CHAPTER 3
PROBLEM 3.1
A foot valve for a pneumatic system is hinged at B. Knowing that
a = 28∞, determine the moment of the 16-N force about Point B
by resolving the force into horizontal and vertical components.
Fx
Solution
Note that
and
θ°
A
q = a - 20∞ = 28∞ - 20∞ = 8∞
Fx = (16 N ) cos 8∞ = 15.8443 N
Fy = (16 N )sin 8∞ = 2.2268 N
y
20 °
16N
Fy
20 ° a = 28 °
x
B
0.17 m
C
x = (0.17 m) cos 20∞ = 0.159748 m
y = (0.17 m)sin 20∞ = 0.058143 m
Noting that the direction of the moment of each force component about B is counterclockwise,
Also
MB = xFy + yFx
= (0.159748 m)(2.2268 N )
+ (0.058143 m)(15.8443 N )
= 1.277 N ◊ m
or M B = 1.277 N ◊ m

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3
PROBLEM 3.2
A foot valve for a pneumatic system is hinged at B. Knowing
that a = 28∞, determine the moment of the 16-N force about Point
B by resolving the force into components along ABC and in a
direction perpendicular to ABC.
Solution
First resolve the 4-lb force into components P and Q, where
Q = (16 N )sin 28∞
= 7.5115 N
Then
Q
A
16 N
α = 28°
P
B
0.17 m
M B = rA/B Q
C
= (0.17 m)(7.5115 N )
= 1.277 N ◊ m or M B = 1.277 N ◊ m

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4
PROBLEM 3.3
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)
D
r
25°
= 271.89 N
Fy = (300 N )sin 25∞
0.2 m
= 126.785 N
F = (271.89 N )i + (126.785 N ) j
A
300 N
uuur
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
0.1 m
0.2 m
D
4
28
28
0.
0.2 m
Fx = (300 N ) cos 25∞
m
45°
B
Q
M D = 41.7 N ◊ m 
uuur
(b)The smallest force Q at B must be perpendicular to DB at
45°
uuur
M D = Q( DB)
41.700 N ◊ m = Q(0.28284 m) Q = 147.4 N
45° 
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5
PROBLEM 3.4
A 300-N force is applied at A as shown. Determine (a) the moment
of the 300-N force about D, (b) the magnitude and sense of the
horizontal force applied at C that creates the same moment about
D, (c) the smallest force applied at C that creates the same moment
about D.
Solution
(a)
See Problem 3.3 for the figure and analysis leading to the determination of MD
M D = 41.7 N ◊ m

0.2 m
D
0.125 m
r
C
(b)
(c)
Since C is horizontal C = C i
C = Ci
uuur
r = DC = (0.2 m)i - (0.125 m) j
M D = r ¥ C i = C (0.125 m)k
41.7 N ◊ m = (0.125 m)(C )
C = 333.60 N
C = 334 N 
The smallest force C must be perpendicular to DC; thus, it forms a with the vertical
0.125 m
tan a =
0.2 m
a = 32.0∞
0.2 m
D
a
0.125 m
M D = C ( DC ); DC = (0.2 m)2 + (0.125 m)2
C
= 0.23585 m
41.70 N ◊ m = C (0.23585 m) a
C = 176.8 N
58.0° 
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6
PROBLEM 3.5
A 150-N force acts on the end of the 1 m lever as shown. Determine the moment of
the force about O.
Solution
First note
Px = 150 cos 30∞
= 129.9 N
Py = 150 sin 30∞
Py
20°
P = 150 N
30°
Px
= 75 N
Noting that the direction of the moment of Px about O is clockwise and Py about O
is counterclockwise,
M B = xPy - yPx
= (0.643)(75) - (0.766)(129.9)
= -51.3 N ◊ m
1m
o
y = 1 sin 50°
= 0.766 m
50°
x = 1 cos 50°
= 0.643 m
or MB = 51.3 N · m

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7
PROBLEM 3.6
For the shift lever shown, determine the magnitude and the direction of the smallest
force P that has a 25 N · m clockwise moment about B.
Solution
For P to be minimum, it must be perpendicular to the line joining Points A and B. Thus,
a =q
0.2 m
= tan
0.6 m
= 18.43∞
-1
and
where
α
= (0.2 m)2 + (0.6 m)2
A
θ
M B = dPmin
d = rA/B
P
0.6 m
B
0.2 m
= 0.632 m
Then
25 N ◊ m
0.632 m
= 39.6 N
Pmin =
Pmin = 39.6 N
18.43° 
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8
PROBLEM 3.7
A 55-N force P is applied to a shift lever. The moment of P about B is clockwise and has
a magnitude of 34 N · m. Determine the value of a.
Solution
By definition
M B = rA/B P sinq
where
q = a + (90∞ - f )
and
f = tan -1
also
Then
or
or
f
P = 55 N
a
θ
A
a
f
0.2 m
= 18.43∞
0.6 m
rA/B = (0.2 m)2 + (0.6 m)2
= 0.632 m
34 N ◊ m = (0.632 m)(55 N )sin(a + 90∞ - 18.43∞)
sin(a + 71.57∞) = 0.978
a + 71.57∞ = 77.96∞
a + 71.57∞ = 102.04∞
rA/B
0.6 m
B
0.2 m
a = 6.39∞, 30.5∞ 
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9
PROBLEM 3.8
It is known that a vertical force of 1000 N 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 a = 10∞, (c) the smallest
force P that creates the same moment about B.
Solution
(a)
A
a
M B = rC /B FN
We have
= (100 mm)(1000 N ) = 105 N ◊ mm
= 100 N ◊ m
P 450 mm
70°
FN = 1000 N
B
C
100 mm
q
(b)
By definition
Then
P
a = 10°
70°
or
rA/B
A
70°
C

M B = rA/B P sin q
q = 10∞ + (180∞ - 70∞)
= 120∞
100 N ◊ m = (0.45 m) ¥ P sin 120∞ or
P = 257 N 
(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
P
a = 20°
MB = 100 N ◊ m
M B = dPmin
d = rA/B
or
100 N ◊ m = (0.45 m)Pmin
or
Pmin = 222 N
Pmin = 222 N
20° 
B
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10
PROBLEM 3.9
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
0.875 m
5
=
1.90 m + 0.2 m 12
12
= (TAB )
13
12
= (1040 N )
13
= 960 N
5
= (1040 N )
13
= 400 N
EC =
TABx
TABy
TABx
C
TAB = 1040 N
13
E
TABy
0.875 m
5
12
D
1.90 m
0.2 m
(a)
MD = 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 MD = 760 N ◊ m

M D = TABx ( y ) + TABx ( x )
C
= (960 N )(0) + ( 400 N )(1.90 m)
= 760 N ◊ m
13
TABx
TAB
12
E
5
D
TABy
1.90 m
(b)
or M D = 760 N ◊ m

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11
PROBLEM 3.10
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
C
25
TABx
E
0.875 m
7
D
TABy
TAB
Slope of line
2.80 m
EC =
0.875 m
7
=
2.80 m + 0.2 m 24
24
TAB
25
7
= TAB
25
Then
TABx =
and
TABy
We have
0.2 m
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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12
PROBLEM 3.11
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
C
0.875 m
(TABmax)x E
TABmax
q
D
(TABmax)y
0.20 m
d
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
Now
M D = 960 N ◊ m
(TAB max ) y = TAB max sin q = (2400 N )sin q
sin q =
0.875 m
( d + 0.20)2 + (0.875)2 m
È
˘
0.875
˙ (d )
960 N ◊ m = 2400 N Í
Í ( d + 0.20)2 + (0.875)2 ˙
Î
˚
or
( d + 0.20)2 + (0.875)2 = 2.1875d
( d + 0.20)2 + (0.875)2 = 4.7852d 2
or
or
3.7852d 2 - 0.40 d - .8056 = 0
Using the quadratic equation, the minimum values of d are 0.51719 m and - .41151 m.
Since only the positive value applies here, d = 0.51719 m
or d = 517 mm 
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13
PROBLEM 3.12
The tailgate of a car is supported by the hydraulic lift BC. If the lift
exerts a 625-N force directed along its centerline on the ball and
socket at B, determine the moment of the force about A.
Solution
First note
dCB = (300 mm)2 + (58 mm)2
= 305.6 mm
Then
and
380 mm
A
y
300 mm
x
300 mm
cos q =
305.6 mm
58 mm
sin q =
305.6 mm
C
+
θ
FCB
300 mm
B
58 mm
FCB = FCB cos q i - FCB sin q j
=
625 N
[(300 mm)i - (58 mm) j]
305.6 mm
Now
MA = rB /A ¥ FCB
where
rB /A = (380 mm)i - (300 mm + 58 mm) j
= (380 mm)i - (358 mm) j
Then
625 N
(300 mm i - 58 mm j)
305.6 mm
= -45075 N ◊ mm
m i + 219650 N ◊ mm k
= 174575 N ◊ mm k
= 175 N ◊ m k
or M A = 175 N ◊ m
M A = [(380 mm)i - (358 mm) j] ¥

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14
PROBLEM 3.13
The tailgate of a car is supported by the hydraulic lift BC. If the lift
exerts a 625-N force directed along its centerline on the ball and
socket at B, determine the moment of the force about A.
Solution
First note
Then
and
512 mm
2
dCB = ( 430 mm) + (190 mm)
2
= 470.1 mm
430 mm
cos q =
470.1 mm
190 mm
sin q =
470.1 mm
B
y
x
110 mm
FCB
190 mm
C +
430 mm
FCB = ( FCB cos q )i - ( FCB sin q ) j
=
625 N
[( 430 mm)i + (190 mm) j]
470.1 mm
Now
MA = rB /A ¥ FCB
where
rB /A = (512 mm)i - (110 mm) j
Then
MA = [(512 mm)i - (110 mm) j] ¥
A
625 N
( 430 mm i + 190 mm j)
( 470.1 mm)
= 129334 N ◊ mm k + 62886 N ◊ mm k
= 192220 N ◊ mm k
= 192 N ◊ m k
or M A = 192 N.m

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15
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.
120 mm
Solution
We have
MC = rB /C ¥ FB
Noting the direction of the moment of each force component about C is
clockwise.
MC = xFBy ¥ yFBx
where
C
90 mm
72 mm
FBx
x = 120 mm - 65 mm = 55 mm
y = 72 mm + 90 mm = 162 mm
FBy
65 mm
and
FBx =
FBy =
65
(65)2 + (72)2
72
(65)2 + (72)2
(485 N) = 325 N (485 N) = 360 N
M C = (55 mm)(360 N) + (162)(325 N)
= 72450 N ◊ mm
= 72.450 N ◊ m
or MC = 72.5 N ◊ m

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16
PROBLEM 3.15
Form the vector products B 3 C and B9 3 C, where B  B9, and use the ­results
obtained to prove the identity
1
1
sin a cos b = sin (a + b ) + sin (a - b ).
2
2
Solution
B = B(cos b i + sin b j)
B ¢ = B(cos b i - sin b j)
C = C (cos a i + sin a j)
Note:
| B ¥ C | = BC sin (a - b ) (1)
| B ¢ ¥ C | = BC sin (a + b ) (2)
By definition:
B ¥ C = B(cos b i + sin b j) ¥ C (cos a i + sin a j)
Now
= BC(cos b sin a - sin b cos a )k (3)
B ¢ ¥ C = B(cos b i - sin b j) ¥ C (cos a i + sin a j)
and
= BC(cos b sin a + sin b cos a ) k (4)
Equating the magnitudes of B ¥ C from Equations (1) and (3) yields:
BC sin(a - b ) = BC (cos b sin a - sin b cos a ) (5)
Similarly, equating the magnitudes of B ¢ ¥ C from Equations (2) and (4) yields:
BC sin(a + b ) = BC (cos b sin a + sin b cos a ) (6)
Adding Equations (5) and (6) gives:
sin(a - b ) + sin(a + b ) = 2 cos b sin a
1
1
or sin a cos b = sin(a + b ) + sin(a - b ) 
2
2
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17
PROBLEM 3.16
A line passes through the Points (20 m, 16 m) and (21 m, 24 m). Determine the perpendicular distance d from
the line to the origin O of the system of coordinates.
y
Solution
+
rB
d AB = [20 m - ( -1 m)]2 + [16 m - ( -4 m)]2
= 29 m
0
Assume that a force F, or magnitude F(N), acts at Point A and is
­directed from A to B.
Then,
d
+
A (–1 m, – 4 m)
x
rB - rA
d AB
1
=
(21i + 20 j)
29
l AB =
By definition
MO = | rA ¥ F | = dF
Then
lAB
F
F = F l AB
where
where
rA
B (20 m, 16 m)
rA = - (1 m)i - ( 4 m) j
MO = [-( -1 m)i - ( 4 m) j] ¥
F
[(21 m)i + (20 m) j]
29 m
F
[-(20)k + (84)k ]
29
Ê 64 ˆ
= Á F˜ k N ◊ m
Ë 29 ¯
=
Finally
Ê 64 ˆ
ÁË F ˜¯ = d ( F )
29
d=
64
m
29
d = 2.21 m 
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18
PROBLEM 3.17
The vectors P and Q are two adjacent sides of a parallelogram. Determine the area of the parallelogram when
(a) P  2 7i  3j 2 3k and Q  2i  2j  5k, (b) P  6i 2 5j 2 2k and Q  2 2i  5j 2 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 in. 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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19
PROBLEM 3.18
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 2 5k and 4i 2 7j 2 5k, (b) 3i 2 3j  2k and 2 2i  6j 2 4k.
Solution
(a)
A¥B
|A¥B|
We have
l=
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
15( -3i - 1j - 1k )
15 11
A¥B
l=
|A ¥ B|
l=
(b)
We have
where
or l =
1
( -3i - j - k ) 
11
A = 3i - 3j + 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
l=
4(2 j + 3k )
4 13
or l =
1
(2 j + 3k ) 
13
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20
PROBLEM 3.19
Determine the moment about the origin O of the force F  4i  5j 2 3k that acts at a Point A. Assume that the
position vector of A is (a) r  2i 2 3j  4k, (b) r  2i  2.5j 2 1.5k, (c) r  2i  5j  6k. The components
of F and r are in N and m respectively.
Solution
(a)
i j k
MO = 2 -3 4
4 5 -3
= (9 - 20)i + (16 + 6) j + (10 + 12)k (b)
i
MO = 2
4
j
k
2.5 - 1.5
5 -3
= ( -7.5 + 7.5)i + ( -6 + 6) j + (10 - 10)k (c)
MO = -11i + 22 j + 22k N ◊ m 
MO = 0 
i j k
MO = 2 5 6
4 5 -3
= ( -15 - 30)i + (24 + 6) j + (10 - 20)k MO = -45i + 30 j - 10k N ◊ m 
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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21
PROBLEM 3.20
Determine the moment about the origin O of the force F 5 2 2i 1 3j 1 5k that acts at a Point A. Assume that
the position vector of A is (a) r  i  j  k, (b) r  2i  3j 2 5k, (c) r  2 4i  6j  10k. The components
of F and r are in N and m respectively.
Solution
(a)
i j k
MO = 1 1 1
-2 3 5
= (5 - 3)i + ( -2 - 5) j + (3 + 2)k (b)
i j k
MO = 2 3 - 5
-2 3 5
= (15 + 15)i + (10 - 10) j + (6 + 6)k (c)
MO = 2i - 7 j + 5k N ◊ m 
MO = 30i + 12k N ◊ m 
i j k
MO = -4 6 10
-2 3 5
= (30 - 30)i + ( -20 + 20) 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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22
PROBLEM 3.21
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
MA = 200 0.06
0.075
0
0
- cos 30∞ sin 30∞
= 200 [(0.075 sin 30∞)i - (0.06 sin 30∞) j - (0.06 cos 30∞)k ]
or MA = (7.50 N ◊ m)i - (6.00 N ◊ m) j - (10.39 N ◊ m)k 
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23
PROBLEM 3.22
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
MA = 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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24
PROBLEM 3.23
A wooden board AB, which is used as a temporary prop to support a
small roof, exerts at Point A of the roof a 300-N force directed along
BA. Determine the moment about C of that force.
Solution
We have
MC = rA/C ¥ FBA
where
rA/C = (1200 mm)i - (150 mm) j + (900 mm)k
and
FBA = l BA FBA
y
È -(125mm)i + (2250 mm) j - (750 mm)k ˘
˙ (300 N )
=Í
2
2
2
Í
˙
+
+
(
125
)
(
2250
)
(
30
)
Î
˚
= -(16.64 N )i + (299.5 N ) j - (99.84 N )k
900 mm
z
C
1200 mm
rA/C
D
150 mm
x
A
FBA
i
j
k
MC = 1200 -150
900 N ◊ m
-16.64 299.5 -99.84
= -(254574 N ◊ mm)i + (104832 N ◊ mm) j + (356904 N ◊ mm)k
or MC = -(255 N ◊ m)i + (104.8 N ◊ m) j + (357 N ◊ m)k 
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25
PROBLEM 3.24
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
MA = rE /A ¥ TDE
where
rE /A = (2.3 m) j
TDE = l DE TDE
(0.6 m)i + (3.3 m) j - (3 m)k
=
(810 N )
(0.6)2 + (3.3)2 + (3)2 m
= (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 MA = -(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 = l CGTCG
=
-(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
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 MA = -(1242 N ◊ m)i + (1458 N ◊ m) j + (1852 N ◊ m)k 
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26
PROBLEM 3.25
A small boat hangs from two davits, one of which is shown in the
­figure. The tension in line ABAD is 410 N. Determine the moment
about C of the resultant force RA exerted on the davit at A.
Solution
We have
R A = 2FAB + FAD
where
FAB = -( 410 N ) j
uuur
(2i - 2.5 j - k )
AD
FAD = FAD
= ( 410 N )
AD
(2)2 + (2.25)2 + (1)2
and
FAD = (258.5N )i - (323.1 N ) j - (129.25 N )k
Thus
R A = 2FAB + FAD = (258.5N )i - (1143.1 N ) j - (129.25 N )k
Also
rA/C = (2.5 m) j + (1 m)k
Using Eq. (3.21):
i
j
k
MC = 0
2.5
1
258.5 - 1143.1 -129.25
= (820 N ◊ m)i + (259 N ◊ m) j - (646 N ◊ m)k
MC = (820 N ◊ m)i + (259 N ◊ m) j - (646 N ◊ m)k 
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27
PROBLEM 3.26
In Sample Problem 3.5, determine the perpendicular distance
from Point O to cable AB.
SAMPLE PROBLEM 3.5 Before the trunk of a large tree
is felled, cables AB and BC are attached as shown. Knowing
that the tensions in cables AB and BC are 555 N and 660 N,
respectively, determine the moment about O of the resultant
force ­exerted on the tree by the cables at B.
Solution
We have
where
| MO | = TBA d
y
d = perpendicular distancefrom O to line AB.
Now
MO = rB /O ¥ TBA
and
rB /O = (7 m) j
TBA
TBA = l BATAB
-(0.75 m)i - (7 m) j + (6 m)k
=
(555 N )
(0.75)2 + (7)2 + (6)2 m
= -( 45.0 N )i - ( 420 N ) j + (360 N )k
i
j
B
k
6m
0.75 m
d
A
rB/O
7m
0
x
z
MO = 0
7
0 N◊m
-45 -420 360
= (2520.0 N ◊ m)i + (315.00 N ◊ m)k
and
| MO | = (2520.0)2 + (315.00)2
= 2539.6 N ◊ m
2539.6 N ◊ m = (555 N)d
or
d = 4.5759 m or d = 4.58 m 
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28
PROBLEM 3.27
In Sample Problem 3.5, determine the perpendicular distance
from Point O to cable BC.
SAMPLE PROBLEM 3.5 Before the trunk of a large tree
is felled, cables AB and BC are attached as shown. Knowing
that the tensions in cables AB and BC are 555 N and 660 N,
respectively, determine the moment about O of the resultant
force ­exerted on the tree by the cables at B.
Solution
We have
where
| MO | = TBC d
y
B
d = perpendicular distance from O to line BC .
TBC
MO = rB /O ¥ TBC
7m
rB /O = 7 mj
rB/O
TBC = l BC TBC
=
( 4.25 m)i - (7 m) j + (1 m)k
2
2
2
( 4.25) + (7) + (1) m
4.25 m
(660 N )
= (340 N )i - (560 N ) j + (80 N )k
i
j
k
MO = 0
7
0
340 -560 80
0
d
z
1m
C
x
= (560 N ◊ m)i - (2380 N ◊ m)k
and
| MO | = (560)2 + (2380)2
= 2445.0 N ◊ m
2445.0 N ◊ m = (660 N)d
d = 3.7045 m or d = 3.70 m 
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29
PROBLEM 3.28
In Problem 3.23, determine the perpendicular distance from Point D
to a line drawn through Points A and B.
PROBLEM 3.23 A wooden board AB, which is used as a temporary
prop to support a small roof, exerts at Point A of the roof a 300 N
force directed along BA. Determine the moment about C of that
force.
y
Solution
We have
where
| M D | = FBA d
900 mm
d = perpendicular distance from D to line AB.
MD = rA/D ¥ FBA
z
rA/D = -(150 mm) j + (900 mm)k
FBA = l BA FBA
C
1200 mm
rA/D
D
150 mm
x
A
FBA
È -(125 mm)i + (2250 mm) j - (750 mm)k ˘
˙ (300 N )
=Í
2
2
2
Í
˙
(
125
)
+
(
2250
)
+
(
30
)
Î
˚
= -(16.64 N )i + (299.5 N ) j - (99.84 N )k
MD =
and
i
j
k
0
-150
900 N ◊ m
-16.64 299.5 -99.84
= -(254574 N ◊ mm)i - (14976 N ◊ mm) j - (2496 N ◊ mm)k
ur
u
| M D | = (254574)2 + (14976)2 + (2496)2
= 255026 N ◊ mm
255026 N ◊ mm = (300 N)d
d = 850 mm or d = 850 mm 
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30
PROBLEM 3.29
In Problem 3.23, determine the perpendicular distance from Point C
to a line drawn through Points A and B.
PROBLEM 3.23 A wooden board AB, which is used as a temporary
prop to support a small roof, exerts at Point A of the roof a 300 N
force directed along BA. Determine the moment about C of that
force.
y
Solution
We have
where
| MC | = FBA d
900 mm
d = perpendicular distance from C to line AB.
z
MC = rA/C ¥ FBA
rA/C = (1200 mm)i - (150 mm) j + (900 mm)k
FBA = l BA FBA
C
1200 mm
rA/C
D
150 mm
x
A
FBA
È -(125 mm)i + (2250 mm) j - (750 mm)k ˘
˙ (300 N )
=Í
2
2
2
Í
˙
+
+
(
125
)
(
2250
)
(
30
)
Î
˚
= -(16.64 N )i + (299.5 N ) j - (99.84 N )k
i
j
k
MC = 1200 -150
900 N ◊ m
-16.64 299.5 -99.84
and
= -(254574 N ◊ mm)i + (104832 N ◊ mm) j + (356904 N ◊ mm)k
ur
u
| MC | = (254574)2 + (104832)2 + (356904)2
= 450753 N ◊ mm
450753 = (300 N )d
d = 1503 mm or d = 1.503 m 
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31
PROBLEM 3.30
In Problem 3.24, determine the perpendicular distance from
Point A to portion DE of cable DEF.
PROBLEM 3.24 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
We have
where
y
| M A | = TDE d
E
2.3 m
d = perpendicular distance from A to line DE.
rE/A
d
M A = rE /A ¥ TDE
A
rE /A = (2.3 m) j
TDE = l DE TDE
(0.6 m)i + (3.3 m) j - (3 m)k
=
(810 N )
(0.6)2 + (3.3)2 + (3)2 m
= (108 N )i + (594 N) j - (540 N )k
TDE
1m
x
D
0.6 m
3m
i
j
k
MA = 0
2.3
0 N◊m
108 594 540
= - (1242.00 N ◊ m)i - (248.00 N ◊ m)k
and
|M A | = (1242.00)2 + (248.00)2
= 1266.52 N ◊ m
1266.52 N ◊ m = (810 N )d
d = 1.56360 m or d = 1.564 m 
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32
PROBLEM 3.31
In Problem 3.24, determine the perpendicular distance from
Point A to a line drawn through Points C and G.
PROBLEM 3.24 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.
y
Solution
We have
where
E
|M A | = TCG d
rG/A
d = perpendicular distance from A to line CG.
M A = rG /A ¥ TCG
z
d
A
B
rG /A = (2.7 m)i + (2.3 m) j
TCG = l CGTCG
=
-(0.6 m) i + (3.3 m) j - (3 m) k
2
2
2
(0.6) + (3.3) + (3) m
(810 N )
= -(108 N ) i + (594 N ) j - (540 N ) k
2.7 m
0.6 m
6
1m x
TCG
C
23 m
3m
i
j
k
0 N◊m
M A = 2.7 2.3
-108 594 -540
= -(1242.00 N ◊ m)i + (1458.00 N ◊ m) j + (1852.00 N ◊ m)k
and
|M A | = (1242.00)2 + (1458.00)2 + (1852.00)2
= 2664.3 N ◊ m
2664.3 N ◊ m = (810 N )d
d = 3.2893 m or d = 3.29 m 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to
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33
PROBLEM 3.32
In Problem 3.25, determine the perpendicular distance from Point C
to portion AD of the line ABAD.
PROBLEM 3.25 A small boat hangs from two davits, one of which
is shown in the figure. The tension in line ABAD is 410 N. 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 3.25:
FDA = - FAD
= -(258.5 N ) i + (323.1 N ) j + (129.25 N )k
MC = rD /C ¥ FDA
= +(2 m)i ¥ [-(258.5 N )i + (323.1 N ) j + (129.25 N )k ]
= -(258.5 N ◊ m) j + (646.2 N ◊ m)k
MC = (258.5)2 + (646.2)2
= 695.99 N ◊ m
Then
MC = FDA d
Since
FDA = 410 N
d=
=
MC
FDA
695.99 N ◊ m
410 N
d = 1.698 m 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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34
PROBLEM 3.33
Determine the value of a that minimises 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,
|MC | = |rA/ C ¥ F| = F ( d )
d = perpendicular distance from C to line AB
where
F = l AB F
=
(8 m)i + (7.32 m) j - (9.05 m)k
(8)2 + (7.32)2 + (9.05)2
= F (0.5664 i + 0.5183j - 0.6407k )
rA/C = (1 m)i - (3.05 m) j - ( a - 3.05 m)k
i
j
k
MC =
1
-3.05 3.05 - a F
0.5664 0.5183 -0.6407
B
y
F
α
C
rA/C
A
= [(0.37332 + 0.5183a)i + (2.36822 - 0.5664a) j + (2.24582)k ] F
|MC | = |rA/C ¥ F2 |
Since
|rA/C ¥ F2 | = ( dF )2
or
(0.3733 + 0.5183a)2 + (2.368 - 0.5664a)2 + (2.246)2 = d 2
Setting
d
(d 2 )
da
= 0 to find a to minimize d
[2(0.5183)(0.3733 + 0.5183a) + 2( -0.5664)(2.368 - 0.5664a)] = 0
Solving
a = 1.947 m or a = 1.947 m 
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35
PROBLEM 3.34
Given the vectors P = 3i - j + 2k , Q = 4i + 5 j - 3k , and S = -2i + 3j - k , compute the scalar products P ? Q,
P ? S, and Q ? S.
Solution
P ◊ Q = (3i - 1j + 2k ) ◊ ( 4i - 5 j - 3k )
= (3)( 4) + ( -1)( -5) + (2)( -3)
=1
or P ◊ Q = 1 
P ◊ S = (3i - 1j + 2k ) ◊ ( -2i + 3j - 1k )
= (3)( -2) + ( -1)(3) + (2)( -1)
= -11
or
Q ◊ S = ( 4i - 5 j - 3k ) ◊ ( -2i + 3j - 1k )
= ( 4)( -2) + (5)(3) + ( -3)( -1)
= 10
or Q ◊ S = 10 
P ◊ S = -11 
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36
PROBLEM 3.35
Form the scalar products B ◊ C and B ¢ × C, where B = B ¢, and use the results
obtained to prove the identity
1
1
cos a cos b = cos (a + b ) + cos (a - b ).
2
2
Solution
y
By definition
C
B = B [(cos b )i + (sin b ) j]
C = C [(cos a )i + (sin a ) j]
where
B
α–β
B ◊ C = BC cos(a - b )
( B cos b )( C cos a ) + ( B sin b )(C sin a ) = BC cos (a - b )
α+β
cos b cos a + sin b sin a = cos(a - b ) or
x
B
By definition
where
(1)
B ¢ ◊ C = BC cos (a + b )
B ¢ = [(cos b )i - (sin b ) j]
( B cos b )(C cos a ) + ( - B sin b )(C sin a ) = BC cos (a + b )
or
cos b cos a - sin b sin a = cos (a + b ) (2)
Adding Equations (1) and (2),
2 cos b cos a = cos (a - b ) + cos (a + b )
1
1
or cos a cos b = cos (a + b ) + cos (a - b ) 
2
2
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37
PROBLEM 3.36
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
uuur
AB = AB(sin 37∞ j - cos 37∞k )
CD = CD( - cos 40∞ cos 55∞ j + sin 40∞ j - cos 40∞ sin 55∞k )
D
Y
(CD)
40°
(CD)y
(CD)xz
(CD)z
55°
(CD)x
Z
C
X
Now
uuur uuur
AB ◊ CD = ( AB)(CD ) cos q
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 q
or
cos q = (sin 37∞)(sin 40∞) + ( - cos 37∞)( - cos 40∞ sin 55∞)
= 0.88799
or q = 27.4∞ 
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38
PROBLEM 3.37
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 EF.
Solution
First note
uuur
AB = AB(sin 37∞ j - cos 37∞k )
uuur
EF = EF (cos 32∞ cos 45∞i + sin 32∞ j - cos 32∞ sin 45∞k )
F
y
(EF)y
(EF)z
EF
E
z
(EF)x
(EF)xz
32°
45°
x
Now
uuur uuur
AB ◊ EF = ( AB)( EF ) cos q
or
AB(sin 37∞ j - cos 37∞k ) ◊ EF (cos 32∞ cos 45∞ j + sin 32∞ j - cos 32∞ sin 45∞k))
= ( AB)( EF ) cos q
or
cos q = (sin 37∞)(sin 32∞) + ( - cos 37∞)( - cos 32∞ sin 45∞)
= 0.79782
or q = 37.1∞ 
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39
PROBLEM 3.38
Consider the volleyball net shown.
­Determine the angle formed by guy
wires AB and AC.
Solution
First note
AB = ( -1.98)2 + ( -2.44)2 + (0.61)2
= 3.2 m
AC = (0)2 + ( -2.44)2 + (2)2
and
By definition
or
= 3.155 m
uuur
AB = -(1.98 m)i - (2.44 m) j + (0.61 m)k
uuur
AC = -(2.44 m) j + (2 m)k
uuur uuur
AB ◊ AC = ( AB)( AC ) cosq
[-(1.98)i - (2.44) j + (0.61)k ] ◊ [-(2.44) j + (2)k ] = (3.2)(3.155) cos q
[( -2.44) ¥ ( -2.44)] + [(0.61)(2)] = 10.096 cos q
cos q = 0.711
or
or q = 44.7∞ 
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40
PROBLEM 3.39
Consider the volleyball net shown.
­Determine the angle formed by guy
wires AC and AD.
Solution
First note
AC = (0)2 + ( -2.44)2 + (2)2
= 3.155 m
AD = (1.22)2 + ( -2.44)2 + (0.3)2
and
By definition
or
= 2.744 m
uuur
AC = ( -2.44 m)j + (2 m)k
uuur
AD = (1.22 m)i - (2.44 m) j + (0.3 m)k
uuur uuur
AC ◊ AD = ( AC )( AD ) cosq
( -2.44 j + 2k ) ◊ (1.22i - 2.44 j + 0.3k ) = (3.155)(2.744) cos q
( -2.44)( -2.44) + 2(0.3) = 8.6573 cos q
or
cos q = 0.757 or q = 40.8∞ 
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41
PROBLEM 3.40
Knowing that the tension in cable AC is 1260 N, determine (a) the
angle between cable AC and the boom AB, (b) the projection on AB
of the force exerted by cable AC at Point A.
Solution
(a)
First note
AC = ( -2.4)2 + (0.8)2 + (1.2)2
= 2.8 m
AB = ( -2.4)2 + ( -1.8)2 + (0)2
= 3.0 m
and
By definition
or
or
uuur
AC = -(2.4 m)i + (0.8 m) j + (1.2 m)k
uuur
AB = -(2.4 m)i - (1.8 m) j
uuur uuur
AC ◊ AB = ( AC )( AB) cosq
( -2.4i + 0.8 j + 1.2k ) ◊ ( -2.4i - 1.8 j) = (2.8)(30) ¥ cos q
( -2.4)( -2.4) + (0.8)( -1.8) + (1.2)(0) = 8.4 cos q
cos q = 0.514 29 or
(b)
We have
or q = 59.0∞ 
(TAC ) AB = TAC ◊ l AB
= TAC cos q
= (1260 N )(0.51429)
or (TAC ) AB = 648 N 
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42
PROBLEM 3.41
Knowing that the tension in cable AD is 405 N, determine (a) the
angle between cable AD and the boom AB, (b) the projection on AB
of the force exerted by cable AD at Point A.
Solution
(a)
First note
AD = ( -2.4)2 + (1.2)2 + ( -2.4)2
= 3.6 m
AB = ( -2.4)2 + ( -1.8)2 + (0)2
and
By definition,
= 3.0 m
&
AD = -(2.4 m)i + (1.2 m) j - (2.4 m)k
AB = -(2.4 m)i - (1.8 m) j
& ◊ AB = ( AD )( AB) cos q
AD
( -2.4i + 1.2 j - 2.4k ) ◊ ( -2.4i - 1.8 j) = (3.6)(3.0) cos q
( -2.4)( -2.4) + (1.2)( -1.8) + ( -2.4)(0) = 10.8 cos q
cosq =
(b)
1
3
q = 70.5∞ 
(TAD ) AB = TAD ◊ l AB
= TAD cos q
Ê 1ˆ
= ( 405 N ) Á ˜ Ë 3¯
(TAD ) AB = 135.0 N 
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43
PROBLEM 3.42
Slider P can move along rod OA. An elastic cord PC is attached
to the slider and to the vertical member BC. Knowing that the
distance from O to P is 150 mm and that the tension in the cord
is 15 N, determine (a) the angle between the elastic cord and the
rod OA, (b) the projection on OA of the force exerted by cord PC
at Point P.
Solution
First note
OA = (300)2 + (300)2 + ( -150)2
= 450 mm
uuur
OA
1
Then
(300i + 300 j - 150k )
l OA =
=
OA 450
1
= (2i + 2 j - k )
3
1
OP = 150 mm. fi OP = (OA)
Now
3
The coordinates of Point P are (100 mm, 100 mm, 2 50 mm)
uuur
PC = (125 mm)i + (275 mm) j + (350 mm)k
so that
PC = (125)2 + (275)2 + (350)2 = 462.33 mm
and
(a)
We have
or
or
uuur
PC ◊ l OA = ( PC ) cosq
1
(125i + 275 j + 350k ) ◊ (2i + 2 j - k ) = 462.33 cos q
3
cos q = 0.324 4
(b)
We have
or q = 71.1∞ 
(TPC ) OA = TPC ◊ l OA
= (TPC l PC ) ◊ l OA
uuur
PC
= TPC
◊ l OA
PC
= TPC cos q
= (15 N )(0.324 44)
or (TPC )OA = 4.87 N 
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44
PROBLEM 3.43
Slider P can move along rod OA. An elastic cord PC is attached to
the slider and to the vertical member BC. Determine the distance
from O to P for which cord PC and rod OA are ­perpendicular.
Solution
First note
OA = (300)2 + (300)2 + ( -150)2
= 450 mm
uuur
OA
1
Then
(300i + 300 j - 150k )
l OA =
=
OA 450
1
= (2i + 2 j - k )
3
Let the coordinates of Point P be (x mm, y mm, z mm). Then
uuur
PC = [(225 - x )i + (375 - y ) j + [(300 - z )k]
uuur
d
Also,
OP = dOP l OA = OP (2i + 2 j - k )
3
uuur
and
OP = xi + yj + zk
2
2
1
x = dOP y = dOP z = dOP
3
3
3
The requirement that OA and PC be perpendicular implies that
uuur
l OA ◊ PC = 0
or
or
1
(2i + 2 j - k ) ◊ [(225 - x )i + (375 - y ) j + (300 - z )k ] = 0
3
2
2
È
ˆ˘
Ê 1
ˆ
Ê
ˆ
Ê
(2) Á 225 - dOP ˜ + (2) Á 375 - dOP ˜ + ( -1) Í300 - Á - dOP ˜ ˙ = 0
¯˚
Ë
¯
Ë
¯
Ë
3
3
3
Î
or dOP = 300 mm 
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45
PROBLEM 3.44
Determine the volume of the parallelepiped of Fig. 3.25 when (a) P  4i 2 3j  2k,
Q  2 2i 2 5j  k, and S  7i  j 2 k, (b) P  5i 2 j  6k, Q  2i  3j  k,
and S  2 3i 2 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
7
1 -1
(b)
= (20 - 21 - 4 + 70 + 6 - 4)
= 67
or Volume = 67.0 
Vol = P ◊ (Q ¥ S)
5 -1 6
= 2 3 1
-3 -2 4
= (60 + 3 - 24 + 54 + 8 + 10)
= 111
or Volume = 111.0 
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46
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 + 10 S x - 6 - 20 + 8 - 12S x = 0 Sx = 7 
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47
PROBLEM 3.46
The 0.61 ¥ 1.00-m lid ABCD of a storage bin is hinged
along side AB and is held open by looping cord DEC over
a frictionless hook at E. If the tension in the cord is 66 N,
determine the moment about each of the ­coordinate axes of
the force exerted by the cord at D.
Solution
First note
z = (0.61)2 - (0.11)2
= 0.60 m
Then
D
0.61 m
0.11 m
z
A
dDE = (0.3)2 + (0.6)2 + ( -0.6)2
= 0.9 m
Now
66 N
(0.3i + 0.6 j - 0.6k )
0.9
= 22[(1 N )i + (2 N ) j - (2 N )k ]
M A = rD /A ¥ TDE
where
rD /A = (0.11 m) j + (0.60 m)k
Then
i
j
k
M A = 22 0 0.11 0.60
1
2
-2
and
TDE =
= 22[( -0.22 - 1.20)i + 0.60 j - 0.11k ]
= - (31.24 N ◊ m)i + (13.20 N ◊ m) j - (2.42 N ◊ m)k
M x = -31.2 N ◊ m, M y = 13.20 N ◊ m, M z = -2.42 N ◊ m 
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48
PROBLEM 3.47
The 0.61 ¥ 1.00-m lid ABCD of a storage bin is hinged alongside AB and is held open by looping cord DEC over a frictionless
hook at E. If the tension in the cord is 66 N, determine the
moment about each of the coordinate axes of the force exerted
by the cord at C.
Solution
First note
z = (0.61)2 - (0.11)2
= 0.60 m
C
0.61 m
0.11 m
B
z
Then
dCE = ( -0.7)2 + (0.6)2 + ( -0.6)2
= 1.1 m
66 N
( -0.7i + 0.6 j - 0.6k )
1.1
= 6[-(7 N )i + (6 N ) j - (6 N )k ]
and
TCE =
Now
M A = rE /A ¥ TCE
where
rE /A = (0.3 m)i + (0.71 m) j
Then
i
j
k
M A = 6 0.3 0.71 0
-7
6
-6
= 6[-4.26i + 1.8 j + (1.8 + 4.97)k ]
= - (25.56 N ◊ m)i + (10.80 N ◊ m) j + ( 40.62 N ◊ m)k
M x = -25.6 N ◊ m, M y = 10.80 N ◊ m, M z = 40.6 N ◊ m 
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49
PROBLEM 3.48
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 2460 N ? m, determine the distance a.
Solution
First note
uuur
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
Then forming the ratio
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
TBA
a
d BA
T
M z = -10.56 BA
d BA
M y = 2.2
( N ◊ m)
( N ◊ m)
My
Mz
T
2.2 dBA ( N ◊ m)
120 N ◊ m
BA
=
-460 N ◊ m -10.56 TBA ( N ◊ m)
d
or a = 1.252 m 
BA
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50
PROBLEM 3.49
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 man applies a 195-N force to
end A of the rope and that the moment of that force about the y axis is
132 N ? m, determine the distance a.
Solution
d BA = (2.2)2 + ( -3.2)2 + ( - a)2
First note
= 15.08 + a2 m
195 N
(2.2i - 3.2 j - a k )
TBA =
d BA
and
Now
M y = j ◊ (rA/D ¥ TBA )
where
rA/ 0 = (2.2 m)i + (1.6 m) j
Then
0
1
0
195
My =
2.2 1.6
0
d BA
2.2 -3.2 - a
=
195
(2.2a) ( N ◊ m)
d BA
Substituting for My and dBA
132 N ◊ m =
or
195
15.08 + a2
(2.2a)
0.30769 15.08 + a2 = a
Squaring both sides of the equation
0.094675(15.08 + a2 ) = a2
or a = 1.256 m 
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51
PROBLEM 3.50
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 470 N · m in
­absolute value. Determine the largest allowable tension in line
ABAD when x = 2 m.
Solution
First note
R A = 2TAB + TAD
Also note that only TAD will contribute to the moment about the z axis.
Now
AD = (2)2 + ( -2.5)2 + ( -1)2
Then,
= 3.354 m
uuur
AD
=T
AD
T
(2i - 2.5 j - k )
=
3.354
Now
where
Then for Tmax ,
TAD
M z = k ◊ (rA/C ¥ TAD )
rA/C = (2.5 m) j + (1 m)k
A
TAD
2TAB
C
2.5 m
z
x
D
x
1m
0
0
1
Tmax
470 =
0 2.5 1
3.354
2 -2.5 -1
=
y
Tmax
(2)(2.25)
3.354
or Tmax = 350 N 
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52
PROBLEM 3.51
For the davit of Problem 3.50, determine the largest allowable
distance x when the tension in line ABAD is 300 N.
Solution
From the solution of Problem 3.50, TAD is now
AD
AD
300 N( xi - 2.5 j - k )
TAD = T
=
x 2 + ( -2.5)2 + ( -1)2
Then M z = k ◊ (rA/ C ¥ TAD ) becomes
300
470 =
470 =
x 2 + ( -2.5)2 + ( -1)2
300
2
x + 7.25
0
0
1
0 2.5 1
x -2.5 -1
| - (2.5)( x ) |
470 x 2 + 7.25 = 750 x
0.627 x 2 + 7.25 = x
Squaring both sides:
0.393x 2 + 2.850 = x 2
x = 2.17 m 
x 2 = 4.695 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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53
PROBLEM 3.52
To loosen a frozen valve, a force F of magnitude 350 N is
applied to the handle of the valve. Knowing that q = 25∞,
Mx = -102 N ◊ m and M z = -72 N ◊ m , determine f and d.
Solution
We have
where
SMO : rA/O ¥ F = MO
rA/O = -(100 mm)i + (275 mm) j - ( d )k
F = F (cos q cos f i - sin q j + cos q sin f k )
For
F = 350 N, q = 25∞
F = (350 N )[(0.90631cos f )i - 0.42262 j + (0.90631sin f )k ]
i
MO = (350 N )
-100
0.90631cos f
j
k
275
-d
mm
-0.42262 0.90631sin f
= (350 N )[(249.2 sin f - 0.42262d ) i + ( -0.90631d cos f + 90.631sin f ) j
+ ( 42.262 - 249.2 cos f )k ] mm
and
M x = (350 N )(249.2 sin f - 0.4226d ) mm = -(102) N ◊ m ¥ 103 mm / m (1)
M z = (350 N )( 42.262 - 249.2 cos f ) mm = -72 ( N ◊ m)(1000 mm / m) (2)
From Equation (2)
Ê 247.98 ˆ
= 5.68∞
f = cos -1 Á
Ë 249.2 ˜¯
or
From Equation (1)
Ê 316.09 ˆ
= 747.97 mm d =Á
Ë 0.4226 ˜¯
or
f = 5.68∞ 
d = 748 mm 
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54
PROBLEM 3.53
When a force F is applied to the handle of the valve shown,
its moments about the x and z axes are, respectively,
M x = -130 N ◊ m and M z = -135 N ◊ m . For d = 675 mm,
determine the moment My of F about the y axis.
Solution
We have
Where
SMO : rA/O ¥ F = MO
rA/O = -(100 mm)i + (275 mm) j - (675 mm)k
F = F (cos q cos f i - sin q j + cos q sin fk )
MO = F
i
j
-100
cos q cos f
275
- sin q
k
-675 N ◊ mm
cos q sin f
= F [(275 cos q sin f - 675 sin q )i + ( -675 cos q cos f + 100 cos q sin f ) j
+ (100 sin q - 275 cos q cos f )k ]N ◊ mm
and
M x = F (275 cos q sin f - 675 sin q )( N ◊ mm) (1)
M y = F ( -675 cos q cos f + 100 cos q sin f ) ( N ◊ mm) (2)
M z = F (100 sin q - 275 cos q cos f ) ( N ◊ mm) (3)
Now, Equation (1)
cos q sin f =
1 Ê Mx
ˆ
+ 675 sin q ˜ ÁË
¯
275 F
(4)
and Equation (3)
cos q cos f =
Mz ˆ
1 Ê
ÁË100 sin q ˜
275
F ¯
(5)
Substituting Equations (4) and (5) into Equation (2),
Ï
Mz ˆ ˘
È 1 Ê Mx
È 1 Ê
ˆ ˘¸
+ 675 sin q ˜ ˙ ˝
M y = F Ì-675 Í
ÁË100 sin q ˜¯ ˙ + 100 Í
ÁË
¯ ˚˛
F ˚
Î 275
Î 275 F
Ó
or
My =
1
(675 M z + 100 M x )
275
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55
PROBLEM 3.53 (continued)
Noting that the ratios 675
and 100
are the ratios of lengths, have
275
275
675
100
( -135 N ◊ m) +
( -130 N ◊ m)
275
275
= (331.36) - ( 47.273)
= -(378.63) N ◊ m
My =
or
M y = -379 N ◊ m 
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56
PROBLEM 3.54
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 = l AD ◊ (rB / A ¥ TBH )
where
and
1
l 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
= 0.5 0
0
5
150 300 -300
1
= [( -3)(0.5)(300)]
5
or
M AD = - 90.0 N ◊ m 
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57
PROBLEM 3.55
In Problem 3.54, determine the moment about the diagonal AD
of the force exerted on the frame by portion BG of the cable.
Solution
MAD = l AD ◊ (rB /A ¥ TBG )
where
and
1
l 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
uv
450 N
T BG =
( -0.5i + 0.925 j - 0.4k )
1.125
= -(200 N )i + (370 N ) j - (160 N)k
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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58
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 AE at A is 55 N, determine
the moment of that force about the line joining Points D and B.
Solution
First note
d AE = (0.9)2 + ( -0.6)2 + (0.2)2 = 1.1 m
Then
TAE =
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 ]
DB = (1.2)2 + ( -0.35)2 + (0)2
l DB
= 1.25 m
uuur
DB
=
DB
1
(1.2i - 0.35 j)
=
1.25
1
= (24i - 7 j)
25
M DB = l DB ◊ (rA/D ¥ TAE )
TDA = -(0.1 m) j + (0.2 m)k
M DB =
24 -7
0
1
(5) 0 -0.1 0.2
25
9
-6
2
1
= ( -4.8 - 12.6 + 28.8)
5
or
MDB = 2.28 N ◊ m 
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59
PROBLEM 3.57
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
dCF = (0.6)2 + ( -0.9)2 + ( -0.2)2 = 1.1 m
Then
TCF =
Also
DB = (1.2)2 + ( -0.35)2 + (0)2
Then
Now
where
Then
33 N
(0.6i - 0.9 j + 0.2k )
1.1
= 3[(6 N )i - (9 N ) j - (2 N )k ]
l DB
= 1.25 m
uuur
DB
=
DB
1
(1.2i - 0.35 j)
=
1.25
1
= (24i - 7 j)
25
MDB = l DB ◊ (rC / D ¥ TCF )
rC / D = (0.2 m) j - (0.4 m)k
MDB
24 -7
0
1
= (3) 0 0.2 -0.4
25
6 -9 -2
=
3
( -9.6 + 16.8 - 86.4)
25
or
MDB = -9.50 N ◊ m 
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60
PROBLEM 3.58
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
(OA)x
M OA = lOA ◊ (rC /O ¥ P)
O
where
From triangle OBC
2
(OA) =
(OA)2x
+ (OA)2y
aÊ 1 ˆ
a
=
2 ÁË 3 ˜¯ 2 3
Ê a ˆ
Ê aˆ
a2 = Á ˜ + (OA)2y + Á
Ë 2¯
Ë 2 3 ˜¯
(OA) y = a2 -
Then
and
6
x
(OA)z
rA/O =
l OA
z
B
+ (OA)2z
2
or
30°
a
(OA) x =
2
(OA) z = (OA) x tan 30∞ =
Since
C
0°
y
2
A
lOA
2
a2 a2
=a
4 12
3
2
a
a
i+a
j+
k
2
3
2 3
1
2
1
= i+
j+
k
2
3 2 3
P = l BC P =
O
rC/O
z
B
C
x
P
( a sin 30∞)i - ( a cos 30∞)k
P
( P ) = (i - 3k )
a
2
rC /O = ai
M OA
1
2
=
1
1
=
2
3
0
1
2 3
Ê Pˆ
( a) Á ˜
Ë 2¯
0
0
- 3
aP Ê 2 ˆ
aP
- ˜ (1)( - 3 ) =
Á
2 Ë 3¯
2
M OA =
aP

2
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61
PROBLEM 3.59
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 3.59 to
determine the perpendicular distance between edges OA and BC.
(OA)x
Solution
(a)
For edge OA to be perpendicular to edge BC,
uuur uuur
OA ◊ BC = 0
C
O
30°
x
(OA)z
60°
where
(OA) x =
From triangle OBC
a
2
z
(OA) z = (OA) x tan 30∞ =
=
so that
(b)
Have M OA
OA
Èa
a
Ê a ˆ ˘
k ˙ ◊ (i - 3k ) = 0
Í i + (OA) y j + ÁË
˜
2
2 3¯ ˚
Î2
A
O
a
a 3
a
ik = (i - 3 k )
2
2
2
2
or
y
uuur Ê a ˆ
Ê a ˆ
OA = Á ˜ i + (OA) y j + Á
k
Ë 2¯
Ë 2 3 ˜¯
uuur
BC = ( a sin 30∞) i - ( a cos 30∞) k
and
Then
B
aÊ 1 ˆ
a
=
2 ÁË 3 ˜¯ 2 3
d
z
C
x
BC
B
2
a
a
+ (OA) y (0) =0
4
4
uuur uuur
OA ◊ BC = 0
uuur
uuur
OA is perpendicular to BC 
uuur uuur
= Pd , with P acting along BC and d the perpendicular distance from OA to BC .
From the results of Problem 3.56
M OA =
Pa
2
Pa
= Pd 2
or d =
a

2
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62
PROBLEM 3.60
A sign erected on uneven ground is guyed
by cables EF and EG. If the force exerted
by cable EF at E is 230 N, determine the
moment of that force about the line joining
points A and D.
Solution
First note that BC = (1200)2 + (900)2 = 1500 mm and that
( 43 ¥ 1200, 2400, 43 ¥ 900)
BE
BC
1125
= 1500
= 43 . The coordinates of Point E are then
or (900 mm, 2400 mm, 675 mm). Then
dEF = ( -15)2 + ( -110)2 + (30)2
= 2875 mm
Then
TEF = ( -375)2 + ( -2750)2 + (750)2
230 N
( -375i - 2750 j + 750k )
2875
= 2[-(15 N )i - (110 N ) j + (30 N )k ]
=
Also
AD = (1200)2 + ( -300)2 + (900)2
= 1529.71 mm
Then
l AD
uuur
AD
=
AD
1
(1200i - 300 j + 900k )
=
1529.71
= (0.7845 i - 0.1961j + 0.5883k)
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63
PROBLEM 3.60 (Continued)
Now
where
Then
M AD = l AD ◊ (rE /A ¥ TEF )
rE /A = (900 mm)i + (2400 mm) j + (675 mm)k
M AD
0.7845 - 0.1961 + 0.5883
= 900
2400
675
-30
-220
60
= 229466 + 14560 - 74126
= 169900 N ◊ mm
MAD = 169.9 N ◊ m 
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64
PROBLEM 3.61
A sign erected on uneven ground is guyed
by cables EF and EG. If the force exerted
by cable EG at E is 270 N, determine the
moment of that force about the line joining
points A and D.
Solution
First note that BC = (1200)2 + (900)2 = 1500 mm and that
then
( 43 ¥ 1200, 2400, 43 ¥ 900)
BE
BC
1125
= 1500
= 43 . The coordinates of Point E are
or (900 mm, 2400 mm, 675 mm) Then
dEG = (275)2 + ( -2200)2 + ( -1100)2
= 2475 mm
270 N
(275i - 2200 j - 1100k )
2475
= (30 N )i - (240 N ) j - (120 N )k
Then
TEG =
Also
AD = (1200)2 + ( -300)2 + (900)2
Then
l AD
= 1529.71 mm
uuur
AD
=
AD
1
(1200i - 300 j + 900k )
=
1529.71
= (0.7845 i - 0.1961j + 0.5883k)
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65
PROBLEM 3.61 (Continued)
Now
where
Then
M AD = l AD ◊ (rE /A ¥ TEF )
rE /A = (900 mm)i + (2400 mm) j + (675 mm)k
M AD
0.7845 -0.1961 +0.5883
= 900
2400
675
30
-240
-120
= -98847 - 25150 - 169430 N ◊ mm
= -293427 N ◊ mm
or MAD = -293 N ◊ m 
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66
PROBLEM 3.62
Two forces F1 and F2 in space have the same magnitude F. Prove that the moment of F1 about the line of
a­ ction of F2 is equal to the moment of F2 about the line of action of F1 .
Solution
F1
A
A
rB/A
rA/B
B
First note that
F1
F2
B
F2
F1 = F1l1 and F2 = F2 l 2
Let M1 = moment of F2 about the line of action of M1 and M 2 = moment of F1 about the line of action
of M 2
Now, by definition
M1 = l1 ◊ (rB /A ¥ F2 )
= l1 ◊ (rB /A ¥ l 2 ) F2
M 2 = l2 ◊ (rA/B ¥ F1 )
= l2 ◊ (rA/B ¥ l 1 ) F1
Since
F1 = F2 = F
and rA/B = -rB /A
M1 = l 1 ◊ (rB /A ¥ l 2 ) F
M 2 = l 2 ◊ ( -rB /A ¥ l 1 ) F
Using Equation (3.38) l 1 ◊ (rB /A ¥ l 2 ) = l 2 ◊ ( -rB /A ¥ l 1 )
so that
M 2 = l 1◊ (rB /A ¥ l 2 ) F M12 = M 21 
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67
PROBLEM 3.63
In Problem 3.54, determine the perpendicular distance between
portion BH of the cable and the diagonal AD.
PROBLEM 3.54 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 3.54:
TBH = 450 N
TBH = (150 N )i + (300 N ) j - (300 N )k
| M AD | = 90.0 N ◊ m
1
l AD = ( 4i - 3k )
5
Based on the discussion of Section
uuur3.11, it follows that only the perpendicular component of TBH will contribute
to the moment of TBH about line AD.
Now
(TBH ) parallel = TBH ◊ l 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 l 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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68
PROBLEM 3.64
In Problem 3.55, determine the perpendicular distance between
portion BG of the cable and the diagonal AD.
PROBLEM 3.55 In Problem 3.54, 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 3.55:
TBG = 450 N
TBG = -(200 N )i + (370 N ) j - (160 N )k
| M AD | = 111 N ◊ m
1
l AD = ( 4i - 3k )
5
Based on the discussion of Sectionuuu3.11,
v it follows that only the perpendicular component of TBG will contribute
to the moment of TBG about line AD.
Now
Also
so that
(TBG ) parallel = TBG ◊ l AD
TBG
1
= ( -200i + 370 j - 160k ) ◊ ( 4i - 3k )
5
1
= [( -200))( 4) + ( -160)( -3)]
5
= -64 N
= (TBG ) parallel + (TBG ) perpendicular
(TBG ) perpendicular = ( 450)2 - ( -64)2 = 445.43 N
Since l 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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69
PROBLEM 3.65
In Problem 3.56, determine the perpendicular distance between
cable AE 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 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 3.56
TAE = 55 N
TAE = 5[(9 N )i - (6 N ) j + (2 N )k ]
| M DB | = 2.28 N ◊ m
1
l DB = (24i - 7 j)
25
Based on the discussion of Section
it follows that only the perpendicular component of TAE will ­contribute
uuu3.11,
r
to the moment of TAE about line DB.
Now
(TAE ) parallel = TAE ◊ l 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 l 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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70
PROBLEM 3.66
In Problem 3.57, determine the perpendicular distance between
cable CF and the line joining Points D and B.
PROBLEM 3.57 The triangular plate ABC is supported by balland-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
From the solution to Problem 3.57
TCF = 33 N
TCF = 3[(6 N )i - (9 N ) j - (2 N )k ]
| M DB | = 9.50 N ◊ m
1
l DB = (24i - 7 j)
25
Based on the discussion of Sectionuuu3.11,
it
follows
that
only the perpendicular component of TCF will contribute
r
to the moment of TCF about line DB.
Now
(TCF ) parallel = TCF ◊ l 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 l 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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or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to
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71
PROBLEM 3.67
In Problem 3.60, determine the perpendicular
distance between cable EF and the line joining
Points A and D.
PROBLEM 3.60 A sign erected on uneven
ground is guyed by cables EF and EG. If
the force exerted by cable EF at E is 230 N,
­determine the moment of that force about the
line joining Points A and D.
Solution
From the solution to Problem 3.61
TEF = 230 N
TEF = ( -30 N )i - (220 N ) j + (60 N )k
| M AD | = 169.9 N ◊ m
l AD = (0.7845i - 0.1961j + 0.5883k )
Based on the discussion of Section
it follows that only the perpendicular component of TEF will ­contribute
uuu3.11,
r
to the moment of TEF about line AD.
Now
Also
so that
(TEF ) parallel = TEF ◊ l AD
TEF
= ( -30i - 220 j + 60k ) ◊ (0.7845i - 0.1961j + 0.58883k)
= 54.905 N
= (TEF ) parallel + (TEF ) perpendicular
(TEF ) perpendicular = (230)2 - (54.905)2 = 223.35 N
Since l AD and (TEF ) perpendicular are perpendicular, it follows that
M AD = d (TEF ) perpendicular
or
169.9 N ◊ m = d ¥ 223.35 N or d = 0.761 m 
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72
PROBLEM 3.68
In Problem 3.61, determine the perpendicular
distance between cable EG and the line joining
Points A and D.
PROBLEM 3.61 A sign erected on uneven
ground is guyed by cables EF and EG. If
the force exerted by cable EG at E is 270 N,
determine the moment of that force about the
line joining Points A and D.
Solution
From the solution to Problem 3.62
TEG = 270 N
TEG = (30 N )i - (240 N ) j - (120 N )k
| M AD | = 293 N ◊ m
l AD = (0.7845i - 0.1961j + 0.5883k )
Based on the discussion of Sectionuuu3.11,
r it follows that only the perpendicular component of TEG will ­contribute
to the moment of TEG about line AD.
Now
Thus,
(TEG ) parallel = TEG ◊ l AD
(TEG ) perpendicular
= (30i - 240 j - 120k ) ◊ (0.7845i - 0.1961j + 0.58883k)
=0
= TEG = 270 N
Since l AD and (TEG ) perpendicular are perpendicular, it follows that
| M AD | = d (TEG ) perpendicular
or
293 N ◊ m = d ¥ 270 N
or
d = 1.085 m 
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73
PROBLEM 3.69
Two parallel 60-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
SM B : - d1C x + d2C y = M
where
d1 = (0.360 m)sin 55∞
= 0.29489 m
d2 = (0.360 m)sin 55∞
= 0.20649 m
(c)
We have
C
60 N
B
Cx
d1
B
20°
Bx
55°
= 56.382 N
C y = (60 N) sin 20∞
We have
d
60 N
C x = (60 N ) cos 20∞
(b)
Cy
d2
20°
A
= 20.521 N
M = -(0.29489 m)(56.382 N)k + (0.20649 m)(20.521 N)k
= -(12.3893 N ◊ m)k
M = Fd( -k )
= 60 N[(0.360 m)sin(55∞ - 20∞)]( -k )
= -(12.3893 N ◊ m)k SM A : S (rA ¥ F) = rB /A ¥ FB + rC /A ¥ FC = M
By
55°
A
or M = 12.39 N ◊ m

or M = 12.39 N ◊ m

or M = 12.39 N ◊ m

i
j
k
M = (0.520 m)(60 N) cos 55∞
sin 55∞ 0 - cos 20∞ - sin 20∞ 0
i
j
k
+ (0.800 m)(60 N) cos 55∞ sin 55∞ 0
cos 20∞ sin 20∞ 0
= (17.8956 N ◊ m - 30.285 N ◊ m)k
= -(12.3892 N ◊ m)k
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74
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 105 N
forces, (b) the perpendicular distance between the 60 N forces if the
resultant of the two couples is zero, (c) the value of a if the resultant
couple is 9 N ◊ m clockwise and d is 1050 mm.
Solution
(a)
60 N
C
D
We have
M1 = d1 F1
where
d1 = 400 mm
F1 = 105 N
M1 = ( 400 mm)(105 N)
= 42000 N ◊ mm
or M1 = 42 N ◊ m 
d2
A
B
60 N
d2
(b)
or
60 N
(c)
D
42 N ◊ m - d2 (60 N) = 0 or
Ê 1050 ˆ
m (sin a )(60 N)
-9 N ◊ m = 42 N ◊ m - Á
Ë 1000 ˜¯
a
sin a = 0.80952
B
A
60 N
d2 = 0.7 m = 700 mm 
M total = M1 + M 2
We have
C
d sin a
M1 + M 2 = 0
We have
a = 54.049∞ and
or a = 54.0∞ 
d
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75
PROBLEM 3.71
A couple M of magnitude 18 N ◊ m is applied to
the handle of a screwdriver to tighten a screw into
a block of wood. Determine the magnitudes of the
two smallest horizontal forces that are equivalent to
M if they are applied (a) at corners A and D, (b) at
corners B and C, (c) anywhere on the block.
Solution
P
(a)
D
M
or
.24 m
M = Pd
We have
18 N ◊ m = P(.24 m)
P = 75.0 N A
P
D
E
d BC = ( BE )2 + ( EC )2
(b)
.08 m
= (.24 m)2 + (.08 m)2
C
M
= 0.25298 m
P
.24 m
18 N ◊ m = P (0.25298 m)
P = 71.152 N
P
.24 m
= (0.24 m)2 + (0.32 m)2
C
M
P
= 0.4 m
P
M = Pd AC
We have
A
or P = 71.2 N 
d AC = ( AD )2 + ( DC )2
(c)
.32 m
D
M = Pd
We have
dBC
B
A
or Pmin = 75.0 N 
18 N ◊ m = P (0.4 m)
P = 45.0 N or P = 45.0 N 
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76
PROBLEM 3.72
Four 25 mm-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
(a)
175 mm
175 N
225 mm
125 N
(b)
A
–F
d
150 mm
q
q
F
D
200 mm
(c)
q
q
With only one string, pegs A and D, or B and C should be used. We have
150
tan q =
q = 36.9∞
90∞ - q = 53.1∞
200
Direction of forces:
With pegs A and D:
q = 53.1∞ 
With pegs B and C:
q = 53.1∞ 
The distance between the centers of the two pegs is
(200)2 + (150)2 = 250 mm
B
F
1 M = (175 N )(175 mm) + (125 N )(225 mm)
= 30625 N ◊ mm + 28125 N ◊ mm
M = 58750 N ◊ mm 
Therefore, the perpendicular distance d between the forces is
ˆ
Ê 25
d = 250 mm + 2 Á mm˜
¯
Ë 2
–F
C
= 275 mm
We must have M = Fd = 58750 N ◊ mm
= F (275 mm)
F = 213.6 N = 214 N 
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77
PROBLEM 3.73
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 60.25 N ◊ m counterclockwise.
Solution
M = d AD FAD + d BC FBC
M = 60.25 N ◊ m = 60250 N ◊ mm
60250 N ◊ mm = [(150 + d ) mm](175 N ) + [(200 + d ) mm](125 N )]
d = 30 mm 
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78
PROBLEM 3.74
The shafts of an angle drive are acted upon by the two couples shown.
Replace the two couples with a single equivalent couple, specifying its
magnitude and the direction of its axis.
Solution
y
Based on
where
M1 = -(12 N ◊ m)j
M 2 = -(9 N ◊ m)k
M = -(12 N ◊ m)j - (9 N ◊ m)k
and
|M| = (12)2 + (9)2 = 15 N ◊ m 143.1°
z
126.9°
M
M = M1 + M 2
or M = 15 N ◊ m 
M
|M|
-(12 N ◊ m)j - (9 N ◊ m)k
=
15 N ◊ m
= -0.8 j - 0.6k
l=
or
M = |M|l = (15 N ◊ m)( -0.8 j - 0.6k )
cos q x = 0
q x = 90∞
cos q y = -0.8
q y = 143.130∞
cos q z = -0.6
q z = 126.870∞
or q x = 90.0∞ q y = 143.1∞ q z = 126.9∞ 
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79
PROBLEM 3.75
If P = 0, replace the two remaining couples with a single
equivalent couple, specifying its magnitude and the direction
of its axis.
Solution
We have
where
y
M = M1 + M 2
.17 m
M1 = rG /C ¥ F1
Also,
.16 m G
F1 = (18 N )k
M1 = -(0.3 m)i ¥ (18 N )k
= (5.4 N ◊ m) j
M 2 = rD /F ¥ F2
z
F1
.15 m
F2
D
rG /C = -(0.3 m)i
.15 m
A
E
rD/F
F1
F
rG/C
F2
B
x
C
rD /F = -(.15 m)i + (.08 m) j
F2 = l ED F2
(.15 m)i + (.08 m) j + (.17 m)k
(34 N )
=
(.15)2 + (.08)2 + (.17)2 m
= 141.421 N ◊ m(.15i + .08j + .17k )
i
j
k
M 2 = 141.421 N ◊ m -.15 .08 0
-.15 .08 .17
= 141.421(.0136i + 0.0255 j) N ◊ m
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80
PROBLEM 3.75 (Continued)
and
M = [(5.4 N ◊ m)j] + [141.421(.0136i + .0255 j) N ◊ m]
= (1.92333 N ◊ m)i + (9.0062 N ◊ m)j
| M | = ( M x )2 + ( M y )2
= (1.92333)2 + (9.0062)2
= 9.2093 N ◊ m
or M = 9.21 N ◊ m 
M (1.92333 N ◊ m)i + (9.0062 N ◊ m) j
=
|M|
9.2093 N ◊ m
= 0.20885 + 0.97795
cos q x = 0.20885
l=
q x = 77.945∞
or q x = 77.9∞ 
cos q y = 0.97795
q y = 12.054∞ or q y = 12.05∞ 
cos q z = 0.0
q z = 90∞ or q z = 90.0∞ 
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81
PROBLEM 3.76
If P = 0, replace the two remaining couples with a single
e­ quivalent couple, specifying its magnitude and the direction
of its axis.
Solution
M = M1 + M 2 ; F1 = 80 N, F2 = 200 N
M1 = rC ¥ F1 = (750 mm)i ¥ ( -80 N ) j = -(60000 N ◊ mm)k
M 2 = rE /B ¥ F2 ; rE /B = (375 mm)i - (125 mm) j
dDE = (0)2 + (125)2 + (250)2 = 125 5 mm
F2 =
200 N
(125 j - 250k )
125 5
= 40 5 ((1 N )j - (2 N )k )
i
j
k
M 2 = 40 5 375 -125 0
0
1 -2
= 40 5[(250 N ◊ mm)i + (750 N ◊ mm) j + (375 N ◊ mm)k ]
M = (22361i + 67082 j + 33541k ) N.mm
= (22.361 N ◊ mm)i + (67082 N ◊ mm)) j - (26459 N ◊ mm)k
M = (22361)2 + (67082)2 + ( -26459)2
= 75499 N ◊ mm
M = 75.5 N ◊ m 
M
= 0.29617i + 0.88852 j - 0.35045k
M
cos q x = 0.29617
l axis =
cos q y = 0.88852
cos q z = -0.35045
q x = 72.8∞ q y = 27.3∞ q z = 110.5∞ 
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82
PROBLEM 3.77
If P = 100 N, replace the three couples with a single equivalent
couple, specifying its magnitude and the direction of its axis.
Solution
From the solution to Problem. 3.76
80 N force:
M1 = -60000 N ◊ mm k
200 N force:
M 2 = 8 5[(22361 N ◊ mm)i + (67082 N ◊ mm) j + (33541 N ◊ mm) k ]
P = 100 N
M 3 = rC ¥ P
= (750 mm)i ¥ (100 N )( -k)
= (75000 N ◊ mm) j
M = M1 + M 2 + M3
= (22361 N ◊ mm)i + (142082 N ◊ mm) j - (26459 N ◊ mm)k
M = (22361)2 + (142082)2 + ( -26459)2
= 146244 N ◊ mm
M = 146.2 N ◊ m 
M
= 0.15290i + 0.97154 j - 0.18092k
M
cos q x = 0.15290
l axis =
cos q y = 0.97154
cos q z = -0.180921
q x = 81.2∞ q y = 13.70∞ q z = 100.4∞ 
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83
PROBLEM 3.78
If P = 20 N, replace the three couples with a single equivalent
couple, specifying its magnitude and the direction of its axis.
Solution
We have
M = M1 + M 2 + M3
where
M1 = rG /C
i
j k
¥ F1 = -0.3 0 0 N ◊ m = (5.4 N ◊ m)j
0
0 18
i
j
k
M 2 = rD /F ¥ F2 = -.15 .08 0 141.421 N ◊ m
-.15 .08 .17
= 141.421(.0136i + .0255 j) N ◊ m
(See Solution to Problem 3.75)
i
j
k
M3 = rC /A ¥ F3 = 0.3 0 0.17 N ◊ m
0 20
0
= -(3.4 N ◊ m)i + (6 N ◊ m)k
M = [(1.92333 - 3.4)i + (5.4 + 3.6062) j + (6)k ] N ◊ m
= -(1.47667 N ◊ m)i + (9.0062 N ◊ m) j + (6 N ◊ m)
| M | = M x2 + M y2 + M z2
= (1.47667) + (9.0062) + (6)2
= 10.9221 N ◊ m
or M 5 10.92 N . m 
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84
Problem 3.78 (Continued)
M -1.47667 + 9.0062 + 6
=
|M|
10.9221
= -0.135200i + 0.82459 j + 0.54934k
l=
cos q x = -0.135200 q x = 97.770 or q x = 97.8∞ 
cos q y = 0.82459
q y = 34.453 or q y = 34.5∞ 
cos q z = 0.54934
q z = 56.678 or q z = 56.7∞ 
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85
PROBLEM 3.79
Shafts A and B connect the gear box to the wheel assemblies of
a tractor, and shaft C connects it to the engine. Shafts A and B
lie in the vertical yz plane, while shaft C is directed along the
x axis. Replace the couples applied to the shafts with a single
equivalent couple, specifying its magnitude and the direction
of its axis.
Solution
Represent the given couples by the following couple vectors:
M A = -1600 sin 20∞ j + 1600 cos 20∞k
= -(547.232 N ◊ m) j + (1503.51 N ◊ m)k
M B = 1200 sin 20∞ j + 1200 cos 20∞k
= ( 410.424 N ◊ m) j + (1127.63 N ◊ m)k
MC = -(1120 N ◊ m)i
The single equivalent couple is
M = M A + M B + MC
= -(1120 N ◊ m)i - (136.808 N ◊ m) j + (2631.1 N ◊ m)k
M = (1120)2 + (136.808)2 + (2631.1)2
= 2862.8 N ◊ m
-1120
cos q x =
2862.8
-136.808
cos q y =
2862.8
2631.1
cos q z =
2862.8
M = 2860 N ◊ m q x = 113.0∞ q y = 92.7∞ q z = 23.2∞ 
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86
PROBLEM 3.80
The tension in the cable attached to the end C of
an adjustable boom ABC is 2.8 k N. Replace the
force exerted by the cable at C with an equivalent
force-couple system (a) at A, (b) at B.
3m
2.44 m
Tcos 50°
30°
20°
40°
(a)
SF : FA = T = 2800 N
Based on
FA = 2800 N
or
B
M A = 11.67 kN ◊ m
or
(b)
A
20°
FA
FB = 2800 N
C
or
A
20° 
SM B : M B = (T sin 50∞)( d B )
= (2800 N)( sin 50∞) (3 m)
= 6435 N ◊ m
M B = 6.43 kN ◊ m
dB
B

SF : FB = T = (2800) N
Based on
or
MB
20° 
SM A : M A = (T sin 50∞)(d A )
= (2800 N )sin 50∞(5.44 m)
= 11668 N ◊ m
C
dA
MA
T
Solution
T sin 50°
B
30°
A
C

20°
FB
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87
PROBLEM 3.81
A 800 N force P is applied at Point A of a structural member.
Replace P with (a) an equivalent force-couple system at C,
(b) an equivalent system consisting of a vertical force at B and
a second force at D.
Solution
(a)
Based on
where
SF : PC = P = 800 N SM C : M C = - Px dcy + Py dCx
or PC = 800 N 60° 
A
Px = (800 N ) cos 60∞
= 400 N
Py = (800 N )sin 60∞
dCx
PC
dCy
60˚
= 692.82 N
= 1.2 m
C
MC
dCx
dCy = 0.83 m
M C = ( 400 N )(0.83 m) + (692.82 N )(1.2 m)
= 1163.4 N ◊ m
(b)
Based on
or MC = 1163 N ◊ m
A
SFx : PDx = P cos 60∞
= (800 N ) cos 60∞
= 400 N
dDA
PB

PDx
PD
q
D PDX
B
SM D : ( P cos 60∞)( dDA ) = PB ( dDB )
[(800 N ) cos 60∞](0.45 m) = PB (1.8 m)
PB = 100 N
dDB
or PB = 100 N ≠ 
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88
Problem 3.81 (Continued)
SFy : P sin 60∞ = PB + PDy
(800 N )sin 60∞ = 100 N + PDy
PDy = 592.82 N
PD = ( PDx )2 + ( PDy )2
= ( 400)2 + (592.82)2
= 715.15 N
-1 Ê
PDy ˆ
q = tan Á
Ë P ˜¯
Dx
Ê 592.82 ˆ
= tan -1 Á
Ë 400 ˜¯
= 55.991∞
or PD = 715 N
56.0° 
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89
PROBLEM 3.82
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
FB
dB
SF : FB = F = 80 N (a) Based on
A
or FB = 80.0 N ¨ 
SM : M B = Fd B
MB
= 80 N (.05 m)
D
C
= 4.0000 N ◊ m
M B = 4.00 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.
A
FC
FD
FB
Then, with FC and FD acting as shown,
SM : M D = FC d
B
C
D
4.0000 N ◊ m = FC (.04 m)
FC = 100.000 N d
or FC = 100.0 N Ø 
SFy : 0 = FD - FC
FD = 100.000 N or FD = 100.0 N ≠ 
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90
PROBLEM 3.83
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 a with the vertical.
A
B
Then for equivalence,
C
30˚
1040 N
SFx : (1040 N )sin 30∞ = FA sin a + FB sin a (1)
SFy : - (1040 N ) cos 30∞ = - FA cos a - FB cos a (2)
Dividing Equation (1) by Equation (2),
A
30˚
B
FA
6.7 m
(1040 N )sin 30∞
( FA + FB )sin a
=
-(1040 N ) cos 30∞ -( FA + FB ) cos a
C
30˚
FC
4m
Simplifying yields a = 30∞
Based on
SM C : [(1040 N ) cos 30∞]( 4 m) = ( FA cos 30∞)(10.7 m)
FA = 388.79 N
FA = 389 N
or
60° 
Based on
SM A : - [(1040 N ) cos 30∞](6.7 m) = ( FC cos 30∞)(10.7 m)
FC = 651.21 N
FC = 651 N
or
60° 
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91
PROBLEM 3.84
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 a = 30∞ and
b = 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)
SF : F = P or F = 250 N
Equivalence requires
60°
SM B : M = -(0.3 m)(250 N ) = -75 N ◊ m
The equivalent force-couple system at B is
F = 250 N
(b)
M = 75.0 N ◊ m 
60°
Require
C
y
60°
B
A
B
250 N
30°
A
φ
x
C
φ
FB
FA
Equivalence then requires
SFx : 0 = FA cos f + FB cos f
FA = - FB
or cosf = 0
SFy : - 250 = - FA sin f - FB sin f
Now if
FA = - FB fi -250 = 0 reject
cos f = 0
or
and
Also
f = 90∞
FA + FB = 250
S M B : - (0.3 m)(250 N ) = (0.2 m) FA
or
FA = -375 N
and
FB = 625 N
FA = 375 N
60∞
FB = 625 N
60∞ 
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92
PROBLEM 3.85
Solve Problem 3.84, assuming a = b = 25∞.
Solution
C
25˚
25˚
B
P = 250 N
A
0.3 m
25˚
(a)
Equivalence requires
SF : FB = P or FB = 250 N
25.0∞
SM 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)
M B = 57.5 N ◊ m
25.0∞ 
Require
C
C
Q
E
MB
–Q
A
B
50˚
B
250 N
A
25˚
25˚
250 N
0.2 m
Equivalence requires
Adding the forces at B:
M B = d AE Q (0.3 m)[(250 N )sin 50∞]
= [(0.2 m)sin 50∞]Q
Q = 375 N
FA = 375 N
25.0∞ FB = 625 N
25.0∞ 
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93
PROBLEM 3.86
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
360 N
(a)
y
A
B
A
x
450 mm
D
and
d = 0.09 m
d
d
+
C
+
C
F
or d = 90.0 mm below ED 
(b)
150 mm
We have from Part a
and
360 N
A
450 mm
F = -(600 N )k 
SM D : -(360 N )(0.15 m) = -(600 N )( d )
d 5 0.09 m
y
B
A
B
or d = 90.0 mm above ED 
z
E
SM D : (360 N )(0.15 m) = (600 N )( d )
D
E
600 N
E
or F = -(600 N )k 
z
360 N
We have SF : F = (360 N ) j - (360 N ) j - (600 N )k
B
F
C
+
x
d
D
600 N
E
D
360 N
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94
PROBLEM 3.87
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
X
250 N
MH
H
0.18 m
C
H
900 N
H
900 N
250 N
900 N
D
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 Ø
x = 50.0 mm 
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95
PROBLEM 3.88
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.
Have FB = 15 N - 13 N = 2 N, where the 13 N force be part of the couple. Combining the two parallel forces,
M couple = (13 N )[(80 mm + 70 mm) cos 25∞]
= 1767.3 N ◊ mm
M couple = 1767.3 N ◊ mm
and
25°
2N
z
B
A
1767.3 N· mm
x
25°
2N
z
A
B
–a
x
A single equivalent force will be located in the negative z-direction
Based on
SM B : -1767.3 N ◊ mm = [(2 N) cos 25∞](a)
a = 975 mm
F9 = (2 N )(cos 25∞i + sin 25∞k )
F9 = (1.813 N )i + (0.845 N )k and is applied on an extension of handle BD at a
distance of 975 mm to the right of B

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96
PROBLEM 3.89
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)
A
F = 240 N
where
θ
60°
First note that the two 100-N forces form A couple. Then
u
55°
q = 180∞ - (60∞ + 55∞) = 65∞
M = SM B
= (750 mm)(240 N ) cos 55∞ - (1750 mm)(100 N ) cos 20∞
= -61202 N ◊ mm
The equivalent force-couple system at B is
and
F = 240 N
65°
M = 61.2 N ◊ m

(b)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.
SM B : M = - aF ¢ cos 55∞
where a is the distance from B to the point of application of F9. Then
-61202 N ◊ mm = - a(240 N ) cos 55∞
or
a = 444.6 mm F ¢ = 240 N
65.0° 
and is applied to the lever 445 mm
To the left of pin B

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97
PROBLEM 3.90
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 SM E = 0 for the given force-couple system. Further, for
Pmin, must require that P be perpendicular to rB /E . Then
Pmin
SM E : (0.2 sin 30∞ + 0.2) m ¥ 300 N
+ (0.2 m)sin 30∞ ¥ 300 N
- (0.4 m) Pmin = 0
or
a =30°
B
300 N
300 N
30°
60°
E
Pmin = 300 N
0.2 m
0.2 m
Pmin = 300 N
30.0° 
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98
PROBLEM 3.91
A rectangular plate is acted upon by the force and couple shown.
This system is to be replaced with a single equivalent force. (a) For
a = 40∞, specify the magnitude and the line of action of the ­equivalent
force. (b) Specify the value of a if the line of action of the equivalent
force is to intersect line CD 300 mm to the right of D.
F
Solution
(a)
M
The given force-couple system (F, M) at B is
F′
d
B
F = 48 N Ø
and
M = S M B = (0.4 m)(15 N ) cos 40∞ + (0.24 m)(15 N )sin 40∞
or
M = 6.9103 N ◊ m
The single equivalent force F9 is equal to F. Further for equivalence
SM B : M = dF ¢
6.9103 N ◊ m = d ¥ 48 N
or
d = 0.14396 m or
and the line of action of F9 intersects line AB 144 mm to the right of A.
(b)
F ¢ = 48 N 

Following the solution to Part a but with d = 0.1 m and a unknown, have
SM B : (0.4 m)(15 N ) cos a + (0.24 m)(15 N )sin a
= (0.1 m)( 48 N )
or
5 cos a + 3 sin a = 4
Rearranging and squaring
25 cos2 a = ( 4 - 3 sin a )2
Using cos2 a = 1 - sin 2 a and expanding
25(1 - sin 2 a ) = 16 - 24 sin a + 9 sin 2 a
or
Then
34 sin 2 a - 24 sin a - 9 = 0
24 ± ( -24)2 - 4(34)( -9)
sin a =
2(34)
sin a = 0.97686 or sin a = -0.27098
a = 77.7∞ or a = -15.72∞

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99
PROBLEM 3.92
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
y
SF : - (1220 N )i = F
0.1 m
G
Also, we have
rN/G
z
1220 N
SMG : rA/G ¥ P = M
x
0.06 m
II
i
j
k
1220 0 -.1 -.06 N ◊ m = M
-1 0
0
y
M
M = (1220 N ◊ m)[( -0.06)( -1) j - ( -0.1)( -1)k ]
G
F
z
F = - (1220 N )i 
x
or M = (73.2 N ◊ m) j - (122 N ◊ m)k 
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100
PROBLEM 3.93
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.
y
Solution
750 mm
We have
C
67 mm
rB/C
SF : PAB = FC
1850 mm
B
where
PAB
PAB = l AB PAB
(33 mm)i + (990 mm) j - (594 mm)k
=
(175 N )
1155.00 mm
990 mm
A
100 mm
O
594 mm
x
z
or FC = (5.00 N )i + (150 N ) j - (90.0 N )k 
We have
SMC : rB/C ¥ PAB = MC
i
j
k
MC = 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 MC = (77.4 N ◊ m)i + (61.5 N ◊ m) j + (106.8 N ◊ m)k 
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101
PROBLEM 3.94
An antenna is guyed by three cables as shown. Knowing that
the tension in cable AB is 1.44 kN, 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 = ( -19)2 + ( -38)2 + (5)2 = 42.778 m
Then
TAB =
Now
1440 N
( -19i - 38 j + 5k )
42.778
= -(639.58 N )i - (1279.16 N ) j + (168.31 N )k
M = MO = rA/ O ¥ TAB
= 38 j ¥ [-(639.58)i - (1279.16) j + (168.31)k )]
= (6396 N ◊ m)i + (24304 N ◊ m)k
The equivalent force-couple system at O is
F = -(640 N )i - (1279 N ) j + (168.3 N )k 
M = (6.40 kN ◊ m)i + (24.3 kN ◊ m)k 
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102
PROBLEM 3.95
An antenna is guyed by three cables as shown. Knowing that
the tension in cable AD is 1.35 kN, 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 = ( -19)2 + ( -38)2 + ( -38)2
= 57 m
Then
Now
1350 N
( -19i - 38 j + 38k )
57
= 450 N( -i - 2 j - 2k )
TAD =
M = MO = rA/O ¥ TAD
= 128 j ¥ 450( -i - 2 j - 2k )
= -(115200 N ◊ m)i + (57600 N ◊ m)k
The equivalent force-couple system at O is
F = -( 450 N )i - (900 N ) j - (900 N )k 
M = -(115.2 kN ◊ m)i + (57.6 kN ◊ m)k 
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103
PROBLEM 3.96
Replace the 150-N force with an equivalent force-couple system
at A.
Solution
Equivalence requires
where
Then
SF : F = (150 N )( - cos 35∞ j - sin 35∞k )
= -(122.873 N ) j - (86.036 N )k
SMA : M = rD /A ¥ F
rD /A = (0.18 m)i - (0.12 m) j + (0.1 m)k
i
j
k
M = 0.18
-0.12
0.1
N◊m
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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104
PROBLEM 3.97
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)
We have
SM Bz : M 2 z = 0
k ◊ (rH /B ¥ F1 ) + M1z = 0 where
(1)
rH /B = (0.31 m)i - (0.0233) j
F1 = l 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 = l EJ M1
=
- 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)
0.31 -0.0233 0 +
=0
2
d
+
0
.
0058
42
42
-49
Solving for d, Equation (1) reduces to
(13.0200 + 0.9786) From which
d = 0.1350 m 2.17 N ◊ m
d 2 + 0.0058
=0
or d = 135.0 mm 
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105
Problem 3.97 (Continued)
(b)
F2 = F1 = ( 42i + 42 j - 49k )N or F2 = ( 42 N )i + ( 42 N ) j - ( 49 N )k 
M 2 = rH /B ¥ F1 + M1
i
j
k
= 0.31 -0.0233 0 +
42
42
-49
(0.1350)i + 0.03j - 0.07k
(31 N ◊ m)
0.155000
= (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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106
PROBLEM 3.98
A 230 N force F and a 265 N?m 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
Then
Now
where
Then
d AJ = ( 450)2 + ( -350)2 + ( -75)2 = 575 mm
230 N
F=
( 450i - 350 j - 75k )
575
= (180 N )i - (140 N ) j - (30 N )k
d AC = ( -1125)2 + (0)2 + ( -700)2 = 1325 mm
265 N ◊ m
M=
( -1125i - 700k )
1325
= -(225 N ◊ m)i - (140 N ◊ m)k
M ¢ = M + rA/H ¥ F
rA/H = (1125 mm)i + (350 mm) j
i
j
k
M ¢ = ( -225i - 140k ) + 1.125 0.35
0 = (1.125 m)i + (350 m) j
180 -140 -30
M ¢ = -225i - 140k - 10.5i + 33.75 j - 220.5k
= -(235.5 N ◊ m)i + (33.75 N ◊ m) j - (360.5 N ◊ m)k
F ¢ = (180 N )i - (140 N ) j - (30 N )k 
The equivalent force-couple system at H is M ¢ = -(235.5 N ◊ m)i + (33.75 N ◊ m) j - (360.5 N ◊ m)k 
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107
PROBLEM 3.99
The handpiece for a miniature industrial grinder weighs 3 N,
and its center of gravity is located on the y axis. The head of the
handpiece is offset in the xz plane in such a way that line BC forms
an angle of 25∞ with the x direction. Show that the weight of the
handpiece and the two couples M1 and M2 can be replaced with a
single equivalent force. Further, assuming that M1 = 85 N?mm and
M2 = 81.25 N?mm, determine (a) the magnitude and the direction
of the equivalent force, (b) the point where its line of action
­intersects the xz plane.
y
Solution
First assume that the given force W and couples M1 and M2 act at the origin.
Now
W = -Wj
and
M = M1 + M 2
= - ( M 2 cos 25∞)i + ( M1 - M 2 sin 25∞)k
0
M1
z
25˚
M2
W
x
Note that since W and M are perpendicular, it follows that they can be replaced
with a single equivalent force.
F = W or F = - Wj = - (3 N ) j or F = - (3N)j 
(a)
We have
(b)
Assume that the line of action of F passes through Point P(x, 0, z). Then for equivalence
M = rP /0 ¥ F
where
y
rP /0 = xi + zk
- ( M 2 cos 25∞)i + ( M1 - M 2 sin 25∞)k
i
j
= x 0
0 -W
Equating the i and k coefficients,
(b)
For
z=
W =3N
k
z = (Wz )i - (Wx )k
0
M
z
y
o
o
x
W
z
rP/O
x
P(x, 0, z)
- M z cos 25∞
Ê M - M 2 sin 25∞ ˆ
and x = - Á 1
˜¯
Ë
W
W
M1 = 85 N ◊ mm M 2 = 81.25 N ◊ mm
Ê 0.85 - 81.25 sin 25∞ ˆ
x = -Á
˜¯ = 16.887 mm -3
Ë
z=
F
- 81.25 cos 25∞
= -24.546 mm 3
or
x = 16.89 mm 
or
z = - 24.5 mm 
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108
PROBLEM 3.100
A 4-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?
R
Solution
(a)
(a)
We have
A
SFy : -400 N - 200 N = Ra
M
or
and
B
4m
R a = 600 N Ø 
SM A : 1800 N ◊ m - (200 N )( 4 m) = M a
or M a = 1000 N ◊ m
(b)
We have
SFy : - 600 N = Rb
or
and
R b = 600 N Ø 
SM A : - 900 N ◊ m = M b
or M b = 900 N ◊ m
(c)
We have

SFy : 300 N - 900 N = Rc
or
and

R c = 600 N Ø 
SM A : 4500 N ◊ m - (900 N )( 4 m) = M c
or M c = 900 N ◊ m

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109
Problem 3.100 (Continued)
(d)
We have
SFy : - 400 N + 800 N = Rd
R d = 400 N ≠ 
or
and
SM A : (800 N)( 4 m) - 2300 N ◊ m = M d
or
(e)
We have
M d = 900 N ◊ m
SFy : - 400 N - 200 N = Re
R e = 600 N Ø 
or
and
SM A : 200 N ◊ m + 400 N ◊ m - (200 N)(4 m) = M e
or
(f)
We have
M e = 200 N ◊ m
R f = 600 N Ø 
SM A : - 300 N ◊ m + 300 N ◊ m + (200 N)(4 m) = M f
or
(g)
We have
M f = 800 N ◊ m
R g = 1000 N Ø 
SM A : 200 N ◊ m + 4000 N ◊ m - (800 N)(4 m) = M g
or M g = 1000 N ◊ m
(h)
We have
R h = 600 N Ø 
SM A : 2400 N ◊ m - 300 N ◊ m - (300 N)(4 m) = M h
or
(b)

SFy : - 300 N - 300 N = Rh
or
and

SFy : - 200 N - 800 N = Rg
or
and

SFy : - 800 N + 200 N = R f
or
and

Therefore, loadings (c) and (h) are equivalent.
M h = 900 N ◊ m 

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110
PROBLEM 3.101
A 4-m-long beam is loaded as shown. Determine the loading of
Problem 3.100 which is equivalent to this loading.
Solution
We have
and
SFy : - 200 N - 400 N = R or
R
SM A : -400 N ◊ m + 2800 N ◊ m - ( 400 N )( 4 m) = M
M = 800 N ◊ m
or
R = 600 N Ø 
A
M
B
4m
Equivalent to case (f) of Problem 3.101 
Problem 3.101 Equivalent force-couples at A
Case
R
(a)
600 N Ø
1000 N ◊ m
(b)
600 N Ø
900 N ◊ m
(c)
600 N Ø
900 N ◊ m
(d)
400 N ≠
900 N ◊ m
(e)
600 N Ø
200 N ◊ m
(f)
600 N Ø
800 N ◊ m
(g)
1000 N Ø
1000 N ◊ m
(h)
600 N Ø
900 N ◊ m
M

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111
PROBLEM 3.102
Determine the single equivalent force and the distance from Point A to its line of action for the beam and
­loading of (a) Problem 3.101b, (b) Problem 3.101d, (c) Problem 3.101e.
PROBLEM 3.101 A 4-m-long beam is subjected to a variety of loadings. (a) Replace each loading with an
equivalent force-couple system at end A of the beam. (b) Which of the loadings are equivalent?
Solution
600 N
A
900 N· m
B
C
4m
R
d
A
(a)
For equivalent single force at distance d from A
SFy : - 600 N = R
We have
or R = 600 N Ø 
B
C
SM C : (600 N )( d ) - 900 N ◊ m = 0
and
or d = 1.500 m 
2300 N· m
400 N
A
C
4m
R
d
A
B
800 N
C
A
400 N
C
4m
R
200 N·m
A
d
C
(b)
SFy : - 400 N + 800 N = R
We have
or R = 400 N ≠ 
SM C : ( 400 N )( d ) + (800 N )( 4 - d )
and
B
200 N
B
400 N· m
B
- 2300 N ◊ m = 0
(c)
or d = 2.25 m 
SFy : - 400 N - 200 N = R
We have
or R = 600 N Ø 
SM C : 200 N ◊ m + ( 400 N)( d )
and
- (200 N )( 4 - d ) + 400 N ◊ m = 0
or d = 0.333 m 
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112
PROBLEM 3.103
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 5 (50 N)i and a couple of moment
M 5 (22.5 N ? m)j 1 (22.5 N ? m)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 N)i]. Next move each of the systems to the origin O; the forces remain
unchanged.
A: M A = SMO = (7.5 N ◊ m) j + (22.5 N ◊ m)k + (0.6 m)k ¥ (50 N )i
= (37.5 N ◊ m) j + (22.5 N ◊ m)k
D : M D = SMO = -(7.5 N ◊ m) j + (37.5 N ◊ m)k
+ [(1.35 m)i + (0.3 m) j + (0.6 m)k ] ¥ (50 N )i
= (22.5 N ◊ m) j + (22.5 N ◊ m)k
G : MG = SMO = (22.5 N ◊ m)i + (22.5 N ◊ m) j
I:
M I = SM I = (22.5 N ◊ m) j - (7.5 N ◊ m)k
+ [(1.35 m)i + (0.3 m) j] ¥ (50 N )i
= (22.5 N ◊ m) j - (22.5 N ◊ m)k
The equivalent force-couple system is the system at corner D.

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113
PROBLEM 3.104
The weights of two children sitting at ends A and B of a seesaw
are 420 N and 320 N, 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) 300 N, (b) 260 N.
Solution
420 N
d
Wc
B
A
1.8 m
(a)
C
1.8 m
For the resultant weight to act at C, SM C = 0 WC = 300 N
Then
( 420 N )(1.8 m) - (300 N)( d ) - (320 N)(1.8 m) = 0
d = 0.6 m to the right of C 
(b)
320 N
For the resultant weight to act at C, SM C = 0 WC = 260 N
Then
( 420 N )(1.8 m) - (260 N)( d ) - (320 N)(1.8 m) = 0
d = 0.692 m to the right of C 
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114
PROBLEM 3.105
Three stage lights are mounted on a pipe as shown. The
lights at A and B each weigh 20.5 N, while the one at
C weighs 17.5 N. (a) If d 5 625 mm, 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
20.5 N
20.5 N
17.5 N
A
B
C
D
L
D
E
R
E
For equivalence
SFy : - 20.5 - 20.5 - 17.5 = - R or R = 58.5 N Ø
M D : - (250 mm)(20.5 N ) - (1100 mm)(20.5 N )
-[(1100 + d ) mm](17.5 N ) = -( L mm)(58.5 N )
46925 + 17.5d = 58.5 L ( d , L in mm)
or
d = 625 mm
(a)
We have
46925 + (17.5)(625) = 58.5L or
L = 989 mm
The resultant passes through a Point 989 mm to the right of D.

L = 1050 mm
(b)
We have
46925 + (17.5)d = (58.5)(1050) or d = 829 mm 
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115
PROBLEM 3.106
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
1300 N
400 N
A
For equivalence
or
400
L
600 N
a
N
b
B
A
B
a
SFy : -1300 + 400 - 400 - 600 = - R
b
aˆ
Ê
R = Á 2300 - 400 ˜ N Ë
b¯
SM A :
Then with
(1)
aÊ
aˆ
ÁË 400 ˜¯ - a( 400) - ( a + b)(600) = - LR
2
b
L=
or
1000 a + 600b - 200
2300 - 400
b = 1.5 m L =
a
b
a2
b
4 2
a
3 8
23 - a
3
10a + 9 -
where a, L are in m
(a)
R
(2)
Find value of a to maximize L
8 ˆÊ
8 ˆ Ê
4 ˆ Ê 8ˆ
Ê
10 - a˜ Á 23 - a˜ - Á10 a + 9 - a2 ˜ Á - ˜
dL ÁË
3 ¯Ë
3 ¯ Ë
3 ¯ Ë 3¯
=
2
da
8 ˆ
Ê
ÁË 23 - a˜¯
3
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116
PROBLEM 3.106 (Continued)
or
184
80
64
80
32
a - a + a2 + a + 24 - a2 = 0
3
3
9
3
9
2
16a - 276a + 1143 = 0
Then
a=
or
a = 10.3435 m and a = 6.9065 m
or
Since
(b)
Using Eq. (1)
and using Eq. (2)
230 -
276 ± ( -276)2 - 4(16)(1143)
2(16)
AB = 9 m, a must be less than 9 m
a = 6.91 m b
6.9065
R = 2300 - 400
or R = 458 N b
1.5
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 b
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117
PROBLEM 3.107
4
60°
3
1
15 m
2
27 m
30
O
m
33 m
3
60 m
45°
– 21.65j F – 20j
2
F1
15i
12.5i
27 m 30
15 m
O
33 m
m
30
30
m
m
21 m
4
– 25j
F3
60 m
30
30
m
m
17.68i
F4
17.68j
Four tugboats are used to bring an ocean liner to its pier. Each tugboat
exerts a 25 kN force in the direction shown. Determine (a) the equivalent
force-couple system at the foremast O, (b) the point on the hull where
a single, more powerful tugboat should push to produce the same effect
as the original four tugboats.
Solution
21 m
(a)Force-Couple System at O. Each of the given forces is resolved
into components in the diagram shown (kip units are used). The
force-couple system at O equivalent to the given system of forces
consists of a force R and a couple MOR defined as follows:
R = SF
MRO = – (1400.9 k)
= (12.5i - 21.65 j) + (15i - 20 j) + ( -25 j) + (17.68i - 17.68 j)
= (45.18
8i - 48.97) j
40.65i
O
47.3°
44.1j
M OR = S(r ¥ F)
= ( -27i + 15 j) ¥ (12.5i - 21.65 j)
+ (30i + 21j) ¥ (15i - 20 j)
+ (120i + 21j) ¥ ( -25 j)
+ (90i + 21j) ¥ (17.68i - 17.68 j)
= (584.55 - 187.5 - 600 - 315 - 3000 + 1591.2 + 371.28)k
= -1555.47k
R
R
44.1j
A
40.65i
o
O
x
21 m
The equivalent force-couple system at O is thus,
R = ( 45.2 kN )i - ( 49.0 kN ) j MOR - (1555 kN ◊ m)k
or
R = 666 kN
MOR = 1400.9 kN ◊ m
47.3°
b
(b) S
ingle Tugboat. The force exerted by a single tugboat must be
equal to R, and its point of application A must be such that the
moment of R about O is equal to MOR. Observing that the position
vector of A is
r = xi + 21j
we write
r ¥ R = MOR
( xi + 21j) ¥ ( 45.18i - 4847 j) = -1555.47k
- x( 48.97)k - 948.78k = -1555.47k
x = 12.39 mb
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118
PROBLEM 3.108
A couple of magnitude M = 7 N?m 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
SF : R = ( -50 j) + (150 cos 60∞)i
+ 150 sin 60∞ j + ( -225i )
= -(150 N )i + (79.9 N)j
or R = 170 N
(b)
28.0° b
First reduce the given forces and couple to an equivalent force-couple system ( R, M B ) at B.
We have
SM B : M B = (7000 N ◊ mm) + (300 mm)(50 N ) - (200 mm)(225 N )
= -23000 N ◊ mm
Then with R at D
SM B : - 23000 N ◊ m = a(79.9 N )
or
and with R at E
or
a = 288 m
SM B : - 23000 N ◊ m = c(150 N )
a
B
A
D
c
R
E
C
c = 153.3 mm
The line of action of R intersects line AB 288 mm to the left of B and intersects line BC 153.3 mm
below B.
b
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119
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, must have M iR = 0
(a)
+ M AB = SM A = M + (300 mm)[(150 N )sin 60∞] - (200 mm)(225 N ) = 0
M = +6028.9 N ◊ mm (b)
b
M = 30 N ◊ m
b
+ M BR = SM B = M + (300 mm)(50 N ) - (200 mm)(225 N ) = 0
M = +30000 N ◊ mm (c)
M = 6.03 N ◊ m
+ M CR = SM C = M + (300 mm)(50 N ) - (200 mm)[(150 N ) cos 60∞] = 0
M = 0
M=0 b
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120
PROBLEM 3.110
Four forces act on a 700 3 375-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)
B
A
R = SF
= ( -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
3
5
17 15
4
8
340 N
500 N
C
D
600 N
E
760 N
Ê 1200 N ˆ
tan q = Á Ë 1000 N ˜¯
= -1.20000
q = -50.194∞
(b)
R = 1562 N
50.2° 
MCR = Sr ¥ 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
– 1000 N
1200 N
y
C
x
Intersection 250 mm to right of C and 300 mm above C 
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121
PROBLEM 3.111
Solve Problem 3.110, assuming that the 760-N force is directed
to the right.
PROBLEM 3.110 Four forces act on a 700 3 375-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.
A
Solution
3
R = SF
= ( -400 N + 160 N + 760 N )i
+ (600 N + 300 N + 300 N ) j
= (520 N )i + (12000 N)j
(a)
R = (520 N )2 + (1200 N )2 = 1307.82 N
B
5
17 15
4
8
500N
340N
C
D
E
600N
760N
Ê 1200 N ˆ
tan q = Á
= 2.3077
Ë 520 N ˜¯
R = 1308 N
q = 66.5714∞
MCR = Sr ¥ F
(b)
= (0.5 m)i ¥ (300 N + 300 N ) j
= (300 N ◊ m)k
or
or
(300 N ◊ m)k = xi ¥ (1200 N ) j
x = 0.25000 m
x = 0.250 mm
(300 N ◊ m)k = [ x ¢i + (0.375 m) j] ¥ [(520 N )i + (1200 N ) j]
= (1200 x ¢ - 1995)k
x ¢ = 0.41250 m
x ¢ = 412.5 mm
x´
66.6° b
1200 N
B
A
0.375m
1200 N
C
520N
E
x
Intersection 412 mm to the right of A and 250 mm to the right of C b
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122
PROBLEM 3.112
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
y
1200 N
70˚
R = SF
R = (1200 N )(cos 70∞i - sin 70∞ j) - (800 N ) j
+ (1500 N )( - cos 40∞i - sin 40∞ j) - (900 N ) j
800 N
D
B
1500 N
F
40˚
G
A
C
E
x
900 N
R = -(738.64 N )i - (3791.8 N ) j
R = Rx2 + Ry2
= -(738.6)2 + (3791.8)2
= 3863 N
Ê Ry ˆ
q = tan -1 Á ˜
Ë Rx ¯
Ê -3791.8 ˆ
= tan -1 Á
Ë -738.6 ˜¯
= 78.977∞
or R = 3860 N
79.0∞ b
y
We have SM A = dRy
where
SM A = -[1200 N cos 70∞](1.8 m) - [1200 N sin 70∞](1.2 m)
- (800 N )(3.6 m) + [1500 N cos 40∞](1.8 m)
- [1500 N sin 40∞](6.0 m) - (900 N )(2.4 m)
= -10848.75 N ◊ m
-10848.75 N ◊ m
d=
-3791.8 N
= 2.861 m
F
D
B
G
A
d
RC
x
E
or d = 2.86 m to the right of A b
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123
PROBLEM 3.113
Pulleys A and B are mounted on bracket CDEF. The tension on
each side of the two belts is as shown. Replace the four forces
with a single equivalent force, and determine where its line of
action intersects the bottom edge of the bracket.
Solution
y
Equivalent force-couple at A due to belts on pulley A
We have
RA = 1400 N
RB = 1800 N
MB = 11250 N· mm
C
SF: - 600 N - 800 N = RA
A
B
F
MA = 10000 N· mm
D
x
E
R A = 1400 N Ø
We have
SM A : -200 N(50 mm) = M A
y
MF
M A = 10000 N ◊ mm
F
Equivalent force-couple at B due to belts on pulley B
We have
SF: (1050 N + 750 N )
25°
SM B : - 300 N(37.5 mm) = M B
E
x
y
D
C
M B = 11250 N ◊ mm
F
d
Equivalent force-couple at F
We have
RF = R
25∞ = R B
R B = 1800 N
We have
D
C
R
E
x
SF: R F = ( -1400 N ) j + (1800 N )(cos 25∞i + sin 25∞ j)
= (1631.35 N )i - (639.285 N ) j R = RF
2
2
= RFx
+ RFy
= (1631.35)2 + ( -639.285)2
= 1752.13 N
Ê RFy ˆ
q = tan -1 Á
Ë RFx ˜¯
Ê -639.285 ˆ
= tan -1 Á
Ë 1631.35 ¯˜
= -21.399∞
or R F = R = 1752 N
21.4° 
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124
PROBLEM 3.113 (Continued)
We have
SM F : M F = - (1400 N )(150 mm) - 10, 000 N ◊ mm
- [(1800 N ) cos 25∞](25 mm)
+ [(1800 N )sin 25∞](300 mm) - 11250 N ◊ mm
M F = - ( 43820 N ◊ mm)k
To determine where a single resultant force will intersect line FE,
M F = dRy
d=
MF
Ry
-43820
-639.285
= 68.545 mm
=
or d = 68.6 mm 
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125
Problem 3.114
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
120 N
200 N
80 N
42 N· m
40 N· m
J
Ry
P
MG
+
Rx
+F
R
b
G
I
+
a
+
H
First replace the applied forces and couples with an equivalent force-couple system at G.
SFx : 200 cos 15∞ - 120 cos 70∞ + P = Rx
Thus
Rx = (152.142 + P ) N
or
SFy : - 200 sin 15∞ - 120 sin 70∞ - 80 = Ry
Ry = -244.53 N
or
SM G : - (0.47 m)(200 N ) cos15∞ + (0.05 m)(200 N )sin 15∞
+ (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
Setting P = 0 in Eq. (1):
Now with R at I
SM G : - 55.544 N ◊ m = - a(244.53 N)
or
and with R at J
a = 0.227 m
SM G : - 55.544 N ◊ m = -b(152.142 N)
(1)
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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126
PROBLEM 3.115
Solve Problem 3.114, assuming that P = 60 N.
PROBLEM 3.115 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 3.115 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
SM G : -63.344 N ◊ m = - a(244.53 N )
a = 0.259 m
or
and with R at J
SM 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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127
PROBLEM 3.116
A 160 N 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
700 sin 30°
We have
y
SF : (300 N )i - (160 N ) j + (700 N )(cos 30∞i + sin 30∞ j) = R
700 cos 30°
50 mm
R = (906.22 N )i + (190 N ) j
160 N
50 mm
300 N
O
50 mm
x
or R = 926 N
We have
11.84° 
SM O : SM O = xRy
- [(700 N ) cos 30∞][(100 + 50 cos 30∞) mm] - [(700 N )sin 30∞]
[(50 mm)sin 30∞] - (300 N )(50 mm) = x(190 N )
1
( -86871 - 8750 - 15000) N ◊ mm
190
x = -582.21 mm
x=
and
Or, resultant intersects the base (x axis) 582 mm to the left of the vertical
centerline (y axis) of the motor.

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128
Problem 3.117
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 the 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
y
F
(a)
The slope of any tangent to the surface of member C is
dy d È Ê
x 2 ˆ ˘ -2b
=
Íb Á 1 - 2 ˜ ˙ = 2 x
dx dx ÍÎ Ë a ¯ ˙˚ a
A
2
C
Since the force F is perpendicular to the surface,
Ê dy ˆ
tana = - Á ˜
Ë dx ¯
-1
=
y = b 1 − x2
a
b
x
D
2
a Ê 1ˆ
Á ˜
2b Ë x ¯
a
For equivalence
a2
SF : F = R
SM D : ( F cos a )( y A ) = M D
where
cos a =
2bx
2bx
2 2
( a ) + (2bx )2
Ê
x2 ˆ
y A = b Á1 - 2 ˜
Ë a ¯
MD =
Ê
x3 ˆ
2 Fb2 Á x - 2 ˜
a ¯
Ë
a 4 + 4b 2 x 2
Therefore, the equivalent force-couple system at D is
R=F
M=
Ê a2 ˆ
tan -1 Á
˜ 
Ë 2bx ¯
Ê
x3 ˆ
2 Fb2 Á x - 2 ˜
a ¯ 
Ë
a 4 + 4b 2 x 2
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129
PROBLEM 3.117 (Continued)
(b)
dM
=0
dx
a = 1 m, b = 2 m
To maximize M, the value of x must satisfy
where, for
M=
8F ( x - x 3 )
1 + 16 x 2
dM
= 8F
dx
È1
˘
1 + 16 x 2 (1 - 3x 2 ) - ( x - x 3 ) Í (32 x )(1 + 16 x 2 ) -1/ 2 ˙
2
Î
˚ =0
(1 + 16 x 2 )
(1 + 16 x 2 )(1 - 3x 2 ) - 16 x( x - x 3 ) = 0
32 x 4 + 3x 2 - 1 = 0
or
-3 ± 9 - 4(32)( -1)
= 0.136011 m2
2(32)
Using the positive value of x2
x = 0.36880 m x2 =
and - 0.22976 m2
or x = 369 mm 
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130
Problem 3.118
Four forces are applied to the machine component
ABDE as shown. Replace these forces by 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
= 0.2 m
j
-120 N
- (50 N) j
0.2 m
A
- (300 N) i
z
B
0.16 m
D
x
0.1m
- (250 N) k
E
- (120 N) i
k
0
-300 N -50 N
i
+ 0.2 m
y
i
j
k
0 + 0.2 m 0 0.16 m
0
0
0 -250 N
j
k
-0.1 m 0.1
16 m
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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131
Problem 3.119
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 force-couple
system at A.
Solution
Equivalent force-couple at each pulley
Pulley B
Pulley C
MB B
B
z
z
20˚
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
145 N
R C = (155 N + 240 N )( - sin 10∞ j - cos10∞k )
RB
215 N
y
y
10˚
C
155 N
z
z
= - (68.591 N ) j - (389.00 N )k
MC = (240 N - 155 N )(0.075 m)i
MC
C
10°
RC
10˚ 240 N
= (6.3750 N ◊ m)i
Then
y
y
R = R B + R C = - ( 419.85 N ) j - (339.41)k or R = ( 420 N ) j - (339 N )k 
M A = M B + MC + rB/ A ¥ R B + rC/ A ¥ R C
i
j
k
= - (5.25 N ◊ m)i + (6.3750 N ◊ m)i + 0.225
0
0 N◊m
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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132
Problem 3.120
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 force-couple system at A
consisting of the force R = (13N)i + Ry j - (3.5N)k and the couple
M RA = M x i + (1 ◊ 5 N ◊ m)j - (1.08 N ◊ m)k. (b) Find the corresponding
values of Ry and M x .
Solution
(a)
From the statement of the problem, equivalence requires
ÂF: B + C = R
or
SFx : Bx + C x = 13 N (1)
SFy : - C y = Ry (2)
SFz : - C z = -3.5 N or C z = 3.5 N
and
 M A : (rB / A ¥ B + M B ) + rC / A ¥ C = M AR
or
 M x : (1500 N ◊ mm) + ( 45mm)(C y ) = M x (3)
SM y : (95 mm)( Bx ) + ( 45 mm)(C x ) + (85 mm)(3 ◊ 5 N ) = 1500 N ◊ mm
or
Using Eq. (1)
95 Bx + 45C x = 1202.5
95 Bx + 45(13 - Bx ) = 1202.5
or
Bx = 12.35 N
and
C x = 0.65 N
SM z : - (85 mm ) (C y ) = -1080 N ◊ mm
C y = 12.706 N
or
B = (12.35 N )i C = (0.65 N )i - (12.71 N ) j - (3.5 N )k 
(b)
Eq. (2) fi
Using Eq. (3)
Ry = -12.71 N 
1500 + ( 45)(12.706) = M x
5 2071.77 N . mm
or M x = 2.07 N ◊ m 
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133
Problem 3.121
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 = (40 N)i + (20 N)k and the couple MC = (45 N ◊ m)i,
determine the forces applied at A and at B when Az = 10 N.
Solution
We have
40 N
SF:
or
y
Fx : Ax + Bx = 40 N
Bx = -( Ax + 40 N ) 45000 N· mm
C
z
A+B=C
20 N
x
By
B
Ay
A
Az =10N
Bx
(1)
SFy : Ay + By = 0
Bz
Ay = - By or
(2)
SFz : 10 N + Bz = 20 N
Ax
Bz = 10 N or
(3)
SMC : rB /C ¥ B + rA/C ¥ A = MC
We have
i
200
j
0
k
i
-50 + 200
j
0
k
200 N ◊ mm = ( 45000 N ◊ mm)i
Bx
By
10
Ay
10
Ax
(50 By - 200 Ay )i + (50 Bx - 2000 + 200 Ax - 2000) j
or
+(200 By + 200 Ay )k = ( 45000 N ◊ mm)i
From
i-coefficient
50 By - 200 Ay = 45000 (4)
j-coefficient
50 Bx + 200 Ax = 4000 (5)
k-coefficient
200 By + 200 Ay = 0 (6)
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134
Problem 3.121 (Continued)
From Equations (2) and (4):
50 By - 200( - By ) = 45000
By = 180 N
From Equations (1) and (5):
Ay = -180 N
50( - Ax - 40) + 200 Ax = 4000
Ax = 40 N
From Equation (1):
Bx = -( 40 + 40) = -80 N
A = ( 40 N)i - (180 N ) j + (10 N )k 
B = -(80 N)i + (180 N ) j + (10 N )k 
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135
PROBLEM 3.122
As an adjustable brace BC is used to bring a wall into plumb, the ­forcecouple system shown is exerted on the wall. Replace this force-couple
system with an equivalent force-couple system at A if R = 106 N and
M = 20 N ◊ m.
Solution
y
We have SF : R = R A = Rl BC
where
l BC
B
R
(1050 mm)i - (2400 mm) j - ( 400 mm)k
=
2650 mm
RA =
106 N
(1050 i - 2400 j - 400 k )
2650
or
R A = ( 42 N )i - (96 N ) j - (16 N )k We have
SM A : rC /A ¥ R + M = M A
where
rC /A = (1050 mm)i + (1200 mm)k
M
A
rC/A
x
z
C

= 1.05 mj + 1.2 m k
R = ( 42 N )i - (96 N ) j - (16 N )k
M = - l BC M
-1050i + 2400 j + 400k
(20 N ◊ m)
2650
= -(7.92 N ◊ m)i + (18.113 N ◊ m) j + (3.019 N ◊ m)k
=
Then
i
j
k
1.05 0
1.2 N ◊ m + ( -7.924 i + 18.113 j + 3.019 k ) N ◊ m
42 -96 -16
M A = (107.28 N ◊ m i ) + (85.313 N ◊ mj) - (97.781 N ◊ m k )
or M A = (107.3 N ◊ m)i + (85.3 N ◊ m)j - (97.8 N ◊ m)k 
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136
Problem 3.123
A mechanic replaces a car’s exhaust system by firmly clamping the catalytic converter FG to its mounting
brackets H and I and then loosely assembling the mufflers and the exhaust pipes. To position the tailpipe AB,
he pushes in and up at A while pulling down at B. (a) Replace the given force system with an equivalent forcecouple system at D. (b) Determine whether pipe CD tends to rotate clockwise or counterclockwise relative to
muffler DE, as viewed by the mechanic.
Solution
(a)
Equivalence requires
and
where
SF : R = A + B
= (100 N )(cos 30∞ j - sin 30∞ k ) - (115 N ) j
= -(28.4 N ) j - (50 N )k
SM D : M D = rA/ D ¥ FA + rB / D ¥ FB
rA/D = -(0.48 m)i - (0.225 m) j + (1.12 m)k
rB /D = -(0.38 m)i + (0.82 m)k
Then
i
j
k
i
j
k
M D = 100 -0.48 -0.225
1.12
+ 115 -0.38 0 0.82
0 cos 30∞ - sin 30∞
0 -1
0
= 100[(0.225 sin 30∞ - 1.12 cos 30∞)i + ( -0.48 sin 30∞) j
+ ( -0.48 cos 30∞)k ] + 115 [(0.82)i + (0.38)k ]
= 8.56i - 24.0 j + 2.13k
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137
Problem 3.123 (Continued)
The equivalent force-couple system at D is
R = -(28.4 N ) j - (50.0 N)k M D = (8.56 N ◊ m)i - (24.0 N ◊ m) j + (2.13 N ◊ m)k (b)
Since ( M D ) z is positive, pipe CD will tend to rotate counterclockwise relative to muffler DE.



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138
Problem 3.124
For the exhaust system of Problem 3.123, (a) replace the given force system with an equivalent force-couple
system at F, where the exhaust pipe is connected to the catalytic converter, (b) determine whether pipe EF tends
to rotate clockwise or counterclockwise, as viewed by the mechanic.
Solution
(a)
Equivalence requires
and
where
SF : R = A + B
= (100 N )(cos 30∞ j - sin 30∞ k ) - (115 N ) j
= -(28.4 N ) j - (50 N )k
M F : M F = rA/ F ¥ A + rB / F ¥ B
rA/F = -(0.48 m)i - (0.345 m) j + (2.10 m)k
rB /F = -(0.38 m)i - (0.12 m) j + (1.80 m)k
Then
i
j
k
i
j
k
M F = 100 -0.48 -0.345
2.10
+ 115 -0.38 0.12 1.80
0 cos 30∞ - sin 30∞
0 -1
0
M F = 100[(0.345 sin 30∞ - 2.10 cos 30∞)i + ( -0.48 sin 30∞) j
+ ( -0.48 cos 30∞)k ] + 115 [(1.80)i + (0.38)k ]
= 42.4i - 24.0 j + 2.13k
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139
Problem 3.124 (Continued)
The equivalent force-couple system at F is
R = - (28.4 N ) j - (50 N)k M F = ( 42.4 N ◊ m)i - (24.0 N ◊ m) j + (2.13 N ◊ m)k (b)
Since ( M F ) z is positive, pipe EF will tend to rotate counterclockwise relative to the mechanic.



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140
Problem 3.125
The head-and-motor assembly of a radial drill press was originally
positioned with arm AB parallel to the z axis and the axis of the
chuck and bit parallel to the y axis. The assembly was then rotated
25∞ about the y axis and 20∞ about the centerline of the horizontal
arm AB, bringing it into the position shown. The drilling process
was started by switching on the motor and rotating the handle to
bring the bit into contact with the workpiece. Replace the force and
couple exerted by the drill press with an equivalent force-couple
system at the center O of the base of the vertical column.
Solution
We have
R = F = (55 N)[( sin 20∞ cos 25∞)]i - (cos 20∞) j - (sin 20∞ sin 25∞)k ]
= (17.0485 N )i - (51.683 N)j - (7.9499 N )k
or
We have
where
R = (17.05 N )i - (51.7) j - (7.95 N )k MO = rB /O ¥ F + MC
c
rB /O = [(350 mm)sin 25∞]i + (375 mm) j + [(350 mm) cos 25∞]k
= (147.9175 mm)i + (375 mm) j + (317.208 mm)k

y
z
Fz
Fx
20°
x
rB /O = [(0.1479 m)i + (0.375 m) j + (0.3172 m)k
MC = (11 N ◊ m)[(sin 20∞ cos 25∞)i - (cos 20∞) j - (sin 20∞ sin 25∞)k ]
Fy
F = 55 N
25°
= (3.4097 N ◊ m)i - (10.337 N ◊ m) j - (1.59 N ◊ m)k
i
j
Mc = 11N·m
k
MO = 0.1479 0.375
0.3172 N ◊ m + (3.0497i - 10..337j - 1.59) N ◊ m
17.048 -51.683 -7.9499
= (16.462 N ◊ m)i - (3.6165 N ◊ m) j - (15.627 N ◊ m)k
or MO = (16.46 N ◊ m)i - (3.62 N ◊ m) j - (15.63 N ◊ m)k 
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141
Problem 3.126
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
y
y
FA
G
B
A
R
FB
O
FC
C
z
O
G
E
x
z
xD
E
D
SF : FA + FB + FC = R
-(375 N ) j - (260 N ) j - ( 400 N ) j = R
-(1035 N ) j = R
R = 1035 N 
or
We have
x
F
F
We have
zD
SM 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)(zD )
z D = 3.0483 m We have
or
z D = 3.05 m 
SM z : FA ( x A ) + FB ( x B ) + 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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142
Problem 3.127
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
y
zD
G
z
y
FA
O
A
R
FB
FD
FC
D
xD
2.5 m
E
O
E
G
x
C
2.5 m
H
z
x
F
F
We have
SF : FA + FB + FC = R
-(375 N ) j - (260 N ) j - ( 400 N ) j - ( 425 N ) j = R
R = -(1460 N ) j
We have
SM 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 or
z D = 1.165 m 
SM z : FA ( x A ) + FB ( x B ) + 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 
We have
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143
Problem 3.128
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 = 0.3 m and b = 4 m.
Solution
We have
x
x
y
y
y
450 N
z
800 N
x
P
R
250 N
z
525 N
Assume that the resultant R is applied at Point P whose coordinates are (x, y, 0).
Equivalence then requires
SFz : - 525 - 450 - 800 - 250 = - R
or R = 2025 N 
SM x : (1.5 m)(525 N) - (0.3 m)( 450 N ) + (0.9 m)(800 N )
or
+ (1.65 m)(250 N ) = - y(2025 N )
y = -0.8815 m
SM y : (1.65 m)(525 N) + ( 4 m)( 450 N ) + ( 4.65 m)(800 N )
+ (7.05 m)(250 N ) = -x(2025 N )
or
x = 4.02 N ◊ m
R acts 4.02 N . m to the right of member AB and 0.881 m below member BC.

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144
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 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 SMG of the applied system of forces also be zero. Then
at
G : SM x : - ( a + 0.9) m ¥ ( 450 N ) + (0.6 m)(525 N )
+ (0.75 m)(250 N ) = 0
or a = 0.217 m 
SM y : - (3 m)(525 N ) - ( 4.65 - b) m ¥ ( 450 N )
+ (2.4 m)(250 N ) = 0
or b = 6.82 m 
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145
Problem 3.130*
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
y
224 N
zL
WL
176 N
A
z
xL
D
C
xR
zR
0.33 m
392 N
0.33 m
y
0.33 m
D
R
x
0.6 m
1.4 m
A
z 1m
B
1m
C
G
B
1.5 m
x
1.8 m
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 centerline 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
SF : - (224 N ) j - (392 N ) j - (176 N ) j = R
R = -(792 N ) j
We have
SM 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
SM 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 centerlines. Thus, SMG = 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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146
Problem 3.130* (continued)
xL = 0.33 m
With
at
G : SM Z : (1 - 0.33) m ¥ WL - (1.29101 - 1) m ¥ (792 N ) = 0
WL = 344.00 N
or
Now must check if this is physically possible,
at
G : SM 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 : SM 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 : SM z : (1 - X L ) m ¥ (358.42 N ) - (1.29101 - 1) m ¥ (792 N ) = 0
or
X L = 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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147
Problem 3.131*
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 3.130* 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 ( X R , 0, Z R )
Equivalence requires
y
SFy : - 224 - 392 - 176 = - R
or
or
or
R = 792 N Ø
SM x : (0.33 m)(224 N) + (0.6 m)(392 N)
+ (2 m)(176 N ) = z R (792 N )
zR = 0.83475 m
SM z : - (0.33 m)(224 N) - (1.67 m)(392 N)
- (1.67 m)(176 N ) = xR (792 N)
zL
zH
A
z
xH
D
xR
R
wH
C
G
cL
B
x
AXLE
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 centerlines. Thus,
SMG = 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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148
PROBLEM 3.131* (Continued)
Now consider these two possibilities
With xH = 0.6 m :
at
G : SM 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 : SM 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 : SM x : (2.7 - 1.5) 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 alongside AD, and the center 2.41 m from side DC.

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149
Problem 3.132
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.
y
Solution
F2 = P j
Force-couple system at O:
F1 = Pi
a
A
R = Pi + Pj + Pk = P(i + j + k )
MOR = aj ¥ Pi + ak ¥ Pj + ai ¥ Pk
= - Pak - Pai - Paj
MOR
a
x
O
F3= Pk
a
z
= - Pa(i + j + k )
Since R and MOR have the same direction, they form a wrench with M1 = MOR . Thus, the axis of the wrench is
the diagonal OA. We note that
R
y
M1 = MO
a
1
cos q x = cos q y = cos q z =
=
a 3
3
R
A
R = P 3 q x = q y = q z = 54.7∞
O
M1 = M OR = - Pa 3
Pitch = p =
(a)
R = P 3 q x = q y = q z = 54.7∞ - a
(b)
(c)
M1 - Pa 3
=
= -a
R
P 3
Axis of the wrench is diagonal OA
x
z



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150
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, MOR at the origin.
We have
F1
SF : - Pj + Pj + Pk = R
z
È
Ê5 ˆ ˘
S MO : - ( aP ) j + Í-( aP )i + Á aP ˜ k ˙ = M OR
Ë2 ¯ ˚
Î
F2
y
R
o
x
z
x
5 ˆ
Ê
MOR = aP Á -i - j + k ˜
Ë
2 ¯
or
(a)
R
MO
F3
R = Pk
or
y
Then for the wrench
R=P 
and
laxis
R
= =k
R
cos q x = 0 cos q y = 0 cos q z = 1
or
(b)
q x = 90∞ q y = 90∞ q z = 0∞ 
Now
M1 = laxis ◊ MOR
5 ˆ
Ê
= k ◊ aP Á -i - j + k ˜
Ë
2 ¯
=
Then
P=
5
aP
2
M1 52 aP
=
R
P
or
P=
5
a
2
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151
Problem 3.133* (Continued)
(c)The components of the wrench are ( R, M1 ), where M1 = M1 l axis , and the axis of the wrench is
­assumed to intersect the xy plane at Point Q whose coordinates are (x, y, 0). Thus require
M z = rQ ¥ R R
Then
y
M z = MO ¥ M1
Where
O
5 ˆ 5
Ê
aP Á -i - j + k ˜ - aPk = ( xi + yj) + Pk
Ë
2 ¯ 2
Equating coefficients
x
rQ
Q
y
R
z
M1
x
i : - aP = yP or y = - a
j: - aP = - xP or 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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152
Problem 3.134*
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
SF : - (20 N )i - (15 N ) j = R
R = 25 N
We have
SMO : S (rO ¥ F) + SMC = MOR
MOR = -20 N(0.1 m)j - ( 4 N ◊ m)i - (1 N ◊ m)j
= - ( 4 N ◊ m )i - ( 3 N ◊ m ) j
(a)
(b)
We have
Pitch
p=
M1 5 N ◊ m
=
= 0.200 m
R
25 N
y
R
4
O
3
x
M1
or p = 0.200 m 
(c)
R = -(20.0 N )i - (15.0 N)j 
R
M1 = l R ◊ MOR l =
R
= ( -0.8i - 0.6 j) ◊ [-( 4 N ◊ m)]i - (3 N ◊ m)j]
= 5 N◊m
From above note that
M1 = MOR
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
3
y = x
4
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153
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
z
y
y
o
o
y
z
M2
z
x
R
R
R
MO
o
Q
x
M1
z
x
R
M1
First, reduce the given force system to a force-couple at the origin.
We have
SF : - (50 N ) j - (55 N ) j = R
R = - (105 N ) j
We have
SMO : S (rO ¥ F) + SMC = MOR
MOR
i
j
k
i
j
k
= 0 0 500 N ◊ mm + 0 0 -375 N ◊ mm - (1.5 ¥ 1000 N ◊ mm) j
0 -50 0
0 -55
0
= (44375 N ◊ mm)i - (1500 N ◊ mm) j
R = - (105 N ) j (a)
(b)
We have
and pitch
or R = - (105 N ) j 
R
R
= ( - j) ◊ [( 4375 N ◊ mm)i - (1500 N ◊ mm) j]
= 1500 N ◊ mm and M1 = -(1500 N ◊ mm) j
M1 = l R ◊ MOR
p=
lR =
M1 1500 N ◊ mm
=
= 14.2857 mm R
105 N
or p = 14.29 mm 
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154
Problem 3.135* (continued)
(c)
We have
MOR = M1 + M 2
M 2 = MOR - M1 = ( 4375 N ◊ mm)i
Require
M 2 = rQ/O ¥ R
( 4375 N ◊ mm)i = ( xi + zk ) ¥ [-(105 N ) j]
4375i = -(105 x )k + (105 z)i
From i:
From k:
4375 = 105 z
z = 41.667 mm.
0 = - 105 x
x=0
The axis of the wrench is parallel to the y axis and intersects the xz plane at x = 0, z = 41.7 mm 
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155
PROBLEM 3.136*
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
y
y MR
O
O
R
M2
y
R
x
z
O
R
z
x
M1
z
o
z
M1
Q
x
x
First, reduce the given force system to a force-couple at the origin.
We have
SF : - (84 N ) j - (80 N )k = R
and
SMO : S (rO ¥ F) + SMC = MOR
R = 116 N
i
j k
i
j
k
0.6 0 0.1 + 0.4 0.3 0 + ( -30 j - 32k ) N ◊ m = MOR
0 84 0
0
0 80
MOR = - (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 = l R ◊ MOR
lR =
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 = M1l R = - ( 40.102 N ◊ m) j - (38.192 N ◊ m)k
p=
M1 55.379 N ◊ m
=
= 0.47741 m R
116 N
or p = 0.477 m 
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156
Problem 3.136* (Continued)
(c)
We have
MOR = M1 + M 2
M 2 = MOR - 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
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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157
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
y
y
x
z
B
R
MBR
x
z
z
B
x
Q
First, reduce the given force system to a force-couple at the origin at B.
(a)
We have
375 ˆ
Ê 200
SF : - (132 N )k - (85 N ) Á
i+
j˜ = R
Ë 425
425 ¯
R = - ( 40 N )i - (75 N ) j - (132 N )k 
and
R = 157 N
We have
SM B : rA/B ¥ FA + M A + M B = M RB
M RB
i
j
k
375 ˆ
Ê 200
0 - 27500k - 30000 Á
= 0 - 250
i+
j˜
Ë 425
425 ¯
0
0
- 132
= 330000i - 27500k - 14117.6i - 26470.6 j
M RB = (18882.4 N ◊ mm)i - (26470.6 N ◊ mm
m)j - (27500 N ◊ mm)k
= M RB = (18.88 N ◊ m)i - (26.5 N ◊ m)j - (27.5 N ◊ m)k
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158
Problem 3.137* (Continued)
(b)
M1 = l R ◊ MOR
and
M1 = M1l R = - (7.885 N ◊ m)i - (14.785 N ◊ m) j - (26.021 N ◊ m)k
Then pitch
(c)
R
R
-40i - 75 j - 132k
=
◊ [(18.88 N ◊ m)i - (26.5 N ◊ m) j - (27.5 N ◊ m)k ]
157
= -4.81 + 12.66 + 23.1 = 30.95 N ◊ m
We have
We have
p=
lR =
M1 30.95 N ◊ m
=
R
157 N
or p = 197.1 mm 
M RB = M1 + M 2
M 2 = M RB - M1 = (18.88i - 26.5 j - 27.5k ) - ( - 7.885i - 14.785 j - 26.021k )
Require
= (26.765 N ◊ m)i - (11.715 N ◊ m) j - (1.479 N.m)k
M 2 = rQ/B ¥ R
i
j
k
26.765i - 11.715 j - 1.479k = x
0
z
-40 -75 -132
= (75 z )i - ( 40 z ) j + (132 x ) j - (75 x )k
From i:
26.765 = 75 z
From k:
- 1.479 = - 75 x
z = 0.357 m = 357 mm
x = 0.01972 m = 19.72 mm
The axis of the wrench intersects the xz plane at
x = 19.72 mm
y = 0 z = 357 mm 
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159
Problem 3.138*
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, MOR 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
SF : R = TAC + TBD
= (900i + 300 j + 1350k )
+(1400i + 200 j + 500k )
= (2300 N )i + (500 N ) j + (1850 N )k
SMO : MOR = 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 + (10800 - 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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160
Problem 3.138* (Continued)
(b)
We have
R = 100 (23)2 + (5)2 + (18.5)2 = 2993.7 N
Let
laxis =
R
1
=
(23i + 5 j + 18.5k )
R 29.937
M1 = laxis ◊ MOR
Then
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
=
P=
Finally
(c)
M1 -601.26 N ◊ m
=
R
2993.7 N
or P = - 0.201 m 
We have
M1 = M1 laxis
or
1
(23i + 5 j + 18.5k )
29.937
M1 = -( 461.93 N ◊ m)i - (100.421 N ◊ m) j - (371.56 N ◊ m)k
Now
M 2 = MOR - M1
= ( -601.26 N ◊ m) ¥
= ( -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
y
y
M1
M1
M2
z
y
x
R
z
R
P
O
O
x
z
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161
Problem 3.138* (Continued)
Thus require
M 2 = rP ¥ R
r = ( yj + zk )
Substituting
i
j
k
-3138.1i + 6400.4 j + 2171.6k = 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 y = -0.944 m
z = 2.78 m
y = - 0.944 m
z = 2.78 m 
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162
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
y
MOR
Q
o
x
o
z
z
z
R
(a)
z
x
o
M1
y
y
M1
M2
x
x
R
R
First reduce the given force system to a force-couple at the origin.
We have
SF : P l BA + P l DC + P l DE = R
3 ˆ Ê 3 4 ˆ Ê -9
4
12 ˆ ˘
ÈÊ 4
R = P ÍÁ j - k ˜ + Á i - j˜ + Á i - j + k ˜ ˙
Ë
¯
Ë
¯
Ë
5
5
5
5
25
5
25 ¯ ˚
Î
R=
R=
We have
3P
(2i - 20 j - k ) 
25
3P
27 5
(2)2 + (20)2 + (1)2 =
P
25
25
SM : S (rO ¥ P ) = MOR
3P ˆ
4P
Ê 3P
Ê -4 P
(24a) j ¥ Á
jk ˜ + (20a) j ¥ Á
iË 5
Ë 5
5 ¯
5
MOR =
4P
12 P ˆ
ˆ
Ê -9 P
ij+
k ˜ = MOR
j˜ + (20a) j ¥ Á
¯
Ë 25
5
25 ¯
24 Pa
( -i - k )
5
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163
Problem 3.139* (Continued)
(b)
We have
M1 = l R ◊ MOR
where
lR =
Then
M1 =
and pitch
p=
R 3P
25
1
=
(2i - 20 j - k )
=
(2i - 20 j - k )
R 25
27 5P 9 5
1
9 5
(2i - 20 j - k ) ◊
M1 -8 Pa Ê 25 ˆ -8a
=
=
R 15 5 ÁË 27 5P ˜¯
81
M1 = M1l R =
(c)
24 Pa
-8Pa
( -i - k ) =
5
15 5
or p = - 0.0988a 
-8Pa Ê 1 ˆ
8Pa
˜¯ (2i - 20 j - k ) = 675 ( -2i + 20 j + k )
ÁË
15 5 9 5
Then
M 2 = MOR - M1 =
Require
M 2 = rQ/O ¥ R
24 Pa
8Pa
8 Pa
( -i - k ) ( -2i + 20 j + k ) =
( -430i - 20 j - 406k)
5
675
675
Ê 3P ˆ
Ê 8Pa ˆ
˜¯ ( -403i - 20 j - 406k ) = ( xi + zk ) ¥ ÁË ˜¯ (2i - 20 j - k )
ÁË
25
675
Ê 3P ˆ
= Á ˜ [20 zi + ( x + 2 z ) j - 20 xk ]
Ë 25 ¯
From i:
8( - 403)
Pa
Ê 3P ˆ
= 20 z Á ˜
Ë 25 ¯
675
From k:
8( -406)
Pa
Ê 3P ˆ
= -20 x Á ˜
Ë 25 ¯
675
z = -1.99012a
x = 2.0049a
The axis of the wrench intersects the xz plane at
x = 2.00a, z = -1.990a 
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164
Problem 3.140*
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
and
We have
SF : FA + FG = R
È (40 mm)i + (60 mm) j - (120 mm)k ˘
R = (50 N )k + 70 N Í
˙
140 mm
Î
˚
= (20 N )i + (30 N ) j - (10 N )k
R = 37.417 N
SMO : S (rO ¥ F) + SMC = MOR
MOR = [(0.12 m) j ¥ (50 N )k ] + {(0.16 m)i ¥ [(20 N )i + (30 N ) j - (60 N )k ]}
È (160 mm)i - (120 mm) j ˘
+ (10 N ◊ m) Í
˙
200 mm
Î
˚
È ( 40 mm)i - (120 mm) j + (60 mm)k ˘
+ (14 N ◊ m) Í
˙
140 mm
Î
˚
MOR = (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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165
Problem 3.140* (Continued)
Then for equivalence
y
y
R
z
R
O
z
Thus require
P
x
y
R
MO
O
x
z
MOR = 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
or z = -0.42 m
k : 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 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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166
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
dDA = -(300)2 + (225)2 + (200)2 = 425 mm
d ED = ( -150)2 + (0)2 + ( -200)2 = 250 mm
Then
170 N
= [( -300)i + (225) j + (200)k ]
425
= ( -120 N )i + (90 N ) j + (80 N )k
FDA =
and
150 N
= ( -150i - 200k )
250
= ( -90 N )i - (120 N )k
FED =
Then
SF : R = FDA + FED = -(210 N)i + (90 N)j - ( 40 N )k
For the applied couple
d AK = ( -150)2 + ( -150)2 + ( 450)2 = 150 11 mm
Then
M=
20000 N ◊ mm
( -150i - 150 j + 450k )
150 11
20000
=
( -i - j + 3k) N ◊ mm
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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167
Problem 3.141* (Continued)
Substituting
20000
?
( -i - j + 3k ) = 0
11
( -210i + 90 j - 40k ) ◊
?
20000
[( -210)( -1) + (90)( -1) + ( -40)(3)] = 0
11
?
0 =0
or
or
R and M are perpendicular so that the given system can be reduced to the single equivalent force
R = -(210 N )i + (90 N ) j - ( 40 N )k

Then for equivalence
y
y
D
O
P
x
=
O
x
y
M
z
z
R
R
z
M = rP /D ¥ R
Thus require
rP /D = -(300 mm)i + [( y - 75)mm]j + ( z mm)k
where
Substituting
i
j
k
20000
( -i - j + 3k ) = -300 ( y - 75)
z
11
-210
90
-40
= [( y - 75)( -40) - z (90)]i
+ [( z )( -210) - ( -300)( -40)]j
+ [( -300)(90) - ( y - 75)( -210)]k
Equating coefficients
j: k:
20000
= - 210 z - 12000
or z = -28.4 mm
11
60000
= -27000 + 210( y - 75) or y = 290 mm
11
The line of action of R intersects the yz plane at x = 0
y = 290 mm
z = -28.4 mm 
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168
Problem 3.142*
Replace the wrench shown with an equivalent system consisting of two
forces perpendicular to the y axis and applied respectively at A and B.
y
Solution
Bz
Express the forces at A and B as
A = Ax i + Az k
b
Az
B = Bx i + Bz k
a
Bx
Ax
O
Then, for equivalence to the given force system
x
z
SFx : Ax + Bx = 0 (1)
SFz : Az + Bz = R (2)
SM x : Az ( a) + Bz ( a + b) = 0 (3)
SM z : - Ax ( a) - Bx ( a + b) = M (4)
Bx = - Ax
From Equation (1),
Substitute into Equation (4)
From Equation (2),
and Equation (3),
- Ax ( a) + Ax ( a + b) = M
M
Ax =
and
b
Bz = R - Az
Bx = -
M
b
Az a + ( R - Az )( a + b) = 0
Ê aˆ
Az = R Á1 + ˜
Ë b¯
Ê aˆ
Bz = R - R Á1 + ˜
Ë b¯
a
Bz = - R
b
and
Then
Ê aˆ
ÊMˆ
A = Á ˜ i + R Á1 + ˜ k 
Ë b¯
Ë b¯
ÊMˆ Êa ˆ
B = - Á ˜ i - Á R˜ k 
Ë b ¯ Ëb ¯
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169
Problem 3.143*
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
y
y
R
o
P
rp
o
x
Given plane
(a)
z
A
M
Given point
B
b
z
P
x
(b)
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 xi + M y j + M zk
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
SFx : Rx = Ax + Bx (1)
SFy : Ry = Ay (2)
SFz : Rz = Az + Bz (3)
SM x : M x = yAz - zAy (4)
SM y : M y = zAx - xAz - bBz (5)
SM z : M z = xAy - yAx (6)
Based on the above six independent equations for the six unknowns ( Ax , Ay , Az , Bx , By , Bz , b), there exists a
unique solution for A and B.
Ay = Ry 
From Equation (2)
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170
Problem 3.143* (Continued)
Equation (6)
Ê 1ˆ
Ax = Á ˜ ( xRy - M z ) 
Ë y¯
Equation (1)
Ê 1ˆ
Bx = Rx - Á ˜ ( xRy - M z ) 
Ë y¯
Equation (4)
Ê 1ˆ
Az = Á ˜ ( M x + zRy ) 
Ë y¯
Equation (3)
Ê 1ˆ
Bz = Rz - Á ˜ ( M x + zRy ) 
Ë y¯
b=
Equation (5)
( xM x + yM y + zM z )
( M x - yRz + zRy )

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171
Problem 3.144*
Show that a wrench can be replaced with two perpendicular forces, one of which is applied at a given point.
Solution
y
y
M
R
a
a
P
P
x
Given point
o
B
z
x
z
z
A
(a)
(b)
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
SFx : 0 = Ax + Bx (1)
SFy : R = Ay + By (2)
SFz : 0 = Az + Bz (3)
SM x : 0 = - zBy (4)
SM y : M = - aAz - xBz + zBx (5)
SM z : 0 = aAy + zBy (6)
Since A and B are made perpendicular,
A ◊ B = 0 or
There are eight unknowns:
Ax Bx + Ay By + Az Bz = 0 (7)
Ax , Ay , Az , Bx , B y , 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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172
Problem 3.144* (Continued)
Since the components of A must be real, a nontrivial solution is not possible. Thus, it is required that By π 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)9
R = Ay + By
(2)
0 = Az + Bz
(3)
M = - aAz - xBz (5)9
0 = aAy + xBy
(6)
Ay By + Az Bz = 0 (7)9
Then Equation (2) can be written
Ay = R - By
Equation (3) can be written
Bz = - Az
aAy
x=By
Equation (6) can be written
Substituting into Equation (5)9,
or
Ê
R - By ˆ
M = - aAz - Á - a
˜ ( - Az )
By ¯
Ë
M
Az = By
aR
(8)
Substituting into Equation (7)9,
ˆ
ˆÊ M
Ê M
( R - By ) By + Á B
B =0
Ë aR y ˜¯ ÁË aR y ˜¯
or
By =
a 2 R3
a2 R2 + M 2
Then from Equations (2), (8), and (3)
Ay = R Az = Bz =
a2 R2
RM 2
=
a2 R2 + M 2 a2 R2 + M 2
M Ê a 2 R3 ˆ
aR2 M
=
Á
˜
aR Ë a2 R2 + M 2 ¯
a2 R2 + M 2
aR2 M
a2 R2 + M 2
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173
Problem 3.144* (Continued)
In summary
A=
RM
( Mj - aRk ) a R2 + M 2

B=
aR2
( 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 By  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.
y
A
B
P
O
x
x
a
z
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174
Problem 3.145*
Show that a wrench can be replaced with two forces, one of which has a prescribed line of action.
Solution
y
y
M
R
a
A′
z
A
x
O
O
z
A
A′
a
B
x
z
x
P
A
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 (AA9). 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 AA9 by
l A = lx i + l y j + lz k
it follows that force A can be expressed as
A = Al A = A( l x i + l y j + l z k )
Force B can be expressed as
B = Bx i + By j + Bz k
Next, observe that since the axis of the wrench and the prescribed line of action AA9 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
SFx : 0 = Al x + Bx (1)
SFy : R = Al y + By (2)
SFz : 0 = Al z + Bz (3)
SM x : 0 = - zBy (4)
SM y : M = - aAl z + zBx - xBz (5)
SM x : 0 = - aAl y + xBy (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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175
Problem 3.145* (Continued)
Case 1: Let z = 0 to satisfy Equation (4)
Now Equation (2)
Equation (3)
Equation (6)
Substitution into Equation (5)
Al y = R - By
Bz = - Al z
x=-
aAl y
By
Ê aˆ
= - Á ˜ ( R - By )
Ë By ¯
È Ê aˆ
˘
M = - aAl z - Í- Á ˜ ( R - By )( - Al z )˙
ÍÎ Ë By ¯
˙˚
1 ÊMˆ
A = - Á ˜ By
l z Ë aR ¯
Substitution into Equation (2)
R=-
1 ÊMˆ
Á ˜ By l y + By
l z Ë aR ¯
l z aR2
l z aR - l y M
MR
R
A= =
l z aR - l y M l - aR l
y
z
M
l x MR
Bx = - Al x =
aR
lz - l y M
By =
Then
Bz = - Al z =
l z MR
l z aR - l y M
A=
In summary
B=
and
Ê
Rˆ
x = a Á1 ˜
By ¯
Ë
P
lA 
aR
ly lz
M
R
( l x Mi + l z aRj + l z M k ) 
l z aR - l y M
È
Ê l z aR - l y M ˆ ˘
= a Í1 - R Á
˜˙
Ë l z aR2 ¯ ˙˚
ÍÎ
or x =
ly M
lz R

Note that for this case, the lines of action of both A and B intersect the x axis.
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176
Problem 3.145* (Continued)
Case 2: Let By = 0 to satisfy Equation (4)
R
Now Equation (2)
A=
ly
Equation (1)
Êl ˆ
Bx = - R Á x ˜
Ë ly ¯
Equation (3)
Êl ˆ
Bz = - R Á z ˜
Ë ly ¯
Equation (6)
aAl y = 0
which requires a = 0
Substitution into Equation (5)
È Ê l ˆ˘
È Ê l ˆ˘
ÊMˆ
M = z Í- R Á x ˜ ˙ - x Í- R Á z ˜ ˙ or l z x - l x z = Á ˜ l y
Ë
R¯
ÍÎ Ë l y ¯ ˙˚
ÍÎ Ë l y ¯ ˙˚
This last expression is the equation for the line of action of force B.
In summary
Ê Rˆ
A = Á ˜ lA Ë ly ¯

Ê Rˆ
B = Á ˜ (-lx i - lx k ) Ë ly ¯

Assuming that l x , l y , l z  0, the equivalent force system is as shown below.
A
y
M λy
R λx
B
x
z
M λy
R λz
Note that the component of A in the xz plane is parallel to B.
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177
Problem 3.146
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
0.25 m
2
W = mg = 80 kg(9.81 m/s ) = 784.8 N
SM E : M E = (784.8 N )(0.25 m)
W
B
A
C
D
E
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
and
A
SM E : 196.2 N ◊ m = FB (0.98615 m)
FB = 198.954 N
-1 Ê 0.85
mˆ
q = tan Á
= 59.534∞
Ë 0.5 m ˜¯
q
B
d
0.5 m
C
D
E
0.85 m
FB = 199.0 N
or
FB
59.5° 
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178
Problem 3.147
It is known that the connecting rod AB exerts on the crank BC a 1.5-kN 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
FAB
24
(FAB)x
FAB
A
ˆ
ˆ
Ê 24
Ê 7
= (0.028 m) Á ¥ 1500 N ˜ + (0.021 m) Á ¥ 1500 N ˜
¯
¯
Ë 25
Ë 25
7
y2
(FAB)y
= 42 N ◊ m
B
B
y1
c
x1
(a)
c
(b)
or
MC = 42.0 N ◊ m

or
MC = 42.0 N ◊ m

Using (b)
M C = y2 ( FAB ) x
ˆ
Ê 7
= (0.1 m) Á ¥ 1500 N ˜ = 42 N ◊ m
¯
Ë 25
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179
Problem 3.148
A 1.8 m-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 30 N. Determine the moment about A of the force exerted by the line at B.
Solution
We have
Then
y
Tz
B TBC = 30 N
Ty
Txz
A
z
Tx
Txz = (30 N ) cos 8∞ = 29.708 N
Tx = Txz sin 30∞ = 14.854 N
Ty = -TBC sin 8∞ = - 4.1752 N
Now
M A = rB /A ¥ TBC
where
rB /A = (1.8 sin 45∞) j - (1.8 cos 45∞)k
1.8 m
2
Then
1.8
MA =
2
=
or
C
x
Tz = -Txz cos 30∞ = -25.728 N
=
8˚
( j - k)
i
j
k
0
1
-1
14.854 -4.1752 -25.728
1.8
1.8
1.8
( -25.728 - 4.1752)i (14.854) j (14.854)k
2
2
2
M A = -(38.1 N ◊ m)i - (18.91 N ◊ m) j - (18.91 N ◊ m)k 
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180
Problem 3.149
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
(a)
TBA
( -3i + 3j - 1.5k )
4.5
T
= BA ( -2i + 2 j - k )
3
BD
1
l BD =
( -0.08i + 0.38 j + 0.16k )
=
BD 0.42
1
= ( -4i + 19 j + 8k )
21
TBA ◊ l BD = TBA cos q
TBA =
Then
We have
or
TBA
1
( -2i + 2 j - k ) - ( -4i + 19 j + 8k ) = TBA cos q
3
21
or
cos q =
1
[( -2)( -4) + (2)(19) + ( -1)(8)]
63
= 0.60317
or (b)
We have
(TBA ) BD = TBA ◊ l BD
= TBA cos q
= (540 N )(0.60317)
or
q = 52.9∞ 
(TBA ) BD = 326 N 
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181
Problem 3.150
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 7092 N.m, determine the magnitude
of TDE when TAB 5 1.275 kN.
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 nor y components produce a
moment about the x axis.
We have
where
y
SM x : (TAB ) z ( y A ) + (TDE ) z ( yD ) = M x
A
(TAB ) z = k ◊ TAB
= k ◊ (TAB l AB ) - 0.3i
TAB
È
Ê - 0.3i - 3.6 j + 4k ˆ ˘
= k ◊ Í1275 N Á
˜¯ ˙
Ë
5.390
Î
˚
= 946.2 N
D
TDE
B
O
C
E
x
z
F
(TDE ) z = k ◊ TDE
= k ◊ (TDE l DE )
È
Ê 0.45i - 4.2 j + 4k ˆ ˘
= k ◊ ÍTDE Á
˜¯ ˙
Ë
5.817
Î
˚
= 0.6876 TDE
y A = 3.6 m
yD = 4.2 m
M x = 7092 N ◊ m
and
(946.2 N )(3.6 m) + (0.6876 TDE )( 4.2 m) = 7092 N ◊ m
TDE = 1276.24 N or
TDE = 1.276 kN 
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182
Problem 3.151
Solve Problem 3.150 when the tension in cable AB is 1.53 kN.
PROBLEM 3.150 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 7092 N·m,
determine the magnitude of TDE when TAB 5 1.275 kN
y
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
SM x : (TAB ) z ( y A ) + (TDE ) z ( yD ) = M x
where
(TAB ) z = k ◊ TAB
= k ◊ (TAB l AB )
A
TAB
D
TDE
B
O
C
E
x
z
F
È
Ê -0.3 i - 3.6 j + 4k ˆ ˘
= k ◊ Í1530 N Á
˜¯ ˙
Ë
5.390
Î
˚
= 1135.44 N
(TDE ) z = k ◊ TDE
= k ◊ (TDE l DE )
È
Ê 0.45i - 4.2 j + 4k ˆ ˘
= k ◊ ÍTDE Á
˜¯ ˙
Ë
5.817
Î
˚
= 0.6876 TDE
y A = 3.6 m
yD = 4.2 m
M x = 7092 N ◊ m
(1135.44 N )(3.6 m) + (0.6876 TDE )( 4.2 m) = 7092 N ◊ m
and
TDE = 1040.34 N or TDE = 1.040 kN 
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183
Problem 3.152
A wiring harness is made by routing either two or
three wires around 50 mm-diameter pegs mounted on
a sheet of plywood. If the force in each wire is 15 N,
determine the resultant couple acting on the plywood
when a 5 450 mm and (a) only wires AB and CD are
in place, (b) all three wires are in place.
Solution
In general, M = SdF, where d is the perpendicular distance between the lines of action of the two forces acting
on a given wire.
(a)
25 mm
25 mm
FAB
FAB
600 mm
600 mm
FAB
FAB25 mm
25 mm
We have
25 mm
25 mm
18
24 18
4
24 3
4
3
FCD
450+250 mm
FCD
450+250 mm
FCD
FCD
25 mm
25 mm
M = d AB FAB + dCD FCD
4
ˆ
Ê
= (50 + 600) mm ¥ 15 N + Á 50 + ¥ 700˜ mm ¥ 15 N
¯
Ë
5
(b)
25 mm
25 mm
or M = 18900 N ◊ mm 
FEF
(450+250)
mm
FEF
(450+250) mm
25 mm
25 mm
FEF
FEF
We have
M = [d AB FAB + dCD FCD ] + dEF FEF
= 18900 N ◊ mm - 700 mm ¥ 15 N or M = 8.400 N ◊ mm 
= 8.40 N . m
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184
Problem 3.153
A worker tries to move a rock by applying a 360-N force to a
steel bar as shown. (a) Replace that force with an equivalent
force-couple system at D. (b) Two workers attempt to move
the same rock by applying a vertical force at A and another
force at D. Determine these two forces if they are to be
equivalent to the single force of Part a.
Solution
(b)
(a)
40°
360 N
F
A
A
B
0.65 m
(a)
FA
A
FO
D
q
M
1.05 m
D
30°
30°
D
30°
We have SF : 360 N(- sin 40∞i - cos 40∞ j) = -(231.40 N )i - (275.78 N ) j = F
or F = 360 N
We have
SM D : rB /D ¥ R = M
where
rB /D = -[(0.65 m) cos 30∞]i + [(0.65 m)sin 30∞]j
50° 
= -(0.56292 m)i + (0.325500 m) j
i
j
k
M = -0.56292 0.32500 0 N ◊ m
-231.40 -275.78 0
= [155.240 + 75.206) N ◊ m]k
= (230.45 N ◊ m)k (b)
We have
where
or M = 230 N ◊ m 
SM D : M = rA/D ¥ FA
rB /D = -[(1.05 m) cos 30∞]i + [(1.05 m)sin 30∞]j
= -(0.90933 m)i + (0.525500 m) j
i
j
k
FA = -0.90933 0.52500 0 N ◊ m
0
-1
0
= [230.45 N ◊ m]k
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185
Problem 3.153 (Continued)
or
(0.90933FA )k = 230.45k
FA = 253.42 N We have
or FA = 253 N Ø 
SF : F = FA + FD
-(231.40 N )i - (275.78 N ) j = -(253.42 N ) j + FD ( - cos q i - sin q j)
From
i: 231.40 N = FD cos q (1)
j: 22.36 N = FD sin q (2)
Equation (2) divided by Equation (1)
tan q = 0.096629
q = 5.5193∞ or q = 5.52∞
Substitution into Equation (1)
231.40
FD =
= 232.48 N cos 5.5193∞
or FD = 232 N
5.52°
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186
PROBLEM 3.154
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 force-couple
system at the origin O of the coordinate system.
Solution
y
110 N
15°
35°
we have
SF : PB = F
PB = 110 N[- (sin 15∞) j + (cos15∞)k ]
= -(28.470 N ) j + (106.252 N )k
where
B
rB/O
o
We have
SM O : rB /O ¥ PB = MO
where
rB /O = [(0.22 cos 35∞)i + (0.15) j - (0.22 sin 35∞)k ]m
x
z
or F = -(28.5 N ) j + (106.3 N )k 
150 mm
= (0.180213 m)i + (0.15 m) j - (0.126187 m)k
y
i
j
k
0.180213 0.15 0.126187 N ◊ m = MO
0
-28.5
106.3
MO = [(12.3487)i - (19.1566) j - (5.1361)k ]N ◊ m
o
15°
or MO = (12.35 N ◊ m)i - (19.16 N ◊ m)j - (5.13 N ◊ m)k 
x
z
F
MO
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187
Problem 3.155
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
a = 30∞, (b) the value of a so that the single equivalent force is
applied at Point B.
Solution
We have
800 N
500 N
A
2000 N
A
d
R
θ B
450 N
(a)
For equivalence
SFx : -500 cos 30∞ + 2000 cos 65∞ + 450 cos 65∞ = Rx
or
Rx = 602.4 N
SFy : 500 sin a + 800 + 2000 sin 65∞ + 450 sin 65∞ = Ry
or
Ry = (3020.45 + 500 sin a ) N With a = 30∞
Ry = 3270.45 N
Then
R = (602.4)2 + (3270.45)2
= 3325.5 N
(1)
3270.45
602.4
or q = 79.6∞
tan q =
Also
SM A : (1.150 m)(800 N ) + (1.650 m)(2000 N ) sin 65∞ + (0.650 m)(2000 N)coos 65∞
+ (1.650 m)( 450 N )sin 65∞ + (0.9 m) (450 N )cos 65∞ = d (3270.45)
or
SM A = 53043 N ◊ m and d = 1.622 m and R is applied 1.622 m. To the right of A.
(b)
R = 3330 N
79.6° 

We have d = 1650 mm
Then
or
Using Eq. (1)
SM A : 5304.3 N ◊ m = (1.65 m)( Ry )
Ry = 3214.7 N
3214.7 = 3020.45 + 500 sin a or a = 22.9∞ 
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188
Problem 3.156
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)
SF : R = B + C
Equivalence requires
-(30 N )i + Ry j + Rz k = - Bk + ( -C x i + C y j + C z k )
or
Equating the i coefficients
i : - 30 N = -C x
Also
SM A : M RA = rB /A ¥ B + rC /A ¥ C
or
or C x = 30 N
-(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 ]
Equating coefficients
i : - 12 N ◊ m = -(0.15 m) B
k : 0 = (0.4 m)C y
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)
or B = 80 N
or C y = 0
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: Ry = 0 k : Rz = -80 + 40 Ry = 0 
or
Rz = -40.0 N 
(c)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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189
Problem 3.157
A concrete foundation mat in the shape of a regular hexagon of
side 12 m 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
SMO = 0 or SM x = 0 SM z = 0
For the applied loads.
2m
E
F
2√3 m
A
O
B
z
D
x
C
4m
Then
SM x = 0 : (2 3 m) FB + (2 3 m)(50 kN ) - (2 3 m)(100 kN )
- (2 3 m) FF = 0
or
FB - FF = 50 (1)
SM z = 0 : ( 4 m)(75 kN ) + (2 m) FB - (2 m)(50 kN )
- ( 4 m)(150 kN ) - (2 m)(100 kN) + (2 m) FF = 0
FB + FF = 300 or
(2)
Then (1) + (2) fi FB = 175 kN Ø 
and
FF = 125 kN Ø 
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190