CH04

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.

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4.1

Moment of a Force—

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

which acts perpendicular to the

handle of the wrench and is located a distance 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

the greater the turning effect.This

tendency for rotation caused by is sometimes called a torque, but most

often it is called the moment of a force or simply the moment

Note

that the moment axis (z) is perpendicular to the shaded plane (x–y) which

contains both and and that this axis intersects the plane at point O.

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.

Force System Resultants

117

)

(a)

Fig. 4–1

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118

H A P T E R

4 F

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Y S T E M

E S U LTA N T S

Now consider applying the force 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,

still creates the tendency for

rotation and so the moment

is produced. As before, the force and

distance

lie in the shaded plane (y–z) which is perpendicular to the

moment axis (x). Lastly, if a force

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

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.

Magnitude

. The magnitude of

(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.,

Direction.

The direction of

will be specified by using the “right-

hand 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,

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 two-

dimensional view of Fig. 4–2a is given in Fig. 4–2b. Here

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.

ft.

= Fd

(c)

Fig. 4–1 (cont.)

Sense of rotation

Moment axis

(a)

(b)

Fig. 4–2

(b)

)

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4.1 M

OMENT OF A

ORCE

—S

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ORMULATION

119

Resultant Moment of a System of Coplanar Forces.

If a

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

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

(4–2)

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

axis, whereas a negative moment is

directed along the

axis.

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

and 4–2.

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

then the moment arm

distance is shorter than when the force is applied perpendicular to the bar,

i.e.,

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.

d¿ 6 d.

u = 90°,

u 6 90°,

-z

d + M

= ©Fd

Fig. 4–3

)

max

$ F d

d¿

% F d¿

% Fd

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

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

Here the support at A

prevents the rotation.

= Fd

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120

H A P T E R

4 F

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E S U LTA N T S

For each case illustrated in Fig. 4–4, determine the moment of the

force about point O.

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

= 17 kN214 m - 1 m2 = 21.0 kN

m g

Ans.

= 160 lb211 sin 45° ft2 = 42.4 lb

ft g

Ans.

= 140 lb214 ft + 2 cos 30° ft2 = 229 lb

ft b

Ans.

= 150 N210.75 m2 = 37.5 N

m b

Ans.

= 1100 N212 m2 = 200 N

m b

Ans.

2 m

(a)

100 N

2 ft

(c)

4 ft

2 cos 30& ft

40 lb

30&

2 m

(e)

4 m

1 m

7 kN

Fig. 4–4

2 m

(b)

50 N

0.75 m

(d)

1 sin 45& ft

60 lb

3 ft

45&

1 ft

EXAMPLE 4.1

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EXAMPLE 4.3

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.

SOLUTION

(SCALAR ANALYSIS)

In general,

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,

The curls indicate the sense of rotation of the moment, which is

defined by the direction the force orbits about each point.

= 800 N10.5 m2 = 400 N

m g

Ans.

= 800 N102 = 0 1line of action of F passes through C2

Ans.

= 800 N11.5 m2 = 1200 N

m b

Ans.

= 800 N12.5 m2 = 2000 N

m b

Ans.

M = Fd,

1.25 m

0.5 m

1.5 m

1 m

% 800 N

Fig. 4–5

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

direction, i.e.,

counterclockwise, we have

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.

= -334 N

m = 334 N

m b

Ans.

-40 N14 m + 3 cos 30° m2

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

d+M

= ©Fd;

30&

50 N

40 N

20 N

3 m

2 m

60 N

Fig. 4–6

4.1 M

OMENT OF A

ORCE

—S

CALAR

ORMULATION

121

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122

H A P T E R

4 F

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Y S T E M

E S U LTA N T S

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

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 between their tails

Thus,

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

(4–3)

where the scalar

defines the magnitude of C and the unit vector

defines the direction of C. The terms of Eq. 4–3 are illustrated

graphically in Fig. 4–8.

AB sin u

C = A * B = 1AB sin u2u

C = AB sin u.

10° … u … 180°2.

C = A * B

C % A $ B

Fig. 4–7

C % AB sin u

Fig. 4–8

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Laws of Operation.

1. The commutative law is not valid; i.e.,

Rather,

This is shown in Fig. 4–9 by using the right-hand rule.The cross product

yields a vector that acts in the opposite direction to C; i.e.,

2. Multiplication by a scalar:

This property is easily shown since the magnitude of the resultant

vector

and its direction are the same in each case.

3. The distributive law:

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

the magnitude of the resultant vector is

and its direction is determined using

the right-hand rule. As shown in Fig. 4–10, the resultant vector points

in the

direction. Thus,

In a similar manner,

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.,

Moving clockwise, a negative unit vector is

obtained; e.g., i * k = -j.

k * i = j.

k * i = j k * j = -i k * k = 0

j * k = i

j * i = -k j * j = 0

i * j = k i * k = -j

i * i = 0

i * j = 112k.

1i21j21sin 90°2 = 112112112 = 1,

i * j,

A * 1B + D2 = 1A * B2 + 1A * D2

1 ƒ a ƒ AB sin u2

a1A * B2 = 1aA2 * B = A * 1aB2 = 1A * B2a

B * A = -C.

B * A

A * B = -B * A

A * B Z B * A

k % i $ j

Fig. 4–10

# C % B $ A

C % A $ B

Fig. 4–9

Fig. 4–11

4.2 C

ROSS

RODUCT

123

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*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.,

124

H A P T E R

4 F

O R C E

Y S T E M

E S U LTA N T S

Consider now the cross product of two general vectors A and B which

are expressed in Cartesian vector form. We have

Carrying out the cross-product operations and combining terms yields

(4–4)

This equation may also be written in a more compact determinant form as

(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 * B = 3

A * B = 1A

- A

2i - 1A

- A

2j + 1A

- A

+ A

1k * i2 + A

1k * j2 + A

1k * k2

+ A

1j * i2 + A

1j * j2 + A

1j * k2

= A

1i * i2 + A

1i * j2 + A

1i * k2

A * B = 1A

i + A

j + A

k2 * 1B

i + B

j + B

For element k:

For element j:

For element i:

% i(A

# A

)

% #j(A

# A

)

% k(A

# A

)

By definition, this notation represents the terms

which is simply the

product of the two elements of the arrow slanting downward to the right

minus

the product of the two elements intersected by the arrow slanting downward to the left

For a

determinant, such as Eq. 4–5, the three minors can be generated in

accordance with the following scheme:

3 * 3

- A

Adding the results and noting that the j element must include the minus sign yields the

expanded form of

given by Eq. 4–4.

A * B

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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,

(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

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

where the angle is measured between the tails of r

and F.To establish this angle,r must be treated as a sliding vector so that can

be constructed properly,Fig. 4–12b.Since the moment arm

then

which agrees with Eq. 4–1.

Direction.

The direction and sense of

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

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.

= rF sin u = F1r sin u2 = Fd

d = r sin u,

= rF sin u,

= r * F

Moment axis

(a)

Fig. 4–12

Moment axis

(b)

4.3 M

OMENT OF A

ORCE

—V

ECTOR

ORMULATION

125

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126

H A P T E R

4 F

O R C E

Y S T E M

E S U LTA N T S

Principle of Transmissibility.

Consider the force F applied at

point A in Fig. 4–13.The moment created by F about O is

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

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.

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

(4–7)

where

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

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

If the determinant is expanded, then like Eq. 4–4 we have

(4–8)

The physical meaning of these three moment components becomes

evident by studying Fig. 4–14a. For example, the i component of

determined from the moments of

and

about the x axis. In

particular, note that

does not create a moment or tendency to cause

= 1r

- r

2i - 1r

- r

2j + 1r

- r

, F

, r

= r * F = 3

= r

* F = r

* F

= r

* F;

Fig. 4–13

(a)

Fig. 4–14

Moment

axis

(b)

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turning about the x axis since this force is parallel to the x axis. The line

of action of

passes through point E, and so the magnitude of the

moment of about point A on the x axis is

By the right-hand rule

this component acts in the negative i direction. Likewise,

contributes

a moment component of

Thus,

as shown in

Eq. 4–8. As an exercise, establish the j and k components of

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

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.

Resultant Moment of a System of Forces.

If a body is acted

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

(4–9)

and is shown in 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

and so

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,

As required, both of these vectors are directed from

point A to a point on the line of action of the force.

= r

* F = r

* F.

= r

= ©1r * F2

= 1r

- r

Fig. 4–15

4.3 M

OMENT OF A

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ORMULATION

127

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EXAMPLE 4.4

128

H A P T E R

4 F

O R C E

Y S T E M

E S U LTA N T S

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.

SOLUTION

(VECTOR ANALYSIS)

As shown in Fig. 4–16b, either one of two position vectors can be used

for the solution, since

The position

vectors are represented as

The force has a magnitude of 60 N and a direction specified by the

unit vector

directed from C to B. Thus,

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

= 5-40i - 20j + 40k6 N

F = 160 N2u

= 160 N2c

11 - 32i + 13 - 42j + 12 - 02k

21-22

+ 1-12

+ 122

= 51i + 3j + 2k6 m and r

= 53i + 4j6 m

= r

* F.

= r

* F

(a)

1 m

3 m

2 m

F % 60 N

4 m

2 m

(c)

Fig. 4–16

= [31402 - 21-202]i - [11402 - 21-402]j + [11-202 - 31-402]k

= r

* F = 3

-40

-20 40

= [41402 - 01-202]i - [31402 - 01-402]j + [31-202 - 41-402]k

= r

* F = 3

-40

-20 40

In both cases,

The magnitude of

is therefore

NOTE:

As expected,

acts perpendicular to the shaded plane

containing vectors F,

and

Fig. 4–16c. (How would you find its

coordinate direction angles

) Had

this problem been worked using a scalar approach, where

notice the difficulty that can arise in obtaining the moment arm d.

= Fd,

g = 63.4°?

b = 122°,

a = 44.3°,

= 211602

+ 1-1202

+ 11002

= 224 N

Ans.

= 5160i - 120j + 100k] N

(b)

B(1 m, 3 m, 2 m)

C(3 m, 4 m, 0)

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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.

SOLUTION

Position vectors are directed from point O to each force as shown in

Fig. 4–17b. These vectors are

The resultant moment about O is therefore

= 54i + 5j - 2k6 ft

= 55j6 ft

(b)

5 ft

4 ft

2 ft

% {80i " 40j # 30k} lb

% {50j} lb

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

(a)

= 530i - 40j + 60k6 lb

Ans.

+ [51-302 - 14021-22]i - [41-302 - 801-22]j + [41402 - 80152]k

= [51202 - 40102]i - [0j] + [01402 - 1-602152]k + [0i - 0j + 0k]

= 3

-60 40 20

3 + 3

0 50 0

3 + 3

-2

80 40

-30

= r

* F

+ r

* F

+ r

* F

= ©1r * F2

The moment axis is directed along the line of action of

Since

the magnitude of this moment is

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

Therefore, the coordinate direction angles of the moment axis are

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.

cos g = 0.7682;

g = 39.8°

Ans.

cos b = -0.5121;

b = 121°

Ans.

cos a = 0.3841;

a = 67.4°

Ans.

u =

30i - 40j + 60k

78.10

= 0.3841i - 0.5121j + 0.7682k

= 21302

+ 1-402

+ 1602

= 78.10 lb

%39.8&

%67.4&

%121&

% {30i # 40j " 60k} lb · ft

(c)

Fig. 4–17

4.3 M

OMENT OF A

ORCE

—V

ECTOR

ORMULATION

129

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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

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.,

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.

= r * F.

= Fd,

130

H A P T E R

4 F

O R C E

Y S T E M

E S U LTA N T S

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

Fig. 4–18. We have

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

If the force is replaced by its two components and

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

will create zero moment about A and so

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

will create zero moment about A, and so M

= F

= Fd.

= r * F

+ r * F

= r * 1F

+ F

2 = r * F

F = F

+ F

Fig. 4–18

C F

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Fig. 4–19a

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.

SOLUTION I

The moment arm d can be found by trigonometry, using the

construction shown in Fig. 4–19b. From the right triangle BCD,

Thus,

According to the right-hand rule,

is directed in the

direction

since the force tends to rotate or orbit counterclockwise about point

A. Hence, reporting the moment as a Cartesian vector, we have

Ans.

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

direction, we can apply Eq. 4–2

in which case

Thus

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.

= 514.1k6 N

= 14.1 N

m g

d+M

= 1200 sin 45° N210.20 m2 - 1200 cos 45° N210.10 m2

= ©Fd2,

= 514.1k6 N

= Fd = 200 N10.070 71 m2 = 14.1 N

m g

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

F % 200 N

45&

100 mm

(b)

F % 200 N

45&

100 mm

(a)

0.2 m

0.1 m

(c)

200 cos 45& N

200 sin 45& N

Fig. 4–19

4.4 P

RINCIPLE OF

OMENTS

131

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132

H A P T E R

4 F

O R C E

Y S T E M

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

direction, we have

Ans.

SOLUTION II

(VECTOR ANALYSIS)

Using a Cartesian vector approach, the force and position vectors

shown in Fig. 4–20c can be represented as

The moment is therefore

NOTE:

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

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.

= 5-98.6k6 N

Ans.

= 0i - 0j + [0.41-346.42 - 1-0.221200.02]k

= r * F = 3

0.4

-0.2

200.0

-346.4 0

= 5200.0i - 346.4j6 N

F = 5400 sin 30°i - 400 cos 30°j6 N

r = 50.4i - 0.2j6 m

= 5-98.6k6 N

= -98.6 N

m = 98.6 N

m b

d+M

= 400 sin 30° N10.2 m2 - 400 cos 30° N10.4 m2

0.4 m

0.2 m

30&

F = 400 N

(a)

0.4 m

0.2 m

(b)

400 cos 30& N

400 sin 30& N

0.4 m

0.2 m

30&

(c)

Fig. 4–20

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ROBLEMS

133

6 in.

3 in.

5 in.

13 in.

40 lb

60 lb

45&

30&

Probs. 4–4/5

300 mm

60&

50 mm

Probs. 4–6/7

4 m

3 m

5 m

2 m

400 N

260 N

2 m

30&

Probs. 4–8/9

P R O B L E M S

4–1. If A, B, and D are given vectors, prove the distributive

law for the vector cross product, i.e.,

4–2. Prove the triple scalar product identity

4–3. Given the three nonzero vectors A, B, and C, show

that if

the three vectors must lie in the

same plane.

*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.

1B * C2 = 0,

1B * C2 = 1A * B2

1A * B2 + 1A * D2.

A * 1B + D2=

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

about point O when u = 30°.

15 N

4–7. If the force

determine the angle

so that the force develops a clockwise

moment about point O of 20 N

10 … u … 90°2

F = 100 N,

*4–8. Determine the magnitude and directional sense of

the resultant moment of the forces about point O.

4–9. Determine the magnitude and directional sense of the

resultant moment of the forces about point P.

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134

H A P T E R

4 F

O R C E

Y S T E M

E S U LTA N T S

30 mm

200 mm

40 N

20&

Prob. 4–10

30&

10 ft

250 lb

4 ft

3 ft

4 ft

6 ft

300 lb

Prob. 4–11

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.

4–11. Determine the magnitude and directional sense of

the resultant moment of the forces about point O.

*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.

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.

4–14. Determine the moment of each force about the bolt

located at A. Take

4–15. If

and

determine the resultant

moment about the bolt located at A.

= 45 lb,

= 30 lb

= 40 lb, F

= 50 lb.

30&

15&

% 4 N

% 6 N

% 8 N

15 mm

20 mm

35 mm

0.2 m

0.35 m

0.25 m

0.15 m

Probs. 4–12/13

Probs. 4–14/15

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ROBLEMS

135

16 ft

15 ft

3 ft

30&

70&

Prob. 4–17

40 lb

Prob. 4–18

20&

60&

80 N

150 mm

Prob. 4–19

2 ft

14 ft

16 ft

800 lb

G¿

Prob. 4–20

*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. for -60° … u … 80°.

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.

4–18. Determine the direction

of the

force

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.

F = 40 lb

u 10° … u … 180°2

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.

*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

determine the maximum load W, having a mass center at

that can be lifted. Take u = 30°.

G¿,

M = 20110

2 lb

ft,

16 mm

2.30 kN

Prob. 4–16

ROBLEMS

135

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136

H A P T E R

4 F

O R C E

Y S T E M

E S U LTA N T S

F % 80 lb

1 ft

4 ft

Prob. 4–22

30&

% 275 lb

85 ft

Prob. 4–23

0.3 m

0.7 m

0.9 m

70 N

Probs. 4–24/25

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?

4–22. Determine the clockwise direction

of the force

so that it produces (a) the maximum

moment about point A and (b) no moment about point A.

Compute the moment in each case.

F = 80 lb

u 10° … u … 180°2

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.

*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

4–25. The 70-N force acts on the end of the pipe at B.

Determine the angles

of the force that

will produce maximum and minimum moments about

point A. What are the magnitudes of these moments?

u 10° … u … 180°2

u = 60°.

400 mm

300 mm

60&

50 N

12 5

Prob. 4–21

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ROBLEMS

137

1.5 m

20 m

P % 4 kN

Probs. 4–26/27

60&

30&

2 m

3 m

4 m

Probs. 4–28/29

% 52 lb

30&

20&

8 in.

6 in.

5 in.

Prob. 4–30

0.5 ft

4.5 ft

80 lb

40&

20&

Prob. 4–31

4–26. The towline exerts a force of

at the end of

the 20-m-long crane boom. If

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

at the end of

the 20-m-long crane boom. If

determine the

position of the boom so that this force creates a maximum

moment about point O. What is this moment?

x = 25 m,

P = 4 kN

u = 30°,

P = 4 kN

*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

4–29. If the resultant moment about point A is

clockwise, determine the magnitude of if

and

= 400 N.

= 300 N

4800 N

= 250 N, F

= 300 N, F

= 500 N.

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.

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?

ROBLEMS

137

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138

H A P T E R

4 F

O R C E

Y S T E M

E S U LTA N T S

20&

3 in.

1.5 in.

60&

14 in.

Prob. 4–32

0.4 m

0.2 m

0.5 m

500 N

Prob. 4–33

3 m

2 m

7 m

6 m

4 m

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

Probs. 4–34/35

6 m

2.5 m

3 m

4 m

8 m

6 m

F % 13 kN

Probs. 4–36/37

*4–32. If it takes a force of

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?

F = 125

4–33. The pipe wrench is activated by pulling on the

cable segment with a horizontal force of 500 N. Determine

the moment

produced by the wrench on the pipe when

Neglect the size of the pulley.

u = 20°.

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.

*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.

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3 m

1 m

2 m

3 m

F % 80 N

45&

Probs. 4–38/39

0.4 m

1.2 m

0.2 m

Prob. 4–40

18 ft

12 ft

30&

Prob. 4–41

1.5 m

3 m

4 m

10.5 m

Prob. 4–42

4–38. The curved rod lies in the x–y plane and has a radius

of 3 m. If a force of

acts at its end as shown,

determine the moment of this force about point O.

4–39. The curved rod lies in the x–y plane and has a radius

of 3 m. If a force of

acts at its end as shown,

determine the moment of this force about point B.

F = 80 N

*4–40. The force

acts at the

end of the beam. Determine the moment of the force about

point A.

F = 5600i + 300j - 600k6 N

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.

4–42. The man pulls on the rope with a force of

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.

F = 20 N.

ROBLEMS

139

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140

H A P T E R

4 F

O R C E

Y S T E M

E S U LTA N T S

5 ft

60&

6 ft

7 ft

Prob. 4–43

400 mm

300 mm

200 mm

250 mm

200 mm

30&

40&

F % 80 N

Prob. 4–44

400 mm

300 mm

200 mm

250 mm

200 mm

30&

40&

F % 80 N

Prob. 4–45

1.5 m

1.2 m

60&

Prob. 4–46

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

to be developed at C.

M = 80 lb

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

Determine the moment of this force about point A.

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.

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

Determine the moment of this force about point B.

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ROBLEMS

141

1 m

4 m

8 m

% {100i # 100j # 60k} N

% { # 500k } N

Prob. 4–47

4–47. Using Cartesian vector analysis, determine the

resultant moment of the three forces about the base of the

column at A. Take F

= 5400i + 300j + 120k6 N.

*4–48. A force of

produces a

moment of

about the origin

of coordinates, point O. If the force acts at a point having an

x coordinate of

determine the y and z coordinates.

x = 1 m,

= 54i + 5j - 14k6 kN

F = 56i - 2j + 1k6 kN

4–49. The force

creates a moment

about point O of

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

determine the perpendicular distance d from

point O to the line of action of F.

= Fd,

= 5-14i + 8j + 2k6 N

F = 56i + 8j + 10k6 N

4–50. A force of

produces a

moment of

about the origin

of coordinates, point O. If the force acts at a point having an

x coordinate of

determine the y and z coordinates.

x = 1 m,

= 54i + 5j - 14k6 kN

F = 56i - 2j + 1k6 kN

1 m

Prob. 4–48

1 m

Prob. 4–49

1 m

Prob. 4–50

ROBLEMS

141

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142

H A P T E R

4 F

O R C E

Y S T E M

E S U LTA N T S

0.4 m

0.5 m

0.3 m

F % 20 N

(a)

Fig. 4–21

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.

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

applied at point A.

The moment of this force about point O has a magnitude of

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

about the y axis,

since this component tends to unscrew

the pipe from the flange at O. From Fig. 4–21a,

has a magnitude of

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

and the direction is determined by the right-

hand 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

(4–10)

= Fd

= 0.3120 N2 = 6 N

110 N

m2 = 6 N

= 120 N210.5 m2 = 10 N

F = 20 N

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Here

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.

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.,

If the

handle is not in the horizontal position, then the moment about the z axis is

determined from

where

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,

then finding the projection or component of this

moment along z, i.e.,

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

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

then

This result, of

course, is to be expected, since it represents the j component of M

= M

= 1-8i + 6j2

j = 6 N

= j,

= r

* F = 10.3i + 0.4j2 * 1-20k2 = 5-8i + 6j6 N

= M

cos u.

= Fd,

d¿

= Fd¿,

max

= Fd.

)

max

0.4 m

0.3 m

F % {#20k} N

(b)

% j

Fig. 4–21 (cont.)

4.5 M

OMENT OF A

ORCE ABOUT A

PECIFIED

XIS

143

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144

H A P T E R

4 F

O R C E

Y S T E M

E S U LTA N T S

% r $ F

a¿

b¿

Moment axis

Axis of projection

Fig. 4–22

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

axis.This tendency

for rotation is measured by the moment component

To determine

we first compute the moment of F about any arbitrary point O that

lies on the

axis. In this case,

is expressed by the cross product

where r is directed from O to A. Here

acts along the

moment axis

and so the component or projection of

onto the

axis is then

The magnitude of

is determined by the dot product,

where

is a unit vector that defines the

direction of the

axis. Combining these two steps as a general expression,

we have

Since the dot product is commutative, we can

also write

In vector algebra, this combination of dot and cross product yielding the

scalar

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

or simply

(4–11)

where

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

defining the direction of the

axis

represent the x, y, z components of the position

vector drawn from any point O on the

axis to

any point A on the line of action of the force

represent the x, y, z components of the force

vector.

aa¿

, u

= u

1r * F2 = 3

= 1u

i + u

j + u

= u

1r * F2

= 1r * F2

aa¿

= M

cos u = M

aa¿

bb¿,

= r * F,

aa¿

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When

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

along

the

axis. If it is positive, then

will have the same sense as

whereas if it is negative, then

will act opposite to

Once

is determined, we can then express

as a Cartesian

vector, namely,

(4–12)

Finally, if the resultant moment of a series of forces is to be computed

about the

axis, then the moment components of each force are added

together algebraically, since each component lies along the same axis.

Thus the magnitude of

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

created about the

axis.The moment

of F about a point A that lies on the axis is

The projection of this

moment along the axis, whose direction is defined by the unit vector

Had this moment been calculated using scalar methods, then

the perpendicular distance from the force line of action to the

axis would have

to be determined, which in this case would be a more difficult task.

a–a

= u

1r * F2.

= r * F.

a–a

= ©[u

1r * F2] = u

aa¿

= M

= [u

1r * F2]u

aa¿

Important Points

•

The moment of a force about a specified axis can be determined

provided the perpendicular distance from both the force line

of action and the axis can be determined.

•

If vector analysis is used,

where

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.

•

is calculated as a negative scalar, then the sense of

direction of

is opposite to

•

The moment

expressed as a Cartesian vector is determined

from M

= M

= u

1r * F2,

= Fd

4.5 M

OMENT OF A

ORCE ABOUT A

PECIFIED

XIS

145

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146

H A P T E R

4 F

O R C E

Y S T E M

E S U LTA N T S

EXAMPLE 4.8

The force

acts at point A shown in

Fig. 4–23a. Determine the moments of this force about the x and a

axes.

SOLUTION I

(VECTOR ANALYSIS)

We can solve this problem by using the position vector

Why? Since

and

then applying Eq. 4–11,

= i,

= 5-3i + 4j + 6k6 m

F = 5-40i + 20j + 10k6 N

(b)

80 N ' m

120 N ' m

(c)

4 m

3 m

6 m

10 N 40 N

20 N

Fig. 4–23

Ans

= -80 N

= 1[41102-61202]-0[1-321102-61-402]+0[1-321202-41-402]

= i

* F2 = 3

-3

-40 20 10

The negative sign indicates that the sense of

is opposite to i.

We can compute

also using

because

extends from a point

on the a axis to the force. Also,

Thus,

= -

i +

Ans.

= -120 N

= -

[41102 - 61202]-

[1-321102-61-402]+0[1-321202-41-402]

= u

* F2 = 3

-3

-40 20 10

What does the negative sign indicate?

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

SOLUTION II

(SCALAR ANALYSIS)

Since the force components and moment arms are easy to determine

for computing

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

Although not required here, note also that

If we were to determine

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.

= 140 N214 m2 - 120 N213 m2 = 100 N

= 110 N213 m2 - 140 N216 m2 = -210 N

= 110 N214 m2 - 120 N216 m2 = -80 N

Ans.

(a)

4 m

3 m

6 m

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

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EXAMPLE 4.9

The rod shown in Fig. 4–24a is supported by two brackets at A and B. De-

termine the moment

produced by

which tends to rotate the rod about the AB axis.

SOLUTION

A vector analysis using

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

defines the direction of the AB axis of the rod,

Fig. 4–24b, where

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 and

are

suitable, Fig. 4–24b. (Although not shown,

can also be used.)

For simplicity, we choose

where

The force is

Substituting these vectors into the determinant form and expanding,

we have

The negative sign indicates that the sense of

is opposite to that of

Expressing

as a Cartesian vector yields

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

would have to be used.This

would lead to

Consequently,

and the above result would again be determined.

= M

1-u

= +53.67 N

-u

= 5-48.0i - 24.0j6 N

Ans.

= M

= 1-53.67 N

m210.894i + 0.447j2

= -53.67 N

+ 0[012002 - 0.21-6002]

= 0.894[0.21-3002 - 012002] - 0.447[01-3002 - 01-6002]

= u

* F2 = 3

0.894 0.447

0.2

-600

200

-300

F = 5-600i + 200j - 300k6 N

= 50.2j6 m

0.4i + 0.2j

210.42

+ 10.22

= 0.894i + 0.447j

= u

1r * F2

F = 5-600i + 200j - 300k6 N,

0.4 m

(a)

0.3 m

0.6 m

0.2 m

(b)

C (0.6 m, 0, 0.3 m)

B (0.4 m, 0.2 m, 0)

D (0, 0.2 m, 0)

Fig. 4–24

4.5 M

OMENT OF A

ORCE ABOUT A

PECIFIED

XIS

147

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148

H A P T E R

4 F

O R C E

Y S T E M

E S U LTA N T S

P R O B L E M S

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

4 m

3 m

6 m

2 m

1 m

Prob. 4–51

6 ft

3 ft

2 ft

4 ft

F % 600 lb

Prob. 4–52

60&

45&

120&

= 50 lb

= 80 lb

6 ft

4 ft

30&

5 ft

Prob. 4–53

3 in.

2 in.

8 in.

Prob. 4–54

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

*4–52. Determine the moment of the force F about the aa

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.

4–54. A force of

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.

F = 58i - 1j + 1k6 lb

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ROBLEMS

149

4–55. The 50-lb force acts on the gear in the direction

shown. Determine the moment of this force about the y axis.

*4–56. 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.

4–57. 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.

4–58. The hood of the automobile is supported by the strut

AB,which exerts a force of

on the hood.Determine

the moment of this force about the hinged axis y.

F = 24 lb

F % 50 lb

3 in.

60&

45&

120&

Prob. 4–55

30&

1.25 in.

13 lb

78 lb

Prob. 4–56

40&

30 mm

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

Prob. 4–57

2 ft

4 ft

2 ft

4 ft

Prob. 4–58

ROBLEMS

149

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150

H A P T E R

4 F

O R C E

Y S T E M

E S U LTA N T S

F % 30 N

0.25 m

0.3 m

0.1 m

0.5 m

Probs. 4–59/60

40 mm

30 mm

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

Prob. 4–61

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

A 30&

Prob. 4–62

2.5 m

4 m

1 m

0.5 m

2 m

1.5 m

2 m

Prob. 4–63

4–59. The lug nut on the wheel of the automobile is to be

removed using the wrench and applying the vertical force of

at A. Determine if this force is adequate, provided

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?

*4–60. 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.

14 N

F = 30 N

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.

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.

4–63. Determine the magnitude of the moment of the force

about the base line CA of the

tripod.

F = 550i - 20j - 80k6 N

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ROBLEMS

151

60&

10 in.

0.75 in.

Probs. 4–64/65

*4–64. The flex-headed ratchet wrench is subjected to a

force of

applied perpendicular to the handle as

shown. Determine the moment or torque this imparts along

the vertical axis of the bolt at A.

4–65. If a torque or moment of

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.

80 lb

in.

P = 16 lb,

4–66. The A-frame is being hoisted into an upright position

by the vertical force of

Determine the moment

of this force about the y axis when the frame is in the position

shown.

F = 80 lb.

4–67. Determine the moment of each force acting on

the handle of the wrench about the a axis. Take

= 53i + 2j - 6k6 lb.

= 5-2i + 4j - 8k6 lb,

30&

15&

6 ft

y¿

x¿

6 ft

Prob. 4–66

6 in.

45&

3.5 in.

4 in.

Prob. 4–68

6 in.

45&

3.5 in.

4 in.

Prob. 4–67

*4–68. Determine the moment of each force acting on

the handle of the wrench about the z axis. Take

= 53i + 2j - 6k6 lb.

= 5-2i + 4j - 8k6 lb,

ROBLEMS

151

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152

H A P T E R

4 F

O R C E

Y S T E M

E S U LTA N T S

Fig. 4–25

Fig. 4–26

Fig. 4–27

4.6

Moment of a Couple

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

and

are directed from point O to points A and B lying on the line of

action of

and F. The couple moment computed about O is therefore

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

about A is zero, and we have

(4–13)

The fact that we obtain the same result in both cases can be demonstrated

by noting that in the first case we can write

and by

the triangle rule of vector addition,

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

and

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

(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.,

(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

M = r * F

M = Fd

r = r

- r

+ r = r

M = 1r

- r

2 * F;

M = r * F

-F

M = r

* 1-F2 + r

* F

-F

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Fig. 4–26, the moment of

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

and

Fig. 4–28b, and then these free vectors may be

moved to the arbitrary point P and added to obtain the resultant couple

moment

shown in Fig. 4–28c.

= M

+ M

-F

(a)

(b)

(c)

Fig. 4–28

4.6 M

OMENT OF A

OUPLE

153

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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)

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

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

An equivalent couple moment of

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

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

couple moment is a free vector.

12-N

F¿ = 40 N.

12 N

m = F¿10.32,

12 N

F = 30 N.

12 N

m = F10.4 m2,

12 N

= ©1r * F2

154

H A P T E R

4 F

O R C E

Y S T E M

E S U LTA N T S

0.3 m

F¿

0.4 m

The frictional forces of the floor on the blades of the concrete finishing machine

create a couple moment

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,

is applied on the handle, although

it could be applied at any other point on the machine.

= Fd,

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Fig. 4–29a

4.6 M

OMENT OF A

OUPLE

155

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,

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.

M = r * F,

0.3 m

0.1 m

40 N

(a)

40 N

0.3 m

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.

(b)

24 N ' m

0.2 m

(c)

Fig. 4–29

SOLUTION

(SCALAR ANALYSIS)

The couple has a magnitude of

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

Ans.

F = 120 N

M = Fd

24 N

m = F10.2 m2

M = Fd = 4010.62 = 24 N

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156

H A P T E R

4 F

O R C E

Y S T E M

E S U LTA N T S

1 ft

(a)

150 lb

2 ft

3 ft

EXAMPLE 4.11

Determine the moment of the couple acting on the member shown in

Fig. 4–30a.

1 ft

(b)

120 lb

90 lb

2 ft

3 ft

(c)

M % 390 lb ' ft

Fig. 4–30

SOLUTION

(SCALAR ANALYSIS)

Here it is somewhat difficult to determine the perpendicular distance

between the forces and compute the couple moment as

Instead, we can resolve each force into its horizontal and vertical

components,

and

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,

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

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

one couple

has a moment of

and the other has a

moment of

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

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.

M = 270 lb

ft + 120 lb

ft = 390 lb

ft g,

= 120 lb 11 ft2 = 120 lb

ft.

= 90 lb 13 ft2 = 270 lb

M = Fd,

= 390 lb

ft g

d+M = 90 lb 13 ft2 + 120 lb 11 ft2

= 390 lb

ft g

Ans.

d+M = 120 lb 10 ft2 - 90 lb 12 ft2 + 90 lb 15 ft2 + 120 lb 11 ft2

1150 lb2 = 90 lb,

1150 lb2 = 120 lb

M = Fd.

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30&

25 lb

8 in.

6 in.

(a)

25 lb

(b)

25 lb

(c)

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.

6 in.

25 lb

(d)

30&

Fig. 4–31

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

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

SOLUTION II

(SCALAR ANALYSIS)

Although this problem is shown in three dimensions, the geometry is

simple enough to use the scalar equation

The perpendicular

distance between the lines of action of the forces is

Fig. 4–31d. Hence, taking moments of the

forces about either point A or B yields

Applying the right-hand rule, M acts in the

direction. Thus,

M = 5-130j6 lb

in.

Ans.

-j

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

in.

d = 6 cos 30° = 5.20 in.,

M = Fd.

= 5-130j6 lb

in.

Ans.

= 16 cos 30°i - 6 sin 30°k2 * 125k2

M = r

* 125k2

= 5-130j6 lb

in.

Ans.

= -200i - 129.9j + 200i

= 18j2 * 1-25k2 + 16 cos 30°i + 8j - 6 sin 30°k2 * 125k2

M = r

* 1-25k2 + r

* 125k2

4.6 M

OMENT OF A

OUPLE

157

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158

H A P T E R

4 F

O R C E

Y S T E M

E S U LTA N T S

EXAMPLE 4.13

Replace the two couples acting on the pipe column in Fig. 4–32a by a

resultant couple moment.

% 37.5 N ' m

% 60 N ' m

(b)

(c)

Fig. 4–32

0.3 m

150 N

125 N

0.4 m

150 N

(a)

SOLUTION

(VECTOR ANALYSIS)

The couple moment

developed by the forces at A and B, can

easily be determined from a scalar formulation.

By the right-hand rule,

acts in the

direction, Fig. 4–32b. Hence,

Vector analysis will be used to determine

caused by forces at C

and D. If moments are computed about point D, Fig. 4–32a,

then

Since

and

are free vectors, they may be moved to some

arbitrary point P and added vectorially, Fig. 4–32c. The resultant

couple moment becomes

NOTE:

Try to establish

by using a scalar formulation, Fig. 4–32b.

= M

+ M

= 560i + 22.5j + 30k6 N

Ans.

= 522.5j + 30k6 N

= 10.3i2 * [100j - 75k] = 301i * j2 - 22.51i * k2

= r

* F

= 10.3i2 *

125

j - 125

= r

* F

= 560i6 N

= Fd = 150 N10.4 m2 = 60 N

publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,

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ROBLEMS

159

3 ft

6 ft

4 ft

2 ft

1.5 ft

Prob. 4–70

150 lb

6 ft

8 ft

Prob. 4–71

1 ft

2 ft

7 ft

6 ft

12 ft

Prob. 4–72

4–70. Determine the magnitude and sense of the couple

moment. Each force has a magnitude of

F = 65 lb.

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

moment.

*4–72. If the couple moment has a magnitude of

determine the magnitude F of the couple forces.

300 lb

ft,

2 m

30&

5 kN

4 m

1 m

0.5 m

Prob. 4–69

P R O B L E M S

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

moment.

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160

H A P T E R

4 F

O R C E

Y S T E M

E S U LTA N T S

4 in.

2 in.

3 in.

Prob. 4–74

% 500 N ' m

% 900 N ' m

45&

Prob. 4–75

0.3 m

Prob. 4–76

4–73. A clockwise couple

is resisted by the

shaft of the electric motor. Determine the magnitude of the

reactive forces

and R which act at supports A and B so

that the resultant of the two couples is zero.

–R

M = 5 N

4–74. The resultant couple moment created by the two

couples acting on the disk is

Determine

the magnitude of force T.

= 510k6 kip

in.

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

Determine the magnitude of

and the bend angle so that

the resultant couple moment is zero.

*4–76. The floor causes a couple moment of

and

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 M

= 0?

= 30 N

= 40 N

150 mm

60&

Prob. 4–73

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200 mm

300 mm

400 mm

150 mm

200 mm

Probs. 4–79/80

1.5 m

1.8 m

45&

30&

2 kN

8 kN

0.3 m

8 kN

Prob. 4–81

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

clockwise.

400 N

4–78. Two couples act on the beam. Determine the

magnitude of F so that the resultant couple moment is

counterclockwise. Where on the beam does the

resultant couple moment act?

450 lb

ft,

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

*4–80. If the couple moment acting on the pipe has a

magnitude of

determine the magnitude F of the

vertical force applied to each wrench.

400 N

F = 525k6 N.

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.

250 N

600 N

1 m

40&

Prob. 4–77

200 lb

2 ft

1.5 ft

1.25 ft

30&

Prob. 4–78

ROBLEMS

161

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162

H A P T E R

4 F

O R C E

Y S T E M

E S U LTA N T S

3 m

2 m

3 m

5 m

4 m

F % {8i – 4j + 10k} N

"F % {–8i + 4j – 10k} N

Prob. 4–86

5 m

4 m

80 N

10 m

6 m

4 m

Prob. 4–87

4–82. Two couples act on the beam as shown. Determine

the magnitude of F so that the resultant couple moment is

counterclockwise. Where on the beam does the

resultant couple act?

300 lb

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.

*4–84. Two couples act on the frame. If

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

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.

d = 4 ft,

4–86. Determine the couple moment. Express the result as

a Cartesian vector.

4–87. Determine the couple moment. Express the result as

a Cartesian vector.

200 lb

1.5 ft

Prob. 4–82

2 ft

1 ft

3 ft

50 lb

80 lb

50 lb

30&

80 lb

3 ft

Probs. 4–83/84/85

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*4–88. If the resultant couple of the two couples acting on

the fire hydrant is

determine the

force magnitude P.

= 5-15i + 30j6 N

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.

4–90. If

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

determine the magnitude F of the forces applied to

the wrenches.

15 N

F = 5100k6 N,

*4–92. The gears are subjected to the couple moments

shown. Determine the magnitude and coordinate direction

angles of the resultant couple moment.

0.150 m

–Pi

0.2 m

{75j} N

{–75j} N

Prob. 4–88

600 mm

150 N

400 mm

500 mm

300 mm

Prob. 4–89

60&

300 mm

150 mm

200 mm

Probs. 4–90/91

% 40 lb ' ft

% 30 lb ' ft

20&

30&

15&

Prob. 4–92

ROBLEMS

163

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164

H A P T E R

4 F

O R C E

Y S T E M

E S U LTA N T S

0.5 m

1.5 m

0.5 m

0.8 m

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

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

Prob. 4–93

0.4 m

0.3 m

0.6 m

F = {–6i + 2j + 3k} N

"F % {6i – 2j – 3k} N

Prob. 4–94

60&

175 mm

35 N

25 N

35 N

175 mm

Prob. 4–95

600 mm

200 mm

150 mm

Probs. 4–96/97

4–93. Express the moment of the couple acting on the rod

in Cartesian vector form.What is the magnitude of the couple

moment?

4–94. 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.

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

Determine the magnitude and coordinate direction angles

of the resultant couple moment.

*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

4–97. If the couple moment acting on the pipe has a

magnitude of

determine the magnitude F of the

forces applied to the wrenches.

300 N

F = 125 N.

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Fig. 4–33a

4.7 E

QUIVALENT

YSTEM

165

(a)

(b)

(c)

Fig. 4–33

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

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.

-F

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166

H A P T E R

4 F

O R C E

Y S T E M

E S U LTA N T S

M $ r # 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

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

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

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.

M = Fd.

M = r * F.

-F

M % Fd

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4.8

Resultants of a Force and

Couple System

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

and

are

applied to the body.Furthermore,the couple moment

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

and the resultant couple

moment is

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 are independent

of the location of point O; however,

depends upon this location since

the moments

and

are determined using the position vectors and

Also note that

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.

(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

plus the moments about point O of all

the forces

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.

(4–18)

Note that the resultant force

is equivalent to the vector sum of its two

components

and F

= ©F

= ©M

+ ©M

©M

= ©F

= ©M

+ ©M

= M

+ M

= F

+ F

= r

* F

= r

* F

(a)

(b)

(c)

% r

$ F

% r

$ F

Fig. 4–35

4.8 R

ESULTANTS OF A

ORCE AND

OUPLE

YSTEM

167

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168

H A P T E R

4 F

O R C E

Y S T E M

E S U LTA N T S

% F

" F

% F

" F

d¿

% F

" F

% F

d¿

" F

d¿

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.

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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Fig. 4–36a

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.

SOLUTION

(SCALAR ANALYSIS)

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

As shown in Fig. 4–36b,

has a magnitude of

and a direction of

Ans.

Moment Summation.

The resultant couple moment

is determined by

summing the moments of the forces about point A. Assuming that positive

moments act counterclockwise, i.e., in the

direction, we have

Ans.

NOTE:

Realize that, when

and

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.

= -551 N

m = 551 N

m b

- 1400 cos 45° N210.3 m2

= 100 N102 - 600 N10.4 m2 - 1400 sin 45° N210.8 m2

d+M

= ©M

;

u = tan

-1

b = tan

-1

882.8
382.8

b = 66.6° ud

= 21F

+ 1F

= 21382.82

+ 1882.82

= 962 N

Ans.

= -600 N - 400 sin 45° N = -882.8 N = 882.8 NT

+ cF

= ©F

;

= -100 N - 400 cos 45° N = -382.8 N = 382.8 N ;

+ F

= ©F

;

45&

100 N

600 N

400 N

0.3 m

0.4 m

(a)

% 551 N ' m

(b)

66.6&

= 962 N

Fig. 4–36

4.8 R

ESULTANTS OF A

ORCE AND

OUPLE

YSTEM

169

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Fig. 4–37a

170

H A P T E R

4 F

O R C E

Y S T E M

E S U LTA N T S

% 800 N

0.1 m

% 300 N

0.15 m

1 m

M % 500 N ' m

(a)

EXAMPLE 4.15

A structural member is subjected to a couple moment M and forces

and

as shown in Fig. 4–37a. Replace this system by an equivalent

resultant force and couple moment acting at its base, point O.

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

Force Summation.

Ans.

Moment Summation.

Ans.

The results are shown in Fig. 4–37b.

= 5-166i - 650j + 300k6 N

= 1-400j + 300k2 + 102 + 1-166.4i - 249.6j2

= 1- 400j + 300k2 + 11k2 * 1- 800k2+ 3

-0.15

0.1

-249.6 166.4 0

= M + r

* F

+ r

* F

= ©M

+ ©M

= 5-249.6i + 166.4j - 800k6 N

= F

+ F

= -800k - 249.6i + 166.4j

= ©F;

M = -500

j + 500

k = 5-400j + 300k6 N

= 300c

-0.15i + 0.1j

21-0.152

+ 10.12

d = 5-249.6i + 166.4j6 N

= 1300 N2u

= 1300 N2a

= 5-800k6 N

(b)

Fig. 4–37

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4.9 F

URTHER

EDUCTION OF A

ORCE AND

OUPLE

YSTEM

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

and

resultant couple moment

which are perpendicular to one

another, Fig. 4–38b. Whenever this occurs, we can further simplify the force

and couple moment system by moving

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

and

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 and the aa

axis. This point is chosen such that the distance d satisfies the scalar equation

With

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

acting through a

specific point.This is because in each of these cases

and

will always be

perpendicular to each other when the force system is simplified at any point O.

d = M

= F

= ©M

= ©F

(a)

(b)

(c)

d %

Fig. 4–38

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172

H A P T E R

4 F

O R C E

Y S T E M

E S U LTA N T S

% (F

Fig. 4–39

(a)

Fig. 4–40a

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

is thus perpendicular to the resultant

force

Fig. 4–41b; and so

can be moved to a point a distance d away

so that it produces the same moment about O.

= ©M

+ ©1r * F2

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

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

direction. The resultant moment

is thus

perpendicular to the resultant force

Fig. 4–40b; and so

can be

positioned a distance d from O so as to create the same moment

about O, Fig. 4–40c.

= ©M + ©1r * F2

% (F

(b)

% (M " (r $ F

Fig. 4–40

% (F

(c)

Fig. 4–40c

publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,

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4.9 F

URTHER

EDUCTION OF A

ORCE AND

OUPLE

YSTEM

173

(a)

(b)

% (F

% (M " ( (r $ F)

(c)

Fig. 4–41

The three parallel forces acting on the stick can be replaced by a single resultant force

acting at a distance d from the grip.To be

equivalent we require the resultant force to equal the sum of the forces,

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, F

d = F

+ F

= F

+ F

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

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

H A P T E R

4 F

O R C E

Y S T E M

E S U LTA N T S

(b)

Fig. 4–42

(c)

Fig. 4–42c

(a)

Fig. 4–42a

Reduction to a Wrench.

In the general case, the force and couple

moment system acting on a body, Fig. 4–35a, will reduce to a single

resultant force

and couple moment

at O which are not

perpendicular. Instead,

will act at an angle from

Fig. 4–35c.As

shown in Fig. 4–42a, however,

may be resolved into two

components: one perpendicular,

and the other parallel

to the

line of action of

As in the previous discussion, the perpendicular

component

may be eliminated by moving

to point P, as shown in

Fig. 4–42b.This point lies on axis bb, which is perpendicular to both

and

In order to maintain an equivalency of loading, the distance from

O to P is

Furthermore, when

is applied at P, the moment

tending to cause rotation of the body about O is in the same

direction as

Fig. 4–42a. Finally, since

is a free vector, it may be

moved to P so that it is collinear with

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.

d = M

(a)

(c)

Fig. 4–35 (Repeated)

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Fig. 4–43a

4.9 F

URTHER

EDUCTION OF A

ORCE AND

OUPLE

YSTEM

175

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.

500 N

60&

200 N

(a)

2 m

1.5 m

1 m

100 N

0.5 m

(b)

350.0 N

233.0 N

% 420.5 N

u % 33.7&

Fig. 4–43

SOLUTION

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

The magnitude and direction of the resultant force are established from the

vector addition shown in Fig. 4–43b. We have

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 (or the

moment of ) about point E to equal the moments of the force system

about E. Assuming positive moments are counterclockwise, we have

Ans.

NOTE:

Using a clockwise sign convention would yield this same result.

Since d is positive,

acts to the left of E as shown. Try to solve this problem

by summing moments about point A and show

measured to

the right of A.

d¿ = 0.927 m,

d =

1182.1

233.0

= 5.07 m

- 1100 N210.5 m2 - 1200 N212.5 m2

= 1500 sin 60° N214 m2 + 1500 cos 60° N2102

233.0 N1d2 + 350.0 N102

d+M

= ©M

;

u = tan

-1

233.0
350.0

b = 33.7° cu

= 21350.02

+ 1233.02

= 420.5 N

= 233.0 NT

= -500 sin 60° N + 200 N = -233.0 N

+ cF

= ©F

;

= 500 cos 60° N + 100 N = 350.0 N :

+ F

= ©F

;

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Fig. 4–44a

176

H A P T E R

4 F

O R C E

Y S T E M

E S U LTA N T S

(x,y)

260 lb

325 lb

260 lb

325 lb

(b)

Fig. 4–44

6 ft

5 ft

175 lb

60 lb

(a)

250 lb

5 4

3 ft

5 ft

3 ft

EXAMPLE 4.17

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

As shown by the vector addition in Fig. 4–44b,

Ans.

Moment Summation.

Moments will be summed about the arbitrary

point A.Assuming the line of action of

intersects AB, Fig. 4–44b, we

require the moment of the components of

in Fig. 4–44b about A to

equal the moments of the force system in Fig. 4–44a about A; i.e.,

Ans.

By the principle of transmissibility,

can also be treated as

intersecting BC, Fig. 4–44b, in which case we have

Ans.

NOTE:

We can also solve for these positions by assuming

acts at

the arbitrary point (x, y) on its line of action, Fig. 4–44b. Summing

moments about point A yields

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

then

and along BC set

then x = 10.9 ft.

y = 11 ft,

y = 2.29 ft,

x = 0,

325y - 260x = 745

= 175 lb 15 ft2 - 60 lb 13 ft2 + 250 lb

111 ft2 - 250 lb

18 ft2

325 lb 1y2 - 260 lb 1x2

d+M

= ©M

;

x = 10.9 ft

+ 250 lb

111 ft2 - 250 lb

18 ft2

= 175 lb 15 ft2 - 60 lb 13 ft2

325 lb 111 ft2 - 260 lb 1x2

d+M

= ©M

;

y = 2.29 ft

= 175 lb 15 ft2 - 60 lb 13 ft2 + 250 lb

111 ft2 - 250 lb

18 ft2

325 lb 1y2 + 260 lb 102

d+M

= ©M

;

u = tan

-1

260
325

b = 38.7° ud

= 213252

+ 12602

= 416 lb

= -250 lb

- 60 lb = -260 lb = 260 lbT

+ cF

= ©F

;

= -250 lb

- 175 lb = -325 lb = 325 lb ;

+ F

= ©F

;

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.

Fig. 4–45a

4.9 F

URTHER

EDUCTION OF A

ORCE AND

OUPLE

YSTEM

177

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.

(b)

P(x, y)

Fig. 4–45

2 m

600 N

500 N

100 N

5 m

400 N

8 m

(a)

SOLUTION

(SCALAR ANALYSIS)

Force Summation.

From Fig. 4–45a, the resultant force is

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

direction, we have

Ans.

In a similar manner, assuming that positive moments act in the

direction, a moment equation can be written about the y axis using

moment arms defined by the x coordinates of each force.

Ans.

NOTE:

A force of

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.

= 1400 N

1400x = 4200

x = 3.00 m

11400 N2x = 600 N18 m2 - 100 N16 m2 + 400 N102 + 500 N102

= ©M

;

-1400y = -3500

y = 2.50 m

-11400 N2y = 600 N102 + 100 N15 m2 - 400 N110 m2 + 500 N102

= ©M

;

= -1400 N = 1400 NT

= -600 N + 100 N - 400 N - 500 N

+ cF

= ©F;

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Fig. 4–46a

178

H A P T E R

4 F

O R C E

Y S T E M

E S U LTA N T S

EXAMPLE 4.19

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.

SOLUTION (

VECTOR ANALYSIS)

Force Summation.

From Fig. 4–46a, the force resultant

Ans.

Moment Summation.

Choosing point O as a reference for

computing moments and assuming that

acts at a point P(x, y),

Fig. 4–46b, we require

Equating the corresponding j and i components yields

(1)
(2)

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

Ans.

The negative sign indicates that it was wrong to have assumed a

position for

as shown in 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

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

= ©M

;

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

= ©M

;

x = 2.39 ft

y = -1.16 ft

-650y = 1600 - 848.5

650x = 2400 - 848.5

650xj - 650yi = 2400j + 1600i - 848.5j - 848.5i

+ 1-8 sin 45°i + 8 cos 45°j2 * 1-150k2

1xi + yj2 * 1-650k2 = 18i2 * 1-300k2 + 1-8j2 * 1-200k2

r * F

= r

* 1-300k2 + r

* 1-200k2 + r

* 1-150k2

= ©M

;

= 5-650k6 lb

= -300k - 200k - 150k

= ©F;

200 lb

300 lb

150 lb

45&

8 ft

(a)

(b)

P (x, y)

Fig. 4–46

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mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department.

30 lb

40 lb

60 lb

1 ft

3 ft

2 ft

Probs. 4–104/105

ROBLEMS

179

2 m

375 N

30&

4 m

1 m

2 m

Probs. 4–98/99

60 lb

4 ft

3 ft

2 ft

85 lb

25 lb

45&

6 ft

5 4

Probs. 4–100/101

P R O B L E M S

4–98. Replace the force at A 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.

*4–100. Replace the force system 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.

4–102. Replace the force system by an equivalent force and

couple moment at point O.

4–103. Replace the force system by an equivalent force and

couple moment at point P.

*4–104. Replace the loading system acting on the post by

an equivalent resultant force and couple moment at point O.

4–105. Replace the loading system acting on the post by an

equivalent resultant force and couple moment at point P.

5 ft

430 lb

60&

2 ft

8 ft

3 ft

2 ft

260 lb

Probs. 4–102/103

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mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department.

180

H A P T E R

4 F

O R C E

Y S T E M

E S U LTA N T S

200 lb

500 lb

260 lb

5 ft

3 ft

2 ft

4 ft

Probs. 4–112/113

120 mm

800 mm

Probs. 4–110/111

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.

*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.

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.

4–110. The forces and couple moments which are exerted

on the toe and heel plates of a snow ski are

and

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

and

respectively. Replace this system by an equivalent force and

couple moment acting at point P. Express the results in

Cartesian vector form.

= 5-20i + 8j + 3k6 N

= 5-20i + 60j - 250k6 N, M

5- 6i + 4j + 2k6 N

= 5-50i + 80j - 158k6 N, M

= 5-20i + 8j + 3k6 N

= 5-20i + 60j - 250k6 N, M

= 5-6i + 4j + 2k6 N

= 5-50i + 80j - 158k6 N, M

*4–112. Replace the three forces acting on the shaft by a

single resultant force. Specify where the force acts, measured

from end A.

4–113. Replace the three forces acting on the shaft by a

single resultant force. Specify where the force acts, measured

from end B.

6 kN

4 kN

60&

5 m

4 m

3 m

8 kN ' m

Probs. 4–106/107

2 ft

6 ft

3 ft

4 ft

850 lb

350 lb

125 lb

Probs. 4–108/109

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.

2 m

300 N

30&

60&

1500 N ' m

4 m

3 m

450 N

700 N

Probs. 4–115/116

1.5 m

200 N ' m

200 N

2 m

450 N

30&

1.5 m

0.2 m

Prob. 4–117

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.

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.

4–117. Replace the loading system acting on the beam by

an equivalent resultant force and couple moment at point O.

4–118. Determine the magnitude and direction 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

4–119. Determine the magnitude and direction 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

50 kN

7 ft

2 ft

4 ft

3 ft

300 lb

200 lb

400 lb

200 lb

600 lb ' ft

Prob. 4–114

3 m

4 m

6 m

5 kN

3 kN

)

Probs. 4–118/119

ROBLEMS

181

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.

182

H A P T E R

4 F

O R C E

Y S T E M

E S U LTA N T S

300 mm

)

r % 80 mm

Probs. 4–128/129

*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.

4–121. Replace the loading on the frame by a single

resultant force. Specify where its line of action intersects

member CD, measured from end C.

4–122. 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.

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.

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.

4–126. 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.

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.

*4–128. The belt passing over the pulley is subjected to forces

and

each having a magnitude of 40 N.

acts in the

direction. Replace these forces by an equivalent force and

couple moment at point A. Express the result in Cartesian

vector form. Set

so that

acts in the

direction.

4–129. The belt passing over the pulley is subjected to two

forces and

each having a magnitude of 40 N.

acts in

the

direction. Replace these forces by an equivalent force

and couple moment at point A. Express the result in

Cartesian vector form. Take u = 45°.

-k

-j

u = 0°

-k

400

250 N

300 N

1 m

2 m

3 m

Probs. 4–120/121

2 ft

4 ft

3 ft

25 lb

2 ft

20 lb

30&

35 lb

Probs. 4–122/123/124

5 m

6 m

4 m

3 m

F % {8i " 6j " 8k} kN

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

6 m

3 m

5 m

Probs. 4–125/126/127

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0.25 m

0.3 m

% {6i # 3j # 10k} N

% {2j # 4k} N

0.15 m

Prob. 4–131

75 mm

45 mm

50 mm 40 mm

30 mm

15 mm

Prob. 4–132

4–130. A force and couple act on the pipe assembly. If

and

replace this system by an

equivalent resultant force and couple moment acting at O.

Express the results in Cartesian vector form.

= 80 N,

= 50 N

4–131. Handle forces and 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.

*4–132. A biomechanical model of the lumbar region of

the human trunk is shown. The forces acting in the four

muscle groups consist of

for the rectus,

for the oblique,

for the lumbar latissimus dorsi,

and

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.

= 32 N

= 23 N

= 45 N

= 35 N

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

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 F

= 20 kN, F

= 50 kN.

= 30 kN, F

= 40 kN.

1.25 m

180 N

0.75 m

0.5 m

Prob. 4–130

20 kN

3 m

2 m

8 m

6 m

4 m

50 kN

Probs. 4–133/134

ROBLEMS

183

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184

H A P T E R

4 F

O R C E

Y S T E M

E S U LTA N T S

% {#60j} lb

% {#40i} lb

% {#80k} lb

12 ft

Prob. 4–138

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.

*4–136. 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.

4–137. 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.

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.

0.25 m

0.8 m

{60k} N

0.6 m

0.7 m

0.5 m

0.3 m

#F i

0.3 m

0.25 m

{#60k} N

{#40i} N

Prob. 4–135

2 ft

6 ft

Prob. 4–136

4 m

6 m

% {500i} N

% {300j} N

% {800k} N

Prob. 4–137

publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,

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4.10 R

EDUCTION OF A

IMPLE

ISTRIBUTED

OADING

185

p % p(x)

(a)

Fig. 4–47a

(b)

w % w(x)

Fig. 4–47b

4.10

Reduction of a Simple Distributed

Loading

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

or pascals

(Pa), where

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,

Pa, is only a function of x since the

pressure is uniform along the y axis. If we multiply

by the

width a m of the plate, we obtain

This loading function, shown in Fig. 4–47b, is a measure of load

distribution along the line

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

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 and its location can be specified, Fig. 4–47c.

Magnitude of Resultant Force.

From Eq. 4–17

the

magnitude of

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,

In other words, the magnitude of dF is

determined from the colored differential area dA under the loading curve.

For the entire plate length,

(4–19)

Hence, the magnitude of the resultant force is equal to the total area A

under the loading diagram

Fig. 4–47c.

w = w1x2,

w1x2 dx =

dA = A

+ TF

= ©F;

dF = w1x2 dx = dA.

= ©F2,

w = w1x2

y = 0

w = [p1x2 N>m

]a m = w1x2 N>m.

p = p1x2

1 Pa = 1 N>m

lb>ft

*The more general case of a nonuniform surface loading acting on a body is considered

in Sec. 9.5.

(c)

Fig. 4–47c

(d)

Fig. 4–47

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186

H A P T E R

4 F

O R C E

Y S T E M

E S U LTA N T S

Location of Resultant Force.

Applying Eq. 4–17

the location of the line of action of

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

about O, Fig. 4–47b, then for the entire plate, Fig. 4–47c,

Solving for using Eq. 4–19, we can write

(4–20)

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 distributed-

loading diagram w(x), Fig. 4–47c.

Once is determined,

by symmetry passes through point

on the surface of the plate, Fig. 4–47d. If we now consider the three-

dimensional 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

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

then the resultant force is determined from the area of the rectangle,

The line of action of this force passes through the centroid or center of

this area,

This resultant is equivalent to the distributed load, and so

both loadings produce the same “external” effects or support reactions on the beam.

x = a + b>2.

= w

p = p1x2

1x, 02

x = L

xw1x2 dx

w1x2 dx

= L

x dA

xw1x2 dx

e + M

= ©M

;

x dF = xw1x2 dx

= ©M

a " b/2

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Fig. 4–48a

4.10 R

EDUCTION OF A

IMPLE

ISTRIBUTED

OADING

187

Important Points

•

Distributed loadings are defined by using a loading function

that indicates the intensity

of the loading along the length of the member.This intensity is measured in

•

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.

lb>ft.

N>m

w = w1x2

EXAMPLE 4.20

Determine the magnitude and location of the equivalent resultant

force acting on the shaft in Fig. 4–48a.

SOLUTION

Since

is given, this problem will be solved by integration.

The colored differential area element

Applying Eq. 4–19, by summing these elements from

we obtain the resultant force

Ans.

Since the element of area dA is located an arbitrary distance x from

O, the location of

measured from O, Fig. 4–48b, is determined

from Eq. 4–20.

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,

3
4

12 m2 = 1.5 m

A =

2 m1240 N>m2

= 160 N and x =

3
4

= 1.5 m

x = L

x dA

= L

x160x

2 dx

160

60c

160

60c

160

= 160 N

dA =

60x

dx = 60c

= 60c

= ©F;

x = 2 m,

x = 0

dA = w dx = 60x

dx.

w = w1x2

w % (60 x

)N/m

(a)

dA % w dx

2 m

240 N/m

(b)

x % 1.5 m

% 160 N

Fig. 4–48

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Fig. 4–49a

188

H A P T E R

4 F

O R C E

Y S T E M

E S U LTA N T S

(a)

7200 Pa

9 m

0.2 m

p = 800x Pa

EXAMPLE 4.21

A distributed loading of

Pa acts over the top surface of the

beam shown in Fig. 4–49a. Determine the magnitude and location of

the equivalent resultant force.

p = 800x

w % 160x N/m

(b)

9 m

1440 N/m

Fig. 4–49b

% 6.48 kN

3 m

x % 6 m

(c)

Fig. 4–49

SOLUTION

The loading function

Pa indicates that the load intensity varies

uniformly from

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

note that

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.

Ans.

The line of action of

passes through the centroid C of the

triangle. Hence,

Ans.

The results are shown in Fig. 4–49c.

NOTE:

We may also view the resultant

as acting through the

centroid of the volume of the loading diagram

Fig. 4–49a. Hence

intersects the x–y plane at the point (6 m, 0).

Furthermore, the magnitude of

is equal to the volume under the

loading diagram; i.e.,

Ans.

= V =

17200 N>m

219 m210.2 m2 = 6.48 kN

p = p1x2

x = 9 m -

19 m2 = 6 m

19 m211440 N>m2 = 6480 N = 6.48 kN

w = 1440 N>m.

x = 9 m,

= 1160x2 N>m

w = 1800x N>m

210.2 m2

x = 9 m.

p = 7200 Pa

x = 0

p = 0

p = 800x

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4.10 R

EDUCTION OF A

IMPLE

ISTRIBUTED

OADING

189

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,

The lines of action of these parallel forces act through the centroid of

their associated areas and therefore intersect the beam at

The two parallel forces

and

can be reduced to a single

resultant

The magnitude of

Ans.

With reference to point A, Fig. 4–50b and 4–50c, we can find the

location of

We require

Ans.

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

NOTE:

Using these results, show that again

and x = 4 ft.

= 675 lb

19 ft2 = 3 ft

19 ft2150 lb>ft2 = 225 lb

19 ft21100 lb>ft2 = 450 lb

x = 4 ft

x16752 = 312252 + 4.514502

e + M

= ©M

;

= 225 + 450 = 675 lb

+ TF

= ©F;

19 ft2 = 4.5 ft

19 ft2 = 3 ft

= 19 ft2150 lb>ft2 = 450 lb

19 ft2150 lb>ft2 = 225 lb

100 lb/ft

50 lb/ft

9 ft

(a)

Fig. 4–50a

50 lb/ft

9 ft

(d)

100 lb/ft

Fig. 4–50

9 ft

(b)

50 lb/ft

Fig. 4–50b

(c)

Fig. 4–50c

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190

H A P T E R

4 F

O R C E

Y S T E M

E S U LTA N T S

P R O B L E M S

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

Determine the magnitude of the equivalent resultant

location, measured from point O.

*4–140. Replace the loading by an equivalent resultant

force and couple moment acting at point A.

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.

2 lb/ft

3.5 lb/ft

2.75 ft

4 ft

1.5 ft

Prob. 4–139

2.5 m

600 N/m

2.5 m

Prob. 4–140

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.

7.5 m

4.5 m

500 kN

6 kN/m

15 kN

Probs. 4–141/142

3000 lb

9 ft

80 lb/ft

200 lb/ft

Prob. 4–143

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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.

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

clockwise.

8 kN

15 kN/m

9 m

5 kN/m

Prob. 4–144

15 ft

30&

800 lb/ft

Prob. 4–145

2.5 kN/m

1 kN/m

3 m

1.5 m

1.5 kN/m

Prob. 4–146

9 m

4 kN/m

2.5 kN/m

Prob. 4–147

*4–144. Replace the loading by an equivalent force and

couple moment acting at point O.

4–145. Replace the distributed loading by an equivalent

resultant force, and specify its location on the beam,

measured from the pin at C.

ROBLEMS

191

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192

H A P T E R

4 F

O R C E

Y S T E M

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.

4–149. 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.

4–150. 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.

4–151. Replace the loading by an equivalent resultant force

and specify its location on the beam, measured from point B.

12 ft

9 ft

100 lb/ft

50 lb/ft

300 lb/ft

Prob. 4–149

6 ft

10 ft

60 lb/ft

40 lb/ft

Prob. 4–150

12 ft

9 ft

500 lb/ft

800 lb/ft

Prob. 4–151

3 m

2 m

800 N/m

200 N/m

Prob. 4–148

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4.5 m

3 m

20 kN/m

7.5 kN/m

Prob. 4–154

w % ( x

) kN/m

3 kN/m

3 m

Prob. 4–155

10 m

500 N/m

w % (0.5x

) N/m

Prob. 4–156

6 m

100 N/m

200 N/m

5 m

Probs. 4–152/153

*4–152. Replace the distributed loading by an equivalent

resultant force and specify where its line of action intersects

member AB, measured from A.

4–153. Replace the distributed loading by an equivalent

resultant force and specify where its line of action intersects

member BC, measured from C.

4–154. Replace the loading by an equivalent resultant force

and couple moment acting at point O.

4–155. Determine the equivalent resultant force and couple

moment 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

Simplify this distributed loading to an

equivalent resultant force and specify the magnitude and

location of the force, measured from A.

w = 10.5x

2 N>m.

ROBLEMS

193

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194

H A P T E R

4 F

O R C E

Y S T E M

E S U LTA N T S

4–157. Determine the equivalent resultant force and its

location, measured from point O.

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.

4–159. 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.

*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 % w

sin x

Prob. 4–157

3 ft

w % (86x

) lb/ft

Prob. 4–158

w % 12(1 " 2x

) lb/ft

0.5 ft

12 lb/ft

Prob. 4–159

5x " (16 " x

)

1/2

kN/m

w %

2 kN/m

5.07 kN/m

3 m

1 m

Prob. 4–160

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HAPTER

EVIEW

195

Chapter Review

Moment axis

Moment of Force—Scalar Definition
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.

The direction of the moment is defined

using the right-hand rule.

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

will be zero.

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.)

= Fd = F

y - F

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.

r is a position vector that extends from

point O to any point A, B, C on the line of

action of F.

= Fd

= r

* F = r

* 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.)

= r * F = 3

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196

H A P T E R

4 F

O R C E

Y S T E M

E S U LTA N T S

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

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.

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.

= d

In three dimensions, the scalar triple

product should be used. Here 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

calculated as a negative scalar, then the

sense of direction of

is opposite to

(See pages 142–145.)

= u

1r * F2 = 3

a¿

Axis of projection

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.

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.)

= Fd

= r * F

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Simplification of a Force and Couple

System
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,

and the resultant

couple moment is equal to the sum of all the

moments and couple moments about the

point, M

= ©M

+ ©M

= ©F,

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.

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.)

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.)

d %

M""

w % w(x)

HAPTER

EVIEW

197

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198

H A P T E R

4 F

O R C E

Y S T E M

E S U LTA N T S

*4–164. Determine the moment of the force about the

door hinge at A. Express the result as a Cartesian vector.

4–165. Determine the magnitude of the moment of the

force about the hinged axis aa of the door.

4–166. A force of

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.

F = 80 N

8 m

4 m

6 m

4 m

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

Prob. 4–163

0.5 m

1 m

30&

2.5 m

1.5 m

% 250 N

Probs. 4–164/165

15&

200 mm

300 mm

100 mm

F % 80 N

15&

Prob. 4–166

10 in.

F % 20 lb

6 in.

8 in.

Probs. 4–161/162

R E V I E W P R O B L E M S

4–161. Determine the coordinate direction angles

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

Express the result as a Cartesian vector.

g = 45°.

b = 120°,

a = 60°,

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.

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4–167. Replace the force F having a magnitude of

and acting at point A by an equivalent force and

couple moment at point C.

F = 50 lb

*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?

30 ft

10 ft

15 ft

20 ft

10 ft

Prob. 4–167

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

of the

moment 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.

50 mm

200 mm

10 mm

30 N

45&

Prob. 4–168

50 mm

200 mm

10 mm

30 N

45&

Prob. 4–169

6 in.

4 in.

3 in.

10 lb

30&

10 lb

Prob. 4–170

EVIEW

ROBLEMS

199

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