Schaum's Outline of Engineering Mechanics: Statics, Seventh Edition

By Merle C. Potter, E. W. Nelson, Charles L. Best and William G. McLean

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• 628 fully solved problems to reinforce knowledge
• 1 final practice exam
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• Extra practice on topics such as orthogonal triad of unit vectors, resultant of distributed force system, noncoplanar force systems, slope of the Shear diagram, and slope of the Moment diagram
• Support for all the major textbooks for statics courses

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• Language: English
• Publisher: McGraw-Hill Education
• Release date: Jan 1, 2021
• ISBN: 9781260462890

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Vectors

1.1 DEFINITIONS

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

Vector quantities possess both magnitude and direction; 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 |P| 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; i.e., P + Q = Q + P.

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

Fig. 1-3 A typical triangle.

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

EXAMPLE 1.1 In a plane, find the resultant of a 1300-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,

Now applying the law of sines,

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

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:

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; i.e., 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 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,

The above equation may be extended to any number of vectors.

Fig. 1-6 Composition of vectors. (When a vector passes through the xy plane or the xz plane, it becomes dashed.)

1.6 MULTIPLICATION OF VECTORS BY SCALARS

(a) The product of vector P and scalar m is a vector mP whose magnitude is |m| 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.

Fig. 1-8 Vector components of P.

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. 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 α with the z axis.

SOLUTION: The components of P in the coordinate directions are

We know that (an application of the Pythagorean theorem)

The component in the z-direction is then

The vector P is now written using the unit vectors as

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 θ (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, for example,

Similarly, the magnitude of the vector component of P along any line L can be written 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

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 magnitudes of the two vectors multiplied by 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 θ. 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 is 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 ΔP in P as time changes from t to (t + Δt) is

Then

If P(t) = Pxi + Pyj + Pzk, 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 a 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 shall 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/s 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 the SI system and will be used to avoid the zeros required when using millimeters. Further, a space should be left between the number and unit symbol; e.g., 2.85 mm, not 2.85mm. When using five or more figures, space them in groups of three 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 seventh edition, all of the quantities are in SI units.

We finish this section with comments on significant figures. In most calculations, a material property or a measurement is involved. The fores of interest in Statics that act on elements of structures and machines are found from such calculations. Material properties and dimensions are seldom known to four significant figures and often only to three, or possibly two. Consequently, the information given in problems is assumed accurate 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/s, only three significant figures. A dimension is stated as 10 mm; it is assumed accurate to three, or at most four significant figures. It is usually acceptable to express answers using 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 for Example 1.1. See 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.

Fig. 1-13

Fig. 1-14

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 α.

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 is

Then, by the law of sines, the angle α 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 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- 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?

SOLUTION

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

Fig. 1-16

Fig. 1-17

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.

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).

SOLUTION

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

The vertical component Pv acts up and is

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

Fig. 1-18

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

acting up the plane. The component normal to the plane is

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.

SOLUTION