acceleration of load D and the average tangential force that gear II exerts on gear I.

Kinetics: Work and Energy

405

Prob. 9.31

9.32 Two identical bars are pin-connected at the ends. They are released from rest at position θ = 60 . Each bar has mass m = 10 kg and length l = 1 m. Determine their angular velocities at the instant they become horizontal. Neglect the friction.

Prob. 9.32

9.33 The block weighs 200 N and is released from rest from h = 12 cm above the top of a board, which is attached to a spring. The spring has stiffness k = 1000 N/cm. Neglecting the mass of the board and the spring, determine the maximum compression dmax of the spring.

Engineering Mechanics

406

Prob. 9.33

9.34 A uniform disk and a rod AB is pin-connected at A. The system is released from rest when h ¼ 45 . If the disk rolls without slipping and the wall is smooth, determine the acceleration of point A at the initial instant.

Prob. 9.34

Kinetics: Work and Energy

407

9.35 Load A of weight P is suspended to an inextensible cord, which passes over pulley O and is attached to the center of disk B. Disk B has weight P1 and radius R and pulley O has weight P2 and radius r. They are treated as uniform disks. The system is released from rest and then load A moves downward, pulley O rotates and disk B rolls without slipping. The weightless cord does not slip on the pulley. Determine the velocity and acceleration of load A when it has moved downward a displacement x.

Prob. 9.35

9.36 The ends of a 30-kg uniform rod AB are pin-connected to two weightless collars that can slide along smooth fixed rods. If the rod is released from rest at θ = 15°, determine the angular velocity of the rod when θ = 90°.

Prob. 9.36

Chapter 10 Kinetics: Impulse and Momentum Objectives
Derive and apply the principle of linear impulse and momentum to solve kinetic problems involving force, time and velocity.
Derive and apply the principle of angular impulse and momentum to solve kinetic problems involving force (couple), time and velocity (angular velocity).
Determine the linear and angular momentum of a system of particles, a rigid body or a system of rigid bodies undergoing planar motion.
Derive and apply conservation of momentum to solve kinetic problems. Chapter 8 relates force and acceleration through the equation of motion. Chapter 9 integrates the equation of motion in the tangential direction with respect to displacement and obtain the principle of work and energy, which directly relates force, displacement and velocity. This chapter will integrate the equation of motion with respect to time and obtain the principle of impulse and momentum, which directly relates force, time and velocity. Both of these principles make the determination of the acceleration unnecessary and provide an easier solution for some kinetic problems involving cumulative effect of force.

10.1

10.1.1

Principle of Linear Impulse and Momentum for a Particle Linear Momentum and Linear Impulse

Consider a particle of mass m that is acted upon by forces motion is X dv Fi ¼ ma ¼ m dt

P

Fi . Its equation of

DOI: 10.1051/978-2-7598-2901-9.c010 © Science Press, EDP Sciences, 2022

Engineering Mechanics

410

Rearranging the terms and integrating throughout the time period t1 to t2 gives Z v2 X Z t2 Fi dt ¼ m dv ¼ mv 2  mv1 ð10:1Þ t1

v1

where each of the two vectors in the form mv ¼ L is the particle’s linear momentum. Linear momentum vector L has the same direction as v because m is a positive scalar. Its unit in SI unit system is kgm/s. Rt The integral of the form t12 Fdt ¼ I 1!2 in equation (10.1) is referred to as the linear impulse of a force during the time period t1 to t2. The linear impulse is also a vector and it measures the cumulative effect of the force during the time interval considered. The calculation of the linear impulse can be divided into the following three cases. (1) For a force of constant magnitude and constant direction during the time interval, the linear impulse is represented by the vector Fðt2  t1 Þ, which has the same direction as F. (2) If the force is only constant in direction and its magnitude is expressed as a function of time, i.e., F ¼ F ðt Þ, the linear impulse has the same direction as F and its magnitude can be determined by the integral as follows: Z t2 I1!2 ¼ F ðt Þdt t1

Apparently, the magnitude of the linear impulse equals the shaded area under the curve of force versus time, as shown in figure 10.1. (3) If the force is variable in both the magnitude and direction, we have to resolve F into rectangular components and calculate the corresponding rectangular components of the linear impulse as follows: Z t2  Z t2  Z t2  Z t2 I 1!2 ¼ Fðt Þdt ¼ Fx ðt Þdt i þ Fy ðt Þdt j þ Fz ðt Þdt k ð10:2Þ t1

t1

t1

t1

FIG. 10.1 – Linear impulse equal to the shaded area.

Kinetics: Impulse and Momentum

411

The calculation of each component of the linear impulse is just the above-mentioned second case. It is clear the unit of the linear impulse is Ns in SI unit system. Since  that  1 Ns = 1 kgm/s2 s = 1 kgm/s, the linear impulse has the same unit as the linear

momentum and equation (10.1) is dimensionally homogeneous. Example 10.1. The magnitude of a horizontal force varies with time in the manner shown in figure 10.2. Determine the linear impulse due to the force.

FIG. 10.2 – Determining linear impulse of a variable force.

Solution: The direction of the force is always horizontal and only its magnitude varies with time. Then the direction of the linear impulse is also horizontal and its magnitude is calculated by the shaded area under the curve of force versus time. Z 0:7 1 F ðt Þdt ¼  0:7  2850 ¼ 997:5 Ns I1!2 ¼ 2 0

10.1.2

Principle of Linear Impulse and Momentum

Equation (10.1) can be rewritten in the form X Z t2 mv 1 þ Fi dt ¼ mv 2 t1

ð10:3Þ

This is the principle of linear impulse and momentum for a particle. It states that the initial linear momentum of the particle plus the sum of all the linear impulses applied to the particle during the time period t1 to t2 equals its final linear momentum. Figure 10.3, called the impulse-and-momentum diagram, is a pictorial representation of this principle. For a complicated problem, it is helpful to draw this diagram to present the problem clearly; however, only a free-body diagram is usually used for simple kinetic problems.

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FIG. 10.3 – The impulse-and-momentum diagram. From the derivation, it can be seen that the principle of linear impulse and momentum is simply an integration of the equation of motion with respect to time. It directly relates force, time and velocity. The principle of work and energy in chapter 9 is also an integration of the equation of motion in the tangential direction with respect to displacement. They both apply only in a Newtonian frame of reference. However, unlike the principle of work and energy, in which kinetic energy and work are scalar quantities, the principle of linear impulse and momentum is a vector equation. In the solution, it is usually replaced by the following three scalar equations. 9 X Z t2 > mv1x þ Fxi dt ¼ mv2x > > > > > t1 > > Z = X t2 ð10:4Þ mv1y þ Fyi dt ¼ mv2y > t1 > > > > X Z t2 > > mv1z þ Fzi dt ¼ mv2z > ; t1

Procedure for analysis The principle of linear impulse and momentum is suitable for solving kinetic problems that involve force, time and velocity. The procedure for analysis is as follows: (1) Establish an inertial coordinate system and analyze the initial and final velocities of the particle. Then calculate the corresponding linear momenta from L ¼ mv. (2) Draw the particle’s impulse-and-momentum diagram to show the initial and final linear momentum vectors, and all the liner impulses applied to the particle. The free-body diagram can also be used to show all the forces that produce impulses on the particle for a simple problem. (3) Apply the principle of linear impulse and momentum in vector form or scalar form to solve for the unknowns. Example 10.2. A variable force F = (150 + 10t) N, where t is in seconds, is applied to the cord to raise the 15-kg block starting from rest, figure 10.4a. Neglecting the mass of the pulley and cord, determine the velocity of the block in 5 s.

Kinetics: Impulse and Momentum

413

FIG. 10.4 – Raising the block by exerting a variable force. Solution 1: The free-body diagram of the block is shown in figure 10.4b. Apply the principle of linear impulse and momentum, Z 5 X Z t2 ð þ "Þ mv1y þ Fyi dt ¼ mv2y ; 15v1 þ ð150 þ 10t  15  9:81Þdt ¼ 15v2 t1

0

Since the block is at rest originally, v1 = 0. Substituting and solving yields v2 ¼ 9:28 m/s Solution 2: From the free-body diagram shown in figure 10.4b, write the equation of motion in the vertical direction, X ð þ "Þ Fyi ¼ may ; mg þ 150 þ 10t ¼ ma a ¼ 0:667t þ 0:19 The acceleration is variable with time. From the kinematic equation a ¼ dv=dt, we have Z v2 Z 5 dv ¼ ð0:667t þ 0:19Þ dt ) v2 ¼ 9:28 m/s 0

0

RETHINK: Unlike two steps in solution 2, the principle of linear impulse and momentum in solution 1 directly relates force, time and velocity and make the determination of the acceleration unnecessary. Therefore, solution 1 is much easier here than solution 2 using the equation of motion.

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Example 10.3. A 2-kg particle originally has a velocity v1 = {3i + 2j − 6k} m/s and thereafter is acted upon by forces F1 = {5i − 2tj + tk}N and F2 = {−t2k}N, where t is in seconds. Determine its velocity after 5 s. Solution: Since the weight is mg ¼ f19:62kgN, the resultant force acting on the particle is X F¼ Fi ¼ ð5i  2tj þ tkÞ þ ðt 2 kÞ þ ð19:62kÞ ¼ 5i  2tj þ ðt  t 2  19:62Þk Apply the principle of linear impulse and momentum in x, y, z directions, respectively, Z 5 XZ mv1x þ Fxi dt ¼ mv2x ; 2  3 þ 5dt ¼ 2v2x ) v2x ¼ 15:5 m/s 0

mv1y þ

XZ

Z Fyi dt ¼ mv2y ;

2  2þ

5

ð2t Þdt ¼ 2v2y

)

v2y ¼ 10:5 m/s

0

mv1z þ

XZ

Z Fzi dt ¼ mv2z ;

5

2  ð6Þ þ

ðt  t 2  19:62Þdt ¼ 2v2z

)

0

v2z ¼ 69:63 m/s So, the velocity of the particle after 5 s is

v ¼ f15:5i  10:5j  69:63kg m/s Example 10.4. The 58-g tennis ball has a speed v1 = 30 m/s at an angle of 20° from the horizontal when it is struck by the racket, figure 10.5a. Just after the strike it flies away at an angle of 30° and then reaches a maximum altitude of 12 m, measured from the striking position. Determine the magnitude of the linear impulse of the racket on the ball. Neglect the weight of the ball during the strike and neglect the air resistance during the flight.

FIG. 10.5 – The tennis ball’s strike and flying.

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415

Solution: The impulse-and-momentum diagram of the ball during the strike is shown in figure 10.5b. Applying the principle of linear impulse and momentum along the x, y axes yields XZ ð þ !Þ mv1x þ ð1Þ Fxi dt ¼ mv2x ; m  30 cos 20 þ Ix ¼ mv2 cos 30

ð þ "Þ mv1y þ

XZ

Fyi dt ¼ mv2y ;

m  30 sin 20 þ Iy ¼ mv2 sin 30

ð2Þ

There are three unknowns in the above equations, i.e., Ix , Iy and v2 ; v2 has to be determined to obtain the linear impulse. The impulse-and-momentum diagram of the ball during the flight until it reaches the maximum altitude is shown in figure 10.5c. At the maximum-altitude position, the direction of the velocity of the ball is horizontal. XZ Fxi dt ¼ mv3x ; mv2 cos 30 þ 0 ¼ mv3 ) v3 ¼ 0:866v2 ð þ !Þ mv2x þ Because flight time t is unknown and the altitude measured from the striking position is given, the theorem of conservation of energy will be used to determine velocity v2. Set the datum at the striking position, thus 1 1 T2 þ V2 ¼ T3 þ V3 ; mv22 þ 0 ¼ m ð0:866v2 Þ2 þ mg  12 2 2 v2 ¼ 30:69 m/s Substituting this into equations (1) and (2) yields Ix ¼ 3:177 Ns; Iy ¼ 1:485 Ns The magnitude of the linear impulse is qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi I ¼ Ix2 þ Iy2 ¼ 3:1772 þ 1:4852 ¼ 3:51 Ns RETHINK: If the strike takes place in 0.1 s, the average force of the racket on the ball would be 35.1 N. During this very short time of strike, this force changes the ball’s momentum drastically and this kind of force is called an impulsive force (Also see §10.4). By comparison, the ball’s weight (0.058*9.81 = 0.569 N) has a negligible effect on the change in momentum during the very short time of strike, and it can be considered nonimpulsive and neglected during the strike. However, during the much longer time of flight after the strike, the impulse of the ball’s weight becomes the fundamental reason to cause the change in the momentum of the ball. Therefore, the weight cannot be neglected during the flight. However, the relatively small air resistance can still be neglected.

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Example 10.5. In the pulley-cord system shown in figure 10.6a, two blocks have masses of mA = 50 kg and mB = 10 kg. The coefficient of kinetic friction between block B and the horizontal ground is 0.3. Neglect the mass of the pulleys and cords. If block A is moving downward with a speed of 3 m/s at a given instant, determine its speed 10 s later.

FIG. 10.6 – The pulley-cord system. Solution: From the dependent motion analysis, we have 2sA þ sB ¼ l;

2vA þ vB ¼ 0 ) vB ¼ 2vA

The negative sign means that when block A has a downward velocity, i.e., in the direction of +sA , it causes a corresponding rightward velocity of block B; i.e., B moves in the direction of –sB . Then when block A is moving downward with a velocity vA1 ¼ 3 m/s, we have vB1 ¼ 6 m/s. The free-body diagram of block B is shown in figure 10.6b. Apply the principle of linear impulse and momentum for block B, Z 10 Z 10 XZ Fi dt ¼ mv2 ; mB vB1  T dt þ FB dt ¼ mB vB2 ðþ Þ mv1 þ 0

Z 10  ð6Þ 

10

0

T dt þ ð10  9:81  0:3Þ  10 ¼ 10vB2

0

Z

10

T dt ¼ 234:3  10vB2 ¼ 234:3  10ð2vA2 Þ

ð1Þ

0

Apply the principle of linear impulse and momentum for block A, figure 10.6c, Z 10 Z 10 XZ ð þ #Þ mv1 þ 2T dt þ mA gdt ¼ mA vA2 Fi dt ¼ mv2 ; mA vA1  0

0

Kinetics: Impulse and Momentum Z 50  3  2

10

417

T dt þ ð50  9:81Þ  10 ¼ 50vA2

0

Substituting equation (1) into the above equation and solving yields vA2 ¼ 50:96 m/s RETHINK: This problem has also been solved in example 8.2, where the equation of motion is used for blocks A and B, respectively, to obtain acceleration aA and then the kinematic equation vA2 ¼ vA1 þ aA t is utilized to get vA2 . Meanwhile, It should be noted that we assume unknown aA and v A2 in the direction of positive sA (downward) and unknown aB and v B2 in the direction of positive sB (to the right). This is necessary to seek a simultaneous solution.

10.2

10.2.1

Principle of Angular Impulse and Momentum for a Particle Angular Momentum

At a given instant, particle P of mass m is moving along a curved path with velocity v with respect to a Newtonian frame of reference Oxyz, figure 10.7. The linear momentum of particle mv is tangent to the path. The moment about O of vector mv is called the moment of momentum, or the angular momentum, of the particle about O at this instant. It is usually denoted by H O. Recalling the definition of the moment of a force, we have H O ¼ r  mv

ð10:5Þ

where r is the position vector from point O to the particle. The direction of H O is defined by the right-hand rule and is perpendicular to the shaded plane containing r and mv, as shown in figure 10.7. The magnitude of H O is HO ¼ rmv sin u ¼ mvr sin u ¼ mvd

ð10:6Þ

where d is the moment arm or perpendicular distance from O to the line of mv. It is clear that the unit of H O in SI unit system is kgm2 /s. From the cross-product operation, the angular momentum H O can also be determined by evaluating the determinant:    i j k   y z  H O ¼ r  mv ¼  x  mvx mvy mvz  ¼ ðmyvz  mzvy Þi þ ðmzvx  mxvz Þj þ ðmxvy  myvx Þk

ð10:7Þ

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FIG. 10.7 – Angular momentum. Example 10.6. A particle of mass m is moving in the x–y plane. Its position is defined by x ¼ a cosðpt Þ and y ¼ b sinðpt Þ, where a, b and p are constant parameters. Determine angular momentum H O of the particle about point O. Solution: From its position function, the velocity is vx ¼ x_ ¼ ap sinðpt Þ; vy ¼ y_ ¼ bp cosðpt Þ From equation (10.7), the angular momentum of the particle about point O is      i j k   i j k   y z  ¼  a cosðpt Þ b sinðpt Þ 0  H O ¼  x  mvx mvy mvz   map sinðpt Þ mbp cosðpt Þ 0   ¼ mabp cos2 ðpt Þ þ mabp sin2 ðpt Þ k ¼ ðmabpÞk From the calculation, it can be seen that for a particle moving in the x–y plane, where z  0 and vz  0, H O only has the k component. It means the angular momentum is always perpendicular to the x–y plane, along the þ z axis or z axis. Thus, the angular momentum can be represented by a scalar for planar motion.

10.2.2

Principle of Angular Impulse and Momentum

P _ we have From the equation of motion of a particle, Fi ¼ ma ¼ m v, X r Fi ¼ r  m v_

ð10:8Þ

where r is the position vector from point O to the particle, figure 10.8. The term on the left side is the sum of the moments of all the forces acting on the particle about P point O and can be denoted by M O. For a particle with constant mass, we have _ O ¼ d ðr  mv Þ ¼ r_  mv þ r  m v_ ¼ v  mv þ r  m v_ H dt Since the cross product of a vector with itself is zero, v  mv ¼ m ðv  v Þ ¼ 0. Therefore, equation (10.8) can be rewritten as

Kinetics: Impulse and Momentum X

419

M O ¼ H_ O ¼

dH O dt

ð10:9Þ

Rearranging the terms and integrating throughout the time period t1 to t2 gives Z H O2 X Z t2 M O dt ¼ dH O ¼ H O2  H O1 ð10:10Þ t1

R t2

H O1

where the integral of the form t1 M O dt, is a vector known as the angular impulse, which is determined by integrating the moment of a force acting on the particle about O over the time period t1 to t2. Since the moment of a force about point O is M O ¼ r  F, the angular impulse can be expressed in vector cross product form as Z t2 Z t2 M O dt ¼ ðr  FÞdt ð10:11Þ t1

t1

Equation (10.10) can be rewritten in the form X Z t2 H O1 þ M O dt ¼ H O2 t1

ð10:12Þ

This is the principle of angular impulse and momentum for a particle. It states that the initial angular momentum about O of the particle plus the sum of all the angular impulses about O applied to the particle from t1 to t2 equals its final angular momentum. Note that point O is the origin of the inertial coordinate system. In application, it is usually chosen as a fixed point.

FIG. 10.8 – Derivation of the principle of angular impulse and momentum for a particle. Example 10.7. The 2-kg ball B of negligible size is attached to the end of a weightless rod, figure 10.9a. The rod can rotate about the axis z due to the bearings at O and A. Determine the speed of the ball in 3 s starting from rest if the rod is subjected to a torque (1) M ¼ 10 Nm; (2) M ¼ ð5 þ t 2 Þ Nm, where t is in seconds.

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420

FIG. 10.9 – The ball’s rotation under the action of a constant or variable couple moment.

Solution 1: Consider the system of both the rod and the ball and draw the free-body diagram shown in figure 10.9b. When applying the principle of angular impulses and momentum about the z axis, the reaction forces at bearings and the weight of ball A can both be eliminated from the analysis, because they are passing through or parallel to the z axis and therefore create zero moment about this axis. Z 3 XZ ð þ xÞ Hz1 þ Mz dt ¼ Hz2 ; 0 þ M dt ¼ 2v2  0:8 0

(1) If M ≡ 10 Nm,

Z

3

0þ

10dt ¼ 1:6v2

)

v2 ¼ 18:8 m/s

0

(2) If M ¼ ð5 þ t 2 ÞNm,

Z

0þ

3



 5 þ t 2 dt ¼ 1:6v2

)

v2 ¼ 15 m/s

0

Solution 2: Write the rotational equation of motion, X

 ð þ xÞ MO ¼ IO a; M ¼ 2  0:82  a ¼ 1:28a

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421

(1) If M ≡ 10 Nm, the angular acceleration is as follows: a¼

M 10 ¼ ¼ 7:812 rad/s2 1:28 1:28

at ¼ ar ¼ 7:812  0:8 ¼ 6:250 m/s2 Tangential acceleration at is constant and from kinematics, we have v2 ¼ v1 þ at t ¼ 0 þ 6.250  3 ¼ 18.8 m/s (2) If M ¼ ð5 þ t 2 Þ Nm, the angular acceleration is as follows: a¼

at ¼ ar ¼

M 5 þ t2 ¼ rad/s2 1:28 1:28

 5 þ t2  0:8 ¼ 0:625 5 þ t 2 m/s2 1:28

Tangential acceleration at is variable and from kinematics, we have at ¼ dv=dt. Since v ¼ 0 at t ¼ 0, integrating and solving yields Z v2 Z 3

 dv ¼ 0:625 5 þ t 2 dt ) v2 ¼ 15 m/s 0

0

RETHINK: The principle of angular impulse and momentum in solution 1 directly relates force, time and (angular) velocity. Solution 2 first get the (angular) acceleration using the equation of motion, then using integration to get (angular) velocity. Therefore, solution 1 is much easier here than solution 2.

10.3

Principle of Impulse and Momentum for a System of Particles

Consider a system of n particles within an enclosed region in space, as shown in figure 10.10. Applying the principle of linear impulse and momentum to the arbitrary i-th particle, which is subjected to an internal force f i and an external force Fi , we obtain Z t2 mi v i1 þ ðFi þ f i Þdt ¼ mi vi2 ð10:13Þ t1

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FIG. 10.10 – Derivation of the principle of impulse and momentum for a system of particles. Applying this principle to each of the other particles of the system, similar equations would be written. Then for the whole system, all n equations vectorially added together gives X X Z t2 X Z t2 X mi v i1 þ Fi dt þ f i dt ¼ mi v i2 ð10:14Þ t1

t1

P R t2 The sum of linear impulses of all the internal forces, t1 f i dt, will be zero, since internal forces always occur in equal but opposite collinear pairs and then their linear impulses cancel out. Therefore, equation (10.14) becomes X X Z t2 X mi v i1 þ Fi dt ¼ mi v i2 ð10:15Þ t1

This is the principle of linear impulse and momentum for a system of particles. It states that the initial linear momentum of the system of particles plus the linear impulses of all the external forces applied to the system from t1 to t2 equals the system’s final linear momentum. From the principle of angular impulse and momentum for a particle, expressed as equation (10.12), we can write the following equation for the i-th particle, Z t2 Z t2 H Oi1 þ M O ðFi Þdt þ M O ðf i Þdt ¼ H Oi2 ð10:16Þ t1

t1

Consider each of the other particles of the system and similar equations can be written. All these equations added together vectorially yields X X Z t2 X Z t2 X H Oi1 þ M O ðFi Þdt þ M O ðf i Þdt ¼ H Oi2 ð10:17Þ t1

t1

P R t2 The sum of angular impulses of all the internal forces, t1 M O ðf i Þdt, will be zero, since internal forces always occur in equal but opposite collinear pairs and then their angular impulses cancel out. Therefore, equation (10.17) becomes

Kinetics: Impulse and Momentum X

H Oi1 þ

423

XZ

t2 t1

M O ðFi Þdt ¼

X

H Oi2

ð10:18Þ

This is the principle of angular impulse and momentum for a system of particles. It states that the initial angular momentum of the system of particles plus the angular impulses of all the external forces applied to the system from t1 to t2 equals the system’s final angular momentum.

10.3.1

Linear Momentum of a System of Particles

The linear momentum of a system of particles is the vector sum of the linear momenta of all the particles of the system, that is X L¼ mi v i ð10:19Þ where mi and v i , are respectively, the mass and velocity of the i-th particle in the system. From the definition of the center of mass, we have P X mi ri or mrC ¼ mi ri ð10:20Þ rC ¼ m P where m ¼ mi is the total mass of the system and the position vector rC locates the center of mass of the system, as shown in figure 10.10. For a system with constant mass, differentiating equation (10.20) with respect to time yields P X mi v i or mv C ¼ vC ¼ mi v i ð10:21Þ m Substituting this result into equation (10.19), we get X L¼ mi v i ¼ mv C

ð10:22Þ

Therefore, the principle of linear impulse and momentum for a system of particles expressed in equation (10.15) can also be rewritten as X Z t2 mv C 1 þ Fi dt ¼ mv C 2 ð10:23Þ t1

Since no restriction is imposed in the way the particles are connected, equation (10.23) can apply to any kind of system of particles, such as a solid, liquid, or gas system. In this book, we mainly apply it to a rigid body and sometimes a system of rigid bodies.

10.3.2

Angular Momentum of a System of Particles

As shown in figure 10.11, we establish a fixed reference frame Oxyz and a translating reference frame Cx 0 y 0 z 0 whose origin is attached to and moves with the mass center C of the system of particles. v C is the mass center’s velocity relative to fixed frame

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Oxyz. v i and vi=C are the velocities of particle Mi relative to fixed frame Oxyz and translating frame Cx 0 y 0 z 0 , respectively. ri and ri=C are the position vectors of particle Mi relative to fixed frame Oxyz and translating frame Cx 0 y 0 z 0 , respectively. From the relative-motion analysis, we have ri ¼ rC þ ri=C ; v i ¼ v C þ vi=C

FIG. 10.11 – Derivation of angular momentum of a system of particles. The angular momentum of the system of particles about fixed point O in equation (10.18) is X X X HO ¼ H Oi ¼ ri  mi v i ¼ ðrC þ ri=C Þ  mi ðv C þ vi=C Þ X  X  X X ¼ rC  mi v C þ rC  mi v i=C þ mi ri=C  v C þ ri=C  mi v i=C ð10:24Þ So, the angular momentum of the system of particles is comprised of four terms. For the first term, X  rC  mi vC ¼ rC  mvC For the second term and the third term, from §§9.2.2, we have X mi vi=C ¼ rC  mvC =C ¼ 0 rC  X For the fourth term

 mi ri=C  vC ¼ mrC =C  v C ¼ 0 X

ri=C  mi v i=C ¼ H C

ð10:25Þ

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425

It represents the angular momentum of the system of particles about C with respect to the translating frame Cx 0 y 0 z 0 . Substituting the four terms into equation (10.24) yields H O ¼ rC  mv C þ H C

ð10:26Þ

It states that the angular momentum of the system of particles about fixed point O is the vector sum of the moment of linear momentum mv C about point O and the angular momentum of the system of particles about C with respect to the translating frame Cx 0 y 0 z 0 . Using the term H O , the principle of angular impulse and momentum for a system of particles expressed by equation (10.18) can be rewritten as X Z t2 H O1 þ M O ðFi Þdt ¼ H O2 ð10:27Þ t1

Referring to equation (10.9), the differential form of the above equation is X dH O ð10:28Þ M O ðF i Þ ¼ dt Substituting equation (10.26) into the above equation yields   X d dH C drC dmv C dH C M O ðFi Þ ¼ ðrC  mv C Þ þ ¼  mv C þ rC  þ dt dt dt dt dt ¼ ðv C  mv C þ rC  maC Þ þ

X  dH C dH C  ¼ 0 þ rC  Fi þ dt dt

It can be rewritten as X X X dH C X ¼ M O ðFi Þ  rC  Fi ¼ ri  F i  rC  F i dt X X X ¼ ðri  rC Þ  Fi ¼ ri=C  Fi ¼ M C ðFi Þ Then we obtain dH C X ¼ M C ðFi Þ dt

ð10:29Þ

Rearranging the terms and integrating throughout the time period t1 to t2 gives X Z t2 HC 1 þ M C ðFi Þdt ¼ H C 2 ð10:30Þ t1

This is the principle of angular impulse and momentum with respect to translating frame Cx0 y0 z0 . This principle is often used for kinetic problems of a rigid body (please see §§10.3.4). However, for an arbitrary system of particles, for example, a system of rigid bodies, the mass center is not a specific point and always changes

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426

with the time, which will make trouble for the integration from t1 to t2 . Therefore, equation (10.30) is inappropriate and alternatively, its differential form equation (10.29) can be used.

10.3.3

Angular Momentum of Rigid Bodies in Planar Motion

As mentioned in §8.5, symmetric rigid bodies undergoing planar motion will be discussed here. A fixed reference frame Oxy and a translating reference frame Cx 0 y 0 whose origin is attached to and moves with mass center C of the body are established in the symmetric plane, as shown in figure 10.12. From example 10.6, the angular momenta H O and H C would always be perpendicular to the x–y plane, along the þ z axis or z axis. Therefore, they can be represented by scalars. Since in general plane motion, v i=C is the tangential velocity due to the relative rotation about mass center C, we have v i=C ¼ x  ri=C and the scalar form of equation (10.25) is  X X

 X 2 HC ¼ ri=C mi vi=C ¼ ri=C mi ri=C x ¼ mi ri=C ð10:31Þ x ¼ IC x From figure 10.13, the scalar form of equation (10.26) is HO ¼ ymvCx þ xmvCy þ IC x

ð10:32Þ

For a rigid body undergoing translation, x ¼ 0 and HC ¼ 0. For a rigid body undergoing rotation about fixed axis O, as shown in figure 10.14, we have vC ¼ rC x. Considering the parallel-axis theorem, the angular momentum about point O becomes

 HO ¼ rC mvC þ IC x ¼ rC m ðrC xÞ þ IC x ¼ mrC2 þ IC x ¼ IO x ð10:33Þ

FIG. 10.12 – Derivation of angular momentum of a body in planar motion.

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427

FIG. 10.13 – Momentum diagram of a body in planar motion.

FIG. 10.14 – Momentum diagram of a rotating body. Example 10.8. The wheel shown in figure 10.15a has mass m and radius of gyration kC about mass center C. At a given instant, the wheel rotates about O with angular velocity x. Determine the linear momentum of the wheel and its angular momenta about point C and point O at this instant.

FIG. 10.15 – Linear momentum and angular momentum of an eccentric wheel. Solution: The momentum diagram of the wheel is shown in figure 10.15b. The velocity of mass center C is vC ¼ xe and its direction is perpendicular to line OC. Then linear momentum L of the wheel has the same direction as v C and its magnitude is

Engineering Mechanics

428 L ¼ mvC ¼ mxe

The angular momenta of the wheel about point C and point O are as follows: ð þ xÞ HC ¼ IC x ¼ mkC2 x

 ð þ xÞ HO ¼ mvC  e þ IC x ¼ meðxeÞ þ mkC2 x ¼ mx e2 þ kC2 From equation (10.33), the angular momentum of the wheel about point O can also be calculated as follows:





 ð þ xÞ HO ¼ IO x ¼ IC þ me2 x ¼ mkC2 þ me2 x ¼ mx e2 þ kC2 Example 10.9. The 2-kg uniform rod shown in figure 10.16a has a length l ¼ 1 m. At a given instant, the velocity of the end point A is vA ¼ 5 m/s. Determine the linear momentum of the rod and its angular momenta about mass center C and fixed point O at this instant.

FIG. 10.16 – Linear momentum and angular momentum of a rod. Solution: The momentum diagram of the rod is shown in figure 10.16b. From the velocities of points A and B, the instantaneous center of rotation (IC) of the rod can be determined and shown in figure 10.16b. From kinematics, we have vA x¼ ¼ 5:774 rad/s; vC ¼ x  ð1  sin 30 Þ ¼ 2:887 m/s 1  cos 30 The direction of v C is perpendicular to line CIC. Then linear momentum L of the wheel has the same direction as v C and its magnitude is L ¼ mvC ¼ 2  2:887 ¼ 5:774 kgm/s The angular momenta of the wheel about mass center C and point O are as follows: ð þ xÞ HC ¼ IC x ¼

ml 2 2  12 x¼  5:774 ¼ 0:962 kgm2 /s 12 12

ð þ xÞ HO ¼ mvC  ð1  sin 30 Þ þ IC x ¼ 2  2:887  0:5 þ 0:962 ¼ 3:849 kgm2 /s

Kinetics: Impulse and Momentum

429

Example 10.10. An inextensible cord of negligible mass is wrapped around two uniform disks, as shown in figure 10.17a. Each of them has weight P and radius R. At a given instant, the angular velocities of two disks are xA and xB , respectively. Determine the linear momentum of the system of two disks and its angular momentum about fixed point A for this instant.

FIG. 10.17 – Linear momentum and angular momentum of a system of rigid bodies.

Solution: The momentum diagram of the system is shown in figure 10.17b. From kinematics, the velocity of the mass center of disk B is vB ¼ RðxA þ xB Þ Its direction is vertically downward. Since the velocity of the mass center of disk A is zero, linear momentum L of the system has the same direction as v B and its magnitude is L ¼ m B vB ¼

PR ðxA þ xB Þ g

The angular momentum about fixed point A of the system is ð þ xÞ HA ¼ IA xA  IB xB  mB vB  2R

¼

PR2 PR2 2PR2 5PR2 xA  xB  ðxA þ xB Þ ¼  ðxA þ xB Þ 2g 2g g 2g

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10.3.4

Principle of Impulse and Momentum for Rigid Bodies in Planar Motion

From impulse and momentum principles expressed in equation (10.23) and equation (10.27) or (10.30), the following three scalar equations can be written for a symmetric rigid body undergoing planar motion: X Z t2 mvC 1x þ Fxi dt ¼ mvC 2x mvC 1y þ IC x1 þ

XZ

XZ t1

t1 t2

t1 t2

ð10:34Þ

Fyi dt ¼ mvC 2y

MC dt ¼ IC x2

or

HO1 þ

XZ

t2 t1

MO dt ¼ HO2

Figure 10.18 is a pictorial representation of these equations and it is called the impulse-and-momentum diagram of a body. Note that in the momentum diagram, the linear momentum, mv C , should be applied at the body’s mass center C; whereas the angular momentum, IC x, can be applied anywhere on the body like a couple moment. The impulse diagram should show the impulses of all the external forces and couples (except nonimpulsive forces, please see section 10.4). It should be pointed out that reaction forces usually do no work, but they create impulses and should be shown in the impulse diagram. Drawing clear impulse-and-momentum diagram is of great help in the analysis and it will allow you to sum components in any direction and to “visualize” the moment arms when summing moments about mass center C or fixed point O in the principle of angular impulse and momentum.

FIG. 10.18 – The impulse-and-momentum diagram of a body. Equation (10.34) can also be applied to a system of connected bodies rather than to each body separately. This eliminates the consideration of interaction impulses which occur at the connections since they are internal to the system. However, it should be noted that since the mass center of the system of rigid bodies is not a P R t2 specific point and always changes with the time, IC x1 þ t1 MC dt ¼ IC x2 is P R t2 inappropriate and only HO1 þ t1 MO dt ¼ HO2 can be used in this case.

Kinetics: Impulse and Momentum

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Procedure for analysis Impulse and momentum principles are suitable for solving kinetic problems that involve force (couple moment), time and velocity (angular velocity) for a symmetric rigid body undergoing general plane motion. The procedure for analysis is as follows: (1) Establish an inertial coordinate system and analyze the velocity of the body’s mass center and the body’s angular velocity at the initial and final instants. (2) Draw the free-body diagram to show all the external forces and couple moments that produce impulses on the body. It should be noted that every external force (except nonimpulsive force, please see section 10.4) acting on the body will create an impulse, although some of these forces do no work (example 10.15). (3) The impulse-and-momentum diagram is strongly recommended for a complicated problem, especially a problem involving a system of rigid bodies. It is particularly helpful to visualize the moment arms used in the principle of angular impulse and momentum. (4) Apply equation (10.34) to solve for the unknowns. Example 10.11. The homogeneous disk has a mass of 10 kg and radius r = 10 cm. It is spinning at the end of the weightless link AB with an angular velocity ω0 = 30 rad/s when it is placed against the wall, figure 10.19a. The coefficient of kinetic friction between the disk and the wall is 0.25. Determine the time needed to stop the disk. Also find the force in the link during this process.

FIG. 10.19 – A rotating disk placed against a rough wall. Solution: The mass moment inertia of the disk about its mass center is IC ¼

1  10  0:12 ¼ 0:05 kgm2 2

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The free-body diagram of the disk is shown in figure 10.19b. From the impulse and momentum principles, we have XZ ðþ"Þ mvC 1y þ Fyi dt ¼ mvC 2y ; 0 þ Fk t þ FAB cos30  t  mgt ¼ 0 ð1Þ

ðþ!Þ mvC 1x þ

ðþxÞ IC x1 þ

XZ XZ

Fxi dt ¼ mvC 2x ;

0 þ FAB sin30  t  Nt ¼ 0

MC dt ¼ IC x2 ; 0:05  30 þ Fk  0:1t ¼ 0

ð2Þ

ð3Þ

Also, the friction equation can be written Fk ¼ lk N ¼ 0:25N

ð4Þ

Substituting and solving the above equations simultaneously yield t ¼ 1:21 s; FAB ¼ 99:0 N Example 10.12. The 10-kg flywheel A has radius of gyration about its center (also mass center) of 10 cm and the homogeneous disk B has a mass of 30 kg, figure 10.20a. The system is initially at rest. Thereafter a motor supplies a torque M = (5t + 10) N·m, where t is in seconds, to the flywheel. How long will it take for disk B to attain an angular velocity of 50 rad/s? The belt does not slip at its contacting surfaces.

FIG. 10.20 – A belt wheel system.

Solution: The mass moment inertia of flywheel A about its center is IA ¼ 10  0:12 ¼ 0:1 kgm2

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When disk B attains an angular velocity of 50 rad/s, the angular velocity of flywheel A is xA ¼

rB 18  50 ¼ 60 rad=s xB ¼ 15 rA

Applying the principle of angular impulse and momentum to flywheel A, figure 10.20b, we have XZ ð þ xÞ IA x1 þ MA dt ¼ IA x2 Z 0þ

t

Z

Z

ð5t þ 10Þdt þ

ðFT2  0:15Þdt 

ðFT 1  0:15Þdt ¼ 0:1  60

0

Z 2:5t þ 10t þ 0:15  2

ðFT 2  FT 1 Þdt ¼ 6

ð1Þ

The mass moment inertia of disk B about its center is IB ¼

1  30  0:182 ¼ 0:486 kgm2 2

Applying the principle of angular impulse and momentum to disk B, figure 10.20c, we have XZ ð þ xÞ IB x1 þ MB dt ¼ IB x2 ; Z

Z 0þ

ðFT 1  0:18Þdt 

ðFT 2  0:18Þdt ¼ 0:486  50

Z ðFT 2  FT 1 Þdt ¼ 135

ð2Þ

Substituting equation (2) into equation (1) and solving yield t ¼ 1:81 s Note that another root, t ¼ 5:81 s, is abandoned, since it does not represent the physical situation. Example 10.13. The uniform disk has a mass of 10 kg and radius r ¼ 0:32 m. It is rotating with an angular velocity ω = 40 rad/s when force F is applied to the brake arm. Magnitude of force F varies with time as shown in figure 10.21a. Neglect the weight of the brake arm and the size of brake pad C. If the coefficient of kinetic friction at brake pad C is 0.3, determine the time required to stop the disk from the instant the brake is applied.

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FIG. 10.21 – The rotating disk and the brake. Solution: The free-body diagrams of the brake arm and the disk are shown in figure 10.21b and c, respectively. For the brake arm, we can write the moment equation of equilibrium, X MO1 ¼ 0; F  ð0:8 þ 0:6Þ  FN  0:6 ¼ 0 ) FN ¼ 2:333F Fk ¼ lk FN ¼ 0:3  2:333F ¼ 0:7F The mass moment of inertia of the disk about its center is 1 IO ¼  10  0:322 ¼ 0:512 kgm2 2 Apply the principle of angular impulse and momentum to the disk, Z t XZ IO x1 þ MO dt ¼ IO x2 ; IO x þ ðFk  0:32Þdt ¼ 0 0

Z

t

0:512  40  0:32  0:7 

 Fdt

¼0

0

Assume t [ 1 s, then Z t 1 Fdt ¼  8  1 þ 8  ðt  1Þ ¼ ð8t  4Þ Ns 2 0

ð1Þ

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Substituting this impulse into equation (1) and solving yield t ¼ 11:9 s Since t ¼ 11:9 s [ 1 s, the assumption is correct and the time needed to stop the disk is 11.9 s. Example 10.14. The tennis racket has mass m, and radius of gyration kC about its mass center C. Assume that the tennis racket is initially at rest and would rotate about point A immediately after it hits the ball, as shown in figure 10.22a. Also assume that the hitting force is approximately normal to the racket face. Determine position P where the ball should be hit such that the force exerted by the racket on the hand would essentially be zero, i.e., the impact on the hand would be almost zero.

FIG. 10.22 – The center of percussion of a tennis racket.

Solution: The impulse-and-momentum diagram of the racket is shown in figure 10.22b. Since the R racket rotates about point A, vC ¼ xrC . Assume the ball exerts an impulse of Fdt on the racket. The following equations can be written from the impulse and momentum principles. Z XZ ð þ Þ mvC 1x þ Fxi dt ¼ mvC 2x ; 0 þ Fdt ¼ mxrC

ð þ xÞ HA1 þ

XZ

Z MA dt ¼ HA2 ;

0þ

 Fdt

 rP ¼ IC x þ m ðxrC Þ  rC

Engineering Mechanics

436 Solving the above two equations yields rP ¼

IC x þ mrC2 x mkC2 x þ mrC2 x kC2 þ rC2 R ¼ ¼ mxrC rC Fdt

RETHINK: By placing the striking point at P, the player would feel little impact to the hand. This point is called the center of percussion, which is widely used in designing many sports rackets, clubs, etc. Example 10.15. The 8-kg gear has radius of gyration kC = 0.25 m about its mass center C. A horizontal force of F = (2t + 10) N, where t is in seconds, is acting on the weightless cord and makes the gear start to move on fixed gear rack from rest, figure 10.23a. Determine the angular velocity of the gear after 5 s.

FIG. 10.23 – Gear and gear rack. Solution 1: The mass moment of inertia of the gear about its mass center is IC ¼ mkC2 ¼ 8  0:252 ¼ 0:5 kgm2 The impulse-and-momentum diagram of the gear is shown in figure 10.23b. We can write the following equations from the impulse and momentum principles, Z 5 Z XZ Fxi dt ¼ mvC 2x ; 0 þ ð2t þ 10Þdt  Ft dt ¼ mvC 2 ð þ !Þ mvC 1x þ 0

Z 75 

Ft dt ¼ 8vC 2

ð1Þ

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XZ ð þ y Þ IC x 1 þ MC dt ¼ IC x2 ; Z 5

Z  0þ ð2t þ 10Þdt  0:2 þ Ft dt  0:4 ¼ IC x2 0

Z 15 þ 0:4

 Ft dt

¼ 0:5x2

ð2Þ

Considering vC 2 ¼ 0:4x2 and solving equations (1) and (2) simultaneously yield x2 ¼ 25:3 rad/s Solution 2: From the impulse-and-momentum diagram of the gear shown in figure 10.23b, we can write the equation from the principle of angular impulse and momentum about a fixed point O on the gear rack, XZ ð þ yÞ HO1 þ MO dt ¼ HO2 ; Z 5

0þ ð2t þ 10Þdt  0:6 ¼ IC x2 þ mvC 2  0:4 0

Note that the angular impulse of the weight and the angular impulse of normal force FN cancel each other out and the angular impulse of tangential reaction force Ft about O is zero. Considering vC 2 ¼ 0:4x2 and solving the above equation yield x2 ¼ 25:3 rad/s Solution 3: The free-body diagram of the gear is shown in figure 10.23c and its kinetic diagram is shown in figure 10.23d. Since the gear appears to be rolling without slipping on the fixed gear rack, we have aC ¼ 0:4a. Write the rotational equation of motion about IC, X X ð þ yÞ MIC ¼ ðMk ÞIC ; ð2t þ 10Þ  0:6 ¼ IC a þ maC  0:4 a ¼ 0:6742t þ 3:371 a is variable with time and from kinematics, we have a ¼ dx=dt. Since x ¼ 0 at t ¼ 0, integrating and solving yield Z x2 Z 5 Z 5 dx ¼ adt ¼ ð0:6742t þ 3:371Þdt ) x2 ¼ 25:3 rad/s 0

0

0

438

Engineering Mechanics

RETHINK: Note that every external force acting on the body will produce an impulse, even though some of these forces do no work. For example, in this problem the tangential and normal reaction forces Ft and FN do no work, but they create impulses. Example 10.16. The bowling ball has a mass of 4.5 kg and radius r = 109 mm. It is cast on the alley with a backspin angular velocity ω0 = 10 rad/s, and a velocity of its mass center vC 0 = 6 m/s, as shown in figure 10.24a. Treat the bowling ball as a uniform sphere. The coefficient of kinetic friction between the ball and the alley is 0.1. How long will it take for the ball to stop back spinning? What is the velocity of its mass center at this instant? Then determine the time for the ball to start rolling without sliding and the velocity of its mass center at this instant.

FIG. 10.24 – Bowling ball cast on the alley.

Solution: The mass moment of inertia of the ball about its mass center is 2 2 IC ¼ mr 2 ¼  4:5  0:1092 ¼ 0:0214 kgm2 5 5 The impulse-and-momentum diagram of the ball from the beginning to the time at which the ball stops back spinning is shown in figure 10.24b. Write the equation in the direction of y from the principle of linear impulse and momentum,

Kinetics: Impulse and Momentum

ð þ "Þ mvC 1y þ

XZ

439

Fyi dt ¼ mvC 2y ;

0 þ Nt1  4:5  9:81t1 ¼ 0

)

N ¼ 44:145 N

Then the kinetic friction force is Fk ¼ lk N ¼ 0:1  44:145 ¼ 4:4145 N Write the other equations from the impulse and momentum principles, XZ ð þ xÞ IC x1 þ MC dt ¼ IC x2 ; IC  10  Fk t1  0:109 ¼ 0 ð þ !Þ mvC 1x þ

XZ

)

t1 ¼ 0:4447 s

Fxi dt ¼ mvC 2x ;

m  6  Fk t1 ¼ mvC 1

)

vC 1 ¼ 5:5637 m/s

Assume after another time interval t2, the ball starts rolling without sliding. The impulse-and-momentum diagram of the ball during this process is shown in figure 10.24c. At the end of the process, the ball’s angular velocity x2 ¼ vC 2 =r. Write the equations from the impulse and momentum principles: XZ ð þ !Þ mvC 1x þ Fxi dt ¼ mvC 2x ; mvC 1  Fk t2 ¼ mvC 2 ð1Þ

ð þ xÞ IC x1 þ

XZ

MC dt ¼ IC x2 ; 0  Fk t2  0:109 ¼ IC 

vC 2 r

ð2Þ

Solving the above equations simultaneously yields t2 ¼ 1:6213 s; vC 2 ¼ 3:9733 m=s RETHINK: We can consider the two phases together and get vC 2 and t ¼ t1 þ t2 directly. From the principle of angular impulse and momentum, the equation can also be written about fixed point O on the alley instead of mass center C. Readers are recommended to finish this by themselves. Example 10.17. The 10-kg spool shown in figure 10.25a has radius of gyration kO ¼ 0:22 m, and R = 0.3 m, r = 0.2 m. Each of the two blocks has a mass of 1 kg. If the system is released from rest, determine the time required for the spool to attain an angular velocity of 10 rad/s.

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FIG. 10.25 – A spool and two blocks. Solution 1: The free-body diagram of each body is shown in 10.25b. The mass moment of inertia of the spool about O is IO ¼ mkO2 ¼ 10  0:222 ¼ 0:484 kgm2 When the spool attains an angular velocity of x ¼ 10 rad/s, the velocities of the two blocks are vA ¼ xR ¼ 10  0:3 ¼ 3 m/s; vB ¼ xr ¼ 10  0:2 ¼ 2 m/s Applying the principle of angular impulse and momentum for the spool, we get XZ ð þ y Þ IO x 1 þ MO dt ¼ IO x2 ; 0 þ ð0:3FT 1  0:2FT 2 Þ  t ¼ 0:484  10 ð1Þ Using the principle of linear impulse and momentum for block A, we get XZ ð þ "Þ mv1y þ Fyi dt ¼ mv2y ; 0 þ ðFT 1  mA gÞ  t ¼ mA vA

ð2Þ

Using the principle of linear impulse and momentum for block B, we get XZ ð þ "Þ mv1y þ Fyi dt ¼ mv2y ; 0 þ ðFT 2  mB gÞ  t ¼ mB vB

ð3Þ

Substituting and solving equations (1)–(3) simultaneously yield t ¼ 6:26 s;

FT 1 ¼ 9:33 N;

FT 2 ¼ 10:1 N

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Solution 2: The impulse-and-momentum diagram of the system is shown in 10.25c. The mass moment of inertia of the spool about O is IO ¼ mkO2 ¼ 10  0:222 ¼ 0:484 kgm2 When the spool attains an angular velocity of x ¼ 10 rad/s, the velocities of the two blocks are vA ¼ xR ¼ 10  0:3 ¼ 3 m/s; vB ¼ xr ¼ 10  0:2 ¼ 2 m/s Applying the principle of angular impulse and momentum for the system, we get XZ ð þ yÞ HO1 þ MO dt ¼ HO2 ; 0 þ mA gtR  mB gtr ¼ IO x þ mA vA R þ mB vB r Substituting and solving this equation yield t ¼ 6:26 s

RETHINK: From solution 2, we can see that by applying the principle of impulse and momentum for the system, the internal impulses acting between the bodies within the system (here FT1 and FT 2 ) are eliminated from the analysis. However, if an internal impulsive force acting on one body of the system is needed, this body has to be isolated and the principle of impulse and momentum is then applied to the body. So, if we want to determine FT 1 and FT 2 in this example, solution 1 should be performed.

10.4

Principle of Conservation of Momentum

From the principle of linear impulse and momentum for a system of particles, expressed by equation (10.15), it can be seen that if the vector sum of all the linear P R t2 impulses applied to the system is zero, i.e., t1 Fi dt ¼ 0, equation (10.15) reduces to the following form: X X mi v i2 or mv C 1 ¼ mv C 2 ð10:35Þ mi vi1 ¼ It means the linear momentum of the system is constant, or conserved. If only the algebraic sum of all the linear impulses in a specific direction is zero, the linear momentum of the system in this direction will be conserved. For example, if P R t2 t1 Fxi dt ¼ 0, we have X X mi vi1x ¼ mi vi2x or mvC 1x ¼ mvC 2x ð10:36Þ

Engineering Mechanics

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Equations (10.35) and (10.36) are referred to as the principle of conservation of linear momentum. From the principle of angular impulse and momentum for a system of particles, expressed by equation (10.18), it can be seen that if the vector sum of all the angular P R t2 impulses about a fixed point O is zero, i.e., t1 M O ðFi Þdt ¼ 0, equation (10.18) reduces to the following form: X X H Oi2 or H O1 ¼ H O2 ð10:37Þ H Oi1 ¼ It means the angular momentum of the system is conserved about point O. If only the algebraic sum of all the angular impulses about a specific axis is zero, the angular momentum of the system about this axis will be conserved. For example, if P R t2 t1 Mz ðFi Þdt ¼ 0, we have X X Hzi1 ¼ Hzi2 ð10:38Þ Equations (10.37) and (10.38) are referred to as the principle of conservation of angular momentum. For a symmetric rigid body undergoing planar motion, from equations (10.30) and (10.31), the conservation of angular momentum can also be expressed as H C 1 ¼ H C 2 or ðIC xÞ1 ¼ ðIC xÞ2

ð10:39Þ

This is the principle of conservation of angular momentum about the body’s mass center C. The conservation of linear momentum and angular momentum is often applied for a collision, strike, or explosion. Since the time period for a collision or explosion is generally very short, some small forces over this very short period of time will produce negligible impulses and may be neglected. This kind of forces causing negligible impulses are called nonimpulsive forces. Typical examples include the weight of a body and the force of a slightly deformed spring. Conversely, forces that are very large and cause a significant change in momentum are called impulsive forces, such as the striking force of one body on another and the interaction force between particles in explosion. Example 10.18. The railway engine of weight P1 ¼ 1100 kN is traveling to the right at v1 ¼ 2 m/s on the horizontal track when it encounters the boxcar at rest, figure 10.26a. They couple and then move to the right together. If the weight of the

FIG. 10.26 – The coupling of a railway engine and a boxcar.

Kinetics: Impulse and Momentum

443

boxcar is P2 ¼ 400 kN, determine (1) their speed just after the coupling; (2) the average force between them if the coupling takes place in 0.6 s. Solution: (1) The free-body diagram of the system is shown in figure 10.26a. Since there is no force in the x direction, linear momentum is conserved in this direction. Assume the railway engine and the boxcar move to the right at v just after the coupling. The equation from the conservation of linear momentum is P1 P1 þ P 2 v1 þ 0 ¼ v g g

ð1Þ

Substituting the values of P1 , P2 and v1 yields v¼

P1 v 1 ¼ 1.467 m/s P1 þ P2

(2) To determine the average coupling force, we have to dismember the system and apply the principle of linear momentum to either the railway engine or the boxcar separately. The free-body diagram of the boxcar is shown in figure 10.26b. Apply the principle of linear impulse and momentum in the x direction, XZ P2 ð þ !Þ mv1x þ v ) Favg ¼ 99:7 kN Fxi dt ¼ mv2x ; 0 þ Favg  0.6 ¼ g

RETHINK: Equation (1) can be rewritten as P1 P2 v ðv 1  v Þ ¼ g g Actually, the left side is the decrease in the linear momentum of the railway engine and the right side is the increase in the linear momentum of the boxcar. It means during the coupling, the railway engine transfers the linear momentum to the boxcar through the forces between them. The decrease in the linear momentum of the railway engine is equal to the increase in the linear momentum of the boxcar and the total linear momentum of the system is conserved. Example 10.19. The 40-N triangular prism is originally at rest. The 20-N block is released from rest at position 1 and slides down 0.5 m to position 2 along the slope, figure 10.27a. Neglecting the mass of the wheels and the friction at all surfaces, determine the prism’s speed when the block reaches position 2.

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FIG. 10.27 – A block moving down a triangular prism. Solution: The free-body diagram of the system at an intermediate position is shown in figure 10.27b. It can be seen that there is no force in the direction of x, therefore the linear momentum is conserved in this direction. Since the system is initially at rest, we have 0¼

20 40 vBx þ vP 9:81 9:81

ð1Þ

Here vB and vP are, respectively, the velocities of the block and the prism with respect to an inertial reference of frame. From the relative-motion analysis, the two velocities can be related. v B ¼ v P þ v B=P The kinematic diagram is shown in figure 10.27c. Therefore, vBx ¼ vP þ vB=P cos 30

ð2Þ

vBy ¼ vB=P sin 30

ð3Þ

There are four unknowns in equations (1)–(3), i.e., vBx , vBy , vP and vB=P ; another equation has to be obtained. During the motion from position 1 to position 2, only the weight of the block does work. Apply the principle of work and energy, T1 þ W1!2 ¼ T2 ,   1 20 2 1 40 2 2 2 vB þ vP ¼ 1:019 vBx 0 þ 20  0:5  sin 30 ¼ þ vBy þ 2:039vP2 ð4Þ 2 9:81 2 9:81 Then the velocity of the prism at position 2 can be solved from equations (1)–(4). vP ¼ 0:738 m=s Example 10.20. A person stands at the front part of a boat, which is also at rest. The masses of the boat and the person are m1 and m2, respectively. The length of the boat is l. Neglecting the resistance of the water, determine the displacement of the boat if the person walks from bow to stern, as shown in figure 10.28a.

Kinetics: Impulse and Momentum

445

FIG. 10.28 – A person walking from bow to stern.

Solution: Since the resistance of the water is neglected, there is no external force in the direction of x for the system of the boat and the man. Then the linear momentum is conserved in this direction. Because the system is originally at rest, from the principle of conservation of linear momentum, we have mvCx1 ¼ mvCx2 ¼ 0 Therefore xC 1 ¼ xC 2 For the initial position shown in figure 10.28b, the x coordinate of the mass center of the system is xC 1 ¼

m1 a þ m2 b m1 þ m2

For the final position shown in figure 10.28b, xC 2 ¼

m1 ða  sÞ þ m2 ðb þ l  sÞ m1 þ m2

Substituting the two expressions into xC 1 ¼ xC 2 and solving, we get the displacement of the boat s¼

m2 l m 2 þ m1

Example 10.21. A 15-g magnet D falls down from rest from the position shown in figure 10.29a. Then magnet D collides and sticks with the 200-g steel uniform bar AB, which is at rest before the impact. Determine the angular velocity of the bar immediately after the impact.

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FIG. 10.29 – A magnet colliding and stick with a bar.

Solution: The velocity of magnet D just before the impact can be calculated from the principle of work and energy. T1 þ W1!2 ¼ T2 ; vD1 ¼

1 2 0 þ mD g  0:32 ¼ mD vD1 2

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2  9:81  0:32 ¼ 2:506 m/s

Since after the impact, the magnet sticks with bar AB at point A, we have vD2 ¼ vA2 . Therefore vA2 vD2 ¼ ) vD2 ¼ 0:3x2 x2 ¼ 0:3 0:3 The impulse-and-momentum diagram of the system through the time period of impact is shown in figure 10.29b. Since this time period is very short, the weights can be considered nonimpulsive. The impulsive forces Ox and Oy pass through pin O. Then the angular momentum of the system about O is conserved. ð þ xÞ HO1 ¼ HO2 ; mD vD1  0:3 ¼ IO x2 þ mD vD2  0:3 0:015  2:506  0:3 ¼

0:2  0:62 x2 þ 0:015  0:3x2  0:3 12

)

x2 ¼ 1:53 rad/s

Example 10.22. Starting from rest, the 30-kg crate moves up the inclined plane with a constant acceleration due to the cable force from motor M, figure 10.30a. The coefficient of kinetic friction between the crate and the incline is 0.2. If the crate moves along the incline a distance 10 m in 4 s, determine the tension developed in the cable.

Kinetics: Impulse and Momentum

447

FIG. 10.30 – A crate moving up an incline due to the cable force.

Solution 1: The free-body diagram of the crate is shown in figure 10.30b. Apply the principle of linear impulse and momentum in the y direction, XZ ð þ -Þ mv1y þ Fyi dt ¼ mv2y ; 0 þ ðN  mg cos 45 Þ  4 ¼ 0 N ¼ mg cos 45 ;

Fk ¼ lk N ¼ lk mg cos 45

Apply the principle of linear impulse and momentum in the x direction, XZ ð þ %Þ mv1x þ Fxi dt ¼ mv2x ; 0 þ ðT  mg sin 45  Fk Þ  4 ¼ mv2

ð1Þ

Using the principle of work and energy, T1 þ W1!2 ¼ T2 , we have 1 2 1 mv1 þ ðT  mg sin 45  Fk Þ  10 ¼ mv22 2 2

ð2Þ

Solving equations (1) and (2) yields T ¼ 287 N Solution 2: Since the crate moves 10 m in 4 s with a constant acceleration, we have 1 s ¼ at 2 2

)

a¼

2s 2  10 ¼ ¼ 1:25 m/s2 t2 42

From the free-body diagram of the crate, figure 10.30b, the following equations of motion can be written. X ð þ -Þ Fyi ¼ may ; N  mg cos 45 ¼ 0 ) N ¼ mg cos 45 Fk ¼ lk N ¼ lk mg cos 45

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ð þ %Þ

X

Fxi ¼ max ;

T  Fk  mg sin 45 ¼ ma

T ¼ Fk þ mg sin 45 þ ma ¼ 287 N

PROBLEMS 10.1 The magnitude of a horizontal force is F = (t2) N, where t is measured in seconds. Calculate the linear impulse due to the force within the first 3 s. 10.2 The world record of service speed of tennis players is 69.7 m/s. If the strike lasts 0.02 s and the ball has a mass of 60 g, determine the average force of tennis racket on the ball. 10.3 As a 4-g bullet passes through the horizontal barrel of a rifle, the horizontal force acting on it varies with time in the manner shown. If the muzzle velocity is 500 m/s when t = 0.8 ms, determine the maximum net force F0.

Prob. 10.3

10.4 The crane cable exerts a constant force of 8 kN to lift a crate of mass 500 kg. Assuming starting from rest, determine the velocity of the crate and the height it goes in 5 s.

Prob. 10.4

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449

10.5 The 10-kg crate is traveling up the slope with a speed of 5 m/s when the cable suddenly fails. If the coefficient of kinetic friction is 0.20, determine the speed of the crate after 5 s.

Prob. 10.5

10.6 The 2-kg block is released from rest when it is 5 m above the plastic floor. Determine the average impulsive force acting on the block by the floor if the block is stopped without rebounding in 0.8 s once the block strikes the floor. Consider the weight of the block as a nonimpulsive force during the impact. 10.7 Calculate the angular momenta of the particle about point O and point P, respectively.

Prob. 10.7

10.8 Calculate the angular momentum HO of each of the two particles about point O. Consider counterclockwise moments as positive.

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Prob. 10.8

10.9 Block A of mass 2 kg is moving on the smooth surface to the right at 5 m/s when it squarely strikes block B, which is at rest and has a mass of 3 kg. If the velocity of block B after the collision is 3 m/s to the right, determine the velocity of block A.

Prob. 10.9

10.10 Motor M delivers a towing force F to lift the 10-kg block from rest. If the magnitude of F varies with time as shown, find the speed of the block when t = 4 s. Neglect the mass of the pulleys and cables.

Prob. 10.10

Kinetics: Impulse and Momentum

451

10.11 Motor M delivers a towing force F to lift the 10-kg block from rest. If the magnitude of F varies with time as shown, find the speed of the block when t = 4 s. Neglect the mass of the pulleys and cables. Hint: First determine the force needed to begin moving the block.

Prob. 10.11

10.12 A ball is attached to cable AOM. Originally the ball moves in a horizontal circular path of radius BM ¼ R with a speed v1 ¼ 4 m/s. Then the cable is pulled down at the end A with a constant rate of 0.5 m/s. Determine the speed of the ball when the radius of the curved path is B1 M1 ¼ R=2.

Prob. 10.12

10.13 The 2-kg drum has radius of gyration kO = 0.3 m about its center O. Starting from rest, it begins to rotate about O when subjected to a moment M ¼ 2ð1  e0:2t Þ Nm, where t is in seconds. Find its angular velocity when t = 5 s. 10.14 The spool has a mass of 30 kg and radius of gyration kO = 0.4 m, and r is 0.3 m. Neglect the mass of the cord. If the 10-kg block A is released from rest, determine the time required for the spool to have an angular velocity x = 10 rad/s. Also find the cord force during the motion.

Engineering Mechanics

452

Prob. 10.14

10.15 The 50-kg drum has radius R = 0.4 m, and radius of gyration kC = 0.25 m about its mass center. If h ¼ 30 and the coefficients of static and kinetic friction are ls ¼ 0:3 and lk ¼ 0:25, respectively, determine the drum’s angular velocity 5 s after it is released from rest. 10.16 The 50-kg drum has radius R = 0.4, and radius of gyration kC = 0.25 mm about its mass center. If h ¼ 30 and the coefficients of static and kinetic friction are ls ¼ 0:15 and lk ¼ 0:1, respectively, determine the drum’s angular velocity 5 s after it is released from rest.

Prob. 10.15/16

10.17 The ring has a mass of 3 kg and radius R = 0.4 m. It is released down the slope with a backspin x0 ¼ 6 rad/s and its center has a velocity vC 0 ¼ 5 m/s as shown. If h ¼ 30 and the coefficient of kinetic friction is 0.25, find the time needed for the ring to stop back spinning. Also find the velocity of its mass center at this instant. 10.18 The ring has a mass of 3 kg and radius R = 0.4 m. It is released down the slope with a backspin x0 ¼ 6 rad/s and its center has a velocity vC 0 ¼ 5 m/s as shown. If h ¼ 30 and the coefficient of kinetic friction is 0.5, find the time needed for the ring to stop slipping and begin to roll without slipping. Also find the velocity of its mass center at this instant.

Prob. 10.17/18

Kinetics: Impulse and Momentum

453

10.19 A rod with mass m1 is suspended from the pin at O. A bullet of mass m2 is fired into the rod with a horizontal velocity vb. Specify the angular velocity of the rod just after the bullet becomes embedded in it.

Prob. 10.19

10.20 Block A of mass 10 kg is connected to block B of mass 20 kg by a spring of stiffness k = 100 N/m. Initially they are held at rest on the smooth surface with the spring stretched 0.3 m. Then they are released. Determine the speed of each block when the spring becomes unstretched.

Prob. 10.20

10.21 Determine the linear momentum and the angular momentum about fixed point O at the instant shown of (a) the rotating uniform disk having a mass m; (b) the rotating uniform bar having a mass m.

Prob. 10.21

Engineering Mechanics

454

10.22 Under the action of force F at the end of the inextensible cord, disk O rotates with an angular velocity ω at a given instant. Each of the two uniform disks has weight P and radius R. Block B also has weight P and it is attached to the center of disk A. Determine the linear momentum of the system and its angular momentum about fixed point O at this instant. Assume the cord does not slip on the disks and neglect the mass of the cord.

Prob. 10.22

Answers

Chapter 1 1.1 (1) 36.9, (2) 255, (3) 44.7, (4) 126, (5) 128 1.2 (1) 0.839 g, (2) 78.7 kN, (3) 35.7 m 1.3 (1) 6.12 g, (2) 2.55 kg, (3) 12.2 Mg, (4) 71.4 Gg 1.4 (1) 8.34 mN, (2) 6.97 kN, (3) 3.59 MN 1.5 94.4 m/s 1.6 54.0 km/h 1.7 (1) 41.2 MN, (2) 4.2 Gg, (3) 6.85 MN 1.8 W = 7.34 kN 1.9 (1) mN
s, (2) Mg/m3, (3) MN/m2, (4) mN
m 1.10 k = 4.29 kN/m 1.11 x ¼ 1.57 rad/s 1.12 n = 57.3 rev/min Chapter 2 2.1 Fx = −5 N, Fy = 9 N 2.2 F = 5.83 kN, Fz = 5.05 kN 2.3 F ¼ f12:1i þ 21:0j þ 97:0kg kN 2.4 FR ¼ 396 N, a ¼ 42:0 ; b ¼ 48:2 ; c ¼ 86:5 2.5 F ¼ f1.11i þ 1:33j þ 1.00kg kN, a ¼ 124 ; b ¼ 48:3 ; c ¼ 60:0 2.6 F1 ¼ f447i þ 224kg N, F2 ¼ f374i  561j þ 187kg N, F3 ¼ f400kg N 2.7 h ¼ 56:2 2.8 FR ¼ 46:9 N, a ¼ 48:4 ; b ¼ 72:8 ; c ¼ 133 2.9 r ¼ f4.5i þ 7:5j  5.8kg m, r ¼ 10:5 m, a ¼ 64:6 ; b ¼ 44:4 ; c ¼ 124 2.10 h ¼ 83:3 , r1
u2 ¼ 1:0 m, r2
u1 ¼ 0:581 m 2.11 h ¼ 79:3 , r1
u2 ¼ 7:28 mm, r2
u1 ¼ 3:84 mm 2.12 FR ¼ f2:11i  15:3j  4.15kg N, uFR ¼ f0.132i  0:957j  0.259kg 2.13 a ¼ 62:9 ; b ¼ 55:3 ; c ¼ 133 2.14 h ¼ 76.1 , FAO ¼ 24:0 N, F? ¼ 97:1 N 2.15 h ¼ 59.3 , FBA ¼ 1:02 kN, F? ¼ 1:72 kN

456

Answers

Chapter 3 3.1 M O ¼ f34:3j þ 103kgðN
mÞ 3.2 M C ¼ f68:6i  103kgðN
mÞ 3.3 MA ðFÞ ¼ 99:0 N
m 3.4 MO ðFA Þ ¼ 86:6 N
m, MO ðFB Þ ¼ 81 N
m; Counterclockwise 3.5 MA ðFÞ ¼ Fb cos a, MB ðFÞ ¼ Fða sin a  b cos aÞ pffiffiffiffiffiffiffiffiffiffiffiffiffi 3.6 (a) MO ðFÞ ¼ Fb, (b) MO ðFÞ ¼ F b2 þ l 2 sin b 3.7 Mx ¼ 0, My ¼ 50 N
m, Mz ¼ 61:2 N
m 3.8 Mx ¼ 2.89 kN
m, My ¼ 2.89 kN
m, Mz ¼ 0 3.9 Ma ðFÞ ¼ 24 N
m, My ðFÞ ¼ 81 N
m 3.10 Mz ¼ 102 N
m 3.11 M R ¼ f413kgkN
m 3.12 F = 50 N, a ¼ 143 3.13 MR ¼ 2:84 kN
m, a ¼ 128 ; b ¼ 106 ; c ¼ 137 3.14 FR ¼ f107i  727jg N 3.15 FR ¼ f3ig kN, yR ¼ 5:33 m 3.16 FR ¼ Fj, xR ¼ 2a 3.17 FR ¼ f183i  461jgkN, xR ¼ 3:68 m 3.18 F 0R ¼ 2Fi  2Fj, MRO ¼ 2Fa; FR ¼ 2Fi  2Fj pffiffiffi 3.19 MR ¼  3Fa=2; The same 3.20 FR ¼ f20kgN, x ¼ 22:3 m, y ¼ 11:3 m 3.21 FR ¼ f20kgN, x ¼ 6 cm; y ¼ 3 cm 3.22 F 0R ¼ f345i þ 250j þ 10:5kgN, M RO ¼ f51:9i  36:6j þ 104kg N
m 3.23 F 0R ¼ f75:4i  31:7j  78:8kgkN, M RO ¼ f192i þ 223j  335kg kN
m pffiffiffi pffiffiffi pffiffiffi pffiffiffi 3.24 F 0R ¼ 2Fj þ 2Fk, M RO ¼  2Faj þ 2Fak 3.25 FR ¼ f80i þ 60jgN, y ¼ 0:125 m, z ¼ 8 m 3.26 F 0R ¼ f80i þ 100jgN, M RP ¼ f156i þ 195jgN
m, y ¼ 0:125 m, z ¼ 6:44 m 3.27 F0R ¼ f30i þ 50kgkN, M RP ¼ f26:49i  44:15jgkN
m, x ¼ 1:1 m, y ¼ 1:47 m 3.28 xC ¼ 0; yC ¼ 0:869 m 3.29 (a) xC ¼ 3a=8; yC ¼ 2h=5 (b) xC ¼ 4 m, yC ¼ 1:786 m (c) xC ¼ 2b=5; yC ¼ 4ab3 =7 (d) xC ¼ 1 m, yC ¼ 1:6 m 3.30 xC ¼ yC ¼ 0; zC ¼ 1:67 m 3.31 xC ¼ 1:875 m, yC ¼ zC ¼ 0 3.32 xC ¼ yC ¼ 0, zC ¼ 0:5kh þ 0:25ah 3 3.33 xC ¼ 8 cm, yC ¼ 23 cm 4R3  6pr 3 3.34 xC ¼ 0, yC ¼ 3 pR2  6 pr 2 3.35 P ¼ 72 N; xC ¼ 2:83 m; yC ¼ 2 m 3.36 xC ¼ 2:017 m; yC ¼ 1:155 m; zC ¼ 0:716 m pffiffiffi 3.37 h ¼ 3r 3.38 FA0 ¼ 5 kN ð#Þ, MRA ¼ 6:67 kN
m 3.39 FR ¼ 4:67 N, xC ¼ 21 mm

Answers

457

Chapter 4 4.3 TAB ¼ 5:2 kN; TBC ¼ 7:3 kN 4.4 F1 ¼ F2 ¼ 1 kN; F3 ¼ 1:41 kN; F4 ¼ 1:58 kN; F5 ¼ 1:15 kN 4.5 33.6 cm 4.6 a ¼ 2 arcsinðP1 =P Þ 4.7 (a) FA ¼ FB ¼ M =l; (b) FA ¼ FB ¼ M =ðl cos aÞ 4.8 FAx ¼ 0:26 kNð Þ; FAy ¼ 4:24 kNð#Þ; MA ¼ 5:26 kN
m (x) 4.9 FAx ¼ 0; FAy ¼ 849 kNð"Þ; FB ¼ 666 kNð"Þ 4.10 FAx ¼ 25 kN, FAy ¼ 27:7 kN, FB ¼ 35:6 kN 4.11 FAx = –1.41 kN, FAy = –1.08 kN, FB = 2.49 kN Pb  Fa  M Fa þ Pb þ M , FB ¼ 4.12 FAx ¼ F, FAy ¼ 2b 2b 4.13 FAx = 0, FAy = 5.99 kN, FB = 6.1 kN 4.14 FAx = 0, FAy = P + ql, MA ¼ lðP þ 0:5qlÞ 4.15 x ¼ 9:17 m 4.16 P ¼ 860 N, yC ¼ 0:419 m, xC ¼ 0:488 m 4.17 TA ¼ 0:889 kN; TB ¼ 1 kN, TC ¼ 0:111 kN 4.18 Ax ¼ 0; Ay ¼ 0:17 kN, Az ¼ 2 kN, By ¼ 6:83 kN, Bz ¼ 6 kN 4.19 FAx ¼ 0; FAy ¼ 10 kN; FAz ¼ 0; MAx ¼ 15 kN
m, MAy ¼ 0; MAz ¼ 20 kN
m 4.20 FOx ¼ 400 N, FOy ¼ 0; FOz ¼ 500 N, MOx ¼ 150 N
m, MOy ¼ 600 N
m, MOz ¼ 120 N
m 4.21 FAx ¼ FDx ¼ M =l, FAy ¼ FDy ¼ 0 4.22 M2 = 3 N
m, FAB = 5 N 4.23 FD ¼ 69:3 kN, FE ¼ FF ¼ 20 kN 4.24 FB = 1.77 kN, FC = 1.77 kN 4.25 FA = 0.35F, FB = 0.79F 4.26 FAx ¼ 0, FAy = –15 kN, FB = 40 kN, FD = 15 kN 4.27 FAx ¼ 0, FAy ¼ 6 kN, MA ¼ 32 kN
m, FC ¼ 18 kN 4.28 FA ¼ 13 kN, FB ¼ 97 kN, FC ¼ 172 kN, FD ¼ 88 kN 4.29 FAx = 0, FAy = –51.3 kN, FB = 105 kN, FD = 6.25 kN 4.30 T ¼ Fa cos  a=2h  r 4.31 P1 min ¼ 2P 1  R 4.33 Pmin ¼ R tan a cotð45  0:5aÞP1 =a 4.34 FAx = –32 kN, FAy = 16 kN, FBx = 32 kN, FBy = –4 kN 4.35 Cx ¼ 150 Nð!Þ; Cy ¼ 150 Nð#Þ 4.36 Bx ¼ 0; By ¼ 150 N( # Þ 4.37 F1 = 14.6 kN, F2 = –8.75 kN, F3 = 11.7 kN 4.38 FCx = –120 kN, FCy = 34 kN, MC = 276 kN
m 4.39 FAx = 0, FAy = 300 kN, MA = 900 kN
m, FDH = 450 kN 4.40 FCDx ¼ 1.5 kN, FCDy ¼ 3 kN 4.41

458

Answers

4.42 FJI ¼ 3:75 kN, FJE ¼ 12:5 kN, FJD ¼ 0 FIL ¼ 1:5F, FJI ¼ F, FJL ¼ 2:24F, FCL ¼ 2:24F F1 ¼ 0:889 kN; F2 ¼ 1.33 kN; F3 ¼ 0 F1 = –0.417F FOA ¼ 1:41 kN, FOB ¼ FOC ¼ 0:707 kN FAC ¼ 50 kN, FA0 C ¼ FAB0 ¼ 83:3 kN, FCC 0 ¼ FAA0 ¼ 66:7 kN. member forces are zero 4.49 F1 ¼ F2 ¼ 1667 N, F3 ¼ 1667 N, F4 ¼ F5 ¼ 0; F6 ¼ 667 N 4.50 F1 ¼ F5 ¼ F, F3 ¼ F, F2 ¼ F4 ¼ F6 ¼ 0 4.43 4.44 4.45 4.46 4.47 4.48

Chapter 5 5.1 Fs ¼ 8:64 Nð%Þ 5.2 Fk ¼ 0:9 kNð.Þ 5.3 Rising: T = 26.1 kN; falling: T = 21.0 kN 5.4 F ¼ 11:25 N 5.5 0.224 5.6 M = 90.6 N
m 5.7 a  72.3 5.8 Fs ¼ 2 Nð"Þ 5.9 Pmax = 500 N rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi rls 5.10 umax ¼ arcsin ð1 þ l2s Þb sin a  ls cos a sin a þ ls cos a F; (b) F1 max ¼ F 5.11 (a) F1 min ¼ cos a þ ls sin a cos a  ls sin a 5.12 11.7 kN 5.13 (a) Fmin ¼ 2650 N; (b) Fmin ¼ 2370 N 5.14 μs = 0.289 5.15 Fs ¼ 289 Nð%Þ 5.16 PB min ¼ 467 N 5.17 PA max ¼ 11.9 N 5.18 Fmin ¼ 200 N d 5.19 ls  2R 5.20 (a) 1:69 m  x  3:02 m; (b) FAx ¼ 0; FAy ¼ 4 kN; MA ¼ 6:67 kN
m 5.21 Pmax ¼ 208 N 5.22 Fs ¼ P1 sin a; FN ¼ P  P1 cos a; M ¼ P1 ðR sin a  rÞ 5.23 Fmax = 0.698 kN Chapter 6 6.1 18 m 6.2 6.93 s 6.3 t = 0.12 h, d = 4.8 km

Other

Answers

6.4 6.5 6.6 6.7 6.8 6.9 6.10 6.11 6.13 6.14 6.15 6.16 6.17 6.18 6.19 6.20 6.21 6.22 6.23 6.24 6.25 6.26 6.27 6.28 6.29 6.30 6.31 6.32

v ¼ 3:85 m/s, x ¼ 6:92 m    c 1 1 v ¼ v0  ebt  1 , x ¼ x0 þ v0 t þ bc t þ ebt  b b b q ¼ 5:60 m x 2 y  52 þ ¼1 4 5 ax ¼ 36 m=s2 ; ay ¼ 36 m=s2 ; at ¼ 0; an ¼ 50:9 m/s2 vx ¼ 4 m/s, vy ¼ 10 m/s, vz ¼ 3:14 m/s, ax ¼ 0; ay ¼ 2 m/s2 ; az ¼ 0 v ¼ 0; a ¼ 2 m/s2 v ¼ 25 m/s, a ¼ 0:708 m/s2 a = 2.83 m/s2, ða; iÞ ¼ 45 v ¼ ð600t Þcm/s, at ¼ 600 cm/s2 ; an ¼ ð4800t 2 Þ cm/s2 pffiffiffiffiffiffiffiffiffiffiffiffiffiffi t ¼ ðR=at Þ v ¼ f2:51i þ 11:5jgm/s, a ¼ f25:9i þ 10:1jgm/s2 , x 2 =9 þ y 2 =16 ¼ 1 sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi gL  v0 ¼  H 1 þ L cot h sin 2h

ymax = 2.25 m, t = 0.678 s v ¼ vt ¼ 756 m/s, vn  0; at ¼ 3:92 m/s2 ; an ¼ 8:99 m/s2 t ¼ 12:97 s, aA ¼ 16:9 m/s2 ; aB ¼ 2:69 m/s2 vB ¼ 1 m/sð"Þ; aB ¼ 0:75 m/s2 ð"Þ vA ¼ 2 m/sð"Þ; aA ¼ 0:5 m/s2 ð"Þ vA ¼ 10 cm/sð%Þ; aA ¼ 2:5 cm/s2 ð.Þ vM ¼ 0:566 m/sð#Þ; aM ¼ 0:255 m/s2 ð"Þ t = 15 s v B=A ¼ f3:38i þ 5:75jg m/s, aB=A ¼ f0:902i  0:608jg m/s2 v B=A ¼ f65:4i þ 20jg km/h, aB=A ¼ f307i  2132jg km/h2 vb = 20 m/min, vw = 12 m/min, l = 200 m v2 = 1.04 m/s, v1=2 = 3.98 m/s xB ¼ 10t 2 cm, yB ¼ ðh  5t 2 Þcm pffiffiffi pffiffiffi vM =A ¼ 3 cm/s, aM =A ¼ 3=3 cm/s2 p ffiffi ffi aK ¼ 2 cm/s2 ; aK =D ¼ 2 3 cm/s2

Chapter 7 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 7.1 vM ¼ xR; aM ¼ R a2 þ x4 7.2 vM ¼ 9:42 m=s; aM ¼ 444 m=s2 7.3 ttotal = 38 min 7.4 x = 2 rad/s, R = 25 cm 7.5 t ¼ 15:1 s 7.6 Z3 ¼ 8 R1 r n1 7.7 vtr ¼ 30R2

459

Answers

460

7.8 7.9 7.10 7.11 7.12 7.13 7.14 7.15 7.16 7.17 7.18 7.19 7.20 7.21 7.22 7.23 7.24 7.25 7.26 7.27 7.28 7.29 7.30 7.31 7.32 7.33

vA ¼ 44 m/sð#Þ; aAt ¼ 18 m/s2 ð#Þ; aAn ¼ 645 m/s2 ð Þ; vB ¼ 29:3 m/sð#Þ; aBt ¼ 12 m/s2 ð#Þ; aBn ¼ 430 m/s2 ð Þ vA ¼ 5:24 m/s vP ¼ 0:307 m/s, aP ¼ 0:778 m/s2 ; vA ¼ 0:307 m/s, aA ¼ 0:62 m/s2 t ¼ 2:88 s v ¼ 10 cm=s, a ¼ 34:6 cm=s2 v ¼ 17:3 cm=sð"Þ; a ¼ 5 cm=s2 ð#Þ pffiffiffi vCD ¼ 2xr=3; aCD ¼ 10 3x2 r=9 vM ¼ 17:3 m=s; aM ¼ 35 cm=s2 x ¼ v0 sin2 h=h; a ¼ 2v02 sin3 h cos h=h 2 x ¼ 48:1  103 rad/s; a ¼ 9:36  103 rad/s2 v ¼ 2p cm/s, a ¼ an ¼ 0:4p2 cm/s2 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x ¼ e cos xt þ r 2  e2 sin2 xt a ¼ 13:7 cm=s2 a ¼ 22:7 rad=s2 x ¼ f5kgrad/s; vC ¼ f0:5ig m/s v A ¼ f12:4igm/s, aA ¼ f2:5i  19:2jgm/s2 ; v B ¼ f7:6igm/s, aB ¼ f0:5i þ 19:2jgm/s2 x ¼ 16 rad=s; a ¼ 10 rad=s2 aB ¼ 66:2 cm/s2 ; aAB ¼ 2:31 rad/s2 Rx2O xA ¼ 2xO ; aA ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi l 2  R2 vB ¼ 12:0 cm=sð"Þ; aB ¼ 0:04 cm=s2 ð#Þ pffiffiffi vC ¼ 3rxO =2; aC ¼ 3 rx2O =12 vCDB ¼ 2:51 m/s xAB ¼ 5:56 rad/s, vM ¼ 667 cm/s xdisk ¼ 7:25 rad/s xAB ¼ 3 rad/s, xO1B ¼ 5:2 rad/s xCD ¼ 0:25 rad/s

Chapter 8 8.1 F1 = 5.41 kN; F2 = 5 kN; F3 = 3.98 kN 8.2 F = 31.1 kN, safe 8.3 fs ¼ tan h þ a=g cos h 8.4 t = 18 s 8.5 vlimit = 4.66 m/s; vlimit = 64.5 m/s pffiffiffiffiffiffiffiffiffiffiffiffi  8.6 x ¼ 3 3 5t þ 1  1 8.7 aA ¼ 1:2 m/s2 ð Þ; aB ¼ 0:8 m/s2 ð Þ 8.8 aA ¼ 8:4 m/s2 ð!Þ; aB ¼ 4:2 m/s2 ð#Þ 8.9 (1) 0.686 s; (2) d = 3.43 m v0 g g 8.10 x ¼ ð1  ekt Þ; y ¼ h  t þ 2 ð1  ekt Þ k k k 8.11 221 m

Answers

Pl Pl ðax2 þ gÞ, FBM ¼ ðax2  gÞ 2ga 2ga (1) F = Pcos h; (2) F = P(3  2cos h) v ¼ 2:81 m=s, T ¼ 186:6 N (1) 8.00 kN; (2) 7.18 kN; (3) 8.82 kN TA = 8.68 N, TB = 7.35 N IO ¼ mðR21 þ R22 Þ=2 Ix ¼ 318:3  103 kg
m2 Iz ¼ mh 2 =6 m 5m 2 ða þ 3ab þ 3b2 Þ (a) Iz ¼ ða 2 þ 3ab þ 4b2 Þ; (b) Iz ¼ 3 6 3mðR5  r 5 Þ Iz ¼ 10ðR3  r 3 Þ 2l 2 sin2 h ð3P þ P1 Þ Iz ¼ 3g  1 8  IC ¼ MR2 þ mr 2 þ mr R2 þ r 2 þ Rr 2 3 sin h  fs cos h amin ¼ g cos h þ fs sin h sin h þ fs cos h amax ¼ g cos h  fs sin h FA = 12.1 kN; NA = 17.8 kN; FB = 7.92 kN; NB = 11.7 kN pffiffiffiffiffiffiffiffi xmax ¼ g=e 0.351 a ¼ at ¼ 4:91 m/s2 , FA ¼ 71:8 N, FB ¼ 268 N Ax ¼ 0; Ay ¼ 327 N IA ¼ 1081 kg
m2 ; Mf ¼ 6:15 N
m  2  gs 1 Iz ¼ mr 2 2h t ¼ 0:204 s a = 20.8 rad/s2 kC ¼ 0:09 m aC ¼ 1:35 m=s2 ; a ¼ 9 rad=s2 aC ¼ 2:37 m=s2 ; a ¼ 36:1 rad=s2 a ¼ 11:85 rad/s2 ; NB ¼ 55:2 N; NA ¼ 88:3 N T ¼ 9:81 N aCx ¼ 1:96 m/s2 ð!Þ; aCy ¼ 7:85 m/s2 ð#Þ

8.12 FAM ¼ 8.13 8.14 8.15 8.16 8.17 8.18 8.19 8.20 8.21 8.22 8.23 8.24 8.25 8.26 8.27 8.28 8.29 8.30 8.31 8.32 8.33 8.34 8.35 8.36 8.37 8.38 8.39 8.40

Chapter 9 9.1 WF ¼ 509 J, WG ¼ 0, WFd ¼ 177 J 9.2 W ¼ 62:9 J 9.3 110 J h  lk h cot h 9.4 s ¼ lk 9.5 WAB ¼ 0:0858ka 2 ; WBC ¼ 0:0858ka 2

461

Answers

462

9.6 9.7 9.8 9.9 9.10 9.11 9.12 9.13 9.14 9.15 9.16 9.17 9.18 9.19 9.20 9.21 9.22

1 W12 ¼  kl 2 2 88 kN 2Pl 2 x2 T¼ 3g 5 G 22 x l T¼ 48 g P T ¼ l 2 x2 sin2 h 6g v2 T ¼ ð2P þ P2 þ 3P1 Þ 4g  Pr 2 Pr 2  2 xA þ x2B þ T¼ ðxA þ xB Þ2 4g 2g v ¼ 8:03 m=s 61.25 m 10.7 m/s 10.4 MN/m2 vA ¼ 4:33 m/s 29.2 kW 52.1 m P ¼ 239 W vB ¼ 2:23 m/s  

h 3 3 þ cos b  h ¼ arccos l 2 2

9.23 FT 1 ¼ 3P cos b  2P cos h, FT 2 ¼ P cos b þ

2Pl ðcos b  cos hÞ l h

9.24 a ¼ an ¼ 2:63 m/s2 , FA ¼ FB ¼ 249 N 9.25 x0 ¼ 3 rad=s sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 M  m2 grðsin h þ lk cos hÞ 2½M  m2 grðsin h þ lk cos hÞ 9.26 x ¼ u, a ¼ r m1 þ 2m2 r 2 ð2m2 þ m1 Þ 9.27 2.91 m, 82.6 N rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi gp sin h 9.28 x ¼ 4 3r 9.29 1.8 m 2g ðM  PR sin hÞ P 9.30 a ¼ ,T¼ ð3M þ P1 R sin hÞ R2 ð3P þ P1 Þ Rð3P þ P1 Þ 2ð2M  r3 PÞ M ð4P2 þ P3 þ 2PÞ þ 2r3 P1 P 9.31 a ¼ g; Ft ¼ ð4P1 þ 4P2 þ P3 þ 2PÞr3 ð4P1 þ 4P2 þ P3 þ 2PÞr1 9.32 5.05 rad/s 9.33 2.4 cm 3P1 9.34 a ¼ g 9Pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi þ 4P1 ffi r 4Pgx 2Pg 9.35 v ¼ ,a¼ 3P1 þ P2 þ 2P 3P1 þ P2 þ 2P 9.36 x ¼ 4:26 rad/s

Answers

Chapter 10 10.1 I ¼ 9 N
s 10.2 209 N 10.3 F0 ¼ 5000 N 10.4 v ¼ 30:95 m/s, h ¼ 77:4 m 10.5 v ¼ 24:4 m/s 10.6 Fave ¼ 24:8 N 10.7 H O ¼ f32:3i þ 42:8j  8:74kgkg
m2 /s, H P ¼ f40:2i þ 34:1j þ 123:2kgkg
m2 /s 10.8 ðHO ÞA ¼ 42:3 kg
m2 /s, ðHO ÞB ¼ 24:8 kg
m2 /s 10.9 vB2 ¼ 0:5 m/s ð!Þ 10.10 v2 ¼ 6:76 m/s 10.11 v2 ¼ 2:58 m/s 10.12 v2 ¼ 8:02 m/s 10.13 x2 ¼ 20:4 rad/s 10.14 t ¼ 1:94 s, T ¼ 82:6 N 10.15 x2 ¼ 44:1 rad/s 10.16 x2 ¼ 27:2 rad/s 10.17 t ¼ 1:13 s; vC ¼ 8:14 m/s 10.18 t ¼ 2:06 s; vC ¼ 6:35 m/s 6m2 vb 10.19 x2 ¼ 4m1 þ 3m2 10.20 vA ¼ 0:774 m/s, vB ¼ 0:387 m/s 3 1 1 10.21 (a) L ¼ mRxð!Þ; HO ¼ mR2 x; (b) L ¼ mlxð!Þ; HO ¼ ml 2 x 2 2 3 P 9P 2 R x(↲) 10.22 L ¼ xRð"Þ; HO ¼  g 4g

463

References

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