CHAPTER 4
PROBLEM 4.1
A gardener uses a 60-N wheelbarrow to transport a 250-N bag
of fertilizer. What force must she exert on each handle?
SOLUTION
Free-Body Diagram:

M A  0: (2 F )(1 m)  (60 N)(0.15 m)  (250 N)(0.3 m)  0
F  42.0 N 

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PROBLEM 4.2
The gardener of Problem 4.1 wishes to transport a second
250-N bag of fertilizer at the same time as the first one.
Determine the maximum allowable horizontal distance from
the axle A of the wheelbarrow to the center of gravity of the
second bag if she can hold only 75 N with each arm.
PROBLEM 4.1 A gardener uses a 60-N wheelbarrow to
transport a 250-N bag of fertilizer. What force must she exert on
each handle?
SOLUTION
Free-Body Diagram:

MA  0: 2(75 N)(1 m)  (60 N)(0.15 m)
 (250 N)(0.3 m)  (250 N) x  0
x  0.264 m 


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PROBLEM 4.3
An 8.4-kN tractor is used to lift 3.6 kN of gravel. Determine
the reaction at each of the two (a) rear wheels A, (b) front
wheels B.
SOLUTION
1.25 m
(a)
Rear wheels
ΣM B = 0: +(8.4 kN)(1 m) − (3.6 kN)(1.25 m) − 2A(1.5 m) = 0
A = +1.3 kN
(b)
Front wheels
ΣM A : −(8.4 kN)(0.5 m) − (3.6 kN)(2.75 m) + 2B(1.5 m) = 0
B = +4.7 kN
Check:
A = 1.3 kN
+ΣFy = 0:
B = 4.7 kN
2A + 2B − 8.4 kN − 3.6 kN = 0
2(1.3 kN) + 2(4.7 kN) − 8.4 kN − 3.6 kN = 0
0 = 0 (Checks)
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60 N
80 N
80 N
140 N
60 N
A
B
PROBLEM 4.4
For the beam and loading shown, determine (a) the reaction at A,
(b) the tension in cable BC.
C
0.15 m
0.2 m
0.2 m
0.15 m
SOLUTION
Free-Body Diagram:
(a)
Reaction at A:
ΣFx = 0: Ax = 0
ΣMB = 0: (60 N)(0.7 m) + (80 N)(0.55 m) + (140 N)(0.35 m)
+(80 N)(0.15 m) − Ay (0.15 m) = 0
Ay = +980 N
(b)
Tension in BC
A = 980 N
ΣM A = 0: (60 N)(0.55 m) + (80 N)(0.4 m) + (140 N)(0.2 m)
−(60 N)(0.15 m) − FBC (0.15 m) = 0
FBC = +560 N
Check:
FBC = 560 N
ΣFy = 0: −60 N − 80 N − 140 N − 80 N + A −60 N − FBC = 0
−420 + 980 N − 560 N = 0
0 = 0 (Checks)
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PROBLEM 4.5
A load of lumber weighing W  25 kN is being raised by a
mobile crane. The weight of the boom ABC and the
combined weight of the truck and driver are as shown.
Determine the reaction at each of the two (a) front wheels
H, (b) rear wheels K.
SOLUTION
Free-Body Diagram:
(a)
 M K  0:
 25 kN  5.4 m  +  3 kN  3.4 m   2FH  2.5 m  +  50 kN  0.5 m   0
2FH  68.080 kN
(b)
 M H  0:
or FH  34.0 kN

 25 kN  2.9 m  +  3 kN  0.9 m    50 kN  2.0 m  + 2FK  2.5 m   0
2 FK  9.9200 kN
or FK  4.96 kN
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
PROBLEM 4.6
A load of lumber weighing W  25 kN is being raised as
shown by a mobile crane. Knowing that the tension is 25
kN in all portions of cable AEF and that the weight of boom
ABC is 3 kN, determine (a) the tension in rod CD, (b) the
reaction at pin B.
SOLUTION
Free-Body Diagram: (boom)
(a)
 M B  0:
 25 kN  2.6 m  +  3 kN  0.6 m    25 kN  0.4 m   TCD  0.7 m   0
TCD  81.143 kN
(b)
Fx  0:
Bx  0 so that B  B y
Fy  0:
 25  3  25  81.143 kN  B
B  134.143 kN
or TCD  81.1 kN 
0
or B  134.1 kN
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
PROBLEM 4. 7
B
A T-shaped bracket supports the four loads shown. Determine the
reactions at A and B (a) if a = 10 cm, (b) if a = 7 cm.
A
40 N
6 cm 6 cm
50 N
a
30 N
10 N
8 cm
SOLUTION
Free-Body Diagram:
ΣFx = 0: Bx = 0
ΣM B = 0: (40 N)(6 cm) − (30 N)a − (10 N)(a + 8 cm) + (12 cm)A = 0
A=
(40a − 160)
12
(1)
ΣM A = 0: −(40 N)(6 cm) − (50 N)(12 cm) − (30 N)(a + 12 cm)
−(10 N)(a + 20 cm) + (12 cm)By = 0
By =
Bx = 0 B =
Since
(a)
(b)
(1400 + 40a)
12
(1400 + 40a)
12
(2)
For a = 10 cm
Eq. (1):
A=
(40 × 10 − 160)
= +20.0 N
12
Eq. (2):
B=
(1400 + 40 × 10)
= +150.0 N
12
B = 150.0 N
Eq. (1):
A=
(40 × 7 − 160)
= +10.00 N
12
A = 10.00 N
Eq. (2):
B=
(1400 + 40 × 7)
= +140.0 N
12
B = 140.0 N
A = 20.0 N
For a = 7 cm
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PROBLEM 4.8
B
For the bracke t and loading of Proble m 4. 7, de te rmine the smalle st
distance a if the bracket is not to move.
A
40 N
6 cm 6 cm
50 N
a
30 N
8 cm
10 N
PROBLEM 4.7 A T-shaped bracket supports the four loads shown.
Determine the reactions at A and B (a) if a = 10 cm, (b) if a = 7 cm.
SOLUTION
Free-Body Diagram:
For no motion, reaction at A must be downward or zero; smallest distance a for no motion
corresponds to A = 0.
ΣM B = 0: (40 N)(6 cm) − (30 N)a − (10 N)(a + 8 cm) + (12 cm)A = 0
A=
(40a − 160)
12
A = 0: (40a − 160) = 0
a = 40 mm
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PROBLEM 4.9
Three loads are applied as shown to a light beam supported by
cables attached at B and D. Neglecting the weight of the beam,
determine the range of values of Q for which neither cable becomes
slack when P  0.
SOLUTION
Free-Body Diagram:
For Qmin , TD  0
 M B  0:
 7.5 kN  0.5 m   Qmin  3 m   0
Qmin  1.250 kN
For Qmax , TB  0
 M D  0:
 7.5 kN  2.75 m   Qmax  0.75 m   0
Qmax  27.5 kN
Therefore:
1.250 kN  Q  27.5 kN 
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PROBLEM 4.10
Three loads are applied as shown to a light beam supported by
cables attached at B and D. Knowing that the maximum allowable
tension in each cable is 12 kN and neglecting the weight of the
beam, determine the range of values of Q for which the loading is
safe when P  0.
SOLUTION
Free-Body Diagram:
 M D  0:
 7.5 kN  2.75 m   TB  2.25 m 
Q  27.5  3TB
 M B  0:
 7.5 kN  0.5 m 
 Q  0.75 m   0
(1)
+ TD  2.25 m   Q  3 m   0
Q  1.25   0.75 TD 
(2)
For the loading to be safe, cables must not be slack and tension must not exceed 12 kN.
Thus, making 0  TB
 12 kN in (1), we have
-8.50 kN
And making 0  TD
1.25
 Q  27.5 kN
(3)
 12 kN in (2), we have
 Q  10.25 kN
(3) and (4) now give:
(4)
1.250 kN
 Q  10.25 kN 
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PROBLEM 4.11
For the beam of Prob. 4.10, determine the range of values of Q for
which the loading is safe when P = 5 kN.
SOLUTION
Free-Body Diagram:
 7.5 kN  2.75 m   TB  2.25 m  +  5 kN 1.5 m   Q  0.75 m   0
(1)
Q   37.5  3 TB  kN
 7.5 kN  0.5 m    5 kN  0.75 m  + TD  2.25 m   Q  3 m   0
(2)
Q   0.75 TD  kN
 M D  0:
 M B  0:
For the loading to be safe, cables must not be slack and tension must not exceed 12 kN.
Thus, making 0  TB
 12 kN in (1), we have
1.500 kN
And making 0  TD
0
 Q  37.5 kN
(3)
 12 kN in (2), we have
 Q  9.00 kN
(3) and (4) now give:
(4)
1.500 kN
 Q  9.00 kN 
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PROBLEM 4.12
For the beam of Sample Prob. 4.2, determine the range of values
of P for which the beam will be safe, knowing that the maximum
allowable value of each of the reactions is 125 kN and that the
reaction at A must be directed upward.
SOLUTION
Free-Body Diagram:
 P(1.5 m)  By (4.5 m)  (30 kN)(5.5 m)  (30 kN)(6.5 m) = 0
M A  0:
P
 3B y  240 kN
(1)
 Ay  4.5 m  P  3 m   (30 kN)(1 m)  (30 kN)(2 m) = 0
M B  0:
P
1.5Ay +30 kN
(2)
For the loading to meet the design criteria, the reactions must not exceed 125 kN.
Thus, making B y  125 kN in (1), we have
P  135 kN
And making 0  Ay
30
(3)
 12̀5 kips in (2), we have
 P  217.5 kN
(3) and (4) now give:
(4)
30 kN
 P  217.5 kN 
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PROBLEM 4.13
The maximum allowable value of each of the reactions is 180 N.
Neglecting the weight of the beam, determine the range of the
distance d for which the beam is safe.
SOLUTION
Fx  0: Bx  0
B  By
M A  0: (50 N)d  (100 N)(0.45 m  d )  (150 N)(0.9 m  d )  B (0.9 m  d )  0
50d  45  100d  135  150d  0.9B  Bd  0
d
180 N  m  (0.9 m) B
300 A  B
(1)
M B  0: (50 N)(0.9 m)  A(0.9 m  d )  (100 N)(0.45 m)  0
45  0.9 A  Ad  45  0
(0.9 m) A  90 N  m
A
(2)
d
180  (0.9)180 18

 0.15 m
300  180
120
d  150.0 mm 
d
(0.9)180  90 72

 0.40 m
180
180
d
Since B  180 N, Eq. (1) yields
Since A  180 N, Eq. (2) yields
Range:
d  400 mm 
150.0 mm  d  400 mm 
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300 N
a
A
300 N
PROBLEM 4.14
0.15 m
D
For the beam and loading shown, determine the range of the
distance a for which the reaction at B does not exceed 100 N
downward or 200 N upward.
B
C
50 N
0.2 m
0.1 m
0.3 m
SOLUTION
Assume B is positive when directed
Sketch showing distance from D to forces.
ΣM D = 0: (300 N)(0.2 m − a) − (300 N)(a − 0.05 m) − (50 N)(0.1 m) + 0.4B = 0
−600a + 70 + 0.4B = 0
a=
(70 + 0.4B)
600
(1)
For B = 100 N = −100 N, Eq. (1) yields:
a
[70 + 0.4(−100)]
30
=
= 0.05 m
600
600
a
50 mm
For B = 200 N = +200 N, Eq. (1) yields:
a
Required range:
[70 + 0.4(200)]
150
=
= 0.25 m
600
600
0.05 m
a
a
250 mm
0.25 m
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PROBLEM 4.15
Two links AB and DE are connected by a bell crank as
shown. Knowing that the tension in link AB is 720 N,
determine (a) the tension in link DE, (b) the reaction
at C.
SOLUTION
Free-Body Diagram:
M C  0: FAB (100 mm)  FDE (120 mm)  0
FDE 
(a)
For
(1)
FAB  720 N
5
FDE  (720 N)
6
(b)
5
FAB
6
FDE  600 N 
3
Fx  0:  (720 N)  Cx  0
5
C x  432 N
4
Fy  0:  (720 N)  C y  600 N  0
5
C y  1176 N
C  1252.84 N
  69.829
C  1253 N
69.8° 
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PROBLEM 4.16
Two links AB and DE are connected by a bell crank as
shown. Determine the maximum force that may be
safely exerted by link AB on the bell crank if the
maximum allowable value for the reaction at C is
1600 N.
SOLUTION
See solution to Problem 4.15 for F.B.D. and derivation of Eq. (1).
FDE 
5
FAB
6
(1)
3
Fx  0:  FAB  Cx  0
5
Fy  0: 
Cx 
3
FAB
5
4
FAB  C y  FDE  0
5
4
5
 FAB  C y  FAB  0
5
6
49
Cy 
FAB
30
C  C x2  C y2
1
(49)2  (18) 2 FAB
30
C  1.74005FAB

For C  1600 N, 1600 N  1.74005 FAB
FAB  920 N 
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PROBLEM 4.17
The required tension in cable AB is 200 N. Determine (a) the
vertical force P that must be applied to the pedal, (b) the
corresponding reaction at C.
SOLUTION
Free-Body Diagram:
BC  7 cm
(a)
M C  0: P (15 cm)  (200 N)(6.062 cm)  0
P  80.83 N
(b)
P  80.8 N 
C x  200 N
 Fy  0: C x  200 N  0
 Fy  0: C y  P  0 C y  80.83 N  0
C y  80.83 N
  22.0
C  215.7 N
C  216 N
22.0 
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PROBLEM 4.18
Determine the maximum tension that can be developed in cable
AB if the maximum allowable value of the reaction at C is 250 N.
SOLUTION
Free-Body Diagram:
BC  7 cm
M C  0: P (15 cm) T (6.062 cm )  0 P  0.40415 T
 Fy  0:  P  C y  0
 0.40415 P  C y  0
C y  0.40415T
Fx  0: T  C x  0
Cx  T
C  Cx2  C y2  T 2  (0.40415T )2
C  1.0786T
For
C  250 N
250 N  1.0786T
T  232 N 
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PROBLEM 4.19
The bracket BCD is hinged at C and attached to a control
cable at B. For the loading shown, determine (a) the
tension in the cable, (b) the reaction at C.
SOLUTION
Ty
At B:
Tx

0.18 m
0.24 m
3
Ty  Tx
4
(1)
M C  0: Tx (0.18 m)  (240 N)(0.4 m)  (240 N)(0.8 m)  0
(a)
Tx  1600 N
From Eq. (1):
Ty 
3
(1600 N)  1200 N
4
T  Tx2  Ty2  16002  12002  2000 N
(b)
T  2.00 kN 
Fx  0: C x  Tx  0
C x  1600 N  0 C x  1600 N
C x  1600 N
 Fy  0: C y  Ty  240 N  240 N  0
C y  1200 N  480 N  0
C y  1680 N
C y  1680 N
  46.4
C  2320 N
C  2.32 kN
46.4° 
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PROBLEM 4.20
Solve Problem 4.19, assuming that a  0.32 m.
PROBLEM 4.19 The bracket BCD is hinged at C and
attached to a control cable at B. For the loading shown,
determine (a) the tension in the cable, (b) the reaction at C.
SOLUTION
Ty
At B:
Tx

0.32 m
0.24 m
4
Ty  Tx
3
M C  0: Tx (0.32 m)  (240 N)(0.4 m)  (240 N)(0.8 m)  0
Tx  900 N
From Eq. (1):
4
Ty  (900 N)  1200 N
3
T  Tx2  Ty2  9002  12002  1500 N
T  1.500 kN 
Fx  0: C x  Tx  0
C x  900 N  0 C x  900 N
C x  900 N
 Fy  0: C y  T y  240 N  240 N  0
C y  1200 N  480 N  0
C y  1680 N
C y  1680 N
  61.8
C  1906 N
C  1.906 kN
61.8° 
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PROBLEM 4.21
The 12-m boom AB weighs 2 kN; the distance from the
axle A to the center of gravity G of the boom is 6 m. For
the position shown, determine (a) the tension T in the cable,
(b) the reaction at A.
SOLUTION
AD  ( AB )sin10  (12 m)sin10
AD  2.0838 m
 M A  0: T ( AD )  2(6cos 30  )  5(12cos 30  )  0
T  29.923 kN
T  29.92 kN 
 Fx  0:
Ax  (29.923) cos 20  0
Ax  28.118 kN
Fy  0: Ay  (29.923) sin 20  2  5  0
Ay  17.234 kN
A  28.118 2  17.2342  32.79 kN
 17.234 

 28.118 
  tan 1 
 31.5
A  32.98 kN
31.5° 
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PROBLEM 4.22
A lever AB is hinged at C and attached to a control cable at A. If
the lever is subjected to a 500-N horizontal force at B,
determine (a) the tension in the cable, (b) the reaction at C.
SOLUTION
Triangle ACD is isosceles with  C  90  30  120  A   D 
1
(180  120)  30.
2
Thus, DA forms angle of 60° with the horizontal axis.
(a)
We resolve FAD into components along AB and perpendicular to AB.
M C  0: ( FAD sin 30)(250 mm)  (500 N)(100 mm)  0
(b)
FAD  400 N 
Fx  0:  (400 N) cos 60  C x  500 N  0
C x  300 N
 Fy  0:  (400 N) sin 60°  C y  0
C y  346.4 N
C  458 N
49.1° 
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PROBLEM 4.23
For each of the plates and loadings shown, determine the reactions at A and B.
4 cm 50 N
4 cm 50 N
40 N
40 N
10 cm
10 cm
B
A
B
A
30°
20 cm
20 cm
(a)
(b)
SOLUTION
(a)
Free-Body Diagram:
ΣM A = 0: B(20 cm) − (50 N)(4 cm) − (40 N)(10 cm) = 0
B = +30 N
B = 30.0 N
ΣFx = 0: Ax + 40 N = 0
Ax = −40 N
A x = 40.0 N
ΣFy = 0: Ay + B − 50 N = 0
Ay +30 N − 50 N = 0
Ay = +20 N
A y = 20.0 N
α = 26.56°
A = 44.72 N
A = 44.7 N
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26.6°
PROBLEM 4.23 (Continued)
(b)
Free-Body Diagram:
ΣM A = 0: (B cos30°)(20 cm) − (40 N)(10 cm) − (50 N)(4 cm) = 0
B = 34.64 N
B = 34.6 N
60.0°
ΣFx = 0: Ax − B sin 30° + 40 N
Ax −(34.64 N) sin 30° + 40 N = 0
Ax = −22.68 N
A x = 22.68 N
Σ Fy = 0: Ay + B cos30° − 50 N = 0
Ay + (34.64 N) cos30° − 50 N = 0
Ay = +20 N
A y = 20.0 N
α = 41.4°
A = 30.24 N
A = 30.2 N
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369
41.4°
PROBLEM 4.24
For each of the plates and loadings shown, determine the reactions at A and B.
4 cm 50 N
4 cm 50 N
40 N
40 N
10 cm
10 cm
B
A
30º
B
A
20 cm
20 cm
(b)
(a)
SOLUTION
(a)
Free-Body Diagram:
ΣM B = 0: − A(20 cm) + (50 N)(16 cm) − (40 N)(10 cm) = 0
A = +20 N
A = 20.0 N
ΣFx = 0: 40 N + Bx = 0
Bx = −40 N
B x = 40 N
ΣFy = 0: A + By − 50 N = 0
20 N + By − 50 N = 0
By = +30 N
α = 36.87°
B = 50 N
B y = 30 N
B = 50.0 N
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36.9°
PROBLEM 4.24 (Continued)
(b)
ΣM B = 0: −(A cos30°)(20 cm) − (40 N)(10 cm) + (50 N)(16 cm) = 0
A = 23.09 N
A = 23.1 N
60.0°
ΣFx = 0: A sin 30° + 40 N + Bx = 0
(23.09 N) sin30° + 40 N + Bx = 0
Bx = −51.55 N
B x = 51.55 N
ΣFy = 0: A cos30° + By − 50 N = 0
(23.09 N) cos30° + By − 50 N = 0
By = +30 N
B y = 30 N
α = 30.2°
B = 59.64 N
B = 59.6 N
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30.2°
PROBLEM 4.25
A rod AB hinged at A and attached at B to cable BD
supports the loads shown. Knowing that d  200 mm,
determine (a) the tension in cable BD, (b) the reaction at A.
SOLUTION
Free-Body Diagram:
(a)
Move T along BD until it acts at Point D.
M A  0: (T sin 45)(0.2 m)  (90 N)(0.1 m)  (90 N)(0.2 m)  0
T  190.919 N
(b)
T  190.9 N 
Fx  0: Ax  (190.919 N) cos 45  0
Ax  135.0 N
 Fy  0:
A x  135.0 N
Ay  90 N  90 N  (190.919 N) sin 45°  0
Ay  45.0 N
A y  45.0 N
A  142.3 N
18.43°
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PROBLEM 4.26
A rod AB, hinged at A and attached at B to cable BD,
supports the loads shown. Knowing that d  150 mm,
determine (a) the tension in cable BD, (b) the reaction at A.
SOLUTION
Free-Body Diagram:
tan  
(a)
10
;   33.690°
15
Move T along BD until it acts at Point D.
M A  0: (T sin 33.690)(0.15 m)  (90 N)(0.1 m)  (90 N)(0.2 m)  0
T  324.50 N
(b)
T  324 N 
Fx  0: Ax  (324.50 N) cos 33.690  0
Ax  270 N
 Fy  0:
A x  270 N
Ay  90 N  90 N  (324.50 N) sin 33.690  0
Ay  0
A  270 N
Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 4. 27
Determine the reactions at A and B when (a) α = 0, (b) α = 90°,
(c) α = 30°.
SOLUTION
Free-Body Diagram:
Equations of equilibrium:
− (330 N)(0.25 m) + B sin α (0.3 m) + B cos α (0.5 m) = 0
ΣM A = 0:
ΣFx = 0:
ΣFy = 0:
(1)
Ax − B sin α = 0
(2)
Ay − (330 N) + B cosα = 0
(3)
(a) Substitution α = 0 into (1), (2), and (3) and solving for A and B:
B = 165.0 N, Ax = 0, Ay = 165.0 N
or A = 165.0 N , B = 165.0 N
t
(b) Substituting α = 90° into (1), (2), and (3) and solving for A and B:
B = 275.00 N, Ax = 275.00 N, Ay = 330.00 N
A=
Ax2 + Ay2 = (275)2 + (330)2 = 429.56 N
θ = tan−1
Ay
330
= tan−1
= 50.194°
Ax
275
∴ A = 429.6 N
50.2°, B = 275 N
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
t
PROBLEM 4. 27 CONTINUED
(c) Substituting α = 30° into (1), (2), and (3) and solving for A and B:
B = 141.506 N, Ax = 70.753 N, Ay = 207.45 N, ⇒
A=
Ax2 + Ay2 = (70.753)2 + (207.45)2 = 219.18 N
θ = tan−1
Ay
207.45
= tan−1
= 71.168°
Ax
70.753
∴ A = 219.2 N
71.2°, B = 141.5 N
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
60° t
PROBLEM 4.28
Determine the reactions at A and C when (a)   0,
(b)   30.
SOLUTION
(a)
 0
From F.B.D. of member ABC:
M C  0: (300 N)(0.2 m)  (300 N)(0.4 m)  A(0.8 m)  0
A  225 N
or
A  225 N 
 Fy  0: C y  225 N  0
C y  225 N
or
C y  225 N
Fx  0: 300 N  300 N  C x  0
C x  600 N or C x  600 N
Then
C  Cx2  C y2  (600)2  (225)2  640.80 N
and
  tan 1 
 Cy 
1  225 
  tan 
  20.556
 600 
 Cx 
or
(b)
C  641 N
20.6 
  30
From F.B.D. of member ABC:
M C  0: (300 N)(0.2 m)  (300 N)(0.4 m)  ( A cos 30)(0.8 m)
 ( A sin 30)(20 in.)  0
A  365.24 N
or
A  365 N
60.0 
Fx  0: 300 N  300 N  (365.24 N)sin 30  C x  0
C x  782.62
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PROBLEM 4.28 (Continued)
Fy  0: C y  (365.24 N) cos 30  0
C y  316.31 N
or
C y  316 N
Then
C  Cx2  C y2  (782.62)2  (316.31)2  884.12 N
and
  tan 1 
or
 Cy 
1  316.31 
  tan 
  22.007
 782.62 
 Cx 
C  884 N
22.0 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 4.29
Rod ABC is bent in the shape of an arc of circle of radius R. Knowing the
  30, determine the reaction (a) at B, (b) at C.
SOLUTION
Free-Body Diagram:
M D  0: C x ( R )  P ( R )  0
Cx   P
Fx  0: C x  B sin   0
P  B sin   0
B  P/sin 
B
P
sin 

 Fy  0: C y  B cos   P  0
C y  ( P/sin  ) cos   P  0
1 

C y  P 1 
 
tan

For   30,
(a)
(b)
B  P/sin 30  2 P
Cx   P
B  2P
60.0° 
Cx  P
C y  P (1  1/tan 30)   0.732/ P
C y  0.7321P
C  1.239P
36.2° 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 4.30
Rod ABC is bent in the shape of an arc of circle of radius R. Knowing that
  60, determine the reaction (a) at B, (b) at C.
SOLUTION
See the solution to Problem 4.33 for the free-body diagram and analysis leading to the following
expressions:
Cx   P
1 

C y  P 1 

tan  

P
B
sin 
For   60,
(a)
(b)
B  P/sin 60  1.1547 P
Cx   P
B  1.155P
30.0° 
Cx  P
C y  P (1  1/tan 60)   0.4226 P
C y  0.4226 P
C  1.086P
22.9° 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 4.31
Neglecting friction, determine the tension in cable ABD and the
reaction at C when   60.
SOLUTION
M C  0: T (2a  a cos  )  Ta  Pa  0
T
P
1  cos 
(1)
Fx  0: C x  T sin   0
Cx  T sin  
P sin 
1  cos 
 Fy  0: C y  T  T cos   P  0
C y  P  T (1  cos  )  P  P
1  cos 
1  cos 
Cy  0
C y  0, C  C x
Since
CP
sin 
1  cos 
(2)
For   60 :
P
P

1  cos 60 1  12
Eq. (1):
T
Eq. (2):
CP
sin 60
0.866
P
1  cos 60
1  12
T
2
P 
3
C  0.577P
Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 4.32
Neglecting friction, determine the tension in cable ABD and the
reaction at C when   45.
SOLUTION
Free-Body Diagram:
Equilibrium for bracket:
M C  0:  T (a )  P ( a )  (T sin 45)(2a sin 45)
 (T cos 45)(a  2a cos 45)  0
T  0.58579
or T  0.586 P 
Fx  0: C x  (0.58579 P ) sin 45  0
C x  0.41422 P
 Fy  0: C y  0.58579 P  P  (0.58579 P ) cos 45  0
Cy  0
or C  0.414P
Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 4.33
A force P of magnitude 90 N is applied to member ACDE, which is
supported by a frictionless pin at D and by the cable ABE. Since the
cable passes over a small pulley at B, the tension may be assumed to
be the same in portions AB and BE of the cable. For the case when
a  3 cm, determine (a) the tension in the cable, (b) the reaction at D.
SOLUTION
Free-Body Diagram:
(a)
M D  0: (90 N)(9 cm) 
5
12
T (9 cm)  T (7 cm)  T (3 cm)  0
13
13
T  117 N
(b)
Fx  0: Dx  117 N 
T  117.0 N 
5
(117 N)  90  0
13
Dx  72 N
Fy  0: Dy 
12
 N)  0
13
D y  108 N
D  129.8 N
56.3° 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 4.34
Solve Problem 4.33 for a  6 cm.
PROBLEM 4.33 A force P of magnitude 90 N is applied to member
ACDE, which is supported by a frictionless pin at D and by the cable
ABE. Since the cable passes over a small pulley at B, the tension may
be assumed to be the same in portions AB and BE of the cable. For
the case when a  3 cm, determine (a) the tension in the cable, (b) the
reaction at D.
SOLUTION
Free-Body Diagram:
(a)
M D  0: (90 N)(6 cm) 
5
12
T (6 cm)  T (7 cm)  T (6 cm)  0
13
13
T  195 N
(b)
T  195.0 N 
Fx  0: Dx  195 N 
5
(195 N)  90  0
13
Dx  180 N
Fy  0: Dy 
12
 N)  0
13
D y  180 N
D  255 N
45.0° 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 4.35
Bar AC supports two 400-N loads as shown. Rollers at A
and C rest against frictionless surfaces and a cable BD is
attached at B. Determine (a) the tension in cable BD, (b) the
reaction at A, (c) the reaction at C.
SOLUTION
Similar triangles: ABE and ACD
AE BE

;
AD CD
(a)
BE
0.15 m

; BE  0.075 m
0.5 m 0.25 m
 0.075 
M A  0: Tx (0.25 m)  
Tx  (0.5 m)  (400 N)(0.1 m)  (400 N)(0.4 m)  0
 0.35 
Tx  1400 N
0.075
(1400 N)
0.35
 300 N
Ty 
T  1432 N 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 4.35 (Continued)
(b)
 Fy  0:
A  300 N  400 N  400 N  0
A   1100 N
(c)
A  1100 N 
Fx  0: C  1400 N  0
C   1400 N
C  1400 N
Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 4.36
A light bar AD is suspended from a cable BE and supports a
20-kg block at C. The ends A and D of the bar are in contact
with frictionless vertical walls. Determine the tension in cable
BE and the reactions at A and D.
SOLUTION
Free-Body Diagram:
W  (20 kg)(9.81 m/s 2 )  196.20 N
Fx  0:
AD
 Fy  0:
TBE  W
TBE  196.2 N 
We note that the forces shown form two couples.
M B  0: A(200 mm)  (196.20 N)(75 mm)  0
A  73.575 N
A  73.6 N
,
D  73.6 N
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
4 cm
4 cm
PROBLEM 4.37
80 N
40 N
A
B
2 cm
C
D
E
The T-shaped bracket shown is supported by a small wheel at E and pegs at C
and D. Neglecting the effect of friction, determine the reactions at C, D, and E
when θ = 30°.
3 cm
3 cm
θ
SOLUTION
Free-Body Diagram:
ΣFy = 0: E cos 30° − 80 − 40 = 0
E=
120 N
= 138.56 N
cos 30°
E = 139 N
60.0°
ΣM D = 0: (40 N)(4 cm) − (80 N)(4 cm)
−C(3 cm) + E sin30°(3 cm) = 0
−160 − 3C + 138.56(0.5)(3) = 0
C = 15.95 N
C = 16 N
ΣFx = 0: E sin 30° + C − D = 0
(138.56 N)(0.5) + 15.95 N − D = 0
D = 85.23 N
D = 85 N
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
4 cm
4 cm
PROBLEM 4.38
80 N
40 N
A
B
2 cm
C
D
E
3 cm
The T-shaped bracket shown is supported by a small wheel at E and pegs
at C and D. Neglecting the effect of friction, determine (a) the smallest
value of θ for which the equilibrium of the bracket is maintained, (b) the
corresponding reactions at C, D, and E.
3 cm
θ
SOLUTION
Free-Body Diagram:
ΣFy = 0: E cos q − 80 − 40 = 0
E=
120
cos θ
(1)
ΣM D = 0: (40 N)(4 cm) − (80 N)(4 cm) − C(3 cm)
+
120
sin θ 3 cm = 0
cos θ
1
C = (360 tan q − 160)
3
(a)
For C = 0,
360 tan θ = 160
tan θ =
(b)
Eq. (1)
E=
4
θ = 23.962°
9
θ = 24.0°
120
= 131. 32 N
cos 23.962°
ΣFx = 0: −D + C + E sin θ = 0
D = (131. 32) sin 23.962 = 53.3 N
C=0
D = 53 N
E = 131 N
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528
66.0°
PROBLEM 4.39
A movable bracket is held at rest by a cable attached at C and by
frictionless rollers at A and B. For the loading shown, determine
(a) the tension in the cable, (b) the reactions at A and B.
SOLUTION
Free-Body Diagram:
(a)
 Fy  0:
T  600 N  0
T  600 N 
(b)
Fx  0: B  A  0
 BA
Note that the forces shown form two couples.
M  0: (600 N)(600 mm)  A(90 mm)  0
A  4000 N
 B  4000 N
A  4.00 kN
; B  4.00 kN
Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 4.40
A light bar AB supports a 15-kg block at its midpoint C. Rollers at
A and B rest against frictionless surfaces, and a horizontal cable
AD is attached at A. Determine (a) the tension in cable AD, (b) the
reactions at A and B.
SOLUTION
Free-Body Diagram:
W  (15 kg)(9.81 m/s 2 )
 147.150 N
(a)
Fx  0:
TAD  105.107 N  0
TAD  105.1 N 
(b)
Fy  0:
A W  0
A  147.150 N  0
A  147.2 N 
M A  0: B (350 mm)  (147.150 N) (250 mm)  0
B  105.107 N
B  105.1 N
Copyright © McGraw-Hill Education. Permission required for reproduction or display.

4 cm
4 cm
PROBLEM 4.41
2 cm
7 cm
Two slots have been cut in plate DEF, and the plate has
been placed so that the slots fit two fixed, frictionless pins A
and B. Knowing that P = 15 N, determine (a) the force each
pin exerts on the plate, (b) the reaction at F.
F
B
P
A
D
30º
E
3 cm
30 N
SOLUTION
Free-Body Diagram:
ΣFx = 0: 15 N − B sin 30° = 0
B = 30.0 N
60.0°
ΣM A = 0: −(30 N)(4 cm) + B sin30°(3 cm) + B cos30°(11 cm) − F(13 cm) = 0
−120 N ⋅ cm + (30 N) sin30° (3 cm) + (30 N) cos30° (11 cm) − F(13 cm) = 0
F = + 16.2145 N
F = 16.2 N
ΣFy = 0: A − 30 N + B cos30° − F = 0
A − 30 N + (30 N) cos30° − 16.2145 N = 0
A = + 20.23 N
A = 20.2 N
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PROBLEM 4.42
For the plate of Problem 4.41 the reaction at F must be
directed downward, and its maximum allowable value
is 20 N. Neglecting friction at the pins, determine the
required range of values of P.
PROBLEM 4.41 Two slots have been cut in plate
DEF, and the plate has been placed so that the slots fit
two fixed, frictionless pins A and B. Knowing that
P  15 N, determine (a) the force each pin exerts on the
plate, (b) the reaction at F.
SOLUTION
Free-Body Diagram:
B  2P
Fx  0: P  B sin 30  0
60°
M A  0:  (30 N)(4 cm)  B sin 30(3 cm)  B cos 30  (11 cm )  F (13 cm)  0
120 N  cm + 2 P sin 30(3 cm)  2 P cos30 (11 cm)  F (13 cm)  0
120  3P  19.0525P  13F  0
P
13F  120
22.0525
For F  0:

P
13(0)  120
 5.442 N
22.0525
For P  20 N:

P
13(20)  120
 17.232 N
22.0525
For
0  F  20 N:
(1)
5.44 N  P  17.23 N 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 4.43
The rig shown consists of a 5.4-kN horizontal member ABC and a vertical
member DBE welded together at B. The rig is being used to raise a
16.2-kN crate at a distance x = 4.8 m from the vertical member DBE. If
the tension in the cable is 18 kN, determine the reaction at E assuming
that the cable is (a) anchored at F as shown in the figure, (b) attached to
the vertical member at a point located 0.4 m above E.
SOLUTION
(a)
Free-Body Diagram:
ΣFx = 0:
Ex = 0
ΣFy = 0:
E y − 16.2 kN − 5.4 kN − 18 kN = 0
or E = 39.6 kN t
Ey = 39.6 kN
ΣM E = 0:
M E + (16.2 kN)(4.8 m) + (5.4 kN)(2.6 m) − (18 kN)(1.5 m) = 0
M E = − 64.8 kN ⋅ m
or ME = 64.8 kN .m
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.
t
PROBLEM 4.43 CONTINUED
(b)
Free-Body Diagram:
1.5 m
7m
2m
2.6 m 2.2 m
16.2 kN
5.4 kN
4m
Ex
ME
EY
+ ΣF = 0:
x
Ex = 0
+ ΣFy = 0:
Ey − 16.2 kN − 5.4 kN = 0
Ey = 21.6 kN
+ ΣME = 0:
or E = 21.6 kN
ME + (16.2 kN)(4.8 m) + (5.4 kN)(2.6 m) = 0
ME = − 91.8 kN · m
or ME = 91.8 kN · m
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PROBLEM 4.44
For the rig and crate of Prob. 4.43, assuming that the cable is anchored at F as
shown, determine (a) the required tension in cable ADCF if the maximum value of
the couple at E is to be as small as possible as x varies from 0.6 m to 7 m, (b) the
corresponding maximum value of the couple.
SOLUTION
Free-Body Diagram:
7m
1.5 m
2m
2.6 m
16.2 KN
T
5.4 KN
4m
X
ME
Ex
Ey
+ ΣME = 0:
ME + (16.2 kN)x + (5.4 kN)(2.6 m) − T(1.5 m) = 0
ME = (1.5 T − 16.2 x − 14.04) kN · m
(1)
For x = 0.6 m, (1) gives:
(ME)1 = (1.5 T − 23.76) kN · m
For x = 7 m, (1) gives:
(ME)2 = (1.5 T − 127.44) kN · m
(a)
The maximum absolute value of ME is obtained when (ME)1 = − (ME)2 and
1.5 T − 23.76 kN = − (1.5 T − 127.44 kN)
T = 50.400 kN
(b)
or T = 50.4 kN
For this value of T:
ME = 1.5(50.400) kN ⋅ m − 23.76 kN ⋅ m
= 51.84 kN ⋅ m
or ME = 51.8 kN ⋅ m
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PROBLEM 4.45
A 175-kg utility pole is used to support at C the end of an electric
wire. The tension in the wire is 600 N, and the wire forms an angle
of 15° with the horizontal at C. Determine the largest and smallest
allowable tensions in the guy cable BD if the magnitude of the
couple at A may not exceed 500 N·m.
SOLUTION
Free-Body Diagram:
Geometry:
Distance BD =
(1.5)2  (3.6)2  3.90 m
Note also that: W  mg  (175 kg)(9.81 m/s2)  1716.75 N
With MA  500 N  m clockwise: (i.e. corresponding to Tmax )
M A  0:
 1.5 

500 N  m – [(600 N) cos 15o](4.5 m) + 
 Tmax  (3.6 m)  0
 3.90 

Tmax  2244.7 N
or Tmax  2240 N 
With MA  500 N  m counter-clockwise: (i.e. corresponding to Tmin )
M A  0:
 1.5 

500 N  m – [(600 N) cos 15o](4.5 m) + 
 Tmin  (3.6 m)  0
 3.90 

Tmin  1522.44 N
or Tmin  1522 N 
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PROBLEM 4.46
Knowing that the tension in wire BD is 1300 N, determine the
reaction at the fixed support C of the frame shown.
SOLUTION
T  1300 N
5
T
13
 500 N
Tx 
12
T
13
 1200 N
Ty 
M x  0: C x  450 N  500 N  0
C x  50 N
C x  50 N
 Fy  0: C y  750 N  1200 N  0 C y  1950 N
C y  1950 N
C  1951 N
88.5° 
M C  0: M C  (750 N)(0.5 m)  (4.50 N)(0.4 m)
 (1200 N)(0.4 m)  0
M C  75.0 N  m
M C  75.0 N  m
Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 4.47
Determine the range of allowable values of the tension in wire
BD if the magnitude of the couple at the fixed support C is not
to exceed 100 N · m.
SOLUTION
 5 
M C  0: (750 N)(0.5 m)  (450 N)(0.4 m)   T  (0.6 m)
 13 
 12 
  T  (0.15 m)  M C  0
 13 
 4.8 
375 N  m  180 N  m  
m T  M C  0
 13 
T
13
(555  M C )
4.8
For
M C  100 N  m: T 
13
(555  100)  1232 N
4.8
For
M C  100 N  m: T 
13
(555  100)  1774 N
4.8
For
|M C |  100 N  m:
1.232 kN  T  1.774 kN 
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PROBLEM 4.48
5m
A
B
200 N
4m
C
E
W
D
Beam AD carries the two 200-N loads shown. The beam is held by a
fixed support at D and by the cable BE that is attached to the
counterweight W. Determine the reaction at D when (a) W = 500 N,
(b) W = 450 N.
200 N
4m
SOLUTION
W = 500 N
(a)
From f.b.d. of beam AD
ΣFx = 0: Dx = 0
ΣFy = 0: Dy −200 N − 200 N + 500 N = 0
Dy = −100 N
or
D = 100 N
ΣM D = 0: MD − (500 N)(5 m) + (200 N)(8 m)
+(200 N)(4 m) = 0
MD = 100 N ⋅ m
or MD = 100 N ⋅ m
W = 450 N
(b)
From f.b.d. of beam AD
ΣFx = 0: Dx = 0
ΣFy = 0: Dy + 450 N − 200 N − 200 N = 0
Dy = −50 N
or
D = 50 N
or
MD = 150 N ⋅ m
ΣM D = 0: MD − (450 N)(5 m) + (200 N)(8 m)
+(200 N)(4 m) = 0
MD = −150 N ⋅ m
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395
PROBLEM 4.49
5m
A
B
200 N
C
E
D
For the beam and loading shown, determine the range of values of W
for which the magnitude of the couple at D does not exceed 200 N ⋅ m.
W
200 N
4m
4m
SOLUTION
For Wmin ,
From f.b.d. of beam AD
MD = −200 N ⋅ m
ΣM D = 0: (200 N)(8 m) − Wmin (5 m)
+(200 N)(4 m) − 200 N ⋅ m = 0
Wmin = 440 N
For Wmax ,
From f.b.d. of beam AD
MD = 200 N ⋅ m
ΣM D = 0: (200 N)(8 m) − Wmax (5 m)
+(200 N)(4 m) + 200 N ⋅ m = 0
Wmax = 520 N
or 440 N
W
520 N
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396
PROBLEM 4.50
An 8-kg mass can be supported in the three different ways shown. Knowing that the pulleys have a
100-mm radius, determine the reaction at A in each case.
SOLUTION

W  mg  (8 kg)(9.81 m/s 2 )  78.480 N

(a)
Fx  0: Ax  0
 Fy  0:

Ay  W  0
A y  78.480 N
M A  0: M A  W (1.6 m)  0

M A   (78.480 N)(1.6 m)

A  78.5 N , M A  125.6 N  m


(b)
Fx  0: Ax  W  0
 Fy  0:
Ay  W  0
M A   (78.480 N)(1.6 m)

A y  78.480
M A  125.568 N  m
A  111.0 N

(c)
45°,
M A  125.6 N  m

Fx  0: Ax  0
 Fy  0:

45°
M A  0: M A  W (1.6 m)  0



A x  78.480
A  (78.480 N) 2  110.987 N


M A  125.568 N  m
Ay  2W  0
Ay  2W  2(78.480 N)  156.960 N
M A  0: M A  2W (1.6 m)  0
M A   2(78.480 N)(1.6 m)
M A  251.14 N  m
A  157.0 N , M A  251 N  m
Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 4.51
A uniform rod AB with a length of l and weight of W is suspended
from two cords AC and BC of equal length. Determine the angle θ
corresponding to the equilibrium position when a couple M is applied
to the rod.
SOLUTION
Free-Body Diagram:
l
Using h  tan 
2
M C  0:
M  (W)(h sin  )  0
or sin  

M
Wh
M 2

 cot  
W l

cot  

or   sin 1  2M
 
Wl 

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PROBLEM 4.52
Rod AD is acted upon by a vertical force P at end A and by two equal
and opposite horizontal forces of magnitude Q at points B and C.
Neglecting the weight of the rod, express the angle  corresponding to
the equilibrium position in terms of P and Q.
SOLUTION
Free-Body Diagram:
M D  0:
P(3a sin  )  Q(a cos  )  Q(2a cos  )  0
3P sin   Q cos 
tan  
Q
3P
 Q 
or   tan 1 
 
 3P 
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PROBLEM 4.53
A slender rod AB, of weight W, is attached to blocks A and B,
which move freely in the guides shown. The blocks are
connected by an elastic cord that passes over a pulley at C.
(a) Express the tension in the cord in terms of W and .
(b) Determine the value of  for which the tension in the cord is
equal to 3W.
SOLUTION
(a)
From f.b.d. of rod AB:
 1 

M C  0: T (l sin  )  W   cos    T (l cos  )  0
 2 

T
W cos 
2(cos   sin  )
Dividing both numerator and denominator by cos ,
W
1

T 
2  1  tan  
(b)
For T  3W ,
or
3W 
or T 
 W2 
(1  tan  )

 W2 
(1  tan  )
1
1  tan  
6
5
 
  tan 1    39.806
6
or
  39.8 
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PROBLEM 4.54
A vertical load P is applied at end B of rod BC. (a) Neglecting the
weight of the rod, express the angle  corresponding to the
equilibrium position in terms of P, l, and the counterweight W. (b)
Determine the value of  corresponding to equilibrium if
P  2W.
SOLUTION
Free-Body Diagram:
(a)
Triangle ABC is isosceles. We have
 
 
CD  ( BC ) cos    l cos  
2
2


M C  0: P (l cos  )  W  l cos   0
2

Setting cos   2cos2

2
 1:




Pl  2 cos 2  1  Wl cos  0
2 
2

cos 2

 1
W 
cos   0


2  2P 
2 2
cos

2


1 W
W2
 
8
2

4 P
P


1 W
  2cos1  
4  P
 


W2
  

8

P2


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PROBLEM 4.54 (Continued)
(b)
For P  2W ,
cos
cos

2

2

2

 1
11
1
 8   1  33
 
 8
42
4

 0.84307 and cos
 32.534
  65.1


2

2

 0.59307
 126.375
  252.75 (discard)
  65.1 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 4.55
A vertical load P is applied at end B of rod BC. (a) Neglecting the weight
of the rod, express the angle  corresponding to the equilibrium position in
terms of P, l, and the counterweight W. (b) Determine the value of 
corresponding to equilibrium if P  2W.
SOLUTION
(a)
Triangle ABC is isosceles. We have
CD  ( BC )cos

2
 l cos

2


M C  0: W  l cos   P (l sin  )  0
2

Setting sin   2sin




W  2 P sin
(b)
For P  2W ,
sin

2

2
or

cos : Wl cos  2 Pl sin cos  0
2
2
2
2
2

2


2
0
  2sin 1 
W 
 
 2P 
W
W

 0.25
2 P 4W
 14.5
  29.0 
 165.5   331 (discard)
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PROBLEM 4.56
A collar B of weight W can move freely along the vertical rod shown. The
constant of the spring is k, and the spring is unstretched when   0.
(a) Derive an equation in , W, k, and l that must be satisfied when the collar
is in equilibrium. (b) Knowing that W  300 N, l  500 mm, and
k  800 N/m, determine the value of  corresponding to equilibrium.
SOLUTION
First note:
T  ks
where
k  spring constant
s  elongation of spring
l
l
cos 
l

(1  cos  )
cos 
kl
(1  cos  )
T
cos 

(a)
From F.B.D. of collar B:
or
(b)
For
Solving numerically,
 Fy  0: T sin   W  0
kl
(1  cos  )sin   W  0
cos 
or tan   sin  
W

kl
W  300 N
l  500 mm  0.5 m
k  800 N /m
300 N
 0.75
tan   sin  

N
800 m  (0.5 m)


  57.957
or
  58.0 
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PROBLEM 4.57
Solve Sample Problem 4.5, assuming that the spring is
unstretched when   90.
SOLUTION
First note:
T  tension in spring  ks
where
s  deformation of spring
 r
F  kr 
From F.B.D. of assembly:
M 0  0: W (l cos  )  F ( r )  0
Wl cos   kr 2   0
or
cos  
kr 2

Wl
k  45 N/mm
For
r  75 mm
l  200 mm
W  1800 N
cos  
(45 N/mm)(75 mm)2

(1800 N)(200 mm)
cos   0.703125
or
Solving numerically,
  0.89245 rad
or
  51.134
Then
  90  51.134  141.134
or   141.1 
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PROBLEM 4.58
A vertical load P is applied at end B of rod BC. The constant of the
spring is k, and the spring is unstretched when θ = 60°. (a) Neglecting
the weight of the rod, express the angle θ corresponding to the
equilibrium position terms of P, k, and l. (b) Determine the value of θ
corresponding to equilibrium if P = 14 kl.
SOLUTION
Free-Body Diagram:
(a)
Triangle ABC is isosceles. We have
 
 
AB  2( AD )  2l sin   ; CD  l cos  
2
2
Elongation of spring:
x  ( AB )  ( AB )  60
 
 2l sin    2l sin 30
2
  1
T  k x  2kl  sin  
2 2



M C  0: T  l cos   P (l sin  )  0
2

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PROBLEM 4.58 (Continued)



  1

2kl  sin   l cos  Pl  2sin cos   0
2
2
2
2
2



cos

2
0
2(kl  P )sin
or
  180 (trivial)
sin

2

2
 kl  0

1
2
kl
kl  P
1
2


  2sin 1  kl /(kl  P )  
(b)
For P 
1
kl ,
4
sin

2

2

1
2
3
4
kl
kl

 41.8
2
3
  83.6 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 4.59
Eight identical 500  750-mm rectangular plates, each of mass m  40 kg, are held in a vertical plane as
shown. All connections consist of frictionless pins, rollers, or short links. In each case, determine whether
(a) the plate is completely, partially, or improperly constrained, (b) the reactions are statically determinate
or indeterminate, (c) the equilibrium of the plate is maintained in the position shown. Also, wherever
possible, compute the reactions.
SOLUTION
1.
Three non-concurrent, non-parallel reactions:
(a)
Plate: completely constrained
(b)
Reactions: determinate
(c)
Equilibrium maintained
A  C  196.2 N
2.
Three non-concurrent, non-parallel reactions:
(a)
Plate: completely constrained
(b)
Reactions: determinate
(c)
Equilibrium maintained
B  0, C  D  196.2 N
3.
Four non-concurrent, non-parallel reactions:
(a)
Plate: completely constrained
(b)
Reactions: indeterminate
(c)
Equilibrium maintained
A x  294 N
,
D x  294 N
( A y  D y  392 N )
4.
Three concurrent reactions (through D):
(a)
Plate: improperly constrained
(b)
Reactions: indeterminate
(c)
No equilibrium
(M D  0)
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PROBLEM 4.59 (Continued)
5.
Two reactions:
(a)
Plate: partial constraint
(b)
Reactions: determinate
(c)
Equilibrium maintained
C  D  196.2 N
6.
Three non-concurrent, non-parallel reactions:
(a)
Plate: completely constrained
(b)
Reactions: determinate
(c)
Equilibrium maintained
B  294 N
7.
8.
, D  491 N
53.1°
Two reactions:
(a)
Plate: improperly constrained
(b)
Reactions determined by dynamics
(c)
No equilibrium
( Fy  0)
Four non-concurrent, non-parallel reactions:
(a)
Plate: completely constrained
(b)
Reactions: indeterminate
(c)
Equilibrium maintained
B  D y  196.2 N
(C  D x  0)
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PROBLEM 4.60
The bracket ABC can be supported in the eight different ways shown. All connections consist of smooth pins,
rollers, or short links. For each case, answer the questions listed in Problem 4.59, and, wherever possible,
compute the reactions, assuming that the magnitude of the force P is 100 N.
B
0.9 m
1
C
A
P
0.6 m 0.6 m
SOLUTION
1.
Three non-concurrent, non-parallel reactions
(a)
Bracket: complete constraint
(b)
Reactions: determinate
(c)
Equilibrium maintained
A = 120.2 N
2.
3.
56.3°,
Four concurrent, reactions (through A)
(a)
Bracket: improper constraint
(b)
Reactions: indeterminate
(c)
No equilibrium
(ΣMA ≠ 0)
Two reactions
(a)
Bracket: partial constraint
(b)
Reactions: determinate
(c)
Equilibrium maintained
A = 50 N ,
4.
B = 66.7 N
C = 50 N
Three non-concurrent, non-parallel reactions
(a)
Bracket: complete constraint
(b)
Reactions: determinate
(c)
Equilibrium maintained
A = 50 N ,
B = 83.3 N
36.9°, C = 66.7 N
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PROBLEM 4.60 (Continued)
5.
6.
Four non-concurrent, non-parallel reactions
(a)
Bracket: complete constraint
(b)
Reactions: indeterminate
(c)
Equilibrium maintained
(ΣM C = 0) A y = 50 N
Four non-concurrent, non-parallel reactions
(a)
Bracket: complete constraint
(b)
Reactions: indeterminate
(c)
Equilibrium maintained
A x = 66.7 N
, Bx = 66.7 N
( A y + B y = 100 N )
7.
Three non-concurrent, non-parallel reactions
(a)
Bracket: complete constraint
(b)
Reactions: determinate
(c)
Equilibrium maintained
A = C = 50 N
8.
Three concurrent, reactions (through A)
(a)
Bracket: improper constraint
(b)
Reactions: indeterminate
(c)
No equilibrium
(ΣM A ≠ 0)
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411
T
PROBLEM 4.61
G
A
0.5 m
2m
A 2-kN cylindrical tank, 2 m in diameter, is to be raised over a 0.5-m
obstruction. A cable is wrapped around the tank and pulled horizontally
as shown. Knowing that the corner of the obstruction at A is rough, find
the required tension in the cable and the reaction at A.
B
SOLUTION
Free-Body Diagram:
T
C
Ө
1m
G
0.5 m
α
D
A
A
2 kN
Force triangle
GD 0.5 m
cos a = _ = _ = 0.5
1m
AG
1
q =_
a = 30°
2
T = (2 kN) tan 30°
2 kN
A=_
cos 30°
a = 60°
(b = 60°)
T = 1.155 kN
A = 2.31 kN
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60.0°
PROBLEM 4.62
Determine the reactions at A and B when a  180 mm.
SOLUTION
Reaction at B must pass through D where A and 300-N load intersect.
BCD :
Free-Body Diagram:
(Three-force member)
240
180
  53.13
tan  
Force triangle
A  (300 N) tan 53.13
 400 N
300 N
cos 53.13°
 500 N

A  400 N 
B
B  500 N
53.1 
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PROBLEM 4.63
For the bracket and loading shown, determine the range of values
of the distance a for which the magnitude of the reaction at B does
not exceed 600 N.
SOLUTION
Reaction at B must pass through D where A and 300-N load intersect.
Free-Body Diagram:
(Three-force member)
a
240 mm
tan 
(1)
Force Triangle
(with B  600 N)
300 N
 0.5
600 N
  60.0
cos  
Eq. (1)
240 mm
tan 60.0°
 138.56 mm
a
For B  600 N a  138.6 mm 
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PROBLEM 4.64
The spanner shown is used to rotate a shaft. A pin fits in a
hole at A, while a flat, frictionless surface rests against the
shaft at B. If a 60-N force P is exerted on the spanner at D,
find the reactions at A and B.
SOLUTION
Free-Body Diagram:
(Three-force body)
The line of action of A must pass through D, where B and P intersect.
3sin 50
3cos50  15
 0.135756
tan  
  7.7310
60 N
sin 7.7310°
 446.02 N
60 N
B
tan 7.7310°
 441.97 N
A
Force triangle
A  446 N

7.73 
B  442 N
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
PROBLEM 4.65
Determine the reactions at B and C when a = 30 mm.
SOLUTION
Since CD is a two-force member, the force it exerts on member ABD is directed along DC.
Free-Body Diagram of ABD: (Three-Force member)
The reaction at B must pass through E, where D and the 250-N load intersect.
Triangle CFD:
Triangle EAD:
Triangle EGB:
Force triangle
60
 0.6
100
  30.964
tan  
AE  200 tan   120 mm
GE  AE  AG  120  30  90 mm
GB 60

GE 90
  33.690
B
D
250 N


sin120.964 sin 33.690 sin 25.346°
tan  
B  500.7 N
D  323.9 N

B  501 N
C  D  324 N
56.3 
31.0 
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PROBLEM 4.66
A 6-m wooden beam weighing 400 N is supported by a pin and
bracket at A and by cable BC. Find the reaction at A and the tension
in the cable.
SOLUTION
Since CB is a two-force member, the force it exerts on member AB is directed along CB.
Free-Body Diagram of AB: (Three-Force member)
The reaction at B must pass through E, where T and the 400 N load intersect.
BG
 AC 
AB
1
DG   3 m  0.75 m
4
DG 
Triangle CFD:
Triangle EAD:
Triangle EGB:
DG 0.75

AG
3

  14.0362
tan  
Force triangle
400 N
A
T


sin 53.13 sin 75.964 sin 50.906°
A  412.5 N

14.04 
T  500 N
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
PROBLEM 4.67
Determine the reactions at B and D when b  60 mm.
SOLUTION
Since CD is a two-force member, the line of action of reaction at D must pass through Points C and D.
Free-Body Diagram:
(Three-force body)
Reaction at B must pass through E, where the reaction at D and the 80-N force intersect.
220 mm
250 mm
  41.348
tan  
Force triangle
Law of sines:
80 N
B
D


sin 3.652° sin 45 sin131.348
B  888.0 N
D  942.8 N
 B  888 N
41.3
D  943 N
45.0 
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PROBLEM 4.68
For the frame and loading shown, determine the reactions at C and D.
SOLUTION
Since BD is a two-force member, the reaction at D must pass through Points B and D.
Free-Body Diagram:
(Three-force body)
Reaction at C must pass through E, where the reaction at D and the 150-N load intersect.
Triangle CEF:
tan  
1.5 m
1m
  56.310
Triangle ABE:
tan  
1
2
  26.565
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PROBLEM 4.68 (Continued)
Force Triangle
Law of sines:
150 N
C
D


sin 29.745 sin116.565 sin 33.690
C  270.42 N,
D  167.704 N
C  270 N
56.3; D  167.7 N
26.6 
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PROBLEM 4.69
A 50-kg crate is attached to the trolley-beam system shown.
Knowing that a  1.5 m, determine (a) the tension in cable CD,
(b) the reaction at B.
SOLUTION
Three-force body: W and TCD intersect at E.
0.7497 m
1.5 m
  26.556
tan  
Three forces intersect at E.
W  (50 kg) 9.81 m/s 2
Force triangle
 490.50 N
Law of sines:
TCD
B
490.50 N


sin 61.556° sin 63.444 sin 55
TCD  498.99 N
B  456.96 N
(a)
(b)
TCD  499 N 
B  457 N
26.6 
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PROBLEM 4.70
One end of rod AB rests in the corner A and the other end is attached to
cord BD. If the rod supports a 150-N load at its midpoint C, find the
reaction at A and the tension in the cord.
SOLUTION
Free-Body Diagram: (Three-force body) Dimensions in mm
The line of action of reaction at A must pass through E, where T and the 150-N load intersect.
Force triangle
EF 460

AF 240
  62.447
tan  
EH 100

DH 240
  22.620
tan  
A
T
150 N


sin 67.380 sin 27.553 sin 85.067°
A  139.0 N
62.4 
T  69.6 N 
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PROBLEM 4.71
For the boom and loading shown, determine (a) the tension in cord
BD, (b) the reaction at C.
SOLUTION
Three-force body: 3-kN load and T intersect at E.
Geometry:
Free-Body Diagram:
BF CF BF 32

;

AG CG 12 48
BF  8 cm
BH  32  BF  32  8  24 cm
JE
DJ JE 48

 ; JE  36 cm
;
BH DH 24 32
EG  JE  JG  36  32  4 cm
4 cm
tan  
48 cm
  4.7636
24 cm
tan  
32 cm
  36.870
Force Triangle
Law of sines:
(a)
(b)
TBD
3 kN
C


sin 94.764 sin 32.106 sin 53.13
TBD  5.63 kN 
C  4.52 kN
4.76 
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PROBLEM 4.72
A 40-N roller, of diameter 80 mm, which is to be used on a tile floor, is
resting directly on the subflooring as shown. Knowing that the thickness
of each tile is 3 mm, determine the force P required to move the roller
onto the tiles if the roller is (a) pushed to the left, (b) pulled to the right.
SOLUTION
Geometry: For each case as roller comes into contact with
tile,
3.7 cm
4 cm
  22.332
  cos 1

(a)
Roller pushed to left (three-force body):
Forces must pass through O.

Force Triangle
Law of sines:
P
40 N

; P  24.87 N
sin 37.668 sin 22.332
P  24.9 N

(b)
Roller pulled to right (three-force body):
Forces must pass through O.
Law of sines:
P
40 N

; P  15.3361 N
sin 97.668 sin 22.332
P  15.34 N

30.0° 
30.0° 


Force Triangle


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PROBLEM 4.73
A T-shaped bracket supports a 300-N load as shown. Determine
the reactions at A and C when α  45°.
SOLUTION
Free-Body Diagram:
(Three-force body)
The line of action of C must pass through E, where A and the 300-N force intersect.
Triangle ABE is isosceles:
EA  AB  400 mm
In triangle CEF:
tan  
CF
CF
150 mm


EF EA  AF 700 mm
  12.0948
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 4.73 (Continued)
Force Triangle
Law of sines:
A
C
300 N


sin 32.905 sin135 sin12.0948
A  778 N ;
C  1012 N
77.9 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 4.74
A T-shaped bracket supports a 300-N load as shown. Determine
the reactions at A and C when α  60°.
SOLUTION
Free-Body Diagram:
EA  (400 mm) tan 30°
 230.94 mm
In triangle CEF:
tan  
CF
CF

EF EA  AF
150
230.94  300
  15.7759
tan  
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PROBLEM 4.74 (Continued)
Force Triangle
Law of sines:
A
C
300 N


sin 44.224 sin120 sin15.7759
A  770 N
C  956 N
A  770 N ;
C  956 N
74.2 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 4.75
Rod AB is supported by a pin and bracket at A and rests against a frictionless
peg at C. Determine the reactions at A and C when a 170-N vertical force is
applied at B.
SOLUTION
Free-Body Diagram:
(Three-force body)
The reaction at A must pass through D where C and the 170-N force intersect.
160 mm
300 mm
  28.07
tan  
We note that triangle ABD is isosceles (since AC  BC) and, therefore,
 CAD    28.07
Also, since CD  CB, reaction C forms angle   28.07 with the horizontal axis.
Force triangle
We note that A forms angle 2 with the vertical axis. Thus, A and C form angle
180  (90   )  2  90  
Force triangle is isosceles, and we have
A  170 N
C  2(170 N)sin 
 160.0 N
A  170.0 N
33.9°;
C  160.0 N
28.1° 
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PROBLEM 4.76
Solve Problem 4.75, assuming that the 170-N force applied at B is horizontal
and directed to the left.
PROBLEM 4.75 Rod AB is supported by a pin and bracket at A and rests
against a frictionless peg at C. Determine the reactions at A and C when a
170-N vertical force is applied at B.
SOLUTION
Free-Body Diagram: (Three-Force body)
The reaction at A must pass through D, where C and the 170-N force intersect.
160 mm
300 mm
  28.07
tan  
We note that triangle ADB is isosceles (since AC  BC). Therefore  A   B  90   .
 ADB  2
Also
Force triangle
The angle between A and C must be 2    
Thus, force triangle is isosceles and
  28.07
A  170.0 N
C  2(170 N) cos   300 N
A  170.0 N
56.1°
C  300 N
28.1° 
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PROBLEM 4.77
D
140 mm
C
A
B
300 N
a = 240 mm
Member ABC is supported by a pin and bracket at B and by an
inextensible cord attached at A and C and passing over a frictionless
pulley at D. The tension may be assumed to be the same in portions AD
and CD of the cord. For the loading shown and neglecting the size of
the pulley, determine the tension in the cord and the reaction at B.
480 mm
SOLUTION
Free-Body Diagram:
Reaction at B must pass through D.
0.14 m
0.24 m
α = 30.256°
0.14 m
tan β =
0.48 m
β = 16.26°
tan α =
Force triangle
Law of sines
T
T − 300 N
B
=
=
sin 59.744° sin13.996° sin106.26
T(sin 13.996°) = (T − 300 N)(sin 59.744°)
T(0.24185) = (T − 300)(0.86378)
T = 416.66 N
B = (416.66 N)
T = 416.7 N
sin 106.26°
sin 59.744
B = 463.1 N
= 463.07 N
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426
30.3°
PROBLEM 4.78
Using the method of Section 4.7, solve Problem 4.22.
PROBLEM 4.22 A lever AB is hinged at C and attached to a
control cable at A. If the lever is subjected to a 500-N horizontal
force at B, determine (a) the tension in the cable, (b) the
reaction at C.
SOLUTION
Free-Body Diagram:
(Three-Force body)
Reaction at C must pass through E, where FAD and 500-N force intersect.
Since AC  CD  250 mm, triangle ACD is isosceles.
We have
 C  90  30  120
and
 A D 
1
(180  120)  30
2
Dimensions in mm
On the other hand, from triangle BCF:
CF  ( BC )sin 30  200 sin 30  100 mm
FD  CD  CF  250  100  150 mm
From triangle EFD, and since  D  30 :
EF  ( FD ) tan 30  150 tan 30  86.60 mm
From triangle EFC:
CF
100 mm

EF 86.60 mm
  49.11
tan  
Force triangle
Law of sines
FAD
C
500 N


sin 49.11 sin 60 sin 70.89°
FAD  400 N, C  458 N
(a)
(b)
FAD  400 N 
C  458 N
49.1° 
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PROBLEM 4.79
Knowing that   30, determine the reaction (a) at B, (b) at C.
SOLUTION
Free-Body Diagram:
(Three-force body)
Reaction at C must pass through D where force P and reaction at B intersect.
In  CDE:
( 3  1) R
R
 3 1
tan  
  36.2
Force Triangle
Law of sines:
P
B
C


sin 23.8 sin126.2 sin 30
B  2.00 P
C  1.239 P
(a)
B  2P
(b)
C  1.239P
60.0° 
36.2° 
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PROBLEM 4.80
Knowing that   60, determine the reaction (a) at B, (b) at C.
SOLUTION
Reaction at C must pass through D where force P and reaction at B intersect.
In CDE:
tan  
R
1
R
3
Free-Body Diagram:
(Three-force body)
R
1
3
  22.9
Force Triangle
Law of sines:
P
B
C


sin 52.9 sin 67.1 sin 60
B  1.155P
C  1.086 P
(a)
B  1.155P
30.0° 
(b)
C  1.086P
22.9° 
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PROBLEM 4.81
Determine the reactions at A and B when   50°.
SOLUTION
Free-Body Diagram: (Three-force body)
Reaction A must pass through Point D where the 100-N force
and B intersect.
In right  BCD:
  90  75  15
BD  250 tan 75  933.01 mm
In right  ABD:
Dimensions in mm
AB
150 mm

BD 933.01 mm
  9.1333
tan  
Force Triangle
Law of sines:
100 N
A
B


sin 9.1333° sin15 sin155.867
A  163.1 N; B  257.6 N
A  163.1 N
74.1° B  258 N
65.0° 
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PROBLEM 4.82
Determine the reactions at A and B when   80.
SOLUTION
Free-Body Diagram:
(Three-force body)
Reaction A must pass through D where the 100-N force and B intersect.
In right triangle BCD:
  90  75  15
BD  BC tan 75  250 tan75
BD  933.01 mm
In right triangle ABD:
tan  
150 mm
AB

BD 933.01 mm
  9.1333
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PROBLEM 4.82 (Continued)
Force Triangle
Law of sines:
100 N
A
B


sin 9.1333° sin15 sin155.867

A  163.1 N
55.9 

B  258 N
65.0 
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PROBLEM 4.83
Rod AB is bent into the shape of an arc of circle and is lodged between two
pegs D and E. It supports a load P at end B. Neglecting friction and the weight
of the rod, determine the distance c corresponding to equilibrium when a  20
mm and R  100 mm.
SOLUTION
Free-Body Diagram:
Since y ED  xED  a,
slope of ED is
45;
slope of HC is
45.
Also
DE  2 a
and
a
1
DH  HE    DE 
2
2
 
For triangles DHC and EHC,
sin  
a
2
R

a
2R
Now
c  R sin(45   )
For
a  20 mm and R  100 mm
sin  
20 mm
2(100 mm)
 0.141421
  8.1301
and
c  (100 mm)sin(45  8.1301)
 60.00 mm
or c  60.0 mm 
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PROBLEM 4.84
A slender rod of length L is attached to collars that can slide freely along
the guides shown. Knowing that the rod is in equilibrium, derive an
expression for the angle  in terms of the angle .
SOLUTION
As shown in the free-body diagram of the slender rod AB, the three forces intersect at C. From the force
geometry:
Free-Body Diagram:
tan  
xGB
y AB
where
y AB  L cos 
and
xGB 
tan  
1
L sin 
2
1
2
L sin 
L cos 
1
 tan 
2
or tan   2 tan  
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PROBLEM 4.85
An 8-kg slender rod of length L is attached to collars that can slide freely
along the guides shown. Knowing that the rod is in equilibrium and that
  30, determine (a) the angle  that the rod forms with the vertical,
(b) the reactions at A and B.
SOLUTION
(a)
As shown in the free-body diagram of the slender rod AB, the three forces intersect at C. From the
geometry of the forces:
Free-Body Diagram:
tan  
xCB
y BC
where
xCB 
and
1
L sin 
2
yBC  L cos 
tan  
1
tan 
2
or
tan   2 tan 
For
  30
tan   2 tan 30
 1.15470
  49.107
  49.1 
or
W  mg  (8 kg)(9.81 m/s 2 )  78.480 N
(b)
From force triangle:
A  W tan 
 (78.480 N) tan 30
 45.310 N
and
B
78.480 N
W

 90.621 N
cos 
cos30
or
A  45.3 N
or B  90.6 N

60.0 
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PROBLEM 4.86
A uniform rod AB of length 2R rests inside a hemispherical bowl of
radius R as shown. Neglecting friction, determine the angle 
corresponding to equilibrium.
SOLUTION
Based on the F.B.D., the uniform rod AB is a three-force body. Point E is the point of intersection of the
three forces. Since force A passes through O, the center of the circle, and since force C is perpendicular to
the rod, triangle ACE is a right triangle inscribed in the circle. Thus, E is a point on the circle.
Note that the angle  of triangle DOA is the central angle corresponding to the inscribed angle  of
triangle DCA.
  2
The horizontal projections of AE , ( x AE ), and AG , ( x AG ), are equal.
x AE  x AG  x A
or
( AE ) cos 2  ( AG ) cos 
and
(2 R) cos 2  R cos 
Now
then
or
cos 2  2 cos 2   1
4 cos 2   2  cos 
4 cos 2   cos   2  0
Applying the quadratic equation,
cos   0.84307 and cos    0.59307
  32.534 and   126.375 (discard)
or   32.5 
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PROBLEM 4.87
A slender rod BC of length L and weight W is held by two cables as
shown. Knowing that cable AB is horizontal and that the rod forms
an angle of 40 with the horizontal, determine (a) the angle  that
cable CD forms with the horizontal, (b) the tension in each cable.
SOLUTION
Free-Body Diagram:
(Three-force body)
(a)
The line of action of TCD must pass through E, where TAB and W intersect.
CF
EF
L sin 40
 1
L cos 40
2
tan  
 2 tan 40
 59.210
  59.2 
(b)
Force Triangle
TAB  W tan 30.790
 0.59588W
TAB  0.596W 


W
cos30.790 
 1.16408W
TCD 
TCD  1.164 W 
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PROBLEM 4.88
A thin ring of mass 2 kg and radius r  140 mm is held against a frictionless
wall by a 125-mm string AB. Determine (a) the distance d, (b) the tension in
the string, (c) the reaction at C.
SOLUTION
Free-Body Diagram:
(Three-force body)
The force T exerted at B must pass through the center G of the ring, since C and W intersect at that point.
Thus, points A, B, and G are in a straight line.
(a) From triangle ACG:
d  ( AG ) 2  (CG )2
 (265 mm)2  (140 mm)2
 225.00 mm
d  225 mm 
Force Triangle
W  (2 kg)(9.81 m/s 2 )  19.6200 N
Law of sines:
(b)
(c)
T
C
19.6200 N


265 mm 140 mm 225.00 mm
T  23.1 N 
C  12.21 N
Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 4.89
A slender rod of length L and weight W is attached to a
collar at A and is fitted with a small wheel at B. Knowing
that the wheel rolls freely along a cylindrical surface of
radius R, and neglecting friction, derive an equation in , L,
and R that must be satisfied when the rod is in equilibrium.
SOLUTION
Free-Body Diagram (Three-force body)
Reaction B must pass through D where B and W intersect.
Note that ABC and BGD are similar.
AC  AE  L cos
In  ABC:
(CE )2  ( BE ) 2  ( BC )2
(2 L cos  )2  ( L sin  ) 2  R 2
2
R
2
2
   4cos   sin 
L
2
R
2
2
   4cos   1  cos 
L
2
R
2
   3cos   1
L
cos 2  
1  R  2 
 1 

3  L 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 4.90
Knowing that for the rod of Problem 4.89, L  15 cm, R  20
cm, and W  10 N, determine (a) the angle  corresponding
to equilibrium, (b) the reactions at A and B.
SOLUTION
See the solution to Problem 4.86 for the free-body diagram and analysis leading to the following
equation:
cos 2  
1  R  2 
 1

3  L 

For L  15 cm, R  20 cm, and W  10 N,
2

1  20 cm
 ;   59.39

cos 2   
1

3  15 cm



(a)
In  ABC:
  59.4 
1
BE
L sin 

 tan 
CE 2 L cos  2
1
tan   tan 59.39  0.8452
2
  40.2
tan  
Force Triangle
A  W tan   (10 N) tan 40.2  8.45 N

B
(b)
(10 N)
W

 13.09 N
cos  cos 40.2
A  8.45 N
B  13.09 N

49.8° 
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PROBLEM 4.91
Two transmission belts pass over a double-sheaved pulley
that is attached to an axle supported by bearings at A and
D. The radius of the inner sheave is 125 mm and the radius
of the outer sheave is 250 mm. Knowing that when the
system is at rest, the tension is 90 N in both portions of
belt B and 150 N in both portions of belt C, determine the
reactions at A and D. Assume that the bearing at D does
not exert any axial thrust.
SOLUTION
We replace TB and TB by their resultant (180 N)j and TC and TC by their resultant (300 N)k.
Dimensions in mm
We have five unknowns and six equations of equilibrium. Axle AD is free to rotate about the x-axis, but
equilibrium is maintained (Mx  0).
M A  0: (150i)  (180 j)  (250i)  (300k )  (450i)  ( Dy j  Dz k )  0
27  103 k  75  103 j  450 Dy k  450 Dz j  0
Equating coefficients of j and k to zero,
j:
75  103  450Dz  0
Dz  166.7 N
k:
 27  103  450 Dy  0
Dy  60.0 N
Fx  0:
Ax  0
Fy  0: Ay  Dy  180 N  0
Ay  180  60  120.0 N
Fz  0: Az  Dz  300 N  0
Az  300  166.7  133.3 N
A  (120.0 N) j  (133.3 N)k ; D  (60.0 N) j  (166.7 N)k 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 4.92
Solve Problem 4.91, assuming that the pulley rotates at a
constant rate and that TB  104 N, T′B  84 N, TC  175 N.
PROBLEM 4.91 Two transmission belts pass over a
double-sheaved pulley that is attached to an axle supported
by bearings at A and D. The radius of the inner sheave is
125 mm and the radius of the outer sheave is 250 mm.
Knowing that when the system is at rest, the tension is 90 N
in both portions of belt B and 150 N in both portions of
belt C, determine the reactions at A and D. Assume that the
bearing at D does not exert any axial thrust.
SOLUTION
Dimensions in mm
We have six unknowns and six equations of equilibrium.
M A  0: (150i  250k )  (104 j)  (150i  250k )  (84 j)
 (250i  125 j)  (175k )  (250i  125 j)  (TC )
 450i  ( Dy j  Dz k )  0
150(104  84)k  250(104  84)i  250(175  TC ) j  125(175  TC )
 450 Dy k  450 Dz j  0
Equating the coefficients of the unit vectors to zero,
i : 250(104  84)  125(175  TC )  0
175  TC  40
j: 250(175  135)  450 Dz  0
Dz  172.2 N
k :  150(104  84)  450 Dy  0
Dy  62.7 N
TC  135;
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 4.92 (Continued)
Fx  0:
Ax  0
Fy  0:
Ay  104  84  62.7  0
Ay  125.3 N
Fz  0:
Az  175  135  172.2  0
Az  137.8 N
A  (125.3 N) j  (137.8 N)k ; D  (62.7 N) j  (172.2 N)k 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 4.93
A small winch is used to raise a 120-N load.
Find (a) the magnitude of the vertical force P
that should be applied at C to maintain
equilibrium in the position shown, (b) the
reactions at A and B, assuming that the bearing at
B does not exert any axial thrust.
SOLUTION
We have six unknowns and six equations of equilibrium.
rC  (32 cm)i  (10 cm) cos 30j  (10 cm)sin 30 k
 32i  8.6603 j  5k
M A  0: (10i  4k )  (120 j)  (20i )  ( By j  Bz k )  (32i  8.6603j  5k )  ( Pj)  0
 1200k  480i  20 By k  20 Bz j  32 Pk  5Pi  0
Equating the coefficients of the unit vectors to zero,
i : 480  5 P )  0
P  96.0 N;
j: 20 Bz  0
Bz  0
(a ) P  96.0 N 
k :  1200  20 By  32(96.0)  0
Fx  0:
Fy  0:
Fz  0:
By  213.6 N
Ax  0
Ay  120  213.6  96.0  0
Az  Bz  0
Ay  2.40 N
Az   Bz  0
A  (2.40 N) j; B  (214 N) j 
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y
PROBLEM 4.94
3m
A 4 × 8-m sheet of plywood weighing 170 N has been temporarily
placed among three pipe supports. The lower edge of the sheet
rests on small collars at A and B and its upper edge leans against
pipe C. Neglecting friction at all surfaces, determine the reactions
at A, B, and C.
4m
4m
B
1m
A
C
5m
z
3.75 m
x
3m
SOLUTION
rG/B =
3.75 1.3919
i+
j+k
2
2
We have five unknowns and six equations of equilibrium.
Plywood sheet is free to move in z direction, but equilibrium is maintained (ΣFz = 0).
ΣM B = 0: rA/B × ( Ax i + Ay j) + rC/B × ( −Ci ) + rG/B × ( − wj) = 0
i
0
Ax
j
0
Ay
k
i
j
k
i
j
k
4 + 3.75 1.3919 2 + 1.875 0.696 1 = 0
0
−C
0
0
0
−170 0
−4Ayi + 4Axj − 2Cj + 1.3919Ck + 170i − 318.75k = 0
Equating coefficients of unit vectors to zero:
i:
−4Ay + 170 = 0
Ay = 42.5 N
j:
− 2C + 4 Ax = 0
Ax =
k : 1.3919C −318.75 = 0
1
1
C = (229) = 114.5 N
2
2
C = 229 N
ΣFx = 0:
Ax + Bx − C = 0:
Bx = 229 − 114.5 = 114.5 N
ΣFy = 0:
Ay + By − W = 0:
By = 170 − 42.5 = 127.5 N
A(114.5 N)i + (42.5 N)j
B = (114.5 N)i + (127.5 N)j C = −(229 N)i
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444
PROBLEM 4.95
A 250 × 400-mm plate of mass 12 kg and a 300-mm-diameter
pulley are welded to axle AC that is supported by bearings at A and
B. For  = 30°, determine (a) the tension in the cable, (b) the
reactions at A and B. Assume that the bearing at B does not exert
any axial thrust.
SOLUTION
Free-Body Diagram:
W  mg  (12 kg)(9.81 m/s2 )  117.720 j N
rB / A  570k ; rD / A  150 j  770k ;
rG / A  (200sin  )i  (200 cos  ) j  285k
M A  0: rB /A  ( Bx i  By j)  rD/A  T i  rG /A  (W ) j  0
i
0
Bx
j
0
By
k
i
j
k
i
570  0 150 770  200sin 
0
T
0
0
0
j
200 cos 
W
k
285  0
0
570By i  570Bx j  770Tj  150Tk  285Wi  (200sin  )Wk  0
Equating the coefficients of the unit vectors to zero:
i:
570By  285(117.72)  0
j:
 570 Bx  770T  0
k : 150T  (200sin 30 )117.720  0
Fy  0:
Fz  0:
Fx  0:
(b)
Ay  By  W  0
By  58.860 N
Bx 
77
T ; Bx  106.017 N
57
T  78.480 N
(a ) T  78.5 N 
Ay  117.720  58.860  58.860 N
Az  0
Ax  Bx  T  0;
Ax  27.537 N
A  (27.5 N)i  (58.9 N) j; B  (106.0 N)i  (58.9 N) j 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 4.96
Solve Prob. 4.95 for  = 60°.
PROBLEM 4.95 A 250 × 400-mm plate of mass 12 kg and a
300-mm-diameter pulley are welded to axle AC that is supported
by bearings at A and B. For  = 30°, determine (a) the tension in
the cable, (b) the reactions at A and B. Assume that the bearing at
B does not exert any axial thrust.
SOLUTION
Free-Body Diagram:
W  mg  (12 kg)(9.81 m/s2 )  117.720 j N
rB / A  570k ; rD / A  150 j  770k ;
rG / A  (200sin  )i  (200 cos  ) j  285k
M A  0: rB /A  ( Bx i  By j)  rD/A  T i  rG /A  (W ) j  0
i
0
Bx
j
0
By
k
i
j
k
i
570  0 150 770  200sin 
0
T
0
0
0
j
200 cos 
W
k
285  0
0
570By i  570Bx j  770Tj  150Tk  285Wi  (200sin  )Wk  0
i:
570By  285(117.72)  0
By  58.860 N
j:
 570 Bx  770T  0
k:
150T  (200sin 60 )117.720  0
Bx 
77
T ; Bx  183.625 N
57
T  135.931 N
(a ) T  135.9 N 
Fy  0:
Fz  0:
Fx  0:
(b)
Ay  By  W  0
Ay  117.720  58.860  58.860 N
Az  0
Ax  Bx  T  0;
Ax  47.694 N
A  (47.7 N)i  (58.9 N) j; B  (183.6 N)i  (58.9 N) j 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 4.97
The 20 × 20-cm square plate shown weighs 56 N and is
supported by three vertical wires. Determine the tension in
each wire.
SOLUTION
Wb j  0 and W  (56.0 N) j
M A  0: rB / A  TB j  rC / A  TC j  rG / A  (Wj)  0
(16i  10k )  TB j  (16i  10k )  TC j  10i  ( Wj)  0
16TB k  10TB i  16TC k  10TC i  10Wk  0
Equate coefficients of unit vectors to zero:
i : 10TB  10TC  0; TB  TC
k : 16TB  16TC  10W  0
16TB  16TB  10(56.0 N)=0
TB  TC  17.50 N
Fy  0: TA  TB  TC  W  0
TA  17.50  17.50  56.0  0
TA  21.0 N
TA  21.0 N 
T
T
17.50 N 
B
C
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PROBLEM 4.98
The 20 × 20-cm square plate shown weighs 56 cm and is
supported by three vertical wires. Determine the weight and
location of the lightest block that should be placed on the
plate if the tensions in the three wires are to be equal.
SOLUTION
Let  Wb j be the weight of block; x and z its coordinates.
Since the tensions in the wires are equal, let TA  TB  TC  T
M O  0: rA  Tj  rB  Tj  rC  Tj  rG  (Wj)  ( xi  zk )  ( Wb j)  0
(10k )  Tj  (16i )  Tj  (16i  20k )  Tj  (10i  10k )  ( Wj)  ( xi  zk )  ( Wb j)  0
10Ti  16Tk  16Tk  20Ti  10Wk  10Wi  Wb xk  Wb zi  0
Equate coefficients of unit vectors to zero:
i :  30T  10W  Wb z  0; (1)
k : 32T  10W  Wb x  0 (2)
Also,
Fy  0: 3T  W  Wb  0 (3)
(1)+10(3): ( z  10)Wb  0 since x and z must be  20 cm z  10 cm
3(2)  32(3) : 2W  (3x  32)Wb  0
Wb
W
2
2
; since x  20 cm b 

W 3 x  32
W 3(20)  32
1
Finally, with W  56 N; Wb  (56 N)
 4.00 N
14
Wb  4.00 N at x  20.0 cm, z  10.00 cm 
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PROBLEM 4.99
An opening in a floor is covered by a 1  1.2-m sheet of
plywood of mass 18 kg. The sheet is hinged at A and B and
is maintained in a position slightly above the floor by a
small block C. Determine the vertical component of the
reaction (a) at A, (b) at B, (c) at C.
SOLUTION
rB/A  0.6i
rC/A  0.8i  1.05k
rG/A  0.3i  0.6k
W  mg  (18 kg)9.81
W  176.58 N
M A  0: rB/A  Bj  rC/A  Cj  rG/A  (Wj)  0
(0.6i )  Bj  (0.8i  1.05k )  Cj  (0.3i  0.6k )  (Wj)  0
0.6 Bk  0.8Ck  1.05Ci  0.3Wk  0.6Wi  0
Equate coefficients of unit vectors to zero:
 0.6 
i : 1.05C  0.6W  0 C  
176.58 N  100.90 N
 1.05 
k : 0.6 B  0.8C  0.3W  0
0.6 B  0.8(100.90 N)  0.3(176.58 N)  0 B  46.24 N
Fy  0: A  B  C  W  0
A  46.24 N  100.90 N  176.58 N  0
A  121.92 N
(a ) A  121.9 N (b) B  46.2 N (c) C  100.9 N 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 4.100
Solve Problem 4.99, assuming that the small block C is
moved and placed under edge DE at a point 0.15 m from
corner E.
PROBLEM 4.99 An opening in a floor is covered by a
1  1.2-m sheet of plywood of mass 18 kg. The sheet is
hinged at A and B and is maintained in a position slightly
above the floor by a small block C. Determine the vertical
component of the reaction (a) at A, (b) at B, (c) at C.
SOLUTION
rB/A  0.6i
rC/A  0.65i  1.2k
rG/A  0.3i  0.6k
W  mg  (18 kg) 9.81 m/s 2
W  176.58 N
M A  0: rB/A  Bj  rC/A  Cj  rG/A  (Wj)  0
0.6i  Bj  (0.65i  1.2k )  Cj  (0.3i  0.6k )  ( Wj)  0
0.6 Bk  0.65Ck  1.2Ci  0.3Wk  0.6Wi  0
Equate coefficients of unit vectors to zero:
i : 1.2C  0.6W  0
 0.6 
C 
176.58 N  88.29 N
 1.2 
k : 0.6 B  0.65C  0.3W  0
0.6 B  0.65(88.29 N)  0.3(176.58 N)  0 B  7.36 N
Fy  0: A  B  C  W  0
A  7.36 N  88.29 N  176.58 N  0
A  95.648 N
(a ) A  95.6 N (b) B  7.36 N (c) C  88.3 N 
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PROBLEM 4.101
Two steel pipes AB and BC, each having a mass per unit
length of 8 kg/m, are welded together at B and supported
by three wires. Knowing that a  0.4 m, determine the
tension in each wire.
SOLUTION
W1  0.6mg
W2  1.2mg
M D  0: rA/D  TA j  rE/D  ( W1 j)  rF/D  ( W2 j)  rC/D  TC j  0
( 0.4i  0.6k )  TA j  ( 0.4i  0.3k )  ( W1 j)  0.2i  ( W2 j)  0.8i  TC j  0
0.4TAk  0.6TA i  0.4W1k  0.3W1i  0.2W2 k  0.8TC k  0
Equate coefficients of unit vectors to zero:
1
1
i :  0.6TA  0.3W1  0; TA  W1  0.6mg  0.3mg
2
2
k :  0.4TA  0.4W1  0.2W2  0.8TC  0
0.4(0.3mg )  0.4(0.6mg )  0.2(1.2mg )  0.8TC  0
TC 
(0.12  0.24  0.24)mg
 0.15mg
0.8
Fy  0: TA  TC  TD  W1  W2  0
0.3mg  0.15mg  TD  0.6mg  1.2mg  0
TD  1.35mg
mg  (8 kg/m)(9.81m/s 2 )  78.480 N/m
TA  0.3mg  0.3  78.480
TA  23.5 N 
TC  0.15mg  0.15  78.480
TC  11.77 N 
TD  1.35mg  1.35  78.480
TD  105.9 N 
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PROBLEM 4.102
For the pipe assembly of Problem 4.101, determine (a) the
largest permissible value of a if the assembly is not to tip,
(b) the corresponding tension in each wire.
SOLUTION
W1  0.6mg
W2  1.2mg
M D  0: rA/D  TA j  rE/D  ( W1 j)  rF/D  ( W2 j)  rC/D  TC j  0
(  ai  0.6k )  TA j  ( ai  0.3k )  (W1 j)  (0.6  a )i  ( W2 j)  (1.2  a )i  TC j  0
TA ak  0.6TA i  W1ak  0.3W1i  W2 (0.6  a )k  TC (1.2  a )k  0
Equate coefficients of unit vectors to zero:
1
1
i :  0.6TA  0.3W1  0; TA  W1  0.6mg  0.3mg
2
2
k :  TA a  W1a  W2 (0.6  a )  TC (1.2  a )  0
0.3mga  0.6mga  1.2mg (0.6  a )  TC (1.2  a )  0
TC 
(a)
0.3a  0.6a  1.2(0.6  a )
1.2  a
For maximum a and no tipping, TC  0.
0.3a  1.2(0.6  a )  0
0.3a  0.72  1.2a  0
1.5a  0.72
a  0.480 m 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 4.102 (Continued)
(b)
Reactions:
mg  (8 kg/m) 9.81 m/s 2  78.48 N/m
TA  0.3mg  0.3  78.48  23.544 N
TA  23.5 N 
Fy  0: TA  TC  TD  W1  W2  0
TA  0  TD  0.6mg  1.2mg  0
TD  1.8mg  TA  1.8  78.48  23.544  117.72
TD  117.7 N 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
y
PROBLEM 4.103
The 240 N square plate shown is supported by three vertical
wires. Determine (a) the tension in each wire when a = 10 cm,
(b) the value of a for which the tension in each wire is 80 N.
A
B
C
a
a
z
30 cm
x
30 cm
SOLUTION
rB/A = ai + 30k
rC/A = 30i + ak
rG/A = 15i + 15k
By symmetry: B = C
ΣM A = 0: rB/A × Bj + rC/A × Cj + rG/A × (−Wj) = 0
(ai + 30k ) × Bj + (30i + ak ) × Bj + (15i + 15k ) × (−Wj) = 0
Bak − 30 Bi + 30 Bk − Bai − 15Wk + 15Wi = 0
Equate coefficient of unit vector i to zero:
i : − 30 B − Ba + 15W = 0
B=
15W
30 + a
C=B=
15W
30 + a
(1)
ΣFy = 0: A + B + C − W = 0
A+2
(a)
For
a = 10 cm
Eq. (1)
C=B=
Eq. (2)
A=
15W
− W = 0;
30 + a
A=
aW
30 + a
(2)
15(240 N)
= 90 N
30 + 10
10(240 N)
= 60 N
30 + 10
A = 60 N
B = C = 90 N
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 4.103 (Continued)
(b)
For tension in each wire = 80 N
Eq. (1)
80 N =
15(240 N)
30 + a
30 cm + a = 45
a = 15.00 cm
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459
PROBLEM 4.104
The table shown weighs 120 N and has a diameter of 1.6 m. It is supported by
three legs equally spaced around the edge. A vertical load P of magnitude
400 N is applied to the top of the table at D. Determine the maximum
value of a if the table is not to tip over. Show, on a sketch, the area of the
table over which P can act without tipping the table.
SOLUTION
r = 0.8 m b = r sin30° = 0.4 m
We shall sum moments about AB.
(b + r )C + (a − b) P − bW = 0
(0.4 + 0.8)C + (a − 0.4)400 − (0.4)120 = 0
C=
If table is not to tip, C
1
[48 − (a − 0.4)400]
1.2
0
[48 − (a − 0.4)400] $ 0
48 $ (a − 0.4)400
a − 0.4 # 0.12 a # 0.52 m
a = 0.52 m
Only ⊥ distance from P to AB matters. Same condition must be satisfied for each leg. P must be located
in shaded area for no tipping
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y
PROBLEM 4.10 5
6m
A 10-m boom is acted upon by the 4-kN force shown. Determine the
tension in each cable and the reaction at the ball-and-socket joint at A.
E
6m
D
7m
B
A
C
6m
4m
z
x
4 kN
SOLUTION
We have five unknowns and six Eqs. of equilibrium but equilibrium is maintained (ΣM x = 0).
Free-Body Diagram:
BD = (−6 m)i + (7 m)j + (6 m)k
BE = (−6 m)i + (7 m)j − (6 m)k
TBD = TBD
TBE = TBE
BD = 11 m
BE = 11 m
BD TBD
(−6i + 7 j + 6k )
=
BD 11
BE TBE
(−6i + 7 j − 6k )
=
BE 11
ΣM A = 0: rB/A × TBD + rB/A × TBE + rC/A × (−4j) = 0
6i ×
TBD
T
(−6i + 7 j + 6k ) + 6i × BE (−6i + 7j − 6k) + 10i × (−4j) = 0
11
11
42
36
42
36
TBD k − TBD j + TBE k + TBE j − 40k
11
11
11
11
Equate coefficients of unit vectors to zero.
j: −
k:
36
36
TBD + TBE = 0 TBE = TBD
11
11
42
42
TBD +
TBE −40 = 0
11
11
2
42
TBD = 40
11
TBD = 5. 38 m
TBD = 5.24 kN
TBE = 5.24 kN
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
461
PROBLEM 4.105 (Continued)
ΣFx = 0: Ax −
6
(5.24 kN) − 6 (5.24 kN) = 0
11
11
Ax = 5.72 kN
ΣFy = 0: Ay +
7
(5.24 kN) + 7 (5.24 kN) − 4 kN = 0
11
11
Ay = −2.67 kN
ΣFz = 0:
Az +
6
(5.24 kN) − 6 (5.24 kN) = 0
11
11
Az = 0
A = (5.72 kN)i − (2.67 kN)j
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462
PROBLEM 4.106
The 6-m pole ABC is acted upon by a 455-N force as shown.
The pole is held by a ball-and-socket joint at A and by two
cables BD and BE. For a  3 m, determine the tension in
each cable and the reaction at A.
SOLUTION
Free-Body Diagram:
Five unknowns and six Eqs. of equilibrium, but equilibrium is maintained
(M AC  0)
rB  3j
rC  6 j

CF  3i  6 j  2k CF  7 m

BD  1.5i  3j  3k BD  4.5 m

BE  1.5i  3j  3k BE  4.5 m

CF P
 (3i  6 j  2k )
PP
CE 7

T
BD TBD
(1.5i  3j  3k )  BD (i  2 j  2k )
TBD  TBD

BD 4.5
3

BE TBD

(i  2 j  2k )
TBE  TBE 
3
BE
M A  0: rB  TBD  rB  TBE  rC  P  0
i
j
k
i j k
i
j k
TBD
TBE
P
 0 3 0
 0 6 0
0
0 3 0
3
3
7
1 2 2
1 2 2
3 6 2
Coefficient of i:
2TBD  2TBE 
12
P0
7
(1)
Coefficient of k:
TBD  TBF 
18
P0
7
(2)
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 4.106 (Continued)
Eq. (1)  2 Eq. (2):
Eq. (2):
 4TBD 

48
12
P  0 TBD  P
7
7
12
18
6
P  TBE  P  0 TBE  P
7
7
7
P  445 N TBD 
Since
TBE 
12
(455)
7
6
(455)
7
TBD  780 N 
TBE  390 N 
F  0: TBD  TBE  P  A  0
780 390 455
(3)  Ax  0


3
3
7
Coefficient of i:
260  130  195  Ax  0
Coefficient of j:

780
390
455
(2) 
(2) 
(6)  Ay  0
3
3
7
520  260  390  Ay  0
Coefficient of k:

Ax  195.0 N
Ay  1170 N
780
390
455
(2) 
(2) 
(2)  Az  0
3
3
7
520  260  130  Az  0
Az  130.0 N
A  (195.0 N)i  (1170 N) j  (130.0 N)k 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 4.107
Solve Problem 4.106 for a  1.5 m.
PROBLEM 4.106 The 6-m pole ABC is acted upon by a
455- N force as shown. The pole is held by a ball-and-socket
joint at A and by two cables BD and BE. For a  3 m,
determine the tension in each cable and the reaction atA .
SOLUTION
Free-Body Diagram:
Five unknowns and six of equilibrium but equilibrium is maintained
(M AC  0)
rB  3j
M A  0: rB  TBD
i
j
rC  6 j

CF  1.5i  6 j  2k CF  6.5 m

BD  1.5i  3j  3k BD  4.5 m

BE  1.5i  3j  3k BE  4.5 m

CF
P
P
(1.5i  6 j  2k )  (3i  12 j  4k )

PP
13
CE 6.5

T
BD TBD
(1.5i  3j  3k )  BD (i  2 j  2k )
TBD  TBD

BD 4.5
3

BE TBD

(i  2 j  2k )
TBE  TBE 
3
BE
 rB  TBE  rC  P  0
k
i j k
i
j
k
TBD
TBE
P
 0 3 0
 0
0
0 3 0
6
0
3
3
13
1 2 2
1 2 2
3 12  4
Coefficient of i:
2TBD  2TBE 
24
P0
13
(1)
Coefficient of k:
TBD  TBE 
18
P0
13
(2)
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 4.107 (Continued)
Eq. (1) 2 Eq. (2):
Eq (2):
4TBD 

60
15
P  0 TBD  P
13
13
15
18
3
P  TBE  P  0 TBE  P
13
13
13
P  445 N TBD 
Since
TBE 
15
(455)
13
3
(455)
13
TBD  525 N 
TBE  105.0 N 
F  0: TBD  TBE  P  A  0
525 105 455
(3)  Ax  0


3
3
13
Coefficient of i:
175  35  105  Ax  0
Coefficient of j:

525
105
455
(2) 
(2) 
(12)  Ay  0
3
3
13
350  70  420  Ay  0
Coefficient of k:
Ax  105.0 N

Ay  840 N
525
105
455
(2) 
(2) 
(4)  Az  0
3
3
13
350  70  140  Az  0
Az  140.0 N
A  (105.0 N)i  (840 N) j  (140.0 N)k 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 4.108
A 2.4-m boom is held by a ball-and-socket joint at C and by
two cables AD and AE. Determine the tension in each cable
and the reaction at C.
SOLUTION
Free-Body Diagram:
(MAC  0).
Five unknowns and six equations of equilibrium, but equilibrium is maintained
rB  1.2k
TAD
TAE
rA  2.4k

AD  0.8i  0.6 j  2.4k
AD  2.6 m

AE  0.8i  1.2 j  2.4k
AE  2.8 m

AD TAD
(0.8i  0.6 j  2.4k )


AD 2.6

AE TAE
(0.8i  1.2 j  2.4k )


AE 2.8
M C  0: rA  TAD  rA  TAE  rB  (3 kN) j  0
i
j
k
i
j
k
TAD
T
0
0
2.4
0
2.4 AE  1.2k  (3.6 kN) j  0
 0
2.6
2.8
0.8 1.2 2.4
0.8 0.6 2.4
Equate coefficients of unit vectors to zero:
i :  0.55385TAD  1.02857TAE  4.32  0
(1)
j :  0.73846TAD  0.68671TAE  0
TAD  0.92857TAE
From Eq. (1):
(2)
0.55385(0.92857)TAE  1.02857TAE  4.32  0
1.54286TAE  4.32
TAE  2.800 kN
TAE  2.80 kN 
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PROBLEM 4.108 (Continued)
From Eq. (2):
TAD  0.92857(2.80)  2.600 kN
TAD  2.60 kN 
0.8
0.8
(2.6 kN) 
(2.8 kN)  0
2.6
2.8
0.6
1.2
(2.6 kN) 
(2.8 kN)  (3.6 kN)  0
Fy  0: C y 
2.6
2.8
2.4
2.4
Fz  0: C z 
(2.6 kN) 
(2.8 kN)  0
2.6
2.8
Fx  0: C x 
Cx  0
C y  1.800 kN
C z  4.80 kN
C  (1.800 kN) j  (4.80 kN)k 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 4.109
Solve Problem 4.108, assuming that the 3.6-kN load is
applied at Point A.
PROBLEM 4.108 A 2.4-m boom is held by a ball-andsocket joint at C and by two cables AD and AE. Determine
the tension in each cable and the reaction at C.
SOLUTION
Free-Body Diagram:
(MAC  0).
Five unknowns and six equations of equilibrium, but equilibrium is maintained

AD  0.8i  0.6 j  2.4k
AD  2.6 m

AE  0.8i  1.2 j  2.4k
AE  2.8 m

AD TAD
TAD 

(0.8i  0.6 j  2.4k )
AD 2.6

AE TAE
TAE 

(0.8i  1.2 j  2.4k )
AE 2.8
M C  0: rA  TAD  rA  TAE  rA  (3.6 kN) j
Factor rA :
rA  (TAD  TAE  (3.6 kN) j)
TAD  TAE  (3 kN) j  0
or
Coefficient of i:

(Forces concurrent at A)
TAD
T
(0.8)  AE (0.8)  0
2.6
2.8
TAD 
Coefficient of j:
2.6
TAE
2.8
(1)
TAD
T
(0.6)  AE (1.2)  3.6 kN  0
2.6
2.8
2.6
 0.6  1.2
TAE 
TAE  3.6 kN  0

2.8
 2.6  2.8
 0.6  1.2 
TAE 
  3.6 kN
 2.8 
TAE  5.600 kN
TAE  5.60 kN 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 4.109 (Continued)
From Eq. (1):
TAD 
2.6
(5.6)  5.200 kN
2.8
0.8
0.8
(5.2 kN) 
(5.6 kN)  0
2.6
2.8
0.6
1.2
(5.2 kN) 
(5.6 kN)  3.6 kN  0
Fy  0: C y 
2.6
2.8
2.4
2.4
Fz  0: C z 
(5.2 kN) 
(5.6 kN)  0
2.6
2.8
Fx  0: C x 
TAD  5.20 kN 
Cx  0
Cy  0
C z  9.60 kN
C  (9.60 kN)k 
Note: Since the forces and reaction are concurrent at A, we could have used the methods of Chapter 2.
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 4.110
The 10-m flagpole AC forms an angle of 30 with the z axis. It
is held by a ball-and-socket joint at C and by two thin braces
BD and BE. Knowing that the distance BC is 3 m, determine
the tension in each brace and the reaction at C.
SOLUTION
Free-Body Diagram:
Express all forces in terms of rectangular components:
rE   3 m i   3 m j
rB   3 m sin 30 j   3 m cos 30 k  1.5 m j   2.598 m k
rD    3 m i   3 m j
or
or
rA  10 m sin 30 j  10 m cos 30 k   5 m j   8.66 m k

BE  rE  rB   3 m i  3 m j  1.5 m j   2.598 m k

BE   3 m i  1.5 m j   2.598 m k, and BE  4.243 m

BD  rD  rB    3 m i  3 m j  1.5 m j   2.598 m k

BD    3 m i  1.5 m j   2.598 m k, and BE  4.243 m
Then


BD
T BD  TBD
 TBD   0.707i  0.3535j  0.6123k 
BD


BE
T BE  TBE
 TBE  0.707i  0.3535j  0.6123k 
BE
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 4.110 (Continued)
rB  TBD  rB  TBE   5 m j   8.66 m k    75 N  j
MC  0:
or
i
j
k
i
j
k
0
1.5
2.598 TBD  0
1.5
2.598  649.5 i  0
0.707 0.3535  0.6123
 0.707 0.3535  0.6123
Equating the coefficients of the unit vectors to zero:
j:
1.837 TBD  1.837 TBE  0
i:
1.837 TBD  1.837 TBE  649.5 N  0
TBD  176.8 N 
TBE  176.8 N 
Force equations:
Cx  176.8  0.707   176.8  0.707   0,
or
Cx  0
C y  176.8  0.3535   176.8  0.3535   75 N  0,
or
Cz  176.8   0.6123  176.8  0.6123  0,
C z  216.5 N
or
C y   50 N
C    50 N  j   216.5 N  k 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 4.111
y
0.5 m
A 1.2-m boom is held by a ball-and-socket joint at C and
by two cables BF and DAE; cable DAE passes around a
frictionless pulley at A. For the loading shown, determine
the tension in each cable and the reaction at C.
0.4 m
E
D
1.2 m
B
A
z
0.5 m
C
F
0.75 m
x
1.6 kN
SOLUTION
Free-Body Diagram:
Five unknowns and six Eqs. of equilibrium but equilibrium is
maintained (ΣMAC = 0).
T = Tension in both parts of cable DAE.
rB/C = 0.75k
rA/C = 1.2k
AD = −0.5i − 1.2k
AD = 1.3 m
AE = 0.5j − 1.2k
AE = 1.3 m
BF = 0.4i − 0.75k
BF = 0.85 m
AD T
T
= (−0.5i − 1.2k) = (−5i − 12k)
AD 1.3
13
AE T
T
(0.5j − 1.2k) = (5j − 12k)
=T
=
AE 1.3
13
T
BF TBF
= TBF
=
(0.4i − 0.75k) = BF (8i − 15k)
17
BF 0.85
TAD = T
TAE
TBF
ΣM C = 0: rA/C × TAD + rA/C × TAE + rB/C × TBF + rB/C × (−1.6 kN)j = 0
i j k
i j k
i j k
T
T
T
+ 0 0 1.2
+ 0 0 0.75 BF + (0.75k) × (−1.6j) = 0
0 0 1.2
13
13
17
−5 0 −12
0 5 −12
8 0 −15
Coefficient of i:
−
6
T + 1.2 = 0
13
T = 2.6 kN
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473
PROBLEM 4.111 (Continued)
Coefficient of j:
−
6
6
T+
TBF = 0
13
17
TBF =
17
17
T = (2.6)
13
13
TBF = 3.4 kN
ΣF = 0: TAD + TAE + TBF − 1.6j + C = 0
Coefficient of i:
−
5
8
(2.6) + (3.4) + Cx = 0
13
17
−1 + 1.6 + Cx = 0
Coefficient of j:
5
(2.6) − 1.6 + Cy = 0
13
1 − 1.6 + Cy = 0
Coefficient of k:
Cx = −0.6 kN
−
Cy = 0.6 kN
12
12
15
(2.6) −
(2.6) −
(3.4) + Cz = 0
13
13
17
−2.4 − 2.4 − 3 + Cz = 0
Cz = 7.8 kN
Answers: TDAE = T
TDAE = 2.6 kN
TBF = 3.4 kN
C = −(0.6 kN)i + (0.6 kN)j + (7.8 kN)k
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
474
y
PROBLEM 4.112
0.5 m
Solve Problem 4.111 , assuming that the 1.6 kN load is
applied at A.
0.4 m
E
1.2 m
B
A
z
PROBLEM 4.11 1 A 1.2-m boom is held by a ball-andsocket joint at C and by two cables BF and DAE; cable
DAE passes around a frictionless pulley at A. For the
loading shown, determine the tension in each cable and
the reaction at C.
D
0.5 m
C
F
0.75 m
x
1.6 kN
SOLUTION
Free-Body Diagram:
Five unknowns and six Eqs. of equilibrium but equilibrium is
maintained (ΣMAC = 0).
T = tension in both parts of cable DAE.
rB/C = 0.75k
rA/C = 1.2k
AD = −0.5i − 1.2k
AD = 1.3 m
AE = 0.5j − 1.2k
AE = 1.3 m
BF = 0.4i − 0.75k
BF = 0.85 m
AD T
T
= (−0.5i − 1.2k) = (−5i − 12k)
AD 1.3
13
AE T
T
(0.5j − 1.2k) = (5j − 12k)
=T
=
AE 1.3
13
T
BF TBF
= TBF
=
(0.4i − 0.75k) = BF (8i − 15k)
17
BF 0.85
TAD = T
TAE
TBF
ΣM C = 0: rA/C × TAD + rA/C × TAE + rB/C × TBF + rA/C × (−1.6 kN)j = 0
i j k
i j k
i j k
T
T
T
+ 0 0 1.2
+ 0 0 30 BF + 1.2k × (−1.6j) = 0
0 0 1.2
13
13
17
−5 0 −12
0 5 −12
8 0 −15
Coefficient of i:
−
6
T + 1.92 = 0
13
T = 4.16 kN
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
475
PROBLEM 4.112 (Continued)
Coefficient of j:
−
6
6
T+
TBF = 0
13
17
TBF =
17
17
T = (4.16) TBF = 5.44 kN
13
13
ΣF = 0: TAD + TAE + TBF − 1.6j + C = 0
−
Coefficient of i:
5
8
(4.16) + (5.44) + Cx = 0
13
17
−1.6 + 2.56 + Cx = 0
5
(4.16) − 1.6 + Cy = 0
13
Coefficient of j:
1.6 − 1.6 + Cy = 0
Coefficient of k:
−
Cy = 0
12
12
15
(4.16) −
(4.16) −
(5.44) + Cz = 0
13
13
17
−3.84 − 3.84 − 4.8 + Cz = 0
Answers:
Cx = −0.96 kN
TDAE = T
Cz = 12.48 kN
TDAE = 4.16 kN
TBF = 5.44 kN
C = −(0.96 kN)i + (12.48 kN)k
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
476
PROBLEM 4.113
A 10-kg storm window measuring 900 × 1500 mm is held by hinges at
A and B. In the position shown, it is held away from the side of the
house by a 600-mm stick CD. Assuming that the hinge at A does not
exert any axial thrust, determine the magnitude of the force exerted by
the stick and the components of the reactions at A and B.
SOLUTION
Free-Body Diagram: Since CD is a two-force member, FCD
is directed along CD and triangle ACD is isosceles. We have
0.3 m
sin  
 0.2;   11.5370 and 2  23.074
1.5 m
W  (10 kg)(9.81 m/s 2 )  (98.10 N) j
rG  (0.75 m)sin 23.074 i  (0.75 m) cos 23.074 j  (0.45 m)k
rG  (0.29394 m)i  (0.690 m)j  (0.45 m)k
FCD  FCD (cos11.5370 i  sin11.5370 j)
FCD  FCD (0.97980i  0.20 j)
M B  0: rA  A  rG  W  rC  FCD  0
0.9k  ( Ax i  Ay j)  (0.29394i  0.690 j  0.45k )  (98.10 j)
 (1.5 j  0.9k )  FCD (0.97980i  0.20 j)  0
0.90 Ax j  0.90 Ay i  28.836 k  44.145 i  1.46970 FCD k  0.88182 FCD j  0.180FCD i  0
Equating the coefficients of the unit vectors to zero,
k:
 28.836  1.46970 FCD  0; FCD  19.6203 N
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 4.113 (Continued)
j:  0.90 Ax  0.88182(19.6203 N)  0;
Ax  19.2240 N
i:  0.90 Ay  44.145  0.180(19.6203 N)  0;
Ay  45.123 N
FCD  19.62 N 
A  (19.22N)i  (45.1 N) j 
Fx  0: Ax  Bx  FCD cos11.5370  0
 19.2240  Bx  19.6203cos11.5370  0
Bx  0
Fy  0: Ay  By  FCD sin11.5370  W  0
45.122  By  19.6230sin11.5370  98.1  0
By  49.053 N
Fz  0: Bz  0
B   (49.1 N) j 
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PROBLEM 4.114
The bent rod ABEF is supported by bearings at C and D and
by wire AH. Knowing that portion AB of the rod is 250 mm
long, determine (a) the tension in wire AH, (b) the reactions
at C and D. Assume that the bearing at D does not exert any
axial thrust.
SOLUTION
ABH is equilateral.
Free-Body Diagram:
Dimensions in mm
rH/C  50i  250 j
rD/C  300i
rF/C  350i  250k
T  T (sin 30) j  T (cos 30)k  T (0.5 j  0.866k )
M C  0: rH/C  T  rD  D  rF/C  ( 400 j)  0
i
j
k
i
j
50 250
0 T  300 0
0
0.5 0.866
0 Dy
Coefficient i:
k
i
j
k
0  350
0
250  0
Dz
0 400 0
216.5T  100  103  0
T  461.9 N
Coefficient of j:
T  462 N 
43.3T  300 Dz  0
43.3(461.9)  300 Dz  0
Dz  66.67 N
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 4.114 (Continued)
Coefficient of k:
25T  300 Dy  140  103  0
25(461.9)  300 Dy  140  103  0
Dy  505.1 N
D  (505 N) j  (66.7 N)k 
F  0: C  D  T  400 j  0
Coefficient i:
Cx  0
Coefficient j:
C y  (461.9)0.5  505.1  400  0 C y  336 N
Coefficient k:
C z  (461.9)0.866  66.67  0
Cx  0
C z  467 N
C  (336 N) j  (467 N)k 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 4.115
The horizontal platform ABCD weighs 60 N and supports a 240-N
load at its center. The platform is normally held in position by
hinges at A and B and by braces CE and DE. If brace DE is
removed, determine the reactions at the hinges and the force
exerted by the remaining brace CE. The hinge at A does not exert
any axial thrust.
SOLUTION
Free-Body Diagram:
Express forces, weight in terms of rectangular components:

EC   3 m i  4 m  j   2 m k
FCE  FCE

EC
 FCE
EC
3i  4 j  2k
 3 2   4  2   2  2
 0.55709 FCE i  0.74278 FCE j  0.37139 FCE k
W   (mg ) j   (300 N)j
M B  0:
or
 4 m k  A  1.5 m i  2 m k     300 N  j
  3 m i  4 m k   FCE  0
  4 m Ay i  4 m Az j  1.5 m 300 N  k  2 m 300 N  i
i
j
k
3
0
4

FCE m  0
0.55709 0.74278 0.37139
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 4.115 (Continued)
Setting the coefficients of the unit vectors equal to zero:
k:
 0.74278 FCE   3 m   300 N1.5 m 
0
FCE  201.94 N
FCE  202 N 
or
j:
Ax  4 m  0.55709  201.94 N    4 m  0.37139 201.94 N  3 m  0
Ax   56.250 N
i:
 Ay 4 m  0.74278 201.94 N  4 m  300 N 2 m  0
Ay  0
Fx  0:
 56.250 N  Bx  0.55709  201.94 N  0
Bx   56.249 N
Fy  0:
0  B y  300 N  0.74278  201.94 N   0
B y  150.003 N
Fz  0:
Bz  0.371391 201.94 N   0
Bz   74.999 N
Therefore:
A    56.3 N i 
B    56.2 N  i  150.0 N  j   75.0 N k 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 4.116
The lid of a roof scuttle weighs 75 N. It is hinged at corners
A and B and maintained in the desired position by a rod CD
pivoted at C; a pin at end D of the rod fits into one of several
holes drilled in the edge of the lid. For the position shown,
determine (a) the magnitude of the force exerted by rod CD,
(b) the reactions at the hinges. Assume that the hinge at B
does not exert any axial thrust.
SOLUTION
Free-Body Diagram:
Geometry:
Using triangle ACD and the law of sines
sin  sin 50

or   20.946
7 cm 15 cm
    20.946  70.946
Expressing FCD in terms of its rectangular coordinates:
FCD  FCD sin  j  FCD cos  k
 FCD sin 70.946 j  FCD cos 70.946k
FCD  0.94521FCD j  0.32646 FCDk
M B  0:
or
  26 cm i  A   13 cm i  16 cm sin 50 j  16 cm cos50 k    75 N  j
 26 cm i  7 cm k  FCD  0
 26 cm Ayk  26 cm Az j  13 cm 75 N  k  16 cm 75 N cos 50 i
 26 cm  0.94521FCD  k  26 cm 0.32646 FCD  j   7 cm  0.94521FCD  i  0
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 4.116 (Continued)
(a) Setting the coefficients of the unit vectors to zero:
i:
 75 N  16 cm cos50   0.94521FCD  7 cm 
0
FCD  116.6 N 
(b) Fx  0:
k:
Ax  0
 0.94521116.580 N    26 cm  75 N  13 cm  Ay  26 cm  0
Ay   72.693 N
j:
Fy  0:
0.32646 116.580 N    26 cm  Az 26 cm  0
Az   38.059 N
 72.693 N  0.94521 116.580 N   75 N  B y  0
By  37.500 N
Fz  0:
 38.059 N  0.32646 116.580 N   Bz  0
Bz  0
Therefore:
A    72.7 N  j   38.1 N  k 
B
 37.5 N  j 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 4.117
A 100-kg uniform rectangular plate is supported in the
position shown by hinges A and B and by cable DCE that
passes over a frictionless hook at C. Assuming that the
tension is the same in both parts of the cable, determine (a)
the tension in the cable, (b) the reactions at A and B.
Assume that the hinge at B does not exert any axial thrust.
SOLUTION
rB/A (960  180)i  780i
Dimensions in mm
 960
 450
 90  i 
rG/A  
k
2
2


 390i  225k
rC/A  600i  450k
T  Tension in cable DCE

CD  690i  675 j  450k

CE  270i  675 j  450k
CD  1065 mm
CE  855 mm
T
(690i  675 j  450k )
1065
T
TCE 
(270i  675 j  450k )
855
W   mgi  (100 kg)(9.81 m/s 2 ) j  (981 N) j
TCD 
M A  0: rC/A  TCD  rC/A  TCE  rG/A  ( Wj)  rB/A  B  0
i
j
k
i
j
k
T
T
600
0
450
450
 600 0
1065
855
690 675 450
270 675 450
i
j
k
i
j
0
225  780 0
 390
0 981 0
0 By
k
0 0
Bz
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 4.117 (Continued)
Coefficient of i:
(450)(675)
T
T
 (450)(675)
 220.73  103  0
1065
855
(a)
Coefficient of j:
( 690  450  600  450)
T  344.64 N T  345 N 
344.64
344.64
 (270  450  600  450)
 780 Bz  0
1065
855
Bz  185.516 N
Coefficient of k: (600)(675)
344.64
344.64
 (600)(675)
 382.59  103  780 By  0 By  113.178 N
1065
855
B  (113.2 N) j  (185.5 N)k 
(b)
F  0: A  B  TCD  TCE  W  0
690
270
(344.64) 
(344.64)  0
1065
855
Coefficient of i:
Ax 
Coefficient of j:
Ay  113.178 
675
675
(344.64) 
(344.64)  981  0
1065
855
Ay  377.30 N
Coefficient of k:
Az  185.516 
450
450
(344.64) 
(344.64)  0
1065
855
Az  141.496 N
Ax  114.454 N
A  (114.4 N)i  (377 N) j  (141.5 N)k 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 4.118
Solve Problem 4.117, assuming that cable DCE is replaced
by a cable attached to Point E and hook C.
PROBLEM 4.117 A 100-kg uniform rectangular plate is
supported in the position shown by hinges A and B and by
cable DCE that passes over a frictionless hook at C.
Assuming that the tension is the same in both parts of the
cable, determine (a) the tension in the cable, (b) the
reactions at A and B. Assume that the hinge at B does not
exert any axial thrust.
SOLUTION
See solution to Problem 4.113 for free-body diagram and analysis leading to the following:
CD  1065 mm
CE  855 mm
T
(690i  675 j  450k )
1065
T
TCE 
(270i  675 j  450k )
855
W  mgi  (100 kg)(9.81 m/s 2 ) j  (981 N)j
TCD 
Now,
M A  0: rC/A  TCE  rG/A  (W j)  rB/A  B  0
i
j
k
i
j
k
i
j
T
600 0
450
 390
0
225  780 0
855
270 675 450
0 981 0
0 By
Coefficient of i:
(450)(675)
k
0 0
Bz
T
 220.73  103  0
855
T  621 N 
T  621.31 N
Coefficient of j:
Coefficient of k:
(270  450  600  450)
(600)(675)
621.31
 780 Bz  0 Bz  364.74 N
855
621.31
 382.59  103  780 By  0 By  113.186 N
855
B  (113.2 N)j  (365 N)k 
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PROBLEM 4.118 (Continued)
F  0: A  B  TCE  W  0
270
(621.31)  0
855
Coefficient of i:
Ax 
Coefficient of j:
Ay  113.186 
Coefficient of k:
Az  364.74 


Ax  196.2 N
675
(621.31)  981  0
855
450
(621.31)  0
855
Ay  377.3 N
Az  37.7 N 
A  (196.2 N)i  (377 N)j  (37.7 N)k 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 4.119
Solve Prob. 4.113, assuming that the hinge at A has been removed and
that the hinge at B can exert couples about axes parallel to the x and y
axes.
PROBLEM 4.113A 10-kg storm window measuring 900 × 1500 mm is
held by hinges at A and B. In the position shown, it is held away from
the side of the house by a 600-mm stick CD. Assuming that the hinge at
A does not exert any axial thrust, determine the magnitude of the force
exerted by the stick and the components of the reactions at A and B.
SOLUTION
Free-Body Diagram: Since CD is a two-force member, FCD
is directed along CD and triangle ACD is isosceles. We have
W  (10 kg)(9.81 m/s 2 )  (98.10 N) j
rG  (0.75 m)sin 23.074 i  (0.75 m) cos 23.074 j  (0.45 m)k
rG  (0.29394 m)i  (0.690 m)j  (0.45 m)k
FCD  FCD (cos11.5370 i  sin11.5370 j)
FCD  FCD (0.97980i  0.20 j)
With hinge at A removed, hinge at B will develop two couples to prevent rotation about the x and z axes.
M B  0: ( M B ) x i  ( M B ) y j  rG  W  rC  FCD  0
( M B ) x i  ( M B ) y j  (0.29394i  0.690 j  0.45k )  ( 98.10) j
 (1.5 j  0.9k )  FCD (0.97980i  0.20 j)  0
( M B ) x i  ( M B ) y j  28.836 k  44.145 i  1.46970 FCD k  0.88182 FCD j  0.180FCD i  0
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 4.119 (Continued)
Equating the coefficients of the unit vectors to zero,
k:
 28.836  1.46970 FCD  0; FCD  19.6203 N
j: ( M B ) y  0.88182(19.6203)  0;
( M B ) y  17.3016 N  m
i: ( M B ) x  44.145  0.180(19.6203)  0;
( M B ) x  40.613 N  m
FCD  19.62 N 
M B  (40.6 N  m)i  (17.30 N  m) j 
Fx  0: Bx  FCD cos11.5370  0
Bx  19.6203cos11.5370  0
Bx  19.22 N
Fy  0: By  FCD sin11.5370  W  0
By  19.6230sin11.5370  98.1  0
By  94.2 N
Fz  0: Bz  0
B  (19.22 N)i  (94.2 N) j 
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PROBLEM 4.120
Solve Prob. 4.115, assuming that the hinge at B has been removed
and that the hinge at A can exert an axial thrust, as well as couples
about axes parallel to the x and y axes.
PROBLEM 4.115 The horizontal platform ABCD weighs 60 N
and supports a 240-N load at its center. The platform is normally
held in position by hinges at A and B and by braces CE and DE. If
brace DE is removed, determine the reactions at the hinges and
the force exerted by the remaining brace CE. The hinge at A does
not exert any axial thrust.
SOLUTION
Free-Body Diagram:
Bx  By  Bz  0
Express forces, weight in terms of rectangular components:

EC  3 m i  4 m j  2 m k
FCE  FCE

EC
 FCE
EC
3i  4 j  2k
 3 2   4  2   2  2
 0.55709 FCE i  0.74278 FCE j  0.37139 FCE k
W   (mg ) j   (300 N)j
M A  0: rG / A  ( W ) j  rC / A  FCE  ( M Ax )i  ( M Ay ) j  0
M A  0:
(1.5i  2k )  (300 j)  3i  FCE (0.55709i  0.74278j  0.37139k )
 (M Ax )i  (M Ay ) j  0
or
 450k  600i  2.2283FCE k  1.11417 FCE j  M Ax i  M Ay j  0
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 4.120 (Continued)
Setting the coefficients of the unit vectors equal to zero:
k:
 450   2.2283FCE   0
FCE  201.95 N
or
j:
FCE  202 N 
 1.11417(201.95)  (M Ay )  0
M Ay  225.00 N  m
i:
 600  M A x  0
M Ax  600 N  m
Fx  0:
Ax  0.55709  201.94 N   0
Ax  112.499 N
Fy  0:
Ay  300 N  0.74278  201.94 N   0
Ay  150.003 N
Fz  0:
Az  0.371391 201.94 N   0
Az   74.999 N
Therefore:
M A   600 N  m i  (225 N  m)j 
A   112.5 N  i  150.0 N  j   75.0 N  k 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 4.121
y
30 N
50 mm B
The assembly shown is used to control the tension T in a tape that passes
around a frictionless spool at E. Collar C is welded to rods ABC and CDE.
It can rotate about shaft FG but its motion along the shaft is prevented
by a washer S. For the loading shown, determine (a) the tension T in the
tape, (b) the reaction at C.
40 mm
A
F
S
C
105 mm
D
z
G
60 mm
E
T
x
T
SOLUTION
Free-Body Diagram:
50
-(30N) j
rA/C = 105j + 50k
mm
B
rE/C = 40i − 60j
(MC)z k
A
CZk
ΣMC = 0:
Cx i
105 mm
Cy j
Z
(MC)y j
rA/C × (−30j) + rE/C × T(i + k) + (MC)y j + (MC)zk = 0
(105j + 50k) × (−30j) + (40i − 60j) × T(i + k) + (MC)y j + (MC)zk = 0
Tk
E
60 mm
Ti
40 mm
Coefficient of i:
1500 − 60T = 0
Coefficient of j:
−40(25 N) + (MC)y = 0
Coefficient of k:
60(25 N) + (MC)z = 0
T = 25 N
(MC)y = 1000 N · mm = 1 N · m
(MC)z = −1500 N · mm = −1.5 N · m
MC = (1.0 N · m)j − (1.5 N · m)k
ΣF = 0: Cxi + Cyj + Czk − (30 N)j + (25 N)i + (25 N)k = 0
Equate coefficients of unit vectors to zero.
Cx = −25 N
Cy = 30 N
Cz = −25 N
C = − (25 N)i + (30 N)j − (25 N)k
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 4.122
The assembly shown is welded to collar A that fits on the
vertical pin shown. The pin can exert couples about the x
and z axes but does not prevent motion about or along the yaxis. For the loading shown, determine the tension in each
cable and the reaction at A.
SOLUTION
Free-Body Diagram:
First note:
(0.08 m)i  (0.06 m) j
TCF  λ CF TCF 
(0.08) 2  (0.06)2 m
TCF
 TCF (0.8i  0.6 j)
TDE  λ DE TDE 
(0.12 m)j  (0.09 m)k
(0.12)2  (0.09) 2 m
TDE
 TDE (0.8 j  0.6k )
(a)
From F.B.D. of assembly:
Fy  0: 0.6TCF  0.8TDE  480 N  0
0.6TCF  0.8TDE  480 N
or
(1)
M y  0:  (0.8TCF )(0.135 m)  (0.6TDE )(0.08 m)  0
TDE  2.25TCF
or
(2)
Substituting Equation (2) into Equation (1),
0.6TCF  0.8[(2.25)TCF ]  480 N
TCF  200.00 N
TCF  200 N 
or

and from Equation (2):
or
TDE  2.25(200.00 N)  450.00
TDE  450 N 
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PROBLEM 4.122 (Continued)
(b)
From F.B.D. of assembly:
Fz  0: Az  (0.6)(450.00 N)  0
Az  270.00 N
Fx  0: Ax  (0.8)(200.00 N)  0
Ax  160.000 N
or A  (160.0 N)i  (270 N)k 
M x  0:
MAx  (480 N)(0.135 m)  [(200.00 N)(0.6)](0.135 m)
 [(450 N)(0.8)](0.09 m)  0
M Ax  16.2000 N  m
M z  0:
MAz  (480 N)(0.08 m)  [(200.00 N)(0.6)](0.08 m)
 [(450 N)(0.8)](0.08 m)  0
M Az  0
or M A  (16.20 N  m)i 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 4.123
The rigid L-shaped member ABC is
supported by a ball-and-socket joint at A and
by three cables. If a 1.8-kN load is applied at
F, determine the tension in each cable.
SOLUTION
Free-Body Diagram:
In this problem:
We have
Thus,
Dimensions in mm
a  210 mm

CD  (240 mm)j  (320 mm)k CD  400 mm

BD  (420 mm)i  (240 mm)j  (320 mm)k BD  580 mm

BE  (420 mm)i  (320 mm)k BE  528.02 mm
TCD
TBD
TBE

CD
 TCD
 TCD (0.6 j  0.8k )
CD

BD
 TBD
 TBD (0.72414i  0.41379 j  0.55172k )
BD

BE
 TBE
 TBE (0.79542i  0.60604k )
BE
M A  0: (rC  TCD )  (rB  TBD )  (rB  TBE )  (rW  W )  0
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 4.123 (Continued)
rC  (420 mm)i  (320 mm)k
Noting that
rB  (320 mm)k
rW  ai  (320 mm)k
and using determinants, we write
i
j
k
i
j
k
420 0 320 TCD 
0
0
320 TBD
0
0.6 0.8
0.72414 0.41379 0.55172
i
j
k
i
j
k

0
0
320 TBE   a
0
320  0
0.79542 0 0.60604
0 1.8 0
Equating to zero the coefficients of the unit vectors,
i:
192TCD  132.413TBD  576  0
(1)
j:
336TCD  231.72TBD  254.53TBE  0
(2)
k:
252TCD  1.8a  0
(3)
Recalling that a  210 mm, Eq. (3) yields
TCD 
From Eq. (1):
1.8(210)
 1.500 kN
252
192(1.5)  132.413TBD  576  0
TBD  2.1751 kN
From Eq. (2):

TCD  1.500 kN 
TBD  2.18 kN 
336(1.5)  231.72(2.1751)  254.53TBE  0
TBE  3.9603 kN 
TBE  3.96 kN 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 4.124
Solve Problem 4.123, assuming that the 1.8kN load is applied at C.
PROBLEM 4.123 The rigid L-shaped
member ABC is supported by a ball-andsocket joint at A and by three cables. If a 1.8kN load is applied at F, determine the
tension in each cable.
SOLUTION
See solution of Problem 4.123 for free-body diagram and derivation of Eqs. (1), (2), and (3):
192TCD  132.413TBD  576  0
(1)
336TCD  231.72TBD  254.53TBE  0
(2)
252TCD  1.8a  0
(3)
In this problem, the 1.8-kN load is applied at C and we have a  420 mm. Carrying into Eq. (3) and
solving for TCD ,
TCD  3.00
From Eq. (1):
From Eq. (2):
(192)(3)  132.413TBD  576  0
336(3)  0  254.53TBE  0 
TCD  3.00 kN 
TBD  0 
TBE  3.96 kN 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
y
PROBLEM 4.12 5
16 cm
9 cm
The rigid L-shaped member ABF is supported by a ball-and-socket
joint at A and by three cables. For the loading shown, determine the
tension in each cable and the reaction at A.
J
16 cm
H
O
F
E
120 N
16 cm
A
G
z
12 cm
D
C
x
8 cm
120 N 8 cm
B
8 cm
8 cm
SOLUTION
Free-Body Diagram:
rB/A = 12i
rC/A = 12 j − 8k
rD/A = 12i − 16k
rE/A = 12i − 24k
rF/A = 12i − 32k
BG = −12i + 9k
BG = 15 cm
BG = −0.8i + 0.6k
DH = −12i + 16 j; DH = 20 cm; λDH = − 0.6i + 0.8 j
FJ = −12i + 16 j;
FJ = 20 cm;
λFJ = −0.6i + 0.8 j
ΣM A = 0: rB/A × TBG λBG + rD/A × TDH λDH + rF/A × TFJ λFJ
+rC/A × (−120j) + rE/A × (−120j) = 0
i
j k
i
j
k
i
j
k
12 0 0 TBG + 12
0 −16 TDH + 12
0 −32 TFJ
−0.8 0 0.6
−0.6 0.8 0
−0.6 0.8 0
i
j
k
i
j
k
+ 12 0 −8 + 12 0 −24 = 0
0 −120 0
0 −120
0
Coefficient of i:
Coefficient of k:
3
4
Eq. (1) − Eq. (2):
+12.8TDH + 25.6TFJ − 960 − 2880 = 0
(1)
+9.6TDH + 9.6TFJ − 1440 − 1440 = 0
9.6TFJ = 0
(2)
TFJ = 0
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 4.125 (Continued)
Eq. (1):
12.8TDH − 3840 = 0
TDH = 300 N
Coefficient of j:
−7.2TBG + (16 × 0.6)(300 N) = 0
TBG = 400 N
ΣF = 0:
A + TBG λ BG + TDH λ DH −120j −120j = 0
Coefficient of i:
Ax + (400)(−0.8) + (300)(−0.6) = 0 Ax = 500 N
Coefficient of j:
Ay + (300)(0.8) − 120 − 120 = 0
Ay = 0
Coefficient of k:
Az + (400)(+0.6) = 0
Az = −240 N
A = (500 N)i − (240 N)j
Note: The value Ay = 0 can be confirmed by considering ΣM BF = 0
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
536
PROBLEM 4.126
y
16 cm
9 cm
Solve Problem 4.125, assuming that the load at C has been removed.
J
16 cm
PROBLEM 4.1525 The rigid L-shaped member ABF is supported
by a ball-and-socket joint at A and by three cables. For the loading
shown, determine the tension in each cable and the reaction at A.
H
O
E
F
120 N
16 cm
A
120 N
G
z
D
C
B
12 cm
x
8 cm
8 cm
8 cm
8 cm
SOLUTION
Free-Body Diagram:
rB/A = 12i
rC/A = 12i − 16k
rE/A = 12i − 24k
rF/A = 12i − 32k
BG = −12i + 9k;
BG = 15 cm; lBG = −0.8i + 0.6k
DH = −12i + 16j; DH = 20 cm;
FJ = −12i + 16j; FJ = 20 cm;
lDH = −0.6i + 0.8j
lFJ = −0.6i + 0.8j
ΣMA= 0: rB/A × TBG lBG + rD/A × TDH lDH + rF/A × TFJ lFJ + rE/A× (−120j) = 0
⎜
i
12
−0.8
j
0
0
⎟ ⎜
⎟ ⎜
j
k
k
i
i
0 TBG + 12
0 −16 TDH + 12
−0.6 0.8
−0.6
0.6
0
j
0
0.8
⎟ ⎜
k
i
−32 TFJ + 12
0
0
j
0
−120
⎟
k
−24 = 0
0
i:
+12.8TDH + 25.6TFJ −2880 = 0
(1)
k:
+9.6TDH + 9.6TFJ −1440 = 0
(2)
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PROBLEM 4.126 (Continued)
Multiply Eq. (1) by
3
4
and subtract Eq. (2):
TFJ = 75 N 
9.6TFJ − 720 = 0
TDH = 75 N 
12.8TDH + 25.6(75) − 2880 = 0
From Eq. (1):
j:
−7.2TBG + (16)(0.6)(75) + (32)(0.6)(75) = 0
TBG = 300 N 
− 7.2TBG + 2160 = 0
ΣF = 0: A + TBG λ BG + TDHλ DH + TFJ λ FJ − 120j = 0
i : Ax + (300)(− 0.8) + (75)(− 0.6) + (75)(− 0.6) = 0
j:
Ay + (75)(0.8) + (75)(0.8) − 120 = 0
k:
Az + (300)(0.6) = 0
Ax = 330 N
Ay = 0
Az = −180 N
A = (300 N) i − (180 N)k 
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PROBLEM 4.127
Three rods are welded together to form a “corner” that is supported
by three eyebolts. Neglecting friction, determine the reactions at
A, B, and C when P = 240 N, a = 12 cm, b = 8 cm, and c = 10 cm.
SOLUTION
From f.b.d. of weldment
ΣM O = 0: rA/O × A + rB/O × B + rC/O × C = 0
i
12
0
j
0
Ay
k
i
0 + 0
Az Bx
j k
i
8 0 + 0
0 Bz
Cx
j k
0 10 = 0
Cy 0
(8 Bz − 10 C y)i + (−12 Az + 10 C x)j + (12 Ay − 8 Bx ) k = 0
From i-coefficient
8 B z − 10 C y = 0
Bz = 1.25C y
or
j-coefficient
−12 Az + 10 C x = 0
C x = 1.2 Az
or
k-coefficient
(2)
12 Ay − 8 Bx = 0
Bx = 1.5 Ay
or
(1)
(3)
ΣF = 0: A + B + C − P = 0
or
( Bx + C x )i + ( Ay + C y − 240 N ) j + ( Az + Bz )k = 0
From i-coefficient
C x = − Bx
or
j-coefficient
or
Bx + C x = 0
(4)
Ay + C y − 240 N = 0
Ay + C y = 240 N
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501
(5)
PROBLEM 4.127 (Continued)
k-coefficient
Az + Bz = 0
Az = − Bz
or
(6)
Substituting C x from Equation (4) into Equation (2)
− Bx = 1.2 Az
(7)
Using Equations (1), (6), and (7)
Cy =
B
Bz
− Az
1 Bx
=
=
= x
1.25 1.25 1.25 1.2
1.5
(8)
From Equations (3) and (8)
Cy =
1.5 Ay
1.5
or C y = Ay
and substituting into Equation (5)
2 Ay = 240 N
Ay = C y = 120 N
(9)
Using Equation (1) and Equation (9)
Bz = 1.25(120 N ) = 150.0 N
Using Equation (3) and Equation (9)
Bx = 1.5(120 N) = 180.0 N
From Equation (4)
C x = −180.0 N
From Equation (6)
Az = −150.0 N
A = (120.0 N)j − (150.0 N)k
Therefore
B = (180.0 N)i + (150.0 N)k
C = −(180.0 N)i + (120.0 N)j
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502
PROBLEM 4.128
Solve Problem 4.127, assuming that the force P is removed and is
replaced by a couple M = +(6 N ⋅ m)j acting at B.
PROBLEM 4.12 7 Three rods are welded together to form a “corner”
that is supported by three eyebolts. Neglecting friction, determine the
reactions at A, B, and C when P = 240 N, a = 12 cm, b = 8 cm, and
c = 10 cm.
SOLUTION
From f.b.d. of weldment
ΣM O = 0: rA/O × A + rB/O × B + rC/O × C + M = 0
i
12
0
j
0
Ay
k
i
0 + 0
Az Bx
j k
i
8 0 + 0
0 Bz
Cx
j k
0 10 +(600 N ⋅ cm)j = 0
Cy 0
(−12 Az j + 12 Ay k ) + (8 Bz i − 8 Bx k ) + ( −10C y i + 10C x j) + (600 N ⋅ cm)j = 0
From i-coefficient
8 Bz − 10 C y = 0
C y = 0.8Bz
or
j-coefficient
(1)
−12 Az + 10 C x + 600 = 0
C x = 1.2 Az − 60
or
k-coefficient
(2)
12 Ay − 8 Bx = 0
Bx = 1.5 Ay
or
(3)
ΣF = 0: A + B + C = 0
( Bx + C x )i + ( Ay + C y ) j + ( Az + Bz )k = 0
From i-coefficient
C x = − Bx
(4)
j-coefficient
C y = − Ay
(5)
k-coefficient
Az = − Bz
(6)
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503
PROBLEM 4.128 (Continued)
Substituting C x from Equation (4) into Equation (2)
Az = 50 −
Bx
1.2
(7)
Using Equations (1), (6), and (7)
C y = 0.8 Bz = − 0.8 Az =
2
Bx − 40
3
(8)
From Equations (3) and (8)
C y = Ay − 40
Substituting into Equation (5)
2 Ay = 40
Ay = 20.0 N
From Equation (5)
C y = −20.0 N
Equation (1)
Bz = −25.0 N
Equation (3)
Bx = 30.0 N
Equation (4)
C x = −30.0 N
Equation (6)
Az = 25.0 N
A = (20.0 N)j + (25.0 N)k
Therefore
B = (30.0 N)i − (25.0 N)k
C = −(30.0 N)i − (20.0 N)j
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504
PROBLEM 4.129
Frame ABCD is supported by a ball-andsocket joint at A and by three cables. For
a  150 mm, determine the tension in
each cable and the reaction at A.
SOLUTION
First note:
TDG   DG TDG 
(0.48 m)i  (0.14 m)j
(0.48)2  (0.14) 2 m
TDG
0.48i  0.14 j
TDG
0.50
T
 DG (24i  7 j)
25

TBE   BE TBE 
(0.48 m)i  (0.2 m)k
(0.48) 2  (0.2)2 m
TBE
0.48i  0.2k
TBE
0.52
T
 BE (12 j  5k )
13

From F.B.D. of frame ABCD:
 7

M x  0:  TDG  (0.3 m)  (350 N)(0.15 m)  0
25


TDG  625 N 
or
 24

 5

M y  0:   625 N  (0.3 m)   TBE  (0.48 m)  0
 25

 13

TBE  975 N 
or
 7

M z  0: TCF (0.14 m)    625 N  (0.48 m)  (350 N)(0.48 m)  0
25


or
TCF  600 N 
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PROBLEM 4.129 (Continued)
Fx  0: Ax  TCF  (TBE ) x  (TDG ) x  0
 12
  24

Ax  600 N    975 N     625 N   0
13
25

 

Ax  2100 N
Fy  0: Ay  (TDG ) y  350 N  0
 7

Ay    625 N   350 N  0
 25

Ay  175.0 N
Fz  0: Az  (TBE ) z  0
 5

Az    975 N   0
 13

Az  375 N
Therefore,
A  (2100 N)i  (175.0 N) j  (375 N)k 
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PROBLEM 4.130
Frame ABCD is supported by a ball-andsocket joint at A and by three cables.
Knowing that the 350-N load is applied at
D (a  300 mm), determine the tension in
each cable and the reaction at A.
SOLUTION
First note
TDG   DG TDG 
(0.48 m)i  (0.14 m)j
(0.48)2  (0.14) 2 m
TDG
0.48i  0.14 j
TDG
0.50
T
 DG (24i  7 j)
25

TBE   BE TBE 
(0.48 m)i  (0.2 m)k
(0.48) 2  (0.2) 2 m
TBE
0.48i  0.2k
TBE
0.52
T
 BE (12i  5k )
13

From f.b.d. of frame ABCD
 7

M x  0:  TDG  (0.3 m)  (350 N)(0.3 m)  0
25


TDG  1250 N 
or
 24

 5

M y  0:   1250 N  (0.3 m)   TBE  (0.48 m)  0
 25

 13

TBE  1950 N 
or
 7

M z  0: TCF (0.14 m)    1250 N  (0.48 m)  (350 N)(0.48 m)  0
25


or
TCF  0 

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PROBLEM 4.130 (Continued)
Fx  0: Ax  TCF  (TBE ) x  (TDG ) x  0
 12
  24

Ax  0    1950 N     1250 N   0 
13
25

 

Ax  3000 N 

Fy  0: Ay  (TDG ) y  350 N  0
 7

Ay    1250 N   350 N  0
 25

Ay  0
Fz  0: Az  (TBE ) z  0
 5

Az    1950 N   0
 13

Az  750 N
Therefore
A  (3000 N)i  (750 N)k 
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PROBLEM 4.131
The assembly shown consists of an
80-mm rod AF that is welded to a cross
consisting of four 200-mm arms. The
assembly is supported by a ball-andsocket joint at F and by three short links,
each of which forms an angle of 45 with
the vertical. For the loading shown,
determine (a) the tension in each link,
(b) the reaction at F.
SOLUTION
rE/F  200 i  80 j
TB  TB (i  j) / 2
rB/F  80 j  200k
TC  TC ( j  k ) / 2 rC/F  200i  80 j
TD  TD ( i  j) / 2
rD/E  80 j  200k
M F  0: rB/F  TB  rC/F  TC  rD/F  TD  rE/F  ( Pj)  0
i
j
k
0 80 200
1 1
0
TB
2
i
j
k
 200 80 0
0 1 1
TC
2
i
j
k
 0 80 200
1 1 0
TD
2
i
j
 200 80
P
0
k
0 0
0
Equate coefficients of unit vectors to zero and multiply each equation by 2.
i:
200 TB  80 TC  200 TD  0
(1)
j:
200 TB  200 TC  200 TD  0
(2)
k:
80 TB  200 TC  80 TD  200 2 P  0
(3)
80 TB  80 TC  80 TD  0
(4)
80
(2):
200
Eqs. (3)  (4):
160TB  280TC  200 2 P  0
Eqs. (1)  (2):
400TB  120TC  0
TB  
(5)
120
TC  0.3TC
400
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
(6)
PROBLEM 4.131 (Continued)
Eqs. (6)
160( 0.3TC )  280TC  200 2 P  0
(5):
232TC  200 2 P  0
TC  1.2191P
TC  1.219 P 
From Eq. (6):
TB  0.3(1.2191P )   0.36574  P
From Eq. (2):
 200(  0.3657 P )  200(1.2191P )  200T D  0
TB  0.366 P 
TD   0.8534 P
F  0:
TD   0.853P 
F  TB  TC  TD  Pj  0
i : Fx 
( 0.36574 P )
2
Fx   0.3448 P
j: Fy 
k : Fz 
2
0
Fx   0.345 P
( 0.36574 P )
2
Fy  P
( 0.8534 P )


(1.2191P)
2

( 0.8534 P)
2
 200  0
Fy  P
(1.2191P )
2
Fz   0.8620 P
0
Fz   0.862 P
F   0.345P i  Pj  0.862Pk 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 4.132
The uniform 10-kg rod AB is supported by a ball-and-socket
joint at A and by the cord CG that is attached to the midpoint G
of the rod. Knowing that the rod leans against a frictionless
vertical wall at B, determine (a) the tension in the cord, (b) the
reactions at A and B.
SOLUTION
Free-Body Diagram:
Five unknowns and six equations of equilibrium, but equilibrium is
maintained (MAB  0).
(a)
W  mg
 (10 kg) 9.81m/s 2
W  98.1 N

GC   300i  200 j  225k GC  425 mm

GC
T
( 300i  200 j  225k )
T T

GC 425
rB/ A   600i  400 j  150 mm
rG/ A   300i  200 j  75 mm
MA  0: rB/ A  B  rG/ A  T  rG/ A  ( W j)  0
i
j
k
i
j
k
i
j
k
T
 600 400 150   300 200
  300 200 75
75
425
B
0
0
 300 200  225
0
 98.1 0
Coefficient of i : (105.88  35.29)T  7357.5  0
T  52.12 N
T  52.1 N 


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PROBLEM 4.132 (Continued)
(b)
Coefficient of j : 150 B  (300  75  300  225)
52.12
0
425
B  73.58 N
B  (73.6 N)i 
F  0: A  B  T  W j  0
Coefficient of i :
Coefficient of j:
Coefficient of k :
300
0
425
Ax  36.8 N 
200
 98.1  0
425
Ay  73.6 N 
Ax  73.58  52.15
Ay  52.15
Az  52.15
225
0
425
Az  27.6 N 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 4.133
The frame ACD is supported by ball-and-socket joints at A
and D and by a cable that passes through a ring at B and is
attached to hooks at G and H. Knowing that the frame
supports at Point C a load of magnitude P  268 N,
determine the tension in the cable.
SOLUTION
Free-Body Diagram:
 AD
 AD
TBG
TBH

AD (1 m)i  (0.75 m)k


1.25 m
AD
 0.8i  0.6k

BG
 TBG
BG
0.5i  0.925 j  0.4k
 TBG
1.125

BH
 TBH
BH
0.375i  0.75 j  0.75k
 TBH
1.125
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 4.133 (Continued)
rB /A  (0.5 m)i; rC/A  (1 m)i; P  (268 N)j
To eliminate the reactions at A and D, we shall write
M AD  0:
 AD  (rB /A  TBG )   AD  (rB /A  TBH )   AD  (rC /A  P )  0
(1)
Substituting for terms in Eq. (1) and using determinants,
0.8
0
0.8
0
0.8
0
0.6
0.6
0.6
TBG
TBH
0.5
0
0
0
0
0
0 0
 0.5
 1
1.125
1.125
0.5 0.925 0.4
0.375 0.75 0.75
0 268
0
Multiplying all terms by (–1.125),
0.27750TBG  0.22500TBH  180.900
For this problem,
(2)
TBG  TBH  T
(0.27750  0.22500)T  180.900
T  360 N 
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PROBLEM 4.134
Solve Prob. 4.133, assuming that cable GBH is replaced by
a cable GB attached at G and B.
PROBLEM 4.133 The frame ACD is supported by balland-socket joints at A and D and by a cable that passes
through a ring at B and is attached to hooks at G and H.
Knowing that the frame supports at Point C a load of
magnitude P  268 N, determine the tension in the cable.
SOLUTION
Free-Body Diagram:
 AD
 AD
TBG
TBH

AD (1 m)i  (0.75 m)k


1.25 m
AD
 0.8i  0.6k

BG
 TBG
BG
0.5i  0.925 j  0.4k
 TBG
1.125

BH
 TBH
BH
0.375i  0.75 j  0.75k
 TBH
1.125
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 4.134 (Continued)
rB /A  (0.5 m)i; rC/A  (1 m)i; P  (268 N)j
To eliminate the reactions at A and D, we shall write
M AD  0:  AD  (rB /A  TBG )   AD  (rB /A  TBH )   AD  (rC /A  P )  0
(1)
Substituting for terms in Eq. (1) and using determinants,
0.8
0
0.8
0
0.8
0
0.6
0.6
0.6
TBG
TBH
0.5
0
0
0
0
0
0 0
 0.5
 1
1.125
1.125
0.5 0.925 0.4
0.375 0.75 0.75
0 268
0
Multiplying all terms by (–1.125),
0.27750TBG  0.22500TBH  180.900
(2)
For this problem, TBH  0.
Thus, Eq. (2) reduces to
0.27750TBG  180.900
TBG  652 N 
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8 cm
PROBLEM 4.135
y
The bent rod ABDE is supported by ball-and-socket joints at A
and E and by the cable DF. If a 300 N load is applied at C as
shown, determine the tension in the cable.
7 cm
F
9 cm
A
B
300 N
11 cm
C
10 cm
E
z
x
14 cm
16 cm
D
SOLUTION
Free-Body Diagram:
DF = −16i + 11j − 8k
DF = 21 c m
DF T
= (−16i + 11j − 8k )
DF 21
= 16 i
T=T
rD/E
rC/E = 16i − 14k
EA
ΣM EA = 0:
EA
=
EA 7i − 24k
=
25
EA
⋅ (rD/E × T) +
EA
⋅ (rC/E ⋅ (−300j)) = 0
7
0 −24
7
0 −24
1
T
16 0
0
+ 16 0 −14
=0
21 × 25
25
0 −300 0
−16 11 −8
−
24 × 16 × 11
−7 × 14 × 300 + 24 × 16 × 300
T+
=0
21 × 25
25
− 201.14 T + 85,800 = 0
T = 426.6 N
T = 426.6 N
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
8 cm
PROBLEM 4.136
y
7 cm
F
Solve Problem 4.13 5, assuming that cable DF is replaced by a
cable connecting B and F.
9 cm
A
B
300 N
11 cm
C
10 cm
E
z
16 cm
x
14 cm
D
SOLUTION
Free-Body Diagram:
rB/ A = 9i
rC/ A = 9i + 10k
BF = −16i + 11j + 16k
BF = 25.16 cm
BF
T
(−16i + 11j + 16k )
=
BF 25.16
AE 7i − 24k
=
=
AE
25
T =T
AE
ΣMAE = 0:
AE
⋅ (rB/ A × T) +
AE
⋅ (rC/ A ⋅ (−300j)) = 0
7
0 −24
7 0 −24
T
1
9
0
0
+9 0
10
=0
25 × 25.16
25
−16 11 16
0 −300 0
−
24 × 9 × 11
24 × 9 × 300 + 7 × 10 × 300
T+
25 × 25.16
25
− 94.436 T + 85,800 = 0
T = 908.6 N
T = 908.6 N
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512
y
PROBLEM 4.137
Two rectangular plates are welded together to form the assembly
shown. The assembly is supported by ball-and-socket joints at B
and D and by a ball on a horizontal surface at C. For the loading
shown, determine the reaction at C.
6 cm
A
200 N
B
9 cm
D
z
C
8 cm
12 cm
x
SOLUTION
λ BD =
First note
−(6 cm)i − (9 cm)j + (12 cm)k
(6) 2 + (9) 2 + (12)2 cm
1
(− 6i − 9 j + 12k )
16.1555
= −(6 cm)i
=
rA/B
P = (200 N)k
rC/D = (8 cm)i
C = (C ) j
From the f.b.d. of the plates
ΣM BD = 0: λ BD ⋅ (rA/B × P ) + λ BD ⋅ ( rC/D × C ) = 0
−6 −9 12
−6 −9 12
C (8)
6(200)
−1 0 0
+ 1 0 0
=0
16.1555
16.1555
0 0 1
0 1 0
(−9)(6)(200) + (12)(8)C = 0
C = 112.5 N
or C = (113 N)j
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515
PROBLEM 4.138
The pipe ACDE is supported by ball-and-socket joints at A and E
and by the wire DF. Determine the tension in the wire when a
640-N load is applied at B as shown.
SOLUTION
Free-Body Diagram:
Dimensions in mm

AE  480i  160 j  240k
AE  560 mm

AE 480i  160 j  240k

λ AE 
AE
560
6i  2 j  3k
λ AE 
7
rB/A  200i
rD/A  480i  160 j

DF  480i  330 j  240k ;
DF  630 mm

DF
480i  330 j  240k
16i  11j  8k
 TDF
 TDF
TDF  TDF
630
21
DF
M AE  λ AE  (rD/A  TDF )  λ AE  (rB/A  (600 j))  0
3
6
2
TDF
1
 200
480 160 0
0
0 0
21  7
7
16 11 8
0 640 0
6
2
3
6  160  8  2  480  8  3  480  11  3  160  16
3  200  640
TDF 
0
21  7
7
1120TDF  384  103  0
TDF  342.86 N
TDF  343 N 
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PROBLEM 4.139
Solve Problem 4.138, assuming that wire DF is replaced by a
wire connecting C and F.
PROBLEM 4.138 The pipe ACDE is supported by ball-andsocket joints at A and E and by the wire DF. Determine the
tension in the wire when a 640-N load is applied at B as shown.
SOLUTION
Free-Body Diagram:
Dimensions in mm

AE  480i  160 j  240k
AE  560 mm

AE 480i  160 j  240k
λ AE 

AE
560
6i  2 j  3k
λ AE 
7
rB/A  200i
rC/A  480i

CF  480i  490 j  240k ; CF  726.70 mm

CE 480i  490 j  240k
TCF  TCF

726.70
CF
MAE  0: λ AE  (rC/A  TCF )  λ AE  (rB/A  ( 600 j))  0
3
3
6
2
6
2
TCF
1
 200
480
0
0
0
0 0
726.7  7
7
480 490 240
0 640 0
2  480  240  3  480  490
3  200  640
TCF 
0
726.7  7
7
653.91TCF  384  103  0
TCF  587 N 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
y
PROBLEM 4.140
x
Two 2 × 4-m plywood panels, each of weight 60 N, are nailed
together as shown. The panels are supported by ball-and-socket
joints at A and F and by the wire BH. Determine (a) the location
of H in the xy plane if the tension in the wire is to be minimum,
(b) the corresponding minimum tension.
H
O
y
C
A
z
B
2m
D
60 N
60 N
E
2m
2m
F
x
2m
2m
SOLUTION
Free-Body Diagram:
AF = 4i − 2 j − 4k
AF
rG1/A
AF = 6 m
1
= (2i − j − 2k )
3
= 2i − j
rG2 /A = 4i − j − 2k
rB/A = 4i
ΣM AF = 0:
AF
⋅ (rG1/A × ( −60 j) +
AF
⋅ (rG2/A × ( −60 j)) +
2 −1 −2
2
1
+ 4
2 −1 0
3
0 −60 0
0
(2 × 2 × 60)
−1 −2
1
−1 −2 +
3
−60 0
AF
AF
1
1
+(−2 × 2 × 60 + 2 × 4 × 60) +
3
3
AF
⋅ (rB/A × T) = −160 or T ⋅ (
⋅ (rB/A × T ) = 0
⋅ (rB/A × T) = 0
AF
A/F
⋅ (rB/A × T) = 0
× rB/A ) = −160
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
(1)
PROBLEM 4.140 (Continued)
Projection of T on (
AF
× rB/A ) is constant. Thus, Tmin is parallel to
AF
Corresponding unit vector is
1
5
1
1
× rB/A = (2i − j − 2k ) × 4i = (−8 j + 4k )
3
3
(−2 j + k )
Tmin = T ( −2 j + k )
Eq. (1):
1
(2)
5
T
1
(−2 j + k ) ⋅ (2i − j − 2k ) × 4i = −160
3
5
T
1
(−2 j + k ) ⋅ ( −8 j + 4k ) = −160
3
5
T
(16 + 4) = −160
3 5
T =−
3 5(160)
= 24 5
20
T = 53.67 N
Eq. (2)
Tmin = T ( −2 j + k )
1
5
= 24 5(−2 j + k )
Tmin
1
5
= −(48 N)j + (24 N)k
Since Tmin has no i component, wire BH is parallel to the yz plane, and x = 4 m
(a)
(b)
x = 4.00 m; y = 8.00 m
Tmin = 50.7 N
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517
y
PROBLEM 4.141
x
Solve Problem 4.140, subject to the restriction that H must lie on
the y axis.
PROBLEM 4.140 Two 2 × 4-m plywo od panels, each of weight 60 N,
are nailed together as shown. The panels are supported by ball-andsocket joints at A and F and by the wire BH. Determine (a) the
location of H in the xy plane if the tension in the wire is to be
minimum, (b) the corresponding minimum tension.
H
O
y
C
A
z
B
2m
D
60 N
60 N
E
2m
2m
x
F
2m
2m
SOLUTION
Free-Body Diagram:
AF = 4i − 2 j − 4k
AF
rG1/A
1
= (2i − j − 2k )
3
= 2i − j
rG2 /A = 4i − j − 2k
rB/A = 4i
ΣMAF = 0:
AF
⋅ (rG1/A × (−60 j) +
AF
⋅ (rG2 /A × (−60j)) +
2 −1 2
2 −1 −2
1
1
2 −1 0 + 4 −1 −2 +
3
3
0 −60 0
0 −60 0
(2 × 2 × 60)
AF
AF
1
1
+(−2 × 2 × 60 + 2 × 4 × 60) +
3
3
AF
⋅ (rB/A × T ) = 0
⋅ (rB/A × T) = 0
AF
⋅ (rB/A × T) = 0
⋅ (rB/A × T) = −160
BH = −4i + yj − 4k
T=T
(1)
BH = (32 + y 2 )1/ 2
BH
−4i + yj − 4k
=T
BH
(32 + y 2 )1/ 2
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
518
PROBLEM 4.141 (Continued)
Eq. (1):
AF
2 −1 −2
T
⋅ (rB/A × T) = 4 0 0
= −160
3(32 + y 2 )1/ 2
−4 y −4
(−16 − 8 y )T = −3 × 160(32 + y 2 )1/ 2
T = 480
(32 + y 2 )1/ 2
8 y + 16
(2)
(8y +16) 12 (32 + y 2 ) −1/ 2 (2 y ) + (32 + y 2 )1/ 2 (8)
dT
= 0: 480
=0
dy
(8 y + 16) 2
Numerator = 0:
(8 y + 16) y = (32 + y 2 )8
8 y 2 + 16 y = 32 × 8 + 8 y 2
Eq. (2):
T = 480
y = 16.00 m
(32 + 162 )1/ 2
= 56.57 N
8 × 16 + 16
Tmin = 56.6 N
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
519
PROBLEM 4.142
A 16-kN forklift truck is used to lift a 8.5-kN crate.
Determine the reaction at each of the two (a) front wheels A,
(b) rear wheels B.
SOLUTION
Free-Body Diagram:
(a)
Front wheels:
M B  0: (8.5 kN)(1.3 m)  (16 kN)(0.3 m)  2 A(0.9 m)  0

A 8.806 kN
(b)
Rear wheels:
A  8.8 kN 
Fy  0:  8.5 kN 16 kN  2(8.806 kN)  2B  0
B  3.444 kN
B  3.4 kN 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 4.143
The lever BCD is hinged at C and attached to a control rod at B. If
P  100 N, determine (a) the tension in rod AB, (b) the reaction at C.
SOLUTION
Free-Body Diagram:
(a)
M C  0: T (5 cm)  (100 N)(7.5 cm)  0
T  150.0 N 
(b)
3
Fx  0: C x  100 N  (150.0 N)  0
5
C x  190 N
Fy  0: C y 
C y  120 N
  32.3
C x  190 N
4
 N)  0
5
C y  120 N
C  225 N
C  225 N
32.3° 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 4.144
Determine the reactions at A and B when (a) h  0,
(b) h  200 mm.
SOLUTION
Free-Body Diagram:
M A  0: ( B cos 60)(0.5 m)  ( B sin 60)h  (150 N)(0.25 m)  0
37.5
B
0.25  0.866h
(a)
(1)
When h  0,
B
From Eq. (1):
37.5
 150 N
0.25
B  150.0 N
30.0° 
Fy  0: Ax  B sin 60  0
Ax  (150) sin 60  129.9 N
A x  129.9 N
Fy  0: Ay  150  B cos 60  0
Ay  150  (150) cos 60  75 N
A y  75 N
  30
A  150.0 N
(b)
A  150.0 N
30.0° 
When h  200 mm  0.2 m,
From Eq. (1):
B
37.5
 488.3 N
0.25  0.866(0.2)
B  488 N
30.0° 
Fx  0: Ax  B sin 60  0
Ax  (488.3) sin 60  422.88 N
A x  422.88 N
Fy  0: Ay  150  B cos 60  0
Ay  150  (488.3) cos 60  94.15 N
A y  94.15 N
  12.55
A  433.2 N
A  433 N
12.55° 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 4.145
Neglecting friction and the radius of the pulley,
determine (a) the tension in cable ADB, (b) the reaction
at C.
SOLUTION
Free-Body Diagram:
Dimensions in mm
Geometry:
Distance:
AD  (0.36) 2  (0.150) 2  0.39 m
Distance:
BD  (0.2)2  (0.15)2  0.25 m
Equilibrium for beam:
M C  0:
(a)
 0.15 
 0.15 
(120 N)(0.28 m)  
T  (0.36 m)  
T  (0.2 m)  0
 0.39 
 0.25 
T  130.000 N
or
T  130.0 N 
 0.36 
 0.2 
Fx  0: C x  
 (130.000 N)  
 (130.000 N)  0
0.39


 0.25 
(b)
C x   224.00 N
 0.15 
 0.15 
Fy  0: C y  
 (130.00 N)   0.25  (130.00 N)  120 N  0
0.39




C y   8.0000 N
Thus,
C  Cx2  C y2  (224)2  ( 8)2  224.14 N
and
  tan 1
Cy
Cx
 tan 1
8
 2.0454
224
C  224 N
2.05 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 4.146
Bar AD is attached at A and C to collars that can move
freely on the rods shown. If the cord BE is vertical (  0),
determine the tension in the cord and the reactions at A and
C.
SOLUTION
Free-Body Diagram:
Fy  0:  T cos 30  (80 N) cos 30  0
T  80 N
T  80.0 N 
M C  0: ( A sin 30)(0.4 m)  (80 N)(0.2 m)  (80 N)(0.2 m)  0
A   160 N
A  160.0 N
30.0° 
C  160.0 N
30.0° 
M A  0: (80 N)(0.2 m)  (80 N)(0.6 m)  (C sin 30)(0.4 m)  0
C   160 N
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 4.147
A slender rod AB, of weight W, is attached to blocks A and B that
move freely in the guides shown. The constant of the spring is k,
and the spring is unstretched when θ = 0. (a) Neglecting the weight
of the blocks, derive an equation in W, k, l, and θ that must be
satisfied when the rod is in equilibrium. (b) Determine the value
of θ when W = 75 N, l = 30 cm, and k = 300 N/m.
SOLUTION
Free-Body Diagram:
Spring force:
Fs = ks = k (l − l cos θ ) = kl (1 − cos θ )
ΣM D = 0: Fs (l sin θ ) − W
(a)
l
cos θ = 0
2
kl (1 − cos θ )l sin θ −
kl (1 − cos θ ) tan θ −
(b)
For given values of
W
l cos θ = 0
2
W
=0
2
or (1 − cos θ ) tan θ =
W
2kl
W = 75 N
l = 30 cm
k = 300 N/m = 3 N/cm.
(1 − cos θ ) tan θ = tan θ − sin θ
75 N
=
2(3 N/cm)(30 cm)
= 0.41667
Solving numerically:
θ = 49.710°
or
θ = 49.7°
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 4.148
Determine the reactions at A and B when a  150 mm.
SOLUTION
Free-Body Diagram:
Force triangle
80 mm 80 mm

a
150 mm
  28.072
tan  
A
320 N
sin 28.072
B
320 N
tan 28.072
A  680 N
28.1 
B  600 N

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

30 N
A
3 cm
PROBLEM 4.149
For the frame and loading shown, determine the reactions at A and C.
C
D
B
4 cm
6 cm
SOLUTION
Since member AB is acted upon by two forces, A and B, they must be colinear, have the same magnitude, and
be opposite in direction for AB to be in equilibrium. The force B acting at B of member BCD will be equal in
magnitude but opposite in direction to force B acting on member AB. Member BCD is a three-force body with
member forces intersecting at E. The f.b.d.’s of members AB and BCD illustrate the above conditions. The
force triangle for member BCD is also shown. The angle β is found from the member dimensions:
β = tan −1
6 cm
= 30.964°
10 cm
Applying of the law of sines to the force triangle for member BCD,
30 N
B
C
=
=
sin( 45° − β ) sin β sin 135°
or
30 N
B
C
=
=
sin 14.036° sin 30.964° sin 135°
A= B=
(30 N) sin 30.964°
= 63.641 N
sin 14.036°
or
and
or
C=
A = 63.6 N
45.0°
C = 87.5 N
59.0°
(30 N ) sin 135°
= 87.466 N
sin 14.036°
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 4.150
A 200-mm lever and a 240-mm-diameter pulley
are welded to the axle BE that is supported by
bearings at C and D. If a 720-N vertical load is
applied at A when the lever is horizontal,
determine (a) the tension in the cord, (b) the
reactions at C and D. Assume that the bearing at
D does not exert any axial thrust.
SOLUTION
Dimensions in mm
We have six unknowns and six equations of equilibrium.
M C  0: (120k )  ( Dx i  D y j)  (120 j  160k )  T i  (80k  200i )  (720 j)  0
120 Dx j  120 Dy i  120T k  160Tj  57.6  103 i  144  103 k  0
Equating to zero the coefficients of the unit vectors:
k:
120T  144  103  0
i:
120 Dy  57.6  103  0
j:  120 Dx  160(1200 N)  0
(b)
Fx  0:
C x  Dx  T  0
Fy  0:
C y  D y  720  0
Fz  0:
Cz  0
(a) T  1200 N 
Dy  480 N
Dx  1600 N
C x  1600  1200  400 N
C y  480  720  1200 N
C  (400 N)i  (1200 N) j; D  (1600 N)i  (480 N) j 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 4.151
The 45-N square plate shown is supported by three vertical
wires. Determine the tension in each wire.
SOLUTION
Free-Body Diagram:
M B  0: rC /B  TC j  rA /B  TA j  rG /B  ( 45 N) j  0

[(20 cm)i  (15 cm)k ]  TC j (20 cm)k  TA j
 [(10 cm)i  (10 cm)k ]  [ (45 N)j]  0
20TC k  15TC i  20TA i  450k  450i  0
Equating to zero the coefficients of the unit vectors,
k:
i:
Fy  0:
20TC  450  0 ; TC  22.5 N
15(22.5)  20TA  450  0 ; TA  5.625 N
TA  TB  TC  45 N  0
5.625 N  TB  22.5 N  45 N  0 ; TB  16.875 N 
TA  5.63 N; TB  16.88 N; TC  22.5 N 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
y
PROBLEM 4.152
H
4 cm
12 cm
F
A
4 cm 25 cm
D
The rectangular plate shown weighs 75 N and is held in the
position shown by hinges at A and B and by cable EF. Assuming
that the hinge at B does not exert any axial thrust, determine
(a) the tension in the cable, (b) the reactions at A and B.
B
z
30 cm
E
x
20 cm
8 cm
C
SOLUTION
rB/A = (38 − 8)i = 30i
rE/A = (30 − 4)i + 20k
rG/A
= 26i + 20k
38
= i + 10k
2
= 19i + 10k
EF = 8i + 25 j − 20k
EF = 33 cm
AE T
= (8i + 25 j − 20k )
AE 33
ΣM A = 0: rE/A × T + rG/A × (−75 j) + rB/A × B = 0
T =T
i
j
k
i
j
k
i
j
T
+ 19 0 10 + 30 0
26 0 20
33
8 25 −20
0 −75 0
0 By
−(25)(20)
Coefficient of i:
Coefficient of j:
Coefficient of k:
(160 + 520)
(26)( 25)
T
+ 750 = 0:
33
k
0 =0
Bz
T = 49.5 N
49.5
− 30 Bz = 0: Bz = 34 N
33
49.5
− 1425 + 30 By = 0: By = 15 N
33
B = (15 N)j + (34 N)k
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484
PROBLEM 4.152 (Continued)
ΣF = 0: A + B + T − (75 N)j = 0
Ax +
Coefficient of i:
Coefficient of j:
Coefficient of k:
Ay + 15 +
8
(49.5) = 0
33
25
(49.5) − 75 = 0
33
Az + 34 −
20
(49.5) = 0
33
Ax = −12.00 N
Ay = 22.5 N
Az = −4.00 N
A = −(12.00 N)i + (22.5 N)j − (4.00 N)k
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485
PROBLEM 4.153
A force P is applied to a bent rod ABC, which may be supported in four different ways as shown. In each
case, if possible, determine the reactions at the supports.
SOLUTION
(a)
M A  0:  Pa  (C sin 45)2a  (cos 45) a  0
3
C
2
C
P
2
P
3
 2  1
Fx  0: Ax  
P
 3  2


Ax 
P
3
 2  1
Fy  0: Ay  P  
P
 3  2


Ay 
2P
3
C  0.471P
A  0.745P
(b)
45° 
63.4° 
M C  0: Pa  ( A cos 30)2a  ( A sin 30) a  0
A(1.732  0.5)  P
A  0.812 P
A  0.812P
Fx  0: (0.812 P ) sin 30  C x  0
60.0° 
C x  0.406 P
Fy  0: (0.812 P ) cos30  P  C y  0 C y  0.297 P
C  0.503P
36.2° 
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PROBLEM 4.153 (Continued)
(c)
M C  0:  Pa  ( A cos 30)2a  ( A sin 30)a  0
A(1.732  0.5)  P
A  0.448 P
A  0.448P
Fx  0:  (0.448 P ) sin 30  C x  0
60.0° 
C x  0.224 P
Fy  0: (0.448 P) cos 30  P  C y  0 C y  0.612 P
C  0.652P
69.9° 

(d )
Force T exerted by wire and reactions A and C all intersect at Point D.
M D  0: Pa  0
Equilibrium is not maintained.
Rod is improperly constrained. 
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PROBLEM 4.F1
Two crates, each of mass 350 kg, are placed as shown in
the bed of a 1400-kg pick-up truck. Draw the free-body
diagram needed to determine the reactions at each of the
two rear wheels A and front wheels B.
SOLUTION
Free-Body Diagram of Truck:

Weights:
W  (350 kg)(9.81 m/s 2 )  3.4335 N
WT  (1400 kg)(9.81 m/s 2 )  13.734 N



Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 4.F2
A lever AB is hinged at C and attached to a control cable at A.
If the lever is subjected to a 200 N vertical force at B, draw the
free-body diagram needed to determine the tension in the cable
and the reaction at C.
SOLUTION
Free-Body Diagram of Lever AB:

Geometry:
x AC  (10 cm) cos 20  9.3969 cm
y AC  (10 cm)sin 20  3.4202 cm
 y AD  12 cm  3.4202 cm  8.5798 cm

 y AD 
1  8.5798 cm

  tan 
  42.398
x
9.3969
cm


 AC 
 = tan 1 
  90  20  42.398  27.602
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PROBLEM 4.F3
A light rod AD is supported by frictionless pegs at B and C and
rests against a frictionless wall at A. A vertical 600 N force is
applied at D. Draw the free-body diagram needed to determine
the reactions at A, B, and C.
SOLUTION
Free-Body Diagram of Rod AD:


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PROBLEM 4.F4
A tension of 20 N is maintained in a tape as it passes
through the support system shown. Knowing that the
radius of each pulley is 10 mm, draw the free-body
diagram needed to determine the reaction at C.
SOLUTION
Free-Body Diagram of Support System:

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PROBLEM 4.F5
Two tape spools are attached to an axle supported by
bearings at A and D. The radius of spool B is 30 mm and
the radius of spool C is 40 mm. Knowing that TB = 80 N and
that the system rotates at a constant rate, draw the freebody diagram needed to determine the reactions at A and
D. Assume that the bearing at A does not exert any axial
thrust and neglect the weights of the spools and axle.
SOLUTION
Free-Body Diagram of Axle-Spool System:

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PROBLEM 4.F6
A 12-m pole supports a horizontal cable CD and is held by a
ball and socket at A and two cables BE and BF. Knowing that
the tension in cable CD is 14 kN and assuming that CD is
parallel to the x axis ( = 0), draw the free-body diagram
needed to determine the tension in cables BE and BF and the
reaction at A.
SOLUTION
Free-Body Diagram of Pole:

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PROBLEM 4.F7
A 20-kg cover for a roof opening is hinged at corners A
and B. The roof forms an angle of 30 with the
horizontal, and the cover is maintained in a horizontal
position by the brace CE. Draw the free-body diagram
needed to determine the magnitude of the force exerted
by the brace and the reactions at the hinges. Assume that
the hinge at A does not exert any axial thrust.
SOLUTION
Geometry of Brace CE:

Free-Body Diagram of Brace CE:

Cover Weight: W  (20 kg)(9.81 m/s 2 )  196.2 N


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