ENGR 106 Introduction To Engineering...

ENGR 106 Introduction To Engineering Problems

Edited and Created by Janet Ash, Jeff McCauley and Jae Hyeuk Suk

September 20, 2011

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Preface

Welcome to “Introduction To Engineering Problems.” This book is intendedfor use in a quarter an introduction to engineering problems which are basedupon the most fundamental engineering class - ‘Statics.’ Because many stu-dents struggled with the Statics as their first engineering course, EngineeringDepartment at Green River Community College offers one introduction coursefor the Statics. We want to make this course interesting and easy for students.

If you have comments or suggestions for us, please contact us, Janet Ash([email protected]), Jeff McCauley ([email protected]), and Jae Suk([email protected]).

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mailto:[email protected]



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Contents

Preface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

I Formation of Engineering Computations 130.1 Format . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

II Concepts of Vectors 19

1 Introduction to Vector 211.1 Vector and Scalar Quantities . . . . . . . . . . . . . . . . . . . . 211.2 Vector Operations . . . . . . . . . . . . . . . . . . . . . . . . . . 23

1.2.1 Addtion . . . . . . . . . . . . . . . . . . . . . . . . . . . . 231.2.2 Subtraction . . . . . . . . . . . . . . . . . . . . . . . . . . 241.2.3 Multiplication of a Vector by a Scalar . . . . . . . . . . . 241.2.4 Summary of Vector Operations . . . . . . . . . . . . . . . 25

1.3 Vector Algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . 251.3.1 Unit Vector . . . . . . . . . . . . . . . . . . . . . . . . . . 26

1.4 Unit Orthogonal Vectors . . . . . . . . . . . . . . . . . . . . . . . 261.4.1 Components of Vector . . . . . . . . . . . . . . . . . . . . 271.4.2 Addition of Vectors in Component Form . . . . . . . . . . 311.4.3 The Unit-Vector Concept . . . . . . . . . . . . . . . . . . 32

1.5 Position and Displacement Vectors . . . . . . . . . . . . . . . . . 351.6 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

Exercise 41

2 Products of Vectors 492.1 Multiplication of Vectors by Vectors . . . . . . . . . . . . . . . . 492.2 The Dot (or Scalar) Product . . . . . . . . . . . . . . . . . . . . 50

2.2.1 Component of Vector . . . . . . . . . . . . . . . . . . . . 522.3 The Cross (or Vector) Product . . . . . . . . . . . . . . . . . . . 542.4 Division by a Vector . . . . . . . . . . . . . . . . . . . . . . . . . 612.5 Vector Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . 612.6 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62

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6 CONTENTS

Exercise 65

III Application of Vectors 75

3 Application of Vector Products 773.1 Moment of a Force about a Point . . . . . . . . . . . . . . . . . . 773.2 Relation Between Angular and Linear Velocity . . . . . . . . . . 823.3 Geometric Applications of Vector Products . . . . . . . . . . . . 853.4 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88

Exercise 89

IV Dimension and Units 97

4 Dimension 994.1 Concept . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 994.2 Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 994.3 Fundamental and Derived Dimensions . . . . . . . . . . . . . . . 1004.4 Algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100

4.4.1 Taking dimensions . . . . . . . . . . . . . . . . . . . . . . 1004.4.2 Commutative relations with multiplication and division . 1014.4.3 Addition . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1024.4.4 Dimensional Homogeneity . . . . . . . . . . . . . . . . . . 102

4.5 Newton’s Second Law . . . . . . . . . . . . . . . . . . . . . . . . 103

Exercise 105

5 Unit Systems 1095.1 Concept . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1095.2 Notation and Derived Units . . . . . . . . . . . . . . . . . . . . . 1095.3 Newton’s Second Law . . . . . . . . . . . . . . . . . . . . . . . . 1105.4 Metric Unit Systems . . . . . . . . . . . . . . . . . . . . . . . . . 1105.5 British System . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1115.6 Definition of Weight-force . . . . . . . . . . . . . . . . . . . . . . 1125.7 Engineering Practice Units . . . . . . . . . . . . . . . . . . . . . 1135.8 Relation between British and Engineering Units . . . . . . . . . . 1145.9 Conversion of Units . . . . . . . . . . . . . . . . . . . . . . . . . 1155.10 Conventional Reference Points . . . . . . . . . . . . . . . . . . . 115

Exercise 117

6 Dimensional Analysis and Similitude 1196.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1196.2 The Pi-theory of Dimensional Analysis . . . . . . . . . . . . . . . 1196.3 First Example: Speed of a Freely Falling Body . . . . . . . . . . 120

CONTENTS 7

6.4 Second Example: Period of a Simple Pendulum . . . . . . . . . . 122

Exercise 127

8 CONTENTS

List of Tables

5.1 Metric Unit System . . . . . . . . . . . . . . . . . . . . . . . . . 1105.2 MKS vs CGS System . . . . . . . . . . . . . . . . . . . . . . . . . 1115.3 MLT Unit Chart . . . . . . . . . . . . . . . . . . . . . . . . . . . 1125.4 Completed Unit Chart . . . . . . . . . . . . . . . . . . . . . . . . 1145.5 Unit Conversion Table . . . . . . . . . . . . . . . . . . . . . . . . 117

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10 LIST OF TABLES

List of Figures

1 An example for format of engineering computations. . . . . . . . 16

1.1 A Vector . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 221.2 ~M and ~N Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . 231.3 Additon Vectors - ~A + ~B = ~C . . . . . . . . . . . . . . . . . . . . 241.4 Additon Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . 241.5 Subtraction Vectors . . . . . . . . . . . . . . . . . . . . . . . . . 251.6 Unit vectors for x− y − z axes . . . . . . . . . . . . . . . . . . . 261.7 Positive Direction of Unit vectors for x− y − z axes . . . . . . . 271.8 Components . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 271.9 Polygon Components . . . . . . . . . . . . . . . . . . . . . . . . . 281.10 x− y − z Components . . . . . . . . . . . . . . . . . . . . . . . . 281.11 x− y − z Components with Unit vectors . . . . . . . . . . . . . . 291.12 x− y − z Components with Unit vectors and angles . . . . . . . 291.13 Vector Addition by Components . . . . . . . . . . . . . . . . . . 311.14 Concept of Unit Vectors . . . . . . . . . . . . . . . . . . . . . . . 321.15 Position Vector . . . . . . . . . . . . . . . . . . . . . . . . . . . . 351.16 Displacement Vector . . . . . . . . . . . . . . . . . . . . . . . . . 361.17 Another expression for Displacement Vector . . . . . . . . . . . . 371.18 Example 1-7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 371.19 Example 1-8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

2.1 Definition of Work . . . . . . . . . . . . . . . . . . . . . . . . . . 502.2 Definition of Moment . . . . . . . . . . . . . . . . . . . . . . . . . 502.3 Dot Product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 512.4 Component of Vector with Dot Product . . . . . . . . . . . . . . 532.5 Component of Vector with Dot Product . . . . . . . . . . . . . . 532.6 Right Hand Rule for Cross Product . . . . . . . . . . . . . . . . . 542.7 Right Hand Rule for Cross Product . . . . . . . . . . . . . . . . . 552.8 Cross Product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 552.9 Cross Product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 562.10 Right Hand Rule for Cross Product . . . . . . . . . . . . . . . . . 562.11 Cross Product with unit vectors of the Cartesian axes . . . . . . 582.12 ~A · ~B = ~A · ~C does not mean ~B = ~C . . . . . . . . . . . . . . . . 62

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12 LIST OF FIGURES

3.1 Application of Vector Product . . . . . . . . . . . . . . . . . . . . 783.2 Computing Vector Product . . . . . . . . . . . . . . . . . . . . . 783.3 Computing Cross Product for Mmment about a force . . . . . . 793.4 Example 3-1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 803.5 Example 3-2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 813.6 Example 3-3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 823.7 Angular Velocity . . . . . . . . . . . . . . . . . . . . . . . . . . . 833.8 Example 3-4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 833.9 Example 3-5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 843.10 Example 3-6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 853.11 Example 3-8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87

6.1 A simple pendulum . . . . . . . . . . . . . . . . . . . . . . . . . . 1226.2 Similarity relation for simple pendulum . . . . . . . . . . . . . . 124

Part I

Formation of EngineeringComputations

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Format

CHAPTER OBJECTIVES

• Learn how to describe the problems with drawing or wording

• Understand how to present/report your solution

0.1 Format

The following represent some commonly accepted practices for preparing engi-neering reports and calculations. There will undoubtedly be some differencesin various industries and areas of engineering. These are the standards for thisparticular class.

1. Always use a PENCIL.

2. All problems will be done on engineering paper .

3. Use horizontal fraction lines, e.g., fractions should be appears as 2 78 .

4. Show exponents and subscripts clearly.

5. Show decimal quantities less then one with a zero before the decimal point,e.g., 0.031.

6. Work from the top of the page DOWN (as opposed to across) in a logicalsequence.

7. Put only one mathematical statement on a line, i.e., don’t skip around.

8. Multiplication is only shown by using parentheses or brackets.

9. Show all units and don’t mix units.

10. Answer will be indicated with a box drawn around them and referencedin the right-hand margin.

11. Answers will be evaluated to a single numerical value when possible.No radical or trig functions left to evaluate.

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16

Figure 1: An example for format of engineering computations.

0.1. FORMAT 17

12. When more then one problem is presented on a page, the problems willbe separated with a continuous double line (drawn with a straight edge)across the page.

13. Do not crowd your work.

14. Label your sketches and make them large enough to display all knownvalues and unknowns represented.

15. Use the problem format that is attached.

16. Indicate all vectors as vectors or unit vectors.

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Part II

Concepts of Vectors

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Chapter 1

Introduction to Vector

CHAPTER OBJECTIVES

• Distinguish the scalar and the vector

• To express force and position in Cartesian vector form and explain howto determine the vector’s magnitude and direction.

• Understand how to operate vector algebras.

• Understand the meaning of the unit vector and how to calculate an unitvector with a given any vector.

1.1 Vector and Scalar Quantities

It is necessary for engineers to establish an efficient notation for the descriptionof the many physical quantities encountered in engineering design and analysis.Many of these quantities require specification of direction as well as magnitudefor their complete identification. For example, to completely define the veloc-ity of a particle one must state not only magnitude but a direction of motionrelative to some reference system. A quantity that has magnitude and direc-tion, will add according to the parallelogram law, and may be presented by astraight-line segment of definite length is called a vector.

Common physical quantities in engineering mechanics that are vectors arevelocity, acceleration, force, and moment of a force. A vector can be representedgraphically by a directed line segment with an arrowhead indicating its directionand the length of the line indicating its magnitude like Figure 1.1.

In this book a vector will be analytically designated by boldface type and ar-row top, such as ~A , with its magnitude or length represented by |~A|. Boldfacetype is, however, impracticable to reproduce at the blackboard and on paper

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22 CHAPTER 1. INTRODUCTION TO VECTOR

Figure 1.1: A Vector

in normal problem computations; thus in problem computations it is recom-mended that a vector quantity be designated by a symbol with a bar above it,

such as ~A. In Figure 1.1, the vector ~A may also be represented as−−→CD, with a

magnitude denoted by |−−→CD|.

Many quantities encountered in mechanics require only a single number fortheir complete definition; these are known as scalar quantities. A scalar is aquantity having magnitude only, such as mass, length, time, temperature, andso on. A scalar may also be considered as a number whose value is the same inall systems of reference. Thus the length or magnitude of a vector is a scalar.The positive and negative numbers of ordinary algebra are typical scalars, andtherefore the basic operations for handling scalars follow the rules of elementaryalgebra. Of course, scalars also may or may not possess dimensions.

Vector quantities can be further categorized into free or fixed vectors. A freevector is one having not specific location in space, which means it is invariantwhen subjected to parallel translation. Vectors representing the physical quan-tities of acceleration and velocity are examples of free vectors. A fixed vector, onthe other hand, has a unique point of application. The vector used to representa force is a fixed, or bound, vector. In addition, we frequently refer to vectorsrepresenting certain other physical quantities as sliding, or transmissible, vec-tors, which means they may be moved anywhere along their line of action. Theline of action refers to the infinite straight line along which the vector acts. Anull vector is a vector of zero magnitude in nay arbitrary direction.

Let us summarize - Scalar and Vector

Scalar: A scalar is any positive or negative physical quantity that can be com-pletely specified by it magnitude. Examples of scalar quantities includelength, mass, and time.

Vector: A vector is any physical quantity that requires both a magnitude and adirection for its complete description. Examples of vectors encountered instatics are fore, position, and moment. A vector is shown graphically byan arrow. The length of the arrow represents the magnitude of the vector,and the angle θ between the vector and a fixed axis defines t

1.2. VECTOR OPERATIONS 23

Initially in our study we shall consider that vectors are not localized in space.For example, the vectors ~M and ~N in Figure 1.2 are equal if they have the samemagnitude and direction and the units in which they are expressed are identical,in spite of the fact that they have different positions. Limitations to this conceptwill be pointed out in subsequent courses in mechanics.

Figure 1.2: ~M and ~N Vectors

Some quantities, such as finite rotations, although possessing both magni-tude and direction, will not add according to the parallelogram law and thereforeare not vectors.

1.2 Vector Operations

1.2.1 Addtion

The parallelogram law establishes the rule for the addition of vectors and in fact,restricts the definition of a vector to those quantities which have magnitude anddirection and will add according to the parallelogram process.

In Figure 1.3, the sum, or resultant, of vectors ~A and ~B is the diagonalvector ~C formed by the diagonal of the parallelogram. Note that the vectors~A and ~B are arranged tail to tail for addition. The vectors ~A and ~B may becalled components of vector ~C.

The vector sum of ~A and ~B can be obtained equally well by using thetriangle rule. To do this we shift the vectors without changing their magnitudesor directions until, as shown in Figure 1.4, the initial point of ~B coincides withthe terminal point of ~A. Then the sum of vectors ~A and ~B can be defined asthe vector ~C represented by the line drawn from the initial point of ~A to theterminal point of ~B. This shifting of vectors in space is justified, of course byour previous consideration that vectors are not localized in space. It is obvious


Figure 1.3: Additon Vectors - ~A + ~B = ~C

that the resultant ~C is the same as both cases. From the geometry we canconclude that ~A + ~B = ~B + ~A, that is, the order of addition is immaterial. Byrepeated application of the parallelogram law any number of vectors may beadded.

Figure 1.4: Additon Vectors

1.2.2 Subtraction

Subtraction of one vector form another vector is the same thing as addition of itsnegative. For example, as shown in Figure 1.5, to subtract vector ~Q from vector~P we reverse the direction of ~Q and then add according to the parallelogramlaw, so that ~R = ~P− ~Q.

To reverse vector ~Q we must multiply it by the scalar (−1). This operation ofmultiplication by a scalar is described following subsection 1.2.3 in more detail.

1.2.3 Multiplication of a Vector by a Scalar

The operation of multiplication of a vector by a scalar conforms to fundamentallaws of scalar algebra. The product of a vector ~A and a positive scalar m is avector having the same direction as ~A but with a magnitude m times as great.If scalar m is negative, the vector m~A has its direction reversed.

1.3. VECTOR ALGEBRA 25

Figure 1.5: Subtraction Vectors

1.2.4 Summary of Vector Operations

The operations of vector - addition, subtraction, and multiplication of a vectorby a scalar, have been briefly presented to show how they conform to certain lawsof scalar algebra. These operations and laws are familiar as the basic processesof ordinary scalar algebra and therefore will not be discussed in detail. Theyare summarized here without proof for concise review:

1. Commutative law of vector addition: ~A + ~B = ~B + ~A

2. Associative law of vector addition:(~A + ~B

)+ ~C = ~A +

(~B + ~C

)3. Commutative law of scalar multiplication: m~A = ~Am

4. Associative law of scalar multiplication: m(n~A)

= n(m~A

)= (mn) ~A

5. Distributive law of scalar multiplication: (m+ n) ~A = m~A + n~A

6. Distributive law of scalar multiplication: m(~A + ~B

)= m~A +m~B

1.3 Vector Algebra

To facilitate handling the directional aspect of vectors, a vector algebra hasbeen devised by means of which vectors may be manipulated in ways consistentwith the physical of geometrical problem in which they occur. This powerfulmathematical tool known as vector algebra is now an essential part of the math-ematical background of engineers. As will be shown later, vector algebra differsin several important respects from the familiar scalar algebra.


1.3.1 Unit Vector

Frequently it is convenient to express vectors as the product of scalar multipliedby a vector. The unit vector, which has a magnitude of 1 in a certain prescribeddirection, plays an important role in this concept. For consistency of notationit is convenient to designate a unit vector having the line of action of anothervector by the lower-case from the capital letter denoting the vector. Thus brepresents a unit vector whose line of action is colinear with vector ~B. Such avector may be expressed as follows:

b =~B

|~B|(1.1)

This expression can be seen to conform with our previous notation concerningthe multiplication of a vector times a scalar, or

~B = |~B|b (1.2)

1.4 Unit Orthogonal Vectors

Since the three orthogonal axes of Cartesian coordinates are used so extensivelyin engineering analysis, unit vectors directed along these axes are of particularimportance. It is customary to denote the unit vectors directed among thepositive x, y, and z axes as i, j, and k, respectively, as shown in Figure 1.61.The orientation of the x, y, and z axes conforms to the so-called “right-hand

Figure 1.6: Unit vectors for x− y − z axes

rule”, which arbitrarily states that when the fingertips of the right hand curlfrom the positive end of the first axis to the positive end of the second, inalphabetical order, the thumb of the right hand points toward the positive endof the third axis. This is shown in Figure 1.7, along with some other right-handCartesian axes. The student should verify these as correct orientation of theaxes for the right-hand rule. For the sake of uniformity, we shall orient the x,y, and z axes as shown in Figure 1.6.

1In this chapter, three colors (red, blue, and green) will be used for x-i, y-j, and z-kaxes-unit vectors.

1.4. UNIT ORTHOGONAL VECTORS 27

Figure 1.7: Positive Direction of Unit vectors for x− y − z axes

1.4.1 Components of Vector

By reversing the process of addition, any vector ~M may be decomposed into twovectors in any arbitrary direction coplanar with ~M as long as the sum of thetwo component vectors equals the original vector. This is shown graphically in

Figure 1.8: Components

Figure 1.8. Indeed, a vector may be decomposed into any number of componentvectors as long as the last component closes the polygon, as shown in Figure1.9. The vectors need not be coplanar, and the polygon does not necessarilyhave to be a plane figure.

The most useful decomposition of a vector into component vectors is alongthe three orthogonal axes of the Cartesian coordinates. Consider a vector ~Vlocated and oriented as the diagonal in the rectangular parallelepiped shownin Figure 1.10. It can be seen that Vx, Vy, and Vz are the vector interceptsalong the x, y, and z axes, respectively. Some confusion exists in terminologyat this point as to whether Vx, Vy, and Vz, we are considering their magnitudesonly, and will call them the orthogonal scalar components of the vector. Thisnotation is necessary for conformity with the previous idea that the product of avector and scalar is a vector having the direction of the vector and a magnitudeequal to the ordinary scalar product of their magnitudes. To convert thesescalar components Vx, Vy, and Vz into vector components, it is necessary only


Figure 1.9: Polygon Components

Figure 1.10: x− y − z Components


to multiply by the three unit orthogonal vectors i, j, and k, respectively (seeFigugre 1.11).

Figure 1.11: x− y − z Components with Unit vectors

The vector components of ~V parallel to the x, y, and z axes are, respectivelyVxi, Vy j, and Vz k. Since a set of component concurrent vectors is equivalent to

the original vector, we may write vector ~V in the following manner:

~V = Vxi+ Vy j + Vz k (1.3)

At this point we would do well to consider some terminology and some of therelationships that exist between a vector and the orthogonal axes. Consider avector ~V with its vector components as shown in Figure 1.12. The direction of~V in space relative to the orthogonal axes is given by the three angles α, β, andγ, measured from the positive ends of the x, y, and z axes, respectively. Theangels α, β, and γ are called the direction angles of the vector. It can be seenthat the direction angels may vary between 0 and 180.

Figure 1.12: x− y − z Components with Unit vectors and angles


From right-triangle trigonometry it is apparent that the cosines of the direc-tion angles, which are called the direction cosines of the vector, are as follows:

cosα =Vx

|~V|≡ l (1.4)

cosβ =Vy

|~V|≡ m (1.5)

cos γ =Vz

|~V|≡ n (1.6)

These direction cosines are frequently denoted by the lower-case letters l, m,and n. rearranging these relationships give the following:

Vx = |~V| cosα = |~V|l (1.7)

Vy = |~V| cosβ = |~V|m (1.8)

Vz = |~V| cos γ = |~V|n (1.9)

Referring to Figure 1.12, and using the Pythagorean theorem, we can now ex-press in terms of its orthogonal components,

|~V|2 = (V ′)2 + (Vz)2

(V ′)2 = (Vx)2 + (Vy)2

The (V ′)2 can be substitued.

|~V|2 = (Vx)2 + (Vy)2 + (Vz)2 (1.10)

or |~V| =√

(Vx)2 + (Vy)2 + (Vz)2 (1.11)

The equation (1.10) in a slightly different manner gives

|~V|2 = (Vx)2 + (Vy)2 + (Vz)2

= |~V|2(cosα)2 + |~V|2(cosβ)2 + |~V|2(cos γ)2

= |~V|2((cosα)2 + (cosβ)2 + (cos γ)2

)Dividing by |~V|2 yields

1 = cos2 α+ cos2 β + cos2 γ (1.12)

or 1 = l2 +m2 + n2 (1.13)

It is apparent form equations (1.12) and (1.13) that the values of the threedirection angles α, β, γ are not independent. Given any two of the angles, wemay at once determine the third by equation (1.12). It is of importance to notethat the values for the direction cosines may be positive or negative, with theproper choice being dependent upon the physical considerations of the problem.


These important relations may also be written in the form

cosα

Vx=

cosβ

Vy=

cos γ

Vz=

1

|~V|(1.14)

The equation (1.14) shows that Vx, Vy, Vz and ~V are respectively proportionalto cosα, cosβ, cos γ and l.

1.4.2 Addition of Vectors in Component Form

As previously explained, the parallelogram law establishes the procedure foraddition of vectors. In three-dimensional work, if it is desired to find the sumof two or more vectors, we first express them in Cartesian-component form sothat we can add the components along the respective orthogonal axes directly.For example, as shown in Figure 1.13, after being arranged tip to tail, vectors

Figure 1.13: Vector Addition by Components

~A and ~B may be decomposed into Cartesian components whose orthogonalcomponents can then be readily added. Thus the sum of any two vectors maybe easily obtained as the sum of the respective orthogonal components:

~A + ~B =(Axi+Ay j +Az k

)+(Bxi+By j +Bz k

)= ~C

or ~C = (Ax +Bx) i+ (Ay +By) j + (Az +Bz) k

We should observe from Figure 1.13 that a vector sum may not necessarilybe uniquely specified in space, unless, of course, the initial point of the vector is


stated. The scalar components of ~C are (Ax +Bx), (Ay +By), and (Az +Bz),respectively, and obey the ordinary laws of scalar algebra.

1.4.3 The Unit-Vector Concept

At this point we shall digress and return to the concept of the unit vector. Theidea of the unit vector may be demonstrated more clearly by referring to Figure1.14, which shows a vector ~B in the yz plane, with scalar components of 3 and4 along the y and z axes, respectively. Thus expressed in Cartesian components

Figure 1.14: Concept of Unit Vectors

becomes ~B = 3j + 4k. Clearly the magnitude of ~B can be computed as follows:

|~B| =√

32 + 42 =√

25 = 5

to obtain a unit vector along vector ~B by using equation (1.1) as previously

explained, all that is necessary to divide by |~B|, or

b =~B

|~B|=

3j + 4k

5

= 0.6j + 0.8k

Checking to be certain that the magnitude of b is unity,

|b| =√

0.62 + 0.82 =√

0.36 + 0.64 =√

1.00 = 1

Thus we see that the equation (1.1) has been verified.

It should be pointed out that the expansion of one unit vector in termsof another, as in the preceding example, is not unique but depends upon thecoordinate system chosen with respect to the vector.

Example 1-1

A force ~P has components Px = −3 lb., Px = 6 lb., and Pz = 2 lb. (a) Express~P in Cartesian coordinates. (b) Find the magnitude of ~P. (c) Determine thedirection angles.


Solution (a) Using the equation (1.3)

~P = −3i+ 6j + 2klb

(b) From the equation (1.11)

|~P| =√

(−3)2 + (6)2 + (2)2 =√

49 = 7lb

(c) By using the equation (1.14)

cosα

Px=

cosβ

Py=

cos γ

Pz=

1

|~P|cosα

−3=

cosβ

6=

cos γ

2=

1

7α = 115.4

β = 31.0

γ = 73.4

Note that α must be greater than 90 because Px = −3. As a check use theequation (1.12):

1 = cos2 α+ cos2 β + cos2 γ

1 = (−0.429)2 + (0.856)2 + (0.286)2

1 = 0.184 + 0.734 + 0.0082 = 1.000γ

Example 1-2

Given vectors ~A = 3i+ 2j + 4k lb, ~B = 2i+ 5j − 3k lb, and ~C = i+ k lb , find(a) ~A + ~B + ~C, (b) |~A + ~B + ~C| and (c) 2~A− 3~B− 4~C

Solution (a)

~A + ~B + ~C = (3i+ 2j + 4k) + (2i+ 5j − 3k) + (i+ k)

= (3 + 2 + 1)i+ (2 + 5 + 0)j + (4− 3 + 1)k

= 6i+ yj + 2klb

(b)

|~A + ~B + ~C| =√

62 + 72 + 22 =√

36 + 49 + 4 =√

89 = 9.43lb

(c)

2~A− 3~B− 4~C = 2(3i+ 2j + 4k)− 3(2i+ 5j − 3k)− 4(i+ k)

= (6i+ 4j + 8k) + (−6i− 15j + 9k) + (−4i− 4k)

= (6− 6− 4)i+ (4− 15 + 0)j + (8 + 9− 4)k

= −4i− 11yj + 13klb


Example 1-3

Find the angle that vector ~B = 5i− 2j + 2.65k makes with coordinates axes.

Solution From the equation (1.11):

|~B| =√

(5)2 + (−2)2 + (2.65)2 =√

25 + 4− 7 =√

36 = 6

By inspection Bx = 5, By = −2, and Bz = 2.65. By the equation (1.14):

cosα

5=

cosβ

−2=

cos γ

2.65=

1

6α = 33.5

β = 109.5

γ = 63.8

Example 1-4

A vector ~D has direction angles α = 45 and β = 70. It is also known thatangleγ is less than 90. Determine angle gamma. From the equation (1.12):

Solution

cos2 45 + cos2 70 + cos2 γ = 1

(0.707)2 + (0.342)2 + cos2 γ = 1

0.5 + 0.117 + cos2 γ = 1

cos2 γ = 1− 0.617 = 0.383

cos γ = ±√

0.383 = ±0.618

Therefore, γ = 51.8. It should be noted that minus value for cos γ is discardedbecause γ is known to be less than 90.

Example 1-5

Define a vector ~E whose magnitude is 10 and whose direction angles are allequal and less than 90.

Solution From the equation (1.13):

l2 +m2 + n2 = 1

Since direction angles are all equal and less than 90,

l2 = m2 = n2.

1.5. POSITION AND DISPLACEMENT VECTORS 35

Therefore,

3l2 = 1

l2 = 0.333

l = ±√

0.333 = ±0.578 = m = n

Ex = 0.578× 10 = 5.78 = Ey = Ez

Finally,

~E = 5.78i+ 5.78j + 5.78k

Example 1-6

Define a unit vector along vector ~F = −5i− 6j + 4.47k.

Solution Using the equation (1.1):

f =~F

|~F|=

−5i− 6j + 4.47k√(−5)2 + (−6)2 + (4.47)2

=−5i− 6j + 4.47k√

81

= −5

9i− 6

9j +

4.47

9k

1.5 Position and Displacement Vectors

To facilitate the application of vector algebra to some analytical transformationsthat will be discussed later, such as the moment of a force about a point, two ormore useful vector quantities should be defined. The first of these is the positionvector. As shown in Figure 1.15, the position vector, usually denoted as ~r, is

Figure 1.15: Position Vector

the directed line segment between some reference point, usually the origin of thecoordinate system, and any point A in space. The scalar components of are the


space coordinates of the point A, then, as previously described, it is a simplematter to express ~r in vector form:

~r = xi+ yj + zk (1.15)

A second quantity, known as the displacement vector, is frequently involved invector operations, particularly in the computation of the moment of a couple.The displacement vector may be defined as a directed line segment between twopoints when neither point is at the origin of the coordinate reference system.A displacement vector between two points P and Q is shown in Figure 1.16.The sense of the displacement vector, frequently denoted as ~e, must be defined

Figure 1.16: Displacement Vector

with utmost care, lest the signs be reversed on the scalar components. In thisinstance we see intuitively that

~e = (x1 − x)i+ (y1 − y)j + (z1 − z)k (1.16)

Another method of evaluating the displacement vector ~e is to draw positionvectors ~r1 and ~r2 to points P and Q, respectively. Since the vectors ~r1 and ~eare arranged tip to tail they can be added according to the triangle rule, and itcan be seen from Figure 1.17 that

~r1 + ~e = ~r2~e = ~r2 −~r1

(1.17)

From the equation (1.15):

~r1 = xi+ yj + zk & ~r2 = x1i+ y1j + z1k

∴ ~e = ~r2 −~r1 = (x1i+ y1j + z1k)− (xi+ yj + zk)

= (x1 − x)i+ (y1 − y)j + (z1 − z)k

which is the same result as shown in the equation (1.16).

Example 1-7

Write the position vector ~r for the point P located as shown n Figure 1.18.

1.5. POSITION AND DISPLACEMENT VECTORS 37

Figure 1.17: Another expression for Displacement Vector

Figure 1.18: Example 1-7


Solution By inspection and using the equation (1.15):

~P = 4i− 3j + 5k

Example 1-8

What is the displacement vector between position (2, 3, 6) and position (4, 5,3)?

Solution By inspection and from Figure 1.19, it can be seen that vector~e starts at x = 2 and terminates at x = 4, an increase of two units; accordingly,it has an x components of +2. By similar reasoning, the y and z components


are +2 and −3, respectively. Thus

~e = 2i+ 2j − 3k

Also, from the equation (1.17),

~e = ~r2 −~r1 = (4i+ 5j + 3k)− (2i+ 3j + 6k)

= 2i+ 2j − 3k

Any or all of these techniques may be used to determine a displacement vector.

1.6 Summary

A quantity that has magnitude and direction, will add according to the paral-lelogram law, and may be represented graphically by a straight-line segment ofdefinite length is called a vector. A scalar is a quantity having magnitude only.A free vector has no fixed position in space, while a fixed vector has a uniquepoint of application. Sliding or transmissible vectors may be moved anywherealong their lines of action. A null vector has a zero magnitude.

1.6. SUMMARY 39

Vectors add according to the parallelogram law or the triangle rule. Onevector is subtracted from another by reversing its direction and adding. Theoperation of multiplication of a vector by a scalar conforms to the fundamentallaws of ordinary scalar algebra, as do the operations of addition and subtraction.

A unit vector has a magnitude of l in any prescribed direction. Any vectormay be resolved into three components along the orthogonal axes of Cartesiancoordinates. Unit vectors along the positive x, y, and z axes are denoted as i,j, and k, respectively. The orientation of the i, j, and k unit vectors conformsto the so-called right-hand rule. Since a set of component concurrent vectors isequivalent to the original vector, any vector ~V may be expressed in the followingmanner:

~V = Vxi+ Vy j + Vz k

where Vx, Vy, and Vz are known as the orthogonal scalar components of ~V.

The direction of a vector in space relative to the orthogonal axes is givenby the three direction angles α, β, and γ are the direction cosines and are fre-quently denoted as l, m, and n, respectively. The values of the three directionangles α, β, and γ are not independent, for if any two are known, the third maybe obtained by the relationship.

1 = cos2 α+ cos2 β + cos2 γ

The sum of two or more vectors can be most expeditiously obtained by firstexpressing the vectors in Cartesian-component form, so that the componentsalong the respective axes may then be added directly.

A position vector is the directed line segment between some reference point,usually the origin of the coordinate system, and any point in space. The dis-placement vector is the directed line segment between two points when neitherpoint is at the origin of the coordinate reference system.


Exercise

1-1 Determine the magnitude and the direction cosines of a force which as x,y, and z components of 8, 4, and 1 lb., respectively.

1-2 Given | ~A| = 500, find the scalar components Ax, Ay, and Az of vector ~Ashown in Prob. 1-2, 3, 4, and 5.

Prob. 1-2, 3, 4, and 5

1-3 Solve for the direction cosines and direction angles of vector ~B in Prob.1-2, 3, 4, and 5.

1-4 The force | ~B|= 617 lb., in Prob. 1-2, 3, 4, and 5, find its scalar componentsBx, By, and Bz.

1-5 In Prob. 1-2, 3, 4, and 5, if force |~C| = 37.4 lb., find its direction cosinesand scalar components Cx, Cy, and Cz.

1-6 Determine whether the following are vectors or scalars: (a) temperature,(b) mass, (c) acceleration, (d) displacement, (e) density, (f) velocity, (g)speed, (h) pressure.

1-7 In Prob. 1-7, what angle does ~OR make with the orthogonal axes, if pointR is at (4, 8, 1)?

1-8 A force ~C (see Prob. 1-8) whose magnitude is 70.7 lb. acts at pointa(3, 0, 0) through point b(0, 5, 4). Express the force in vector form.

41


Prob. 1-7

Prob. 1-8

1-9 A force ~M has a magnitude of 34 lb., emanates from the origin, and passesthrough the point (12, 8, 9). Write the force in vector notation.

1-10 The tension, ~T , in the cable of Prob. 1-10 is 140 lb. Write the tensionforce in terms of the unit vectors i, j, and k.

Prob. 1-10

1-11 A force ~P = 5Bi+3Bj+3Bk has a positive x component of 125 lb. Write

1.6. SUMMARY 43

the force ~P in vector notation, showing the numeric scalar coefficient.

1-12 Express in vector form the vector ~N whose magnitude is 98 and has adirection parallel to the vector ~M = 3i− 2j + 6k.

1-13 A vector ~V has a magnitude of 50 lb., with components Vx = 13 andVy = −20. Express ~V in Cartesian components and find its directionangles.

1-14 If ~A = 5i+7j−9k and ~B = i−4j+3k, find (a) ~A+ ~B, (b) ~A− ~B, (c) ~B− ~A,

and (d)2 ~A− 3 ~B.

1-15 Find the length of the following vectors: (a)~R = 7i − 5j + 6k ft., (b)~S =

3i+ 4j − 5k ft., and (c)~R− ~S.

1-16 Find the magnitude and direction angles each of the following vectors: (a)~A = −10k, (b) ~B = −3i+ 4j, and (c)~C = 9i+ 3j − 5.57k.

1-17 What is the sum of the following three vectors? What angle does theresultant vector make with the z axis? (a)~P = 4i − 5j + 6k, (b) ~Q is avector whose magnitude is 10, is in the xz plane, makes an angle of 60

with the positive end of the z axis, and is directed away from the origin.(c) ~R = 3i− 7j − 3k.

1-18 Write vector ~P of Prob. 1-18 in terms of the unit vectors i, j, and k.

Prob. 1-18

1-19 If an airplane steering a course due east with an air speed of 200 knotsencounters a wind blowing N30E at 50 knots, how far an in what directionfrom the starting point will the aircraft be in 1 hour? Note: 1 knot = 1nautical mile per hour.

1-20 In Prob. 1-20, if ~P is 100 lb., determine Px, Py, and Pz.


Prob. 1-20

1-21 A destroyer is steaming due north at 25 knots. Ocean currents are knownto be moving S45W at 4 knots. What is the course and speed that theship is actually making?

1-22 Determine a vector in the yz plane that is equally inclined to both the yand z axes and has the same length as ~P = 5i− 7j + 6k.

1-23 Find a vector in the xy plane having the same length and the same pro-jection along the x axis as ~R = 7i− 5j + 9k.

1-24 Resolve the 100 lb. Force in Prob. 1-24 into component vectors withdirections indicated by the dashed lines.

Prob. 1-24

1-25 Given vectors ~A = 10i + 7j − 5k lb., and ~B = 2i + 3j − 5.8k lb, find theangle that the resultant of these two vectors makes with the z axis.

1-26 A vector ~F has a direction angle α = 45. It is also known that cosβ istwice as large a cos γ. Find angles β and γ.

1-27 Given vectors ~R = 5i+ 5j + 6k, ~S = 0.5i+ 7j + 5k and ~T = −13i+ 10j −3k, find (a) the resultant, (b) the equilibriant, (c) the magnitude of theresultant, (d) the angle the resultant makes with the y axis.

1.6. SUMMARY 45

1-28 A force of 15 lb., has a line of action whose direction cosines are l = −1/3,m = 2/3, and n = 2/3. Another force of 49 lb. act in a direction withdirection cosines of l = 2/7, m = 6/7, and n = 3/7. Find the resultant ofthese two forces.

1-29 Find the magnitude and direction cosines for the vector ~A = 8i− 4j − k.

1-30 Solve for the magnitude and direction angles of the vector ~M in Prob.1-30.

Prob. 1-30

1-31 An automobile is moving at 30 mph when a man jumps from it with avelocity of 12 ft per sec in a direction making an angle of 60 with the car’sdirection of motion. Find the man’s velocity with respect to the ground.

1-32 Multiply the vector ~V = 10i− 5j + 2k by the scalar m = 3.

1-33 Add vectors ~V1, ~V2, and ~V3 as shown in Prob. 1-33.

Prob. 1-33


1-34 A vector in space is 10 units long and makes angles of 60 and 35 withthe y and z axes, respectively. Find its component along the positive xdirection.

1-35 Determine the resultant of the two forces shown in Prob. 1-35. Force ~Phas direct cosines of l = 0.5, m = 0.5, and n = 0.707, and force ~Q passesthrough the origin and the point (0, 9, 4).

Prob. 1-35

1-36 The bolt in Prob. 1-36 is acted upon by the three forces ~A, ~B, and ~C.Determine the magnitude and direction angles of the resultant force if~A = 100 lb., ~B = 200 lb., and ~C = 350 lb.

Prob. 1-36

1-37 Determine a unit vector parallel to the resultant of vectors ~A = 2i+3j−4k,~B = −3i+ 2j − 2k, and ~C = −4i+ j + 2k.

1-38 Define a unit vector along vector ~M = −7i+ 8j + 10.6k.

1-39 Determine a unit vector parallel to the sum of ~C = 4i − 3j + 5k and~D = 2i+ 5j − 3k.

1-40 A unit vector in space has a component of 0.75 along the y axis and makesan angle of 55 with the positive x axis. Find the zcomponent.

1.6. SUMMARY 47

1-41 Determine a unit vector with a direction opposite to that of the vector~B = 5.91i− 7j + 4k.

1-42 Find a unit vector that is parallel to the line joining point B(4, 7,−5) topoint C(1, 3,−5).

1-43 Write a position vector to the point (5,−10, 3).

1-44 A position vector whose length is 52 makes equal angles with the positivex axis, the negative y axis, and the positive z axis. Write the vector incomponent forms.

1-45 If the position vectors ~OA = 3i+ 4j+ 5k and ~OB = −2i+ 2j− 5k extendfrom the origin to the points A and B, respectively, find the vector fromthe origin to the middle point of ~AB.

1-46 Determine the position vector to a point P2 located 10 units from P1(3,−2, 5).The direction cosines of the line P1P2 are l = −0.5, m = +0.5, andn = 0.707.

1-47 If the position vectors to points M and N are 2i−4j+5k and 5i−7j+2k,respectively, find the length of displacement vector ~MN and its directioncosines.

1-48 If the position vectors to points A and B are 2i+3j+5k and 6i−3j+7k,respectively, find ~AB and determine its direction cosines.

1-49 Determine the magnitude and direction cosines of the displacement vectorfrom the point (4, 0, 5) to the point (2, 4, 2).

1-50 If the position vectors to points M and N are 3i+5j+9k and −5i+7j+4k,respectively, determine the displacement vector ~MN . What angle does~MN make with the y axis?

1-51 A displacement vector ~A starts at a point with the coordinates (3, 4, 6).It has a magnitude of 9 and direction cosines of l = −0.788, m = 0.444,and n = −0.444. What are the coordinates of the terminal point?

1-52 What angle does a displacement vector from the point (7, 5, 8) to the point(3,−2, 3) make with the z axis?

1-53 A displacement vector whose length is 9 starts from a point with thecoordiantes (3, 4,−5) in a direction defined by direction cosines l = 0.333,m = −0.667, and n = 0.667. Find the coordinates of the terminal point.

1-54 A handball player hits a ball so that it travels along the straight line paths~AB and ~BC as shown in Prob. 1-54. The space coordinates of point A

are (25, 15, 3) ft., and the direction angles of vector ~AB are α = 133, β =

133, and γ = 74.6; the direction cosines for vector ~BC are l = −0.625,m = 0.625, and n = −0.467.


Prob. 1-54

(a) What are the coordinates of point C?

(b) What is the smallest angle between the lines of action of ~AB and~BC?

(c) What is the total distance traveled by the handball?

Chapter 2

Products of Vectors

CHAPTER OBJECTIVES

• Understand the algebraic calculations with vectors

• Be able to calculate the dot product

• Be able to calculate the cross product

2.1 Multiplication of Vectors by Vectors

We have not yet discussed the multiplication of one vector by another vector,but by now it should be clear that the usual concept of a product in scalaralgebra, which is restricted to the multiplication of scalar magnitudes only, can-not apply to vectors because of their directional properties. Definitions for theproducts of vectors have been devised that are consistent with the way in whichsuch products occur in certain physical phenomena. Engineering mechanics isinherently a subject that deals with vector quantities, and, as we shall see, thereare numerous instances in which obtaining their products makes certain three-dimensional solutions much simpler and more direct. The field of electricityalso contains many situations involving vectors which submit to problem anal-ysis using these vector products. Let us now turn our attention to the form theproduct of two vectors should take. First, consider the product of the force ~Fand a linear displacement s, with an angle θ between their directions, as shownin Figure 2.1. Work is defined as the product of a displacement by a compo-nent of the force in the direction of the displacement; this can be evaluated asFs cos θ, using scalar notation. Also by definition, work is a scalar quantity.Thus we see that in this instance the product of the vector ~F and vector ~s isthe scalar quantity called work.

Work = ~F ·~s = Fs cos θ

49

50 CHAPTER 2. PRODUCTS OF VECTORS

Figure 2.1: Definition of Work

Figure 2.2: Definition of Moment

Second, consider a rigid body constrained and acted upon by a force, asshown in Figure 2.2. While the body cannot translate, there is the tendency forrotation about the pin at A. Recall that the moment of a force with respectto a point is defined as the product of the force itself and the perpendiculardistance from the point to the line of action of the force. This moment can beevaluated as Fd, a vector quantity by definition. Thus, in the second case theproduct of the vector ~F and the vector ~d is the vector quantity known as themoment. Since the product of two vectors is a scalar in the case of work anda vector in the case of a moment, it is necessary to define two sorts of vectormultiplication, the scalar product and the vector product.

2.2 The Dot (or Scalar) Product

The dot, or scalar, product of two vectors is defined as the product of themagnitudes of ~A and ~B and the cosine of the smallest angle between theirdirections. It is expressed symbolically as ~A · ~B and read as “ ~A dot ~B”. Thevectors must be arranged tail to tail, as shown in Figure 2.3, before taking thedot product. As the name “scalar product” implies, the resulting product is ascalar quantity. In symbols,

~A · ~B = |~A||~B| cos θ (2.1)

2.2. THE DOT (OR SCALAR) PRODUCT 51

Figure 2.3: Dot Product

Note that the dot product may be positive or negative, depending upon weatherthe angle θ is greater or less than 90. The dot product is useful in mechanicsto find the work done by a force moving through a displacement, to find thecomponent or projection of a vector in a given direction, and to find the anglebetween two vectors.

The following laws are valid for the dot product:

1. Commutative Law: ~A · ~B = ~B · ~A

2. Distributive Law: ~A · (~B + ~C) = ~A · ~B + ~A · ~C

3. Associative Law: m(~A · ~B) = (m~A) · ~B = ~A · (m~B) = (~A · ~B)m

The scalar product of two vectors that are not themselves zero is zero whenthe vectors are at right angles. In fact, the condition for perpendicularly oftwo vectors ~A and ~B neither of which is a null vector is ~A · ~B = 0. This isan important difference between the laws of vector algebra and those of scalaralgebra. From the definition of the dot product it should be apparent that thefollowing relations hold for unit orthogonal vectors:

i · i = |i||i| cos 0 = (1)(1)(1) = 1

i · j = |i||j| cos 90 = (1)(1)(0) = 0

Thus,

i · i = j · j = k · k = 1

and

i · j = j · k = k · i = j · i = i · k = k · j = 0

From this we can conclude that the dot product of like unit orthogonal vectorsis unity and that of unlike unit orthogonal vectors is zero.

The scalar product of a vector multiplied by itself is the square of the mag-nitude of the vector:

~A · ~A = |~A||~A| cos 0 = |~A||~A| = |~A|2


If two vectors ~A and ~B are expressed in vector form, the scalar product is

~A · ~B = (Axi+Ay j +Az k) · (Bxi+By j +Bz k)

= (Axi) · (Bxi+By j +Bz k)

+(Ay j) · (Bxi+By j +Bz k)

+(Az k) · (Bxi+By j +Bz k)

which yields

~A · ~B = AxBx +AyBy +AzBz (2.2)

In other words, the scalar product of two vectors expressed in Cartesian coor-dinates is the sum of the products of the respective scalar components.

Example 2-1

Given vectors ~A = 3i− 4j + 5k, ~B = −2i− 5j + 3k, ~C = i+ 2j + 3k, compute(a) ~A · ~B, (b) (~A + ~B) · ~C, (c) (~A− ~C) · ~B.

Solution (a)

~A · ~B = (3i− 4j + 5k) · (−2i− 5j + 3k)

= 3(−2) + (−4)(−5) + (5)(3) = 29

(b)

(~A + ~B) · ~C =(

(3i− 4j + 5k) + (−2i− 5j + 3k))· (i+ 2j + 3k)

= (i− 9j + 8k) · (i+ 2j + 3k)

= 1− 18 + 24 = 7

(c)

(~A− ~C) · ~B =(

(3i− 4j + 5k)− (i+ 2j + 3k))· (−2i− 5j + 3k)

= (2i+ 6j + 3k) · (−2i− 5j + 3k)

= −4 + 30 + 6 = 32

2.2.1 Component of Vector

A previously mentioned, one of the important uses of the scalar product is tofind the component of a vector in a given direction. This geometric concept isof such importance that we shall consider it in detail. Consider the vector ~Amaking an angle θ with a line of known direction, as shown in Figure 2.4. If itis desired to obtain the component or projection of a vector ~A along the givenline, by right-triangle trigonometry it is seen to be |~A| cos θ. When vector ~A

2.2. THE DOT (OR SCALAR) PRODUCT 53

Figure 2.4: Component of Vector with Dot Product

Figure 2.5: Component of Vector with Dot Product

is expressed in Cartesian components, the projection along a given line cannotbe obtained with such ease without resorting to the dot product. As shown inFigure 2.5, a unit vector b is directed along the line of known direction. If thedot product between vector ~A and vector b is evaluated, the result is

~A · b = |~A||b| cos θ = |~A|(1) cos θ = |~A| cos θ

Since |b| is unity, it is apparent that the required component or projection alongthe line is obtained. From this we can conclude that if one of the vectors is aunit vector, the scalar product of it and another vector represents the projectionor components of the vector on an axis that is coincident with the unit vector.

Example 2-2

Find the scalar component of vector ~P = 7i − 14j + 21k along the vector~V = 3i+ 6j − 2k.

Solution From the unit vector

v =~V

|~V|=

3i+ 6j − 2k√9 + 36 + 4

=3i+ 6j − 2k

7

The projection of ~P along ~V can be obtained by

~P · v = (7i− 14j + 21k) ·

(3i+ 6j − 2k

7

)= 3− 12− 6

= −15


2.3 The Cross (or Vector) Product

A second product involving two vectors is known as the cross, or vector, product,which, as the name implies, produces a vector instead of a scalar. Its physicalsignificance is of fundamental importance in statics and relates to the momentof a force about a point. In dynamics the cross product is of great significancein relating angular and linear motion. It also has important geometric use indetermining the perpendicular distance form a point to a line, in finding theperpendicular distance from a point to a plane, and in calculating the shortestperpendicular distance between any two non-intersecting nonparallel lines.

The cross, or vector, product of two vectors ~A and ~B is defined as the vector~C whose magnitude equals the product of the magnitudes of ~A and ~B and thesine of the smaller angle between their directions when they are placed tail totail. The direction of ~C is perpendicular to the plane containing ~A and ~B, witha sense defined by the “right-hand-rule” (the direction of advancement for a

right-hand screw revolved from ~A to ~B through the smaller angle) in Figure2.6.

Figure 2.6: Right Hand Rule for Cross Product

Figure 2.7 shows us the right-hand rule with different position.The line of action ~C is not defined by the cross product, but rather depends

upon the physical problem.

~A× ~B = |~A||~B| sin θu = ~C (2.3)

where u is a unit vector perpendicular to the plane containing ~A and ~B andindicates the direction of ~C. The cross product of vectors ~A and ~B, read “~A

2.3. THE CROSS (OR VECTOR) PRODUCT 55


cross ~B”, is ~A× ~B = ~C, shown graphically in Figure 2.8.

Figure 2.8: Cross Product

Another important difference between vector and scalar algebra occurs inthe cross product, where the operation is not commutative. In other words, achange in the order of the factors in the cross product changes the sign of theproduct. The fact that ~A× ~B 6= ~B× ~A can be deduced by comparing Figures2.8 and 2.9. Note from Figure 2.9 that ~B × ~A = −~C, so the cross products of~A× ~B and ~B× ~A have the same magnitudes but opposite senses. Similarly, it


Figure 2.9: Cross Product

is of importance to observe that

~A× ~B = −~B× ~A

Figure 2.10 shows us the right-hand rule for results of ~A× ~B and ~B× ~A.


Although the commutative law fails for the cross product, the laws that arevalid and for which we have immediate use are as follows:


1. Distributive Law: ~A×(~B + ~C

)= ~A× ~B + ~A× ~C

2. Associative Law: m(~A× ~B

)= m~A× ~B = ~A×m~B =

(~A× ~B

)m

If ~A× ~B = 0 when neither vector ~A nor ~B is zero, then sin θ must be zeroand θ = 0 or θ = 180, indicating that ~A and ~B have the same or oppositesenses and are parallel. This vanishing of the cross product not only when ei-ther of the vectors is zero, but also when the vectors are parallel, is anotherimportant dissimilarity between vector and scalar algebra.

Applying the cross product to the unit orthogonal vector i, j, and k, itfollows that

i× i = |i||i| sin 0u = (1)(1)(0)u = 0

and

i× i = j × j = k × k = 0

For this we conclude that the cross product of like unit orthogonal vectorsis zero, because angle θ is either 0 or 180; consequently, sin θ is zero.

Also, by using the cross product we find that

i× j = |i||j| sin 90u = (1)(1)(1)k = k.

Since vector k is perpendicular to the plan containing vector i and j. Consistentwith the right-handed system, we can write

i× j = k

j × i = −kj × k = i

k × j = −ik × i = j

i× k = −j

Figure 2.11 shows us the results of cross product of unit vectors. The directionof arrows means the result of cross product and the sign of the circle centermeans the direction of result.

Expressing the two vectors ~A and ~B in Cartesian coordinate form and car-


Figure 2.11: Cross Product with unit vectors of the Cartesian axes

rying out the cross product gives

~A× ~B = (Axi+Ay j +Az k)× (Bxi+By j +Bz k)

= Axi× (Bxi+By j +Bz k)

+Ay j × (Bxi+By j +Bz k)

+Az k × (Bxi+By j +Bz k)

= AxBx(i× i) +AxBy (i× j) +AxBz (i× k)

+AyBx(j × i) +AyBy(j × j) +AyBz(j × k)

+AzBx(k × i) +AzBy(k × j) +AzBz(k × k)

= AxByk −AxBz j −AyBxk +AyBz i+AzBxj −AzBy i= (AyBz −AzBy )i+ (AzBx −AxBz)j + (AxBy −AyBx)k.(2.4)

A more convenient way to solve the cross product of is to write it as athree-by three determinant:

~A× ~B =

∣∣∣∣∣∣i j kAx Ay AzBx By Bz

∣∣∣∣∣∣Carefully writing the terms in the sequence conforming to the order of readingof the vectors, in this case ~A cross ~B, we can evaluate this determinant in anumber of ways. One method is to rewrite the first two rows and then take theordinary scalar products along the diagonals, being careful to apply the correctsigns as shown:

(+)

~A× ~B =

∣∣∣∣∣∣∣∣∣∣i j kAx Ay AzBx By Bzi j kAx Ay Az

∣∣∣∣∣∣∣∣∣∣(−)

= (AyBz −AzBy )i+ (AzBx −AxBz)j + (AxBy −AyBx)k


The result is identical to the equation (2.4).

A second method involves rewriting the first two columns and taking theordinary scalar products along the diagonals, again being careful with the signs.

(+)

~A× ~B =

∣∣∣∣∣∣i j k i jAx Ay Az Ax AyBx By Bz Bx By

∣∣∣∣∣∣(−)


A third method produces the same result by use of a two-by-two determinantas follows:

~A× ~B =


∣∣∣∣∣∣(+)

= i

∣∣∣∣ Ay AzBy Bz

∣∣∣∣− j ∣∣∣∣ Ax AzBx Bz

∣∣∣∣+ k

∣∣∣∣ Ax AyBx By

∣∣∣∣(−)


Example 2-4

Given vectors ~P = 3i − 4j + 5k and ~Q = −2i + 3j, find (a) ~P × ~Q, (b) (~P +~Q)× (~P− ~Q).

Solution (a) First, from the equation (2.3)

~P× ~Q = (3i− 4j + 5k)× (−2i+ 3j)

= 3i× (−2i+ 3j)− 4j × (−2i+ 3j) + 5k × (−2i+ 3j)

= 0 + 9k − 8k − 0− 10j − 15i

= −15i− 10j + k

(b)

(~P + ~Q)× (~P− ~Q) =(

(3i− 4j + 5k) + (−2i+ 3j))×(

(3i− 4j + 5k)− (−2i+ 3j))

= (i− j + 5k)× (5i− 7j + 5k)

=

∣∣∣∣∣∣i j k i j1 −1 5 1 −15 −7 5 5 −7

∣∣∣∣∣∣= 30i+ 20j − 2k


Example 2-5

Determine a unit vector c perpendicular to a plane containing vectors ~M =−3i+ 4j + 5k and ~N = 2i− 3j − 4k.

Solution First, determine a vector perpendicular to ~M and ~N by takingthe cross product:

~M× ~N = (−3i+ 4j + 5k)× (2i− 3j − 4k)

=

∣∣∣∣∣∣i j k i j−3 4 5 −3 42 −3 −4 2 −3

∣∣∣∣∣∣= −i− 2j + k

Second, find the unit vector c whose line of action is co-linear with ~C:

c =~C

|~C|

=−i− 2j + k√

(−1)2 + (−2)2 + (1)2

=−i− 2j + k√

6=−i− 2j + k

2.46

Therefore, c = ±(−0.406i − 0.813j + 0.406k). Note that Vector ~C is shown as+ or −, since the sense is not specified.

Example 2-6

If ~A = 10i − 5j − 6k and ~B = 5i − 2j − 2k, what are the direction angles of~A× ~B?

Solution

~A× ~B =

∣∣∣∣∣∣i j k i j

10 −5 −6 10 −55 −2 −2 5 −2

∣∣∣∣∣∣= −2i− 10j + 5k

|~A× ~B| = ±√

(−2)2 + (−10)2 + (5)2

= ±√

4 + 100 + 25 = ±11.4

Finally, we can obtain cosα:

cosα =−2

11.4= 0.175

2.4. DIVISION BY A VECTOR 61

Therefore,

α = 100

cosβ =−10

11.4= −0.878

Therefore,

β = 151.2

cos γ =5

11.4= 0.438

Therefore,

γ = 64.0

To check,

cos2 α+ cosβ + cos2 γ = 1

(−0.175)2 + (−0.878)2 + (0.438)2 = 0.995 ≈ 1

2.4 Division by a Vector

So far our discussion of the vector operations has been confined to addition,subtraction, and multiplication. It may seem strange that there has been nomention of division of a vector. Actually, there is no operation in vector algebracorresponding to division by a vector, and therefore it is a meaningless, orundefined, operation. The reason is that the ordinary cancellation law of scalaralgebra does not hold for either the dot or the cross product of two vectors. Forexample, the equation

~A · ~B = ~A · ~C

does not ensure that ~B = ~C. This is shown graphically in Figure 2.12. Theprojections of vectors ~B and ~C along vector ~A are equal, but obviously ~B and~C need not be equal. Note, however, that division of a vector quantity by ascalar quantity is a legitimate operation.

2.5 Vector Equations

A valid vector equation must express an equality between two vectors, not be-tween a vector and a scalar. Thus equations such as

5~A + 3~B = 15 and ~A× ~B = (~C× ~D) · ~E

are meaningless, because the left side is a vector and the right side is a scalar. Asalways, dimensional homogeneity must also be considered. Furthermore, since


Figure 2.12: ~A · ~B = ~A · ~C does not mean ~B = ~C

any vector may be resolved into its i, j, and k components, the coefficients oflike vectors can be equated. Hence, the vector equation

Axi+Ay j +Azk = Bz i+By j +Bz k

is tantamount to

Ax = Bx

Ay = By

Az = Bz

2.6 Summary

The dot, or scalar, product of two vectors is defined as the product of their mag-nitudes and the cosine of the smallest angle between their directions. Expressedin symbols,

~A · ~B = |~A||~B| cos θ

The condition for perpendicularity of two vectors neither of which is a nullvector is that their dot product be zero. The dot product of like unit orthog-onal vectors is unity, and that of unlike unit orthogonal vectors is zero. Whentwo vectors are expressed in vector form, the scalar product is the sum of theproducts of the respective scalar components; hence,

~A · ~B = AxBx +AyBy +AzBz

An important function of the scalar product is to find the component of a vec-tor in a given direction. The required component of a vector along a given linecan be obtained by evaluating the scalar product between the vector and a unitvector coincident with the line.

2.6. SUMMARY 63

The cross, or vector, product of two vectors is defined as a vector whosemagnitude equals the product of the magnitudes of the two vectors and thesine of the smaller angle between their directions. The direction of the vectorproduct is perpendicular to the plane containing the two vectors, with a sensedefined by the right-hand-rule. In symbols,

~A× ~B = |~A||~B| sin θu = ~C

where u is a unit vector perpendicular to the plane containing ~A and ~B andindicates the direction of ~C. The line of action of ~C is not defined by the crossproduct. The commutative law fails for the cross product. Parallelism of twovectors neither of which is a null vector is indicated when their vector productis zero. The vector product of like unit orthogonal vectors is zero. The vectorproduct of two unlike unit orthogonal vectors results in the third unit orthog-onal vector, with a sense determined by the right-hand-screw rule. Expressingtwo vectors in Cartesian-component form and carrying out the cross productgives the determinant.

~A× ~B =


∣∣∣∣∣∣Division by a vector is an undefined operation, but division of a vector quan-

tity by a scalar is a legitimate operation.

A valid vector equation must express an equality between two vectors, notbetween a vector and a scalar.


Exercise

2-1 Given the vectors ~A = 10i+4j, ~B = 3i−2j+5k, and ~C = 3j+ k, evaluate(a) ~A · ~B, (b) ~B · ( ~A+ ~C), (c) ( ~A · ~C) ~B, (d) ( ~A · ~B)~C.

2-2 For vectors ~A = 17i − 34j + 51k and ~B = 8i + 9j + 12k find the scalarcomponent of ~A along ~B.

2-3 Find the angle between the following vectors: (a) ~R = 2i + 3j − 7k and~S = 4i− 4j + 3k; (b) Two lines with direction cosines l = 1/3, m = 2/3,and n = −2/3 and l = −2/3, m = 1/3, and n = −2/3.

2-4 Determine the component of the force ~F = 24i+21j−30k in the direction~N = 1.5i− 2j + 1.66k.

2-5 Find the work done by a force ~P = 5i + 6j + 7k tons moving through adisplacement of ~S = 2i− 3j + 4k ft.

2-6 A vector ~A has a magnitude of 10 and direction cosines which are all equaland positive. Vector ~B has a magnitude of 5 and lies in the yz plane withdirection angles β = 60 and γ = 30. Find the angle between ~A and ~B.

2-7 A force ~P has x, y, and z components of 2, 8, and 4 lb., respectively. Findthe component of this force along an intersecting line whose directioncosines are l =

√3/4, m = −3/4 and n = 1/2.

2-8 Find the angle between (a) the position vectors to the pointers with coor-dinates (3, 4, 5) and (4, 0, 6), (b) the lines AB and AC as shown in Prob.2-8.

2-9 What is the component of ~A = 0.7i−2.8j+14k parallel to ~B = 3i−6j+2k.

2-10 A vector ~P whose magnitude is 5 has direction angles of 45, 60, and120, respectively. Find the component of ~P parallel with ~Q = 0.3i−0.4j.

2-11 If vectors ~A = 5i−3j+ zk and ~B = 3i+ 2j+ 5k are perpendicular, find z.

2-12 If vectors ~A = 3xi+ 4j − 2k and ~B = 2xi− 3xj + 12k are perpendicular,solve for the value of x.

65


Prob. 2-8

2-13 Define a vector or vectors that have a y component of 5, lie in the xyplane, and make an angle of 60 with the vector ~V = 3i+ 4j + 5k.

2-14 Compute the work done in moving a particle along a vector ~S = 4i+3j−2kft if the force is ~F = 5i− 2j − k lb.

2-15 Compute the projection of ~V = 7i+14j+3.5k on the line passing throughthe points (8, 6, 5) and (2, 8, 8).

2-16 Find a force acting in the direction i + j + k that will produce a 30 lb.component in the direction 2i+ 5j + 4k.

2.6. SUMMARY 67

2-17 In Prob. 2-17, if | ~A| = 10 and | ~B| = 20, compute ~A · ~B.

Prob. 2-17

2-18 Determine the scalar product of the vector ~P and the vector ~R as shownin Prob. 2-18.

Prob. 2-18

2-19 Compute the work done by the 50 lb. force in moving the block fromposition A to position B in Prob. 2-19.

Prob. 2-19


2-20 Find the angle between line in the xy plane with direction angles α = 30

and β = 60 and a line making equal angles with the positive x, y, and zaxes.

2-21 What is the angle between the diagonal of a cube and one of its edges?

2-22 Given two vectors ~A = i+ 5j + 10k and ~B = 4i− 3j + 2k, find the scalarcomponent of ~A along ~B.

2-23 Show that the vectors ~P = 5i − 4j + 10k and ~Q = 2i − 5j − 3k areperpendicular.

2-24 Find the angle between the vectors that are represented by the diagonalsAF and BE in Prob. 2-24.

Prob. 2-24

2-25 Find the magnitude of the force acting in the direction i+ 2j + 3k that isrequired to produce a 15 lb. component in the direction 2i− 3j + 5k.

2-26 Find a vector in the xy plane that has the same length and same x pro-jection as 6i+ 2j − 3k.

2-27 What is the angle between a line making equal angle with the negativex, y, and z axes and a line in the xy plane making equal angles with thepositive x and y axes?

2-28 Find a vector or vectors in the xy plane that are perpendicular to a vectormaking equal angles with the positive coordinate axes.

2-29 A force ~A = 2i − 5j + 3k lb. and a force ~B = −3i + 2j − k lb. act on aparticle, giving it a displacement ~S = −3i+7j−3k ft. Compute the workdone.

2.6. SUMMARY 69

2-30 Using the dot product, prove the law of cosines for the triangle shown inProb. 2-30.

Prob. 2-30

2-31 Given the points A(4, 3, 2), B(2, 4, 4), and C(5, 0,−3), find (a) ~AB, (b)

| ~AB|, (c) the direction angles of ~AB, (d) ~AC, (e) the angle between ~AB

and ~AC.

2-32 Find the sum ~A+ ~B+ ~C if ~A has x, y, and z components of 5, 10, and 15,respectively, ~B = 4i+ 8j + 16k, and ~C is in the yz plane at an inclinationof 60 to the positive end of the y axis with a magnitude of 10.

2-33 What is the component of ~A+ ~B+ ~C of Problem 2-32 along a line havingdirection cosines of l = 0.5 and m = 0.378, with n known to be negative?

2-34 If ~E = 2i+ 5j + 4k and ~F = 4i+ 7j − 3k, find the component of ~E along~F .

2-35 For the unit cube shown in Prob. 2-35, find the angle between the diago-nals ~AB and ~CD.

Prob. 2-35


2-36 Show that vectors ~A = 3i+5j−2k and ~B = 10i−4j+5k are perpendicular.

2-37 What is the component of ~A+ ~B+ ~C of Problem 2-32 along a line havingdirection cosines of l = 0.5 and m = 0.378, with n known to be negative?

2-38 If ~E = 2i+ 5j + 4k and ~F = 4i+ 7j − 3k, find the component of ~E along~F .

2-39 For the unit cube shown in Prob. 2-39, find the angle between the diago-nals ~AB and ~CD.

Prob. 2-39

2-40 Show that vectors ~A = 3i+5j−2k and ~B = 10i−4j+5k are perpendicular.

2-41 Find the work done on a particle by the force ~F1 = 5i + 6j + 7k lb. and~F2 = 3i+ 2j + 4k lb. during a displacement of 7i− 4j + 3k ft.

2-42 A particle acted upon by the forces ~F1 = 3i + 5j + 8k tons and ~F2 =2i + 7j − 2k tons is displaced from the point (−1, 5, 7) ft. to (7, 8, 2) ft.Determine the total work done by the forces.

2-43 The vector~= 2i+P2k makes an angle of 62 with ~Q = i+ 2j + 2k. FindP2.

2-44 If ~A = 0.5i+ 2j−5k and ~B = 4i+ 3j+ 2k, evaluate (a) ~A× ~B, (b) ~B× ~A,

(c) ( ~A× ~B)× ~B, (d) [( ~A+ ~B)× ~A] · ~B.

2-45 If ~A = 3i− 4j+ 5k, ~B = i+ 2j− 3k and ~C = 2i+ 5j+ 2k, find (a) ~A× ~B,

(b) ~A× ~C, (c) ~B × ~A, (d) ~A× ( ~B − ~C), (e) ~A× ( ~B × ~C).

2.6. SUMMARY 71

2-46 In Prob 2-46, vectors ~P and ~Q lie in the xy plane. It is also known that|~P | = 5 and | ~Q| = 10. Find ~P × ~Q.

Prob. 2-46

2-47 Determine a unit vector perpendicular to the vectors ~S = 10i + 5j − 2kand ~T = 5i+ 2j + 2k.

2-48 If ~M = i + 2j + 3k, ~N = 5i − 3j + 2k, | ~O| = 45, and ~O is perpendicular

to both ~M and ~N , write ~O in vector form.

2-49 Given ~A = 3i−5j+4k, ~B = i+2j−3k, |~C| = 36.8, with ~C perpendicular

to both ~A and ~B, write ~C in vector form.

2-50 If vector ~C = 10i − 5j + 2k and vector ~D = 4i + 3j + 2k, (a) determine~C × ~D; (b) using the cross product find the sine of the angle between ~C

and ~D.

2-51 Find a unit vector with a positive x component perpendicular to the planecontaining ~A = 5i− 4j + 6k and ~B = 3i− 2j + 4k.

2-52 The vectors 5i− 2j + 4k and i+ yj + zk are known to be parallel. Find yand z.

2-53 Find a unit vector orthogonal to the plane determined by the pointsA(1, 2, 3), B(4, 5,−1), and C(3,−2, 1).


2-54 Given vectors ~P , ~Q, and ~R emanating from the origin to the points asshown in Prob. 2-54, (a) determine a vector ~S such that the resultant ofthe system is zero; (b) define a vector which is perpendicular to the plane

containing ~Q, and ~R and directed upward.

Prob. 2-54

2-55 Prove that the magnitude of ~P × ~Q equals the area of the parallelogramshown in Prob. 2-55.

Prob. 2-55

2-56 The two vectors ~P = 5i−3j+7k and ~Q = 3i−yj+zk are parallel vectors.Find y and z.

2-57 Determine a unit vector parallel to the xz plane and perpendicular tovector ~P = 5i+ 4j − 3k.

2-58 If ~P × ~Q = ~P × ~R, does ~Q have to equal ~R? Prove your answer.

2-59 If ~M = 3i+ 4j − 5k and ~N = 2i− 5j − 3k, evaluate ~M × (− ~N).

2-60 Find the area of a parallelogram determined by the vectors ~A = 4i−3j−2kand ~B = 3i+ 4j + 5k. Dimensions are in inches.

2.6. SUMMARY 73

2-61 Determine a unit vector orthogonal to the plane defined by the three pointsP (3, 4, 5), Q(−2, 7, 3), and F (0,−2,−4).

2-62 Determine the area of a triangle with vertices A(3, 4, 5), B(−1,−2, 2), andC(−7,−3,−4).


Part III

Application of Vectors

75

Chapter 3

Application of VectorProducts

CHAPTER OBJECTIVES

• Understand how to calculate/use the cross product for applications

• Understand the meaning of the cross product and the dot product forphysical and geometric applications

• Apply the cross product to physical applications

• Apply the geometric application for the cross product

3.1 Moment of a Force about a Point

Previously we stated that the moment of a force with respect to a point is anaction which has a tendency to produce rotation. The moment of a force abouta point was also defined as the product of the force itself and the perpendiculardistance from the point to the line of action of the force. When the perpendiculardistance (moment arm) is not readily available, it is convenient to use Varignon’stheorem to evaluate the moment. Varignon’s theorem states that about anypoint the algebraic sum of the moments of the components of a force equalsthe moment of the force. For example, as shown in Figure 3.1 if it is desiredto compute the moment of force ~F about point A, the force ~F would first bedecomposed into its x and y components. The moment would then be ~M =r~F sin θ, using scalar notation. It can readily be seen that the cross productbetween the position vector ~r and the force vector ~F would give the same resultand thus fits the requirement for the evaluation of the moment of a force abouta point. Written in vector notation, it would be:

~M = ~r× ~F = |~r||~F| sin θu

77

78 CHAPTER 3. APPLICATION OF VECTOR PRODUCTS

Figure 3.1: Application of Vector Product

It can now be shown that the force ~F in forming the moment vector may beconsidered transmissible and that the position vector r may extend to any pointon the line of action of the force ~F. Consider the force ~F oriented as shown inFigure 3.2. To compute the moment about the origin, the position vectors r1, r2,

Figure 3.2: Computing Vector Product

and r, which are directed to any point on the line of action of the force, makingangles of θ, φ, and 90, respectively, with the force ~F, are crossed with the force.Remember that the vectors must be arranged tail to tail before making the crossproduct. Again looking at Figure 3.2 and considering the shaded triangle formedby the vectors r, r1, and the line of action of ~F, it is apparent that

sin θ =|~r||~r1|

or |~r1| =|~r|

sin θ(3.1)

Figure 3.3 shows us that the moment about a force vector, ~F, is same if thevector ~r is from the origin to any point under the line of action of the forcevector, ~F.

3.1. MOMENT OF A FORCE ABOUT A POINT 79

Figure 3.3: Computing Cross Product for Mmment about a force

Now if the cross product of ~r1 and ~F is evaluated, it will give

~r1 × ~F = |~r1||~F| sin θu

Substituting form the equation (3.1) yields

~r1 × ~F =|~r|

sin θ|~F| sin θu = |~r||~F|u (3.2)

Carrying out the same operation using the position vector ~r results in

sinφ =|~r||~r2|

or|~r2| =|~r|

sinφ

and

~r2 × ~F = |~r2||~F| sinφu =|~r|

sinφ|~F| sinφu = |~r||~F|u (3.3)

Finally, using position vector ~r, which is at right angles to the force ~F clearlygives the magnitude of the moment of the force ~F about the origin by simplescalar multiplication. However, using the cross products gives

~r× ~F = |~r||~F| sin 90u = |~r||~F|u (3.4)

We see that the results of equations (3.2), (3.3), and (3.4) are identical.

From this we can conclude that the moment of a force about any point maybe computed by employing the cross product between the force and a positionvector directed from the point to any point on the line of action of the force.Hence, we have the general expression

~M = ~r× ~F (3.5)

Since the moment is a vector quantity, it possesses both magnitude and direc-tion. The direction of the moment vector is perpendicular to the plane of ~r and~F, with a sense defined by the right-hand-screw rule.


Example 3-1

Determine the moment about the origin of force ~F = 10i + 5j + 2k lb. actingat the point (4, 5, 6) ft. in Figure 3.4. What are the direction angles of theresulting moment vector?


Solution By inspection, ~r = 4i+ 5j + 6k ft. by the equation 3.5,

~Mo = ~r× ~F= (4i+ 5j + 6k)× (10i+ 5j + 2k)

= −20i+ 52j − 30k

| ~Mo| =√

(−20)2 + (52)2 + (−30)2

=√

400 + 2704 + 900 = 63.1ft.lb.

For the direction of cosine,

cosα =−20

63.1= −0.317

α = 108.5

cosβ =52

63.1= 0.822

β = 34.8

cos γ =−30

63.1= −0.475

γ = 118.4

Example 3-2

Find the moment of the force shown in Figure 3.5 with respect to the origin andwith respect to point B. Finding a unit vector along the line of action of force~F,

3.1. MOMENT OF A FORCE ABOUT A POINT 81


Solution

u =−4i+ 5j + 3k√(−4)2 + 52 + 32

=−4i+ 5j + 3k√

50=−4i+ 5j + 3k

7.07

~F = |~F|u = 70.7−4i+ 5j + 3k

7.07= −40i+ 50j + 30k (3.6)

The moment with respect to the origin is, by the equation (3.5),

~Mo = ~r× ~F = 4i× (−40i+ 50j + 30k) = −120j + 200kft.lb.

The moment with respect to point B is

~MB = ~r× ~F = (6i+ 3j − 2k)× (−40i+ 50j + 30k)

= 190i− 100j + 420kft.lb.

Example 3-3

The bracket in Figure 3.6 is acted upon by the force ~F = 10i + 5j + 4k lb. asshown. (a) Compute the moment of the force about the origin. (b) Computethe moment of the force about point A.

Solution (a) First find a position vector from the origin to any point on

the line of action of the force ~F:

~r = 2i+ 9j − 6k

Compute the moment ~Mo by the equation (3.5)

~Mo = ~r× ~F = (2i+ 9j − 6k)× (10i+ 5j + 4k)

= 66i− 68j − 80kin.lb.



(b) Determine a position vector from point A to any point on the line of action

of the force ~F:

~r = −2i+ 9j − 10k

Then

~MA = ~r× ~F = (−2i+ 9j − 10k)× (10i+ 5j + 4k)

= 86i− 92j − 100kin.lb

3.2 Relation Between Angular and Linear Ve-locity

Another important use for the cross product between two vectors is in the con-sideration of the circular motion of a particle about an axis. Consider the pointP in a rigid body rotating about an axis with a constant angular velocity ~ω, asshown in Figure 3.7. The angular velocity ~ω is a vector which, in accordancewith the right-hand-screw rule, is directed normal to the plane described by themotion of particle P and with a sense such that when the thumb of the righthand points in the direction of ~ω, the fingers curl in the direction of rotationabout the axis. From any point O on the axis of rotation, a position vector ~r isdirected to the point P . Since the linear, or tangential, velocity of any point in acircle is equal to the product of the angular velocity and the radius of the circle,the linear velocity ~V is clearly proportional to the distance b from the axis ofrotation. From trigonometry we see that radius b of the circular trajectory ofparticle P has a magnitude of |~r sin θ, and therefore the magnitude of vector ~Vwill be

|~V| = |~ω||~r| sin θ. (3.7)

3.2. RELATION BETWEEN ANGULAR AND LINEAR VELOCITY 83

Figure 3.7: Angular Velocity

We see that the right side of the equation (3.7) is the cross product ~ω × ~r.

The direction of ~V, of course, is perpendicular to the plane of ~ω and ~r, witha sense designated by the right-hand-screw rule. It should be apparent fromFigure 3.7 that the particular location of point O has no significance as long asit is on the axis of rotation. Thus we may write the important general vectorequation.

~V = ~ω ×~r (3.8)

This shows that the linear velocity of a particle that is in pure rotation aboutan axis can be computed by the cross product ~ω × ~r, where ~ω is the angularvelocity (expressed in radians per unit of time) and~r is a position vector directedfrom any point on the axis of rotation to the particle.

Example 3-4

The particle P has an angular velocity of 15 rad. per sec. About the z axis, asshown in Figure 3.8. What is the linear tangential velocity of point P?



Solution Direct a position vector from any point on the axis of rotationto the particle.

~r = 5i+ 6j + 7k

By the right-hand-screw rule, ~ω = 15k rad. per sec. By the equation (3.8),

~V = ~ω ×~r = 15k × (5i+ 6j + 7k)

= −90i+ 75jfps. (3.9)

Example 3-5

The rigid body is turning about the axis of rotation that passes through theorigin and the point (1, 4, 8) with angular speed of 18 rad. per sec., as shown inFigure 3.9. What is the linear velocity of point P?


Solution First determine the angular velocity ~ω by finding the directioncosines of the axis of rotation.

d =√

(dx)2 + (dy)2 + (dz)2

=√

(1)2 + (4)2 + (8)2 =√

1 + 16 + 64

=√

81 = 9

and

~ω = |~ω|(cosαi+ cosβj + cos γk)

= 18

(1

9i+

4

9j +

8

9k

)= 2i+ 8j + 16k

3.3. GEOMETRIC APPLICATIONS OF VECTOR PRODUCTS 85

Then direct a position vector from any point on the axis of rotation to the pointin question:

~r = 2i− 4j + 1k~V = ~ω ×~r = (2i+ 8j + 16k)× (2i− 4j + 1k)

= (72i+ 30j − 24k)ft./sec.

3.3 Geometric Applications of Vector Products

There are numerous applications of vector algebra which greatly simplify certainthree-dimensional geometric problems. Our discussion will cover the operationsof determining the perpendicular distance from a point to a line, finding theperpendicular distance from a point to a plane, and calculating the shortestperpendicular distance between any two non-intersecting non-parallel lines. Weshall describe these three specific operations in the following examples.

Example 3-6

To find the perpendicular distance from a point to a line, determine the per-pendicular distance from point P (3, 5, 4) ft. to the line of action of vector~A = 3i+ 6j + 2k acting at point Q(2,−3, 2) ft., as shown in Figure 3.10.


Solution First determine a displacement vector from point Q to point P :

~QP = i+ 8j + 2k

Now find a unit vector along ~A:

a =~A

|~A|=

3i+ 6j + 2k√32 + 62 + 22

=3

7i+

6

7j +

2

7k

Next use the cross product to obtain the perpendicular distance from point Pto the line of action of A as follows:

~QP × a = | ~QP ||a| sin θu


From trigonometry we see that the magnitude of ~QP × a is the perpendiculardistance form point P to the line of action of A. Call this distance D. Then

~D = ~QP × a

= (i+ 8j + 2k)× (3

7i+

6

7j +

2

7k)

=4

7i+

4

7j − 18

7k

Therefore,

|~D| =√

(4

7)2 + (

4

7)2 + (−18

7)2 = 2.7ft.

Example 3-7

To find the perpendicular distance from a point to a plane, find the perpendic-ular distance from the point P (5, 7, 4) to a plane defined by the three pointsQ(2, 3, 2), R(−2,−1,−3), and S(7, 8,−1). All dimensions are expressed in feet.

Solution First find displacement vectors ~QR and ~QS:

~QR = −4i− 4j − 5k~QS = 5i+ 5j − 3k

Now obtain a perpendicular to the plane QRS by the cross product ~QS × ~QR:

~QS × ~QR = (5i+ 5j − 3k)× (−4i− 4j − 5k)

= −37i+ 37j

Call this vector ~N. Next find a unit vector uN :

uN =~N

|~N|

=−37i+ 37j√

(−37)2 + (37)2=−37i+ 37j

52.2

Determine a displacement vector from any point in the plane QRS to point P .In this case,

~QP = 3i+ 4j + 2k

Finally, obtain the projection of vector ~QP on vector uN :

~QP · uN = (3i+ 4j + 2k) · −37i+ 37j

52.2= −2.12 + 2.83 = 0.71ft

3.3. GEOMETRIC APPLICATIONS OF VECTOR PRODUCTS 87

Example 3-8

Find the shortest perpendicular distance between two non-intersecting nonpar-allel vectors, given vector ~A = 5i + 3j + 4k acting at point P and vector~B = 2i− 5j − 6k acting at point Q. As shown in Figure 3.11.


Solution First determine a perpendicular to both A and B:

~A× ~B = (5i+ 3j + 4k)× (2i− 5j − 6k)

= −38i+ 38j + 19k

Next find any displacement vector form a point on the line of action A to anypoint on B. The most convenient in this case is

~QP = −i+ 5j − k

Now obtain the projection of ~QP on ~A× ~B. To do this, first determine a unitvector along (~A× ~B):

u~A×~B =~A× ~B

|~A× ~B|=

−38i+ 38j + 19k√(−38)2 + (38)2 + (19)2

=−38i+ 38j + 19k

57

Finally, compute the shortest perpendicular distance:

~QP · u~A×~B = (−i+ 5j − k) · −38i38j + 19k

57= 0.667 + 3.33− 0.333

= 3.666ft


3.4 Summary

The moment of a force about a point may be evaluated by employing the crossproduct between the force and a position vector directed from the point to anypoint on the line of action of the force.

The linear velocity of a particle that is in pure rotation about an axis maybe computed by the cross product of the angular velocity and a position vectordirected from any point on the axis of rotation to the particle.

Some applications of vector products to three dimensional geometric prob-lems are determining the perpendicular distance from a point to a line, findingthe perpendicular distance from a point to a plane, and computing the shortestperpendicular distance between any two non-intersecting nonparallel lines.

Exercise

3-1 A force F = 10i+5j−8k lb. acts at a point (5,−7, 3). What is its momentabout the origin? About the point (−3,−5, 4)?

3-2 Determine the moment of the 35 lb. force about point A. The force passesthrough the points (0, 0, 0) ft. and (2, 3, 6) ft. as shown in Figure 3-2.

Prob. 3-2

3-3 Determine the moment about the origin of the 10 lb. force whose line ofaction passes through points A and B as shown in Figure 3-3.

Prob. 3-3

3-4 Calculate the moment about a point (5, 4, 3) ft. caused by the forces

P = 3i − 2j − 4k lb. and Q = 5i + 7j − 2k lb. acting at the points(7, 2,−2) ft. and (−3,−2, 5) ft., respectively.

89


3-5 Find the moment about the point (2, 3, 4) ft. of a force F = i + 2k lb.acting through the point (5,−1, 2) ft.

3-6 A force of 26 lb. acts through a point (4, 5,−7) ft. and is equally inclinedto the positive ends of the orthogonal axes. What is the moment of thisforce about the origin?

3-7 A force F = 7i+ 6j − 5k lb. acts at the origin. Find the magnitude anddirection cosines of the moment of this force about the point (−4, 5, 6) ft.

3-8 A force F = 5i + 10k lb. acts at a point (2, 4,−6) ft. It has a moment

M = 20i− 55j − 10k ft. lb. about the point (1, 2, z) ft. Find z.

3-9 Find the moment of the force shown in Figure 3-9with respect to the originand with respect to point B.

Prob. 3-9

3-10 A force F = 3i−5j+7k lb. acts at a point whose coordinates are (−4, 5, 2)ft. Compute the moment of the force about the origin.

3-11 A force F = 10i+5j−3k lb. acts at a point whose coordinates are (3,−5, 2)ft. Compute the moment of the force about a point whose coordinates are(−2, 1, 5) ft.

3-12 Find the magnitude and direction cosines of the moment about the point(3,−4, 5) ft. of a force F = −5i+ 2j − 4k lb. whose line of action passesthrough the point (−1,−1, 2) ft.

3-13 The force F = −2i+ 4j + zk lb. acts at a point (5, 4,−2) ft. and creates

a moment of M = 36i− 31j + 28k ft. lb. about the origin. Determine z.

3-14 A force F = −4i + 5j − 2k lb. acting at a point (−7, 5, 3) ft. creates a

moment of M = −7i− 6j − k ft. lb. about a point (x,−1, 4). Determinex.

3-15 A force F = xi+yj+4k lb. acting at a point (−5,−3, 2) creates a moment

of M = −6i+ 30j + 30k ft. lb. about the origin. Determine x and y.

3.4. SUMMARY 91

3-16 A force P = −10i + 20j + 30k lb. produces a moment with scalar com-ponents Mx = 120, My = −150, Mz = 140 ft. lb. about the origin.Determine where the force P intersects the xy plane.

3-17 The angular velocity of a rigid body rotating about an axis of rotation isω = 5i + 2j − 4k rad. per. sec. Determine the linear velocity of point Bon the body with a position vector r = 3i − j + 3k ft. directed from apoint on the axis of rotation.

3-18 A rigid body is rotating about an axis with an angular velocity ω =10i + 5j − 2k rad. per. sec. What is the linear velocity of a point P onthe body with a position vector r = 2i− 4j + 5k ft. directed from a pointon the axis of rotation?

3-19 Particle P is on a rigid body spinning about an axis having directioncosines of l = 0.5, m = −0.707, and n = 0.5; as shown in Figuire 3-19. The angular speed is 10 rad. per. sec., and the linear velocity isV = −62.4i− 5j + 55.4k. Determine y and z.

Prob. 3-19


3-20 In Figure 3-20, a belt is being used to drive a pulley 12 in. in diameter.If the pulley rotates at 120 rpm, what is the linear speed of the pulley?

Prob. 3-20

3-21 The diesel locomotive wheel shown in Figure 3-21has a diameter of 60 in.If the locomotive is moving at 60 mph, what is the angular velocity of thewheel?

Prob. 3-21

3.4. SUMMARY 93

3-22 The weight W attached to the string is whirled around in a circle that liesin the xy plane, as shown in Figure 3-22. If W has an angular speed of10 rad. per. sec. and a linear speed of 70 ft. per. sec., what must be thelength of the string?

Prob. 3-22

3-23 In Figure 3-23, a particle P in a rigid body is spinning about an axis whichpasses through the point B(6, 8, 4) ft. If point P has an angular velocity

ω = 3i+ 4j + 5k rad. per. sec., find its linear velocity.

Prob. 3-23


3-24 In Figure 3-24, an emery wheel is rotating about the axis with an angularspeed of 100 rad. per. sec. The axis of rotation has direction cosines ofl = −0.072, m = −0.758, and n = 0.650. A particle P locates as shownis dislodged from the wheel. Determine the linear velocity and anglesthe path of the particle makes with the orthogonal axes at the instant itbecomes dislodged.

Prob. 3-24

3-25 Find the perpendicular distance from point P to the line AB in Figure3-25.

Prob. 3-25

3-26 Find the perpendicular distance from the point P (4, 5, 7) to a plane definedby three points A(−1, 2, 3), B(5, 3,−2), and C(2,−4, 1).

3.4. SUMMARY 95

3-27 Given vectors A = 3i+ 7j − 2k and B = 2i− 6j + 5k acting at the pointsshown in Figure 3-27, find the shortest perpendicular distance betweentheir lines of action.

Prob. 3-27

3-28 Determine the shortest distance between the lines CD and EF given bythe points C(3, 2, 1), D(5,−4, 3), E(−1, 4,−3), and F (−4, 2, 5).

3-29 Two non-intersecting pipe lines AB and CD are shown in Figure 3-29.Determine the shortest possible connection between them.

Prob. 3-29

3-30 How close does the vector V = 3i−2j+5k acting through a point (4, 5, 6)in. come to the y axis?


3-31 In Figure 3-31, the lines AB and CD are the center lines of two conduits1 ft. and 2 ft. in diameter respectively. Determine the maximum value ofz so that the two may pass without interference. Conduit CD must passunder AB.

Prob. 3-31

3-32 If the lines AB and CD in Figure 3-32 represent two control cables in anairplane, find the minimum clearance between the two cables.

Prob. 3-32

Part IV

Dimension and Units

97

Chapter 4

Dimension

4.1 Concept

The dimension of the quantity tells what kind of stuff it is made of. For anexample, consider the quantity a = 5.0 seconds. a is made of time, and we say“a has the dimension of time.” Put more formally, a is a measure in the timedimension.

4.2 Notation

In the above example, [a] = T . The square brackets are read as “dimensionof” and the upper case letter T here designates the dimensional value of time.Similarly,

[b] = L says that b has dimension of length[c] = M says that c has dimension of mass[d] = F says that d has dimension of force[e] = θ says that e has dimension of temperature[f ] = Q says that f has dimension of electric charge[g] = N says that g has null dimension (it is a dimensionless number).

(4.1)

The symbols a, b, · · · used here as arguments are arbitrary. The symbols T , L,M , used for the dimensional values are conventional and should be used consis-tently.

The [ ] brackets may be regarded as an operator which extracts certaininformation (the dimension) from its argument and ignores its other properties(magnitude). This idea may be clarified by the example of a similar kind ofoperator, the arithmetic sign function:

sgn(x) =x

|x|(4.2)

99

100 CHAPTER 4. DIMENSION

A little consideration reveals that this operator extracts the sign from any realnumber x (tells whether the argument x is made of positive or negative stuff).However, the [ ] operator, unlike the sgn operator , cannot be expressed in moreelementary terms.

The reader is cautioned to remember that the dimensional values T , L, M ,· · · are not quantities (like a, b, c, · · · ) of time, length, mass. . . but are valuesof a much more general nature. T , L, M , · · · may be considered as sets. Thenthe statement [a] = T is interpreted as a is a member of the set of all quantitieswhich are measures of time. Notice that the sets T , L, M , · · · are disjoint: nophysical quantity can be at once a member of more than one set. For example, ifp is a mass (really, a quantity, or measure of mass) then p cannot simultaneouslybe a length. This reflects the disjoint property of M and L.

4.3 Fundamental and Derived Dimensions

Of course not every physical quantity has one of the dimensions T , L, M , · · ·as listed above. But some, such as the dimension of velocity may be expressedas a compound of those given dimensions; if v is a velocity then [v] = L/T .Notice that we could add a dimensional value V to our list above, to designatethe dimension of velocity, but the dimension of velocity is expressible as thequotient of the dimensions length and time:

V =L

T(4.3)

and it is therefore not necessary to establish V as a new dimension. However itis often convenient to utilize such derived dimensions. We say that the dimen-sion of velocity is derived from the fundamental dimensions of length and time.

The distinction between the fundamental and derived dimensions is in factarbitrary. If our senses were better suited for determining velocities than lengths,then we would probably consider V as a fundamental dimension, and we wouldwrite the equation (4.3) as L = V T , which expresses L as a derived dimensionin terms of the fundamentals V and T . But it is customary, as well as naturalfor us, to consider length and time more fundamental than velocity or volume(is it obvious that the dimension of volume may be expressed as L3?).

4.4 Algebra

4.4.1 Taking dimensions

Two physical quantities are said to be equal only when they are equal measuresin the same dimension (at this point we will concentrate on the sameness ofdimensions). It follows that if p is a mass and q is a velocity then they cannot

4.4. ALGEBRA 101

be equal quantities. Then if

p = q (4.4)

it is also true that

[p] = [q] (4.5)

and we may execute the extraction of dimensions as though it were an algebraicoperation on either side of the equation. The equation remains valid, but isreduced to a relation between dimensional values. We say that the equation(4.5) is obtained by “taking dimensions” of the equation (4.4).

4.4.2 Commutative relations with multiplication and di-vision

Using quantities from the list in Section 4.2, we have a velocity b/a and we mayexpress (some of) our knowledge of b/a by the equation,[

b

a

]=L

T(4.6)

But it is basic that

[b]

[a]=L

T(4.7)

and we see that the dimensional value of

[b

a

]and

[b]

[a]are equal: the outcome of

successive use of the operations ÷ and [ ] is independent of the order in whichthey are taken. The operations ÷ and [ ] are then said to be commutative.

For another example, consider the formula for area of a rectangle:

A = lw (4.8)

Then

[A] = [lw]. (4.9)

But [A] = L2 (an intuitive observation), and we have

L2 = [lw] (4.10)

It is clear that since l and w are lengths,

[l][w] = LL = L2 (4.11)

and we have

[lw] = [l][w]. (4.12)

It is apparent that the [ ] and x operations are commutative.


4.4.3 Addition

Consider the addition A+B where [A] = [B] = T . Clearly, the sum of the twoparticular times A and B is also a time(the dimensional sets T , L, M , · · · areclosed with respect to addition), and we have

[A+B] = [A] + [B] = T + T = T. (4.13)

The strange-looking equation,

T + T = T (4.14)

reflects the fact that T is not a measure (measures add numerically), but thatthe sum of a time and a time is just a time.

Now consider the addition of the time A with the mass C. A + C is neithera time nor a mass. It is a heterogeneous (lumpy) mixture of time and mass andsuch mixtures are not useful in engineering problems. Such an addition is, forour purposes, undefined and we say that the sum A + C does not exist.

The additive relations for dimensional values are summarized:

• Addition of like dimensional values leaves their value unchanged (equation(4.14)), and

• Addition of unlike dimensional values is undefined.

4.4.4 Dimensional Homogeneity

In order to be a physical equation (and therefore properly describe a physicalrelation), an equation must be dimensionally homogeneous. This notion em-bodies the ideas of subsection 4.4.1 and 4.4.3 above. Specifically an equation issaid to be dimensionally homogeneous whenever

4-1 all sums include only terms of like dimension, and

4-2 dimensions of right and left-hand members are the same.

As an example for dimensional homogeneity, consider the formula for speed ofsound in a perfect gas,

c =√γRT (4.15)

in which

c = the speed, [c]=L

T.

γ = ratio of specific heats of the gas, [γ]= N

R = Universal Gas Constant, [R]=FL

MθT = gas temperature, [T ]= θ

4.5. NEWTON’S SECOND LAW 103

The reader should recognize the dimensions written for c and T , but furtherfamiliarity with the particular quantities would be required in order to deducethe dimensions of and R. It is well to note in passing that physical quantitiescalled ratios are ordinarily dimensionless, as in this case. For convenience incomputation, square equation (4.15) and then take dimensions:

[c2] = [γ][R][T ]L2

T 2= N

FL

Mθθ = F

L

M

(4.16)

Notice that the null dimension N may be ignored when it appears as a factor,like the number 1. To exhibit the dimensional homogeneity of this relation, itwill be necessary to borrow a dimensional relation which is given in the nextsection: F = MLT−2. Substituting this into equation (4.16) gives

L2

T 2=ML

T 2· LM

=L2

T 2

and the dimensional character of equation (4.15) is proved to be correct; it isdimensionally homogeneous.

4.5 Newton’s Second Law

The basic law of classical dynamics (distinct from the relativistic and quantummechanics of modern physics) is the relationship given by Isaac Newton in 1685,which may be stated: The product of the mass of a body with its accelerationgives the value of the disturbing force on the body. This is concisely stated bythe equation

f = ma (4.17)

where f is the disturbing force, m is the mass of the body in question, and a isits acceleration.

Now if we take dimensions of equation (4.17),

[f ] = [m] [a]

F = ML

T 2= MLT−2. (4.18)

we find that F , L, M , T are interdependent dimensions. Either F or M isordinarily subordinated (considered to be a derived dimension, based on equa-tion (4.18), and this choice forms the basis of the two rather different kinds ofunit systems in common usage: MLT and FLT systems. This point will bediscussed in the following section.


Exercise

4-1 The dimensional values T , L, M , F , θ, Q given in section 4.2, serve as abasic vocabulary of dimensions. The dimensions of other familiar physicalquantities may be expressed in terms of these few values. Give a nameassociated with each of the following:

a.F

L2: Answer “pressure”

b.L

T 2:

c. L3:

d.M

L3:

e.Q

T:

4-2 Translate the following into dimensional notation.

a. speed- answer:L

T

b. arc length1-

Prob. 4-2

c. angle2-

1Remember that the angle α in radians, is given by the ratio of arc length to radius: α =S

r2Remember that the angle α in radians, is given by the ratio of arc length to radius: α =

S

r

105


d. rotation rate (angular velocity)-

e. electric charge density

4-3 Given the following as sets of fundamental dimensions, derive expressionsfor L, T , M , F in terms of the given dimensions.

a. area (A), velocity (V), mass-density(D) answer: L = A1/2, T =A1/2

V, M = DA3/2, F = DAV 2

b. area (A), velocity (V), pressure(P)

c. volume (V), acceleration (A), pressure(P)

It should be understood that the notations A, V , D, · · · introduced forthe purpose of this problem, are not ordinarily used. For example, thedimension of area is ordinarily written asL2.

4-4 Determine whether the following equations are dimensionally homoge-neous. h and l represent lengths, p is a pressure, m is a mass, g is anacceleration, t is a time, f is a force.

a. h = 12gt

2

b. h = gt2

c. t =pl

mf

4-5 whether the following equations are dimensionally homogeneous in accor-dance with data from problem 4.

a. p =mg

l2+

f

h2

b. f =2ph3

l+mh

3t2

c. gt =

√plh2

m

4-6 a. What is the dimension of tan ? (Remember its definition.)

b. The exponential function ex may be represented by the infinite series:

ex = 1 + x+x2

2!+x3

3!+ · · ·

What must be the dimension of x?

c. What must be the dimension of ex?

4-7 In the quadratic equation

a · x2 + b · x+ c = 0

x represents a time and c represents a velocity. What are the dimensionsof a and b?

4.5. NEWTON’S SECOND LAW 107

4-8 In the cubic equation

a · y3 + b · y2 + c · y + d = 0

y represents a length, and a is mass-density. What are the dimensions ofb, c, and d?

4-9 What is the dimension of the bi-linear form,

a · x+ b

c · x+ d

when a = 1 yard, b = 1 acre, c = 1 gallon.

4-10 What is the dimension of the bi-quadratic form,

p · z2 + q · z + r

s · z2 + t · z + u

when r is a velocity, and q and s are accelerations?

4-11 Is it possible that v is a velocity in the equation

v = a · t+ v√

2 + t2

where a is an acceleration and t is a time? Is the equation dimensionallyhomogeneous?

4-12 Equivalence between mass and energy is given by Einsteins formula,

E = mc2

where E is the energy equivalent of the mass m and c is the speed of light.Determine the dimension of energy in terms of F , L, and T .

4-13 The energy (see previous problem 4-12) stored in a spring of stiffnessk is given by 1

2ke2, where e denotes elongation of the spring: [e] = L.

Determine the dimension of ke in terms of F , L, T . What is the dimensionof k in terms of M , L, T?

4-14 The power P produced by a pump in a uniform pipeline is

P = Q∆p

where Q is the volume flow [Q] = L3T−1, through the pump, and ∆p isthe pressure increase produced by the pump. Give the dimension of powerin terms of F , L, T .

4-15 The power produced at the output shaft of an engine is given by

p = TΩ

where T is torque and Ω is the shaft rotation rate. What is the dimensionof the torque T in terms of dimensions F , L, T?


4-16 The average curvature in the interval (P1, P2), in the drawing below, isgiven by the increase in the slope angle ∆θ, divided by the arc length,∆S:

Caverage =∆θ

∆S

What is the dimension of curvature?

Prob. 4-16

Chapter 5

Unit Systems

5.1 Concept

In order to describe physical (dimensional) quantities it is necessary to establisha unit for each dimension. This is a standard quantity for the dimension, andwill serve as a basis for measurement or description of any quantity in thesame dimension. In our example, a = 5.0 seconds, the second is used as theunit for the time dimension and we are told that if a is cut up into one-secondchunks, then there will be five of them.

5.2 Notation and Derived Units

To say that the second is established as the unit of time, one may write,

UT = 1second. (5.1)

Notice that this equation has the dimensional content [UT ] = [1 sec] = T , andthe numerical content 1 = 1; the magnitude of UT , by its definition, is 1. Simi-larly, we will write UL to denote the unit of length, etc.

Supposing that the foot is adopted as UL, we ask whether this places anyrestriction on the unit chosen for area. In view of the dimensional relation

[area] = L2 (5.2)

it is natural to write

Uarea = U2L, (5.3)

and in fact this is the only definition of Uarea which satisfies the required di-mensional relation, equation (5.2), as well as the associated numerical equation:l = l2. Evidently,

Uarea = (1ft)2 = 1ft2,

109

110 CHAPTER 5. UNIT SYSTEMS

and we say that the unit of area is a derived unit, 1 ft2 based on the fundamentalunit, 1 ft. The subscript “area” is here written out in order to avoid the in-troduction of too-numerous dimensional symbols. Dimensional expressions aresimplest when written in fundamental terms and on this account it is wise toavoid the use of special symbols for derived dimensions.

5.3 Newton’s Second Law

From the second law we have the dimensional relation F = MLT−2. Now anyconsistent system of units must satisfy the equation,

UF =UMULU2T

(5.4)

It is clear that the dimensional requirement is satisfied, and because all elementshave unit numerical value, the numerical requirement is also satisfied: 1 =1× 1

12. We will also have occasion to use this relation in the transposed (but, of

course, equivalent) form,

UM =UFU

2T

UL(5.5)

In MLT unit systems, those based on M , L, T as fundamental dimensions,equation (5.4) will be used to derive UF . For FLT unit systems, equation (5.5)will determine UM .

5.4 Metric Unit Systems

We begin our discussion of specific unit systems with the MKS (meter, kilo-gram, second) system. These units may be tabulated: In the MKS system,

Table 5.1: Metric Unit SystemDimension MKS Unit

M UM = kilogram(kg)L UL = meter (m)T UT = second (sec.)F UF = ?

these fundamental units are defined and agreed upon by the International Com-

5.5. BRITISH SYSTEM 111

mittee on Weights and Measures1;

1kilogram = mass of the Prototype Kilogram, a platinum-iridiumcylinder having height and diameter of 39 mm,which is maintained by the International Committeein a vault at Sevres, France;

1meter = 1,650,763.73 wavelengths of the red-orange lightemitted by gaseous Krypton 86;

1second = 1/31,556,925.9747 of the year 1900

But there is no such definition required (or allowable) for UF in the MKSsystem because the value of UF is now dictated by equation (5.4);

UF =UMULU2T

=(1kg)(1m)

(1sec)2= 1

kg ·msec2

(5.6)

Once UM , UL, UT have been established this is the only consistent definitionfor UF . For brevity this unit of force is given the name Newton so that

1kg ·msec2

= 1newton(N) (5.7)

Now let us expand the units chart to include the CGS system: The definitions

Table 5.2: MKS vs CGS SystemXXXXXXXXXXUnit

SystemMKS CGS

UM kg gram(gm)UL m centimeter(cm)UT sec sec

UFkg ·msec2

= Ngm · cmsec2

= dyne

of the gram and centimeter, based on the kilogram and meter, are familiar.Notice that the unit force is derived for the CGS system exactly as it was forthe MKS system, and it is given the name “dyne.”

5.5 British System

The units of mass, length, and time for the British system are defined (by theInternational Committee, and ratified by the Congress of the United States)

1At the time of this writing, the International Committee on Weights and Measures hastaken steps to redefine the second on the basis of transition time of a certain magnetic momentin the atom of Cesium 133. However, the definition has not yet been formally accepted. SeeNBS Technical News Bulletin, 48:12:209 Dec. 1964.


relative to the fundamental MKS units:

UM = 1 pound-mass (lbm)≡ 0.45359237kgUL = 1 foot (ft.)≡ 0.3048mUT = 1 second

Continuing to expand the units chart, we have

Table 5.3: MLT Unit ChartXXXXXXXXXXUnitSystem MLT System

MKS CGS British

UM kg gram(gm) lbmUL m centimeter(cm) ftUT sec sec sec

UFkg ·msec2

= Ngm · cmsec2

= dynelbm · ftsec2

= pdl

Notice that the evaluation of UF follows (via equation (5.4)) directly fromthe values adopted for UM , UL, UT in all three systems, reflecting their MLTbasis. The British unit of force,

1lbm · ftsec2

≡ 1poundl(pdl.) (5.8)

is seldom used in engineering practice.

5.6 Definition of Weight-force

Before we expand the units chart finally to include the Engineering Practicesystem of units, it will be useful to pause and consider the notion of “weight”(weight-force). The weight of a body is defined to be that force on the bodydue to gravitation. The weight force which you experience presently (yes, that’swhat keeps you in your chair) is due to the gravitational field of the earth actingon the mass of your body. (The sun and moon also affect your weight, but onlyvery slightly because they are relatively far away.)

The force produced by a gravitational field acting on a body will not serveto characterize the field since the force also depends on the properties (mass)of the particular body. However, any freely falling body will experience exactlythe same acceleration in a given gravitational field, and that acceleration willcharacterize the field. The acceleration due to the earths gravity is given the

symbol g and has the approximate value g = 32.2ft

sec2at sea level.

Just as for any other force, the acceleration produced by the weight-force w,on a body of mass m is given by Newton’s second law, F = ma:

5.7. ENGINEERING PRACTICE UNITS 113

(weight-force, w) = (mass, m) x (acceleration caused by weight-force).

But the acceleration caused by the weight-force is just g = 32.2ft

sec2and we

have

w = mg (5.9)

This relation, a special case of Newton’s law, may be taken as a quantitativedefinition of weight. This equation is valid for whatever value of g prevails andproperly predicts weightlessness in the absence of gravity.

5.7 Engineering Practice Units

The Engineering Practice system of units is an FLT system. This says the unitsof force, length, and time are the fundamental units of the system, and the unitof mass is derived according to equation (5.5). The familiar British units oflength and time are employed, but now the unit of force is defined to be thepound-force (lbf.):

UF = 1lbf ≡ theweight− force(weight)ofonepound−mass(lbm) (5.10)

in a standard gravitational field.

This definition requires the further definition of the standard gravitationalfield, and such a field is defined as one in which a body (subject to no otherforces) will fall with an acceleration of exactly

gstandard = 32.174ft

sec2. (5.11)

Of course this represents the value of gravitational acceleration at the earth’ssurface; but since that acceleration varies from point to point on the surface,and also from time to time, it has been necessary to establish this standardgravity in order that the unit 1lbf be independent of those variations. Exceptin cases where very precise calculation is necessary, the value g = 32.2 ft

sec2 isused to represent standard gravity.

Now since UL = 1ft, UT = 1sec, UF = 1lbf are fundamental units in theEngineering Practice system, UM must follow from equation (5.5):

UM =UFU

2T

UL=

(1lbf)(1sec)2

(1ft)= 1

lbf · sec2

ft(5.12)

and this derived unit is given the not-too-romantic name slug, so that

1lbf · sec2

ft≡ 1slug (5.13)


Table 5.4: Completed Unit Chart

SystemMLT System (absolute) FLT(gravitational)

MKS CGS British Engineering

UM kg gram(gm) lbmlbf · sec2

ft= slug

UL m centimeter(cm) ft ftUT sec sec sec sec

UFkg ·msec2

= Ngm · cmsec2

= dynelbm · ftsec2

= pdl lbf

Then the completed unit chart is

The Engineering system is divided apart from the MLT Systems, under theheading FLT to emphasize the structural difference between it and the MLTsystems. The Engineering system is sometimes called a gravitational systembecause of the role played by gravity in the definition of the pound-force. TheMLT systems are gravity-independent and so are called absolute systems.

5.8 Relation between British and EngineeringUnits

Just as the mass units 1kg and 1lbm are related by 1lbm = 1.4536kg, theremust be such a relation between the lbm and the slug. For this let us appeal toequations (5.9) and (5.11 ) and write

1lbf = (1lbm)

(32.174

ft

sec2

). (5.14)

By rearranging the coefficients,

1lbf = (32.174lbm)

(1ft

sec2

)(5.15)

and we have

32.174lbm = 1lbf · sec2

ft. (5.16)

And using equation (5.13) we see that

32.174lbm = 1slug (5.17)

which is the desired relation. It is left as a problem to derive the correspondingrelation between the pound-force and poundal.

5.9. CONVERSION OF UNITS 115

5.9 Conversion of Units

For the purpose of minimizing confusion in conversions among the several unitsystems we will adopt the formalism of the conversion factor. A conversionfactor is a quotient of equal measures, which therefore has only the numericalvalue 1. Taking an example from the preceding discussion, we have the relations

1 lbm = 0.4536 kg, and 1 slug = 32.175 lbm.

These will serve as a basis for the conversion of 1kg into slugs:

1kg = 1kg

(1lbm

0.4539kg

)(1slug

32.174lbm

)=

1slug

0.4536× 32.174= 0.06852slug

Notice that each of the bracketed quotients (the conversion factors) is equal tounity, and so these terms do not alter value; they only change the form. Otherconversion factors are, for example(

1gallon

0.1337ft3

),

(5280ft

1mile

),

(1revolution

2πradians

).

5.10 Conventional Reference Points

It is important to observe that certain scales of measurement not only establisha unit for their dimension, but also establish a reference point (origin) fromwhich to measure. For example, the Celsius (centigrade) scale establishes, as aunit of temperature, one one-hundredth of the temperature interval between thefreezing point and boiling point of pure water at standard atmosphere pressure.That unit is called the Celsius degree, 1C (note position of degree symbol).And the freezing point is established as the origin of the Celsius scale, 0C(note position of degree symbol). Notice that the Celsius degree is a measure(interval) of temperature, just as the second is a measure of time. But unlikethe second, use of the Celsius degree implies measurement from a conventionalreference point. The specification of a temperature as 30C means that thetemperature is thirty Celsius degrees above the freezing point of water.

Similarly, atmospheric pressure (14.7lbf

in2= 14.7psi) is often used as a refer-

ence point when measuring pressure. When it is said that the pressure in a tireis 30.0psi, it is ordinarily meant that the air pressure inside the tire is 30.0psihigher than the air pressure outside (atmospheric). Such a specification is called“gage pressure” and it would be written as 30.0psig. If the actual pressure werespecified, then it would be 14.7psi higher, and would be written as 44.7psia(“absolute pressure”).


While many measurements in engineering are independent of conventionalreference points, it is important to recognize the role of such references wherethey are used.

Exercise

“weight” here means weight in standard gravity unless otherwise specified.When converting units in the following problems, do not refer to any tablesother than the following:

Table 5.5: Unit Conversion Table1 acre = 43560 ft2 1 mile =5280 ft1 foot =0.3048 m 1 pound-mass = 0.4536

1 gallon = 0.1337 ft3 1 radian = 57.3 degrees1 horsepower = 550 ft-lbf/sec 1 slug =32.174 lbm

1 inch = 2.54 m

5-1 What must be the unit of the length in a system which has the acre as itsunit of area?

5-2 What must be the unit of the length in a system having the acre-foot asits unit of volume?

5-3 What is the unit of mass in a system using the units 1inch, 1sec, 1lbf?This unit is sometimes convenient in engineering and has been dubbed the“slinch.”

5-4 What is the unit of length in a system having 1mile

houras its unit of velocity,

and 1day as its unit of time?

5-5 Describe the way in which the definition of the pound-force depends onthe kilogram, foot and second.

5-6 Beginning with their definitions, derive a relation between the pound-forceand poundal.

5-7 What is the weight of a kilograms, expressed in lbf?

5-8 What is the equivalent of 1newton, expressed in pounds-force?

5-9 What is the weight of 1slug, expressed in pounds-force?

117


5-10 Convert 1poundal to dynes.

5-11 Convert 1acre · foot to gallons.

5-12 The joule is a unit of work, or energy, defined by

1 joule = 1 newton ·meter

Convert 1 joule to ft · lbf .

5-13 The watt is a unit of power defined by

1watt = 1joule

sec= 1

nt ·msec

Convert 1 watt to horsepower(HP).

5-14 The liter is defined as 1000 cubic centimeters. Show that

1liter = 0.264gallon.

5-15 The specific gravity of a substance is the ratio of its mass-density to thatof water. It is numerically equal with the mass-density of the substance,expressed in grams per cubic centimeter. What is the specific gravity ofsteel, which weighs 0.283lbf/in3?

5-16 The Kelvin temperature scale utilizes the Celsius degree, so that 1K =1C, but takes absolute zero as its origin so that 0K = −273.15C or0C = 273.15K. What is the boiling point of water (at standard atmo-spheric pressure), on the Kelvin scale?

5-17 Like the Kelvin scale, the Rankine temperature scale utilizes the Fahren-heit degree with is 5/9 as large as the Kelvin (or Celsius) degree: 1F =5/9C. What is the freezing point of water (at standard atmosphere pres-sure), on the Rankine scale?

5-18 Express the pressure in a perfect vacuum in psia, and also in psig.

Chapter 6

Dimensional Analysis andSimilitude

6.1 Introduction

From the preceding sections it is apparent that the dimensional variables in aphysical relation must appear in such an arrangement that the condition ofdimensional homogeneity is satisfied. And we can learn something about themathematical form of the relation by considering the dimensions of the severalvariables in the light of that condition. The method of deriving informationabout the mathematical form of a physical relation by consideration of the di-mensions involved is known as dimensional analysis. A dimensional analysisleads us directly to the elements of similitude: the relation which exists be-tween physical processes which are so similar that they may be described byidentical mathematical equations. Similitude is the central principle in the the-ory of models, which is the basis of such diverse engineering methods as thesimulation of full-scale flight in the laboratory by means of a wind tunnel, orsimulation of the erosive action of the tides by means of a laboratory-scale model.

These ideas will be clarified by an example following the next section. Nowlet us set down the basis of dimensional analysis, the Buckingham Pi-theorem.

6.2 The Pi-theory of Dimensional Analysis

If a physical process involves m physical quantities a, b, c, · · · , x then the gov-erning relation for the process will be a mathematical equation involving thosequantities. In other words, the process places a restriction on the quantitiesinvolved so that not all of them are freely variable. We represent the governingrelation formally by writing,

f(a, b, c, ·, x) = 0. (6.1)

119

120 CHAPTER 6. DIMENSIONAL ANALYSIS AND SIMILITUDE

It must be understood that equation (6.1) only states that there is some relationamong the quantities a, b, c, · · · , x; the specific form of the function f is notknow.

Now the dimensions of a, b, c, · · · , x will include some number n of fun-damental dimensions. This means, for example, that if the list a, b, c, · · · , xcomprises quantities having the dimensions of mass, velocity, area, and time,then n = 3 since these dimensions may all be expressed as compounds of threedimensions (such as FLT or MLT ), but not less than three.

The Pi-theorem says that we can findm−n dimensionless groups, P1, P2, · · · ,Pm−n, involving the quantities a, b, c, · · · , x and that the governing equation(6.1) may be rewritten in terms of those m − n groups, which are called pi-factors1,

F (P1, P2, · · · , Pm−n) = 0 (6.2)

Like the function f in equation (6.1), the form of the function F in equation(6.2) is unknown. But it is important to observe that the latter equation issimpler in the sense that it involves fewer variables.

6.3 First Example: Speed of a Freely FallingBody

Suppose we wish to find a relation which describes the speed of a freely fallingbody. In order to apply the dimensional analysis, it is necessary as a first stepto assemble a list of the physical quantities which relate to the problem, in thiscase

(1) the speed, v, since it is the desired quantity,

(2) gravity, g, since without it the body would not fall at all,

(3) we suspect the mass, m, of the body is important, and,

(4) the distance h which the body has fallen (we wish to express v in termsof h rather than the time of falling, t, and so t is excluded from the list).

(5) all the other physical quantities such as temperature of the body and thetime of day appear to be unrelated to the problem, and so the list isterminated.

1Traditionally, the notation π1, π2, · · · has been used for these dimensionless groups,following Buckingham. However, that notation sometimes causes confusion, due to the usualassociation of the value 3.1416 with π.

6.3. FIRST EXAMPLE: SPEED OF A FREELY FALLING BODY 121

Notice that this step is purely intuitive, based on some feeling or qualitativeknowledge, gained in previous experience with the kind of system in question.The governing relation may be represented by

f(v, g,m, h) = 0 (6.3)

We observe that there are four physical quantities involved in the governingrelation, so that m = 4.

Our second step is to enumerate the dimensions of the quantities involved:

[v] = LT−1

[g] = LT−2

[m] = M(or, alternatively, FT 2L−1)[h] = L

(6.4)

We observe that three fundamental dimensions L, T , M(or L, T , F if [m] is writ-ten in the alternative form) are represented among these quantities. Therefore,n = 3, and the Pi-theorem tells us that the governing relation for this systemis expressible in terms of a single (m− n = 4− 3 = 1) pi-factor (dimensionlessgroup). The dimensionless form of the governing relation is then

F (P1) = 0 (6.5)

The problem remains (as our third step) to find the required dimension-less group, P1. There are systematic methods for finding dimensionless groupsamong a given system of dimensional quantities2, but in the present example wecan continue the discussion on an intuitive basis. It is clear that the compoundhg has the dimension L2T−2, and that this is also the dimension of v2. Therequired dimensionless group is then

P1 =v2

gh. (6.6)

Notice that it is impossible to fabricate any other dimensionless group (notcounting functions of the same group as “others”) involving the quantities ofthis system. Particularly, it is impossible for the mass m to be involved in thepi-factor. This tells us that the speed v must be independent of the mass.

The fourth step in our dimensional analysis will be to extract an expressionfor v equation (6.5). That equation expresses a condition on the single variableP1, and restricts its value to be a root of that equation. Let us denote the valueof that root (constant) by α2. If the character of the function F were known,then the value of its root could be determined by solving equation (6.5), but

2See, for example, Dimensional Analysis and Theory of Models by H.L. Langhaar, JohnWiley and Sons, 1951.


dimensional analysis never tells us the nature of the function F . We can onlyknow that it requires P1 to have a certain value:

P1 = α2 (6.7)

Now we have, from the equation (6.6),

v2

gh= α2,

or v = α√gh

(6.8)

This is the desired mathematical from relating v with the other quantities listedfor the system. You might recognize, from your physics course, that the requiredvalue of α is

√2, so that

v =√

2gh (6.9)

Without such prior knowledge,our a complete mathematical solution of the pro-gram, it would be necessary to determine α experimentally.

6.4 Second Example: Period of a Simple Pen-dulum

A simple pendulum is a pendulum in which all mass is concentrated in a pointat the end of an inextensible string fixed without friction to an immovable sup-port. (No real pendulum can be a simple pendulum, but many systems are soconstructed that they differ only slightly from a simple pendulum and so aregoverned by the theory of the simple pendulum.) The period of the pendulum

Figure 6.1: A simple pendulum

is the time which it requires to go from an end-point of its swing through a cycle

6.4. SECOND EXAMPLE: PERIOD OF A SIMPLE PENDULUM 123

of its oscillation and return to that end-point.

We list the quantities which would seem to influence the period:

(1) The period, p is the desired quantity,

(2) the amplitude, θo of the oscillation,

(3) gravitational acceleration, g,

(4) the pendulum mass, m.

(5) the pendulum length, l,

(6) all the other innumrable physical properties, such as temperature, color,etc., appear to be quite unrelated.

The governing relation is then

f(p, θo, g,m, l) = 0.

We list the dimensions:

[p] = T[θo] = N(remember: angles are dimensionless)[g] = LT−2

[m] = M[l] = L

(6.10)

Observe that m = 5, and n = 3. Note that the null dimension of θo is notcounted in n. The dimensionless governing equation is then

F (P1, P2) = 0. (6.11)

One of the pi-factors is clearly the amplitude θo, since it is itself already dimen-sionless:

P1 = θo (6.12)

It is easy to confirm that p

√g

lis another dimensionless group, and it will serve

as the second pi-factor:

P2 = p

√g

l(6.13)

Equation (6.11) may be solved formally for P2:

P2 = H(P1) (6.14)


where H, like F , is an unknown function. Substituting equations (6.12) and(6.13) into (6.14), we obtain,

p

√g

l= H(θo) (6.15)

By analogy with equation (6.8), it tis evident that the amplitude-function H(θo)here corresponds to the role of the constant α2 in the previous example, whichinvolved only a single pi-factor.

By measuring the period of a certain pendulum of known length in a knowngravitational field for a sequence of known amplitudes, the function H(θo) canbe determined experimentally, using equation (6.15). The result of such anexperimental program is represented at the below. This plot exhibits a similarity

Figure 6.2: Similarity relation for simple pendulum

relation for simple pendulum and may be used in computations for any simplependulum whatever.

6.4. SECOND EXAMPLE: PERIOD OF A SIMPLE PENDULUM 125

Example

Determine the period of a simple pendulum of length 10 ft swinging with anamplitude of 75 in a standard gravitational field.

Solution From the equation (6.15),

p =

√l

gH(θo)

=

√10ft · sec2

32.2ftH(75)

using Figure 6.2, H(75) = 7, and

p =y√3.22

sec. = 3.9sec.


Exercise

6-1 Astronauts arriving a new planet wish to measure its gravity. They ob-serve that a simple pendulum of length 1 ft. requires 5 seconds for acycle of its oscillation, at an amplitude of 2. What is the acceleration ofgravity?

6-2 A jet of water from a firehose is directed squarely against a brick wall.The velocity of the stream is v, its cross-section area is A, and the mass-density of the water is ρ. Find an expression for the force F exerted onthe wall by the stream.

6-3 A mass m slides in a circular path on a frictionless horizontal surface,tethered by a string of length r. If the mass slides with velocity v incircles at the end of its tether, what will be the tension force in the string(as far as you can determine by a dimensional analysis.)?

6-4 It is observed that the acoustical speed c in a solid depends only on theelastic modulus E, [E] = FL−2, and the mass-density ρ of the material.How is c related to E and ρ?

6-5 The pitch (frequency of vibration) of a guitar string is determined by

(a) its length l, by fingering on the frets,

(b) its tension t, by tuning the string, and

(c) its lineal mass-density m, the “size” of the string.

Lineal mass-density is just the mass of the string per unit length. Howdoes the frequency f depend on t, l, and m? How much must one increasethe tension in order to raise the pitch by one octave (exactly double thefrequency)? How much must one shorten the string in order to raise itspitch by one octave?

127


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129

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