Schaum's Outline of Engineering Mechanics Dynamics, Seventh Edition

Preface

This book is designed to supplement standard texts, primarily to assist students of engineering and science in acquiring a more thorough knowledge and proficiency in dynamics, the course that follows statics in the mechanics sequence. It is based on the authors’ conviction that numerous solved problems constitute one of the best means for clarifying and fixing in mind the basic principles. While this book will not mesh precisely with any one text, the authors feel that it can be a very valuable adjunct to all.

The previous editions of this book have been very favorably received. This edition incorporates SI units only. This eliminates the problems encountered when mixing units and allows students to focus on the subject being studied.

The authors attempt to use the best mathematical tools available to students at the sophomore level. Thus the vector approach is applied in those chapters where its techniques provide an elegance and simplicity in theory and problems. On the other hand, we have not hesitated to use scalar methods elsewhere, since they provide entirely adequate solutions to many of the problems. Chapter 1 is a complete review of the minimum number of vector definitions and operations necessary for the entire book, and applications of this introductory chapter are made throughout the book.

Chapter topics correspond to material usually covered in a standard introductory dynamics course. Most chapters contain the appropriate derivations along with examples that illustrate the basic principles. The text material is followed by graded sets of solved and supplementary problems. The solved problems serve to illustrate and amplify the theory, present methods of analysis, provide practical examples, and bring into sharp focus those fine points that enable the student to apply the basic principles correctly and confidently. Numerous derivations of formulas are also included among the solved problems. The many supplementary problems serve as a review of the material covered in each chapter.

In the first edition the authors gratefully acknowledged their indebtedness to Paul B. Eaton and J. Warren Gillon. In the second edition the authors received helpful suggestions and criticism from Charles L. Best and John W. McNabb. Also in that edition Larry Freed and Paul Gary checked the solutions to the problems. In the third and fourth editions, computer solutions were added to numerous problems; these solutions have been eliminated in this seventh edition since several software packages have been developed that allow students to perform such solutions. For typing the manuscripts of the third and fourth editions we are indebted to Elizabeth Bullock.

M. C. POTTER

E. W. NELSON

C. L. BEST

W. G. MCLEAN

Vectors

1.1 DEFINITIONS

Scalar quantities possess only magnitude; examples are time, volume, energy, mass, density, and work. Scalars are added by ordinary algebraic methods, for example, 2 s + 7 s = 9 s and 14 kg − 5 kg = 9 kg.

Vector quantities possess both magnitude and direction; direction is understood to include both the angle that the line of action makes with a given reference line and the sense of the vector along the line of action. Examples are force, displacement, and velocity. A vector is represented by an arrow at the given angle. The head of the arrow indicates the sense, and the length usually represents the magnitude of the vector. The symbol for a vector is shown in print in boldface type, such as P. The magnitude is represented by or P. Often, when writing by hand, we would use , rather than P.

A free vector may be moved anywhere in space provided it maintains the same direction and magnitude.

A sliding vector may be applied at any point along its line of action. By the principle of transmissibility, the external effects of a sliding vector remain the same.

A bound or fixed vector must remain at the same point of application.

A unit vector is a vector one unit in length. It is represented by i, n, or in written form by

The negative of a vector P is the vector -P that has the same magnitude and angle but is of the opposite sense, that is, it acts in the opposite direction.

The resultant of a system of vectors is the least number of vectors that will replace the given system.

1.2 ADDITION OF TWO VECTORS

(a)   The parallelogram law states that the resultant R of two vectors P and Q is the diagonal of the parallelogram for which P and Q are adjacent sides. All three vectors P, Q, and R are concurrent as shown in Fig. 1-1(a). P and Q are also called the components of R.

Fig. 1-1 The addition of vectors.

(b)   If the sides of the parallelogram in Fig. 1-1(a) are perpendicular, the vectors P and Q are said to be rectangular components of the vector R. The rectangular components are illustrated in Fig. 1-1(b). The magnitudes of the rectangular components are given by

(c)   Triangle law. Place the tail end of either vector at the head end of the other. The resultant is drawn from the tail end of the first vector to the head end of the other. The triangle law follows from the parallelogram law because opposite sides of the parallelogram are free vectors, as shown in Fig. 1-2.

Fig. 1-2 The triangle law.

(d)   Vector addition is commutative; that is, P + Q = Q + P.

(e)   The law of cosines (refer to Fig. 1-3) is

The law of sines (refer to Fig. 1-3) is

Fig. 1-3 A typical triangle.

EXAMPLE 1.1 In a plane, find the resultant of a 300-N force at 30° and a -250-N force at 90°, using the parallelogram method. Refer to Fig. 1-4(a). Also, find the angle α between the resultant and the y axis. (Angles are always measured counterclockwise from the positive x axis.)

Fig. 1-4

SOLUTION: Draw a sketch of the problem, not necessarily to scale. The negative sign indicates that the 250-N force acts along the 90° line downward toward the origin. This is equivalent to a positive 250-N force along the 270° line, according to the principle of transmissibility.

As in Fig. 1-4(b), place the tail ends of the two vectors at a common point. Complete the parallelogram. Consider the triangle, one side of which is the y axis, in Fig. 1-4(b). The sides of this triangle are R, 250, and 300. The angle between the 250 and 300 sides is 60°. Applying the law of cosines gives

Now applying the law of sines, we get

Note: If the forces and angles are drawn to scale, the magnitude of R and the angle a could be measured from the drawing.

1.3 SUBTRACTION OF A VECTOR

Subtraction of a vector is accomplished by adding the negative of the vector:

Note also that

-(P + Q) = -P - Q

EXAMPLE 1.2 In a plane, subtract 130 N at 60° from 280 N at 320° (see Fig. 1-5).

Fig. 1-5

SOLUTION: To the 280-N, 320° force add the negative of the 130-N, 60° force. The resultant is found by applying the law of cosines as follows:

The law of sines allows us to find α:

Thus, R makes an angle of -62.9° with the x axis.

1.4 ZERO VECTOR

A zero vector is obtained when a vector is subtracted from itself; that is, P - P = 0. This is also called a null vector.

1.5 COMPOSITION OF VECTORS

Composition of vectors is the process of determining the resultant of a system of vectors. A vector polygon is drawn by placing the tail end of each vector in turn at the head end of the preceding vector, as shown in Fig. 1-6. The resultant is drawn from the tail end of the first vector to the head end (terminus) of the last vector. As will be shown later, not all vector systems reduce to a single vector. Since the order in which the vectors are drawn is immaterial, it can be seen that for three given vectors P, Q, and S,

Fig. 1-6 Composition of vectors.

Equation (1.5) may be extended to any number of vectors.

1.6 MULTIPLICATION OF VECTORS BY SCALARS

(a)   The product of vector P and scalar m is a vector mP whose magnitude is times as great as the magnitude of P and that is similarly or oppositely directed to P, depending on whether m is positive or negative.

(b)   Other operations with scalars m and n are

1.7 ORTHOGONAL TRIAD OF UNIT VECTORS

An orthogonal triad of unit vectors i, j, and k is formed¹ by drawing unit vectors along the x, y, and z axes, respectively. A right-handed set of axes is shown in Fig. 1-7.

Fig. 1-7 Unit vectors i, j, k.

A vector P is written as

where Pxi, Py j, and Pzk are the vector components of P along the x, y, and z axes, respectively, as shown in Fig. 1-8.

Fig. 1-8 Vector components of P.

Note that

EXAMPLE 1.3 Using the triad of unit vectors, write the vector P that has magnitude of 100 which makes an angle 50° with the negative x axis, 80° with the y axis, and angle a with the z axis.

SOLUTION: The components of P in the coordinate directions are

Px = -100 cos 50 = -64.3,    Py = 100 cos 80° = 17.4,    Pz = l00 cos a

We know that (an application of the Pythagorean theorem)

100² = (-64.3)² + 17.4² + (l00 cos α)². ∴ a = 4l.8°

The component in the z direction is then

Pz = 100 cos 4l.8 = 74.5

The vector P is now written using the unit vectors as

P = -64.3i + 17.4j + 74.5k

1.8 POSITION VECTOR

The position vector r of a point (x, y, z) in space is written

where (see Fig. 1-9).

Fig. 1-9 The position vector r.

EXAMPLE 1.4 A position vector r has a magnitude of 40 cm. Its known components are rx = 20 cm and rz = -30 cm. Write r using the triad of unit vectors.

SOLUTION: Using the components, we can write

The vector r is written as

1.9 DOT OR SCALAR PRODUCT

The dot or scalar product of two vectors P and Q, written P · Q, is a scalar quantity and is defined as the product of the magnitudes of the two vectors and the cosine of their included angle q (see Fig. 1-10). Thus,

Fig. 1-10 The included angle θ between two vectors.

The following laws hold for dot products, where m is a scalar:

Since i, j, and k are orthogonal,

Also, if and then

The magnitudes of the vector components of P along the rectangular axes can be written

since, e.g.,

Similarly, the magnitude of the vector component of P along any line L can be written P · eL, where eL is the unit vector along the line L. (Some authors use u as the unit vector.) Figure 1-11 shows a plane through the tail end A of vector P and a plane through the head B, both planes being perpendicular to line L. The planes intersect line L at points C and D. The vector CD is the component of P along L, and its magnitude equals P · eL = PeL cos θ.

Fig. 1-11 The component of P along a line.

EXAMPLE 1.5 Two vectors are given as P = 20i + 40j − 30k and Q = 20i − 40j + 30k. Determine the angle between the two vectors.

SOLUTION: Use the definition of the dot product:

EXAMPLE 1.6 Determine the unit vector eL for a line L that originates at point (2, 3, 0) and passes through point (−2, 4, 6). Next determine the projection of the vector P = 2i + 3j − k along the line L.

SOLUTION: The line L changes from +2 to -2 in the x direction, or a change of -4. The change in the y direction is 4 − 3 = 1. The change in the z direction is 6 − 0 = 6. The unit vector is

The projection of P is then

1.10 THE CROSS OR VECTOR PRODUCT

The cross or vector product of two vectors P and Q, written P × Q, is a vector R whose magnitude is the product of the magnitudes of the two vectors and the sine of their included angle. The vector R = P × Q is normal to the plane of P and Q and points in the direction of advance of a right-handed screw when turned in the direction from P to Q through the smaller included angle q. Thus if e is the unit vector that gives the direction of R = P × Q, the cross product can be written

Figure 1-12 indicates that P × Q = -Q × P (not commutative).

Fig. 1-12 The cross product of two vectors.

The following laws hold for cross products, where m is a scalar:

Since i, j, and k are orthogonal,

Also, if and then

The proof of this cross-product determinant is the objective of Example 1.7.

EXAMPLE 1.7 Show that the cross product of two vectors P and Q can be written as

SOLUTION: Write the given vectors in component form and expand the cross product to obtain

But i × i = j × j = k × k = 0; and i × j = k and j × i = -k, etc. Hence,

These terms can be grouped as

or in determinant form as

Be careful to observe that the scalar components of the first vector P in the cross product are written in the middle row of the determinant.

1.11 VECTOR CALCULUS

(a)   Differentiation of a vector P that varies with respect to a scalar quantity such as time t is performed as follows.

Let P = P(t); that is, P is a function of time t. A change ∆P in P as time changes from t to t + ∆t is

Then

If where Px, Py, and Pz are functions of time t, we have

The following operations are valid:

(b)   Integration of a vector P that varies with respect to a scalar quantity, such as time t, is performed as follows. Let P = P(t); that is, P is a function of time t. Then

1.12 DIMENSIONS AND UNITS

In the study of mechanics, the characteristics of a body and its motion can be described in terms of a set of fundamental quantities called dimensions. In the United States, engineers have been accustomed to a gravitational system using the dimensions of force, length, and time (with units of lb, ft, and s). Most countries throughout the world use an absolute system in which the selected dimensions are mass, length, and time (with units of kg, m, and s). There is a growing trend to use this second system in the United States.

Both systems derive from Newton’s second law of motion, which is often written as

where R is the resultant of all forces acting on an object, a is the acceleration of the object, and m is its mass.

The International System (SI)

In the International System (SI),* the unit of mass is the kilogram (kg), the unit of length is the meter (m), and the unit of time is the second (s). The unit of force is the newton (N) and is defined as the force that will accelerate a mass of one kilogram one meter per second squared (m/s²). Thus,

A mass of 1 kg falling freely near the surface of the earth has an acceleration of gravity g that varies very slightly from place to place. In this book we assume an average value of 9.80 m/s². Thus the force of gravity acting on a 1-kg mass becomes

Of course, problems in statics involve forces; but, in a problem, a mass given in kilograms is not a force. The gravitational force acting on the mass, referred to as the weight W, must be used. In all work involving mass, the student must remember to multiply the mass in kilograms by 9.80 m/s² to obtain the gravitational force in newtons. A 5-kg mass has a gravitational force of 5 × 9.8 = 49 N acting on it.

In solving statics problems, the mass may not be mentioned. It is important to realize that the mass in kilograms is a constant for a given body. On the surface of the moon, this same given mass will have acting on it a force of gravity approximately one-sixth of that on the earth.

The student should also note that, in SI, the millimeter (mm) is the standard linear dimension unit for engineering drawings. Centimeters are tolerated in SI and can be used to avoid the zeros required when using millimeters. Further, a space should be left between the number and unit symbol, for example, 2.85 mm, not 2.85mm. When using five or more figures, space them in groups of 3 starting at the decimal point as 12 830 000. Do not use commas in SI. A number with four figures can be written without the space unless it is in a column of quantities involving five or more figures.

Tables of SI units, SI prefixes, and conversion factors for the modern metric system (SI) are included in Appendix A. In this edition, all the quantities are in SI units.

We finish this section with comments on significant figures. In most calculations, a material property or a measured quantity is involved. The quantities of interest in dynamics involve dimensions, mass, gravity, velocity, and acceleration, to name a few, and all of these quantities are seldom known to four significant figures and often only two or three. Consequently, the information given in a problem is assumed known to three, possibly four significant figures. Thus, it is not appropriate to express answers to five or six significant figures. Our calculations are only as accurate as the least significant figure. For example, we use gravity as 9.80 m/s², only three significant figures. A dimension is stated as 10 mm; it is assumed accurate to three and at most four significant figures. It is usually acceptable to express answers using at most four significant figures, but not five or six. The use of calculators may even provide eight. The engineer does not, in general, work with five or six significant figures.

SOLVED PROBLEMS

1.1.   Use the triangle law and solve Example 1.1 (see Fig. 1-13).

Fig. 1-13

SOLUTION

It is immaterial which vector is chosen first. Take the 300-N force. To the head of this vector attach the tail end of the 250-N force. Sketch the resultant from the tail end of the 300-N force to the head end of the 250-N force. Using the triangle shown, the results are the same as in Example 1.1.

1.2.   The resultant of two forces in a plane is 400 N at 120°, as shown in Fig. 1-14. One of the forces is 200 N at 20°. Determine the missing force F and the angle a.

Fig. 1-14

SOLUTION

Select a point through which to draw the resultant and the given 200-N force. Draw the force connecting the head ends of the given force and the resultant. This represents the missing force F.

The result is obtained by the laws of trigonometry. The angle between R and the 200-N force is 100°, and hence, by the law of cosines, the unknown force F follows

Then, by the law of sines, the angle a is found:

1.3.   Determine the resultant of the following coplanar system of forces: 26 N at 10°; 39 N at 114°; 63 N at 183°; 57 N at 261° (see Fig. 1-15).

Fig. 1-15

SOLUTION

This problem can be solved by using the idea of rectangular components. Resolve each force in Fig. 1-15 into x and y components. Since all the x components are collinear, they can be added algebraically, as can the y components. Now, if the x components and y components are added, the two sums form the x and y components of the resultant. Thus,

1.4.   In Fig. 1-16 the rectangular component of the force F is 10 N in the direction of OH. The force F acts at 60° to the positive x axis. What is the magnitude of the force?

Fig. 1-16

SOLUTION

The component of F in the direction of OH is Fcosθ. Hence,

1.5.   An 80-kg block is positioned on a board inclined 20° with the horizontal. What is the gravitational component (a) normal to the board and (b) parallel to the board? See Fig. 1-17.

Fig. 1-17

SOLUTION

(a)   The normal component is at an angle of 20° with the gravitational force vector (the weight), which has a magnitude of 80(9.8) = 784 N. The normal component is

(b)   The parallel component is

1.6.   A force P of 235 N acts at an angle of 60° with the horizontal on a block resting on a 22° inclined plane. Determine (a) the horizontal and vertical components of P and (b) the components of P perpendicular to and along the plane. Refer to Fig. 1-18(a).

Fig. 1-18

SOLUTION

(a)   The horizontal component Ph acts to the left and is

The vertical component acts up and is

as shown in Fig. 1-18(b).

(b)   The component P|| parallel to the plane

acting up the plane. The component P⊥ normal to the plane

as shown in Fig. 1-18(c).

1.7.   The three forces shown in Fig. 1-19 produce a resultant force of 20 N acting upward along the y axis. Determine the magnitudes of F and P.

Fig. 1-19

SOLUTION

For the resultant to be a force of 20 N upward along the y axis, and As the sum of the x components must be equal to the x component of the resultant

Similarly,

1.8.   Refer to Fig. 1-20. The x, y, and z edges of a rectangular parallelepiped are 4, 3, and 2 m, respectively. If the diagonal OP drawn from the origin represents a 50-N force, determine the x, y, and z components of the force. Express the force as a vector in terms of the unit vectors i, j, and k.

Fig. 1-20

SOLUTION

Let represent, respectively, the angles between the diagonal OP and the x, y, z axes. Then

Length of Hence,

Since each component in the sketch is in the positive direction of the axis along which it acts,

The vector P is written as

1.9.   Determine the x, y, and z components of a 100-N force passing from the origin through the point (2, –4, 1). Express the vector in terms of the unit vectors i, j, and k.

SOLUTION

The direction cosines of the force line are

Hence, The vector P is

1.10.   A force F = 2.63i + 4.28j − 5.92k N acts through the origin. What is the magnitude of this force and what angles does it make with the x, y, and z axes?

SOLUTION

1.11.   Find the dot product of and Q = -2.81i − 6.09j + 1.12k m.

SOLUTION

1.12.   Determine the projection of the force P = 10i − 8j + 14k N on the directed line L which originates at point (2, -5, 3) and passes through point (5, 2, -4).

SOLUTION

The unit vector along L is

The projection of P on L is

The minus sign indicates that the projection is directed opposite to the direction of L.

1.13.   Find the cross product of P = 2.85i + 4.67j − 8.09k and Q = 28.3i + 44.6j + 53.3k.

SOLUTION

1.14.   Determine the time derivative of the position vector where i, j, and k are fixed vectors.

SOLUTION

The time derivative is

1.15. Determine the time integral from time t1 = 1 s to time t2 = 3 s of the velocity vector

where i, j, and k are fixed vectors.

SOLUTION

SUPPLEMENTARY PROBLEMS

1.16. Determine the resultant of the coplanar forces 100 N at 0° and 200 N at 90°.

Ans.   224 N,

1.17. Determine the resultant of the coplanar forces 32 N at 20° and 64 N at 190°.

Ans.   33.0 N,

1.18. Find the resultant of the coplanar forces 80 N at -30° and 60 N at 60°.

Ans.   100 N,

1.19. Find the resultant of the concurrent coplanar forces 120 N at 78° and 70 N at 293°.

Ans.   74.7 N,

1.20. The resultant of two coplanar forces is 18 N at 30°. If one of the forces is 28 N at 0°, determine the other.

Ans.   15.3 N, 144°

1.21. The resultant of two coplanar forces is 36 N at 45°. If one of the forces is 24 N at 0°, find the other force.

Ans.   25.5 N, 87°

1.22. The resultant of two coplanar forces is 50 N at 143°. One of the forces is 120 N at 238°. Determine the missing force.

Ans.   134 N,

1.23. The resultant of two forces, one in the positive x direction and the other in the positive y direction, is 100 N at 50° counterclockwise from the positive x direction. What are the two forces?

Ans.   

1.24. A force of 120 N has a rectangular component of 84 N acting along a line making an angle of 20° counterclockwise from the positive x axis. What angle does the 120-N