CHAPTER 3
PROBLEM 3.1
A 20-lb force is applied to the control rod AB as shown. Knowing that the length of
the rod is 9 in. and that α = 25°, determine the moment of the force about Point B
by resolving the force into horizontal and vertical components.
SOLUTION
Free-Body Diagram of Rod AB:
x = (9 in.) cos 65°
= 3.8036 in.
y = (9 in.)sin 65°
= 8.1568 in.
F = Fx i + Fy j
rA/B
= (20 lb cos 25°)i + ( −20 lb sin 25°) j
= (18.1262 lb)i − (8.4524 lb) j

= BA = ( −3.8036 in.)i + (8.1568 in.)j
M B = rA /B × F
= (−3.8036i + 8.1568 j) × (18.1262i − 8.4524 j)
= 32.150k − 147.852k
= −115.702 lb-in.
M B = 115.7 lb-in.

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161
PROBLEM 3.2
A 20-lb force is applied to the control rod AB as shown. Knowing that the length of
the rod is 9 in. and that α = 25°, determine the moment of the force about Point B by
resolving the force into components along AB and in a direction perpendicular to AB.
SOLUTION
Free-Body Diagram of Rod AB:
θ = 90° − (65° − 25°)
= 50°
Q = (20 lb) cos 50°
= 12.8558 lb
M B = Q (9 in.)
= (12.8558 lb)(9 in.)
= 115.702 lb-in.
M B = 115.7 lb-in.

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162
PROBLEM 3.3
A 20-lb force is applied to the control rod AB as shown. Knowing that the length of the
rod is 9 in. and that the moment of the force about B is 120 lb · in. clockwise, determine
the value of α.
SOLUTION
Free-Body Diagram of Rod AB:
α = θ − 25°
Q = (20 lb) cos θ
and
Therefore,
M B = (Q )(9 in.)
120 lb-in. = (20 lb)(cos θ )(9 in.)
120 lb-in.
cos θ =
180 lb-in.
or
θ = 48.190°
Therefore,
α = 48.190° − 25°
α = 23.2° 
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163
PROBLEM 3.4
A crate of mass 80 kg is held in the position shown. Determine
(a) the moment produced by the weight W of the crate about E,
(b) the smallest force applied at B that creates a moment of
equal magnitude and opposite sense about E.
SOLUTION
(a)
By definition,
We have
W = mg = 80 kg(9.81 m/s 2 ) = 784.8 N
ΣM E : M E = (784.8 N)(0.25 m)
M E = 196.2 N ⋅ m
(b)

For the force at B to be the smallest, resulting in a moment (ME) about E, the line of action of force FB
must be perpendicular to the line connecting E to B. The sense of FB must be such that the force
produces a counterclockwise moment about E.
Note:
We have
d = (0.85 m) 2 + (0.5 m) 2 = 0.98615 m
ΣM E : 196.2 N ⋅ m = FB (0.98615 m)
FB = 198.954 N
and
 0.85 m 
 = 59.534°
 0.5 m 
θ = tan −1 
FB = 199.0 N
or
59.5° 
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164
PROBLEM 3.5
A crate of mass 80 kg is held in the position shown. Determine
(a) the moment produced by the weight W of the crate about E,
(b) the smallest force applied at A that creates a moment of equal
magnitude and opposite sense about E, (c) the magnitude, sense,
and point of application on the bottom of the crate of the smallest
vertical force that creates a moment of equal magnitude and
opposite sense about E.
SOLUTION
First note. . .
W = mg = (80 kg)(9.81 m/s 2 ) = 784.8 N
(a)
We have
M E = rH /EW = (0.25 m)(784.8 N) = 196.2 N ⋅ m
(b)
(c)
or M E = 196.2 N ⋅ m

For FA to be minimum, it must be perpendicular to the line
joining Points A and E. Then with FA directed as shown,
we have (− M E ) = rA/E ( FA )min .
Where
rA /E = (0.35 m)2 + (0.5 m)2 = 0.61033 m
then
196.2 N ⋅ m = (0.61033 m)( FA )min
or
( FA ) min = 321 N
Also
tan φ =
0.35 m
0.5 m
or
φ = 35.0°
(FA ) min = 321 N
35.0° 
For Fvertical to be minimum, the perpendicular distance from its line of action to Point E must be
maximum. Thus, apply (Fvertical)min at Point D, and then
(− M E ) = rD / E ( Fvertical ) min
or (Fvertical )min = 231 N
196.2 N ⋅ m = (0.85 m)( Fvertical ) min
at Point D 
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165
PROBLEM 3.6
A 300-N force P is applied at Point A of the bell crank shown.
(a) Compute the moment of the force P about O by resolving
it into horizontal and vertical components. (b) Using the
result of part (a), determine the perpendicular distance from
O to the line of action of P.
SOLUTION
x = (0.2 m) cos 40°
= 0.153209 m
y = (0.2 m)sin 40°
= 0.128558 m
∴ rA /O = (0.153209 m)i + (0.128558 m) j
(a)
Fx = (300 N)sin 30°
= 150 N
Fy = (300 N) cos 30°
= 259.81 N
F = (150 N)i + (259.81 N) j
M O = rA/ O × F
= (0.153209i + 0.128558 j) m × (150i + 259.81j) N
= (39.805k − 19.2837k ) N ⋅ m
= (20.521 N ⋅ m)k
(b)
M O = 20.5 N ⋅ m

M O = Fd
20.521 N ⋅ m = (300 N)(d )
d = 0.068403 m
d = 68.4 mm 
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166
PROBLEM 3.7
A 400-N force P is applied at Point A of the bell crank shown. (a) Compute
the moment of the force P about O by resolving it into components along
line OA and in a direction perpendicular to that line. (b) Determine the
magnitude and direction of the smallest force Q applied at B that has the
same moment as P about O.
SOLUTION
(a)
Portion OA of crank:
θ = 90° − 30° − 40°
θ = 20°
S = P sin θ
= (400 N) sin 20°
= 136.81 N
M O = rO /A S
= (0.2 m)(136.81 N)
= 27.362 N ⋅ m
(b)
M O = 27.4 N ⋅ m

Smallest force Q must be perpendicular to OB.
Portion OB of crank:
M O = rO /B Q
M O = (0.120 m)Q
27.362 N ⋅ m = (0.120 m)Q
Q = 228 N
42.0° 
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167
PROBLEM 3.8
It is known that a vertical force of 200 lb is required to remove the nail
at C from the board. As the nail first starts moving, determine (a) the
moment about B of the force exerted on the nail, (b) the magnitude of
the force P that creates the same moment about B if α = 10°, (c) the
smallest force P that creates the same moment about B.
SOLUTION
(a)
M B = rC/B FN
We have
= (4 in.)(200 lb)
= 800 lb ⋅ in.
or MB = 800 lb ⋅ in. 
(b)
By definition,
M B = rA/B P sin θ
θ = 10° + (180° − 70°)
= 120°
Then
800 lb ⋅ in. = (18 in.) × P sin120°
or P = 51.3 lb 
(c)
For P to be minimum, it must be perpendicular to the line joining
Points A and B. Thus, P must be directed as shown.
Thus
M B = dPmin
d = rA/B
or
or
800 lb ⋅ in. = (18 in.)Pmin
Pmin = 44.4 lb
Pmin = 44.4 lb
20° 
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168
PROBLEM 3.9
It is known that the connecting rod AB exerts on the crank BC a 500-lb force
directed down and to the left along the centerline of AB. Determine the moment of
the force about C.
SOLUTION
Using (a):
M C = y1 ( FAB ) x + x1 ( FAB ) y
 7

 24

= (2.24 in.)  × 500 lb  + (1.68 in.)  × 500 lb 
 25

 25

= 1120 lb ⋅ in.
(a)
MC = 1.120 kip ⋅ in.

Using (b):
M C = y2 ( FAB ) x
 7

= (8 in.)  × 500 lb 
 25

= 1120 lb ⋅ in.
(b)
M C = 1.120 kip ⋅ in.

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169
PROBLEM 3.10
It is known that the connecting rod AB exerts on the crank BC a 500-lb force directed
down and to the left along the centerline of AB. Determine the moment of the force
about C.
SOLUTION
Using (a):
M C = − y1 ( FAB ) x + x1 ( FAB ) y
 7

 24

= −(2.24 in.)  × 500 lb  + (1.68 in.)  × 500 lb 
 25

 25

= +492.8 lb ⋅ in.
(a)
MC = 493 lb ⋅ in.

Using (b):
M C = y2 ( FAB ) x
 7

= (3.52 in.)  × 500 lb 
 25

= +492.8 lb ⋅ in.
(b)
M C = 493 lb ⋅ in.

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170
PROBLEM 3.11
A winch puller AB is used to straighten a fence post. Knowing that the tension in cable BC is 1040 N and
length d is 1.90 m, determine the moment about D of the force exerted by the cable at C by resolving that
force into horizontal and vertical components applied (a) at Point C, (b) at Point E.
SOLUTION
(a)
Slope of line:
Then
and
Then
EC =
0.875 m
5
=
1.90 m + 0.2 m 12
12
(TAB )
13
12
= (1040 N)
13
= 960 N
5
TABy = (1040 N)
13
= 400 N
TABx =
(a)
M D = TABx (0.875 m) − TABy (0.2 m)
= (960 N)(0.875 m) − (400 N)(0.2 m)
= 760 N ⋅ m
(b)
We have
or M D = 760 N ⋅ m
M D = TABx ( y ) + TABx ( x)
= (960 N)(0) + (400 N)(1.90 m)
= 760 N ⋅ m

(b)
or M D = 760 N ⋅ m

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171
PROBLEM 3.12
It is known that a force with a moment of 960 N · m about D is required to straighten the fence post CD. If
d = 2.80 m, determine the tension that must be developed in the cable of winch puller AB to create the
required moment about Point D.
SOLUTION
Slope of line:
EC =
0.875 m
7
=
2.80 m + 0.2 m 24
Then
TABx =
24
TAB
25
and
TABy =
7
TAB
25
We have
M D = TABx ( y ) + TABy ( x)
24
7
TAB (0) + TAB (2.80 m)
25
25
= 1224 N
960 N ⋅ m =
TAB
or TAB = 1224 N 
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172
PROBLEM 3.13
It is known that a force with a moment of 960 N · m about D is required to straighten the fence post CD. If the
capacity of winch puller AB is 2400 N, determine the minimum value of distance d to create the specified
moment about Point D.
SOLUTION
The minimum value of d can be found based on the equation relating the moment of the force TAB about D:
M D = (TAB max ) y (d )
where
M D = 960 N ⋅ m
(TAB max ) y = TAB max sin θ = (2400 N) sin θ
Now
sin θ =
0.875 m
(d + 0.20)2 + (0.875)2 m

0.875
960 N ⋅ m = 2400 N 
 ( d + 0.20) 2 + (0.875) 2

or
(d + 0.20) 2 + (0.875) 2 = 2.1875d
or
(d + 0.20) 2 + (0.875) 2 = 4.7852d 2
or

 (d )


3.7852d 2 − 0.40d − 0.8056 = 0
Using the quadratic equation, the minimum values of d are 0.51719 m and − 0.41151 m.
Since only the positive value applies here, d = 0.51719 m
or d = 517 mm 
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173
PROBLEM 3.14
A mechanic uses a piece of pipe AB as a lever when tightening an
alternator belt. When he pushes down at A, a force of 485 N is
exerted on the alternator at B. Determine the moment of that force
about bolt C if its line of action passes through O.
SOLUTION
We have
M C = rB/C × FB
Noting the direction of the moment of each force component about C is
clockwise,
M C = xFBy + yFBx
where
and
x = 120 mm − 65 mm = 55 mm
y = 72 mm + 90 mm = 162 mm
FBx =
FBy =
65
(65) + (72) 2
2
72
(65) + (72) 2
2
(485 N) = 325 N
(485 N) = 360 N
M C = (55 mm)(360 N) + (162)(325 N)
= 72450 N ⋅ m
= 72.450 N ⋅ m
or M C = 72.5 N ⋅ m

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174
PROBLEM 3.15
Form the vector products B × C and B′ × C, where B = B′, and use the results
obtained to prove the identity
sin α cos β =
1
1
sin (α + β ) + sin (α − β ).
2
2
SOLUTION
Note:
B = B(cos β i + sin β j)
B′ = B(cos β i − sin β j)
C = C (cos α i + sin α j)
By definition,
Now
| B × C | = BC sin (α − β )
(1)
| B′ × C | = BC sin (α + β )
(2)
B × C = B(cos β i + sin β j) × C (cos α i + sin α j)
= BC (cos β sin α − sin β cos α )k
and
(3)
B′ × C = B(cos β i − sin β j) × C (cos α i + sin α j)
= BC (cos β sin α + sin β cos α ) k
(4)
Equating the magnitudes of B × C from Equations (1) and (3) yields:
BC sin(α − β ) = BC (cos β sin α − sin β cos α )
(5)
Similarly, equating the magnitudes of B′ × C from Equations (2) and (4) yields:
BC sin(α + β ) = BC (cos β sin α + sin β cos α )
(6)
Adding Equations (5) and (6) gives:
sin(α − β ) + sin(α + β ) = 2cos β sin α
or sin α cos β =
1
1
sin(α + β ) + sin(α − β ) 
2
2
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175
PROBLEM 3.16
The vectors P and Q are two adjacent sides of a parallelogram. Determine the area of the parallelogram when
(a) P = −7i + 3j − 3k and Q = 2i + 2j + 5k, (b) P = 6i − 5j − 2k and Q = −2i + 5j − k.
SOLUTION
(a)
We have
A = |P × Q|
where
P = −7i + 3j − 3k
Q = 2i + 2 j + 5k
Then
i j k
P × Q = −7 3 −3
2 2 5
= [(15 + 6)i + ( −6 + 35) j + ( −14 − 6)k ]
= (21)i + (29) j(−20)k
A = (20) 2 + (29) 2 + (−20) 2
(b)
We have
A = |P × Q|
where
P = 6i − 5 j − 2k
or A = 41.0 
Q = −2i + 5 j − 1k
Then
i
j k
P × Q = 6 −5 −2
−2 5 −1
= [(5 + 10)i + (4 + 6) j + (30 − 10)k ]
= (15)i + (10) j + (20)k
A = (15) 2 + (10) 2 + (20) 2
or A = 26.9 
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176
PROBLEM 3.17
A plane contains the vectors A and B. Determine the unit vector normal to the plane when A and B are equal
to, respectively, (a) i + 2j − 5k and 4i − 7j − 5k, (b) 3i − 3j + 2k and −2i + 6j − 4k.
SOLUTION
(a)
A×B
|A × B|
We have
λ=
where
A = 1i + 2 j − 5k
B = 4i − 7 j − 5k
Then
i j k
A × B = 1 +2 −5
4 −7 −5
= (−10 − 35)i + (20 + 5) j + (−7 − 8)k
= 15(3i − 1j − 1k )
and
|A × B | = 15 (−3)2 + (−1) 2 + (−1)2 = 15 11
λ=
(b)
15(−3i − 1j − 1k )
15 11
or λ =
1
11
(−3i − j − k ) 
A×B
|A × B|
We have
λ=
where
A = 3i − 3 j + 2k
B = −2i + 6 j − 4k
Then
i
j k
A × B = 3 −3 2
−2 6 −4
= (12 − 12)i + (−4 + 12) j + (18 − 6)k
= (8 j + 12k )
and
|A × B| = 4 (2) 2 + (3) 2 = 4 13
λ=
4(2 j + 3k )
4 13
or λ =
1
13
(2 j + 3k ) 
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177
PROBLEM 3.18
A line passes through the Points (20 m, 16 m) and (−1 m, −4 m). Determine the perpendicular distance d from
the line to the origin O of the system of coordinates.
SOLUTION
d AB = [20 m − ( −1 m)]2 + [16 m − ( −4 m)]2
= 29 m
Assume that a force F, or magnitude F(N), acts at Point A and is
directed from A to B.
Then
where
F = F λ AB
λ AB =
=
By definition,
where
Then
rB − rA
d AB
1
(21i + 20 j)
29
M O = | rA × F | = dF
rA = −(1 m)i − (4 m) j
M O = [ −(−1 m)i − (4 m) j] ×
F
[(21 m)i + (20 m) j]
29 m
F
[−(20)k + (84)k ]
29
 64 
=  F k N ⋅ m
 29 
=
Finally,
 64 
 29 F  = d ( F )


64
d=
m
29
d = 2.21 m 
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178
PROBLEM 3.19
Determine the moment about the origin O of the force F = 4i − 3j + 5k that acts at a Point A. Assume that the
position vector of A is (a) r = 2i + 3j − 4k, (b) r = −8i + 6j − 10k, (c) r = 8i − 6j + 5k.
SOLUTION
MO = r × F
(a)
i j k
M O = 2 3 −4
4 −3 5
= (15 − 12)i + (−16 − 10) j + (−6 − 12)k
(b)
i
j
k
M O = −8 6 −10
4 −3 5
= (30 − 30)i + ( −40 + 40) j + (24 − 24)k
(c)
M O = 3i − 26 j − 18k 
MO = 0 
i j k
M O = 8 −6 5
4 −3 5
= (−30 + 15)i + (20 − 40) j + (−24 + 24)k
M O = −15i − 20 j 
Note: The answer to Part (b) could have been anticipated since the elements of the last two rows of the
determinant are proportional.
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179
PROBLEM 3.20
Determine the moment about the origin O of the force F = 2i + 3j − 4k that acts at a Point A. Assume that the
position vector of A is (a) r = 3i − 6j + 5k, (b) r = i − 4j − 2k, (c) r = 4i + 6j − 8k.
SOLUTION
MO = r × F
(a)
i j k
M O = 3 −6 5
2 3 −4
= (24 − 15)i + (10 + 12) j + (9 + 12)k
(b)
i j k
M O = 1 −4 −2
2 3 −4
= (16 + 6)i + (−4 + 4) j + (3 + 8)k
(c)
M O = 9i + 22 j + 21k 
M O = 22i + 11k 
i j k
M O = 4 6 −8
2 3 −4
= (−24 + 24)i + ( −16 + 16) j + (12 − 12)k
MO = 0 
Note: The answer to Part (c) could have been anticipated since the elements of the last two rows of the
determinant are proportional.
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180
PROBLEM 3.21
The wire AE is stretched between the corners A and E of a bent
plate. Knowing that the tension in the wire is 435 N, determine the
moment about O of the force exerted by the wire (a) on corner A,
(b) on corner E.
SOLUTION

AE = (0.21 m)i − (0.16 m) j + (0.12 m)k
(a)
AE = (0.21 m) 2 + (−0.16 m) 2 + (0.12 m) 2 = 0.29 m

AE
FA = FA λ AE = F
AE
0.21i − 0.16 j + 0.12k
= (435 N)
0.29
= (315 N)i − (240 N) j + (180 N)k
rA/O = −(0.09 m)i + (0.16 m) j
i
j
k
M O = −0.09 0.16
0
315 −240 180
= 28.8i + 16.20 j + (21.6 − 50.4)k
(b)
M O = (28.8 N ⋅ m)i + (16.20 N ⋅ m) j − (28.8 N ⋅ m)k 
FE = −FA = −(315 N)i + (240 N) j − (180 N)k
rE / O = (0.12 m)i + (0.12 m)k
i
j
k
M O = 0.12
0
0.12
−315 240 −180
= −28.8i + (−37.8 + 21.6) j + 28.8k
M O = −(28.8 N ⋅ m)i − (16.20 N ⋅ m) j + (28.8 N ⋅ m)k 
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181
PROBLEM 3.22
A small boat hangs from two davits, one of which is shown in the
figure. The tension in line ABAD is 82 lb. Determine the moment
about C of the resultant force RA exerted on the davit at A.
SOLUTION
We have
R A = 2FAB + FAD
where
and
FAB = −(82 lb) j

AD
6i − 7.75 j − 3k
FAD = FAD
= (82 lb)
AD
10.25
FAD = (48 lb)i − (62 lb) j − (24 lb)k
Thus
R A = 2FAB + FAD = (48 lb)i − (226 lb) j − (24 lb)k
Also
rA/C = (7.75 ft) j + (3 ft)k
Using Eq. (3.21):
i
j
k
M C = 0 7.75
3
48 − 226 −24
= (492 lb ⋅ ft)i + (144.0 lb ⋅ ft) j − (372 lb ⋅ ft)k
M C = (492 lb ⋅ ft)i + (144.0 lb ⋅ ft) j − (372 lb ⋅ ft)k 
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182
PROBLEM 3.23
A 6-ft-long fishing rod AB is securely anchored in the sand of a beach. After a fish takes the bait, the resulting
force in the line is 6 lb. Determine the moment about A of the force exerted by the line at B.
SOLUTION
We have
Then
Txz = (6 lb) cos 8° = 5.9416 lb
Tx = Txz sin 30° = 2.9708 lb
Ty = TBC sin 8° = − 0.83504 lb
Tz = Txz cos 30° = −5.1456 lb
Now
M A = rB/A × TBC
where
rB/A = (6sin 45°) j − (6cos 45°)k
=
Then
or
6 ft
2
(j − k)
i
j
k
MA =
0
1
−1
2
2.9708 −0.83504 −5.1456
6
6
6
(−5.1456 − 0.83504)i −
(2.9708) j −
(2.9708)k
=
2
2
2
6
M A = −(25.4 lb ⋅ ft)i − (12.60 lb ⋅ ft) j − (12.60 lb ⋅ ft)k

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183
PROBLEM 3.24
A precast concrete wall section is temporarily held by two
cables as shown. Knowing that the tension in cable BD is 900
N, determine the moment about Point O of the force exerted by
the cable at B.
SOLUTION
F=F

BD
BD
where F = 900 N

BD = −(1 m)i − (2 m) j + (2 m)k
BD = (−1 m) 2 + ( −2 m) 2 + (2 m) 2
=3m
−i − 2 j + 2k
3
= −(300 N)i − (600 N) j + (600 N)k
= (2.5 m)i + (2 m) j
F = (900 N)
rB /O
M O = rB /O × F
i
j
k
= 2.5
2
0
−300 −600 600
= 1200i − 1500 j + ( −1500 + 600)k
MO = (1200 N ⋅ m)i − (1500 N ⋅ m) j − (900 N ⋅ m)k 
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184
PROBLEM 3.25
A 200-N force is applied as shown to the bracket ABC. Determine
the moment of the force about A.
SOLUTION
We have
M A = rC/A × FC
where
rC/A = (0.06 m)i + (0.075 m) j
FC = −(200 N) cos 30° j + (200 N)sin 30°k
Then
i
j
k
M A = 200 0.06
0.075
0
0
− cos 30° sin 30°
= 200[(0.075sin 30°)i − (0.06sin 30°) j − (0.06 cos 30°)k ]
or M A = (7.50 N ⋅ m)i − (6.00 N ⋅ m) j − (10.39 N ⋅ m)k 
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185
PROBLEM 3.26
The 6-m boom AB has a fixed end A. A steel cable is stretched from
the free end B of the boom to a Point C located on the vertical wall.
If the tension in the cable is 2.5 kN, determine the moment about A of
the force exerted by the cable at B.
SOLUTION
First note
d BC = (−6)2 + (2.4) 2 + (−4) 2
= 7.6 m
2.5 kN
(−6i + 2.4 j − 4k )
7.6
Then
TBC =
We have
M A = rB/A × TBC
where
rB/A = (6 m)i
Then
M A = (6 m)i ×
2.5 kN
(−6i + 2.4 j − 4k )
7.6
or M A = (7.89 kN ⋅ m) j + (4.74 kN ⋅ m)k 
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186
PROBLEM 3.27
In Prob. 3.21, determine the perpendicular distance from point O to
wire AE.
PROBLEM 3.21 The wire AE is stretched between the corners A
and E of a bent plate. Knowing that the tension in the wire is 435 N,
determine the moment about O of the force exerted by the wire (a) on
corner A, (b) on corner E.
SOLUTION
From the solution to Prob. 3.21
M O = (28.8 N ⋅ m)i + (16.20 N ⋅ m) j − (28.8 N ⋅ m)k
M O = (28.8) 2 + (16.20) 2 + (28.8) 2
= 43.8329 N ⋅ m
But
M O = FA d
or
MO
FA
43.8329 N ⋅ m
d=
435 N
= 0.100765 m
d=
d = 100.8 mm 
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187
PROBLEM 3.28
In Prob. 3.21, determine the perpendicular distance from point B to
wire AE.
PROBLEM 3.21 The wire AE is stretched between the corners A
and E of a bent plate. Knowing that the tension in the wire is 435
N, determine the moment about O of the force exerted by the wire
(a) on corner A, (b) on corner E.
SOLUTION
From the solution to Prob. 3.21
FA = (315 N)i − (240 N) j + (180 N)k
rA/B = −(0.210 m)i
M B = rA /B × FA = −0.21i × (315i − 240 j + 180k )
= 50.4k + 37.8 j
M B = (50.4) 2 + (37.8) 2
= 63.0 N ⋅ m
M B = FA d
or
MB
FA
63.0 N ⋅ m
d=
435 N
= 0.144829 m
d=
d = 144.8 mm 
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188
PROBLEM 3.29
In Problem 3.22, determine the perpendicular distance from
point C to portion AD of the line ABAD.
PROBLEM 3.22 A small boat hangs from two davits, one of
which is shown in the figure. The tension in line ABAD is
82 lb. Determine the moment about C of the resultant force
RA exerted on the davit at A.
SOLUTION
First compute the moment about C of the force FDA exerted by the line on D:
From Problem 3.22:
FDA = − FAD
= −(48 lb) i + (62 lb) j + (24 lb)k
M C = rD/C × FDA
= + (6 ft)i × [−(48 lb)i + (62 lb) j + (24 lb)k ]
= −(144 lb ⋅ ft) j + (372 lb ⋅ ft)k
M C = (144) 2 + (372)2
= 398.90 lb ⋅ ft
Then
M C = FDA d
Since
FDA = 82 lb
d=
=
MC
FDA
398.90 lb ⋅ ft
82 lb
d = 4.86 ft 
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189
PROBLEM 3.30
In Prob. 3.23, determine the perpendicular distance from point A to a line drawn through points B and C.
PROBLEM 3.23 A 6-ft-long fishing rod AB is securely anchored in the sand of a beach. After a fish takes the
bait, the resulting force in the line is 6 lb. Determine the moment about A of the force exerted by the line at B.
SOLUTION
From the solution to Prob. 3.23:
M A = −(25.4 lb ⋅ ft)i − (12.60 lb ⋅ ft) j − (12.60 lb ⋅ ft)k
M A = (−25.4) 2 + (−12.60) 2 + (−12.60)2
= 31.027 lb ⋅ ft
M A = TBC d
MA
TBC
31.027 lb ⋅ ft
=
6 lb
= 5.1712 ft
d=
or
d = 5.17 ft 
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190
PROBLEM 3.31
In Prob. 3.23, determine the perpendicular distance from point D to a line drawn through points B and C.
PROBLEM 3.23 A 6-ft-long fishing rod AB is securely anchored in the sand of a beach. After a fish takes the
bait, the resulting force in the line is 6 lb. Determine the moment about A of the force exerted by the line at B.
SOLUTION
AB = 6 ft
TBC = 6 lb
We have
| M D | = TBC d
where d = perpendicular distance from D to line BC.
M D = rB /D × TBC
rB /D = (6sin 45° ft) j = (4.2426 ft)
TBC : (TBC ) x = (6 lb) cos8° sin 30° = 2.9708 lb
(TBC ) y = −(6 lb) sin 8° = −0.83504 lb
(TBC ) z = −(6 lb) cos8° cos 30° = −5.1456 lb
TBC = (2.9708 lb)i − (0.83504 lb) j − (5.1456 lb)k
i
j
k
MD =
0
4.2426
0
2.9708 −0.83504 −5.1456
= −(21.831 lb ⋅ ft)i − (12.6039 lb ⋅ ft)
| M D | = ( −21.831) 2 + (−12.6039) 2 = 25.208 lb ⋅ ft
25.208 lb ⋅ ft = (6 lb)d
d = 4.20 ft 
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191
PROBLEM 3.32
In Prob. 3.24, determine the perpendicular distance from point O
to cable BD.
PROBLEM 3.24 A precast concrete wall section is temporarily
held by two cables as shown. Knowing that the tension in cable
BD is 900 N, determine the moment about Point O of the force
exerted by the cable at B.
SOLUTION
From the solution to Prob. 3.24 we have
M O = (1200 N ⋅ m)i − (1500 N ⋅ m) j − (900 N ⋅ m)k
M O = (1200) 2 + (−1500) 2 + ( −900)2 = 2121.3 N ⋅ m
M O = Fd
MO
F
2121.3 N ⋅ m
=
900 N
d=
d = 2.36 m 
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192
PROBLEM 3.33
In Prob. 3.24, determine the perpendicular distance from point C
to cable BD.
PROBLEM 3.24 A precast concrete wall section is
temporarily held by two cables as shown. Knowing that the
tension in cable BD is 900 N, determine the moment about
Point O of the force exerted by the cable at B.
SOLUTION
From the solution to Prob. 3.24 we have
F = −(300 N)i − (600 N) j + (600 N)k
rB /C = (2 m) j
M C = rB /C × F = (2 m) j × ( −300 Ni − 600 Nj + 600 Nk )
= (600 N ⋅ m)k + (1200 N ⋅ m)i
M C = (600)2 + (1200) 2 = 1341.64 N ⋅ m
M C = Fd
MC
F
1341.64 N ⋅ m
=
900 N
d=
d = 1.491 m 
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193
PROBLEM 3.34
Determine the value of a that minimizes the
perpendicular distance from Point C to a section
of pipeline that passes through Points A and B.
SOLUTION
Assuming a force F acts along AB,
|M C | = |rA / C × F| = F ( d )
d = perpendicular distance from C to line AB
where
F = λ AB F
=
(24 ft) i + (24 ft) j − (28 ft) k
(24) 2 + (24) 2 + (28) 2 ft
F
F
(6) i + (6) j − (7) k
11
= (3 ft)i − (10 ft) j − (a − 10 ft)k
=
rA/C
i
j
k
F
M C = 3 −10 10a
11
6 6
−7
= [(10 + 6a)i + (81 − 6a) j + 78 k ]
|M C | = |rA/C × F 2 |
Since
or
F
11
|rA/C × F 2 | = ( dF ) 2
1
(10 + 6a) 2 + (81 − 6a) 2 + (78)2 = d 2
121
Setting
d
da
(d 2 ) = 0 to find a to minimize d:
1
[2(6)(10 + 6a) + 2(−6)(81 − 6a)] = 0
121
Solving
a = 5.92 ft
or a = 5.92 ft 
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194
PROBLEM 3.35
Given the vectors P = 3i − j + 2k, Q = 4i + 5j − 3k, and S = −2i + 3j − k, compute the scalar products P · Q,
P · S, and Q · S.
SOLUTION
P ⋅ Q = (3i − j + 2k ) ⋅ (4i + 5 j − 3k )
= (3)(4) + (−1)(5) + (2)(−3)
= 12 − 5 − 6
P ⋅ Q = +1 
P ⋅ S = (3i − j + 2k ) ⋅ (−2i + 3j − k )
= (3)(−2) + (−1)(3) + (2)( −1)
= −6 − 3 − 2
P ⋅ S = −11 
Q ⋅ S = (4i + 5 j − 3k ) ⋅ (−2i + 3j − k )
= (4)(−2) + (5)(3) + ( −3)(−1)
= −8 + 15 + 3
Q ⋅ S = +10 
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195
PROBLEM 3.36
Form the scalar product B · C and use the result obtained to prove the
identity
cos (α − β) = cos α cos β + sin α sin β .
SOLUTION
B = B cos α i + B sin α j
(1)
C = C cos β i + C sin β j
(2)
By definition:
B ⋅ C = BC cos(α − β )
(3)
From (1) and (2):
B ⋅ C = ( B cos α i + B sin α j) ⋅ (C cos β i + C sin β j)
= BC (cos α cos β + sin α sin β )
(4)
Equating the right-hand members of (3) and (4),
cos(α − β ) = cos α cos β + sin α sin β 
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196
PROBLEM 3.37
Consider the volleyball net shown.
Determine the angle formed by guy
wires AB and AC.
SOLUTION
First note:
AB = (−6.5)2 + (−8)2 + (2) 2 = 10.5 ft
AC = (0) 2 + (−8) 2 + (6)2 = 10 ft
and
By definition,
or
or

AB = −(6.5 ft)i − (8 ft) j + (2 ft)k

AC = −(8 ft) j + (6 ft)k
 
AB ⋅ AC = ( AB )( AC ) cos θ
(−6.5i − 8 j + 2k ) ⋅ (−8 j + 6k ) = (10.5)(10) cos θ
(−6.5)(0) + ( −8)( −8) + (2)(6) = 105cos θ
cos θ = 0.72381
or θ = 43.6° 
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197
PROBLEM 3.38
Consider the volleyball net shown.
Determine the angle formed by guy
wires AC and AD.
SOLUTION
First note:
AC = (0)2 + (−8) 2 + (6) 2
= 10 ft
AD = (4) 2 + ( −8) 2 + (1) 2
and
By definition,
or
= 9 ft

AC = −(8 ft)j + (6 ft)k

AD = (4 ft)i − (8 ft) j + (1 ft)k
 
AC ⋅ AD = ( AC )( AD ) cos θ
(−8 j + 6k ) ⋅ (4i − 8 j + k ) = (10)(9) cos θ
(0)(4) + ( −8)( −8) + (6)(1) = 90 cos θ
or
cos θ = 0.77778
or θ = 38.9° 
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198
PROBLEM 3.39
Three cables are used to support a container as shown. Determine
the angle formed by cables AB and AD.
SOLUTION
First note:
AB = (450 mm)2 + (600 mm)2
= 750 mm
AD = (−500 mm) 2 + (600 mm) 2 + (360 mm) 2
and
By definition,
= 860 mm

AB = (450 mm)i + (600 mm) j

AD = (−500 mm)i + (600 mm) j + (360 mm)k
 
AB ⋅ AD = ( AB)( AD ) cos θ
(450i + 600 j) ⋅ (−500i − 600 j + 360k ) = (750)(860) cos θ
(450)(−500) + (600)(600) + (0)(360) = (750)(860) cos θ
cos θ = 0.20930
or
θ = 77.9° 
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199
PROBLEM 3.40
Three cables are used to support a container as shown. Determine
the angle formed by cables AC and AD.
SOLUTION
First note:
AC = (600 mm) 2 + (−320 mm) 2
= 680 mm
AD = (−500 mm) 2 + (600 mm) 2 + (360 mm) 2
and
By definition,
= 860 mm

AC = (600 mm)j + (−320 mm)k

AD = (−500 mm)i + (600 mm) j + (360 mm)k
 
AC ⋅ AD = ( AC )( AD ) cos θ
(600 j − 320k ) ⋅ (−500i + 600 j + 360k ) = (680)(860) cos θ
0(−500) + (600)(600) + (−320)(360) = (680)(860) cos θ
cos θ = 0.41860
θ = 65.3° 
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200
PROBLEM 3.41
The 20-in. tube AB can slide along a horizontal rod. The ends A and B
of the tube are connected by elastic cords to the fixed point C. For the
position corresponding to x = 11 in., determine the angle formed by
the two cords (a) using Eq. (3.32), (b) applying the law of cosines to
triangle ABC.
SOLUTION
(a)
Using Eq. (3.32):

CA = 11i − 12 j + 24k
CA = (11) 2 + ( −12) 2 + (24) 2 = 29 in.

CB = 31i − 12 j + 24k
CB = (31) 2 + ( −12) 2 + (24) 2 = 41 in.
 
CA ⋅ CB
cos θ =
(CA)(CB)
(11i − 12 j + 24k ) ⋅ (31i − 12 j + 24k )
=
(29)(41)
(11)(31) + (−12)(−12) + (24)(24)
=
(29)(41)
= 0.89235
(b)
θ = 26.8° 
Law of cosines:
( AB) 2 = (CA) 2 + (CB) 2 − 2(CA)(CB ) cos θ
(20) 2 = (29)2 + (41) 2 − 2(29)(41) cos θ
cos θ = 0.89235
θ = 26.8° 
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201
PROBLEM 3.42
Solve Prob. 3.41 for the position corresponding to x = 4 in.
PROBLEM 3.41 The 20-in. tube AB can slide along a horizontal
rod. The ends A and B of the tube are connected by elastic cords
to the fixed point C. For the position corresponding to x = 11 in.,
determine the angle formed by the two cords (a) using Eq. (3.32),
(b) applying the law of cosines to triangle ABC.
SOLUTION
(a)
Using Eq. (3.32):

CA = 4i − 12 j + 24k
CA = (4) 2 + (−12) 2 + (24) 2 = 27.129 in.

CB = 24i − 12 j + 24k
CB = (24) 2 + (−12)2 + (24)2 = 36 in.
 
CA ⋅ CB
cos θ =
(CA)(CB)
(4i − 12 j + 24k ) ⋅ (24i − 12 j + 24k )
=
(27.129)(36)
= 0.83551
(b)
θ = 33.3° 
Law of cosines:
( AB) 2 = (CA) 2 + (CB )2 − 2(CA)(CB ) cos θ
(20) 2 = (27.129)2 + (36) 2 − 2(27.129)(36) cos θ
cos θ = 0.83551
θ = 33.3° 
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202
PROBLEM 3.43
Ropes AB and BC are two of the ropes used to
support a tent. The two ropes are attached to a
stake at B. If the tension in rope AB is 540 N,
determine (a) the angle between rope AB and
the stake, (b) the projection on the stake of the
force exerted by rope AB at Point B.
SOLUTION
First note:
BA = (−3) 2 + (3)2 + (−1.5) 2 = 4.5 m
BD = (−0.08) 2 + (0.38)2 + (0.16) 2 = 0.42 m
λ BD
(a)
TBA
(−3i + 3j − 1.5k )
4.5
T
= BA (−2i + 2 j − k )
3

1
BD
( −0.08i + 0.38 j + 0.16k )
=
=
BD 0.42
1
= (−4i + 19 j + 8k )
21
TBA =
Then
We have
or
or
TBA ⋅ λ BD = TBA cos θ
TBA
1
( −2i + 2 j − k ) ⋅ (−4i + 19 j + 8k ) = TBA cos θ
3
21
1
[(−2)( −4) + (2)(19) + (−1)(8)]
63
= 0.60317
cos θ =
or θ = 52.9° 
(b)
We have
(TBA ) BD = TBA ⋅ λ BD
= TBA cos θ
= (540 N)(0.60317)
or (TBA ) BD = 326 N 
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203
PROBLEM 3.44
Ropes AB and BC are two of the ropes used to
support a tent. The two ropes are attached to a
stake at B. If the tension in rope BC is 490 N,
determine (a) the angle between rope BC and
the stake, (b) the projection on the stake of the
force exerted by rope BC at Point B.
SOLUTION
First note:
BC = (1) 2 + (3) 2 + (−1.5) 2 = 3.5 m
BD = (−0.08) 2 + (0.38)2 + (0.16) 2 = 0.42 m
TBC
(i + 3j − 1.5k )
3.5
T
= BC (2i + 6 j − 3k )
7

1
BD
( −0.08i + 0.38 j + 0.16k )
=
=
BD 0.42
1
= (−4i + 19 j + 8k )
21
TBC =
λBD
(a)
TBC ⋅ λBD = TBC cos θ
TBC
1
(2i + 6 j − 3k ) ⋅ (−4i + 19 j + 8k ) = TBC cos θ
7
21
1
[(2)( −4) + (6)(19) + (−3)(8)]
147
= 0.55782
cos θ =
θ = 56.1° 
(b)
(TBC ) BD = TBC ⋅ λBD
= TBC cos θ
= (490 N)(0.55782)
(TBC ) BD = 273 N 
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204
PROBLEM 3.45
Given the vectors P = 4i − 2 j + 3k , Q = 2i + 4 j − 5k , and S = S x i − j + 2k , determine the value of S x for
which the three vectors are coplanar.
SOLUTION
If P, Q, and S are coplanar, then P must be perpendicular to (Q × S).
P ⋅ (Q × S) = 0
(or, the volume of a parallelepiped defined by P, Q, and S is zero).
Then
or
4
2
Sx
−2 3
4 −5 = 0
−1 2
32 + 10S x − 6 − 20 + 8 − 12S x = 0
Sx = 7 
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205
PROBLEM 3.46
Determine the volume of the parallelepiped of Fig. 3.25 when
(a) P = 4i − 3j + 2k, Q = −2i − 5j + k, and S = 7i + j − k,
(b) P = 5i − j + 6k, Q = 2i + 3j + k, and S = −3i − 2j + 4k.
SOLUTION
Volume of a parallelepiped is found using the mixed triple product.
(a)
Vol. = P ⋅ (Q × S)
4 −3 2
= −2 −5 1 in.3
7
1 −1
= (20 − 21 − 4 + 70 + 6 − 4)
= 67
or Volume = 67.0 
(b)
Vol. = P ⋅ (Q × S)
5 −1 6
= 2
3 1 in.3
−3 −2 4
= (60 + 3 − 24 + 54 + 8 + 10)
= 111
or Volume = 111.0 
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206
PROBLEM 3.47
Knowing that the tension in cable AB is 570 N, determine the
moment about each of the coordinate axes of the force exerted
on the plate at B.
SOLUTION
BA = (−900 mm)i + (600 mm) j + (360 mm)k
BA = (−900) 2 + (600)2 + (360) 2 = 1140 mm

BA
FB = FB
BA
−900i + 600 j + 360k
= (570 N)
1140
= −(450 N)i + (300 N) j + (180 N)k
rB = (0.9 m)i
M O = rB × FB = 0.9i × (−450i + 300 j + 180k )
= 270k − 162 j
MO = M x i + M y j + M z k
= −(162 N ⋅ m) j + (270 N ⋅ m)k
M x = 0, M y = −162.0 N ⋅ m, M z = +270 N ⋅ m 
Therefore,
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207
PROBLEM 3.48
Knowing that the tension in cable AC is 1065 N, determine the
moment about each of the coordinate axes of the force exerted
on the plate at C.
SOLUTION

CA = (−900 mm)i + (600 mm) j + (−920 mm)k
CA = ( −900)2 + (600) 2 + (−920) 2 = 1420 mm

CA
FC = FC
CA
−900i + 600 j − 920k
= (1065 N)
1420
= −(675 N)i + (450 N) j − (690 N)k
rC = (0.9 m)i + (1.28 m)k
Using Eq. (3.19):
i
j
k
0.9
0 1.28
M O = rC × FC =
−675 450 −690
M O = −(576 N ⋅ m)i − (243 N ⋅ m) j + (405 N ⋅ m)k
But
Therefore,
MO = M x i + M y j + M z k
M x = −576 N ⋅ m, M y = −243 N ⋅ m, M z = +405 N ⋅ m 
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208
PROBLEM 3.49
A small boat hangs from two davits, one of which is shown in
the figure. It is known that the moment about the z-axis of the
resultant force RA exerted on the davit at A must not exceed
279 lb⋅ft in absolute value. Determine the largest allowable
tension in line ABAD when x = 6 ft.
SOLUTION
First note:
R A = 2TAB + TAD
Also note that only TAD will contribute to the moment about the z-axis.
Now
AD = (6) 2 + (−7.75) 2 + (−3) 2
= 10.25 ft

AD
=T
AD
T
(6i − 7.75 j − 3k )
=
10.25
Then
TAD
Now
M z = k ⋅ (rA/C × TAD )
where
rA/C = (7.75 ft) j + (3 ft)k
Then for Tmax ,
0
0
1
Tmax
279 =
0 7.75 3
10.25
6 −7.75 −3
=
Tmax
| − (1)(7.75)(6)|
10.25
or Tmax = 61.5 lb 
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209
PROBLEM 3.50
For the davit of Problem 3.49, determine the largest allowable
distance x when the tension in line ABAD is 60 lb.
SOLUTION
From the solution of Problem 3.49, TAD is now

AD
TAD = T
AD
=
60 lb
x 2 + ( −7.75) 2 + (−3)2
( xi − 7.75 j − 3k )
Then M z = k ⋅ (rA / C × TAD ) becomes
279 =
279 =
60
x 2 + (−7.75) 2 + ( −3) 2
60
x 2 + 69.0625
0
0
1
0 7.75 3
x −7.75 −3
| − (1)(7.75)( x) |
279 x 2 + 69.0625 = 465 x
0.6 x 2 + 69.0625 = x
Squaring both sides:
0.36 x 2 + 24.8625 = x 2
x 2 = 38.848
x = 6.23 ft 
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210
PROBLEM 3.51
A farmer uses cables and winch pullers B and E to plumb
one side of a small barn. If it is known that the sum of the
moments about the x-axis of the forces exerted by the
cables on the barn at Points A and D is equal to 4728 lb ⋅ ft,
determine the magnitude of TDE when TAB = 255 lb.
SOLUTION
The moment about the x-axis due to the two cable forces can be found using the z components of each
force acting at their intersection with the xy plane (A and D). The x components of the forces are parallel to
the x-axis, and the y components of the forces intersect the x-axis. Therefore, neither the x or y components
produce a moment about the x-axis.
We have
ΣM x : (TAB ) z ( y A ) + (TDE ) z ( yD ) = M x
where
(TAB ) z = k ⋅ TAB
= k ⋅ (TAB λ AB )

 − i − 12 j + 12k  
= k ⋅  255 lb 

17



= 180 lb
(TDE ) z = k ⋅ TDE
= k ⋅ (TDE λDE )

 1.5i − 14 j + 12k  
= k ⋅ TDE 

18.5



= 0.64865TDE
y A = 12 ft
yD = 14 ft
M x = 4728 lb ⋅ ft
(180 lb)(12 ft) + (0.64865TDE )(14 ft) = 4728 lb ⋅ ft
and
TDE = 282.79 lb
or TDE = 283 lb 
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211
PROBLEM 3.52
Solve Problem 3.51 when the tension in cable AB is 306 lb.
PROBLEM 3.51 A farmer uses cables and winch pullers B
and E to plumb one side of a small barn. If it is known that the
sum of the moments about the x-axis of the forces exerted by
the cables on the barn at Points A and D is equal to 4728 lb ⋅ ft,
determine the magnitude of TDE when TAB = 255 lb.
SOLUTION
The moment about the x-axis due to the two cable forces can be found using the z components of each force
acting at the intersection with the xy plane (A and D). The x components of the forces are parallel to the x-axis,
and the y components of the forces intersect the x-axis. Therefore, neither the x or y components produce a
moment about the x-axis.
We have
ΣM x : (TAB ) z ( y A ) + (TDE ) z ( yD ) = M x
Where
(TAB ) z = k ⋅ TAB
= k ⋅ (TAB λAB )

 − i − 12 j + 12k  
= k ⋅ 306 lb 

17



= 216 lb
(TDE ) z = k ⋅ TDE
= k ⋅ (TDE λDE )

 1.5i − 14 j + 12k  
= k ⋅ TDE 

18.5



= 0.64865TDE
y A = 12 ft
yD = 14 ft
M x = 4728 lb ⋅ ft
(216 lb)(12 ft) + (0.64865TDE )(14 ft) = 4728 lb ⋅ ft
and
or TDE = 235 lb 
TDE = 235.21 lb
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212
PROBLEM 3.53
A single force P acts at C in a direction perpendicular to the
handle BC of the crank shown. Knowing that Mx = +20 N ·
m and My = −8.75 N · m, and Mz = −30 N · m, determine the
magnitude of P and the values of φ and θ.
SOLUTION
rC = (0.25 m)i + (0.2 m) sin θ j + (0.2 m) cos θ k
P = − P sin φ j + P cos φ k
i
j
M O = rC × P = 0.25 0.2sin θ
0
− P sin φ
k
0.2 cos θ
P cos φ
Expanding the determinant, we find
M x = (0.2) P(sin θ cos φ + cos θ sin φ )
Dividing Eq. (3) by Eq. (2) gives:
M x = (0.2) P sin(θ + φ )
(1)
M y = −(0.25) P cos φ
(2)
M z = −(0.25) P sin φ
(3)
tan φ =
Mz
My
(4)
−30 N ⋅ m
−8.75 N ⋅ m
φ = 73.740
tan φ =
φ = 73.7° 
Squaring Eqs. (2) and (3) and adding gives:
M y2 + M z2 = (0.25) 2 P 2
or
P = 4 M y2 + M z2
(5)
P = 4 (8.75) 2 + (30) 2
= 125.0 N
P = 125.0 N 
Substituting data into Eq. (1):
(+20 N ⋅ m) = 0.2 m(125.0 N) sin(θ + φ )
(θ + φ ) = 53.130° and (θ + φ ) = 126.87°
θ = −20.6° and θ = 53.1°
Q = 53.1° 
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213
PROBLEM 3.54
A single force P acts at C in a direction perpendicular to the
handle BC of the crank shown. Determine the moment Mx of P
about the x-axis when θ = 65°, knowing that My = −15 N · m
and Mz = −36 N · m.
SOLUTION
See the solution to Prob. 3.53 for the derivation of the following equations:
M x = (0.2) P sin(θ + φ )
M
tan φ = z
My
P = 4 M y2 + M z2
(1)
(4)
(5)
Substituting for known data gives:
tan φ =
−36 N ⋅ m
−15 N ⋅ m
φ = 67.380°
P = 4 ( −15) 2 + (−36) 2
P = 156.0 N
M x = 0.2 m(156.0 N) sin(65° + 67.380°)
= 23.047 N ⋅ m
M x = 23.0 N ⋅ m 
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214
PROBLEM 3.55
The triangular plate ABC is supported by ball-and-socket joints
at B and D and is held in the position shown by cables AE and
CF. If the force exerted by cable AE at A is 55 N, determine the
moment of that force about the line joining Points D and B.
SOLUTION
First note:
TAE = TAE

AE
AE
AE = (0.9)2 + (−0.6) 2 + (0.2) 2 = 1.1 m
Then
Also,
Then
Now
where
Then
55 N
(0.9i − 0.6 j + 0.2k )
1.1
= 5[(9 N)i − (6 N) j + (2 N)k ]
TAE =
DB = (1.2) 2 + ( −0.35) 2 + (0)2
λ DB
= 1.25 m

DB
=
DB
1
(1.2i − 0.35 j)
=
1.25
1
(24i − 7 j)
=
25
M DB = λ DB ⋅ (rA/D × TAE )
rA/D = −(0.1 m) j + (0.2 m)k
M DB
24 −7
0
1
(5) 0 −0.1 0.2
=
25
9
2
−6
1
= (−4.8 − 12.6 + 28.8)
5
or M DB = 2.28 N ⋅ m 
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215
PROBLEM 3.56
The triangular plate ABC is supported by ball-and-socket joints
at B and D and is held in the position shown by cables AE and
CF. If the force exerted by cable CF at C is 33 N, determine the
moment of that force about the line joining Points D and B.
SOLUTION
First note:
TCF = TCF

CF
CF
CF = (0.6)2 + (−0.9) 2 + (−0.2) 2 = 1.1 m
33 N
(0.6i − 0.9 j + 0.2k )
1.1
= 3[(6 N)i − (9 N) j − (2 N)k ]
Then
TCF =
Also,
DB = (1.2) 2 + ( −0.35) 2 + (0)2
= 1.25 m

DB
=
DB
1
(1.2i − 0.35 j)
=
1.25
1
(24i − 7 j)
=
25
Then
λ DB
Now
M DB = λ DB ⋅ (rC/D × TCF )
where
rC/D = (0.2 m) j − (0.4 m)k
Then
M DB
24 −7
0
1
(3) 0 0.2 −0.4
=
25
6 −9 −2
=
3
(−9.6 + 16.8 − 86.4)
25
or M DB = −9.50 N ⋅ m 
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216
PROBLEM 3.57
The 23-in. vertical rod CD is welded to the midpoint C of the
50-in. rod AB. Determine the moment about AB of the 235-lb
force P.
SOLUTION

AB = (32 in.)i − (30 in.) j − (24 in.)k
AB = (32)2 + ( −30) 2 + ( −24) 2 = 50 in.

AB
= 0.64i − 0.60 − 0.48k
λ AB =
AB
We shall apply the force P at Point G:
rG /B = (5 in.)i + (30 in.)k

DG = (21 in.)i − (38 in.) j + (18 in.)k
DG = (21) 2 + (−38)2 + (18) 2 = 47 in.

DG
21i − 38 j + 18k
P=P
= (235 lb)
47
DG
P = (105 lb)i − (190 lb) j + (90 lb)k
The moment of P about AB is given by Eq. (3.46):
M AB
0.64
−0.60 −0.48
0
30 in.
= λ AB ⋅ (rG /B × P) = 5 in.
105 lb −190 lb 90 lb
M AB = 0.64[0 − (30 in.)(−190 lb)]
− 0.60[(30 in.)(105 lb) − (5 in.)(90 lb)]
− 0.48[(5 in.)( −190 lb) − 0]
= +2484 lb ⋅ in.
M AB = +207 lb ⋅ ft 
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217
PROBLEM 3.58
The 23-in. vertical rod CD is welded to the midpoint C of the
50-in. rod AB. Determine the moment about AB of the 174-lb
force Q.
SOLUTION

AB = (32 in.)i − (30 in.) j − (24 in.)k
AB = (32)2 + ( −30) 2 + ( −24) 2 = 50 in.

AB
= 0.64i − 0.60 j − 0.48k
λ AB =
AB
We shall apply the force Q at Point H:
rH /B = −(32 in.)i + (17 in.) j

DH = −(16 in.)i − (21 in.) j − (12 in.)k
DH = (16) 2 + (−21) 2 + (−12)2 = 29 in.

DH
−16i − 21j − 12k
Q=
= (174 lb)
DH
29
Q = −(96 lb)i − (126 lb) j − (72 lb)k
The moment of Q about AB is given by Eq. (3.46):
M AB
0.64
−0.60
−0.48
= λ AB ⋅ (rH /B × Q) = −32 in. 17 in.
0
−96 lb −126 lb −72 lb
M AB = 0.64[(17 in.)(−72 lb) − 0]
− 0.60[(0 − (−32 in.)( −72 lb)]
− 0.48[(−32 in.)(−126 lb) − (17 in.)(−96 lb)]
= −2119.7 lb ⋅ in.
M AB = 176.6 lb ⋅ ft 
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218
PROBLEM 3.59
The frame ACD is hinged at A and D and is supported by a
cable that passes through a ring at B and is attached to hooks
at G and H. Knowing that the tension in the cable is 450 N,
determine the moment about the diagonal AD of the force
exerted on the frame by portion BH of the cable.
SOLUTION
M AD = λ AD ⋅ (rB/A × TBH )
Where
and
1
λ AD = (4i − 3k )
5
rB/A = (0.5 m)i
d BH = (0.375)2 + (0.75) 2 + (−0.75) 2
= 1.125 m
Then
Finally,
450 N
(0.375i + 0.75 j − 0.75k )
1.125
= (150 N)i + (300 N) j − (300 N)k
TBH =
MAD
4
0
−3
1
0
= 0.5 0
5
150 300 −300
1
= [(−3)(0.5)(300)]
5
or M AD = − 90.0 N ⋅ m 
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219
PROBLEM 3.60
In Problem 3.59, determine the moment about the diagonal AD
of the force exerted on the frame by portion BG of the cable.
PROBLEM 3.59 The frame ACD is hinged at A and D and is
supported by a cable that passes through a ring at B and is
attached to hooks at G and H. Knowing that the tension in the
cable is 450 N, determine the moment about the diagonal AD
of the force exerted on the frame by portion BH of the cable.
SOLUTION
M AD = λ AD ⋅ (rB/A × TBG )
Where
and
1
λ AD = (4i − 3k )
5
rB/A = (0.5 m) j
BG = (−0.5) 2 + (0.925)2 + (−0.4)2
= 1.125 m
Then
Finally,
450 N
(−0.5i + 0.925 j − 0.4k )
1.125
= −(200 N)i + (370 N) j − (160 N)k
TBG =
MAD
4
0
−3
1
=
0.5
0
0
5
−200 370 −160
1
= [(−3)(0.5)(370)]
5
M AD = −111.0 N ⋅ m 
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220
PROBLEM 3.61
A regular tetrahedron has six edges of length a. A force P is
directed as shown along edge BC. Determine the moment of P
about edge OA.
SOLUTION
We have
From triangle OBC:
M OA = λOA ⋅ (rC /O × P)
(OA) x =
a
2
(OA) z = (OA) x tan 30° =
Since
a 1 
a

=
2 3  2 3
(OA) 2 = (OA) 2x + (OA) 2y + (OAz )2
2
or
 a 
a
a =   + (OA) 2y + 

2
2 3
2
2
(OA) y = a 2 −
2
a2 a2
−
=a
4 12
3
a
2
a
i+a
j+
k
2
3
2 3
Then
rA/O =
and
λOA = i +
1
2
2
1
j+
k
3
2 3
P = λ BC P =
(a sin 30°)i − (a cos 30°)k
P
( P) = (i − 3k )
a
2
rC /O = ai
M OA
1
2
=
1
2
3
0
2 3
P
(a)  
0
2
1
0
− 3
=
1
aP  2 
aP
 −
 (1)(− 3) =
2  3
2
M OA =
aP
2

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221
PROBLEM 3.62
A regular tetrahedron has six edges of length a. (a) Show that two
opposite edges, such as OA and BC, are perpendicular to each
other. (b) Use this property and the result obtained in Problem 3.61
to determine the perpendicular distance between edges OA and BC.
SOLUTION
(a)
For edge OA to be perpendicular to edge BC,
 
OA ⋅ BC = 0
From triangle OBC:
(OA) x =
a
2
a 1 
a

=
2 3  2 3
  a 
 a 
OA =   i + (OA) y j + 
k
2
2 3

BC = ( a sin 30°) i − (a cos 30°) k
(OA) z = (OA) x tan 30° =
and
=
Then
or
so that
(b)
We have M OA
a
a 3
a
i−
k = (i − 3 k )
2
2
2
a
 a  
a
 i + (OA) y j + 
 k  ⋅ (i − 3k ) = 0
2
2 3 
2
a2
a2
+ (OA) y (0) −
=0
4
4
 
OA ⋅ BC = 0


OA is perpendicular to BC. 
 
= Pd , with P acting along BC and d the perpendicular distance from OA to BC.
From the results of Problem 3.57,
M OA =
Pa
2
Pa
2
= Pd
or d =
a
2

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222
PROBLEM 3.63
Two forces F1 and F2 in space have the same magnitude F. Prove that the moment of F1 about the line of
action of F2 is equal to the moment of F2 about the line of action of F1 .
SOLUTION
First note that
F1 = F1λ1 and F2 = F2 λ 2
Let M1 = moment of F2 about the line of action of F1 and M 2 = moment of F1 about the line of
action of F2 .
Now, by definition,
M1 = λ 1 ⋅ (rB /A × F2 )
= λ 1 ⋅ (rB /A × λ 2 ) F2
M 2 = λ 2 ⋅ (rA/B × F1 )
= λ 2 ⋅ (rA/B × λ 1 ) F1
Since
F1 = F2 = F
and rA /B = −rB /A
M1 = λ 1 ⋅ (rB /A × λ 2 ) F
M 2 = λ 2 ⋅ (−rB /A × λ 1 ) F
Using Equation (3.39):
so that
λ 1 ⋅ (rB /A × λ 2 ) = λ 2 ⋅ (−rB /A × λ 1 )
M 2 = λ 1⋅ (rB /A × λ 2 ) F 
M12 = M 21 
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223
PROBLEM 3.64
In Problem 3.55, determine the perpendicular distance between
cable AE and the line joining Points D and B.
PROBLEM 3.55 The triangular plate ABC is supported by
ball-and-socket joints at B and D and is held in the position
shown by cables AE and CF. If the force exerted by cable AE
at A is 55 N, determine the moment of that force about the line
joining Points D and B.
SOLUTION
From the solution to Problem 3.55:
ΤAE = 55 N
TAE = 5[(9 N)i − (6 N) j + (2 N)k ]
| M DB | = 2.28 N ⋅ m
λ DB =
1
(24i − 7 j)
25
Based on the discussion of Section 3.11,
it follows that only the perpendicular component of TAE will
contribute to the moment of TAE about line DB.
Now
(TAE )parallel = TAE ⋅ λ DB
= 5(9i − 6 j + 2k ) ⋅
1
(24i − 7 j)
25
1
= [(9)(24) + (−6)(−7)]
5
= 51.6 N
Also,
so that
TAE = (TAE ) parallel + (TAE ) perpendicular
(TAE )perpendicular = (55) 2 + (51.6)2 = 19.0379 N
Since λ DB and (TAE )perpendicular are perpendicular, it follows that
M DB = d (TAE ) perpendicular
or
2.28 N ⋅ m = d (19.0379 N)
d = 0.119761
d = 0.1198 m 
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224
PROBLEM 3.65
In Problem 3.56, determine the perpendicular distance
between cable CF and the line joining Points D and B.
PROBLEM 3.56 The triangular plate ABC is supported by
ball-and-socket joints at B and D and is held in the position
shown by cables AE and CF. If the force exerted by cable CF at
C is 33 N, determine the moment of that force about the line
joining Points D and B.
SOLUTION
ΤCF = 33 N
From the solution to Problem 3.56:
TCF = 3[(6 N)i − (9 N) j − (2 N)k ]
| M DB | = 9.50 N ⋅ m
λ DB =
1
(24i − 7 j)
25
Based on the discussion of Section 3.11,
it follows that only the perpendicular component of TCF will
contribute to the moment of TCF about line DB.
Now
(TCF )parallel = TCF ⋅ λ DB
= 3(6i − 9 j − 2k ) ⋅
1
(24i − 7 j)
25
3
[(6)(24) + (−9)( −7)]
25
= 24.84 N
=
Also,
so that
TCF = (TCF ) parallel + (TCF ) perpendicular
(TCF )perpendicular = (33) 2 − (24.84) 2
= 21.725 N
Since λ DB and (TCF )perpendicular are perpendicular, it follows that
| M DB | = d (TCF ) perpendicular
or
9.50 N ⋅ m = d × 21.725 N
or d = 0.437 m 
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225
PROBLEM 3.66
In Prob. 3.57, determine the perpendicular distance between
rod AB and the line of action of P.
PROBLEM 3.57 The 23-in. vertical rod CD is welded to the
midpoint C of the 50-in. rod AB. Determine the moment about
AB of the 235-lb force P.
SOLUTION

AB = (32 in.)i − (30 in.) j − (24 in.)k
AB = (32)2 + (−30) 2 + ( −24) 2 = 50 in.

AB
= 0.64i − 0.60 j − 0.48k
λ AB =
AB
λP =
P 105i − 190 j + 90k
=
P
235
Angle θ between AB and P:
cos θ = λ AB ⋅ λ P
= (0.64i − 0.60 j − 0.48k ) ⋅
105i − 190 j + 90k
235
= 0.58723
∴ θ = 54.039°
The moment of P about AB may be obtained by multiplying the projection of P on a plane perpendicular to
AB by the perpendicular distance d from AB to P:
M AB = ( P sin θ )d
From the solution to Prob. 3.57: M AB = 207 lb ⋅ ft = 2484 lb ⋅ in.
We have
2484 lb ⋅ in. = (235 lb)(sin 54.039) d
d = 13.06 in. 
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226
PROBLEM 3.67
In Prob. 3.58, determine the perpendicular distance between
rod AB and the line of action of Q.
PROBLEM 3.58 The 23-in. vertical rod CD is welded to the
midpoint C of the 50-in. rod AB. Determine the moment about
AB of the 174-lb force Q.
SOLUTION

AB = (32 in.)i − (30 in.) j − (24 in.)k
AB = (32)2 + (−30) 2 + ( −24) 2 = 50 in.

AB
λ AB =
= 0.64i − 0.60 j − 0.48k
AB
λQ =
Q −96i − 126 j − 72k
=
Q
174
Angle θ between AB and Q:
cos θ = λ AB ⋅ λQ
= (0.64i − 0.60 j − 0.48k ) ⋅
(−96i − 126 j − 72k )
174
= 0.28000
∴ θ = 73.740°
The moment of Q about AB may be obtained by multiplying the projection of Q on a plane perpendicular to
AB by the perpendicular distance d from AB to Q:
M AB = (Q sin θ )d
From the solution to Prob. 3.58: M AB = 176.6 lb ⋅ ft = 2119.2 lb ⋅ in.
2119.2 lb ⋅ in. = (174 lb)(sin 73.740°) d
d = 12.69 in. 
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227
PROBLEM 3.68
In Problem 3.59, determine the perpendicular distance between
portion BH of the cable and the diagonal AD.
PROBLEM 3.59 The frame ACD is hinged at A and D and is
supported by a cable that passes through a ring at B and is
attached to hooks at G and H. Knowing that the tension in the
cable is 450 N, determine the moment about the diagonal AD
of the force exerted on the frame by portion BH of the cable.
SOLUTION
From the solution to Problem 3.59:
TBH = 450 N
TBH = (150 N)i + (300 N) j − (300 N)k
| M AD | = 90.0 N ⋅ m
1
λ AD = (4i − 3k )
5
Based on the discussion of Section 3.11, 
it follows that only the perpendicular component of TBH will
contribute to the moment of TBH about line AD.
Now
(TBH )parallel = TBH ⋅ λ AD
1
= (150i + 300 j − 300k ) ⋅ (4i − 3k )
5
1
= [(150)(4) + (−300)(−3)]
5
= 300 N
Also,
so that
TBH = (TBH ) parallel + (TBH )perpendicular
(TBH )perpendicular = (450) 2 − (300) 2 = 335.41 N
Since λ AD and (TBH )perpendicular are perpendicular, it follows that
M AD = d (TBH ) perpendicular
or
90.0 N ⋅ m = d (335.41 N)
d = 0.26833 m
d = 0.268 m 
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228
PROBLEM 3.69
In Problem 3.60, determine the perpendicular distance between
portion BG of the cable and the diagonal AD.
PROBLEM 3.60 In Problem 3.59, determine the moment
about the diagonal AD of the force exerted on the frame by
portion BG of the cable.
SOLUTION
From the solution to Problem 3.60:
ΤBG = 450 N
TBG = −(200 N)i + (370 N) j − (160 N)k
| M AD | = 111 N ⋅ m
1
λ AD = (4i − 3k )
5
Based on the discussion of Section 3.11, 
it follows that only the perpendicular component of TBG will
contribute to the moment of TBG about line AD.
Now
(TBG ) parallel = TBG ⋅ λ AD
1
= ( −200i + 370 j − 160k ) ⋅ (4i − 3k )
5
1
= [(−200)(4) + (−160)(−3)]
5
= −64 N
Also,
so that
TBG = (TBG ) parallel + (TBG )perpendicular
(TBG ) perpendicular = (450) 2 − ( −64) 2 = 445.43 N
Since λ AD and (TBG ) perpendicular are perpendicular, it follows that
M AD = d (TBG ) perpendicular
or
111 N ⋅ m = d (445.43 N)
d = 0.24920 m
d = 0.249 m 
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229
PROBLEM 3.70
A plate in the shape of a parallelogram is acted upon by two
couples. Determine (a) the moment of the couple formed by the
two 21-lb forces, (b) the perpendicular distance between the 12-lb
forces if the resultant of the two couples is zero, (c) the value of α
if the resultant couple is 72 lb ⋅ in. clockwise and d is 42 in.
SOLUTION
(a)
M1 = d1 F1
We have
d1 = 16 in.
where
F1 = 21 lb
M1 = (16 in.)(21 lb)
= 336 lb ⋅ in.
(b)
(c)
336 lb ⋅ in. − d 2 (12 lb) = 0
d 2 = 28.0 in. 
M total = M1 + M 2
We have
or

M1 + M 2 = 0
We have
or
or M1 = 336 lb ⋅ in.
−72 lb ⋅ in. = 336 lb ⋅ in. − (42 in.)(sin α )(12 lb)
sin α = 0.80952
α = 54.049°
and
or α = 54.0° 
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230
PROBLEM 3.71
Four 1-in.-diameter pegs are attached to a board as shown. Two
strings are passed around the pegs and pulled with the forces
indicated. (a) Determine the resultant couple acting on the board.
(b) If only one string is used, around which pegs should it pass
and in what directions should it be pulled to create the same
couple with the minimum tension in the string? (c) What is the
value of that minimum tension?
SOLUTION
M = (35 lb)(7 in.) + (25 lb)(9 in.)
= 245 lb ⋅ in. + 225 lb ⋅ in.
(a)
M = 470 lb ⋅ in.
(b)

With only one string, pegs A and D, or B and C should be used. We have
6
8
tan θ =
θ = 36.9°
90° − θ = 53.1°
Direction of forces:
(c)
With pegs A and D:
θ = 53.1° 
With pegs B and C:
θ = 53.1° 
The distance between the centers of the two pegs is
82 + 62 = 10 in.
Therefore, the perpendicular distance d between the forces is
1 
d = 10 in. + 2  in. 
2 
= 11 in.
M = Fd
We must have
470 lb ⋅ in. = F (11 in.)
F = 42.7 lb 
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231
PROBLEM 3.72
Four pegs of the same diameter are attached to a board as shown.
Two strings are passed around the pegs and pulled with the
forces indicated. Determine the diameter of the pegs knowing
that the resultant couple applied to the board is 485 lb·in.
counterclockwise.
SOLUTION
M = d AD FAD + d BC FBC
485 lb ⋅ in. = [(6 + d ) in.](35 lb) + [(8 + d ) in.](25 lb)
d = 1.250 in. 
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232
PROBLEM 3.73
A piece of plywood in which several holes are being drilled
successively has been secured to a workbench by means of two
nails. Knowing that the drill exerts a 12-N·m couple on the piece of
plywood, determine the magnitude of the resulting forces applied to
the nails if they are located (a) at A and B, (b) at B and C, (c) at A
and C.
SOLUTION
(a)
M = Fd
12 N ⋅ m = F (0.45 m)
F = 26.7 N 
(b)
M = Fd
12 N ⋅ m = F (0.24 m)
F = 50.0 N 
(c)
M = Fd
d = (0.45 m)2 + (0.24 m)2
= 0.510 m
12 N ⋅ m = F (0.510 m)
F = 23.5 N 
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233
PROBLEM 3.74
Two parallel 40-N forces are applied to a lever as shown.
Determine the moment of the couple formed by the two forces
(a) by resolving each force into horizontal and vertical components
and adding the moments of the two resulting couples, (b) by using
the perpendicular distance between the two forces, (c) by summing
the moments of the two forces about Point A.
SOLUTION
(a)
We have
where
ΣM B : − d1C x + d 2 C y = M
d1 = (0.270 m) sin 55°
= 0.22117 m
d 2 = (0.270 m) cos 55°
= 0.154866 m
C x = (40 N) cos 20°
= 37.588 N
C y = (40 N) sin 20°
= 13.6808 N
M = −(0.22117 m)(37.588 N)k + (0.154866 m)(13.6808 N)k
= −(6.1946 N ⋅ m)k
(b)
We have
We have

or M = 6.19 N ⋅ m

or M = 6.19 N ⋅ m

M = Fd (−k )
= 40 N[(0.270 m)sin(55° − 20°)](−k )
= −(6.1946 N ⋅ m)k
(c)
or M = 6.19 N ⋅ m
ΣM A : Σ(rA × F ) = rB/A × FB + rC/A × FC = M
i
j
k
sin 55° 0
M = (0.390 m)(40 N) cos 55°
− cos 20° − sin 20° 0
i
j
k
+ (0.660 m)(40 N) cos 55° sin 55° 0
cos 20° sin 20° 0
= (8.9478 N ⋅ m − 15.1424 N ⋅ m)k
= −(6.1946 N ⋅ m)k
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234
PROBLEM 3.75
The two shafts of a speed-reducer unit are subjected to
couples of magnitude M1 = 15 lb·ft and M2 = 3 lb·ft,
respectively. Replace the two couples with a single equivalent
couple, specifying its magnitude and the direction of its axis.
SOLUTION
M1 = (15 lb ⋅ ft)k
M 2 = (3 lb ⋅ ft)i
M = M12 + M 22
= (15) 2 + (3) 2
= 15.30 lb ⋅ ft
15
tan θ x =
=5
3
θ x = 78.7°
θ y = 90°
θ z = 90° − 78.7°
= 11.30°
M = 15.30 lb ⋅ ft; θ x = 78.7°, θ y = 90.0°, θ z = 11.30° 
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235
PROBLEM 3.76
Replace the two couples shown with a single equivalent couple,
specifying its magnitude and the direction of its axis.
SOLUTION
Replace the couple in the ABCD plane with two couples P and Q shown:
P = (50 N)
CD
 160 mm 
= (50 N) 
 = 40 N
CG
 200 mm 
Q = (50 N)
CF
 120 mm 
= (50 N) 
 = 30 N
CG
 200 mm 
Couple vector M1 perpendicular to plane ABCD:
M1 = (40 N)(0.24 m) − (30 N)(0.16 m) = 4.80 N ⋅ m
Couple vector M2 in the xy plane:
M 2 = −(12.5 N)(0.192 m) = −2.40 N ⋅ m
144 mm
θ = 36.870°
192 mm
M1 = (4.80 cos 36.870°) j + (4.80 sin 36.870°)k
= 3.84 j + 2.88k
tan θ =
M 2 = −2.40 j
M = M1 + M 2 = 1.44 j + 2.88k
M = 3.22 N ⋅ m; θ x = 90.0°, θ y = 53.1°, θ z = 36.9° 
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236
PROBLEM 3.77
Solve Prob. 3.76, assuming that two 10-N vertical forces have
been added, one acting upward at C and the other downward at B.
PROBLEM 3.76 Replace the two couples shown with a single
equivalent couple, specifying its magnitude and the direction of
its axis.
SOLUTION
Replace the couple in the ABCD plane with two couples P and Q shown.
P = (50 N)
CD
 160 mm 
= (50 N) 
 = 40 N
CG
 200 mm 
Q = (50 N)
CF
 120 mm 
= (50 N) 
 = 30 N
CG
 200 mm 
Couple vector M1 perpendicular to plane ABCD.
M1 = (40 N)(0.24 m) − (30 N)(0.16 m) = 4.80 N ⋅ m
144 mm
θ = 36.870°
192 mm
M1 = (4.80cos 36.870°) j + (4.80sin 36.870°)k
= 3.84 j + 2.88k
tan θ =
M 2 = −(12.5 N)(0.192 m) = −2.40 N ⋅ m
= −2.40 j
M 3 = rB /C × M 3 ; rB /C = (0.16 m)i + (0.144 m) j − (0.192 m)k
= (0.16 m)i + (0.144 m) j − (0.192 m)k × (−10 N) j
= −1.92i − 1.6k
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237
PROBLEM 3.77 (Continued)
M = M1 + M 2 + M 3 = (3.84 j + 2.88k ) − 2.40 j + (−1.92i − 1.6k )
= −(1.92 N ⋅ m)i + (1.44 N ⋅ m) j + (1.28 N ⋅ m)k
M = ( −1.92) 2 + (1.44) 2 + (1.28) 2 = 2.72 N ⋅ m
M = 2.72 N ⋅ m 
cos θ x = −1.92/2.72
cos θ y = 1.44/2.72
θ x = 134.9° θ y = 58.0° θ z = 61.9° 
cos θ z = 1.28/2.72
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238
PROBLEM 3.78
If P = 0, replace the two remaining couples with a single
equivalent couple, specifying its magnitude and the direction
of its axis.
SOLUTION
M = M1 + M 2 ; F1 = 16 lb, F2 = 40 lb
M1 = rC × F1 = (30 in.)i × [−(16 lb) j] = −(480 lb ⋅ in.)k
M 2 = rE/B × F2 ; rE/B = (15 in.)i − (5 in.) j
d DE = (0) 2 + (5) 2 + (10) 2 = 5 5 in.
F2 =
40 lb
5 5
(5 j − 10k )
= 8 5[(1 lb) j − (2 lb)k ]
i
j k
M 2 = 8 5 15 −5 0
0 1 −2
= 8 5[(10 lb ⋅ in.)i + (30 lb ⋅ in.) j + (15 lb ⋅ in.)k ]
M = −(480 lb ⋅ in.)k + 8 5[(10 lb ⋅ in.)i + (30 lb ⋅ in.) j + (15 lb ⋅ in.)k ]
= (178.885 lb ⋅ in.)i + (536.66 lb ⋅ in.) j − (211.67 lb ⋅ in.)k
M = (178.885) 2 + (536.66) 2 + (−211.67) 2
= 603.99 lb ⋅ in
M = 604 lb ⋅ in. 
M
= 0.29617i + 0.88852 j − 0.35045k
M
cos θ x = 0.29617
λ axis =
cos θ y = 0.88852
θ x = 72.8° θ y = 27.3° θ z = 110.5° 
cos θ z = −0.35045
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239
PROBLEM 3.79
If P = 20 lb, replace the three couples with a single
equivalent couple, specifying its magnitude and the
direction of its axis.
SOLUTION
From the solution to Problem. 3.78:
16-lb force:
M1 = −(480 lb ⋅ in.)k
40-lb force:
M 2 = 8 5[(10 lb ⋅ in.)i + (30 lb ⋅ in.) j + (15 lb ⋅ in.)k ]
P = 20 lb
M 3 = rC × P
= (30 in.)i × (20 lb)k
= (600 lb ⋅ in.) j
M = M1 + M 2 + M 3
= −(480)k + 8 5 (10i + 30 j + 15k ) + 600 j
= (178.885 lb ⋅ in.)i + (1136.66 lb ⋅ in.) j − (211.67 lb ⋅ in.)k
M = (178.885) 2 + (113.66) 2 + (211.67)2
M = 1170 lb ⋅ in. 
= 1169.96 lb ⋅ in.
M
= 0.152898i + 0.97154 j − 0.180921k
M
cos θ x = 0.152898
λ axis =
cos θ y = 0.97154
θ x = 81.2° θ y = 13.70° θ z = 100.4° 
cos θ z = −0.180921
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240
PROBLEM 3.80
In a manufacturing operation, three holes are drilled
simultaneously in a workpiece. If the holes are perpendicular to the
surfaces of the workpiece, replace the couples applied to the drills
with a single equivalent couple, specifying its magnitude and the
direction
of
its axis.
SOLUTION
M = M1 + M 2 + M 3
= (1.5 N ⋅ m)(− cos 20° j + sin 20°k ) − (1.5 N ⋅ m) j
+ (1.75 N ⋅ m)(− cos 25° j + sin 25°k )
= −(4.4956 N ⋅ m) j + (0.22655 N ⋅ m)k
M = (0) 2 + (−4.4956) 2 + (0.22655) 2
= 4.5013 N ⋅ m
M = 4.50 N ⋅ m 
M
= −(0.99873j + 0.050330k )
M
cos θ x = 0
λaxis =
cos θ y = −0.99873
θ x = 90.0°, θ y = 177.1°, θ z = 87.1° 
cos θ z = 0.050330
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241
PROBLEM 3.81
A 260-lb force is applied at A to the rolled-steel section shown. Replace that
force with an equivalent force-couple system at the center C of the section.
SOLUTION
AB = (2.5 in.) 2 + (6.0 in.)2 = 6.50 in.
2.5 in. 5
=
6.5 in. 13
6.0 in. 12
cos α =
=
6.5 in. 13
sin α =
α = 22.6°
F = − F sin α i − F cos α j
5
12
= −(260 lb) i − (260 lb) j
13
13
= −(100.0 lb)i − (240 lb) j
M C = rA /C × F
= (2.5i + 4.0 j) × (−100.0i − 240 j)
= 400k − 600k
= −(200 lb ⋅ in.)k
F = 260 lb
67.4°; M C = 200 lb ⋅ in.

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242
PROBLEM 3.82
A 30-lb vertical force P is applied at A to the bracket shown, which is held by
screws at B and C. (a) Replace P with an equivalent force-couple system at B.
(b) Find the two horizontal forces at B and C that are equivalent to the couple
obtained in part a.
SOLUTION
(a)
M B = (30 lb)(5 in.)
= 150.0 lb ⋅ in.
F = 30.0 lb , M B = 150.0 lb ⋅ in.
(b)
B=C =

150 lb ⋅ in.
= 50.0 lb
3.0 in.
B = 50.0 lb
; C = 50.0 lb

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243
PROBLEM 3.83
The force P has a magnitude of 250 N and is applied at the end C of
a 500-mm rod AC attached to a bracket at A and B. Assuming α = 30°
and β = 60°, replace P with (a) an equivalent force-couple system at B,
(b) an equivalent system formed by two parallel forces applied at
A and B.
SOLUTION
(a)
ΣF : F = P or F = 250 N
Equivalence requires
60°
ΣM B : M = −(0.3 m)(250 N) = −75 N ⋅ m
The equivalent force-couple system at B is
FB = 250 N
(b)
M B = 75.0 N ⋅ m
60°

We require
Equivalence then requires
ΣFx : 0 = FA cos φ + FB cos φ
FA = − FB
or cos φ = 0
ΣFy : − 250 = − FA sin φ − FB sin φ
Now if
FA = − FB  −250 = 0, reject.
cos φ = 0
or
and
Also,
φ = 90°
FA + FB = 250
ΣM B : − (0.3 m)(250 N) = (0.2m) FA
or
FA = −375 N
and
FB = 625 N
FA = 375 N
60°
FB = 625 N
60.0° 
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244
PROBLEM 3.84
Solve Problem 3.83, assuming α = β = 25°.
PROBLEM 3.83 The force P has a magnitude of 250 N and is applied
at the end C of a 500-mm rod AC attached to a bracket at A and B.
Assuming α = 30° and β = 60°, replace P with (a) an equivalent forcecouple system at B, (b) an equivalent system formed by two parallel
forces applied at A and B.
SOLUTION
(a)
Equivalence requires
ΣF : FB = P or FB = 250 N
25.0°
ΣM B : M B = −(0.3 m)[(250 N)sin 50°] = −57.453 N ⋅ m
The equivalent force-couple system at B is
FB = 250 N
(b)
25.0°
M B = 57.5 N ⋅ m

We require
Equivalence requires
M B = d AE Q (0.3 m)[(250 N) sin 50°]
= [(0.2 m) sin 50°]Q
Q = 375 N
Adding the forces at B:
FA = 375 N
25.0°
FB = 625 N
25.0° 
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245
PROBLEM 3.85
The 80-N horizontal force P acts on a bell crank as shown.
(a) Replace P with an equivalent force-couple system at B.
(b) Find the two vertical forces at C and D that are equivalent
to the couple found in part a.
SOLUTION
(a)
ΣF : FB = F = 80 N
Based on
or
FB = 80.0 N

M B = 4.00 N ⋅ m

ΣM : M B = Fd B
= 80 N (0.05 m)
= 4.0000 N ⋅ m
or
(b)
If the two vertical forces are to be equivalent to MB, they must be
a couple. Further, the sense of the moment of this couple must be
counterclockwise.
Then with FC and FD acting as shown,
ΣM : M D = FC d
4.0000 N ⋅ m = FC (0.04 m)
FC = 100.000 N
or FC = 100.0 N 
ΣFy : 0 = FD − FC
FD = 100.000 N
or FD = 100.0 N 
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246
PROBLEM 3.86
A dirigible is tethered by a cable attached to its cabin at B. If
the tension in the cable is 1040 N, replace the force exerted by
the cable at B with an equivalent system formed by two
parallel forces applied at A and C.
SOLUTION
Require the equivalent forces acting at A and C be parallel and at an
angle of α with the vertical.
Then for equivalence,
ΣFx : (1040 N)sin 30° = FA sin α + FB sin α
(1)
ΣFy : −(1040 N) cos 30° = − FA cos α − FB cos α
(2)
Dividing Equation (1) by Equation (2),
( FA + FB ) sin α
(1040 N) sin 30°
=
−(1040 N) cos 30° −( FA + FB ) cos α
Simplifying yields α = 30°.
Based on
ΣM C : [(1040 N) cos 30°](4 m) = ( FA cos 30°)(10.7 m)
FA = 388.79 N
FA = 389 N
or
60.0° 
Based on
ΣM A : − [(1040 N) cos 30°](6.7 m) = ( FC cos 30°)(10.7 m)
FC = 651.21 N
FC = 651 N
or
60.0° 
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247
PROBLEM 3.87
Three control rods attached to a lever ABC exert on it the forces
shown. (a) Replace the three forces with an equivalent force-couple
system at B. (b) Determine the single force that is equivalent to the
force-couple system obtained in part a, and specify its point of
application on the lever.
SOLUTION
(a)
First note that the two 90-N forces form a couple. Then
F = 216 N
where
and
θ
θ = 180° − (60° + 55°) = 65°
M = ΣM B
= (0.450 m)(216 N) cos 55° − (1.050 m)(90 N) cos 20°
= −33.049 N ⋅ m
The equivalent force-couple system at B is
F = 216 N
(b)
65.0°; M = 33.0 N ⋅ m

The single equivalent force F′ is equal to F. Further, since the sense of M is clockwise, F′ must be
applied between A and B. For equivalence,
ΣM B : M = aF ′ cos 55°
where a is the distance from B to the point of application of F′. Then
−33.049 N ⋅ m = − a(216 N) cos 55°
a = 0.26676 m
or
F ′ = 216 N
65.0° applied to the lever 267 mm to the left of B 
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248
PROBLEM 3.88
A hexagonal plate is acted upon by the force P and the couple shown.
Determine the magnitude and the direction of the smallest force P for
which this system can be replaced with a single force at E.
SOLUTION
From the statement of the problem, it follows that ΣM E = 0 for the given force-couple system. Further,
for Pmin, we must require that P be perpendicular to rB/E . Then
ΣM E : (0.2 sin 30° + 0.2)m × 300 N
+ (0.2 m)sin 30° × 300 N
− (0.4 m) Pmin = 0
or
Pmin = 300 N
Pmin = 300 N
30.0° 
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249
PROBLEM 3.89
A force and couple act as shown on a square plate of side a = 25 in. Knowing
that P = 60 lb, Q = 40 lb, and α = 50°, replace the given force and couple by
a single force applied at a point located (a) on line AB, (b) on line AC. In each
case determine the distance from A to the point of application of the force.
SOLUTION
Replace the given force-couple system with an equivalent forcecouple system at A.
Px = (60 lb)(cos 50°) = 38.567 lb
Py = (60 lb)(sin 50°) = 45.963 lb
M A = Py a − Qa
= (45.963 lb)(25 in.) − (40 lb)(25 in.)
= 149.075 lb ⋅ in.
(a)
Equating moments about A gives:
149.075 lb ⋅ in. = (45.963 lb) x
x = 3.24 in.
P = 60.0 lb
(b)
50.0°; 3.24 in. from A 
149.075 lb ⋅ in. = (38.567 lb) y
y = 3.87 in.
P = 60.0 lb
50.0°; 3.87 in. below A 
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250
PROBLEM 3.90
The force and couple shown are to be replaced by an equivalent single force.
Knowing that P = 2Q, determine the required value of α if the line of action of
the single equivalent force is to pass through (a) Point A, (b) Point C.
SOLUTION
(a)
We must have M A = 0
( P sin α )a − Q (a ) = 0
sin α =
Q Q 1
=
=
P 2Q 2
α = 30.0° 
(b)
MC = 0
We must have
( P sin α )a − ( P cos α ) a − Q (a) = 0
sin α − cos α =
Q Q 1
=
=
P 2Q 2
sin α = cos α +
1
2
(1)
1
4
1
1 − cos 2 α = cos 2 α + cos α +
4
sin 2 α = cos 2 α + cos α +
2 cos 2 α + cos α − 0.75 = 0
(2)
Solving the quadratic in cos α :
cos α =
−1 ± 7
4
α = 65.7° or 155.7°
Only the first value of α satisfies Eq. (1),
therefore α = 65.7° 
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251
PROBLEM 3.91
The shearing forces exerted on the cross section of a steel channel can
be represented by a 900-N vertical force and two 250-N horizontal
forces as shown. Replace this force and couple with a single force F
applied at Point C, and determine the distance x from C to line BD.
(Point C is defined as the shear center of the section.)
SOLUTION
Replace the 250-N forces with a couple and move the 900-N force to Point C such that its moment about H is
equal to the moment of the couple
M H = (0.18)(250 N)
= 45 N ⋅ m
Then
or
M H = x(900 N)
45 N ⋅ m = x(900 N)
x = 0.05 m
F = 900 N
x = 50.0 mm 
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252
PROBLEM 3.92
A force and a couple are applied as shown to the end of a
cantilever beam. (a) Replace this system with a single force F
applied at Point C, and determine the distance d from C to a
line drawn through Points D and E. (b) Solve part a if the
directions of the two 360-N forces are reversed.
SOLUTION
(a)
We have
ΣF : F = (360 N) j − (360 N) j − (600 N)k
or F = −(600 N)k 
ΣM D : (360 N)(0.15 m) = (600 N)(d )
and
d = 0.09 m
or d = 90.0 mm below ED 
(b)
We have from part a:
F = −(600 N)k 
ΣM D : − (360 N)(0.15 m) = −(600 N)(d )
and
d = 0.09 m
or d = 90.0 mm above ED 
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253
PROBLEM 3.93
An antenna is guyed by three cables as shown. Knowing
that the tension in cable AB is 288 lb, replace the force
exerted at A by cable AB with an equivalent force-couple
system at the center O of the base of the antenna.
SOLUTION
We have
d AB = (−64)2 + (−128) 2 + (16) 2 = 144 ft
Then
TAB =
Now
288 lb
( −64i − 128 j + 16k )
144
= (32 lb)( −4i − 8 j + k )
M = M O = rA / O × TAB
= 128 j × 32(−4i − 8 j + k )
= (4096 lb ⋅ ft)i + (16,384 lb ⋅ ft)k
The equivalent force-couple system at O is
F = −(128.0 lb)i − (256 lb) j + (32.0 lb)k 
M = (4.10 kip ⋅ ft)i + (16.38 kip ⋅ ft)k 
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254
PROBLEM 3.94
An antenna is guyed by three cables as shown. Knowing
that the tension in cable AD is 270 lb, replace the force
exerted at A by cable AD with an equivalent force-couple
system at the center O of the base of the antenna.
SOLUTION
We have
d AD = ( −64) 2 + (−128)2 + (−128) 2
= 192 ft
Then
Now
270 lb
(−64i − 128 j + 128k )
192
= (90 lb)(−i − 2 j − 2k )
TAD =
M = M O = rA/O × TAD
= 128 j × 90(−i − 2 j − 2k )
= −(23, 040 lb ⋅ ft)i + (11,520 lb ⋅ ft)k
The equivalent force-couple system at O is
F = −(90.0 lb)i − (180.0 lb) j − (180.0 lb)k 
M = −(23.0 kip ⋅ ft)i + (11.52 kip ⋅ ft)k 
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255
PROBLEM 3.95
A 110-N force acting in a vertical plane parallel to the yz-plane
is applied to the 220-mm-long horizontal handle AB of a
socket wrench. Replace the force with an equivalent forcecouple system at the origin O of the coordinate system.
SOLUTION
We have
ΣF : PB = F
where
PB = 110 N[− (sin15°) j + (cos15°)k ]
= −(28.470 N) j + (106.252 N)k
or F = −(28.5 N) j + (106.3 N)k 
We have
where
ΣM O : rB/O × PB = M O
rB /O = [(0.22 cos 35°)i + (0.15) j − (0.22sin 35°)k ] m
= (0.180213 m)i + (0.15 m) j − (0.126187 m)k
i
j
k
0.180213 0.15 0.126187 N ⋅ m = M O
0
−28.5
106.3
M O = [(12.3487)i − (19.1566) j − (5.1361)k ] N ⋅ m
or M O = (12.35 N ⋅ m)i − (19.16 N ⋅ m)j − (5.13 N ⋅ m)k 
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256
PROBLEM 3.96
An eccentric, compressive 1220-N force P is applied to the end
of a cantilever beam. Replace P with an equivalent force-couple
system at G.
SOLUTION
We have
ΣF : − (1220 N)i = F
F = − (1220 N)i 
Also, we have
ΣM G : rA/G × P = M
i
j
k
1220 0 −0.1 −0.06 N ⋅ m = M
−1 0
0
M = (1220 N ⋅ m)[(−0.06)(−1) j − ( −0.1)( −1)k ]
or M = (73.2 N ⋅ m) j − (122 N ⋅ m)k 
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257
PROBLEM 3.97
To keep a door closed, a wooden stick is wedged between the
floor and the doorknob. The stick exerts at B a 175-N force directed
along line AB. Replace that force with an equivalent force-couple
system at C.
SOLUTION
We have
ΣF : PAB = FC
where
PAB = λAB PAB
=
(33 mm)i + (990 mm) j − (594 mm)k
(175 N)
1155.00 mm
or FC = (5.00 N)i + (150.0 N) j − (90.0 N)k 
We have
ΣM C : rB/C × PAB = M C
i
j
k
M C = 5 0.683 −0.860 0 N ⋅ m
1
30
−18
= (5){(− 0.860)(−18)i − (0.683)(−18) j
+ [(0.683)(30) − (0.860)(1)]k}
or M C = (77.4 N ⋅ m)i + (61.5 N ⋅ m) j + (106.8 N ⋅ m)k 
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258
PROBLEM 3.98
A 46-lb force F and a 2120-lb⋅in. couple M are
applied to corner A of the block shown. Replace
the given force-couple system with an equivalent
force-couple system at corner H.
SOLUTION
We have
Then
Also
d AJ = (18) 2 + (−14) 2 + (−3) 2 = 23 in.
46 lb
(18i − 14 j − 3k )
23
= (36 lb)i − (28 lb) j − (6 lb)k
F=
d AC = (−45) 2 + (0) 2 + (−28) 2 = 53 in.
2120 lb ⋅ in.
( −45i − 28k )
53
= −(1800 lb ⋅ in.)i − (1120 lb ⋅ in.)k
Then
M=
Now
M ′ = M + rA/H × F
where
Then
rA/H = (45 in.)i + (14 in.) j
i
j
k
M ′ = (−1800i − 1120k ) + 45 14
0
36 −28 −6
= (−1800i − 1120k ) + {[(14)(−6)]i + [−(45)( −6)]j + [(45)(−28) − (14)(36)]k}
= (−1800 − 84)i + (270) j + (−1120 − 1764)k
= −(1884 lb ⋅ in.)i + (270 lb ⋅ in.)j − (2884 lb ⋅ in.)k
= −(157 lb ⋅ ft)i + (22.5 lb ⋅ ft) j − (240 lb ⋅ ft)k
F′ = (36.0 lb)i − (28.0 lb) j − (6.00 lb)k 
The equivalent force-couple system at H is
M ′ = −(157.0 lb ⋅ ft)i + (22.5 lb ⋅ ft) j − (240 lb ⋅ ft)k 
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259
PROBLEM 3.99
A 77-N force F1 and a 31-N ⋅ m couple M1 are
applied to corner E of the bent plate shown. If
F1 and M1 are to be replaced with an equivalent
force-couple system (F2, M2) at corner B and if
(M2)z = 0, determine (a) the distance d, (b) F2
and M2.
SOLUTION
(a)
ΣM Bz : M 2 z = 0
We have
k ⋅ (rH /B × F1 ) + M 1z = 0
where
(1)
rH /B = (0.31 m)i − (0.0233) j
F1 = λ EH F1
(0.06 m)i + (0.06 m) j − (0.07 m)k
(77 N)
0.11 m
= (42 N)i + (42 N) j − (49 N)k
=
M1z = k ⋅ M1
M1 = λEJ M 1
=
− di + (0.03 m) j − (0.07 m)k
d 2 + 0.0058 m
(31 N ⋅ m)
Then from Equation (1),
0
0
1
( −0.07 m)(31 N ⋅ m)
=0
0.31 −0.0233 0 +
2
+
0.0058
d
−49
42
42
Solving for d, Equation (1) reduces to
(13.0200 + 0.9786) −
from which
2.17 N ⋅ m
d 2 + 0.0058
d = 0.1350 m
=0
or d = 135.0 mm 
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260
PROBLEM 3.99 (Continued)
(b)
F2 = F1 = (42i + 42 j − 49k ) N
or F2 = (42.0 N)i + (42.0 N) j − (49.0 N)k 
M 2 = rH /B × F1 + M1
i
j
k
(0.1350)i + 0.03j − 0.07k
(31 N ⋅ m)
= 0.31 −0.0233 0 +
0.155000
42
42
−49
= (1.14170i + 15.1900 j + 13.9986k ) N ⋅ m
+ (−27.000i + 6.0000 j − 14.0000k ) N ⋅ m
M 2 = − (25.858 N ⋅ m)i + (21.190 N ⋅ m) j
or M 2 = − (25.9 N ⋅ m)i + (21.2 N ⋅ m) j 
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261
PROBLEM 3.100
A 2.6-kip force is applied at Point D of the cast iron post shown.
Replace that force with an equivalent force-couple system at the
center A of the base section.
SOLUTION

DE = −(12 in.) j − (5 in.)k; DE = 13.00 in.

DE
F = (2.6 kips)
DE
F = (2.6 kips)
−12 j − 5k
13
F = −(2.40 kips) j − (1.000 kip)k 
M A = rD /A × F
where
rD /A = (6 in.)i + (12 in.) j
i
j
k
12 in.
0
M A = 6 in.
0
−2.4 kips −1.0 kips
M A = −(12.00 kip ⋅ in.)i + (6.00 kip ⋅ in.) j − (14.40 kip ⋅ in.)k 
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262
PROBLEM 3.101
A 3-m-long beam is subjected to a variety of loadings. (a) Replace each loading with an equivalent forcecouple system at end A of the beam. (b) Which of the loadings are equivalent?
SOLUTION
(a)
(a) We have
ΣFY : − 300 N − 200 N = Ra
or R a = 500 N 
and
ΣM A : − 400 N ⋅ m − (200 N)(3 m) = M a
or M a = 1000 N ⋅ m
(b) We have

ΣFY : 200 N + 300 N = Rb
or R b = 500 N 
and
ΣM A : − 400 N ⋅ m + (300 N)(3 m) = M b
or M b = 500 N ⋅ m
(c) We have

ΣFY : − 200 N − 300 N = Rc
or R c = 500 N 
and
ΣM A : 400 N ⋅ m − (300 N)(3 m) = M c
or M c = 500 N ⋅ m

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263
PROBLEM 3.101 (Continued)
(d) We have
ΣFY : − 500 N = Rd
or R d = 500 N 
and
ΣM A : 400 N ⋅ m − (500 N)(3 m) = M d
or M d = 1100 N ⋅ m
(e) We have

ΣFY : 300 N − 800 N = Re
or R e = 500 N 
and
ΣM A : 400 N ⋅ m + 1000 N ⋅ m − (800 N)(3 m) = M e
or M e = 1000 N ⋅ m
(f ) We have

ΣFY : − 300 N − 200 N = R f
or R f = 500 N 
and
ΣM A : 400 N ⋅ m − (200 N)(3 m) = M f
or M f = 200 N ⋅ m
(g) We have

ΣFY : − 800 N + 300 N = Rg
or R g = 500 N 
and
ΣM A : 1000 N ⋅ m + 400 N ⋅ m + (300 N)(3 m) = M g
or M g = 2300 N ⋅ m
(h) We have

ΣFY : − 250 N − 250 N = Rh
or R h = 500 N 
and
ΣM A : 1000 N ⋅ m + 400 N ⋅ m − (250 N)(3 m) = M h
or M h = 650 N ⋅ m
(b)

Therefore, loadings (a) and (e) are equivalent.
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264
PROBLEM 3.102
A 3-m-long beam is loaded as shown. Determine the loading
of Prob. 3.101 that is equivalent to this loading.
SOLUTION
We have
ΣFY : − 200 N − 300 N = R
R = 500 N
or
and
ΣM A : 500 N ⋅ m + 200 N ⋅ m − (300 N)(3 m) = M
M = 200 N ⋅ m
or
Problem 3.101 equivalent force-couples at A:
Case

R

M
(a)
500 N
1000 N⋅m
(b)
500 N
500 N⋅m
(c)
500 N
500 N⋅m
(d)
500 N
1100 N⋅m
(e)
500 N
1000 N⋅m
(f )
500 N
200 N⋅m
(g)
500 N
2300 N⋅m
(h)
500 N
650 N⋅m
Equivalent to case (f ) of Problem 3.101 
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265
PROBLEM 3.103
Determine the single equivalent force and the distance from Point A to its line of action for the beam and
loading of (a) Prob. 3.101a, (b) Prob. 3.101b, (c) Prob. 3.102.
SOLUTION
For equivalent single force at distance d from A:
(a)
We have
ΣFY : − 300 N − 200 N = R
or R = 500 N 
ΣM C : − 400 N ⋅ m + (300 N)(d )
and
− (200 N)(3 − d ) = 0
or d = 2.00 m 
(b)
We have
ΣFY : 200 N + 300 N = R
or R = 500 N 
ΣM C : − 400 N ⋅ m − (200 N)(d )
and
+ (300 N)(3 − d ) = 0
or d = 1.000 m 
(c)
We have
ΣFY : − 200 N − 300 N = R
or R = 500 N 
ΣM C : 500 N ⋅ m + 200 N ⋅ m
and
+ (200 N)( d ) − (300 N)(3 − d ) = 0
or d = 0.400 m 
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266
PROBLEM 3.104
Five separate force-couple systems act at the corners of a piece of sheet metal, which has been bent into the
shape shown. Determine which of these systems is equivalent to a force F = (10 lb)i and a couple of moment
M = (15 lb ⋅ ft)j + (15 lb ⋅ ft)k located at the origin.
SOLUTION
First note that the force-couple system at F cannot be equivalent because of the direction of the force [The
force of the other four systems is (10 lb)i]. Next, move each of the systems to the origin O; the forces remain
unchanged.
A: M A = ΣM O = (5 lb ⋅ ft) j + (15 lb ⋅ ft)k + (2 ft)k × (10 lb)i
= (25 lb ⋅ ft) j + (15 lb ⋅ ft)k
D : M D = ΣM O = −(5 lb ⋅ ft) j + (25 lb ⋅ ft)k
+ [(4.5 ft)i + (1 ft) j + (2 ft)k ] × 10 lb)i
= (15 lb ⋅ ft)i + (15 lb ⋅ ft)k
G : M G = ΣM O = (15 lb ⋅ ft)i + (15 lb ⋅ ft) j
I : M I = ΣM I = (15 lb ⋅ ft) j − (5 lb ⋅ ft)k
+ [(4.5 ft)i + (1 ft) j] × (10 lb) j
= (15 lb ⋅ ft) j − (15 lb ⋅ ft)k
The equivalent force-couple system is the system at corner D.

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267
PROBLEM 3.105
Three horizontal forces are applied as shown to a vertical cast iron arm.
Determine the resultant of the forces and the distance from the ground to
its line of action when (a) P = 200 N, (b) P = 2400 N, (c) P = 1000 N.
SOLUTION
(a)
RD = +200 N − 600 N − 400 N = −800 N
M D = −(200 N)(0.450 m) + (600 N)(0.300 m) + (400 N)(0.1500 m)
= +150.0 N ⋅ m
y=
M D 150 N ⋅ m
=
= 0.1875 m
R
800 N
R = 800 N
; y = 187.5 mm 
(b)
RD = +2400 N − 600 N − 400 N = +1400 N
M D = −(2400 N)(0.450 m) + (600 N)(0.300 m) + (400 N)(0.1500 m)
= −840 N ⋅ m
y=
M D 840 N ⋅ m
=
= 0.600 m
R
1400 N
R = 1400 N
; y = 600 mm 
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268
PROBLEM 3.105 (Continued)
(c)
RD = +1000 − 600 − 400 = 0
M D = −(1000 N)(0.450 m) + (600 N)(0.300 m) + (400 N)(0.1500 m)
= −210 N ⋅ m
∴ y = ∞ System reduces to a couple.
M D = 210 N ⋅ m

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269
PROBLEM 3.106
Three stage lights are mounted on a pipe as shown.
The lights at A and B each weigh 4.1 lb, while the one
at C weighs 3.5 lb. (a) If d = 25 in., determine the
distance from D to the line of action of the resultant
of the weights of the three lights. (b) Determine the
value of d so that the resultant of the weights passes
through the midpoint of the pipe.
SOLUTION
For equivalence,
ΣFy : − 4.1 − 4.1 − 3.5 = − R or R = 11.7 lb
ΣFD : − (10 in.)(4.1 lb) − (44 in.)(4.1 lb)
−[(4.4 + d ) in.](3.5 lb) = −( L in.)(11.7 lb)
375.4 + 3.5d = 11.7 L (d , L in in.)
or
d = 25 in.
(a)
We have
375.4 + 3.5(25) = 11.7 L or
L = 39.6 in.
The resultant passes through a point 39.6 in. to the right of D.

L = 42 in.
(b)
We have
375.4 + 3.5d = 11.7(42)
or d = 33.1 in. 
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270
PROBLEM 3.107
The weights of two children sitting at ends A and B of a seesaw
are 84 lb and 64 lb, respectively. Where should a third child sit
so that the resultant of the weights of the three children will
pass through C if she weighs (a) 60 lb, (b) 52 lb.
SOLUTION
(a)
For the resultant weight to act at C,
Then
ΣM C = 0 WC = 60 lb
(84 lb)(6 ft) − 60 lb(d ) − 64 lb(6 ft) = 0
d = 2.00 ft to the right of C 
(b)
For the resultant weight to act at C,
Then
ΣM C = 0 WC = 52 lb
(84 lb)(6 ft) − 52 lb(d ) − 64 lb(6 ft) = 0
d = 2.31 ft to the right of C 
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271
PROBLEM 3.108
A couple of magnitude M = 54 lb ⋅ in. and the three forces shown are
applied to an angle bracket. (a) Find the resultant of this system of
forces. (b) Locate the points where the line of action of the resultant
intersects line AB and line BC.
SOLUTION
(a)
We have
ΣF : R = (−10 j) + (30 cos 60°)i
+ 30 sin 60° j + (−45i )
= −(30 lb)i + (15.9808 lb) j
or R = 34.0 lb
(b)
28.0° 
First reduce the given forces and couple to an equivalent force-couple system (R , M B ) at B.
We have
ΣM B : M B = (54 lb ⋅ in) + (12 in.)(10 lb) − (8 in.)(45 lb)
= −186 lb ⋅ in.
Then with R at D,
or
and with R at E,
or
ΣM B : −186 lb ⋅ in = a(15.9808 lb)
a = 11.64 in.
ΣM B : −186 lb ⋅ in = C (30 lb)
C = 6.2 in.
The line of action of R intersects line AB 11.64 in. to the left of B and intersects line BC 6.20 in.
below B.

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272
PROBLEM 3.109
A couple M and the three forces shown are applied to an angle bracket.
Find the moment of the couple if the line of action of the resultant of the
force system is to pass through (a) Point A, (b) Point B, (c) Point C.
SOLUTION
In each case, we must have M1R = 0
(a)
M AB = ΣM A = M + (12 in.)[(30 lb) sin 60°] − (8 in.)(45 lb) = 0
M = +48.231 lb ⋅ in.
(b)

M = 240 lb ⋅ in.

M BR = ΣM B = M + (12 in.)(10 lb) − (8 in.)(45 lb) = 0
M = +240 lb ⋅ in.
(c)
M = 48.2 lb ⋅ in.
M CR = ΣM C = M + (12 in.)(10 lb) − (8 in.)[(30 lb) cos 60°] = 0
M =0
M=0 
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273
PROBLEM 3.110
A 32-lb motor is mounted on the floor. Find the resultant of the
weight and the forces exerted on the belt, and determine where
the line of action of the resultant intersects the floor.
SOLUTION
We have
ΣF : (60 lb)i − (32 lb) j + (140 lb)(cos 30°i + sin 30° j) = R
R = (181.244 lb)i + (38.0 lb) j
or R = 185.2 lb
We have
11.84° 
ΣM O : ΣM O = xRy
− [(140 lb) cos 30°][(4 + 2 cos 30°)in.] − [(140 lb) sin 30°][(2 in.)sin 30°]
− (60 lb)(2 in.) = x(38.0 lb)
x=
and
1
(− 694.97 − 70.0 − 120) in.
38.0
x = −23.289 in.
Or resultant intersects the base (x-axis) 23.3 in. to the left of
the vertical centerline (y-axis) of the motor.

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274
PROBLEM 3.111
A machine component is subjected to the forces and couples
shown. The component is to be held in place by a single rivet
that can resist a force but not a couple. For P = 0, determine the
location of the rivet hole if it is to be located (a) on line FG,
(b) on line GH.
SOLUTION
We have
First replace the applied forces and couples with an equivalent force-couple system at G.
ΣFx : 200cos 15° − 120 cos 70° + P = Rx
Thus,
Rx = (152.142 + P) N
or
ΣFy : − 200sin 15° − 120sin 70° − 80 = Ry
Ry = −244.53 N
or
ΣM G : − (0.47 m)(200 N) cos15° + (0.05 m)(200 N)sin15°
+ (0.47 m)(120 N) cos 70° − (0.19 m)(120 N)sin 70°
− (0.13 m)( P N) − (0.59 m)(80 N) + 42 N ⋅ m
+ 40 N ⋅ m = M G
M G = −(55.544 + 0.13P) N ⋅ m
or
(1)
Setting P = 0 in Eq. (1):
Now with R at I,
ΣM G : − 55.544 N ⋅ m = − a(244.53 N)
a = 0.227 m
or
and with R at J,
ΣM G : − 55.544 N ⋅ m = −b(152.142 N)
b = 0.365 m
or
(a)
The rivet hole is 0.365 m above G.

(b)
The rivet hole is 0.227 m to the right of G.

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275
PROBLEM 3.112
Solve Problem 3.111, assuming that P = 60 N.
PROBLEM 3.111 A machine component is subjected to the
forces and couples shown. The component is to be held in
place by a single rivet that can resist a force but not a couple.
For P = 0, determine the location of the rivet hole if it is to
be located (a) on line FG, (b) on line GH.
SOLUTION
See the solution to Problem 3.111 leading to the development of Equation (1):
M G = −(55.544 + 0.13P) N ⋅ m
Rx = (152.142 + P) N
and
P = 60 N
For
we have
Rx = (152.142 + 60)
= 212.14 N
M G = −[55.544 + 0.13(60)]
= −63.344 N ⋅ m
Then with R at I,
ΣM G : −63.344 N ⋅ m = −a(244.53 N)
a = 0.259 m
or
and with R at J,
ΣM G : −63.344 N ⋅ m = −b(212.14 N)
b = 0.299 m
or
(a)
The rivet hole is 0.299 m above G.

(b)
The rivet hole is 0.259 m to the right of G.

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276
PROBLEM 3.113
A truss supports the loading shown. Determine the equivalent
force acting on the truss and the point of intersection of its
line of action with a line drawn through Points A and G.
SOLUTION
We have
R = ΣF
R = (240 lb)(cos 70°i − sin 70° j) − (160 lb) j
+ (300 lb)(− cos 40°i − sin 40° j) − (180 lb) j
R = −(147.728 lb)i − (758.36 lb) j
R = Rx2 + Ry2
= (147.728) 2 + (758.36) 2
= 772.62 lb
 Ry 

 Rx 
 −758.36 
= tan −1 

 −147.728 
= 78.977°
θ = tan −1 
or
We have
ΣM A = dRy
where
ΣM A = −[240 lb cos 70°](6 ft) − [240 lbsin 70°](4 ft)
R = 773 lb
79.0° 
− (160 lb)(12 ft) + [300 lb cos 40°](6 ft)
− [300 lb sin 40°](20 ft) − (180 lb)(8 ft)
= −7232.5 lb ⋅ ft
−7232.5 lb ⋅ ft
−758.36 lb
= 9.5370 ft
d=
or
d = 9.54 ft to the right of A 
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277
PROBLEM 3.114
Four ropes are attached to a crate and exert the forces shown.
If the forces are to be replaced with a single equivalent force
applied at a point on line AB, determine (a) the equivalent
force and the distance from A to the point of application of
the force when α = 30°, (b) the value of α so that the single
equivalent force is applied at Point B.
SOLUTION
We have
(a)
For equivalence,
ΣFx : −100 cos 30° + 400 cos 65° + 90 cos 65° = Rx
or
Rx = 120.480 lb
ΣFy : 100 sin α + 160 + 400 sin 65° + 90 sin 65° = Ry
or
Ry = (604.09 + 100sin α ) lb
With α = 30°,
Ry = 654.09 lb
Then
R = (120.480) 2 + (654.09) 2
= 665 lb
(1)
654.09
120.480
or θ = 79.6°
tan θ =
ΣM A : (46 in.)(160 lb) + (66 in.)(400 lb) sin 65°
Also
or
+ (26 in.)(400 lb) cos 65° + (66 in.)(90 lb)sin 65°
+ (36 in.)(90 lb) cos 65° = d (654.09 lb)
ΣM A = 42, 435 lb ⋅ in. and d = 64.9 in.
79.6° 

and R is applied 64.9 in. to the right of A.
(b)
R = 665 lb
We have d = 66 in.
Then
or
Using Eq. (1):
ΣM A : 42, 435 lb ⋅ in = (66 in.) Ry
Ry = 642.95 lb
642.95 = 604.09 + 100sin α
or α = 22.9° 
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278
PROBLEM 3.115
Solve Prob. 3.114, assuming that the 90-lb force is removed.
PROBLEM 3.114 Four ropes are attached to a crate and exert
the forces shown. If the forces are to be replaced with a single
equivalent force applied at a point on line AB, determine
(a) the equivalent force and the distance from A to the point of
application of the force when α = 30°, (b) the value of α so
that the single equivalent force is applied at Point B.
SOLUTION
(a)
For equivalence,
ΣFx : − (100 lb) cos 30° + (400 lb)sin 25° = Rx
or
Rx = 82.445 lb
ΣFy : 160 lb + (100 lb) sin 30° + (400 lb) cos 25° = Ry
or
Ry = 572.52 lb
R = (82.445)2 + (572.52) = 578.43 lb
tan θ =
572.52
82.445
or θ = 81.806°
ΣM A : (46 in.)(160 lb) + (66 in.)(400 lb) cos 25° + (26 in.)(400 lb)sin 25°
= d (527.52 lb)
d = 62.3 in.
R = 578 lb
(b)
81.8° and is applied 62.3 in. to the right of A.

We have d = 66.0 in. For R applied at B,
ΣM A : Ry (66 in.) = (160 lb)(46 in.) + (66 in.)(400 lb) cos 25° + (26 in.)(400 lb)sin 25°
Ry = 540.64 lb
ΣFY : 160 lb + (100 lb)sin α + (400 lb) cos 25° = 540.64 lb
α = 10.44° 
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279
PROBLEM 3.116
Four forces act on a 700 × 375-mm plate as shown. (a) Find
the resultant of these forces. (b) Locate the two points where
the line of action of the resultant intersects the edge of the
plate.
SOLUTION
(a)
R = ΣF
= (−400 N + 160 N − 760 N)i
+ (600 N + 300 N + 300 N) j
= −(1000 N)i + (1200 N) j
R = (1000 N) 2 + (1200 N)2
= 1562.09 N
 1200 N 
tan θ =  −

 1000 N 
= −1.20000
θ = −50.194°
(b)
R = 1562 N
50.2° 
M CR = Σr × F
= (0.5 m)i × (300 N + 300 N) j
= (300 N ⋅ m)k
(300 N ⋅ m)k = xi × (1200 N) j
x = 0.25000 m
x = 250 mm
(300 N ⋅ m) = yj × ( −1000 N)i
y = 0.30000 m
y = 300 mm
Intersection 250 mm to right of C and 300 mm above C 
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280
PROBLEM 3.117
Solve Problem 3.116, assuming that the 760-N force is directed
to the right.
PROBLEM 3.116 Four forces act on a 700 × 375-mm plate as
shown. (a) Find the resultant of these forces. (b) Locate the two
points where the line of action of the resultant intersects the edge
of the plate.
SOLUTION
R = ΣF
(a)
= ( −400N + 160 N + 760 N)i
+ (600 N + 300 N + 300 N) j
= (520 N)i + (1200 N) j
R = (520 N) 2 + (1200 N) 2 = 1307.82 N
 1200 N 
tan θ = 
 = 2.3077
 520 N 
θ = 66.5714°
R = 1308 N
66.6° 
M CR = Σr × F
(b)
= (0.5 m)i × (300 N + 300 N) j
= (300 N ⋅ m)k
(300 N ⋅ m)k = xi × (1200 N) j
x = 0.25000 m
or
x = 0.250 mm
(300 N ⋅ m)k = [ x′i + (0.375 m) j] × [(520 N)i + (1200 N) j]
= (1200 x′ − 195)k
x′ = 0.41250 m
or
x′ = 412.5 mm
Intersection 412 mm to the right of A and 250 mm to the right of C 
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281
PROBLEM 3.118
As follower AB rolls along the surface of member C, it exerts
a constant force F perpendicular to the surface. (a) Replace F
with an equivalent force-couple system at Point D obtained
by drawing the perpendicular from the point of contact to the
x-axis. (b) For a = 1 m and b = 2 m, determine the value of x
for which the moment of the equivalent force-couple system
at D is maximum.
SOLUTION
(a)
The slope of any tangent to the surface of member C is
dy d  
x2
=
b 1 − 2
dx dx   a
  −2b
  = 2 x
  a
Since the force F is perpendicular to the surface,
 dy 
tan α = −  
 dx 
−1
=
a2  1 
2b  x 
For equivalence,
ΣF : F = R
ΣM D : ( F cos α )( y A ) = M D
where
cos α =
2bx
(a ) + (2bx)2
2 2

x2 
y A = b 1 − 2 
 a 

x3 
2 Fb 2  x − 2 
a 

MD =
4
2 2
a + 4b x
Therefore, the equivalent force-couple system at D is
R=F
 a2 
tan −1 
 
 2bx 

x3 
2 Fb2  x − 2 
a 

M=

a 4 + 4b 2 x 2
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282
PROBLEM 3.118 (Continued)
(b)
To maximize M, the value of x must satisfy
dM
=0
dx
a = 1 m, b = 2 m
where for
M=
8F ( x − x3 )
1 + 16 x 2
dM
= 8F
dx
1

1 + 16 x 2 (1 − 3x 2 ) − ( x − x3 )  (32 x)(1 + 16 x 2 ) −1/ 2 
2
 =0
(1 + 16 x 2 )
(1 + 16 x 2 )(1 − 3x 2 ) − 16 x( x − x3 ) = 0
32 x 4 + 3x 2 − 1 = 0
or
x2 =
−3 ± 9 − 4(32)(−1)
= 0.136011 m 2
2(32)
Using the positive value of x2:
x = 0.36880 m
and − 0.22976 m 2
or x = 369 mm 
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283
PROBLEM 3.119
As plastic bushings are inserted into a 60-mm-diameter cylindrical sheet
metal enclosure, the insertion tools exert the forces shown on the enclosure.
Each of the forces is parallel to one of the coordinate axes. Replace these
forces with an equivalent force-couple system at C.
SOLUTION
For equivalence,
ΣF :
R = FA + FB + FC + FD
= −(17 N) j − (12 N) j − (16 N)k − (21 N)i
= −(21 N)i − (29 N) j − (16 N)k
ΣM C : M = rA /C × FA + rB /C × FB + rD /C × FD
M = [(0.11 m) j − (0.03 m)k ] × [−(17 N)] j
+ [(0.02 m)i + (0.11 m) j − (0.03 m)k ] × [ −(12 N)]j
+ [(0.03 m)i + (0.03 m) j − (0.03 m)k ] × [ −(21 N)]i
= −(0.51 N ⋅ m)i + [−(0.24 N ⋅ m)k − (0.36 N ⋅ m)i]
+ [(0.63 N ⋅ m)k + (0.63 N ⋅ m) j]
∴
The equivalent force-couple system at C is
R = −(21.0 N)i − (29.0 N) j − (16.00 N)k 
M = −(0.870 N ⋅ m)i + (0.630 N ⋅ m) j + (0.390 N ⋅ m)k 
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284
PROBLEM 3.120
Two 150-mm-diameter pulleys are mounted on
line shaft AD. The belts at B and C lie in
vertical planes parallel to the yz-plane. Replace
the belt forces shown with an equivalent forcecouple system at A.
SOLUTION
Equivalent force-couple at each pulley:
Pulley B:
R B = (145 N)(− cos 20° j + sin 20°k ) − 215 Nj
= − (351.26 N) j + (49.593 N)k
M B = − (215 N − 145 N)(0.075 m)i
= − (5.25 N ⋅ m)i
Pulley C:
R C = (155 N + 240 N)(− sin10° j − cos10°k )
= − (68.591 N) j − (389.00 N)k
M C = (240 N − 155 N)(0.075 m)i
= (6.3750 N ⋅ m)i
Then
R = R B + R C = − (419.85 N) j − (339.41)k
or R = (420 N) j − (339 N)k 
M A = M B + M C + rB/ A × R B + rC/ A × R C
i
j
k
0
0
N⋅m
= − (5.25 N ⋅ m)i + (6.3750 N ⋅ m)i + 0.225
0
−351.26 49.593
i
j
k
+ 0.45
0
0
N⋅m
0
−68.591 −389.00
= (1.12500 N ⋅ m)i + (163.892 N ⋅ m) j − (109.899 N ⋅ m)k
or M A = (1.125 N ⋅ m)i + (163.9 N ⋅ m) j − (109.9 N ⋅ m)k 
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285
PROBLEM 3.121
Four forces are applied to the machine component
ABDE as shown. Replace these forces with an
equivalent force-couple system at A.
SOLUTION
R = −(50 N) j − (300 N)i − (120 N)i − (250 N)k
R = −(420 N)i − (50 N)j − (250 N)k
rB = (0.2 m)i
rD = (0.2 m)i + (0.16 m)k
rE = (0.2 m)i − (0.1 m) j + (0.16 m)k
M RA = rB × [−(300 N)i − (50 N) j]
+ rD × (−250 N)k + r × ( − 120 N)i
i
j
k
i
j
k
= 0.2 m
0
0 + 0.2 m 0 0.16 m
−300 N −50 N 0
0
0 −250 N
i
j
k
+ 0.2 m −0.1 m 0.16 m
−120 N
0
0
= −(10 N ⋅ m)k + (50 N ⋅ m) j − (19.2 N ⋅ m) j − (12 N ⋅ m)k
Force-couple system at A is
R = −(420 N)i − (50 N) j − (250 N)k M RA = (30.8 N ⋅ m) j − (220 N ⋅ m)k 
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286
PROBLEM 3.122
While using a pencil sharpener, a student applies
the forces and couple shown. (a) Determine the
forces exerted at B and C knowing that these
forces and the couple are equivalent to a forcecouple system at A consisting of the force
R = (2.6 lb)i + Ry j − (0.7 lb)k and the couple
M RA = M x i + (1.0 lb · ft)j − (0.72 lb · ft)k.
(b) Find the corresponding values of Ry and M x .
SOLUTION
(a)
From the statement of the problem, equivalence requires
ΣF : B + C = R
or
ΣFx : Bx + C x = 2.6 lb
(1)
ΣFy : − C y = R y
(2)
ΣFz : − C z = −0.7 lb or C z = 0.7 lb
and
ΣM A : (rB/A × B + M B ) + rC/A × C = M AR
or
 1.75 
ΣM x : (1 lb ⋅ ft) + 
ft  (C y ) = M x
 12

(3)
 3.75 
 1.75 
 3.5 
ΣM y : 
ft  ( Bx ) + 
ft  (C x ) + 
ft  (0.7 lb) = 1 lb ⋅ ft
 12 
 12 
 12 
or
Using Eq. (1):
3.75Bx + 1.75C x = 9.55
3.75Bx + 1.75(2.6 Bx ) = 9.55
or
Bx = 2.5 lb
and
C x = 0.1 lb
 3.5 
ΣM z : − 
ft  (C y ) = −0.72 lb ⋅ ft
 12 
C y = 2.4686 lb
or
B = (2.50 lb)i C = (0.1000 lb)i − (2.47 lb) j − (0.700 lb)k 
(b)
Eq. (2) 
Using Eq. (3):
Ry = −2.47 lb 
 1.75 
1+ 
 (2.4686) = M x
 12 
or M x = 1.360 lb ⋅ ft 
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287
PROBLEM 3.123
A blade held in a brace is used to tighten a screw at A.
(a) Determine the forces exerted at B and C, knowing
that these forces are equivalent to a force-couple
system at A consisting of R = −(30 N)i + Ry j + Rz k
and M RA = − (12 N · m)i. (b) Find the corresponding
values of Ry and Rz . (c) What is the orientation of the
slot in the head of the screw for which the blade is least
likely to slip when the brace is in the position shown?
SOLUTION
(a)
Equivalence requires
or
Equating the i coefficients:
Also,
or
Equating coefficients:
ΣF : R = B + C
−(30 N)i + Ry j + Rz k = − Bk + (−C x i + C y j + C z k )
i : − 30 N = −C x
or C x = 30 N
ΣM A : M RA = rB/A × B + rC/A × C
−(12 N ⋅ m)i = [(0.2 m)i + (0.15 m)j] × (− B)k
+(0.4 m)i × [−(30 N)i + C y j + C z k ]
i : − 12 N ⋅ m = −(0.15 m) B
k : 0 = (0.4 m)C y
or
or
B = 80 N
Cy = 0
j: 0 = (0.2 m)(80 N) − (0.4 m)C z or
C z = 40 N
B = −(80.0 N)k C = −(30.0 N)i + (40.0 N)k 
(b)
Now we have for the equivalence of forces
−(30 N)i + Ry j + Rz k = −(80 N)k + [(−30 N)i + (40 N)k ]
Equating coefficients:
j: R y = 0
Ry = 0 
k : Rz = −80 + 40
(c)
or
Rz = −40.0 N 
First note that R = −(30 N)i − (40 N)k. Thus, the screw is best able to resist the lateral force Rz
when the slot in the head of the screw is vertical.

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288
PROBLEM 3.124
In order to unscrew the tapped faucet A, a plumber uses two
pipe wrenches as shown. By exerting a 40-lb force on each
wrench, at a distance of 10 in. from the axis of the pipe and in a
direction perpendicular to the pipe and to the wrench, he
prevents the pipe from rotating, and thus avoids loosening or
further tightening the joint between the pipe and the tapped
elbow C. Determine (a) the angle θ that the wrench at A should
form with the vertical if elbow C is not to rotate about the
vertical, (b) the force-couple system at C equivalent to the two
40-lb forces when this condition is satisfied.
SOLUTION
We first reduce the given forces to force-couple systems at A and B, noting that
| M A | = | M B | = (40 lb)(10 in.)
= 400 lb ⋅ in.
We now determine the equivalent force-couple system at C.
R = (40 lb)(1 − cos θ )i − (40 lb) sin θ j
(1)
M CR = M A + M B + (15 in.)k × [−(40 lb) cos θ i − (40 lb)sin θ j]
+ (7.5 in.)k × (40 lb)i
= + 400 − 400 − 600cos θ j + 600sin θ i + 300 j
= (600 lb ⋅ in.)sin θ i + (300 lb ⋅ in.)(1 − 2 cos θ ) j
(a)
(2)
For no rotation about vertical, y component of M CR must be zero.
1 − 2cos θ = 0
cos θ = 1/2
θ = 60.0° 
(b)
For θ = 60.0° in Eqs. (1) and (2),
R = (20.0 lb)i − (34.641 lb) j; M CR = (519.62 lb ⋅ in.)i
R = (20.0 lb)i − (34.6 lb) j; M CR = (520 lb ⋅ in.)i 
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289
PROBLEM 3.125
Assuming θ = 60° in Prob. 3.124, replace the two 40-lb forces
with an equivalent force-couple system at D and determine
whether the plumber’s action tends to tighten or loosen the
joint between (a) pipe CD and elbow D, (b) elbow D and
pipe DE. Assume all threads to be right-handed.
PROBLEM 3.124 In order to unscrew the tapped faucet A,
a plumber uses two pipe wrenches as shown. By exerting a
40-lb force on each wrench, at a distance of 10 in. from the
axis of the pipe and in a direction perpendicular to the pipe
and to the wrench, he prevents the pipe from rotating, and
thus avoids loosening or further tightening the joint between
the pipe and the tapped elbow C. Determine (a) the angle θ
that the wrench at A should form with the vertical if elbow C
is not to rotate about the vertical, (b) the force-couple system
at C equivalent to the two 40-lb forces when this condition
is satisfied.
SOLUTION
The equivalent force-couple system at C for θ = 60° was obtained in the solution to Prob. 3.124:
R = (20.0 lb)i − (34.641 lb) j
M CR = (519.62 lb ⋅ in.)i
The equivalent force-couple system at D is made of R and M RD where
M RD = M CR + rC /D × R
= (519.62 lb ⋅ in.)i + (25.0 in.) j × [(20.0 lb)i − (34.641 lb) j]
= (519.62 lb ⋅ in.)i − (500 lb ⋅ in.)k
Equivalent force-couple at D:
R = (20.0 lb)i − (34.6 lb) j; M CR = (520 lb ⋅ in.)i − (500 lb ⋅ in.)k 
(a)
(b)
Since M RD has no component along the y-axis, the plumber’s action will neither loosen
nor tighten the joint between pipe CD and elbow.

, the plumber’s action will tend to tighten
Since the x component of M RD is
the joint between elbow and pipe DE.

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290
PROBLEM 3.126
As an adjustable brace BC is used to bring a wall into plumb, the
force-couple system shown is exerted on the wall. Replace this
force-couple system with an equivalent force-couple system at A
if R = 21.2 lb and M = 13.25 lb · ft.
SOLUTION
We have
where
or
We have
where
ΣF : R = R A = RλBC
λBC =
(42 in.)i − (96 in.) j − (16 in.)k
106 in.
RA =
21.2 lb
(42i − 96 j − 16k )
106
R A = (8.40 lb)i − (19.20 lb) j − (3.20 lb)k

ΣM A : rC/A × R + M = M A
rC/A = (42 in.)i + (48 in.)k =
1
(42i + 48k )ft
12
= (3.5 ft)i + (4.0 ft)k
R = (8.40 lb)i − (19.50 lb) j − (3.20 lb)k
M = −λBC M
−42i + 96 j + 16k
(13.25 lb ⋅ ft)
106
= −(5.25 lb ⋅ ft)i + (12 lb ⋅ ft) j + (2 lb ⋅ ft)k
=
Then
i
j
k
3.5
0
4.0 lb ⋅ ft + (−5.25i + 12 j + 2k ) lb ⋅ ft = M A
8.40 −19.20 −3.20
M A = (71.55 lb ⋅ ft)i + (56.80 lb ⋅ ft)j − (65.20 lb ⋅ ft)k
or M A = (71.6 lb ⋅ ft)i + (56.8 lb ⋅ ft)j − (65.2 lb ⋅ ft)k 
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291
PROBLEM 3.127
Three children are standing on a 5 × 5-m raft. If the weights
of the children at Points A, B, and C are 375 N, 260 N, and
400 N, respectively, determine the magnitude and the point
of application of the resultant of the three weights.
SOLUTION
We have
ΣF : FA + FB + FC = R
−(375 N) j − (260 N) j − (400 N) j = R
−(1035 N) j = R
or
We have
R = 1035 N 
ΣM x : FA ( z A ) + FB ( z B ) + FC ( zC ) = R ( z D )
(375 N)(3 m) + (260 N)(0.5 m) + (400 N)(4.75 m) = (1035 N)(z D )
z D = 3.0483 m
We have
or
z D = 3.05 m 
ΣM z : FA ( x A ) + FB ( xB ) + FC ( xC ) = R ( xD )
375 N(1 m) + (260 N)(1.5 m) + (400 N)(4.75 m) = (1035 N)( xD )
xD = 2.5749 m
or
xD = 2.57 m 
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292
PROBLEM 3.128
Three children are standing on a 5 × 5-m raft. The weights of
the children at Points A, B, and C are 375 N, 260 N, and 400 N,
respectively. If a fourth child of weight 425 N climbs onto the
raft, determine where she should stand if the other children
remain in the positions shown and the line of action of the
resultant of the four weights is to pass through the center of
the raft.
SOLUTION
We have
ΣF : FA + FB + FC = R
−(375 N) j − (260 N) j − (400 N) j − (425 N) j = R
R = −(1460 N) j
We have
ΣM x : FA ( z A ) + FB ( z B ) + FC ( zC ) + FD ( z D ) = R ( z H )
(375 N)(3 m) + (260 N)(0.5 m) + (400 N)(4.75 m)
+ (425 N)(z D ) = (1460 N)(2.5 m)
z D = 1.16471 m
We have
or
z D = 1.165 m 
ΣM z : FA ( x A ) + FB ( xB ) + FC ( xC ) + FD ( xD ) = R ( xH )
(375 N)(1 m) + (260 N)(1.5 m) + (400 N)(4.75 m)
+ (425 N)(xD ) = (1460 N)(2.5 m)
xD = 2.3235 m
or
xD = 2.32 m 
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293
PROBLEM 3.129
Four signs are mounted on a frame spanning a highway,
and the magnitudes of the horizontal wind forces acting on
the signs are as shown. Determine the magnitude and the
point of application of the resultant of the four wind forces
when a = 1 ft and b = 12 ft.
SOLUTION
We have
Assume that the resultant R is applied at Point P whose coordinates are (x, y, 0).
Equivalence then requires
ΣFz : − 105 − 90 − 160 − 50 = − R
or R = 405 lb 
ΣM x : (5 ft)(105 lb) − (1 ft)(90 lb) + (3 ft)(160 lb)
+ (5.5 ft)(50 lb) = − y (405 lb)
or
y = −2.94 ft
ΣM y : (5.5 ft)(105 lb) + (12 ft)(90 lb) + (14.5 ft)(160 lb)
+ (22.5 ft)(50 lb) = − x(405 lb)
x = 12.60 ft
or
R acts 12.60 ft to the right of member AB and 2.94 ft below member BC.

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294
PROBLEM 3.130
Four signs are mounted on a frame spanning a highway,
and the magnitudes of the horizontal wind forces acting on
the signs are as shown. Determine a and b so that the point
of application of the resultant of the four forces is at G.
SOLUTION
Since R acts at G, equivalence then requires that ΣM G of the applied system of forces also be zero. Then at
G : ΣM x : − (a + 3) ft × (90 lb) + (2 ft)(105 lb)
+ (2.5 ft)(50 lb) = 0
or a = 0.722 ft 

ΣM y : − (9 ft)(105 ft) − (14.5 − b) ft × (90 lb)
+ (8 ft)(50 lb) = 0

or b = 20.6 ft 
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295
PROBLEM 3.131*
A group of students loads a 2 × 3.3-m flatbed trailer with two
0.66 × 0.66 × 0.66-m boxes and one 0.66 × 0.66 × 1.2-m box. Each
of the boxes at the rear of the trailer is positioned so that it is
aligned with both the back and a side of the trailer. Determine the
smallest load the students should place in a second 0.66 × 0.66 ×
1.2-m box and where on the trailer they should secure it, without
any part of the box overhanging the sides of the trailer, if each
box is uniformly loaded and the line of action of the resultant of
the weights of the four boxes is to pass through the point of
intersection of the centerlines of the trailer and the axle.
(Hint: Keep in mind that the box may be placed either on its
side or on its end.)
SOLUTION
For the smallest weight on the trailer so that the resultant force of the four weights acts over the axle at the
intersection with the center line of the trailer, the added 0.66 × 0.66 × 1.2-m box should be placed adjacent to
one of the edges of the trailer with the 0.66 × 0.66-m side on the bottom. The edges to be considered are based
on the location of the resultant for the three given weights.
We have
ΣF : − (224 N) j − (392 N) j − (176 N) j = R
R = −(792 N) j
We have
ΣM z : − (224 N)(0.33 m) − (392 N)(1.67 m) − (176 N)(1.67 m) = ( −792 N)( x)
xR = 1.29101 m
We have
ΣM x : (224 N)(0.33 m) + (392 N)(0.6 m) + (176 N)(2.0 m) = (792 N)( z )
z R = 0.83475 m
From the statement of the problem, it is known that the resultant of R from the original loading and the
lightest load W passes through G, the point of intersection of the two center lines. Thus, ΣM G = 0.
Further, since the lightest load W is to be as small as possible, the fourth box should be placed as far from G
as possible without the box overhanging the trailer. These two requirements imply
(0.33 m ≤ x ≤ 1 m) (1.5 m ≤ z ≤ 2.97 m)
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296
PROBLEM 3.131* (Continued)
xL = 0.33 m
With
at
G : ΣM z : (1 − 0.33) m × WL − (1.29101 − 1) m × (792 N) = 0
WL = 344.00 N
or
Now we must check if this is physically possible,
at
G : ΣM x : ( z L − 1.5) m × 344 N) − (1.5 − 0.83475) m × (792 N) = 0
z L = 3.032 m
or
which is not acceptable.
z L = 2.97 m:
With
at
G : ΣM x : (2.97 − 1.5) m × WL − (1.5 − 0.83475) m × (792 N) = 0
WL = 358.42 N
or
Now check if this is physically possible,
at
G : ΣM z : (1 − xL ) m × (358.42 N) − (1.29101 − 1) m × (792 N) = 0
or
xL = 0.357 m ok!
WL = 358 N 
The minimum weight of the fourth box is
And it is placed on end A (0.66 × 0.66-m side down) along side AB with the center of the box 0.357 m
from side AD.

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297
PROBLEM 3.132*
Solve Problem 3.131 if the students want to place as much
weight as possible in the fourth box and at least one side of
the box must coincide with a side of the trailer.
PROBLEM 3.131* A group of students loads a 2 × 3.3-m
flatbed trailer with two 0.66 × 0.66 × 0.66-m boxes and one
0.66 × 0.66 × 1.2-m box. Each of the boxes at the rear of the
trailer is positioned so that it is aligned with both the back
and a side of the trailer. Determine the smallest load the
students should place in a second 0.66 × 0.66 × 1.2-m box and
where on the trailer they should secure it, without any part
of the box overhanging the sides of the trailer, if each box is
uniformly loaded and the line of action of the resultant of
the weights of the four boxes is to pass through the point of
intersection of the centerlines of the trailer and the axle.
(Hint: Keep in mind that the box may be placed either on its
side or on its end.)
SOLUTION
First replace the three known loads with a single equivalent force R applied at coordinate ( xR , 0, z R ).
Equivalence requires
ΣFy : − 224 − 392 − 176 = − R
or
R = 792 N
ΣM x : (0.33 m)(224 N) + (0.6 m)(392 N)
+ (2 m)(176 N) = z R (792 N)
or
z R = 0.83475 m
ΣM z : − (0.33 m)(224 N) − (1.67 m)(392 N)
− (1.67 m)(176 N) = xR (792 N)
or
xR = 1.29101 m
From the statement of the problem, it is known that the resultant of R and the heaviest loads WH passes
through G, the point of intersection of the two center lines. Thus,
ΣM G = 0
Further, since WH is to be as large as possible, the fourth box should be placed as close to G as possible while
keeping one of the sides of the box coincident with a side of the trailer. Thus, the two limiting cases are
xH = 0.6 m or
z H = 2.7 m
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298
PROBLEM 3.132* (Continued)
Now consider these two possibilities.
With xH = 0.6 m
at
G : ΣM z : (1 − 0.6) m × WH − (1.29101 − 1) m × (792 N) = 0
WH = 576.20 N
or
Checking if this is physically possible
at
or
G : ΣM x : ( z H − 1.5) m × (576.20 N) − (1.5 − 0.83475) m × (792 N) = 0
z H = 2.414 m
which is acceptable.
With z H = 2.7 m
at
or
G : ΣM x : (2.7 − 1.5) m × WH − (1.5 − 0.83475) m × (792 N) = 0
WH = 439 N
Since this is less than the first case, the maximum weight of the fourth box is
WH = 576 N 
and it is placed with a 0.66 × 1.2-m side down, a 0.66-m edge along side AD, and the center 2.41 m
from side DC.

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299
PROBLEM 3.133*
A piece of sheet metal is bent into the shape shown and is acted
upon by three forces. If the forces have the same magnitude P,
replace them with an equivalent wrench and determine (a) the
magnitude and the direction of the resultant force R, (b) the pitch
of the wrench, (c) the axis of the wrench.
SOLUTION
(
)
First reduce the given forces to an equivalent force-couple system R, M OR at the origin.
We have
ΣF : − Pj + Pj + Pk = R
R = Pk
or

5  
ΣM O : − (aP ) j +  −( aP )i +  aP  k  = M OR
2  

5 

M OR = aP  −i − j + k 
2 

or
(a)
Then for the wrench,
R=P 
and
λ axis =
R
=k
R
cos θ x = 0 cos θ y = 0 cos θ z = 1
or
(b)
θ x = 90° θ y = 90° θ z = 0°

Now
M1 = λ axis ⋅ M OR
5 

= k ⋅ aP  −i − j + k 
2 

5
= aP
2
Then
P=
M1 25 aP
=
R
P
or P =
5
a 
2
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300
PROBLEM 3.133* (Continued)
(c)
The components of the wrench are (R , M1 ), where M1 = M1λ axis , and the axis of the wrench is
assumed to intersect the xy-plane at Point Q, whose coordinates are (x, y, 0). Thus, we require
M z = rQ × R R
where
M z = M O × M1
Then
5  5

aP  −i − j + k  − aPk = ( xi + yj) + Pk
2  2

Equating coefficients:
i : − aP = yP
or
j: − aP = − xP or
y = −a
x=a
The axis of the wrench is parallel to the z-axis and intersects the xy-plane at
x = a, y = −a. 
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301
PROBLEM 3.134*
Three forces of the same magnitude P act on a cube of side a as
shown. Replace the three forces by an equivalent wrench and
determine (a) the magnitude and direction of the resultant force
R, (b) the pitch of the wrench, (c) the axis of the wrench.
SOLUTION
Force-couple system at O:
R = Pi + Pj + Pk = P(i + j + k )
M OR = aj × Pi + ak × Pj + ai × Pk
= − Pak − Pai − Paj
M OR = − Pa (i + j + k )
Since R and M OR have the same direction, they form a wrench with M1 = M OR . Thus, the axis of the wrench
is the diagonal OA. We note that
cos θ x = cos θ y = cos θ z =
a
a 3
=
1
3
R = P 3 θ x = θ y = θ z = 54.7°
M1 = M OR = − Pa 3
Pitch = p =
M 1 − Pa 3
=
= −a
R
P 3
(a)
R = P 3 θ x = θ y = θ z = 54.7° 
(b)
– a
(c)
Axis of the wrench is diagonal OA.
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302
PROBLEM 3.135*
The forces and couples shown are applied to two screws as a piece
of sheet metal is fastened to a block of wood. Reduce the forces and
the couples to an equivalent wrench and determine (a) the resultant
force R, (b) the pitch of the wrench, (c) the point where the axis of
the wrench intersects the xz-plane.
SOLUTION
First, reduce the given force system to a force-couple at the origin.
We have
ΣF : − (10 lb) j − (11 lb) j = R
R = − (21 lb) j
We have
ΣM O : Σ(rO × F ) + ΣM C = M OR
M OR
i
j
k
i
j
k
= 0 0 20 lb ⋅ in. + 0 0 −15 lb ⋅ in. − (12 lb ⋅ in) j
0 −10 0
0 −11 0
= (35 lb ⋅ in.)i − (12 lb ⋅ in.) j
R = − (21 lb) j
(a)
(b)
We have
or R = − (21.0 lb) j 
R
R
= (− j) ⋅ [(35 lb ⋅ in.)i − (12 lb ⋅ in.) j]
M1 = λ R ⋅ M OR
λR =
= 12 lb ⋅ in. and M1 = −(12 lb ⋅ in.) j
and pitch
p=
M 1 12 lb ⋅ in.
=
= 0.57143 in.
R
21 lb
or
p = 0.571 in. 
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303
PROBLEM 3.135* (Continued)
(c)
We have
M OR = M1 + M 2
M 2 = M OR − M1 = (35 lb ⋅ in.)i
We require
M 2 = rQ/O × R
(35 lb ⋅ in.)i = ( xi + zk ) × [ −(21 lb) j]
35i = −(21x)k + (21z )i
From i:
From k:
35 = 21z
z = 1.66667 in.
0 = − 21x
z=0
The axis of the wrench is parallel to the y-axis and intersects the xz-plane at x = 0, z = 1.667 in.

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304
PROBLEM 3.136*
The forces and couples shown are applied to two screws as a piece
of sheet metal is fastened to a block of wood. Reduce the forces and
the couples to an equivalent wrench and determine (a) the resultant
force R, (b) the pitch of the wrench, (c) the point where the axis of
the wrench intersects the xz-plane.
SOLUTION
First, reduce the given force system to a force-couple system.
We have
ΣF : − (20 N)i − (15 N) j = R
We have
ΣM O : Σ(rO × F ) + ΣM C = M OR
R = 25 N
M OR = −20 N(0.1 m)j − (4 N ⋅ m)i − (1 N ⋅ m)j
= −(4 N ⋅ m)i − (3 N ⋅ m) j
R = −(20.0 N)i − (15.00 N)j 
(a)
(b)
We have
R
R
= (−0.8i − 0.6 j) ⋅ [−(4 N ⋅ m)i − (3 N ⋅ m)j]
M1 = λR ⋅ M OR
λ=
= 5 N⋅m
Pitch:
p=
M1 5 N ⋅ m
=
= 0.200 m
R
25 N
or p = 0.200 m 
(c)
From above, note that
M1 = M OR
Therefore, the axis of the wrench goes through the origin. The line of action of the wrench lies in the
xy-plane with a slope of
y=
3
x 
4
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305
PROBLEM 3.137*
Two bolts at A and B are tightened by applying the forces
and couples shown. Replace the two wrenches with a
single equivalent wrench and determine (a) the resultant
R, (b) the pitch of the single equivalent wrench, (c) the
point where the axis of the wrench intersects the xz-plane.
SOLUTION
First, reduce the given force system to a force-couple at the origin.
We have
ΣF : − (84 N) j − (80 N)k = R
and
ΣM O : Σ(rO × F ) + ΣM C = M OR
R = 116 N
i
j k
i
j k
0.6 0 0.1 + 0.4 0.3 0 + (−30 j − 32k ) N ⋅ m = M OR
0 84 0
0
0 80
M OR = − (15.6 N ⋅ m)i + (2 N ⋅ m) j − (82.4 N ⋅ m)k
R = − (84.0 N) j − (80.0 N)k 
(a)
(b)
We have
M1 = λ R ⋅ M OR
λR =
R
R
−84 j − 80k
⋅ [− (15.6 N ⋅ m)i + (2 N ⋅ m) j − (82.4 N ⋅ m)k ]
116
= 55.379 N ⋅ m
=−
and
Then pitch
M1 = M1λR = − (40.102 N ⋅ m) j − (38.192 N ⋅ m)k
p=
M 1 55.379 N ⋅ m
=
= 0.47741 m
R
116 N
or p = 0.477 m 
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306
PROBLEM 3.137* (Continued)
(c)
We have
M OR = M1 + M 2
M 2 = M OR − M1 = [(−15.6i + 2 j − 82.4k ) − (40.102 j − 38.192k )] N ⋅ m
= − (15.6 N ⋅ m)i + (42.102 N ⋅ m) j − (44.208 N ⋅ m)k
We require
M 2 = rQ/O × R
(−15.6i + 42.102 j − 44.208k ) = ( xi + zk ) × (84 j − 80k )
= (84 z )i + (80 x) j − (84 x)k
From i:
or
From k:
or
−15.6 = 84 z
z = − 0.185714 m
z = − 0.1857 m
−44.208 = −84 x
x = 0.52629 m
x = 0.526 m
The axis of the wrench intersects the xz-plane at
x = 0.526 m
y = 0 z = − 0.1857 m 
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307
PROBLEM 3.138*
Two bolts at A and B are tightened by applying the forces and
couples shown. Replace the two wrenches with a single equivalent
wrench and determine (a) the resultant R, (b) the pitch of the
single equivalent wrench, (c) the point where the axis of the
wrench intersects the xz-plane.
SOLUTION
First, reduce the given force system to a force-couple at the origin at B.
(a)
We have
15 
 8
ΣF : − (26.4 lb)k − (17 lb)  i + j  = R
 17 17 
R = − (8.00 lb)i − (15.00 lb) j − (26.4 lb)k 
and
We have
R = 31.4 lb
ΣM B : rA/B × FA + M A + M B = M BR
M RB
i
j
k
15 
 8
0 − 220k − 238  i + j  = 264i − 220k − 14(8i + 15 j)
= 0 −10
 17 17 
0 0 − 26.4
M RB = (152 lb ⋅ in.)i − (210 lb ⋅ in.)j − (220 lb ⋅ in.)k
(b)
We have
R
R
−8.00i − 15.00 j − 26.4k
=
⋅ [(152 lb ⋅ in.)i − (210 lb ⋅ in.) j − (220 lb ⋅ in.)k ]
31.4
= 246.56 lb ⋅ in.
M1 = λ R ⋅ M OR
λR =
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308
PROBLEM 3.138* (Continued)
and
Then pitch
(c)
We have
M1 = M1λR = − (62.818 lb ⋅ in.)i − (117.783 lb ⋅ in.) j − (207.30 lb ⋅ in.)k
p=
M 1 246.56 lb ⋅ in.
=
= 7.8522 in.
31.4 lb
R
or p = 7.85 in. 
M RB = M1 + M 2
M 2 = M RB − M1 = (152i − 210 j − 220k ) − ( − 62.818i − 117.783j − 207.30k )
= (214.82 lb ⋅ in.)i − (92.217 lb ⋅ in.) j − (12.7000 lb ⋅ in.)k
We require
M 2 = rQ/B × R
i
j
k
214.82i − 92.217 j − 12.7000k = x
0
z
−8 −15 −26.4
= (15 z )i − (8 z ) j + (26.4 x) j − (15 x)k
From i:
214.82 = 15 z
z = 14.3213 in.
From k:
−12.7000 = −15 x
x = 0.84667 in.
The axis of the wrench intersects the xz-plane at
x = 0.847 in. y = 0 z = 14.32 in. 
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309
PROBLEM 3.139*
A flagpole is guyed by three cables. If the tensions in the
cables have the same magnitude P, replace the forces exerted
on the pole with an equivalent wrench and determine (a) the
resultant force R, (b) the pitch of the wrench, (c) the point
where the axis of the wrench intersects the xz-plane.
SOLUTION
(a)
First reduce the given force system to a force-couple at the origin.
We have
ΣF : Pλ BA + P λDC + P λDE = R
 4
3  3
4   −9
4
12  
R = P  j − k  +  i − j  +  i − j + k  
5
5
5
5
25
5
25  
 
 

R=
R=
We have
3P
(2i − 20 j − k ) 
25
3P
27 5
P
(2) 2 + (20) 2 + (1)2 =
25
25
ΣM : Σ(rO × P) = M OR
3P 
4P 
4P
12 P 
 −4 P
 3P
 −9 P
(24a) j × 
j−
k  + (20a) j × 
i−
j  + (20a) j × 
i−
j+
k  = M OR
5
5
5
5
25
5
25






M OR =
24 Pa
( −i − k )
5
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310
PROBLEM 3.139* (Continued)
(b)
We have
M1 = λ R ⋅ M OR
where
λR =
Then
M1 =
25
1
R 3P
(2i − 20 j − k )
(2i − 20 j − k )
=
=
R 25
27 5 P 9 5
p=
and pitch
1
9 5
(2i − 20 j − k ) ⋅
M 1 −8Pa  25  −8a
=

=
R 15 5  27 5 P  81
M1 = M1λ R =
(c)
Then
M 2 = M OR − M1 =
24 Pa
−8 Pa
(−i − k ) =
5
15 5
or p = − 0.0988a 
−8Pa  1 
8 Pa
( −2i + 20 j + k )

 (2i − 20 j − k ) =
675
15 5  9 5 
24 Pa
8Pa
8 Pa
( −i − k ) −
(−2i + 20 j + k ) =
(−430i − 20 j − 406k )
5
675
675
M 2 = rQ/O × R
We require
 8Pa 
 3P 
 675  (−403i − 20 j − 406k ) = ( xi + zk ) ×  25  (2i − 20 j − k )




 3P 
=
 [20 zi + ( x + 2 z ) j − 20 xk ]
 25 
From i:
8(− 403)
Pa
 3P 
= 20 z 

675
 25 
From k:
8(−406)
Pa
 3P 
= −20 x 
 x = 2.0049a
675
 25 
z = −1.99012a
The axis of the wrench intersects the xz-plane at
x = 2.00a, z = −1.990a 
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311
PROBLEM 3.140*
Two ropes attached at A and B are used to move the trunk of a
fallen tree. Replace the forces exerted by the ropes with an
equivalent wrench and determine (a) the resultant force R,
(b) the pitch of the wrench, (c) the point where the axis of the
wrench intersects the yz-plane.
SOLUTION
(a)
(
)
First replace the given forces with an equivalent force-couple system R, M OR at the origin.
We have
d AC = (6) 2 + (2) 2 + (9) 2 = 11 m
d BD = (14) 2 + (2)2 + (5) 2 = 15 m
Then
1650 N
= (6i + 2 j + 9k )
11
= (900 N)i + (300 N) j + (1350 N)k
TAC =
and
1500 N
= (14i + 2 j + 5k )
15
= (1400 N)i + (200 N) j + (500 N)k
TBD =
Equivalence then requires
ΣF : R = TAC + TBD
= (900i + 300 j + 1350k )
+(1400i + 200 j + 500k )
= (2300 N)i + (500 N) j + (1850 N)k
ΣM O : M OR = rA × TAC + rB × TBD
= (12 m)k × [(900 N)i + (300 N)j + (1350 N)k ]
+ (9 m)i × [(1400 N)i + (200 N)j + (500 N)k ]
= −(3600)i + (10,800 − 4500) j + (1800)k
= −(3600 N ⋅ m)i + (6300 N ⋅ m)j + (1800 N ⋅ m)k
The components of the wrench are (R , M1 ), where
R = (2300 N)i + (500 N) j + (1850 N)k

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312
PROBLEM 3.140* (Continued)

(b)
We have
R = 100 (23)2 + (5) 2 + (18.5) 2 = 2993.7 N
Let
λ axis =
Then
1
R
(23i + 5 j + 18.5k )
=
R 29.937
M1 = λ axis ⋅ M OR
1
(23i + 5 j + 18.5k ) ⋅ (−3600i + 6300 j + 1800k )
29.937
1
[(23)( −36) + (5)(63) + (18.5)(18)]
=
0.29937
= −601.26 N ⋅ m
=
Finally,
P=
M1 −601.26 N ⋅ m
=
R
2993.7 N
or P = − 0.201 m 
(c)
We have
M1 = M 1 λ axis
= (−601.26 N ⋅ m) ×
1
(23i + 5 j + 18.5k )
29.937
or
M1 = −(461.93 N ⋅ m)i − (100.421 N ⋅ m) j − (371.56 N ⋅ m)k
Now
M 2 = M OR − M1
= (−3600i + 6300 j + 1800k )
− ( −461.93i − 100.421j − 371.56k )
= − (3138.1 N ⋅ m)i + (6400.4 N ⋅ m)j + (2171.6 N ⋅ m)k
For equivalence:
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313
PROBLEM 3.140* (Continued)
Thus, we require
M 2 = rP × R
r = ( yj + zk )
Substituting:
−3138.1i + 6400.4 j + 2171.6k =
i
j
k
0
y
z
2300 500 1850
Equating coefficients:
j : 6400.4 = 2300 z
or
k : 2171.6 = −2300 y or
The axis of the wrench intersects the yz-plane at
z = 2.78 m
y = − 0.944 m
y = −0.944 m
z = 2.78 m

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314
PROBLEM 3.141*
Determine whether the force-and-couple system shown can be
reduced to a single equivalent force R. If it can, determine R
and the point where the line of action of R intersects the
yz-plane. If it cannot be so reduced, replace the given system
with an equivalent wrench and determine its resultant, its pitch,
and the point where its axis intersects the yz-plane.
SOLUTION
First determine the resultant of the forces at D. We have
d DA = (−12) 2 + (9) 2 + (8) 2 = 17 in.
d ED = (−6) 2 + (0)2 + (−8)2 = 10 in.
Then
34 lb
= (−12i + 9 j + 8k )
17
= −(24 lb)i + (18 lb) j + (16 lb)k
FDA =
and
30 lb
= (−6i − 8k )
10
= −(18 lb)i − (24 lb)k
FED =
Then
ΣF : R = FDA + FED
= (−24i + 18 j + 16k + ( −18i − 24k )
= −(42 lb)i + (18 lb)j − (8 lb)k
For the applied couple
d AK = ( −6) 2 + (−6) 2 + (18)2 = 6 11 in.
Then
M=
160 lb ⋅ in.
( −6i − 6 j + 18k )
6 11
160
=
[−(1 lb ⋅ in.)i − (1 lb ⋅ in.)j + (3 lb ⋅ in.)k ]
11
To be able to reduce the original forces and couple to a single equivalent force, R and M
must be perpendicular. Thus
?
R ⋅ M =0
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315
PROBLEM 3.141* (Continued)
Substituting
(−42i + 18 j − 8k ) ⋅
160
or
11
160
11
?
( −i − j + 3k ) = 0
?
[(−42)(−1) + (18)(−1) + (−8)(3)] = 0

0 =0
or
R and M are perpendicular so that the given system can be reduced to the single equivalent force.
R = −(42.0 lb)i + (18.00 lb) j − (8.00 lb)k

Then for equivalence,
M = rP/D × R
Thus, we require
where
rP/D = −(12 in.)i + [( y − 3)in.] j + ( z in.)k
Substituting:
i
j
k
( −i − j + 3k ) = −12 ( y − 3) z
11
18
−42
−8
= [( y − 3)( −8) − ( z )(18)]i
160
+ [( z )(−42) − (−12)(−8)]j
+ [( −12)(18) − ( y − 3)(−42)]k
Equating coefficients:
j: −
k:
160
= − 42 z − 96
or
11
480
= −216 + 42( y − 3) or
11
The line of action of R intersects the yz-plane at
x=0
z = −1.137 in.
y = 11.59 in.
y = 11.59 in. z = −1.137 in.

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316
PROBLEM 3.142*
Determine whether the force-and-couple system shown can be
reduced to a single equivalent force R. If it can, determine R
and the point where the line of action of R intersects the
yz-plane. If it cannot be so reduced, replace the given system
with an equivalent wrench and determine its resultant, its
pitch, and the point where its axis intersects the yz-plane.
SOLUTION
First, reduce the given force system to a force-couple at the origin.
We have
ΣF : FA + FG = R
 (40 mm)i + (60 mm) j − (120 mm)k 
R = (50 N)k + 70 N 

140 mm


= (20 N)i + (30 N) j − (10 N)k
and
We have
R = 37.417 N
ΣM O : Σ(rO × F) + ΣM C = M OR
M OR = [(0.12 m) j × (50 N)k ] + {(0.16 m)i × [(20 N)i + (30 N) j − (60 N)k ]}
 (160 mm)i − (120 mm) j 
+ (10 N ⋅ m) 

200 mm


 (40 mm)i − (120 mm) j + (60 mm)k 
+ (14 N ⋅ m) 

140 mm


M 0R = (18 N ⋅ m)i − (8.4 N ⋅ m) j + (10.8 N ⋅ m)k
To be able to reduce the original forces and couples to a single equivalent force, R and M must be
perpendicular. Thus, R ⋅ M = 0.
Substituting
?
(20i + 30 j − 10k ) ⋅ (18i − 8.4 j + 10.8k ) = 0
?
or
(20)(18) + (30)(−8.4) + (−10)(10.8) = 0
or
0=0

R and M are perpendicular so that the given system can be reduced to the single equivalent force.
R = (20.0 N)i + (30.0 N) j − (10.00 N)k 
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317
PROBLEM 3.142* (Continued)

Then for equivalence,
Thus, we require
M OR = rp × R rp = yj + zk
Substituting:
i
j
k
18i − 8.4 j + 10.8k = 0 y
z
20 30 −10
Equating coefficients:
j: − 8.4 = 20 z
k:
or
z = −0.42 m
10.8 = −20 y or
y = −0.54 m
The line of action of R intersects the yz-plane at
x=0
y = −0.540 m z = −0.420 m

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318
PROBLEM 3.143*
Replace the wrench shown with an equivalent system consisting of two
forces perpendicular to the y-axis and applied respectively at A and B.
SOLUTION
Express the forces at A and B as
A = Ax i + Az k
B = Bx i + Bz k
Then, for equivalence to the given force system,
ΣFx : Ax + Bx = 0
(1)
ΣFz : Az + Bz = R
(2)
ΣM x : Az ( a) + Bz ( a + b) = 0
(3)
ΣM z : − Ax (a) − Bx (a + b) = M
(4)
Bx = − Ax
From Equation (1),
Substitute into Equation (4):
− Ax ( a) + Ax ( a + b) = M
M
M
and Bx = −
Ax =
b
b
From Equation (2),
Bz = R − Az
and Equation (3),
Az a + ( R − Az )(a + b) = 0
 a
Az = R 1 + 
 b
and
a

Bz = R − R 1 + 
 b
a
Bz = − R
b
M
A=
 b
Then
a


 i + R 1 + b  k 



M
B = −
 b
 a 
i −  b Rk 
 

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319
PROBLEM 3.144*
Show that, in general, a wrench can be replaced with two forces chosen in such a way that one force passes
through a given point while the other force lies in a given plane.
SOLUTION
First, choose a coordinate system so that the xy-plane coincides with the given plane. Also, position the
coordinate system so that the line of action of the wrench passes through the origin as shown in Figure a.
Since the orientation of the plane and the components (R, M) of the wrench are known, it follows that the
scalar components of R and M are known relative to the shown coordinate system.
A force system to be shown as equivalent is illustrated in Figure b. Let A be the force passing through the
given Point P and B be the force that lies in the given plane. Let b be the x-axis intercept of B.
The known components of the wrench can be expressed as
R = Rx i + Ry j + Rz k and M = M x i + M y j + M z k
while the unknown forces A and B can be expressed as
A = Ax i + Ay j + Az k and B = Bx i + Bz k
Since the position vector of Point P is given, it follows that the scalar components (x, y, z) of the
position vector rP are also known.
Then, for equivalence of the two systems,
ΣFx : Rx = Ax + Bx
(1)
ΣFy : Ry = Ay
(2)
ΣFz : Rz = Az + Bz
(3)
ΣM x : M x = yAz − zAy
(4)
ΣM y : M y = zAx − xAz − bBz
(5)
ΣM z : M z = xAy − yAx
(6)
Based on the above six independent equations for the six unknowns ( Ax , Ay , Az , Bx , Bz , b), there
exists a unique solution for A and B.
Ay = Ry 
From Equation (2),
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320
PROBLEM 3.144* (Continued)
1
Ax =   ( xRy − M z ) 
 y
Equation (6):
1
Bx = Rx −   ( xRy − M z ) 
 y
Equation (1):
Equation (4):
1
Az =   ( M x + zRy ) 
 y
Equation (3):
1
Bz = Rz −   ( M x + zRy ) 
 y
b=
Equation (5):
( xM x + yM y + zM z )
( M x − yRz + zRy )

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321
PROBLEM 3.145*
Show that a wrench can be replaced with two perpendicular forces, one of which is applied at a given point.
SOLUTION
First, observe that it is always possible to construct a line perpendicular to a given line so that the constructed
line also passes through a given point. Thus, it is possible to align one of the coordinate axes of a rectangular
coordinate system with the axis of the wrench while one of the other axes passes through the given point.
See Figures a and b.
We have
R = Rj and M = Mj
and are known.
The unknown forces A and B can be expressed as
A = Ax i + Ay j + Az k and B = Bx i + By j + Bz k
The distance a is known. It is assumed that force B intersects the xz-plane at (x, 0, z). Then for equivalence,
ΣFx :
0 = Ax + Bx
(1)
ΣFy :
R = Ay + By
(2)
ΣFz :
0 = Az + Bz
(3)
0 = − zBy
(4)
ΣM x :
ΣM y : M = − aAz − xBz + zBx
(5)
ΣM z :
(6)
0 = aAy + xBy
Since A and B are made perpendicular,
A ⋅ B = 0 or
There are eight unknowns:
Ax Bx + Ay B y + Az Bz = 0
(7)
Ax , Ay , Az , Bx , By , Bz , x, z
But only seven independent equations. Therefore, there exists an infinite number of solutions.
0 = − zBy
Next, consider Equation (4):
If By = 0, Equation (7) becomes
Ax Bx + Az Bz = 0
Ax2 + Az2 = 0
Using Equations (1) and (3), this equation becomes
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322
PROBLEM 3.145* (Continued)
Since the components of A must be real, a nontrivial solution is not possible. Thus, it is required that By ≠ 0,
so that from Equation (4), z = 0.
To obtain one possible solution, arbitrarily let Ax = 0.
(Note: Setting Ay , Az , or Bz equal to zero results in unacceptable solutions.)
The defining equations then become
0 = Bx
(1)′
R = Ay + By
(2)
0 = Az + Bz
(3)
M = − aAz − xBz
(5)′
0 = aAy + xBy
(6)
Ay By + Az Bz = 0
(7)′
Then Equation (2) can be written
Ay = R − By
Equation (3) can be written
Bz = − Az
Equation (6) can be written
x=−
aAy
By
Substituting into Equation (5)′,

R − By
M = − aAz −  − a

By

M
Az = −
By
aR
or

 ( − Az )


(8)
Substituting into Equation (7)′,
 M
 M

( R − By ) By +  −
By 
By  = 0
 aR  aR 
or
By =
a 2 R3
a2 R2 + M 2
Then from Equations (2), (8), and (3),
a2 R2
RM 2
= 2 2
2
2
a R +M
a R + M2

M  a 2 R3
aR 2 M
Az = −
=
−


aR  a 2 R 2 + M 2 
a2 R2 + M 2
Ay = R −
Bz =
2
aR 2 M
a2 R2 + M 2
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323
PROBLEM 3.145* (Continued)
In summary,
A=
RM
( Mj − aRk )
a R2 + M 2

B=
aR 2
(aRj + Mk )
a2 R2 + M 2

2
Which shows that it is possible to replace a wrench with two perpendicular forces, one of which is applied at
a given point.
Lastly, if R > 0 and M > 0, it follows from the equations found for A and B that Ay > 0 and B y > 0.
From Equation (6), x < 0 (assuming a > 0). Then, as a consequence of letting Ax = 0, force A lies in a plane
parallel to the yz-plane and to the right of the origin, while force B lies in a plane parallel to the yz-plane but
to the left to the origin, as shown in the figure below.

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324
PROBLEM 3.146*
Show that a wrench can be replaced with two forces, one of which has a prescribed line of action.
SOLUTION
First, choose a rectangular coordinate system where one axis coincides with the axis of the wrench and
another axis intersects the prescribed line of action (AA′). Note that it has been assumed that the line of action
of force B intersects the xz-plane at Point P(x, 0, z). Denoting the known direction of line AA′ by
λ A = λx i + λ y j + λz k
it follows that force A can be expressed as
A = Aλ A = A(λx i + λ y j + λz k )
Force B can be expressed as
B = Bx i + B y j + Bz k
Next, observe that since the axis of the wrench and the prescribed line of action AA′ are known, it follows that
the distance a can be determined. In the following solution, it is assumed that a is known.
Then for equivalence,
ΣFx : 0 = Aλx + Bx
(1)
ΣFy : R = Aλ y + By
(2)
ΣFz : 0 = Aλz + Bz
(3)
ΣM x : 0 = − zBy
(4)
ΣM y : M = − aAλz + zBx − xBz
(5)
ΣM x : 0 = − aAλ y + xB y
(6)
Since there are six unknowns (A, Bx, By, Bz, x, z) and six independent equations, it will be possible to
obtain a solution.
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325
PROBLEM 3.146* (Continued)
Case 1: Let z = 0 to satisfy Equation (4).
Now Equation (2):
Equation (3):
Equation (6):
Aλ y = R − B y
Bz = − Aλz
x=−
aAλ y
By
 a
= −
 By


 ( R − By )


Substitution into Equation (5):
  a
M = − aAλz −  − 
  By
1 M 
A=− 
B
λz  aR  y


 ( R − By )(− Aλz ) 



Substitution into Equation (2):
R=−
By =
Then
1 M 
B λ + By
λz  aR  y y
λz aR 2
λz aR − λ y M
MR
R
=
λz aR − λ y M λ − aR λ
y
z
M
λx MR
Bx = − Aλx =
λz aR − λ y M
A=−
Bz = − Aλz =
λz MR
λz aR − λ y M
A=
In summary,
B=
and

R 
x = a 1 −


By 


 λz aR − λ y M
= a 1 − R 
 λ aR 2

z

P
λA 
aR
λy −
λz
M
R
(λ M i + λz aRj + λz M k ) 
λz aR − λ y M x

 
 
or x =
λy M
λz R

Note that for this case, the lines of action of both A and B intersect the x-axis.
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326
PROBLEM 3.146* (Continued)
Case 2: Let By = 0 to satisfy Equation (4).
Now Equation (2):
A=
R
λy
Equation (1):
λ
Bx = − R  x
 λy





Equation (3):
λ
Bz = − R  z
 λy





Equation (6):
aAλ y = 0
which requires a = 0
Substitution into Equation (5):
 λ
M = z − R  x
  λ y
 λ

 − x − R  z

  λ y


M
  or λz x − λx z = 

 R


 λy

This last expression is the equation for the line of action of force B.
In summary,
 R
A=
 λy


 λA



 R
B=
 λy


 ( −λ x i − λx k )



Assuming that λx , λ y , λz ⬎ 0, the equivalent force system is as shown below.
Note that the component of A in the xz-plane is parallel to B.
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327
PROBLEM 3.147
A 300-N force is applied at A as shown. Determine (a) the moment
of the 300-N force about D, (b) the smallest force applied at B that
creates the same moment about D.
SOLUTION
(a)
Fx = (300 N) cos 25°
= 271.89 N
Fy = (300 N) sin 25°
= 126.785 N
F = (271.89 N)i + (126.785 N) j

r = DA = −(0.1 m)i − (0.2 m) j
MD = r × F
M D = [−(0.1 m)i − (0.2 m) j] × [(271.89 N)i + (126.785 N) j]
= −(12.6785 N ⋅ m)k + (54.378 N ⋅ m)k
= (41.700 N ⋅ m)k
M D = 41.7 N ⋅ m
(b)

The smallest force Q at B must be perpendicular to

DB at 45°

M D = Q ( DB)
41.700 N ⋅ m = Q (0.28284 m)
Q = 147.4 N
45.0° 
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328
PROBLEM 3.148
The tailgate of a car is supported by the hydraulic lift BC. If the
lift exerts a 125-lb force directed along its centerline on the ball
and socket at B, determine the moment of the force about A.
SOLUTION
First note
dCB = (12.0 in.) 2 + (2.33 in.) 2
= 12.2241 in.
Then
and
12.0 in.
12.2241 in.
2.33 in.
sin θ =
12.2241 in.
cos θ =
FCB = FCB cos θ i − FCB sin θ j
=
125 lb
[(12.0 in.) i − (2.33 in.) j]
12.2241 in.
Now
M A = rB/A × FCB
where
rB/A = (15.3 in.) i − (12.0 in. + 2.33 in.) j
= (15.3 in.) i − (14.33 in.) j
Then
M A = [(15.3 in.)i − (14.33 in.) j] ×
125 lb
(12.0i − 2.33j)
12.2241 in.
= (1393.87 lb ⋅ in.)k
= (116.156 lb ⋅ ft)k
or M A = 116.2 lb ⋅ ft

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329
PROBLEM 3.149
The ramp ABCD is supported by cables at corners C and D.
The tension in each of the cables is 810 N. Determine the
moment about A of the force exerted by (a) the cable at D,
(b) the cable at C.
SOLUTION
(a)
We have
M A = rE/A × TDE
where
rE/A = (2.3 m) j
TDE = λ DE TDE
=
(0.6 m)i + (3.3 m) j − (3 m)k
(0.6) 2 + (3.3)2 + (3) 2 m
(810 N)
= (108 N)i + (594 N) j − (540 N)k
i
j
k
MA = 0
2.3
0 N⋅m
108 594 −540
= −(1242 N ⋅ m)i − (248.4 N ⋅ m)k
or M A = −(1242 N ⋅ m)i − (248 N ⋅ m)k 
(b)
We have
M A = rG/A × TCG
where
rG/A = (2.7 m)i + (2.3 m) j
TCG = λ CGTCG
=
−(.6 m)i + (3.3 m) j − (3 m)k
(.6) 2 + (3.3) 2 + (3) 2 m
(810 N)
= −(108 N)i + (594 N) j − (540 N)k
i
j
k
M A = 2.7 2.3
0 N⋅m
−108 594 −540
= −(1242 N ⋅ m)i + (1458 N ⋅ m) j + (1852 N ⋅ m)k
or M A = −(1242 N ⋅ m)i + (1458 N ⋅ m) j + (1852 N ⋅ m)k 
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330
PROBLEM 3.150
Section AB of a pipeline lies in the yz-plane and forms an angle
of 37° with the z-axis. Branch lines CD and EF join AB as
shown. Determine the angle formed by pipes AB and CD.
SOLUTION
First note

AB = AB(sin 37° j − cos 37°k )
CD = CD(− cos 40° cos 55° j + sin 40° j − cos 40° sin 55°k )
Now
 
AB ⋅ CD = ( AB)(CD ) cos θ
or
AB(sin 37° j − cos 37°k ) ⋅ CD (− cos 40° cos 55°i + sin 40° j − cos 40° sin 55°k )
= (AB)(CD) cos θ
or
cos θ = (sin 37°)(sin 40°) + (− cos 37°)(− cos 40° sin 55°)
= 0.88799
or θ = 27.4° 
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331
PROBLEM 3.151
To lift a heavy crate, a man uses a block and tackle attached to the
bottom of an I-beam at hook B. Knowing that the moments about the y
and the z axes of the force exerted at B by portion AB of the rope are,
respectively, 120 N ⋅ m and −460 N ⋅ m, determine the distance a.
SOLUTION
First note

BA = (2.2 m)i − (3.2 m) j − ( a m)k
Now
M D = rA/D × TBA
where
rA/D = (2.2 m)i + (1.6 m) j
TBA =
Then
i
j
k
TBA
MD =
2.2 1.6 0
d BA
2.2 −3.2 − a
=
Thus
TBA
(2.2i − 3.2 j − ak ) (N)
d BA
TBA
{−1.6a i + 2.2a j + [(2.2)(−3.2) − (1.6)(2.2)]k}
d BA
M y = 2.2
TBA
a
d BA
M z = −10.56
Then forming the ratio
TBA
d BA
(N ⋅ m)
(N ⋅ m)
My
Mz
T
2.2 dBA (N ⋅ m)
120 N ⋅ m
BA
=
−460 N ⋅ m −10.56 TdBA (N ⋅ m)
or a = 1.252 m 
BA
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332
PROBLEM 3.152
To loosen a frozen valve, a force F of magnitude 70 lb is
applied to the handle of the valve. Knowing that θ = 25°,
Mx = −61 lb ⋅ ft, and M z = − 43 lb ⋅ ft, determine φ and d.
SOLUTION
We have
ΣM O : rA/O × F = M O
where
rA/O = −(4 in.)i + (11 in.) j − (d )k
F = F (cos θ cos φ i − sin θ j + cos θ sin φ k )
For
F = 70 lb, θ = 25°
F = (70 lb)[(0.90631cos φ )i − 0.42262 j + (0.90631sin φ )k ]
i
M O = (70 lb)
−4
−0.90631cos φ
j
k
11
−d
in.
−0.42262 0.90631sin φ
= (70 lb)[(9.9694sin φ − 0.42262d ) i + (−0.90631d cos φ + 3.6252sin φ ) j
+ (1.69048 − 9.9694cos φ )k ] in.
and
M x = (70 lb)(9.9694sin φ − 0.42262d ) in. = −(61 lb ⋅ ft)(12 in./ft)
(1)
M y = (70 lb)(−0.90631d cos φ + 3.6252sin φ ) in.
(2)
M z = (70 lb)(1.69048 − 9.9694 cos φ ) in. = − 43 lb ⋅ ft(12 in./ft)
(3)
 634.33 
 = 24.636°
 697.86 
From Equation (3):
φ = cos −1 
or
From Equation (1):
 1022.90 
d =
 = 34.577 in.
 29.583 
or d = 34.6 in. 
φ = 24.6° 
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333
PROBLEM 3.153
The tension in the cable attached to the end C of an
adjustable boom ABC is 560 lb. Replace the force
exerted by the cable at C with an equivalent forcecouple system (a) at A, (b) at B.
SOLUTION
(a)
Based on
ΣF : FA = T = 560 lb
FA = 560 lb
or
20.0° 
ΣM A : M A = (T sin 50°)(d A )
= (560 lb)sin 50°(18 ft)
= 7721.7 lb ⋅ ft
M A = 7720 lb ⋅ ft
or
(b)
Based on
ΣF : FB = T = 560 lb
FB = 560 lb
or

20.0° 
ΣM B : M B = (T sin 50°)(d B )
= (560 lb) sin 50°(10 ft)
= 4289.8 lb ⋅ ft
M B = 4290 lb ⋅ ft
or

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334
PROBLEM 3.154
While tapping a hole, a machinist applies the horizontal forces
shown to the handle of the tap wrench. Show that these forces
are equivalent to a single force, and specify, if possible, the
point of application of the single force on the handle.
SOLUTION
Since the forces at A and B are parallel, the force at B can be replaced with the sum of two forces with one of
the forces equal in magnitude to the force at A except with an opposite sense, resulting in a force-couple.
We have FB = 2.9 lb − 2.65 lb = 0.25 lb, where the 2.65-lb force is part of the couple. Combining the two
parallel forces,
M couple = (2.65 lb)[(3.2 in. + 2.8 in.) cos 25°]
= 14.4103 lb ⋅ in.
and
M couple = 14.4103 lb ⋅ in.
A single equivalent force will be located in the negative z direction.
Based on
ΣM B : −14.4103 lb ⋅ in. = [(0.25 lb) cos 25°](a )
a = 63.600 in.
F′ = (0.25 lb)(cos 25°i + sin 25°k )
F′ = (0.227 lb)i + (0.1057 lb)k and is applied on an extension of handle BD at a
distance of 63.6 in. to the right of B.

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335
PROBLEM 3.155
Replace the 150-N force with an equivalent force-couple
system at A.
SOLUTION
Equivalence requires
ΣF : F = (150 N)(− cos 35° j − sin 35°k )
= −(122.873 N) j − (86.036 N)k
ΣMA : M = rD/A × F
where
Then
rD/A = (0.18 m)i − (0.12 m) j + (0.1 m)k
i
j
k
−0.12
0.1 N ⋅ m
M = 0.18
0
−122.873 −86.036
= [( −0.12)(−86.036) − (0.1)(−122.873)]i
+ [−(0.18)(−86.036)]j
+ [(0.18)(−122.873)]k
= (22.6 N ⋅ m)i + (15.49 N ⋅ m) j − (22.1 N ⋅ m)k
The equivalent force-couple system at A is
F = −(122.9 N) j − (86.0 N)k 
M = (22.6 N ⋅ m)i + (15.49 N ⋅ m) j − (22.1 N ⋅ m)k 
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336
PROBLEM 3.156
A beam supports three loads of given magnitude and a fourth load whose magnitude is a function of position.
If b = 1.5 m and the loads are to be replaced with a single equivalent force, determine (a) the value of a so that
the distance from support A to the line of action of the equivalent force is maximum, (b) the magnitude of the
equivalent force and its point of application on the beam.
SOLUTION
For equivalence,
ΣFy : −1300 + 400
a
− 400 − 600 = − R
b
a

R =  2300 − 400  N
b


or
ΣM A :
a
a
400  − a (400) − (a + b)(600) = − LR
2 
b
L=
or
Then with
(1)
1000a + 600b − 200
2300 − 400
b = 1.5 m
L=
a2
b
a
b
10a + 9 −
4 2
a
3
8
23 − a
3
(2)
where a, L are in m.
(a)
Find value of a to maximize L.
8 
8  
4  8 

10 − a  23 − a  − 10a + 9 − a 2  − 
dL 
3 
3  
3  3 
=
2
da
8 

 23 − 3 a 


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337
PROBLEM 3.156 (Continued)
230 −
or
16a 2 − 276a + 1143 = 0
276 ± (−276) 2 − 4(16)(1143)
2(16)
Then
a=
or
a = 10.3435 m and a = 6.9065 m
Since
(b)
184
80
64 2 80
32
a− a+
a + a + 24 − a 2 = 0
3
3
9
3
9
or
AB = 9 m, a must be less than 9 m
a = 6.91 m 
6.9065
1.5
or R = 458 N 
Using Eq. (1),
R = 2300 − 400
and using Eq. (2),
4
10(6.9065) + 9 − (6.9065)2
3
L=
= 3.16 m
8
23 − (6.9065)
3
R is applied 3.16 m to the right of A. 
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338
PROBLEM 3.157
A mechanic uses a crowfoot wrench to loosen a bolt at C. The
mechanic holds the socket wrench handle at Points A and B
and applies forces at these points. Knowing that these forces
are equivalent to a force-couple system at C consisting of the
force C = (8 lb)i + (4 lb)k and the couple M C = (360 lb · in.)i,
determine the forces applied at A and at B when Az = 2 lb.
SOLUTION
We have
ΣF :
A+B=C
or
Fx : Ax + Bx = 8 lb
Bx = −( Ax + 8 lb)
(1)
ΣFy : Ay + By = 0
Ay = − By
or
(2)
ΣFz : 2 lb + Bz = 4 lb
Bz = 2 lb
or
(3)
ΣM C : rB/C × B + rA/C × A = M C
We have
i
8
Bx
j
0
By
k
i
2 + 8
2
Ax
j
0
Ay
k
8
2
lb ⋅ in. = (360 lb ⋅ in.)i
(2 By − 8 Ay )i + (2 Bx − 16 + 8 Ax − 16) j
or
+ (8By + 8 Ay )k = (360 lb ⋅ in.)i
From
i-coefficient:
j-coefficient:
k-coefficient:
2 By − 8 Ay = 360 lb ⋅ in.
−2 Bx + 8 Ax = 32 lb ⋅ in.
8 By + 8 Ay = 0
(4)
(5)
(6)
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339
PROBLEM 3.157 (Continued)
From Equations (2) and (4):
2 By − 8(− By ) = 360
By = 36 lb
From Equations (1) and (5):
Ay = 36 lb
2(− Ax − 8) + 8 Ax = 32
Ax = 1.6 lb
From Equation (1):
Bx = −(1.6 + 8) = −9.6 lb
A = (1.600 lb)i − (36.0 lb) j + (2.00 lb)k 
B = −(9.60 lb)i + (36.0 lb) j + (2.00 lb)k 
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340
PROBLEM 3.158
A concrete foundation mat in the shape of a regular hexagon of side
12 ft supports four column loads as shown. Determine the magnitudes
of the additional loads that must be applied at B and F if the resultant
of all six loads is to pass through the center of the mat.
SOLUTION
From the statement of the problem, it can be concluded that the six applied loads are equivalent to the
resultant R at O. It then follows that
ΣM O = 0 or ΣM x = 0 ΣM z = 0
For the applied loads:
Then
ΣM x = 0: (6 3 ft) FB + (6 3 ft)(10 kips) − (6 3 ft)(20 kips)
− (6 3 ft) FF = 0
FB − FF = 10
or
(1)
ΣM z = 0: (12 ft)(15 kips) + (6 ft) FB − (6 ft)(10 kips)
− (12 ft)(30 kips) − (6 ft)(20 kips) + (6 ft) FF = 0
FB + FF = 60
or
(2)
Then Eqs. (1) + (2) 
FB = 35.0 kips 
and
FF = 25.0 kips 
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341