MECÂNICA

Application of forces to the handle of this wrench will produce a tendency to rotate the wrench about its end. It is important to calculate this effect and, in some cases, to be able to simplify this system to a single resultant force and specify where this resultant acts on the wrench.

Unpublished Work Š 2007 by R. C. Hibbeler. To be published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River, New Jersey. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department.

Force System Resultants CHAPTER OBJECTIVES

• To discuss the concept of the moment of a force and show how to calculate it in two and three dimensions.

• • • •

To provide a method for finding the moment of a force about a specified axis.

To define the moment of a couple.

To present methods for determining the resultants of nonconcurrent force systems.

To indicate how to reduce a simple distributed loading to a resultant force having a specified location.

4.1 Moment of a Force—

z

Scalar Formulation

The moment of a force about a point or axis provides a measure of the tendency of the force to cause a body to rotate about the point or axis. For example, consider the horizontal force Fx, which acts perpendicular to the handle of the wrench and is located a distance dy from point O, Fig. 4–1a.

It is seen that this force tends to cause the pipe to turn about the z axis.

The larger the force or the distance dy , the greater the turning effect. This tendency for rotation caused by Fx is sometimes called a torque, but most often it is called the moment of a force or simply the moment 1M O2z. Note that the moment axis (z) is perpendicular to the shaded plane (x–y) which contains both Fx and dy and that this axis intersects the plane at point O.

(MO)z O y

x

dy

Fx (a)

Fig. 4–1 117 Unpublished Work © 2007 by R. C. Hibbeler. To be published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River, New Jersey. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department.

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z

Now consider applying the force Fz to the wrench, Fig. 4–1b. This force will not rotate the pipe about the z axis. Instead, it tends to rotate it about the x axis. Keep in mind that although it may not be possible to actually “rotate” or turn the pipe in this manner, Fz still creates the tendency for rotation and so the moment 1M O2x is produced. As before, the force and distance dy lie in the shaded plane (y–z) which is perpendicular to the moment axis (x). Lastly, if a force Fy is applied to the wrench, Fig. 4–1c, no moment is produced about point O. This results in a lack of turning since the line of action of the force passes through O and therefore no tendency for rotation is possible.

We will now generalize the above discussion and consider the force F and point O which lie in a shaded plane as shown in Fig. 4–2a. The moment M O about point O, or about an axis passing through O and perpendicular to the plane, is a vector quantity since it has a specified magnitude and direction.

O y

(MO)x

x

dy

Fz

(b) z

O

Fy

dy

x

(c)

Fig. 4–1 (cont.)

Moment axis

MO

F

d

O

(a)

d F

MO O

(b)

Fig. 4–2

Sense of rotation

Magnitude. The magnitude of MO is y

MO = Fd

(4–1)

where d is referred to as the moment arm or perpendicular distance from the axis at point O to the line of action of the force. Units of moment magnitude consist of force times distance, e.g., N # m or lb # ft.

Direction. The direction of M O will be specified by using the “righthand rule.”To do this, the fingers of the right hand are curled such that they follow the sense of rotation, which would occur if the force could rotate about point O, Fig. 4–2a. The thumb then points along the moment axis so that it gives the direction and sense of the moment vector, which is upward and perpendicular to the shaded plane containing F and d.

In three dimensions, M O is illustrated by a vector arrow with a curl on it to distinguish it from a force vector, Fig. 4–2a. Many problems in mechanics, however, involve coplanar force systems that may be conveniently viewed in two dimensions. For example, a twodimensional view of Fig. 4–2a is given in Fig. 4–2b. Here M O is simply represented by the (counterclockwise) curl, which indicates the action of F. The arrowhead on this curl is used to show the sense of rotation caused by F. Using the right-hand rule, however, realize that the direction and sense of the moment vector in Fig. 4–2b are specified by the thumb, which points out of the page since the fingers follow the curl. In particular, notice that this curl or sense of rotation can always be determined by observing in which direction the force would “orbit” about point O (counterclockwise in Fig. 4–2b). In two dimensions we will often refer to finding the moment of a force “about a point” (O).

Keep in mind, however, that the moment always acts about an axis which is perpendicular to the plane containing F and d, and this axis intersects the plane at the point (O), Fig. 4–2a.

Unpublished Work © 2007 by R. C. Hibbeler. To be published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River, New Jersey. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department.

119

4.1 MOMENT OF A FORCE—SCALAR FORMULATION

Resultant Moment of a System of Coplanar Forces. If a

z

system of forces lies in an x–y plane, then the moment produced by each force about point O will be directed along the z axis, Fig. 4–3. Consequently, the resultant moment M RO of the system can be determined by simply adding the moments of all forces algebraically since all the moment vectors are collinear. We can write this vector sum symbolically as d + MRO = ©Fd

M RO F2 d2

(4–2)

d3 F3

Here the counterclockwise curl written alongside the equation indicates that, by the scalar sign convention, the moment of any force will be positive if it is directed along the +z axis, whereas a negative moment is directed along the -z axis.

The following examples illustrate numerical application of Eqs. 4–1 and 4–2.

O d1 F1

Fig. 4–3

By pushing down on the pry bar the load on the ground at A can be lifted. The turning effect, caused by the applied force, is due to the moment about A.To produce this moment with minimum effort we instinctively know that the force should be applied to the end of the bar; however, the direction in which this force is applied is also important.This is because a moment is the product of the force and the moment arm. Notice that when the force is at an angle u 6 90°, then the moment arm distance is shorter than when the force is applied perpendicular to the bar, u = 90°, i.e., d¿ 6 d. Hence the greatest moment is produced when the force is farthest from point A and applied perpendicular to the axis of the bar so as to maximize the moment arm.

F

F

u

d

MA % F d¿ A

d¿

A

(MA)max $ F d

In the photo to the right, the moment of a force does not always cause a rotation.

For example, the force F tends to rotate the beam clockwise about its support at A with a moment MA = FdA. The actual rotation would occur if the support at B were removed. In the same manner, F creates a tendency to rotate the beam counterclockwise about B with a moment MB = FdB. Here the support at A prevents the rotation.

F MA % FdA A

dA

MB % FdB dB

B

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EXAMPLE 4.1 For each case illustrated in Fig. 4–4, determine the moment of the force about point O. 100 N

SOLUTION (SCALAR ANALYSIS) The line of action of each force is extended as a dashed line in order to establish the moment arm d. Also illustrated is the tendency of rotation of the member as caused by the force. Furthermore, the orbit of the force is shown as a colored curl. Thus,

O 2m

Fig. 4–4a Fig. 4–4b Fig. 4–4c Fig. 4–4d Fig. 4–4e

(a)

MO MO MO MO MO

= = = = =

1100 N212 m2 = 200 N # m b 150 N210.75 m2 = 37.5 N # m b 140 lb214 ft + 2 cos 30° ft2 = 229 lb # ft b 160 lb211 sin 45° ft2 = 42.4 lb # ft g 17 kN214 m - 1 m2 = 21.0 kN # m g

Ans. Ans. Ans. Ans. Ans.

2 ft 2m

30& 40 lb

O

O 0.75 m

4 ft 50 N

2 cos 30& ft

(c)

(b)

2m 1m

7 kN

3 ft 4m

O

1 ft

45&

1 sin 45& ft 60 lb O

(d)

(e)

Fig. 4–4e

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121

4.1 MOMENT OF A FORCE—SCALAR FORMULATION

EXAMPLE 4.2 Determine the moments of the 800-N force acting on the frame in Fig. 4–5 about points A, B, C, and D. 1.25 m

SOLUTION (SCALAR ANALYSIS) In general, M = Fd, where d is the moment arm or perpendicular distance from the point on the moment axis to the line of action of the force. Hence, MA = 800 N12.5 m2 = 2000 N # m b MB = 800 N11.5 m2 = 1200 N # m b

Ans.

MC = 800 N102 = 0 1line of action of F passes through C2 MD = 800 N10.5 m2 = 400 N # m g

Ans. Ans.

D

0.5 m F % 800 N

C

1.5 m

Ans.

B 1m A

The curls indicate the sense of rotation of the moment, which is defined by the direction the force orbits about each point.

Fig. 4–5

EXAMPLE 4.3 Determine the resultant moment of the four forces acting on the rod shown in Fig. 4–6 about point O. SOLUTION Assuming that positive moments act in the +k direction, i.e., counterclockwise, we have

y 50 N 2m

2m

d+MRO = ©Fd;

60 N

MRO = -50 N12 m2 + 60 N102 + 20 N13 sin 30° m2 MRO

-40 N14 m + 3 cos 30° m2 = -334 N # m = 334 N # m b

30&

O

x 20 N

3m

Ans.

For this calculation, note how the moment-arm distances for the 20-N and 40-N forces are established from the extended (dashed) lines of action of each of these forces.

40 N

Fig. 4–6

Unpublished Work © 2007 by R. C. Hibbeler. To be published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River, New Jersey. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department.

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4.2 Cross Product The moment of a force will be formulated using Cartesian vectors in the next section. Before doing this, however, it is first necessary to expand our knowledge of vector algebra and introduce the cross-product method of vector multiplication.

The cross product of two vectors A and B yields the vector C, which is written C = A * B and is read “C equals A cross B.”

Magnitude. The magnitude of C is defined as the product of the magnitudes of A and B and the sine of the angle u between their tails 10° … u … 180°2. Thus, C = AB sin u.

C%A$B

Direction. Vector C has a direction that is perpendicular to the plane containing A and B such that C is specified by the right-hand rule; i.e., curling the fingers of the right hand from vector A (cross) to vector B, the thumb then points in the direction of C, as shown in Fig. 4–7.

Knowing both the magnitude and direction of C, we can write A

u

C = A * B = 1AB sin u2uC B

Fig. 4–7

(4–3)

where the scalar AB sin u defines the magnitude of C and the unit vector uC defines the direction of C. The terms of Eq. 4–3 are illustrated graphically in Fig. 4–8.

C uC

A

C % AB sin u

u

B

Fig. 4–8 Unpublished Work © 2007 by R. C. Hibbeler. To be published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River, New Jersey. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department.

123

4.2 CROSS PRODUCT

Laws of Operation.

C%A$B

1. The commutative law is not valid; i.e., A * B Z B * A

B

Rather, A

A * B = -B * A

B

This is shown in Fig. 4–9 by using the right-hand rule.The cross product B * A yields a vector that acts in the opposite direction to C; i.e., B * A = -C.

A

2. Multiplication by a scalar: a1A * B2 = 1aA2 * B = A * 1aB2 = 1A * B2a

#C%B$A

This property is easily shown since the magnitude of the resultant vector 1 ƒ a ƒ AB sin u2 and its direction are the same in each case.

Fig. 4–9

3. The distributive law: A * 1B + D2 = 1A * B2 + 1A * D2

z

The proof of this identity is left as an exercise (see Prob. 4–1). It is important to note that proper order of the cross products must be maintained, since they are not commutative.

Cartesian Vector Formulation. Equation 4–3 may be used to find the cross product of a pair of Cartesian unit vectors. For example, to find i * j, the magnitude of the resultant vector is 1i21j21sin 90°2 = 112112112 = 1, and its direction is determined using the right-hand rule. As shown in Fig. 4–10, the resultant vector points in the +k direction. Thus, i * j = 112k. In a similar manner,

k%i$j

j i x

Fig. 4–10

i * j = k i * k = -j i * i = 0 j * k = i j * i = -k j * j = 0 k * i = j k * j = -i k * k = 0 These results should not be memorized; rather, it should be clearly understood how each is obtained by using the right-hand rule and the definition of the cross product. A simple scheme shown in Fig. 4–11 is helpful for obtaining the same results when the need arises. If the circle is constructed as shown, then “crossing” two unit vectors in a counterclockwise fashion around the circle yields the positive third unit vector; e.g., k * i = j. Moving clockwise, a negative unit vector is obtained; e.g., i * k = -j.

y

i " j

#

k

Fig. 4–11

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Consider now the cross product of two general vectors A and B which are expressed in Cartesian vector form. We have A * B = = + +

1A x i + A y j + A z k2 * 1Bx i + By j + Bz k2 A xBx1i * i2 + A xBy1i * j2 + A xBz1i * k2 A yBx1j * i2 + A yBy1j * j2 + A yBz1j * k2 A zBx1k * i2 + A zBy1k * j2 + A zBz1k * k2

Carrying out the cross-product operations and combining terms yields A * B = 1A yBz - A zBy2i - 1A xBz - A zBx2j + 1A xBy - A yBx2k (4–4) This equation may also be written in a more compact determinant form as i 3 A * B = Ax Bx

j Ay By

k Az 3 Bz

(4–5)

Thus, to find the cross product of any two Cartesian vectors A and B, it is necessary to expand a determinant whose first row of elements consists of the unit vectors i, j, and k and whose second and third rows represent the x, y, z components of the two vectors A and B, respectively.* *A determinant having three rows and three columns can be expanded using three minors, each of which is multiplied by one of the three terms in the first row. There are four elements in each minor, e.g., A11 A21

A12 A22

By definition, this notation represents the terms 1A 11A 22 - A 12A 212, which is simply the product of the two elements of the arrow slanting downward to the right 1A 11A 222 minus the product of the two elements intersected by the arrow slanting downward to the left 1A 12A 212. For a 3 * 3 determinant, such as Eq. 4–5, the three minors can be generated in accordance with the following scheme: For element i:

i Ax Bx

j Ay By

k Az % i(AyBz # AzBy) Bz

For element j:

i Ax Bx

j Ay By

k Az % #j(AxBz # AzBx) Bz

For element k:

i Ax Bx

j Ay By

k Az % k(AxBy # AyBx) Bz

Adding the results and noting that the j element must include the minus sign yields the expanded form of A * B given by Eq. 4–4.

Unpublished Work © 2007 by R. C. Hibbeler. To be published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River, New Jersey. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department.

4.3 MOMENT OF A FORCE—VECTOR FORMULATION

125

4.3 Moment of a Force—Vector Formulation

The moment of a force F about point O, or actually about the moment axis passing through O and perpendicular to the plane containing O and F, Fig. 4–12a, can be expressed using the vector cross product, namely, MO = r * F

(4–6)

Here r represents a position vector drawn from O to any point lying on the line of action of F. We will now show that indeed the moment M O, when determined by this cross product, has the proper magnitude and direction.

Magnitude. The magnitude of the cross product is defined from Eq. 4–3

as MO = rF sin u, where the angle u is measured between the tails of r and F.To establish this angle, r must be treated as a sliding vector so that u can be constructed properly, Fig. 4–12b. Since the moment arm d = r sin u, then MO = rF sin u = F1r sin u2 = Fd which agrees with Eq. 4–1.

Direction. The direction and sense of M O in Eq. 4–6 are determined

by the right-hand rule as it applies to the cross product. Thus, extending r to the dashed position and curling the right-hand fingers from r toward F, “r cross F,” the thumb is directed upward or perpendicular to the plane containing r and F and this is in the same direction as M O, the moment of the force about point O, Fig. 4–12b. Note that the “curl” of the fingers, like the curl around the moment vector, indicates the sense of rotation caused by the force. Since the cross product is not commutative, it is important that the proper order of r and F be maintained in Eq. 4–6.

Moment axis

Moment axis

MO

MO

r

A

O

r

u u

A

d r

O

F

F (a)

(b)

Fig. 4–12 Unpublished Work © 2007 by R. C. Hibbeler. To be published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River, New Jersey. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department.

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Principle of Transmissibility. Consider the force F applied at

z

point A in Fig. 4–13. The moment created by F about O is M O = rA * F; however, it was shown that “r” can extend from O to any point on the line of action of F. Consequently, F may be applied at point B or C, and the same moment M O = rB * F = rC * F will be computed. As a result, F has the properties of a sliding vector and can therefore act at any point along its line of action and still create the same moment about point O. We refer to this as the principle of transmissibility, and we will discuss this property further in Sec. 4.7.

C rC

B F rB

MO

A rA y

O

Cartesian Vector Formulation. If we establish x, y, z coordinate axes, then the position vector r and force F can be expressed as Cartesian vectors, Fig. 4–14. Applying Eq. 4–5 we have

x

i M O = r * F = 3 rx Fx

Fig. 4–13

where

j ry Fy

k rz 3 Fz

(4–7)

rx, ry, rz

represent the x, y, z components of the position vector drawn from point O to any point on the line of action of the force

Fx, Fy, Fz

represent the x, y, z components of the force vector

If the determinant is expanded, then like Eq. 4–4 we have M O = 1ryFz - rzFy2i - 1rxFz - rzFx2j + 1rxFy - ryFx2k

(4–8)

The physical meaning of these three moment components becomes evident by studying Fig. 4–14a. For example, the i component of M O is determined from the moments of Fx, Fy, and Fz about the x axis. In particular, note that Fx does not create a moment or tendency to cause z

Fz F

C

O

F

Fy

r rx

z

MO

rz

E

ry B

Fx

A x

Moment axis

r

y

O

y

D x (a)

(b)

Fig. 4–14 Unpublished Work © 2007 by R. C. Hibbeler. To be published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River, New Jersey. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department.

127

4.3 MOMENT OF A FORCE—VECTOR FORMULATION

turning about the x axis since this force is parallel to the x axis. The line of action of Fy passes through point E, and so the magnitude of the moment of Fy about point A on the x axis is rzFy, By the right-hand rule this component acts in the negative i direction. Likewise, Fz contributes a moment component of ryFzi. Thus, 1MO2x = 1ryFz - rzFy2 as shown in Eq. 4–8. As an exercise, establish the j and k components of M O in this manner and show that indeed the expanded form of the determinant, Eq. 4–8, represents the moment of F about point O. Once M O is determined, realize that it will always be perpendicular to the shaded plane containing vectors r and F, Fig. 4–14b.

It will be shown in Example 4.4 that the computation of the moment using the cross product has a distinct advantage over the scalar formulation when solving problems in three dimensions. This is because it is generally easier to establish the position vector r to the force, rather than determining the moment-arm distance d that must be directed perpendicular to the line of action of the force.

z F3

F1

r3

r1

F2

MRO

r2

Resultant Moment of a System of Forces. If a body is acted

O

upon by a system of forces, Fig. 4–15, the resultant moment of the forces about point O can be determined by vector addition resulting from successive applications of Eq. 4–6. This resultant can be written symbolically as M RO = ©1r * F2

(4–9)

and is shown in Fig. 4–15.

x

Fig. 4–15

If we pull on cable BC with a force F at any point along the cable, the moment of this force about the base of the utility pole at A will always be the same. This is a consequence of the principle of transmissibility. Note that the moment arm, or perpendicular distance from A to the cable, is rd, and so MA = rdF. In three dimensions this distance is often difficult to determine, and so we can use the vector cross product to obtain the moment in a more direct manner. For example, M A = rAB * F = rAC * F. As required, both of these vectors are directed from point A to a point on the line of action of the force.

B

F rAB

MA A

rd rAC

C

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y

128

CHAPTER 4

F O R C E S Y S T E M R E S U LTA N T S

EXAMPLE 4.4 z

The pole in Fig. 4–16a is subjected to a 60-N force that is directed from C to B. Determine the magnitude of the moment created by this force about the support at A.

B F % 60 N 2m

A

y

1m 3m

2m

C

4m x

SOLUTION (VECTOR ANALYSIS) As shown in Fig. 4–16b, either one of two position vectors can be used for the solution, since M A = rB * F or M A = rC * F. The position vectors are represented as rB = 51i + 3j + 2k6 m and rC = 53i + 4j6 m

(a) z B(1 m, 3 m, 2 m)

rB

The force has a magnitude of 60 N and a direction specified by the unit vector uF, directed from C to B. Thus, F = 160 N2uF = 160 N2c

F A

y uF

rC

C(3 m, 4 m, 0) x

11 - 32i + 13 - 42j + 12 - 02k 21-222 + 1-122 + 1222

d

= 5- 40i - 20j + 40k6 N Substituting into the determinant formulation, Eq. 4–7, and following the scheme for determinant expansion as stated in the footnote on page 124, we have

(b)

MA

i j k 3 = rB * F = 1 3 2 3 -40 -20 40 = [31402 - 21-202]i - [11402 - 21-402]j + [11 -202 - 31-402]k

or MA

i j k 3 = rC * F = 3 4 0 3 -40 -20 40 = [41402 - 01-202]i - [31402 - 01-402]j + [31 -202 - 41-402]k

z d

MA g a

rB

In both cases,

B F

b

y

A rC C

x

(c)

Fig. 4–16

M A = 5160i - 120j + 100k] N # m

The magnitude of M A is therefore

MA = 2116022 + 1-12022 + 110022 = 224 N # m

Ans.

NOTE: As expected, M A acts perpendicular to the shaded plane

containing vectors F, rB , and rC , Fig. 4–16c. (How would you find its coordinate direction angles a = 44.3°, b = 122°, g = 63.4°?) Had this problem been worked using a scalar approach, where MA = Fd, notice the difficulty that can arise in obtaining the moment arm d.

Unpublished Work © 2007 by R. C. Hibbeler. To be published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River, New Jersey. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department.

129

4.3 MOMENT OF A FORCE—VECTOR FORMULATION

EXAMPLE 4.5 Three forces act on the rod shown in Fig. 4–17a. Determine the resultant moment they create about the flange at O and determine the coordinate direction angles of the moment axis.

z F1 % {#60i " 40j " 20k} lb

x 5 ft

F3 % {80i " 40j # 30k} lb (a)

M RO = ©1r * F2 = rA * F1 + rA * F2 + rB * F3 i j k i j k i j k 3 3 3 3 3 = 0 5 0 + 0 5 0 + 4 5 -2 3 -60 40 20 0 50 0 80 40 -30 = [51202 - 40102]i - [0j] + [01402 - 1-602152]k + [0i - 0j + 0k] + [51 -302 - 14021 -22]i - [41 -302 - 801-22]j + [41402 - 80152]k = 530i - 40j + 60k6 lb # ft Ans. The moment axis is directed along the line of action of M RO. Since the magnitude of this moment is MRO = 213022 + 1-4022 + 16022 = 78.10 lb # ft

Therefore, the coordinate direction angles of the moment axis are a = 67.4° b = 121° g = 39.8°

F1 rA

A

F2

y

rB B

30i - 40j + 60k = = 0.3841i - 0.5121j + 0.7682k 78.10

cos a = 0.3841; cos b = - 0.5121; cos g = 0.7682;

z

O

x

the unit vector which defines the direction of the moment axis is MRO

4 ft

B

The resultant moment about O is therefore

u =

y

2 ft

rA = 55j6 ft rB = 54i + 5j - 2k6 ft

M RO

F2 % {50j} lb

A

O

SOLUTION Position vectors are directed from point O to each force as shown in Fig. 4–17b. These vectors are

Ans. Ans. Ans.

NOTE: These results are shown in Fig. 4–17c. Realize that the three

forces tend to cause the rod to rotate about this axis in the manner shown by the curl indicated on the moment vector.

F3 (b) z MRO % {30i # 40j " 60k} lb · ft

g %39.8&

a%67.4&

b %121& O

x (c)

Fig. 4–17

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y

130

CHAPTER 4

F1

F

F2

r

O

Fy

F

B

Fx h

F

C Fx

d

b

4.4 Principle of Moments A concept often used in mechanics is the principle of moments, which is sometimes referred to as Varignon’s theorem since it was originally developed by the French mathematician Varignon (1654–1722). It states that the moment of a force about a point is equal to the sum of the moments of the force’s components about the point. The proof follows directly from the distributive law of the vector cross product. To show this, consider the force F and two of its rectangular components, where F = F1 + F2, Fig. 4–18. We have M O = r * F1 + r * F2 = r * 1F1 + F22 = r * F

Fig. 4–18

Fy

F O R C E S Y S T E M R E S U LTA N T S

MA

This concept has important applications to the solution of problems and proofs of theorems that follow, since it is often easier to determine the moments of a force’s components rather than the moment of the force itself.

The guy cable exerts a force F on the pole and this creates a moment about the base at A of MA = Fd. If the force is replaced by its two components Fx and Fy at point B where the cable acts on the pole, then the sum of the moments of these two components about A will yield the same resultant moment. For the calculation Fy will create zero moment about A and so MA = Fxh. This is an application of the principle of moments. In addition we can apply the principle of transmissibility and slide the force to where its line of action intersects the ground at C. In this case Fx will create zero moment about A, and so MA = Fyb.

A

Important Points • The moment of a force indicates the tendency of a body to turn about an axis passing through a specific point O.

• Using the right-hand rule, the sense of rotation is indicated by the fingers, and the thumb is directed • • •

along the moment axis, or line of action of the moment.

The magnitude of the moment is determined from MO = Fd, where d is the perpendicular or shortest distance from point O to the line of action of the force F. In three dimensions use the vector cross product to determine the moment, i.e., M O = r * F. Remember that r is directed from point O to any point on the line of action of F. The principle of moments states that the moment of a force about a point is equal to the sum of the moments of the force’s components about the point. This is a very convenient method to use in two dimensions.

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131

4.4 PRINCIPLE OF MOMENTS

EXAMPLE 4.6 A 200-N force acts on the bracket shown in Fig. 4–19a. Determine the moment of the force about point A. F % 200 N 45& B

100 mm A

100 mm 100 mm y

(a)

SOLUTION I Fig. 4–19a The moment arm d can be found by trigonometry, using the construction shown in Fig. 4–19b. From the right triangle BCD,

F % 200 N C

D

CB = d = 100 cos 45° = 70.71 mm = 0.070 71 m

45& B

Thus,

MA = Fd = 200 N10.070 71 m2 = 14.1 N # m g According to the right-hand rule, M A is directed in the +k direction since the force tends to rotate or orbit counterclockwise about point A. Hence, reporting the moment as a Cartesian vector, we have M A = 514.1k6 N # m

A

100 mm

(b)

Ans.

d+MA = 1200 sin 45° N210.20 m2 - 1200 cos 45° N210.10 m2 = 14.1 N # m g M A = 514.1k6 N # m

Ans.

NOTE: By comparison, it is seen that Solution II provides a more

convenient method for analysis than Solution I since the moment arm for each component force is easier to establish.

x

d

SOLUTION II The 200-N force may be resolved into x and y components, as shown in Fig. 4–19c. In accordance with the principle of moments, the moment of F computed about point A is equivalent to the sum of the moments produced by the two force components. Assuming counterclockwise rotation as positive, i.e., in the +k direction, we can apply Eq. 4–2 1MA = ©Fd2, in which case

Thus

45&

y 200 sin 45& N B 0.1 m

200 cos 45& N A

x 0.2 m (c)

Fig. 4–19c

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F O R C E S Y S T E M R E S U LTA N T S

EXAMPLE 4.7 The force F acts at the end of the angle bracket shown in Fig. 4–20a.

Determine the moment of the force about point O. SOLUTION I (SCALAR ANALYSIS) The force is resolved into its x and y components as shown in Fig. 4–20b, and the moments of the components are computed about point O. Taking positive moments as counterclockwise, i.e., in the +k direction, we have

O 0.2 m

0.4 m 30&

d+MO = 400 sin 30° N10.2 m2 - 400 cos 30° N10.4 m2 = -98.6 N # m = 98.6 N # m b

F = 400 N

(a)

or M O = 5-98.6k6 N # m

y

O

x 0.2 m 400 sin 30& N 0.4 m 400 cos 30& N (b)

Ans.

SOLUTION II (VECTOR ANALYSIS) Using a Cartesian vector approach, the force and position vectors shown in Fig. 4–20c can be represented as r = 50.4i - 0.2j6 m F = 5400 sin 30°i - 400 cos 30°j6 N = 5200.0i - 346.4j6 N The moment is therefore

y

O

x 0.2 m

0.4 m

Fig. 4–20

Ans.

NOTE: By comparison, it is seen that the scalar analysis (Solution I)

30& (c)

i j k M O = r * F = 3 0.4 -0.2 03 200.0 -346.4 0 = 0i - 0j + [0.41 -346.42 - 1-0.221200.02]k = 5-98.6k6 N # m

F

provides a more convenient method for analysis than Solution II since the direction of the moment and the moment arm for each component force are easy to establish. Hence, this method is generally recommended for solving problems displayed in two dimensions. On the other hand, Cartesian vector analysis is generally recommended only for solving three-dimensional problems, where the moment arms and force components are often more difficult to determine.

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133

PROBLEMS

PROBLEMS 4–1. If A, B, and D are given vectors, prove the distributive law for the vector cross product, i.e., A * 1B + D2= 1A * B2 + 1A * D2.

4–2. Prove the triple A # 1B * C2 = 1A * B2 # C.

scalar

product

"4–7.

If the force F = 100 N, determine the angle u 10 … u … 90°2 so that the force develops a clockwise moment about point O of 20 N # m. y

identity

F

4–3. Given the three nonzero vectors A, B, and C, show that if A # 1B * C2 = 0, the three vectors must lie in the same plane.

u 50 mm

*4–4. Determine the magnitude and directional sense of the resultant moment of the forces at A and B about point O. 4–5. Determine the magnitude and directional sense of the resultant moment of the forces at A and B about point P. 300 mm 60& x

O

y

Probs. 4–6/7 *4–8. Determine the magnitude and directional sense of the resultant moment of the forces about point O.

40 lb 6 in.

B

P

4–9. Determine the magnitude and directional sense of the resultant moment of the forces about point P.

3 in.

O 3 in.

45& x

13 in.

6 in.

y

5 in.

4m

260 N

A

3m

13

12 5

30& 60 lb

Probs. 4–4/5

P

O

2m

B

x

5m 400 N A

30&

2m

4–6. Determine the magnitude of the force F that should be applied at the end of the lever such that this force creates a clockwise moment of 15 N # m about point O when u = 30°.

Probs. 4–8/9

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4–10. A force of 40 N is applied to the wrench. Determine the moment of this force about point O. Solve the problem using both a scalar analysis and a vector analysis. y

4–13. To correct a birth defect, the tibia of the leg is straightened using three wires that are attached through holes made in the bone and then to an external brace that is worn by the patient. Determine the moment of each wire force about joint B.

30 mm

A

20 mm 0.15 m

200 mm

O

0.25 m

35 mm

40 N

A

20&

x

Prob. 4–10

15& F1 % 4 N F2 % 8 N

0.35 m

F3 % 6 N

15 mm

30&

0.2 m B

4–11. Determine the magnitude and directional sense of the resultant moment of the forces about point O. Probs. 4–12/13 y A 10 ft 30& O

4 5

3

250 lb

4–15. If FB = 30 lb and FC = 45 lb, determine the resultant moment about the bolt located at A.

x

4 ft

6 ft B 3 ft

4–14. Determine the moment of each force about the bolt located at A. Take FB = 40 lb, FC = 50 lb.

4 ft

P

300 lb 30&

Prob. 4–11 *4–12. To correct a birth defect, the tibia of the leg is straightened using three wires that are attached through holes made in the bone and then to an external brace that is worn by the patient. Determine the moment of each wire force about joint A.

Probs. 4–14/15

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135

PROBLEMS *4–16. The elbow joint is flexed using the biceps brachii muscle, which remains essentially vertical as the arm moves in the vertical plane. If this muscle is located a distance of 16 mm from the pivot point A on the humerus, determine the variation of the moment capacity about A if the constant force developed by the muscle is 2.30 kN. Plot these results of M vs. u for -60° … u … 80°.

4–18. Determine the direction u 10° … u … 180°2 of the force F = 40 lb so that it produces (a) the maximum moment about point A and (b) the minimum moment about point A. Compute the moment in each case.

40 lb 2.30 kN

Prob. 4–18 16 mm

4–19. The rod on the power control mechanism for a business jet is subjected to a force of 80 N. Determine the moment of this force about the bearing at A.

u A

20&

60&

80 N

Prob. 4–16

150 mm

A

4–17. The Snorkel Co. produces the articulating boom platform that can support a weight of 550 lb. If the boom is in the position shown, determine the moment of this force about points A, B, and C.

A

Prob. 4–19 16

3 ft

ft

30&

B

*4–20. The boom has a length of 30 ft, a weight of 800 lb, and mass center at G. If the maximum moment that can be developed by the motor at A is M = 2011032 lb # ft, determine the maximum load W, having a mass center at G¿, that can be lifted. Take u = 30°.

800 lb

15 ft C

70&

14 ft

G¿

16 ft

A

M

G

u

W 2 ft

Prob. 4–17

Prob. 4–20

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F O R C E S Y S T E M R E S U LTA N T S

4–21. The tool at A is used to hold a power lawnmower blade stationary while the nut is being loosened with the wrench. If a force of 50 N is applied to the wrench at B in the direction shown, determine the moment it creates about the nut at C. What is the magnitude of force F at A so that it creates the opposite moment about C?

A

4–23. The Y-type structure is used to support the high voltage transmission cables. If the supporting cables each exert a force of 275 lb on the structure at B, determine the moment of each force about point A. Also, by the principle of transmissibility, locate the forces at points C and D and determine the moments.

B

13

F

12

5

F1 % 275 lb 400 mm

85 ft

C

F2 % 275 lb 30&

30&

50 N C B

A

60&

300 mm

D

Prob. 4–23

Prob. 4–21

*4–24. The 70-N force acts on the end of the pipe at B. Determine (a) the moment of this force about point A, and (b) the magnitude and direction of a horizontal force, applied at C, which produces the same moment. Take u = 60°.

4–22. Determine the clockwise direction u 10° … u … 180°2 of the force F = 80 lb so that it produces (a) the maximum moment about point A and (b) no moment about point A. Compute the moment in each case.

4–25. The 70-N force acts on the end of the pipe at B. Determine the angles u 10° … u … 180°2 of the force that will produce maximum and minimum moments about point A. What are the magnitudes of these moments?

A

4 ft 0.9 m A

1 ft

70 N u F % 80 lb

Prob. 4–22

u C

0.3 m

0.7 m

B

Probs. 4–24/25

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137

PROBLEMS 4–26. The towline exerts a force of P = 4 kN at the end of the 20-m-long crane boom. If u = 30°, determine the placement x of the hook at A so that this force creates a maximum moment about point O. What is this moment?

4–27. The towline exerts a force of P = 4 kN at the end of the 20-m-long crane boom. If x = 25 m, determine the position u of the boom so that this force creates a maximum moment about point O. What is this moment?

4–30. The flat-belt tensioner is manufactured by the Daton Co. and is used with V-belt drives on poultry and livestock fans. If the tension in the belt is 52 lb, when this pulley is not turning, determine the moment of each of these forces about the pin at A. F1 % 52 lb

B

30& P % 4 kN

20 m

6 in.

8 in.

u

O

1.5 m

A

A

5 in.

x

Prob. 4–30

Probs. 4–26/27 *4–28. Determine the resultant moment of the forces about point A. Solve the problem first by considering each force as a whole, and then by using the principle of moments. Take F1 = 250 N, F2 = 300 N, F3 = 500 N. 4–29. If the resultant moment about point A is 4800 N # m clockwise, determine the magnitude of F3 if F1 = 300 N and F2 = 400 N. F1

30&

4–31. The worker is using the bar to pull two pipes together in order to complete the connection. If he applies a horizontal force of 80 lb to the handle of the lever, determine the moment of this force about the end A. What would be the tension T in the cable needed to cause the opposite moment about point A?

F2 60&

A 2m

F2 % 52 lb

20&

80 lb

3m 40& 4m 0.5 ft A

4.5 ft 20& T

5

4 3

Probs. 4–28/29

F3

Prob. 4–31

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*4–32. If it takes a force of F = 125 lb to pull the nail out, determine the smallest vertical force P that must be applied to the handle of the crowbar. Hint: This requires the moment of F about point A to be equal to the moment of P about A. Why?

4–34. Determine the moment of the force F at A about point O. Express the result as a Cartesian vector.

4–35. Determine the moment of the force F at A about point P. Express the result as a Cartesian vector.

z

A 60&

F 4m

O 20& P

F % {60i # 30j # 20k} N 3 in.

3m

A

14 in.

1.5 in.

y

O

7m

4m

6m

2m

x

P

Prob. 4–32

Probs. 4–34/35

"4–33. The pipe wrench is activated by pulling on the cable segment with a horizontal force of 500 N. Determine the moment MA produced by the wrench on the pipe when u = 20°. Neglect the size of the pulley.

*4–36. Determine the moment of the force F at A about point O. Express the result as a Cartesian vector.

4–37. Determine the moment of the force F at A about point P. Express the result as a Cartesian vector. z

B

0.2 m A

500 N

F % 13 kN 8m

6m 0.5 m

2.5 m

A

u

4m O

3m

y

3m

MA

6m 0.4 m

Prob. 4–33

8m x

P

Probs. 4–36/37

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139

PROBLEMS 4–38. The curved rod lies in the x–y plane and has a radius of 3 m. If a force of F = 80 N acts at its end as shown, determine the moment of this force about point O.

4–41. The pole supports a 22-lb traffic light. Using Cartesian vectors, determine the moment of the weight of the traffic light about the base of the pole at A. z

4–39. The curved rod lies in the x–y plane and has a radius of 3 m. If a force of F = 80 N acts at its end as shown, determine the moment of this force about point B.

12 ft

z

O y B 3m 45&

3m 18 ft

A

1m

A

F % 80 N

x

2m

30&

C

y

x

Probs. 4–38/39

Prob. 4–41

*4–40. The force F = 5600i + 300j - 600k6 N acts at the end of the beam. Determine the moment of the force about point A.

4–42. The man pulls on the rope with a force of F = 20 N. Determine the moment that this force exerts about the base of the pole at O. Solve the problem two ways, i.e., by using a position vector from O to A, then O to B. z

A

z F A

10.5 m x

O

F

1.2 m

B

B 0.4 m

1.5 m O

0.2 m

Prob. 4–40

y

3m x

4m y

Prob. 4–42

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4–43. Determine the smallest force F that must be applied along the rope in order to cause the curved rod, which has a radius of 5 ft, to fail at the support C.This requires a moment of M = 80 lb # ft to be developed at C.

4–45. The pipe assembly is subjected to the 80-N force.

Determine the moment of this force about point B. z

A

z

400 mm C 5 ft

B x

60&

300 mm

A

5 ft

200 mm

y F

200 mm

6 ft

C

250 mm

B

40&

7 ft

x

y

30& F % 80 N

Prob. 4–43

Prob. 4–45 *4–44. The pipe assembly is subjected to the 80-N force.

Determine the moment of this force about point A. z

4–46. The x-ray machine is used for medical diagnosis. If the camera and housing at C have a mass of 150 kg and a mass center at G, determine the moment of its weight about point O when it is in the position shown.

A

z 400 mm B

x

300 mm

C

y

G

200 mm 200 mm C

O

250 mm 40&

60& 1.5 m

x

1.2 m

y

30& F % 80 N

Prob. 4–44

Prob. 4–46

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141

PROBLEMS 4–47. Using Cartesian vector analysis, determine the resultant moment of the three forces about the base of the column at A. Take F1 = 5400i + 300j + 120k6 N. z

4–49. The force F = 56i + 8j + 10k6 N creates a moment about point O of M O = 5-14i + 8j + 2k6 N # m. If the force passes through a point having an x coordinate of 1 m, determine the y and z coordinates of the point.Also, realizing that MO = Fd, determine the perpendicular distance d from point O to the line of action of F.

F1

B F2 % {100i # 100j # 60k} N 4m

z

F

P MO

F3 % { # 500k } N

z

d

8m 1m

E

y

O y

y

A

1m

x

Prob. 4–49 x

Prob. 4–47 *4–48. A force of F = 56i - 2j + 1k6 kN produces a moment of M O = 54i + 5j - 14k6 kN # m about the origin of coordinates, point O. If the force acts at a point having an x coordinate of x = 1 m, determine the y and z coordinates.

z

4–50. A force of F = 56i - 2j + 1k6 kN produces a moment of M O = 54i + 5j - 14k6 kN # m about the origin of coordinates, point O. If the force acts at a point having an x coordinate of x = 1 m, determine the y and z coordinates.

z

F

P

P

F

MO z

d O

y

y

O

1m

MO

y x

z

d

1m

y x

Prob. 4–48

Prob. 4–50

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4.5 Moment of a Force about a Specified Axis

Recall that when the moment of a force is computed about a point, the moment and its axis are always perpendicular to the plane containing the force and the moment arm. In some problems it is important to find the component of this moment along a specified axis that passes through the point. To solve this problem either a scalar or vector analysis can be used. z b

5 3

O

x

MO

4

y

My 0.3 m

0.5 m A

0.4 m

F % 20 N

Fig. 4–21

(a)

Scalar Analysis. As a numerical example of this problem, consider

the pipe assembly shown in Fig. 4–21a, which lies in the horizontal plane and is subjected to the vertical force of F = 20 N applied at point A. The moment of this force about point O has a magnitude of MO = 120 N210.5 m2 = 10 N # m, and a direction defined by the right-hand rule, as shown in Fig. 4–21a. This moment tends to turn the pipe about the Ob axis. For practical reasons, however, it may be necessary to determine the component of M O about the y axis, M y , since this component tends to unscrew the pipe from the flange at O. From Fig. 4–21a, M y has a magnitude of My = 53110 N # m2 = 6 N # m and a sense of direction shown by the vector resolution. Rather than performing this two-step process of first finding the moment of the force about point O and then resolving the moment along the y axis, it is also possible to solve this problem directly.To do so, it is necessary to determine the perpendicular or moment-arm distance from the line of action of F to the y axis. From Fig. 4–21a this distance is 0.3 m. Thus the magnitude of the moment of the force about the y axis is again My = 0.3120 N2 = 6 N # m, and the direction is determined by the righthand rule as shown.

In general, then, if the line of action of a force F is perpendicular to any specified axis aa, the magnitude of the moment of F about the axis can be determined from the equation Ma = Fda

(4–10)

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143

4.5 MOMENT OF A FORCE ABOUT A SPECIFIED AXIS

Here da is the perpendicular or shortest distance from the force line of action to the axis. The direction is determined from the thumb of the right hand when the fingers are curled in accordance with the direction of rotation as produced by the force. In particular, realize that a force will not contribute a moment about a specified axis if the force line of action is parallel to the axis or its line of action passes through the axis.

z (Mz)max d

A

F

If a horizontal force F is applied to the handle of the flex-headed wrench, it tends to turn the socket at A about the z axis. This effect is caused by the moment of F about the z axis.The maximum moment is determined when the wrench is in the horizontal plane so that full leverage from the handle can be achieved, i.e., 1Mz2max = Fd. If the handle is not in the horizontal position, then the moment about the z axis is determined from Mz = Fd¿, where d¿ is the perpendicular distance from the force line of action to the axis. We can also determine this moment by first finding the moment of F about A, MA = Fd, then finding the projection or component of this moment along z, i.e., Mz = MA cos u.

z

u

Mz

MA

d¿ A

F

Vector Analysis. The previous two-step solution of first finding the

moment of the force about a point on the axis and then finding the projected component of the moment about the axis can also be performed using a vector analysis,Fig. 4–21b.Here the moment about point O is first determined from M O = rA * F = 10.3i + 0.4j2 * 1-20k2 = 5-8i + 6j6 N # m. The component or projection of this moment along the y axis is then determined from the dot product (Sec. 2.9). Since the unit vector for this axis (or line) is ua = j, then My = M O # ua = 1-8i + 6j2 # j = 6 N # m. This result, of course, is to be expected, since it represents the j component of M O.

z

MO ua % j y My

O

0.3 m

rA

x

A

0.4 m (b)

F % {#20k} N

Fig. 4–21 (cont.)

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a

b Ma O Moment axis b¿

MO % r $ F

u

r

ua

A

Axis of projection a¿

Fig. 4–22

F

A vector analysis such as this is particularly advantageous for finding the moment of a force about an axis when the force components or the appropriate moment arms are difficult to determine. For this reason, the above two-step process will now be generalized and applied to a body of arbitrary shape. To do so, consider the body in Fig. 4–22, which is subjected to the force F acting at point A. Here we wish to determine the effect of F in tending to rotate the body about the aa¿ axis. This tendency for rotation is measured by the moment component M a. To determine M a we first compute the moment of F about any arbitrary point O that lies on the aa¿ axis. In this case, M O is expressed by the cross product M O = r * F, where r is directed from O to A. Here M O acts along the moment axis bb¿, and so the component or projection of M O onto the aa¿ axis is then M a. The magnitude of M a is determined by the dot product, Ma = MO cos u = M O # ua where ua is a unit vector that defines the direction of the aa¿ axis. Combining these two steps as a general expression, we have Ma = 1r * F2 # ua. Since the dot product is commutative, we can also write Ma = ua # 1r * F2 In vector algebra, this combination of dot and cross product yielding the scalar Ma is called the triple scalar product. Provided x, y, z axes are established and the Cartesian components of each of the vectors can be determined, then the triple scalar product may be written in determinant form as i Ma = 1uaxi + uayj + uazk2 # 3 rx Fx

j ry Fy

k rz 3 Fz

or simply

Ma = ua # 1r * F2 =

where

uax 3 rx Fx

uay ry Fy

uaz rz 3 Fz

(4–11)

uax, uay, uaz represent the x, y, z components of the unit vector defining the direction of the aa¿ axis rx, ry, rz

Fx, Fy, Fz

represent the x, y, z components of the position vector drawn from any point O on the aa¿ axis to any point A on the line of action of the force represent the x, y, z components of the force vector.

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145

4.5 MOMENT OF A FORCE ABOUT A SPECIFIED AXIS

When Ma is evaluated from Eq. 4–11, it will yield a positive or negative scalar. The sign of this scalar indicates the sense of direction of M a along the aa¿ axis. If it is positive, then M a will have the same sense as ua, whereas if it is negative, then M a will act opposite to ua.

Once Ma is determined, we can then express M a as a Cartesian vector, namely, M a = Maua = [ua # 1r * F2]ua

(4–12)

Finally, if the resultant moment of a series of forces is to be computed about the aa¿ axis, then the moment components of each force are added together algebraically, since each component lies along the same axis.

Thus the magnitude of M a is Ma = ©[ua # 1r * F2] = ua # ©1r * F2 The examples which follow illustrate a numerical application of the above concepts.

Wind blowing on the face of this traffic sign creates a resultant force F that tends to tip the sign over due to the moment M A created about the a–a axis.The moment of F about a point A that lies on the axis is M A = r * F. The projection of this moment along the axis, whose direction is defined by the unit vector ua, is Ma = ua # 1r * F2. Had this moment been calculated using scalar methods, then the perpendicular distance from the force line of action to the a–a axis would have to be determined, which in this case would be a more difficult task.

F

r ua a

Ma

A

a

MA

Important Points • The moment of a force about a specified axis can be determined •

provided the perpendicular distance da from both the force line of action and the axis can be determined. Ma = Fda. If vector analysis is used, Ma = ua # 1r * F2, where ua defines the direction of the axis and r is directed from any point on the axis to any point on the line of action of the force.

• If Ma is calculated as a negative scalar, then the sense of •

direction of M a is opposite to ua.

The moment M a expressed as a Cartesian vector is determined from M a = Maua.

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EXAMPLE 4.8 The force F = 5-40i + 20j + 10k6 N acts at point A shown in Fig. 4–23a. Determine the moments of this force about the x and a axes.

F % {#40i " 20j " 10k} N

z

A

rA

a

ua 3m

O x

SOLUTION I (VECTOR ANALYSIS) We can solve this problem by using the position vector rA. Why? Since rA = 5-3i + 4j + 6k6 m and ux = i, then applying Eq. 4–11,

6m

Mx = i # 1rA * F2 =

y

4m (a)

F A

0 6 3 10

a O

y

120 N ' m (b)

z "

The negative sign indicates that the sense of M x is opposite to i.

We can compute Ma also using rA because rA extends from a point on the a axis to the force. Also, ua = - 35 i + 54 j. Thus, Ma = ua # 1rA

80 N ' m

20 N

3m " (c)

Fig. 4–23

4 20

0 6 3 10

= - 35[41102 - 61202]- 45[1-321102 -61-402]+0[1-321202 -41-402] = -120 N # m Ans. What does the negative sign indicate?

The moment components are shown in Fig. 4–23b.

Ans.

Although not required here, note also that

a

4m

4 5

Mx = 110 N214 m2 - 120 N216 m2 = -80 N # m

6m

O

- 53 * F2 = 3 -3 -40

SOLUTION II (SCALAR ANALYSIS) Since the force components and moment arms are easy to determine for computing Mx, a scalar analysis can be used to solve this problem.

Referring to Fig. 4–23c, only the 10-N and 20-N forces contribute moments about the x axis. (The line of action of the 40-N force is parallel to this axis and hence its moment about the x axis is zero.)

Using the right-hand rule, the algebraic sum of the moment components about the x axis is therefore

10 N 40 N A

" x

0 4 20

= 1[41102-61202]-0[1-321102 -61-402]+0[1-321202 -41-402] = -80 N # m Ans.

z

x

1 3 -3 -40

y

My = 110 N213 m2 - 140 N216 m2 = -210 N # m Mz = 140 N214 m2 - 120 N213 m2 = 100 N # m

If we were to determine Ma by this scalar method, it would require much more effort since the force components of 40 N and 20 N are not perpendicular to the direction of the a axis. The vector analysis yields a more direct solution.

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147

4.5 MOMENT OF A FORCE ABOUT A SPECIFIED AXIS

EXAMPLE 4.9 z

The rod shown in Fig. 4–24a is supported by two brackets at A and B. Determine the moment M AB produced by F = 5-600i + 200j - 300k6 N, which tends to rotate the rod about the AB axis.

0.6 m

SOLUTION A vector analysis using MAB = uB # 1r * F2 will be considered for the solution since the moment arm or perpendicular distance from the line of action of F to the AB axis is difficult to determine. Each of the terms in the equation will now be identified.

Unit vector uB defines the direction of the AB axis of the rod, Fig. 4–24b, where uB =

0.4i + 0.2j rB = = 0.894i + 0.447j rB 210.422 + 10.222

0.3 m A

C

F

D

y

0.4 m B

0.2 m

(a)

x

Vector r is directed from any point on the AB axis to any point on the line of action of the force. For example, position vectors rC and rD are suitable, Fig. 4–24b. (Although not shown, rBC or rBD can also be used.)

For simplicity, we choose rD, where

z

rD = 50.2j6 m The force is F = 5-600i + 200j - 300k6 N

rC

Substituting these vectors into the determinant form and expanding, we have MAB = uB # 1rD

0.894 * F2 = 3 0 -600

0.447 0.2 200

0 0 3 -300

F C (0.6 m, 0, 0.3 m)

A rD D (0, 0.2 m, 0)y MAB uB B (0.4 m, 0.2 m, 0)

x

= 0.894[0.21-3002 - 012002] - 0.447[01-3002 - 01-6002] + 0[012002 - 0.21-6002] = - 53.67 N # m

(b)

Fig. 4–24

The negative sign indicates that the sense of M AB is opposite to that of uB. Expressing M AB as a Cartesian vector yields M AB = MABuB = 1-53.67 N # m210.894i + 0.447j2 = 5-48.0i - 24.0j6 N # m

Ans.

The result is shown in Fig. 4–24b.

NOTE: If axis AB is defined using a unit vector directed from B toward A, then in the above formulation -uB would have to be used. This would lead to MAB = + 53.67 N # m. Consequently, M AB = MAB1-uB2, and the above result would again be determined.

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PROBLEMS 4–51. Determine the moment of the force F about the Oa axis. Express the result as a Cartesian vector.

4–53. Determine the resultant moment of the two forces about the Oa axis. Express the result as a Cartesian vector.

z

z

F % {50i # 20j " 20k} N

F1 = 80 lb

45& 4m

a

B F2 = 50 lb

3m

6m

O

120& 6 ft

y

1m

60&

O

2m

5 ft 30&

x

4 ft

y 30&

a

Prob. 4–51

x

Prob. 4–53 *4–52. Determine the moment of the force F about the aa axis. Express the result as a Cartesian vector.

4–54. A force of F = 58i - 1j + 1k6 lb is applied to the handle of the box wrench. Determine the component of the moment of this force about the z axis which is effective in loosening the bolt.

z

a z

6 ft

3 ft

4 ft

a 2 ft

x

2 ft

4 ft

F % 600 lb

y F

2 in.

3 in.

y

8 in. x

Prob. 4–52

Prob. 4–54

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149

PROBLEMS 4–55. The 50-lb force acts on the gear in the direction shown. Determine the moment of this force about the y axis.

4–57. The cutting tool on the lathe exerts a force F on the shaft in the direction shown. Determine the moment of this force about the y axis of the shaft.

z

z

60&

F % 50 lb

120& 45&

F % {6i # 4j # 7k} kN

30 mm

40& 3 in.

x

y

y x

Prob. 4–55

Prob. 4–57

*4–56. The Rollerball skate is an in-line tandem skate that uses two large spherical wheels on each skate, rather than traditional wafer-shape wheels. During skating the two forces acting on the wheel of one skate consist of a 78-lb normal force and a 13-lb friction force. Determine the moment of both of these forces about the axle AB of the wheel.

4–58. The hood of the automobile is supported by the strut AB, which exerts a force of F = 24 lb on the hood. Determine the moment of this force about the hinged axis y.

z B

y F

4 ft

30& A 1.25 in.

B

78 lb

x

x

A 2 ft

2 ft

4 ft

y

13 lb

Prob. 4–56

Prob. 4–58

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4–59. The lug nut on the wheel of the automobile is to be removed using the wrench and applying the vertical force of F = 30 N at A. Determine if this force is adequate, provided 14 N # m of torque about the x axis is initially required to turn the nut. If the 30-N force can be applied at A in any other direction, will it be possible to turn the nut?

4–62. The wooden shaft is held in a lathe. The cutting tool exerts a force F on the shaft in the direction shown.

Determine the moment of this force about the x axis of the shaft. Express the result as a Cartesian vector. The distance OA is 25 mm.

*4–60. Solve Prob. 4–59 if the cheater pipe AB is slipped over the handle of the wrench and the 30-N force can be applied at any point and in any direction on the assembly.

z

z F % 30 N B O 0.25 m

A

0.3 m

A

30&

x F % {#5i # 3j " 8k} N

0.5 m

y

Prob. 4–62

0.1 m x

y

4–63. Determine the magnitude of the moment of the force F = 550i - 20j - 80k6 N about the base line CA of the tripod.

Probs. 4–59/60

4–61. The bevel gear is subjected to the force F which is caused from contact with another gear. Determine the moment of this force about the y axis of the gear shaft.

z F D

z

4m F % {20i " 8j # 15k} N

y 2m

40 mm 1.5 m

30 mm

C

A y

x

x

Prob. 4–61

2.5 m

2m 0.5 m

B

1m

Prob. 4–63

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151

PROBLEMS *4–64. The flex-headed ratchet wrench is subjected to a force of P = 16 lb, applied perpendicular to the handle as shown. Determine the moment or torque this imparts along the vertical axis of the bolt at A.

4–67. Determine the moment of each force acting on the handle of the wrench about the a axis. Take F1 = 5-2i + 4j - 8k6 lb, F2 = 53i + 2j - 6k6 lb.

4–65. If a torque or moment of 80 lb # in. is required to loosen the bolt at A, determine the force P that must be applied perpendicular to the handle of the flex-headed ratchet wrench.

z a

F2

C

P

60&

10 in.

A

6 in. y

F1 x

4 in.

45&

A

3.5 in.

B B

0.75 in.

Prob. 4–67

Probs. 4–64/65 4–66. The A-frame is being hoisted into an upright position by the vertical force of F = 80 lb. Determine the moment of this force about the y axis when the frame is in the position shown.

*4–68. Determine the moment of each force acting on the handle of the wrench about the z axis. Take F1 = 5-2i + 4j - 8k6 lb, F2 = 53i + 2j - 6k6 lb.

z

z a

F

F2

C

C A

6 ft

x¿

15&

A

6 ft

30& B

x

x y¿

Prob. 4–66

4 in.

6 in. y

F1

y

45&

B

3.5 in.

B

Prob. 4–68

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4.6 Moment of a Couple F d "F

Fig. 4–25

A couple is defined as two parallel forces that have the same magnitude, have opposite directions, and are separated by a perpendicular distance d, Fig. 4–25. Since the resultant force is zero, the only effect of a couple is to produce a rotation or tendency of rotation in a specified direction.

The moment produced by a couple is called a couple moment. We can determine its value by finding the sum of the moments of both couple forces about any arbitrary point. For example, in Fig. 4–26, position vectors rA and rB are directed from point O to points A and B lying on the line of action of -F and F. The couple moment computed about O is therefore M = rA * 1-F2 + rB * F

B

r

A

F

"F

Rather than sum the moments of both forces to determine the couple moment, it is simpler to take moments about a point lying on the line of action of one of the forces. If point A is chosen, then the moment of -F about A is zero, and we have M = r * F

rB

rA

O

Fig. 4–26

M

The fact that we obtain the same result in both cases can be demonstrated by noting that in the first case we can write M = 1rB - rA2 * F; and by the triangle rule of vector addition, rA + r = rB or r = rB - rA, so that upon substitution we obtain Eq. 4–13. This result indicates that a couple moment is a free vector, i.e., it can act at any point since M depends only upon the position vector r directed between the forces and not the position vectors rA and rB, directed from the arbitrary point O to the forces. This concept is therefore unlike the moment of a force, which requires a definite point (or axis) about which moments are determined.

Scalar Formulation. The moment of a couple, M, Fig. 4–27, is defined as having a magnitude of M = Fd

"F

d F

Fig. 4–27

(4–13)

(4–14)

where F is the magnitude of one of the forces and d is the perpendicular distance or moment arm between the forces. The direction and sense of the couple moment are determined by the right-hand rule, where the thumb indicates the direction when the fingers are curled with the sense of rotation caused by the two forces. In all cases, M acts perpendicular to the plane containing these forces.

Vector Formulation. The moment of a couple can also be expressed by the vector cross product using Eq. 4–13, i.e., M = r * F

(4–15)

Application of this equation is easily remembered if one thinks of taking the moments of both forces about a point lying on the line of action of one of the forces. For example, if moments are taken about point A in Unpublished Work © 2007 by R. C. Hibbeler. To be published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River, New Jersey. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department.

4.6 MOMENT OF A COUPLE

153

Fig. 4–26, the moment of -F is zero about this point, and the moment or F is defined from Eq. 4–15. Therefore, in the formulation r is crossed with the force F to which it is directed.

Equivalent Couples. Two couples are said to be equivalent if they

produce the same moment. Since the moment produced by a couple is always perpendicular to the plane containing the couple forces, it is therefore necessary that the forces of equal couples lie either in the same plane or in planes that are parallel to one another. In this way, the direction of each couple moment will be the same, that is, perpendicular to the parallel planes.

Resultant Couple Moment. Since couple moments are free vectors, they may be applied at any point P on a body and added vectorially. For example, the two couples acting on different planes of the body in Fig. 4–28a may be replaced by their corresponding couple moments M 1 and M 2, Fig. 4–28b, and then these free vectors may be moved to the arbitrary point P and added to obtain the resultant couple moment M R = M 1 + M 2, shown in Fig. 4–28c.

$

(a)

M1

(b)

$

M2

MR

M1

P

M2

(c)

Fig. 4–28 Unpublished Work © 2007 by R. C. Hibbeler. To be published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River, New Jersey. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department.

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If more than two couple moments act on the body, we may generalize this concept and write the vector resultant as (4–16)

M R = ©1r * F2

These concepts are illustrated numerically in the examples which follow. In general, problems projected in two dimensions should be solved using a scalar analysis since the moment arms and force components are easy to compute.

A moment of 12 N # m is needed to turn the shaft connected to the center of the wheel. To do this it is efficient to apply a couple since this effect produces a pure rotation. The couple forces can be made as small as possible by placing the hands on the rim of the wheel, where the spacing is 0.4 m. In this case 12 N # m = F10.4 m2, F = 30 N. An equivalent couple moment of 12 N # m can be produced if one grips the wheel within the inner hub, although here much larger forces are needed. If the distance between the hands becomes 0.3 m, then 12 N # m = F¿10.32, F¿ = 40 N. Also, realize that if the wheel was connected to the shaft at a point other than at its center, the wheel would still turn when the forces are applied since the 12-N # m couple moment is a free vector.

F F¿ 0.4 m

0.3 m F¿

F

Mc

#F d

F

The frictional forces of the floor on the blades of the concrete finishing machine create a couple moment M c on the machine that tends to turn it. An equal but opposite couple moment must be applied by the hands of the operator to prevent the turning. Here the couple moment, Mc = Fd, is applied on the handle, although it could be applied at any other point on the machine.

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155

4.6 MOMENT OF A COUPLE

Important Points • A couple moment is produced by two noncollinear forces that are equal but opposite. Its effect is to produce • • • •

pure rotation, or tendency for rotation in a specified direction.

A couple moment is a free vector, and as a result it causes the same effect of rotation on a body regardless of where the couple moment is applied to the body.

The moment of the two couple forces can be computed about any point. For convenience, this point is often chosen on the line of action of one of the forces in order to eliminate the moment of this force about the point.

In three dimensions the couple moment is often determined using the vector formulation, M = r * F, where r is directed from any point on the line of action of one of the forces to any point on the line of action of the other force F. A resultant couple moment is simply the vector sum of all the couple moments of the system.

EXAMPLE 4.10 A couple acts on the gear teeth as shown in Fig. 4–29a. Replace it by an equivalent couple having a pair of forces that act through points A and B. 40 N 24 N ' m

0.3 m B

A

0.3 m

0.1 m 40 N

(b)

(a)

SOLUTION (SCALAR ANALYSIS) Fig. 4–29a The couple has a magnitude of M = Fd = 4010.62 = 24 N # m and a direction that is out of the page since the forces tend to rotate counterclockwise. M is a free vector, and so it can be placed at any point on the gear, Fig. 4–29b. To preserve the counterclockwise rotation of M, vertical forces acting through points A and B must be directed as shown in Fig. 4–29c. The magnitude of each force is M = Fd 24 N # m = F10.2 m2 F = 120 N

Ans.

"F

F B

A 0.2 m (c)

Fig. 4–29

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EXAMPLE 4.11 Determine the moment of the couple acting on the member shown in Fig. 4–30a.

150 lb

5

1 ft

A

D

4

90 lb

3

150 lb 5

2 ft

3 ft

B

4

3

120 lb A

D

1 ft

2 ft

3 ft

(a)

B 120 lb 90 lb

(b)

M % 390 lb ' ft

(c)

Fig. 4–30c

SOLUTION (SCALAR ANALYSIS) Here it is somewhat difficult to determine the perpendicular distance between the forces and compute the couple moment as M = Fd. Instead, we can resolve each force into its horizontal and vertical components, Fx = 451150 lb2 = 120 lb and Fy = 531150 lb2 = 90 lb, Fig. 4–30b, and then use the principle of moments. The couple moment can be determined about any point. For example, if point D is chosen, we have for all four forces, d+M = 120 lb 10 ft2 - 90 lb 12 ft2 + 90 lb 15 ft2 + 120 lb 11 ft2 = 390 lb # ft g Ans. It is easier, however, to determine the moments about point A or B in order to eliminate the moment of the forces acting at the moment point. For point A, Fig. 4–30b, we have d+M = 90 lb 13 ft2 + 120 lb 11 ft2 = 390 lb # ft g

Ans.

NOTE: Show that one obtains the same result if moments are summed about point B. Notice also that the couple in Fig. 4–30a can be replaced by two couples in Fig. 4–30b. Using M = Fd, one couple has a moment of M1 = 90 lb 13 ft2 = 270 lb # ft and the other has a moment of M2 = 120 lb 11 ft2 = 120 lb # ft. By the right-hand rule, both couple moments are counterclockwise and are therefore directed out of the page. Since these couples are free vectors, they can be moved to any point and added, which yields M = 270 lb # ft + 120 lb # ft = 390 lb # ft g, the same result determined above. M is a free vector and can therefore act at any point on the member, Fig. 4–30c. Also, realize that the external effects, such as the support reactions on the member, will be the same if the member supports the couple, Fig. 4–30a, or the couple moment, Fig. 4–30c.

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157

4.6 MOMENT OF A COUPLE

EXAMPLE 4.12 Determine the couple moment acting on the pipe shown in Fig. 4–31a.

Segment AB is directed 30° below the x–y plane.

z

z O O

x

8 in.

25 lb A

rB

A y

30&

25 lb

rA

x

25 lb

y 25 lb

6 in.

B

B

(b) z

(a)

SOLUTION I (VECTOR ANALYSIS) The moment of the two couple forces can be found about any point. If point O is considered, Fig. 4–31b, we have M = = = =

O

25 lb

rA * 1-25k2 + rB * 125k2 x 18j2 * 1-25k2 + 16 cos 30°i + 8j - 6 sin 30°k2 * 125k2 -200i - 129.9j + 200i 5-130j6 lb # in.

Ans.

A y 25 lb

It is easier to take moments of the couple forces about a point lying on the line of action of one of the forces, e.g., point A, Fig. 4–31c. In this case the moment of the force A is zero, so that M = rAB * 125k2 = 16 cos 30°i - 6 sin 30°k2 * 125k2 = 5-130j6 lb # in.

B (c)

z

Ans.

x SOLUTION II (SCALAR ANALYSIS) Although this problem is shown in three dimensions, the geometry is simple enough to use the scalar equation M = Fd. The perpendicular distance between the lines of action of the forces is d = 6 cos 30° = 5.20 in., Fig. 4–31d. Hence, taking moments of the forces about either point A or B yields

M = Fd = 25 lb 15.20 in.2 = 129.9 lb # in.

O

25 lb A y 25 lb

6 in.

30& d

B

Applying the right-hand rule, M acts in the -j direction. Thus, M = 5-130j6 lb # in.

rAB

(d)

Ans.

Fig. 4–31

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EXAMPLE 4.13 Replace the two couples acting on the pipe column in Fig. 4–32a by a resultant couple moment. z 125 N

5 5 3

M2 % 37.5 N ' m

4

3

4

D

C

MR

0.3 m 150 N

x A

B 150 N

0.4 m

(a)

125 N 5

3

M1

y

M2 p

4

M1 % 60 N ' m

(b)

(c)

Fig. 4–32

SOLUTION (VECTOR ANALYSIS) The couple moment M 1, developed by the forces at A and B, can easily be determined from a scalar formulation.

M1 = Fd = 150 N10.4 m2 = 60 N # m

By the right-hand rule, M 1 acts in the +i direction, Fig. 4–32b. Hence, M 1 = 560i6 N # m

Vector analysis will be used to determine M 2, caused by forces at C and D. If moments are computed about point D, Fig. 4–32a, M 2 = rDC * FC, then M 2 = rDC * FC = 10.3i2 * C 125 A 45 B j - 125 A 35 B k D = 10.3i2 * [100j - 75k] = 301i * j2 - 22.51i * k2 = 522.5j + 30k6 N # m Since M 1 and M 2 are free vectors, they may be moved to some arbitrary point P and added vectorially, Fig. 4–32c. The resultant couple moment becomes M R = M 1 + M 2 = 560i + 22.5j + 30k6 N # m

Ans.

NOTE: Try to establish M 2 by using a scalar formulation, Fig. 4–32b.

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159

PROBLEMS

PROBLEMS 4–69. Determine the magnitude and sense of the couple moment.

4–71. Determine the magnitude and sense of the couple moment.

y 0.5 m

5 kN

y

A A

5 kN 30&

30&

150 lb

4m

8 ft

6 ft

6 ft x

B

6 ft

1m

8 ft

8 ft

150 lb

x

O

B

2m

Prob. 4–69

Prob. 4–71

4–70. Determine the magnitude and sense of the couple moment. Each force has a magnitude of F = 65 lb..

*4–72. If the couple moment has a magnitude of 300 lb # ft, determine the magnitude F of the couple forces.

y y B

2 ft

12 ft 3 ft

2 ft

"F 1.5 ft

"F

x

o

2 ft 6 ft

4 ft F 6 ft

Prob. 4–70

1 ft

7 ft x

F 12 ft

A

Prob. 4–72

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4–73. A clockwise couple M = 5 N # m is resisted by the shaft of the electric motor. Determine the magnitude of the reactive forces – R and R which act at supports A and B so that the resultant of the two couples is zero.

4–75. Three couple moments act on the pipe assembly.

Determine the magnitude of M 3 and the bend angle u so that the resultant couple moment is zero.

M3 M u 150 mm

M1 % 900 N ' m

45& 60&

A

60&

B M2 % 500 N ' m

"R

R

Prob. 4–75

Prob. 4–73

4–74. The resultant couple moment created by the two couples acting on the disk is M R = 510k6 kip # in. Determine the magnitude of force T.

*4–76. The floor causes a couple moment of MA = 40 N # m and MB = 30 N # m on the brushes of the polishing machine.

Determine the magnitude of the couple forces that must be developed by the operator on the handles so that the resultant couple moment on the polisher is zero. What is the magnitude of these forces if the brush at B suddenly stops so that MB = 0?

z F "T A

"T

0.3 m

3 in.

2 in.

4 in.

T

MA

B

T

MB

"F y

x

Prob. 4–74

Prob. 4–76

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161

PROBLEMS 4–77. The ends of the triangular plate are subjected to three couples. Determine the magnitude of the force F so that the resultant couple moment is 400 N # m clockwise.

4–79. Express the moment of the couple acting on the pipe assembly in Cartesian vector form. Solve the problem (a) using Eq. 4–13, and (b) summing the moment of each force about point O. Take F = 525k6 N. *4–80. If the couple moment acting on the pipe has a magnitude of 400 N # m, determine the magnitude F of the vertical force applied to each wrench.

F

600 N

z

O

"F

600 N 40&

300 mm

40&

200 mm

y

F

150 mm

1m

B

250 N

250 N x

Prob. 4–77

"F

400 mm

200 mm A

Probs. 4–79/80

4–78. Two couples act on the beam. Determine the magnitude of F so that the resultant couple moment is 450 lb # ft, counterclockwise. Where on the beam does the resultant couple moment act?

4–81. Determine the resultant couple moment acting on the beam. Solve the problem two ways: (a) sum moments about point O; and (b) sum moments about point A.

8 kN 200 lb

"F 30&

1.5 ft

1.25 ft

30&

45&

F

30&

O

B 0.3 m A

200 lb 2 ft

Prob. 4–78

1.8 m

1.5 m 2 kN

30&

45&

2 kN

8 kN

Prob. 4–81

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4–82. Two couples act on the beam as shown. Determine the magnitude of F so that the resultant couple moment is 300 lb # ft counterclockwise. Where on the beam does the resultant couple act?

4–86. Determine the couple moment. Express the result as a Cartesian vector.

z 200 lb 1.5 ft

"F % {–8i + 4j – 10k} N

F % {8i – 4j + 10k} N

200 lb A

B 3m

5m 2m

3m

y

4m

Prob. 4–82 4–83. Two couples act on the frame. If the resultant couple moment is to be zero, determine the distance d between the 80-lb couple forces.

x

Prob. 4–86

*4–84. Two couples act on the frame. If d = 4 ft, determine the resultant couple moment. Compute the result by resolving each force into x and y components and (a) finding the moment of each couple (Eq. 4–13) and (b) summing the moments of all the force components about point A. 4–85. Two couples act on the frame. If d = 4 ft, determine the resultant couple moment. Compute the result by resolving each force into x and y components and (a) finding the moment of each couple (Eq. 4–13) and (b) summing the moments of all the force components about point B.

4–87. Determine the couple moment. Express the result as a Cartesian vector.

z

y 2 ft B

30&

80 N

3 ft 50 lb

B 6m

1 ft 5

3

d

4

80 lb

30&

50 lb

10 m 4m

10 m

y

5m 5 4

80 lb

3

3 ft

A

Probs. 4–83/84/85

x

4m

x

A 80 N

Prob. 4–87

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163

PROBLEMS *4–88. If the resultant couple of the two couples acting on the fire hydrant is M R = 5-15i + 30j6 N # m, determine the force magnitude P.

4–90. If F = 5100k6 N, determine the couple moment that acts on the assembly. Express the result as a Cartesian vector.

Member BA lies in the x–y plane.

4–91. If the magnitude of the resultant couple moment is 15 N # m, determine the magnitude F of the forces applied to the wrenches.

z

z O Pi {75j} N

B

0.150 m y

0.2 m

x

–Pi x

60&

F 200 mm y

300 mm

{–75j} N

"F 150 mm 200 mm

Prob. 4–88

A

Probs. 4–90/91 4–89. If the resultant couple of the three couples acting on the triangular block is to be zero, determine the magnitude of forces F and P.

*4–92. The gears are subjected to the couple moments shown. Determine the magnitude and coordinate direction angles of the resultant couple moment.

z

20&

C

F

z

M1 % 40 lb ' ft

"F

M2 % 30 lb ' ft

150 N B 300 mm

x

400 mm

D 500 mm

150 N A

"P

Prob. 4–89

P

y

30&

15&

y

600 mm

x

Prob. 4–92

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4–93. Express the moment of the couple acting on the rod in Cartesian vector form.What is the magnitude of the couple moment?

4–95. A couple acts on each of the handles of the dual valve.

Determine the magnitude and coordinate direction angles of the resultant couple moment.

z z

"F % {#14i " 8j " 6k} N

0.5 m

35 N

y

25 N

0.8 m 1.5 m

x

60&

0.5 m

y 175 mm

x F % {14i # 8j # 6k} N

175 mm 25 N

35 N

Prob. 4–93

Prob. 4–95 *4–96. Express the moment of the couple acting on the pipe in Cartesian vector form.What is the magnitude of the couple moment? Take F = 125 N.

4–94. Express the moment of the couple acting on the pipe assembly in Cartesian vector form. Solve the problem (a) using Eq. 4–13, and (b) summing the moment of each force about point O.

4–97. If the couple moment acting on the pipe has a magnitude of 300 N # m, determine the magnitude F of the forces applied to the wrenches.

z

z y O x

O 5F

y

150 mm

0.3 m

600 mm

A F = {–6i + 2j + 3k} N 0.4 m "F % {6i – 2j – 3k} N

0.6 m

B x

200 mm

150 mm F

Prob. 4–94

Probs. 4–96/97

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4.7 EQUIVALENT SYSTEM

165

4.7 Equivalent System A force has the effect of both translating and rotating a body, and the amount by which it does so depends upon where and how the force is applied. In the next section we will discuss the method used to simplify a system of forces and couple moments acting on a body to a single resultant force and couple moment acting at a specified point O.To do this, however, it is necessary that the force and couple moment system produce the same “external” effects of translation and rotation of the body as their resultants.

When this occurs these two sets of loadings are said to be equivalent.

In this section we wish to show how to maintain this equivalency when a single force is applied to a specific point on a body and when it is located at another point O. Two cases for the location of point O will now be considered.

Point O Is On the Line of Action of the Force. Consider the

body shown in Fig. 4–33a, which is subjected to the force F applied to point A. In order to apply the force to point O without altering the external effects on the body, we will first apply equal but opposite forces F and -F at O, as shown in Fig. 4–33b. The two forces indicated by the slash across them can be canceled, leaving the force at point O as required, Fig. 4–33c. By using this construction procedure, an equivalent system has been maintained between each of the diagrams, as shown by the equal signs. Note, however, that the force has simply been “transmitted” along its line of action, from point A, Fig. 4–33a, to point O, Fig. 4–33c. In other words, the force can be considered as a sliding vector since it can act at any point O along its line of action. In Sec. 4.3 we referred to this concept as the principle of transmissibility. It is important to realize that only the external effects, such as the body’s motion or the forces needed to support the body if it is stationary, remain unchanged after F is moved. Certainly the internal effects depend on where F is located. For example, when F acts at A, the internal forces in the body have a high intensity around A; whereas movement of F away from this point will cause these internal forces to decrease.

A

O (a)

Fig. 4–33a

F

A

$

F O

F

$

A F O

"F (b)

(c)

Fig. 4–33b

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166

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F O R C E S Y S T E M R E S U LTA N T S

A

O

M$r#F

F

F

$

A

F r O

P

$

A

F O

"F (a)

(b)

(c)

Fig. 4–34

Point O Is Not On the Line of Action of the Force. This case is shown in

Fig. 4–34a, where F is to be moved to point O without altering the external effects on the body.

Following the same procedure as before, we first apply equal but opposite forces F and -F at point O, Fig. 4–34b. Here the two forces indicated by a slash across them form a couple which has a moment that is perpendicular to F and is defined by the cross product M = r * F. Since the couple moment is a free vector, it may be applied at any point P on the body as shown in Fig. 4–34c. In addition to this couple moment, F now acts at point O as required.

To summarize these concepts, when the point on the body is on the line of action of the force, simply transmit or slide the force along its line of action to the point. When the point is not on the line of action of the force, then move the force to the point and add a couple moment anywhere to the body. This couple moment is found bytaking the moment of the force about the point.

When these rules are carried out, equivalent external effects will be produced.

Photo A: Consider the effects on the hand when a stick of negligible weight supports a force F at its end.When the force is applied horizontally, the same force is felt at the grip, regardless of where it is applied along its line of action. This is a consequence of the principle of transmissibility.

Photo B: When the force is applied vertically it causes both a downward force F to be felt at the grip and a clockwise couple moment or twist of M = Fd. These same effects are felt if F is applied at the grip and M is applied anywhere on the stick. In both cases the systems are equivalent.

F

$

F

A F d

$

F M % Fd

B Unpublished Work © 2007 by R. C. Hibbeler. To be published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River, New Jersey. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department.

167

4.8 RESULTANTS OF A FORCE AND COUPLE SYSTEM

4.8 Resultants of a Force and Couple System

FR = ©F M RO = ©M c + ©M O

r1

O

(a)

r2

$

Mc

M2 % r2 $ F2

F2 Mc

F1

(b) O M1 % r1 $ F1

FR

(4–17)

The first equation states that the resultant force of the system is equivalent to the sum of all the forces; and the second equation states that the resultant couple moment of the system is equivalent to the sum of all the couple moments ©M c, plus the moments about point O of all the forces ©M O. If the force system lies in the x–y plane and any couple moments are perpendicular to this plane, that is along the z axis, then the above equations reduce to the following three scalar equations.

FRx = ©Fx FRy = ©Fy MRo = ©Mc + ©MO

F1

F2

$

When a rigid body is subjected to a system of forces and couple moments, it is often simpler to study the external effects on the body by replacing the system by an equivalent single resultant force acting at a specified point O and a resultant couple moment.To show how to determine these resultants we will consider the rigid body in Fig. 4–35a and use the concepts discussed in the previous section. Since point O is not on the line of action of the forces, an equivalent effect is produced if the forces are moved to point O and the corresponding couple moments M 1 = r1 * F1 and M 2 = r2 * F2 are applied to the body. Furthermore, the couple moment M c is simply moved to point O since it is a free vector. These results are shown in Fig. 4–35b. By vector addition, the resultant force is FR = F1 + F2, and the resultant couple moment is M RO = M c + M 1 + M 2, Fig. 4–35c. Since equivalency is maintained between the diagrams in Fig. 4–35, each force and couple system will cause the same external effects, i.e., the same translation and rotation of the body. Note that both the magnitude and direction of FR are independent of the location of point O; however, M RO depends upon this location since the moments M 1 and M 2 are determined using the position vectors r1 and r2.

Also note that M RO is a free vector and can act at any point on the body, although point O is generally chosen as its point of application.

The above method of simplifying any force and couple moment system to a resultant force acting at point O and a resultant couple moment can be generalized and represented by application of the following two equations.

M RO

u

(c)

O

Fig. 4–35

(4–18)

Note that the resultant force FR is equivalent to the vector sum of its two components FRx and FRy. Unpublished Work © 2007 by R. C. Hibbeler. To be published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River, New Jersey. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department.

168

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If the two forces acting on the stick are replaced by an equivalent resultant force and couple momen t at point A, or by the equivalent resultant force and couple moment at point B, then in each case the hand must provide the same resistance to translation and rotation in order to keep the stick in the horizontal position. In other words, the external effects on the stick are the same in each case.

d1

d¿1 F1

A

d2

FR % F1 " F2 F2 d¿2

B

$

A

MRA % F1d1 " F2d2

FR % F1 " F2

$

B MRB % F1d¿1 " F2d¿2

Procedure for Analysis The following points should be kept in mind when applying Eqs. 4–17 or 4–18.

• Establish the coordinate axes with the origin located at point O and the axes having a selected orientation.

Force Summation.

• If the force system is coplanar, resolve each force into its x and y components. If a component is directed •

along the positive x or y axis, it represents a positive scalar; whereas if it is directed along the negative x or y axis, it is a negative scalar.

In three dimensions, represent each force as a Cartesian vector before summing the forces.

Moment Summation.

• When determining the moments of a coplanar force system about point O, it is generally advantageous •

to use the principle of moments, i.e., determine the moments of the components of each force rather than the moment of the force itself.

In three dimensions use the vector cross product to determine the moment of each force about the point.

Here the position vectors extend from point O to any point on the line of action of each force.

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4.8 RESULTANTS OF A FORCE AND COUPLE SYSTEM

169

EXAMPLE 4.14 Replace the forces acting on the brace shown in Fig. 4–36a by an equivalent resultant force and couple moment acting at point A. y

MRA % 551 N ' m 100 N

A

A

x 66.6&

400 N

0.3 m

45&

FR = 962 N

0.4 m

0.4 m

(b)

Fig. 4–36b

600 N (a)

Fig. 4–36a ANALYSIS) SOLUTION (SCALAR The principle of moments will be applied to the 400-N force, whereby the moments of its two rectangular components will be considered.

Force Summation. The resultant force has x and y components of + F = ©F ; F = - 100 N - 400 cos 45° N = - 382.8 N = 382.8 N ; : Rx

x

Rx

+ c FRy = ©Fy; FRy = - 600 N - 400 sin 45° N = - 882.8 N = 882.8 NT As shown in Fig. 4–36b, FR has a magnitude of FR = 21FRx22 + 1FRy22 = 21382.822 + 1882.822 = 962 N and a direction of u = tan-1 a

FRy FRx

b = tan-1 a

882.8 b = 66.6° ud 382.8

Ans.

Ans.

Moment Summation. The resultant couple moment M RA is determined by summing the moments of the forces about point A. Assuming that positive moments act counterclockwise, i.e., in the +k direction, we have d+MRA = ©MA; MRA = 100 N102 - 600 N10.4 m2 - 1400 sin 45° N210.8 m2 - 1400 cos 45° N210.3 m2 # # = - 551 N m = 551 N m b Ans. NOTE: Realize that, when M RA and FR act on the brace at point A, Fig. 4–36b, they will produce the same external effects or reactions at the supports as those produced by the force system in Fig. 4–36a.

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170

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F O R C E S Y S T E M R E S U LTA N T S

EXAMPLE 4.15 M % 500 N ' m 3

F1 % 800 N 0.1 m

5 4

A structural member is subjected to a couple moment M and forces F1 and F2 as shown in Fig. 4–37a. Replace this system by an equivalent resultant force and couple moment acting at its base, point O.

z

C

F2 % 300 N

B 0.15 m

rC rB

SOLUTION (VECTOR ANALYSIS) The three-dimensional aspects of the problem can be simplified by using a Cartesian vector analysis. Expressing the forces and couple moment as Cartesian vectors, we have F1 = 5-800k6 N

1m

F2 = 1300 N2uCB = 1300 N2a

O x

y

= 300c

(a)

-0.15i + 0.1j

rCB b rCB

21-0.1522 + 10.122

d = 5-249.6i + 166.4j6 N

M = - 500 A 45 B j + 500 A 35 B k = 5- 400j + 300k6 N # m

Fig. 4–37a z

Force Summation.

FR = ©F;

FR = F1 + F2 = - 800k - 249.6i + 166.4j = 5-249.6i + 166.4j - 800k6 N

MRO

Ans.

Moment Summation.

O x

FR (b)

Fig. 4–37b

y

M RO = ©M C + ©M O M RO = M + rC * F1 + rB * F2 M RO

i j 3 = 1- 400j + 300k2 + 11k2 * 1- 800k2+ - 0.15 0.1 -249.6 166.4 = 1- 400j + 300k2 + 102 + 1-166.4i - 249.6j2 = 5-166i - 650j + 300k6 N # m

k 13 0 Ans.

The results are shown in Fig. 4–37b.

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4.9 FURTHER REDUCTION OF A FORCE AND COUPLE SYSTEM

171

4.9 Further Reduction of a Force and Couple System

Simplification to a Single Resultant Force. Consider now a special case for which the system of forces and couple moments acting on a rigid body, Fig. 4–38a, reduces at point O to a resultant force FR = ©F and resultant couple moment M RO = ©M O, which are perpendicular to one another, Fig. 4–38b. Whenever this occurs, we can further simplify the force and couple moment system by moving FR to another point P, located either on or off the body so that no resultant couple moment has to be applied to the body, Fig. 4–38c. In other words, if the force and couple moment system in Fig. 4–38a is reduced to a resultant system at point P, only the force resultant will have to be applied to the body, Fig. 4–38c.

The location of point P, measured from point O, can always be determined provided FR and M RO are known, Fig. 4–38b.As shown in Fig. 4–38c, P must lie on the bb axis, which is perpendicular to both the line of action of FR and the aa axis. This point is chosen such that the distance d satisfies the scalar equation MRO = FRd or d = MRO>FR. With FR so located, it will produce the same external effects on the body as the force and couple moment system in Fig. 4–38a, or the force and couple moment resultants in Fig. 4–38b.

If a system of forces is either concurrent, coplanar, or parallel, it can always be reduced, as in the above case, to a single resultant force FR acting through a specific point. This is because in each of these cases FR and M RO will always be perpendicular to each other when the force system is simplified at any point O. F2

M2 F1

F4

r2 r4

O

FR

$

r1 r3

a

b

MRO

F3

O a

M1

b

(a) d%

FR

$

a

b P

MRO FR

O

a

b

(b)

Fig. 4–38

(c)

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172

CHAPTER 4 F3

F2 P

F O R C E S Y S T E M R E S U LTA N T S

FR % (F

$

P

Concurrent Force Systems. A concurrent force system has been treated in detail in Chapter 2. Obviously, all the forces act at a point for which there is no resultant couple moment, so the point P is automatically specified, Fig. 4–39.

Coplanar Force Systems. Coplanar force systems, which may include

F1

couple moments directed perpendicular to the plane of the forces as shown in Fig. 4–40a, can be reduced to a single resultant force, because when each force in the system is moved to any point O in the x–y plane, it produces a couple moment that is perpendicular to the plane, i.e., in the ;k direction. The resultant moment M RO = ©M + ©1r * F2 is thus perpendicular to the resultant force FR, Fig. 4–40b; and so FR can be positioned a distance d from O so as to create the same moment M RO about O, Fig. 4–40c.

Fig. 4–39

y

y

M2

F3

F2

r2 O

x

r1

MRO % (M " (r $ F

$

r3 M1

x

O

F1 (a)

Fig. 4–40a

y

FR % (F (b)

$

d

O

FR % (F x

(c)

Fig. 4–40c

Fig. 4–40b

Parallel Force Systems. Parallel force systems, which can include

couple moments that are perpendicular to the forces, as shown in Fig. 4–41a, can be reduced to a single resultant force because when each force is moved to any point O in the x–y plane, it produces a couple moment that has components only about the x and y axes. The resultant moment M RO = ©M O + ©1r * F2 is thus perpendicular to the resultant force FR, Fig. 4–41b; and so FR can be moved to a point a distance d away so that it produces the same moment about O.

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4.9 FURTHER REDUCTION OF A FORCE AND COUPLE SYSTEM z

F1 r1

z F2 r2

O

F3

y

r3

%

z FR % (F

MRO % (M " ( (r $ F)

173

O

FR % (F y

%

d

O

y

M1 M2

x

x

x (b)

(a)

(c)

Fig. 4–41 The three parallel forces acting on the stick can be replaced by a single resultant force FR acting at a distance d from the grip. To be equivalent we require the resultant force to equal the sum of the forces, FR = F1 + F2 + F3, and to find the distance d the moment of the resultant force about the grip must be equal to the moment of all the forces about the grip, FRd = F1d1 + F2d2 + F3d3.

FR F2

F1

d1

F3

$

d

d2 d3

Procedure for Analysis The technique used to reduce a coplanar or parallel force system to a single resultant force follows a similar procedure outlined in the previous section.

• Establish the x, y, z, axes and locate the resultant force FR an arbitrary distance away from the origin of the coordinates.

Force Summation.

• The resultant force is equal to the sum of all the forces in the system.

• For a coplanar force system, resolve each force into its x and y components. Positive components are directed along the positive x and y axes, and negative components are directed along the negative x and y axes.

Moment Summation.

• The moment of the resultant force about point O is equal to the sum of all the couple moments in the •

system plus the moments about point O of all the forces in the system.

This moment condition is used to find the location of the resultant force from point O.

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174

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F O R C E S Y S T E M R E S U LTA N T S

Reduction to a Wrench. In the general case, the force and couple

F1

F2 r1

O

(a)

moment system acting on a body, Fig. 4–35a, will reduce to a single resultant force FR and couple moment M RO at O which are not perpendicular. Instead, FR will act at an angle u from M RO, Fig. 4–35c. As shown in Fig. 4–42a, however, M RO may be resolved into two components: one perpendicular, M " , and the other parallel M 7, to the line of action of FR. As in the previous discussion, the perpendicular component M " may be eliminated by moving FR to point P, as shown in Fig. 4–42b. This point lies on axis bb, which is perpendicular to both M RO and FR. In order to maintain an equivalency of loading, the distance from O to P is d = M" >FR. Furthermore, when FR is applied at P, the moment of FR tending to cause rotation of the body about O is in the same direction as M " , Fig. 4–42a. Finally, since M 7 is a free vector, it may be moved to P so that it is collinear with FR, Fig. 4–42c. This combination of a collinear force and couple moment is called a wrench or screw. The axis of the wrench has the same line of action as the force. Hence, the wrench tends to cause both a translation along and a rotation about this axis. Comparing Fig. 4–42a to Fig. 4–42c, it is seen that a general force and couple moment system acting on a body can be reduced to a wrench.

The axis of the wrench and the point through which this axis passes can always be determined.

r2

$

Mc

FR

u

M RO (c)

O

Fig. 4–35 (Repeated)

u

b

a

M""

M""

MRO

a O b

M"

FR

FR

FR

$

M"" a

b

O a

d

(a)

(b)

Fig. 4–42a

Fig. 4–42b

P b

$

a

b

O a

d

P b

(c)

Fig. 4–42c

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175

4.9 FURTHER REDUCTION OF A FORCE AND COUPLE SYSTEM

EXAMPLE 4.16 The beam AE in Fig. 4–43a is subjected to a system of coplanar forces.

Determine the magnitude, direction, and location on the beam of a resultant force which is equivalent to the given system of forces measured from E. y 200 N

500 N

A

60&

2m

y D B

100 N 0.5 m

1.5 m

1m (a)

d E

x

A

350.0 N u % 33.7& 233.0 N

1.5 m

FR % 420.5 N

Fig. 4–43b

SOLUTION Fig. 4–43a The origin of coordinates is located at point E as shown in Fig. 4–43a.

Force Summation. Resolving the 500-N force into x and y components and summing the force components yields + F = ©F ; : F = 500 cos 60° N + 100 N = 350.0 N : Rx

x

E

(b)

Rx

+ c FRy = ©Fy;

FRy = - 500 sin 60° N + 200 N = - 233.0 N = 233.0 NT

The magnitude and direction of the resultant force are established from the vector addition shown in Fig. 4–43b. We have FR = 21350.022 + 1233.022 = 420.5 N 233.0 u = tan-1 a b = 33.7° cu 350.0

Ans. Ans.

Moment Summation. Moments will be summed about point E. Hence, from Figs. 4–43a and 4–43b, we require the moments of the components of FR (or the moment of FR) about point E to equal the moments of the force system about E. Assuming positive moments are counterclockwise, we have d+MRE = ©ME ; 233.0 N1d2 + 350.0 N102 = 1500 sin 60° N214 m2 + 1500 cos 60° N2102 - 1100 N210.5 m2 - 1200 N212.5 m2 1182.1 = 5.07 m Ans. 233.0 NOTE: Using a clockwise sign convention would yield this same result.

Since d is positive, FR acts to the left of E as shown. Try to solve this problem by summing moments about point A and show d¿ = 0.927 m, measured to the right of A. d =

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x

176

CHAPTER 4

F O R C E S Y S T E M R E S U LTA N T S

EXAMPLE 4.17 y B

3 ft

5 ft

3 ft C

5 4 3

6 ft 60 lb

175 lb

250 lb

The jib crane shown in Fig. 4–44a is subjected to three coplanar forces. Replace this loading by an equivalent resultant force and specify where the resultant’s line of action intersects the column AB and boom BC. SOLUTION Force Summation. Resolving the 250-lb force into x and y components and summing the force components yields + F = ©F ; F = - 250 lb A 3 B - 175 lb = - 325 lb = 325 lb ; : Rx x Rx 5

5 ft

+ c FRy = ©Fy; FRy = - 250 lb A 45 B - 60 lb = - 260 lb = 260 lb T As shown by the vector addition in Fig. 4–44b,

x

A

FR = 2132522 + 126022 = 416 lb 260 u = tan-1 a b = 38.7° ud 325

(a)

Fig. 4–44a

d +MRA = ©MA;

x B

C

325 lb FR 325 lb

260 lb

(x,y)

325 lb u 260 lb

325 lb 1y2 + 260 lb 102

= 175 lb 15 ft2 - 60 lb 13 ft2 + 250 lb A 35 B 111 ft2 - 250 lb A 45 B 18 ft2 Ans. y = 2.29 ft By the principle of transmissibility, FR can also be treated as intersecting BC, Fig. 4–44b, in which case we have d +MRA = ©MA;

260 lb

A

Ans.

Moment Summation. Moments will be summed about the arbitrary point A. Assuming the line of action of FR intersects AB, Fig. 4–44b, we require the moment of the components of FR in Fig. 4–44b about A to equal the moments of the force system in Fig. 4–44a about A; i.e.,

y

FR

Ans.

325 lb 111 ft2 - 260 lb 1x2

= 175 lb 15 ft2 - 60 lb 13 ft2 + 250 lb A 35 B 111 ft2 - 250 lb A 45 B 18 ft2 x = 10.9 ft

y x (b)

Fig. 4–44b

Ans.

NOTE: We can also solve for these positions by assuming FR acts at

the arbitrary point (x, y) on its line of action, Fig. 4–44b. Summing moments about point A yields d +MRA = ©MA; 325 lb 1y2 - 260 lb 1x2 = 175 lb 15 ft2 - 60 lb 13 ft2 + 250 lb A 35 B 111 ft2 - 250 lb A 45 B 18 ft2 325y - 260x = 745 which is the equation of the colored dashed line in Fig. 4–44b. To find the points of intersection with the crane along AB, set x = 0, then y = 2.29 ft, and along BC set y = 11 ft, then x = 10.9 ft.

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177

4.9 FURTHER REDUCTION OF A FORCE AND COUPLE SYSTEM

EXAMPLE 4.18 The slab in Fig. 4–45a is subjected to four parallel forces. Determine the magnitude and direction of a resultant force equivalent to the given force system and locate its point of application on the slab. z

600 N

z 500 N 100 N

400 N 5m

5m

C

O B

2m

"

y

"

O

8m

x

P(x, y)

"

x

FR

y

"

(a)

x

y (b)

Fig. 4–45a

Fig. 4–45b

SOLUTION (SCALAR ANALYSIS) Force Summation. From Fig. 4–45a, the resultant force is + c FR = ©F;

FR = -600 N + 100 N - 400 N - 500 N = -1400 N = 1400 NT

Ans.

Moment Summation. We require the moment about the x axis of the resultant force, Fig. 4–45b, to be equal to the sum of the moments about the x axis of all the forces in the system, Fig. 4–45a. The moment arms are determined from the y coordinates since these coordinates represent the perpendicular distances from the x axis to the lines of action of the forces. Using the right-hand rule, where positive moments act in the +i direction, we have MRx = ©Mx; -11400 N2y = 600 N102 + 100 N15 m2 - 400 N110 m2 + 500 N102 -1400y = -3500 y = 2.50 m Ans. In a similar manner, assuming that positive moments act in the +j direction, a moment equation can be written about the y axis using moment arms defined by the x coordinates of each force.

MRy = ©My; 11400 N2x = 600 N18 m2 - 100 N16 m2 + 400 N102 + 500 N102 1400x = 4200 x = 3.00 m Ans. NOTE: A force of FR = 1400 N placed at point P(3.00 m, 2.50 m) on the slab, Fig. 4–45b, is therefore equivalent to the parallel force system acting on the slab in Fig. 4–45a.

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178

CHAPTER 4

F O R C E S Y S T E M R E S U LTA N T S

EXAMPLE 4.19

z

Three parallel bolting forces act on the rim of the circular cover plate in Fig. 4–46a. Determine the magnitude and direction of a resultant force equivalent to the given force system and locate its point of application, P, on the cover plate.

150 lb

200 lb

B

rC

300 lb rB

O rA

C 45& 8 ft

y

SOLUTION (VECTOR ANALYSIS) Force Summation. From Fig. 4–46a, the force resultant FR is FR = ©F;

A

FR = -300k - 200k - 150k = 5-650k6 lb

Ans.

x

Moment Summation. Choosing point O as a reference for computing moments and assuming that FR acts at a point P(x, y), Fig. 4–46b, we require (a)

M RO = ©M O; r * FR = rA * 1-300k2 + rB * 1-200k2 + rC * 1-150k2 1xi + yj2 * 1-650k2 = 18i2 * 1-300k2 + 1-8j2 * 1-200k2 + 1-8 sin 45°i + 8 cos 45°j2 * 1-150k2 650xj - 650yi = 2400j + 1600i - 848.5j - 848.5i

Fig. 4–46a z FR

x

O

r

y

Equating the corresponding j and i components yields y

P (x, y)

650x = 2400 - 848.5

(1)

-650y = 1600 - 848.5

(2)

Solving these equations, we obtain the coordinates of point P, x

x = 2.39 ft

y = -1.16 ft

Ans.

The negative sign indicates that it was wrong to have assumed a +y position for FR as shown in Fig. 4–46b. (b)

Fig. 4–46b

NOTE: It is also possible to establish Eqs. 1 and 2 directly by summing

moments about the y and x axes. Using the right-hand rule we have MRy = ©My;

650x = 300 lb 18 ft2 - 150 lb 18 sin 45° ft2

MRx = ©Mx;

-650y = 200 lb 18 ft2 - 150 lb 18 cos 45° ft2

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179

PROBLEMS

PROBLEMS 4–98. Replace the force at A by an equivalent force and couple moment at point O.

4–102. Replace the force system by an equivalent force and couple moment at point O.

4–99. Replace the force at A by an equivalent force and couple moment at point P.

4–103. Replace the force system by an equivalent force and couple moment at point P.

y

y

2m

A

2 ft 60&

30& 4m

13

P

O

1m

O

260 lb

8 ft

430 lb

375 N

3 ft

x

x

5 ft

P

2m

12 5

2 ft

Probs. 4–98/99

Probs. 4–102/103

*4–100. Replace the force system by an equivalent resultant force and couple moment at point O.

*4–104. Replace the loading system acting on the post by an equivalent resultant force and couple moment at point O.

4–101. Replace the force system by an equivalent resultant force and couple moment at point P.

4–105. Replace the loading system acting on the post by an equivalent resultant force and couple moment at point P. 30 lb P

y

1 ft 85 lb 5

60 lb

25 lb

4 3

4 3

3 ft

60 lb 3 ft

2 ft O 6 ft

5

45&

P 4 ft

Probs. 4–100/101

40 lb x 2 ft O

x

Probs. 4–104/105

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180

CHAPTER 4

F O R C E S Y S T E M R E S U LTA N T S

4–106. Replace the force and couple system by an equivalent force and couple moment at point O. 4–107. Replace the force and couple system by an equivalent force and couple moment at point P.

y 3m

8 kN ' m

P

O 6 kN 4m

5m 12

13

x

3m 4 kN

4–110. The forces and couple moments which are exerted on the toe and heel plates of a snow ski are Ft = 5-50i + 80j - 158k6N, M t = 5-6i + 4j + 2k6 N # m, and Fh = 5-20i + 60j - 250k6 N, M h = 5-20i + 8j + 3k6 N # m, respectively. Replace this system by an equivalent force and couple moment acting at point O. Express the results in Cartesian vector form.

4–111. The forces and couple moments that are exerted on the toe and heel plates of a snow ski are Ft = 5-50i + 80j - 158k6 N, M t = 5- 6i + 4j + 2k6 N # m, and Fh = 5-20i + 60j - 250k6 N, M h = 5-20i + 8j + 3k6 N # m, respectively. Replace this system by an equivalent force and couple moment acting at point P. Express the results in Cartesian vector form.

60&

5

z P

A

Fh

4m Ft

O

Mh 800 mm

Mt

Probs. 4–106/107

120 mm

y

x

*4–108. Replace the force system by a single force resultant and specify its point of application, measured along the x axis from point O.

Probs. 4–110/111

4–109. Replace the force system by a single force resultant and specify its point of application, measured along the x axis from point P.

*4–112. Replace the three forces acting on the shaft by a single resultant force. Specify where the force acts, measured from end A.

y

4–113. Replace the three forces acting on the shaft by a single resultant force. Specify where the force acts, measured from end B.

850 lb

5 ft

350 lb

125 lb

P

O 2 ft

6 ft

3 ft

Probs. 4–108/109

4 ft

x

3 ft

2 ft

4

500 lb

4 ft

5

200 lb

260 lb

Probs. 4–112/113

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PROBLEMS 4–114. Replace the loading on the frame by a single resultant force. Specify where its line of action intersects member AB, measured from A. 300 lb

200 lb 3 ft

181

4–117. Replace the loading system acting on the beam by an equivalent resultant force and couple moment at point O.

400 lb 4 ft

y B

A

450 N 30&

2 ft

600 lb ' ft

0.2 m

200 N ' m

200 lb O

x 1.5 m

7 ft

1.5 m

2m 200 N

C

Prob. 4–117

Prob. 4–114 4–115. Replace the loading acting on the beam by a single resultant force. Specify where the force acts, measured from end A. *4–116. Replace the loading acting on the beam by a single resultant force. Specify where the force acts, measured from B.

4–118. Determine the magnitude and direction u of force F and its placement d on the beam so that the loading system is equivalent to a resultant force of 12 kN acting vertically downward at point A and a clockwise couple moment of 50 kN # m.

4–119. Determine the magnitude and direction u of force F and its placement d on the beam so that the loading system is equivalent to a resultant force of 10 kN acting vertically downward at point A and a clockwise couple moment of 45 kN # m.

5 kN

450 N

700 N 30&

300 N

24

25 7

A

F

3 kN

)

60& d

A

B 2m

4m

3m

1500 N ' m 3m

Probs. 4–115/116

4m

6m

Probs. 4–118/119

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*4–120. Replace the loading on the frame by a single resultant force. Specify where its line of action intersects member AB, measured from A.

4–125. Replace the force and couple-moment system by an equivalent resultant force and couple moment at point O. Express the results in Cartesian vector form.

4–121. Replace the loading on the frame by a single resultant force. Specify where its line of action intersects member CD, measured from end C.

4–126. Replace the force and couple-moment system by an equivalent resultant force and couple moment at point P. Express the results in Cartesian vector form.

300 N 1m C

2m

250 N

3m

5

B

D

4

3

4–127. Replace the force and couple-moment system by an equivalent resultant force and couple moment at point Q. Express the results in Cartesian vector form. z F % {8i " 6j " 8k} kN

M % {#20i # 70j " 20k} kN ' m

400 3m 3m 4m A

x

4–123. Replace the force system acting on the frame by an equivalent resultant force and specify where the resultant’s line of action intersects member BC, measured from point B. *4–124. Replace the force system acting on the frame by an equivalent resultant force and couple moment acting at point A.

A

y

5m

6m P

Probs. 4–125/126/127

4–122. Replace the force system acting on the frame by an equivalent resultant force and specify where the resultant’s line of action intersects member AB, measured from point A.

30&

6m

O

Probs. 4–120/121

35 lb

5m

Q

*4–128. The belt passing over the pulley is subjected to forces F1 and F2, each having a magnitude of 40 N. F1 acts in the -k direction. Replace these forces by an equivalent force and couple moment at point A. Express the result in Cartesian vector form. Set u = 0° so that F2 acts in the -j direction.

4–129. The belt passing over the pulley is subjected to two forces F1 and F2, each having a magnitude of 40 N. F1 acts in the -k direction. Replace these forces by an equivalent force and couple moment at point A. Express the result in Cartesian vector form. Take u = 45°. z

20 lb 4 ft

)

B

r % 80 mm

y

2 ft 300 mm

3 ft 25 lb

A

x

2 ft C

Probs. 4–122/123/124

F2 F1

Probs. 4–128/129

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183

PROBLEMS 4–130. A force and couple act on the pipe assembly. If F1 = 50 N and F2 = 80 N, replace this system by an equivalent resultant force and couple moment acting at O. Express the results in Cartesian vector form.

z

*4–132. A biomechanical model of the lumbar region of the human trunk is shown. The forces acting in the four muscle groups consist of FR = 35 N for the rectus, FO = 45 N for the oblique, FL = 23 N for the lumbar latissimus dorsi, and FE = 32 N for the erector spinae. These loadings are symmetric with respect to the y–z plane. Replace this system of parallel forces by an equivalent force and couple moment acting at the spine, point O. Express the results in Cartesian vector form. z

180 N O

1.25 m

FR

y

FO FE

FL

FE FO

F2 x

FR

FL

0.5 m F1

O 0.75 m

75 mm

15 mm 45 mm

Prob. 4–130

50 mm

30 mm 40 mm

y

x

Prob. 4–132 4–131. Handle forces F1 and F2 are applied to the electric drill. Replace this system by an equivalent resultant force and couple moment acting at point O. Express the results in Cartesian vector form.

4–133. The building slab is subjected to four parallel column loadings. Determine the equivalent resultant force and specify its location (x, y) on the slab.Take F1 = 30 kN, F2 = 40 kN. 4–134. The building slab is subjected to four parallel column loadings. Determine the equivalent resultant force and specify its location (x, y) on the slab.Take F1 = 20 kN, F2 = 50 kN.

F2 % {2j # 4k} N

z z

0.15 m

F1 % {6i # 3j # 10k} N 20 kN

F1

50 kN

0.25 m F2 0.3 m

x

O y

x

Prob. 4–131

3m

4m 8m

6m 2m

Probs. 4–133/134

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y

184

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4–135. The pipe assembly is subjected to the action of a wrench at B and a couple at A. Determine the magnitude F of the couple forces so that the system can be simplified to a wrench acting at point C.

4–137. Replace the three forces acting on the plate by a wrench. Specify the magnitude of the force and couple moment for the wrench and the point P(x, y) where its line of action intersects the plate.

z #F i A

C 0.6 m

FA % {500i} N

A

z

FB % {800k} N

0.3 m 0.3 m

0.8 m {60k} N 0.25 m

x

B 0.25 m {#40i} N

Fi

0.5 m

D

0.7 m

x

B

x

P

y

y

y

6m

4m C

{#60k} N

FC % {300j} N

Prob. 4–135

Prob. 4–137

*4–136. The three forces acting on the block each have a magnitude of 10 lb. Replace this system by a wrench and specify the point where the wrench intersects the z axis, measured from point O.

4–138. Replace the three forces acting on the plate by a wrench. Specify the magnitude of the force and couple moment for the wrench and the point P(y, z) where its line of action intersects the plate.

z

12 ft

z

B F % {#60j} lb B

F2 2 ft

O F3

y

y F1

6 ft

P

x

6 ft

FA % {#80k} lb

x

Prob. 4–136

12 ft

z

A

C FC % {#40i} lb

y

Prob. 4–138

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185

4.10 REDUCTION OF A SIMPLE DISTRIBUTED LOADING

4.10 Reduction of a Simple Distributed

p % p(x)

Loading

p x

In many situations a very large surface area of a body may be subjected to distributed loadings such as those caused by wind, fluids, or simply the weight of material supported over the body’s surface. The intensity of these loadings at each point on the surface is defined as the pressure p (force per unit area), which can be measured in units of lb>ft2 or pascals (Pa), where 1 Pa = 1 N>m2.

In this section we will consider the most common case of a distributed pressure loading, which is uniform along one axis of a flat rectangular body upon which the loading is applied.* An example of such a loading is shown in Fig. 4–47a. The direction of the intensity of the pressure load is indicated by arrows shown on the load-intensity diagram. The entire loading on the plate is therefore a system of parallel forces, infinite in number and each acting on a separate differential area of the plate. Here the loading function, p = p1x2 Pa, is only a function of x since the pressure is uniform along the y axis. If we multiply p = p1x2 by the width a m of the plate, we obtain w = [p1x2 N>m2]a m = w1x2 N>m.

This loading function, shown in Fig. 4–47b, is a measure of load distribution along the line y = 0 which is in the plane of symmetry of the loading, Fig. 4–47a. As noted, it is measured as a force per unit length, rather than a force per unit area. Consequently, the load-intensity diagram for w = w1x2 can be represented by a system of coplanar parallel forces, shown in two dimensions in Fig. 4–47b. Using the methods of Sec. 4.9, this system of forces can be simplified to a single resultant force FR and its location x can be specified, Fig. 4–47c.

L y a 2

FR =

LL

w1x2 dx =

LA

dA = A

(a)

Fig. 4–47a w w % w(x)

x

O

L (b)

Fig. 4–47b FR A C

Magnitude of Resultant Force. From Eq. 4–17 1FR = ©F2, the magnitude of FR is equivalent to the sum of all the forces in the system.

In this case integration must be used since there is an infinite number of parallel forces dF acting along the plate, Fig. 4–47b. Since dF is acting on an element of length dx and w(x) is a force per unit length, then at the location x, dF = w1x2 dx = dA. In other words, the magnitude of dF is determined from the colored differential area dA under the loading curve.

For the entire plate length, + TFR = ©F;

a 2

O x

L (c)

Fig. 4–47c FR

(4–19)

Hence, the magnitude of the resultant force is equal to the total area A under the loading diagram w = w1x2, Fig. 4–47c. *The more general case of a nonuniform surface loading acting on a body is considered in Sec. 9.5.

x

y a 2

(d)

Fig. 4–47d

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x

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Location of Resultant Force. Applying Eq. 4–17 1MRO = ©MO2, the location x of the line of action of FR can be determined by equating the moments of the force resultant and the force distribution about point O (the y axis). Since dF produces a moment of x dF = xw1x2 dx about O, Fig. 4–47b, then for the entire plate, Fig. 4–47c, e + MRO = ©MO;

xFR =

xw1x2 dx LL Solving for x, using Eq. 4–19, we can write

x =

LL

xw1x2 dx

LL

= w1x2 dx

x dA LA LA

(4–20)

dA

This equation represents the x coordinate for the geometric center or centroid of the area under the distributed-loading diagram w(x).

Therefore, the resultant force has a line of action which passes through the centroid C (geometric center) of the area defined by the distributedloading diagram w(x), Fig. 4–47c.

Once x is determined, FR by symmetry passes through point 1x, 02 on the surface of the plate, Fig. 4–47d. If we now consider the threedimensional pressure loading p(x), Fig. 4–47a, we can therefore conclude that the resultant force has a magnitude equal to the volume under the distributed-loading curve p = p1x2 and a line of action which passes through the centroid (geometric center) of this volume. Detailed treatment of the integration techniques for computing the centroids of volumes or areas is given in Chapter 9. In many cases, however, the distributed-loading diagram is in the shape of a rectangle, triangle, or some other simple geometric form. The centroids for such common shapes do not have to be determined from Eq. 4–20; rather, they can be obtained directly from the tabulation given on the inside back cover.

The beam supporting this stack of lumber is subjected to a uniform distributed loading, and so the load-intensity diagram has a rectangular shape. If the load intensity is w0, then the resultant force is determined from the area of the rectangle, FR = w0b. The line of action of this force passes through the centroid or center of this area, x = a + b>2. This resultant is equivalent to the distributed load, and so both loadings produce the same “external” effects or support reactions on the beam.

w0

a

b

FR

$ a " b/2

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187

4.10 REDUCTION OF A SIMPLE DISTRIBUTED LOADING

Important Points • Distributed loadings are defined by using a loading function w = w1x2 that indicates the intensity • •

of the loading along the length of the member. This intensity is measured in N>m or lb>ft.

The external effects caused by a coplanar distributed load acting on a body can be represented by a single resultant force.

The resultant force is equivalent to the area under the distributed loading diagram, and has a line of action that passes through the centroid or geometric center of this area.

EXAMPLE 4.20 Determine the magnitude and location of the equivalent resultant force acting on the shaft in Fig. 4–48a.

SOLUTION Since w = w1x2 is given, this problem will be solved by integration.

The colored differential area element dA = w dx = 60x2 dx.

Applying Eq. 4–19, by summing these elements from x = 0 to x = 2 m, we obtain the resultant force FR. FR = ©F; 2 x3 2 23 03 FR = dA = 60x2 dx = 60c d = 60c d 3 0 3 3 LA L0 = 160 N Ans. Since the element of area dA is located an arbitrary distance x from O, the location x of FR measured from O, Fig. 4–48b, is determined from Eq. 4–20.

2

x dA

x =

LA

dA LA = 1.5 m

=

L0

x160x22 dx 160

w w % (60 x2)N/m 240 N/m dA % w dx

x

dx

2m (a)

Fig. 4–48a

w

x4 2 24 04 d 60c d 4 0 4 4 = 160 160

60c =

x

O

FR % 160 N C

Ans.

NOTE: These results may be checked by using the table on the inside back cover, where it is shown that for an exparabolic area of length a, height b, and shape shown in Fig. 4–48a,

2 m1240 N>m2 ab 3 3 A = = = 160 N and x = a = 12 m2 = 1.5 m 3 3 4 4

x

O x % 1.5 m (b)

Fig. 4–48b

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188

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EXAMPLE 4.21 A distributed loading of p = 800x Pa acts over the top surface of the beam shown in Fig. 4–49a. Determine the magnitude and location of the equivalent resultant force.

7200 Pa p = 800x Pa

x

p y

x

9m

0.2 m (a)

w

w % 160x N/m

1440 N/m

x x 9m (b)

Fig. 4–49b FR % 6.48 kN x%6m

3m C

(c)

Fig. 4–49c

SOLUTION Fig. 4–49a The loading function p = 800x Pa indicates that the load intensity varies uniformly from p = 0 at x = 0 to p = 7200 Pa at x = 9 m. Since the intensity is uniform along the width of the beam (the y axis), the loading may be viewed in two dimensions as shown in Fig. 4–49b. Here w = 1800x N>m2210.2 m2 = 1160x2 N>m At x = 9 m, note that w = 1440 N>m. Although we may again apply Eqs. 4–19 and 4–20 as in Example 4.20, it is simpler to use the table on the inside back cover.

The magnitude of the resultant force is equivalent to the area under the triangle.

FR = 2119 m211440 N>m2 = 6480 N = 6.48 kN Ans. The line of action of FR passes through the centroid C of the triangle. Hence, x = 9 m - 1319 m2 = 6 m Ans. The results are shown in Fig. 4–49c.

NOTE: We may also view the resultant FR as acting through the

centroid of the volume of the loading diagram p = p1x2 in Fig. 4–49a. Hence FR intersects the x–y plane at the point (6 m, 0).

Furthermore, the magnitude of FR is equal to the volume under the loading diagram; i.e., FR = V = 1217200 N>m2219 m210.2 m2 = 6.48 kN

Ans.

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189

4.10 REDUCTION OF A SIMPLE DISTRIBUTED LOADING

EXAMPLE 4.22 The granular material exerts the distributed loading on the beam as shown in Fig. 4–50a. Determine the magnitude and location of the equivalent resultant of this load.

SOLUTION The area of the loading diagram is a trapezoid, and therefore the solution can be obtained directly from the area and centroid formulas for a trapezoid listed on the inside back cover. Since these formulas are not easily remembered, instead we will solve this problem by using “composite” areas. In this regard, we can divide the trapezoidal loading into a rectangular and triangular loading as shown in Fig. 4–50b. The magnitude of the force represented by each of these loadings is equal to its associated area, F1 = 2119 ft2150 lb>ft2 = 225 lb F2 = 19 ft2150 lb>ft2 = 450 lb The lines of action of these parallel forces act through the centroid of their associated areas and therefore intersect the beam at x1 = x2 =

1 3 19 1 2 19

100 lb/ft 50 lb/ft B

A 9 ft (a)

Fig. 4–50a F1

50 lb/ft 50 lb/ft A

B x1

ft2 = 3 ft ft2 = 4.5 ft

FR = 225 + 450 = 675 lb

FR x A

B (c)

Ans.

Fig. 4–50c

NOTE: The trapezoidal area in Fig. 4–50a can also be divided into two

triangular areas as shown in Fig. 4–50d. In this case

and

F1 = 2119 ft21100 lb>ft2 = 450 lb F2 = 1219 ft2150 lb>ft2 = 225 lb x1 = x2 =

1 3 19 1 3 19

9 ft

Fig. 4–50b

Ans.

With reference to point A, Fig. 4–50b and 4–50c, we can find the location of FR. We require e + MRA = ©MA; x16752 = 312252 + 4.514502 x = 4 ft

x2

(b)

The two parallel forces F1 and F2 can be reduced to a single resultant FR. The magnitude of FR is + TFR = ©F;

F2

x1

F1

F2

100 lb/ft

ft2 = 3 ft ft2 = 3 ft

NOTE: Using these results, show that again FR = 675 lb and x = 4 ft.

x2 50 lb/ft

9 ft (d)

Fig. 4–50d

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190

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PROBLEMS 4–139. The loading on the bookshelf is distributed as shown.

Determine the magnitude of the equivalent resultant location, measured from point O.

4–141. Replace the loading by an equivalent force and couple moment acting at point O. 4–142. Replace the loading by a single resultant force, and specify the location of the force on the beam measured from point O.

6 kN/m / 2 lb/ft

3.5 lb/ft

15 kN

500 kN

O

A 2.75 ft 4 ft

7.5 m

1.5 ft

4.5 m

Prob. 4–139

Probs. 4–141/142

*4–140. Replace the loading by an equivalent resultant force and couple moment acting at point A.

4–143. The column is used to support the floor which exerts a force of 3000 lb on the top of the column. The effect of soil pressure along its side is distributed as shown. Replace this loading by an equivalent resultant force and specify where it acts along the column, measured from its base A.

3000 lb

2.5 m 600 N/m

80 lb/ft

2.5 m

B

A

9 ft

600 N/m

200 lb/ft A

Prob. 4–140

Prob. 4–143

Unpublished Work © 2007 by R. C. Hibbeler. To be published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River, New Jersey. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department.

191

PROBLEMS *4–144. Replace the loading by an equivalent force and couple moment acting at point O.

4–146. The beam supports the distributed load caused by the sandbags. Determine the resultant force on the beam and specify its location measured from point A.

15 kN/m

O

2.5 kN/m

1.5 kN/m

5 kN/m

1 kN/m

A 9m

B 3m

3m

1.5 m

Prob. 4–144

Prob. 4–146

4–145. Replace the distributed loading by an equivalent resultant force, and specify its location on the beam, measured from the pin at C.

4–147. Determine the length b of the triangular load and its position a on the beam such that the equivalent resultant force is zero and the resultant couple moment is 8 kN # m clockwise.

a

b 4 kN/m 30&

A

C

B A 2.5 kN/m

800 lb/ft 15 ft

15 ft

Prob. 4–145

9m

Prob. 4–147

Unpublished Work © 2007 by R. C. Hibbeler. To be published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River, New Jersey. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department.

192

CHAPTER 4

F O R C E S Y S T E M R E S U LTA N T S

*4–148. Replace the distributed loading by an equivalent resultant force and specify its location, measured from point A.

4–150. The beam is subjected to the distributed loading.

Determine the length b of the uniform load and its position a on the beam such that the resultant force and couple moment acting on the beam are zero.

800 N/m

b 200 N/m

A

40 lb/ft

a

B

60 lb/ft 2m

3m

10 ft

6 ft

Prob. 4–148

Prob. 4–150

4–149. The distribution of soil loading on the bottom of a building slab is shown. Replace this loading by an equivalent resultant force and specify its location, measured from point O.

4–151. Replace the loading by an equivalent resultant force and specify its location on the beam, measured from point B.

800 lb/ft 500 lb/ft O 50 lb/ft

100 lb/ft A 12 ft

300 lb/ft 9 ft

Prob. 4–149

B 12 ft

9 ft

Prob. 4–151

Unpublished Work © 2007 by R. C. Hibbeler. To be published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River, New Jersey. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department.

193

PROBLEMS *4–152. Replace the distributed loading by an equivalent resultant force and specify where its line of action intersects member AB, measured from A.

4–155. Determine the equivalent resultant force and couple moment at point O.

4–153. Replace the distributed loading by an equivalent resultant force and specify where its line of action intersects member BC, measured from C. w

200 N/m

3 kN/m

100 N/m B

w % ( 13 x2) kN/m

C

6m

O

5m

x

3m

Prob. 4–155 200 N/m

A

Probs. 4–152/153 4–154. Replace the loading by an equivalent resultant force and couple moment acting at point O.

*4–156. Wind has blown sand over a platform such that the intensity of the load can be approximated by the function w = 10.5x32 N>m. Simplify this distributed loading to an equivalent resultant force and specify the magnitude and location of the force, measured from A.

3m 7.5 kN/m w 3m

500 N/m

w % (0.5x3) N/m 4.5 m x

A 20 kN/m O

Prob. 4–154

10 m

Prob. 4–156

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194

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F O R C E S Y S T E M R E S U LTA N T S

4–157. Determine the equivalent resultant force and its location, measured from point O.

4–159. Currently eighty-five percent of all neck injuries are caused by rear-end car collisions. To alleviate this problem, an automobile seat restraint has been developed that provides additional pressure contact with the cranium.

During dynamic tests the distribution of load on the cranium has been plotted and shown to be parabolic. Determine the equivalent resultant force and its location, measured from point A.

w % w0 sin pL x

w

A O

x

0.5 ft

12 lb/ft w w % 12(1 " 2x2) lb/ft

B L

Prob. 4–157 x

Prob. 4–159

4–158. Determine the equivalent resultant force acting on the bottom of the wing due to air pressure and specify where it acts, measured from point A.

*4–160. Determine the equivalent resultant force of the distributed loading and its location, measured from point A. Evaluate the integral using Simpson’s rule.

w 3 ft

w % 5x " (16 " x2)1/2 kN/m

x

5.07 kN/m

2 kN/m A w % (86x2) lb/ft

A 3m

w

Prob. 4–158

x

B 1m

Prob. 4–160

Unpublished Work © 2007 by R. C. Hibbeler. To be published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River, New Jersey. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department.

195

CHAPTER REVIEW

Chapter Review Moment of Force—Scalar Definition

Moment axis

A force produces a turning effect about a point O that does not lie on its line of action. In scalar form, the moment magnitude is the product of the force and the moment arm or perpendicular distance from point O to the line of action of the force.

MO = Fd MO

F

d

O

The direction of the moment is defined using the right-hand rule. M O always acts along an axis perpendicular to the plane containing F and d, and passes through the point O. If the force acts through the point O, the moment M O will be zero. y

Rather than finding d, it is normally easier to resolve the force into its x and y components, determine the moment of each component about the point, and then sum the results. This is called the principle of moments.

(See pages 118–119.)

F

Fy x Fx

MO = Fd = Fxy - Fyx

y d

x

O

Moment of a Force—Vector Definition Since three-dimensional geometry is generally more difficult to visualize, the vector cross product can be used to determine the moment.

M O = rA * F = rB * F = rC * F z C

r is a position vector that extends from point O to any point A, B, C on the line of action of F. If the position vector r and force F can be expressed as Cartesian vectors, then the cross product can be expressed by the expansion of a determinant.

(See pages 125–127.)

rC

MO

i 3 = r * F = rx Fx

j ry Fy

k rz 3 Fz

B rB

MO

rA O

F

A y

x

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Moment about an Axis If the moment of a force F is to be determined about an arbitrary axis, then the projection of the moment onto the axis must be obtained. Provided the distance da that is perpendicular to both the line of action of the force and the axis can be found, then the moment of the force about the axis can be determined from a scalar equation.

Ma a

Ma = daF

F

da

a

It’s a scalar equation when the line of action of F intersects an axis and the moment of F about the axis is zero. Also, when the line of action of F is parallel to the axis, the moment of F about the axis is zero.

In three dimensions, the scalar triple product should be used. Here ua is the unit vector that specifies the direction of the axis, and r is a position vector that is directed from any point on the axis to any point on the line of action of the force. If Ma is calculated as a negative scalar, then the sense of direction of M a is opposite to ua.

(See pages 142–145.)

a

Ma

r

ua

uax Ma = ua # 1r * F2 = 3 rx Fx

uay ry Fy

uz rz 3 Fz

F

Axis of projection a¿

Couple Moment A couple consists of two equal but opposite forces that act a perpendicular distance d apart. Couples tend to produce a rotation without translation.

Mc = Fd

#F

d F

The moment of the couple has a direction that is established using the right-hand rule.

If the vector cross product is used to determine the moment of a couple, then r extends from any point on the line of action of one of the forces to any point on the line of action of the other force F that is used in the cross product.

(See pages 152 and 153.)

Mc = r * F

B F

r

A

"F

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197

CHAPTER REVIEW

Simplification of a Force and Couple System

FR

Any system of forces and couples can be reduced to a single resultant force and resultant couple moment acting at a point.

The resultant force is the sum of all the forces in the system, FR = ©F, and the resultant couple moment is equal to the sum of all the moments and couple moments about the point, MRO = ©Mc + ©MO

M RO

O r1 r2

O

$

Mc

FR

FR a

b

MRO O

a

Equating the moment of a resultant force about a point to the moment of the forces and couples in the system about the same point, for any type of force system that is not concurrent, coplanar, or parallel, would yield a wrench, which consists of the resultant force and a resultant collinear couple moment.

(See pages 167, 171–172, and 174.)

b

$

a

b d%

MRO FR

P

O

a

FR M"" M RO

b

O

b

FR

$

Further simplification to a single resultant force is possible provided the force system is concurrent, coplanar, or parallel. For this case, to find the location of the resultant force from a point, it is necessary to equate the moment of the resultant force about the point to the moment of the forces and couples in the system about the same point.

F1

F2

a

O

P

d

a

b

Coplanar Distributed Loading A resultant force can replace a simple distributed loading, which is equivalent to the area under the loading curve.

This resultant has a line of action that passes through the centroid or geometric center of the area or volume under the loading diagram.

(See pages 185 and 186.)

w

FR

w % w(x)

A C

x

O L

O

x

L

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REVIEW PROBLEMS 4–161. Determine the coordinate direction angles a, b, g of F, which is applied to the end A of the pipe assembly, so that the moment of F about O is zero.

4–162. Determine the moment of the force F about point O. The force has coordinate direction angles of a = 60°, b = 120°, g = 45°. Express the result as a Cartesian vector.

*4–164. Determine the moment of the force Fc about the door hinge at A. Express the result as a Cartesian vector.

4–165. Determine the magnitude of the moment of the force Fc about the hinged axis aa of the door. z

F % 20 lb

C 1.5 m

2.5 m FC % 250 N

a

z

y

O

10 in.

30&

6 in.

A 8 in.

x

a

6 in.

Probs. 4–161/162

A B 1m

x

4–163. Replace the force at A by an equivalent resultant force and couple moment at point P. Express the results in Cartesian vector form.

0.5 m

y

Probs. 4–164/165

4–166. A force of F = 80 N acts vertically downward on the Z-bracket. Determine the moment of this force about the bolt axis (z axis), which is directed at 15° from the vertical.

z

z

F % {#300i " 200j # 500k} N

F % 80 N A 200 mm

4m

A 15&

8m 4m

6m

O

300 mm

y y

P x

x

Prob. 4–163

15&

100 mm

Prob. 4–166

Unpublished Work © 2007 by R. C. Hibbeler. To be published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River, New Jersey. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department.

199

REVIEW PROBLEMS 4–167. Replace the force F having a magnitude of F = 50 lb and acting at point A by an equivalent force and couple moment at point C.

4–169. The horizontal 30-N force acts on the handle of the wrench. Determine the moment of this force about point O. Specify the coordinate direction angles a, b, g of the moment axis.

z A

C

z

F

30 ft

A

30 N 45& 45& 10 mm

50 mm

O

O

10 ft

y

x

20 ft

x

200 mm

B

15 ft

Prob. 4–169 10 ft

B

y

Prob. 4–167

*4–168. The horizontal 30-N force acts on the handle of the wrench. What is the magnitude of the moment of this force about the z axis?

4–170. If the resultant couple moment of the three couples acting on the triangular block is to be zero, determine the magnitudes of forces F and P.

z

z

B

"F 200 mm

A

30 N 45& 45&

F 3 in.

10 mm

50 mm O

6 in.

10 lb

3 in. x

Prob. 4–168

P

30& y

x

10 lb

4 in.

"P

Prob. 4–170

Unpublished Work © 2007 by R. C. Hibbeler. To be published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River, New Jersey. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department.

y