PROBLEM 4.1
The boom on a 4300-kg truck is used to unload a pallet of shingles of
mass 1600 kg. Determine the reaction at each of the two (a) rear
wheels B, (b) front wheels C.
SOLUTION
WA
1600 kg 9.81 m/s 2
mA g
15696 N
WA
or
WG
15.696 kN
4300 kg 9.81 m/s 2
mG g
42 183 N
or
WG
42.183 kN
(a) From f.b.d. of truck with boom
MC
0:
15.696 kN
0.5
0.4
42.183 kN 0.5 m
2 FB
6 cos15 m
2 FB
0.5
0.4
4.3 m
0
126.185
5.2
24.266 kN
or FB
12.13 kN
or FC
16.81 kN
(b) From f.b.d. of truck with boom
MB
0:
15.696 kN
2 FC
4.3
6 cos15
4.3 m
0.9 m
0
2 FC
Check:
Fy
0:
42.183 kN
174.786
5.2
4.3
0.4 m
33.613 kN
33.613
42.183
24.266
57.879
57.879 kN
15.696 kN
0 ok
0?
PROBLEM 4.2
Two children are standing on a diving board of mass 65 kg. Knowing that
the masses of the children at C and D are 28 kg and 40 kg, respectively,
determine (a) the reaction at A, (b) the reaction at B.
SOLUTION
WG
mG g
65 kg 9.81 m/s 2
637.65 N
WC
mC g
28 kg 9.81 m/s 2
274.68 N
WD
mD g
40 kg 9.81 m/s 2
392.4 N
(a) From f.b.d. of diving board
MB
0:
Ay 1.2 m
637.65 N 0.48 m
274.68 N 1.08 m
1418.92
1.2
Ay
392.4 N 2.08 m
0
1182.43 N
or A y
1.182 kN
(b) From f.b.d. of diving board
MA
0: By 1.2 m
637.65 N 1.68 m
By
274.68 N 2.28 m
2984.6
1.2
392.4 N 3.28 m
0
2487.2 N
or B y
Check:
Fy
0:
1182.43
2487.2
2487.2
637.65
2487.2 N
0 ok
274.68
392.4 N
0?
2.49 kN
PROBLEM 4.3
Two crates, each weighing 250 lb, are placed as shown in the bed of a
3000-lb pickup truck. Determine the reactions at each of the two
(a) rear wheels A, (b) front wheels B.
SOLUTION
(a) From f.b.d. of truck
MB
0:
250 lb 12.1 ft
250 lb 6.5 ft
2 FA
16350
9.8
3000 lb 3.9 ft
2FA 9.8 ft
0
1668.37 lb
834 lb
FA
(b) From f.b.d. of truck
MA
0:
2 FB 9.8 ft
3000 lb 5.9 ft
2 FB
17950
9.8
250 lb 3.3 ft
250 lb 2.3 ft
1831.63 lb
FB
Check:
Fy
0:
250
1668.37
3500
3500 lb
0
250
0 ok
3000
1831.63 lb
0?
916 lb
PROBLEM 4.4
Solve Problem 4.3 assuming that crate D is removed and that the position
of crate C is unchanged.
P4.3 The boom on a 4300-kg truck is used to unload a pallet of
shingles of mass 1600 kg. Determine the reaction at each of the two
(a) rear wheels B, (b) front wheels C
SOLUTION
(a) From f.b.d. of truck
MB
0:
3000 lb 3.9 ft
2 FA
2FA 9.8 ft
14725
9.8
250 lb 12.1 ft
0
1502.55 lb
or FA
751 lb
(b) From f.b.d. of truck
MA
0:
2 FB 9.8 ft
2 FB
3000 lb 5.9 ft
17125
9.8
250 lb 2.3 ft
0
1747.45 lb
or FB
Check:
Fy
0:
2 751
3250
874
3250 lb
3000
250 lb
0 ok
0?
874 lb
PROBLEM 4.5
A T-shaped bracket supports the four loads shown. Determine the reactions
at A and B if (a) a 100 mm, (b) a 70 mm.
SOLUTION
(a)
From f.b.d. of bracket
MB
0:
10 N 0.18 m
30 N 0.1 m
2.400
0.12
A
MA
0: B 0.12 m
40 N 0.06 m
40 N 0.06 m
B
A 0.12 m
20 N
or A
50 N 0.12 m
18.000
0.12
0
30 N 0.22 m
150 N
20.0 N
10 N 0.3 m
or B
0
150.0 N
(b)
From f.b.d. of bracket
MB
0:
10 N 0.15 m
30 N 0.07 m
1.200
0.12
A
MA
0: B 0.12 m
40 N 0.06 m
10 N 0.27 m
40 N 0.06 m
A 0.12 m
10 N
50 N 0.12 m
0
or A
10.00 N
or B
140.0 N
30 N 0.19 m
0
B
16.800
0.12
140 N
PROBLEM 4.6
For the bracket and loading of Problem 4.5, determine the smallest
distance a if the bracket is not to move.
P4.5 A T-shaped bracket supports the four loads shown. Determine the
reactions at A and B if (a) a 100 mm, (b) a 70 mm.
SOLUTION
The amin value will be based on A
0
From f.b.d. of bracket
MB
0:
40 N 60 mm
a
30 N a
1600
40
10 N a
80 mm
40 mm
or amin
40.0 mm
0
PROBLEM 4.7
A hand truck is used to move two barrels, each weighing 80 lb.
Neglecting the weight of the hand truck, determine (a) the vertical force
P which should be applied to the handle to maintain equilibrium when
35o , (b) the corresponding reaction at each of the two wheels.
SOLUTION
a1
20 in. sin
8 in. cos
a2
32 in. cos
20 in. sin
b
64 in. cos
From f.b.d. of hand truck
0: P b
MB
0: P
Fy
For
(a)
2w
2B
0
W a1
(1)
0
(2)
35
a1
20sin 35
8cos 35
a2
32cos 35
20sin 35
b
64cos 35
52.426 in.
4.9183 in.
14.7413 in.
From Equation (1)
P 52.426 in.
80 lb 14.7413 in.
P
(b)
W a2
80 lb 4.9183 in.
14.9896 lb
0
or P
14.99 lb
or B
72.5 lb
From Equation (2)
14.9896 lb
B
2 80 lb
72.505 lb
2B
0
PROBLEM 4.8
40o.
Solve Problem 4.7 when
P4.7 A hand truck is used to move two barrels, each weighing 80 lb.
Neglecting the weight of the hand truck, determine (a) the vertical force P
which should be applied to the handle to maintain equilibrium when
35o , (b) the corresponding reaction at each of the two wheels.
SOLUTION
a1
20 in. sin
8 in. cos
a2
32 in. cos
20 in. sin
b
64 in. cos
From f.b.d. of hand truck
0: P b
MB
Fy
0: P
For
(a)
W a2
2w
W a1
2B
0
0
(2)
40
a1
20sin 40
8cos 40
a2
32cos 40
20sin 40
b
64cos 40
49.027 in.
6.7274 in.
11.6577 in.
From Equation (1)
P 49.027 in.
80 lb 11.6577 in.
P
80 lb 6.7274 in.
0
8.0450 lb
or P
(b)
(1)
8.05 lb
From Equation (2)
8.0450 lb
B
2 80 lb
2B
0
75.9775 lb
or B
76.0 lb
PROBLEM 4.9
Four boxes are placed on a uniform 14-kg wooden plank which rests
on two sawhorses. Knowing that the masses of boxes B and D are
4.5 kg and 45 kg, respectively, determine the range of values of the
mass of box A so that the plank remains in equilibrium when box C is
removed.
SOLUTION
WA
WB
For m A
min
,E
mA g
mB g
WD
4.5 g
mD g
WG
0:
mA g 2.5 m
14 g
4.5 g 1.6 m
14 g 1 m
mA
, F
0:
ME
0: mA g 0.5 m
max
mG g
0
MF
For m A
45 g
0
2.32 kg
4.5g 0.4 m
45g 2.6 m
mA
45g 0.6 m
14 g 1 m
0
265.6 kg
or 2.32 kg
mA
266 kg
PROBLEM 4.10
A control rod is attached to a crank at A and cords are attached at B and
C. For the given force in the rod, determine the range of values of the
tension in the cord at C knowing that the cords must remain taut and that
the maximum allowed tension in a cord is 180 N.
SOLUTION
For
TC
MO
0:
TC
TC
MO
0:
TC
max
min
min
, TB
400 N 0.060 m
180.0 N
180 N
Tmax
0.120 m
Therefore,
min
0
180 N
Tmax
180 N 0.040 m
400 N 0.060 m
TC
0
TB
200 N
max
TC
For
,
0.120 m
max
TC
max
0
140.0 N
140.0 N
TC
180.0 N
PROBLEM 4.11
The maximum allowable value of each of the reactions is 360 N.
Neglecting the weight of the beam, determine the range of values of the
distance d for which the beam is safe.
SOLUTION
From f.b.d. of beam
Fx
0: Bx
0
Fy
0: A
B
100
A
B
600 N
or
so that
B
200
By
300 N
0
Therefore, if either A or B has a magnitude of the maximum of 360 N,
the other support reaction will be
MA
0:
100 N d
B 1.8
or
200 N 0.9
d
360 N
d
240 N .
300 N 1.8
d
0
d
Since B
360 N 600 N
720 1.8B
600 B
360 N,
720 1.8 360
600 360
d
MB
0:
0.300 m
100 N 1.8
or
d
Since A
d
or
A 1.8
d
1.8 A
360
d
300 mm
200 N 0.9
0
A
360 N,
1.8 360 360
360
0.800 m
or
d
or 300 mm
800 mm
d
800 mm
PROBLEM 4.12
Solve Problem 4.11 assuming that the 100-N load is replaced by a 160-N
load.
P4.11 The maximum allowable value of each of the reactions is 360 N.
Neglecting the weight of the beam, determine the range of values of the
distance d for which the beam is safe.
SOLUTION
From f.b.d of beam
Fx
0: Bx
0
Fy
0: A
B
160
A
B
660 N
or
so that
B
200
By
300 N
0
Therefore, if either A or B has a magnitude of the maximum of 360 N,
the other support reaction will be 360 N 660 360 300 N .
MA
0: 160 N d
200 N 0.9
B 1.8
or
d
d
720 1.8 360
660 360
0
0.240 m
0: 160 N 1.8
or
or
A 1.8
d
d
d
360 N,
MB
Since A
300 N 1.8
720 1.8B
660 B
d
Since B
d
d
1.8 A
d
240 mm
200 N 0.9
0
468
A
360 N,
1.8 360
468
360
0.500 m
or
d
or 240 mm
500 mm
d
500 mm
PROBLEM 4.13
For the beam of Sample Problem 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 45 kips and that the
reaction at A must be directed upward.
SOLUTION
For the force of P to be a minimum, A = 0.
With A
0,
MB
0: Pmin 6 ft
6 kips 2 ft
Pmin
For the force P to be a maximum, A
With A
A max
6 kips 4 ft
0
6.00 kips
45 kips
45 kips,
MB
0:
45 kips 9 ft
Pmax 6 ft
Pmax
6 kips 2 ft
6 kips 4 ft
0
73.5 kips
A check must be made to verify the assumption that the maximum value of P is based on the reaction force at
A. This is done by making sure the corresponding value of B is 45 kips.
Fy
B
0: 45 kips
40.5 kips
73.5 kips
45 kips
B
ok
6 kips
or Pmax
6 kips
0
73.5 kips
and 6.00 kips
P
73.5 kips
PROBLEM 4.14
For the beam and loading shown, determine the range of values of the
distance a for which the reaction at B does not exceed 50 lb
downward or 100 lb upward.
SOLUTION
To determine amax the two 150-lb forces need to be as close to B without
having the vertical upward force at B exceed 100 lb.
From f.b.d. of beam with B
MD
0:
100 lb
150 lb amax
25 lb 2 in.
or
amax
4 in.
150 lb amax
100 lb 8 in.
1 in.
0
5.00 in.
To determine amin the two 150-lb forces need to be as close to A without
having the vertical downward force at B exceed 50 lb.
From f.b.d. of beam with B
MD
0:
50 lb
150 lb 4 in.
25 lb 2 in.
or
Therefore,
amin
150 lb amin
amin
50 lb 8 in.
1 in.
0
1.00 in.
or 1.00 in.
a
5.00 in.
PROBLEM 4.15
A follower ABCD is held against a circular cam by a stretched spring,
which exerts a force of 21 N for the position shown. Knowing that the
tension in rod BE is 14 N, determine (a) the force exerted on the roller
at A, (b) the reaction at bearing C.
SOLUTION
Note: From f.b.d. of ABCD
A
2
A cos 60
Ax
A sin 60
Ay
3
2
A
(a) From f.b.d. of ABCD
A
2
0:
MC
40 mm
21 N 40 mm
14 N 20 mm
A
0
28 N
or A
28.0 N
60
(b) From f.b.d. of ABCD
Fx
Cx
Fy
Cy
Then
and
C
0: C x
14 N
28 N
or
Cx
0:
21 N
28 N sin 60
Cy
3.2487 N
C x2
tan
C y2
1
Cy
Cx
28 N cos 60
28
tan
28.0 N
Cy
or
2
1
0
3.2487
3.2487
28
or C
0
3.25 N
2
28.188 N
6.6182
28.2 N
6.62
PROBLEM 4.16
A 6-m-long pole AB is placed in a hole and is guyed by three cables.
Knowing that the tensions in cables BD and BE are 442 N and 322 N,
respectively, determine (a) the tension in cable CD, (b) the reaction
at A.
SOLUTION
Note:
DB
2.8
2
5.25
2
5.95 m
DC
2.8
2
2.10
2
3.50 m
(a) From f.b.d. of pole
MA
0:
2.8 m
5.95 m
322 N 6 m
2.8 m
TCD
3.50 m
442 N
2.85 m
6m
0
300 N
TCD
or TCD
300 N
(b) From f.b.d. of pole
Fx
2.8 m
5.95 m
0: 322 N
2.8 m
3.50 m
Ax
Fy
and
A
Ax
or
570 N
Ax2
Ax
300 N
5.25 m
5.95 m
0: Ay
Ay
Then
126 N
442 N
Ay2
tan
126 N
2.10 m
3.50 m
442 N
Ay
or
126
1
2
570 N
126 N
0
300 N
0
570 N
570
2
583.76 N
77.535
or A
584 N
77.5
PROBLEM 4.17
Determine the reactions at A and C when (a)
30o.
0, (b)
SOLUTION
(a)
(a)
0
From f.b.d. of member ABC
MC
0:
80 lb 10 in.
80 lb 20 in.
A
A 40 in.
60 lb
or A
Fy
Cy
Fx
C
and
C x2
tan
60 lb
60 lb
or
160 lb
Cx
Then
0: C y
0: 80 lb
C y2
1
Cy
tan
Cx
Cy
60 lb
Cx
0
Cx
or
2
60
160 lb
2
60
160
1
170.880 lb
20.556
or C
(b)
(b)
60.0 lb
0
80 lb
160
0
170.9 lb
20.6
A cos 30
40 in.
30
From f.b.d. of member ABC
MC
0:
80 lb 10 in.
A sin 30
80 lb 20 in.
20 in.
A
0
97.399 lb
or A
97.4 lb
60
PROBLEM 4.17 CONTINUED
Fx
0: 80 lb
0: C y
and
C
C x2
tan
or
97.399 lb cos 30
84.350 lb
Cy
Then
97.399 lb sin 30
208.70 lb
Cx
Fy
80 lb
C y2
1
Cy
Cx
tan
1
2
Cx
209 lb
Cy
84.4 lb
0
0
or
208.70
Cx
84.350
84.350
208.70
or C
2
225.10 lb
22.007
225 lb
22.0
PROBLEM 4.18
Determine the reactions at A and B when (a) h
0, (b) h
8 in.
SOLUTION
(a)
(a) h
0
From f.b.d. of plate
MA
0:
B sin 30
20 in.
B
40 lb 10 in.
40 lb
or B
Fx
Fy
0: Ax
40 lb cos 30
Then
A
or
0: Ay
40 lb
40 lb sin 30
Ax2
and
(b)
Ay2
tan
(b) h
Ax
34.641 lb
20 lb
1
Ay
Ax
tan
1
30
2
20
34.6 lb
0
Ay
or
34.641
40.0 lb
0
Ax
Ay
0
20.0 lb
2
39.999 lb
20
34.641
30.001
or A
40.0 lb
B cos 30
8 in.
30
8 in.
From f.b.d. of plate
MA
0:
B sin 30
20 in.
40 lb 10 in.
B
0
130.217 lb
or B
130.2 lb
30.0
PROBLEM 4.18 CONTINUED
Fx
0: Ax
Ax
Fy
0: Ay
130.217 lb cos30
112.771 lb
40 lb
Then
and
A
Ax2
tan
Ay2
1
130.217 lb sin 30
112.771
Ay
Ax
Ax
or
25.108 lb
Ay
0
tan
1
or
2
Ay
25.108
25.108
112.771
or A
2
112.8 lb
0
25.1 lb
115.532 lb
12.5519
115.5 lb
12.55
PROBLEM 4.19
The lever BCD is hinged at C and is attached to a control rod at B. If
P 200 N, determine (a) the tension in rod AB, (b) the reaction at C.
SOLUTION
(a) From f.b.d. of lever BCD
MC
0: TAB 50 mm
200 N 75 mm
0
TAB
300 N
(b) From f.b.d. of lever BCD
Fx
Fy
0: 200 N
Cx
Cx
380 N
0: C y
and
C
C x2
tan
240 N
C y2
1
or
0.8 300 N
Cy
Then
0.6 300 N
380
Cy
Cx
tan
240
1
Cx
380 N
Cy
240 N
0
or
2
0
2
240
380
or C
449.44 N
32.276
449 N
32.3
PROBLEM 4.20
The lever BCD is hinged at C and is attached to a control rod at B.
Determine the maximum force P which can be safely applied at D if the
maximum allowable value of the reaction at C is 500 N.
SOLUTION
From f.b.d. of lever BCD
MC
0: TAB 50 mm
0: 0.6TAB
P
0: 0.8TAB
0
0.6TAB
0.6 1.5P
Cy
Cy
(2)
1.9 P
0
0.8TAB
Cy
From Equation (1)
Cx
P
Cx
Fy
(1)
P
Cx
From Equation (1)
0
1.5P
TAB
Fx
P 75 mm
(3)
0.8 1.5P
1.2P
From Equations (2) and (3)
C
Since Cmax
C x2
C y2
1.9P
1.2 P
2
2.2472 P
500 N,
500 N
or
2
Pmax
2.2472Pmax
222.49 lb
or P
222 lb
PROBLEM 4.21
The required tension in cable AB is 800 N. Determine (a) the vertical
force P which must be applied to the pedal, (b) the corresponding
reaction at C.
SOLUTION
(a) From f.b.d. of pedal
MC
0: P 0.4 m
800 N
P
0.18 m sin 60
0
311.77 N
or P
312 N
(b) From f.b.d. of pedal
Fx
0: Cx
800 N
0
Cx
800 N
or
Cx
Fy
0: C y
311.77 N
or
and
311.77 N
Cy
C
C x2
tan
C y2
1
800
Cy
Cx
0
311.77 N
Cy
Then
800 N
tan
2
311.77
1
2
311.77
800
or C
858.60 N
21.291
859 N
21.3
PROBLEM 4.22
Determine the maximum tension which can be developed in cable AB
if the maximum allowable value of the reaction at C is 1000 N.
SOLUTION
Have
Cmax
C2
Now
Cy
1000 N
C x2
C y2
2
Cx2
1000
(1)
From f.b.d. of pedal
Fx
0: C x
Tmax
Cx
MD
0: C y 0.4 m
Cy
0
Tmax
Tmax
(2)
0.18 m sin 60
0
0.38971Tmax
(3)
Equating the expressions for C y in Equations (1) and (3), with Cx
Tmax
from Equation (2)
1000
and
2
2
Tmax
0.389711Tmax
2
Tmax
868,150
Tmax
931.75 N
or Tmax
932 N
PROBLEM 4.23
A steel rod is bent to form a mounting bracket. For each of the mounting
brackets and loadings shown, determine the reactions at A and B.
SOLUTION
(a) From f.b.d. of mounting bracket
(a)
ME
0: A 8 in.
80 lb in.
20 lb 12 in.
A
10 lb 6 in.
0
47.5 lb
or A
Fx
0: Bx
10 lb
Bx
or
Bx
Bx2
B
and
tan
0
By
20 lb
By
20.0 lb
By2
1
37.5 lb
20 lb
or
Then
0
37.5 lb
0: By
Fy
47.5 lb
47.5 lb
37.5
By
tan
Bx
2
20.0
2
20
37.5
1
42.5 lb
28.072
or B
42.5 lb
28.1
(b) From f.b.d. of mounting bracket
(b)
MB
0:
A cos 45
8 in.
10 lb 6 in.
A
80 lb in.
20 lb 12 in.
67.175 lb
or A
Fx
0: Bx
Bx
or
Bx
0
10 lb
67.175cos 45
37.500 lb
37.5 lb
67.2 lb
0
45
PROBLEM 4.23 CONTINUED
Fy
0: By
20 lb
By
or
Then
and
B
Bx2
tan
67.175sin 45
27.500 lb
By
27.5 lb
By2
37.5
2
tan
1
1
By
Bx
0
27.5
27.5
37.5
or B
2
46.503 lb
36.254
46.5 lb
36.3
PROBLEM 4.24
A steel rod is bent to form a mounting bracket. For each of the
mounting brackets and loadings shown, determine the reactions at A
and B.
SOLUTION
(a)
(a) From f.b.d. of mounting bracket
MA
0:
B 8 in.
20 lb 12 in.
10 lb 2 in.
B
80 lb in.
37.5 lb
or B
Fx
0:
37.5 lb
Ax
or
0:
20 lb
or
Ax2
and
tan
20 lb
Ay2
1
20.0 lb
47.5
Ay
tan
Ax
1
2
20
20
47.5
2
51.539 lb
22.834
or A
(b)
0
0
Ay
Ay
A
Ax
47.5 lb
Ay
Then
10 lb
37.5 lb
47.5 lb
Ax
Fy
0
51.5 lb
22.8
(b) From f.b.d. of mounting bracket
MA
0:
B cos 45
80 lb in.
B
8 in.
20 lb 2 in.
10 lb 2 in.
53.033 lb
or B
Fx
0: Ax
Ax
or
53.033 lb cos 45
47.500 lb
Ax
0
47.5 lb
45
53.0 lb
10
0
PROBLEM 4.24 CONTINUED
Fy
0: Ay
53.033 lb sin 45
Ay
Then
and
A
Ax2
tan
Ay2
1
Ay
Ax
0
17.500 lb
Ay
or
20
17.50 lb
47.5
tan
1
2
17.5
17.5
47.5
or A
2
50.621 lb
20.225
50.6 lb
20.2
PROBLEM 4.25
A sign is hung by two chains from mast AB. The mast is hinged at A and
is supported by cable BC. Knowing that the tensions in chains DE and
FH are 50 lb and 30 lb, respectively, and that d 1.3 ft, determine
(a) the tension in cable BC, (b) the reaction at A.
SOLUTION
First note
BC
2
8.4
1.3
2
8.5 ft
(a) From f.b.d. of mast AB
8.4
TBC
8.5
0:
MA
2.5 ft
30 lb 7.2 ft
50 lb 2.2 ft
0
131.952 lb
TBC
or TBC
132.0 lb
(b) From f.b.d. of mast AB
Fx
8.4
131.952 lb
8.5
0: Ax
or
Fy
Ax
130.400 lb
Ax
130.4 lb
1.3
131.952 lb
8.5
0: Ay
Ay
Then
and
A
Ax2
Ay2
tan
1
Ax
50 lb
59.819 lb
130.4
Ay
30 lb
59.819 lb
Ay
or
0
tan
2
1
59.819
59.819
130.4
or A
2
143.466 lb
24.643
143.5 lb
24.6
0
PROBLEM 4.26
A sign is hung by two chains from mast AB. The mast is hinged at A
and is supported by cable BC. Knowing that the tensions in chains DE
and FH are 30 lb and 20 lb, respectively, and that d 1.54 ft,
determine (a) the tension in cable BC, (b) the reaction at A.
SOLUTION
First note
BC
8.4
2
2
1.54
8.54 ft
(a) From f.b.d. of mast AB
MA
8.4
TBC
8.54
0:
2.5 ft
30 lb 2.2 ft
TBC
20 lb 7.2 ft
0
85.401 lb
or TBC
85.4 lb
(b) From f.b.d. of mast AB
Fx
8.4
8.54
0: Ax
Ax
Fy
Ay
and
A
Ax2
Ay2
tan
1
85.401 lb
Ax
30 lb
34.600 lb
84.001
Ay
20 lb
34.600 lb
Ay
or
Then
84.001 lb
1.54
8.54
0: Ay
0
84.001 lb
Ax
or
85.401 lb
tan
2
1
34.600
2
90.848 lb
34.6
84.001
22.387
or A
90.8 lb
22.4
0
PROBLEM 4.27
For the frame and loading shown, determine the reactions at A and E
when (a)
30o , (b)
45o.
SOLUTION
(a) Given
(a)
30
From f.b.d. of frame
MA
0:
90 N 0.2 m
E cos 60
90 N 0.06 m
0.160 m
E
E sin 60
0: Ax
90 N
or
Fy
0: Ay
19.7730 N
Ax
19.7730 N
90 N
Ay
Then
Ax2
Ay2
60
0
140.454 N sin 60
0
31.637 N
Ay
A
140.5 N
140.454 N cos 60
Ax
or
0
140.454 N
or E
Fx
0.100 m
31.6 N
19.7730
2
31.637
2
37.308 lb
and
tan
1
Ay
Ax
tan
1
31.637
19.7730
57.995
or A
37.3 N
58.0
PROBLEM 4.27 CONTINUED
(b)
(b) Given
45
From f.b.d. of frame
MA
0:
90 N 0.2 m
E cos 45
90 N 0.06 m
0.160 m
E
E sin 45
0: Ax
90
127.279 N cos 45
Ax
Fy
0: Ay
90
0
127.279 N
or E
Fx
0.100 m
45
0
0
127.279 N sin 45
Ay
127.3 N
0
0
or A
0
PROBLEM 4.28
A lever AB is hinged at C and is attached to a control cable at A. If the
lever is subjected to a 300-N vertical force at B, determine
(a) the tension in the cable, (b) the reaction at C.
SOLUTION
First
x AC
0.200 m cos 20
0.187 939 m
y AC
0.200 m sin 20
0.068 404 m
Then
yDA
0.240 m
y AC
0.240 m
0.068404 m
0.171596 m
yDA
x AC
tan
and
0.171 596
0.187 939
42.397
and
90
20
42.397
27.603
(a) From f.b.d. of lever AB
MC
0: T cos 27.603 0.2 m
300 N 0.3 m cos 20
T
477.17 N
0
or T
477 N
(b) From f.b.d. of lever AB
Fx
0: C x
477.17 N cos 42.397
Cx
Cx
or
Fy
or
0: C y
0
352.39 N
352.39 N
300 N
477.17 N sin 42.397
Cy
621.74 N
Cy
621.74 N
0
PROBLEM 4.28 CONTINUED
Then
and
C
C x2
tan
C y2
1
Cy
Cx
352.39
tan
1
2
621.74
621.74
352.39
or C
2
714.66 N
60.456
715 N
60.5
PROBLEM 4.29
Neglecting friction and the radius of the pulley, determine the tension
in cable BCD and the reaction at support A when d 80 mm.
SOLUTION
First
tan
1
60
280
12.0948
60
80
36.870
1
tan
From f.b.d. of object BAD
MA
0:
40 N 0.18 m
T sin
0.18 m
T sin
0.08 m
T cos
0.18 m
7.2 N m
0.056061
T
T cos
0.08 m
0
128.433 N
or T
Fx
0:
128.433 N cos
Ax
Fy
0: Ay
Then
128.433 N sin
and
Ax2
Ay2
tan
1
Ax
40 N
143.970 N
22.836
Ay
sin
143.970 N
Ay
A
0
22.836 N
Ay
or
Ax
22.836 N
Ax
or
cos
128.4 N
tan
1
2
143.970
143.970
22.836
or A
2
145.770 N
80.987
145.8 N
81.0
0
PROBLEM 4.30
Neglecting friction and the radius of the pulley, determine the tension in
cable BCD and the reaction at support A when d 144 mm.
SOLUTION
First note
1
tan
tan
1
60
216
15.5241
60
144
22.620
From f.b.d. of member BAD
MA
0:
40 N 0.18 m
T cos
T sin
0.18 m
T cos
T sin
0.18 m
0
7.2 N m
0.0178199 m
T
0.08 m
0.08 m
404.04 N
or T
Fx
0: Ax
or
Fy
404.04 N cos
Ax
16.3402 N
Ax
16.3402 N
0: Ay
404.04 N sin
Then
Ay
and
A
Ax2
tan
Ay2
1
Ax
0
sin
40 N
0
303.54 N
16.3402
Ay
cos
303.54 N
Ay
or
404 N
tan
1
2
303.54
303.54
16.3402
or A
2
303.98 N
86.919
304 N
86.9
PROBLEM 4.31
Neglecting friction, determine the tension in cable ABD and the reaction
at support C.
SOLUTION
From f.b.d. of inverted T-member
MC
0: T 25 in.
T 10 in.
T
30 lb 10 in.
20 lb
or T
Fx
0: Cx
20 lb
Cx
Fy
0: C y
Cy
Then
and
or
Cy
C
C x2
C y2
tan
1
20 lb
20.0 lb
20 lb
or
20.0 lb
0
Cx
or
0
30 lb
10 lb
10.00 lb
20
Cy
Cx
0
2
tan
10
1
2
10
20
22.361 lb
26.565
C
22.4 lb
26.6
PROBLEM 4.32
Rod ABC is bent in the shape of a circular arc of radius R. Knowing
35o , determine the reaction (a) at B, (b) at C.
that
SOLUTION
For
35
(a) From the f.b.d. of rod ABC
MD
0: Cx R
Cx
0
P
Cx
or
PR
P
Fx
0: P
B
P
sin 35
B sin 35
0
1.74345P
or B
1.743P
55.0
(b) From the f.b.d. of rod ABC
Fy
0: C y
1.74345P cos 35
Cy
Cy
or
Then
and
C
C x2
tan
C y2
1
Cy
Cx
P
0
0.42815P
0.42815P
P
2
tan
1
0.42815P
0.42815P
P
or C
2
1.08780 P
23.178
1.088P
23.2
PROBLEM 4.33
Rod ABC is bent in the shape of a circular arc of radius R. Knowing
50o , determine the reaction (a) at B, (b) at C.
that
SOLUTION
For
50
(a) From the f.b.d. of rod ABC
MD
0: Cx R
Cx
Cx
or
P R
0
P
P
Fx
0: P
B
P
sin 50
B sin 50
0
1.30541P
or B
1.305P
40.0
(b) From the f.b.d. of rod ABC
Fy
0: C y
P
Cy
Cy
or
Then
and
C
C x2
tan
C y2
1
Cy
Cx
1.30541P cos 50
0
0.160900P
0.1609 P
P
2
tan
1
0.1609P
0.1609P
P
or C
2
1.01286 P
9.1405
1.013P
9.14
PROBLEM 4.34
Neglecting friction and the radius of the pulley, determine (a) the
tension in cable ABD, (b) the reaction at C.
SOLUTION
First note
tan
1
tan
1
15
36
22.620
15
20
36.870
(a) From f.b.d. of member ABC
MC
0:
T sin 22.620
30 lb 28 in.
T sin 36.870
T
36 in.
20 in.
0
32.500 lb
or T
32.5 lb
(b) From f.b.d. of member ABC
Fx
0: Cx
32.500 lb cos 22.620
Cx
Cx
or
Fy
0: C y
30 lb
Cy
or
and
C
C x2
tan
C y2
1
Cy
Cx
0
56.000 lb
56.000 lb
32.500 lb sin 22.620
Cy
Then
cos 36.870
sin 36.870
0
2.0001 lb
2.0001 lb
56.0
tan
1
2
2.001
2.0
56.0
or C
2
56.036 lb
2.0454
56.0 lb
2.05
PROBLEM 4.35
Neglecting friction, determine the tension in cable ABD and the
reaction at C when
60o.
SOLUTION
From f.b.d. of bent ACD
MC
0:
T cos30
T a
2a sin 60
P a
T sin 30
a
2a cos 60
0
T
P
1.5
or T
Fx
0: C x
Cx
2P
cos 30
3
3
P
3
0.57735P
Cx
or
Fy
0: C y
Cy
2
P
3
0
P
0.577 P
2P
cos 60
3
0
0
or C
0.577P
2P
3
PROBLEM 4.36
Neglecting friction, determine the tension in cable ABD and the
30o.
reaction at C when
SOLUTION
From f.b.d. of bent ACD
MC
0:
T cos 60
P a
2a sin 30
T a
T
T sin 60 a
2a cos 30
0
P
1.86603
0.53590P
or T
Fx
0: C x
or
Fy
0: C y
0.53590P cos 60
0
Cx
0.26795P
Cx
0.268P
0.53590 P
Cy
P
0.536 P
0.53590 P sin 60
0
0
or C
0.268P
PROBLEM 4.37
Determine the tension in each cable and the reaction at D.
SOLUTION
First note
BE
20
CF
10
2
2
8
8
2
2
in.
21.541 in.
in.
12.8062 in.
From f.b.d. of member ABCD
MC
0:
8
TBE 10 in.
21.541
120 lb 20 in.
TBE
0
646.24 lb
or TBE
Fy
0:
120 lb
8
646.24 lb
21.541
TCF
8
TCF
12.8062
0
192.099 lb
or TCF
Fx
0:
20
646.24 lb
21.541
D
646 lb
10
192.099 lb
12.8062
D
192.1 lb
0
750.01 lb
or D
750 lb
PROBLEM 4.38
Rod ABCD is bent in the shape of a circular arc of radius 80 mm and
rests against frictionless surfaces at A and D. Knowing that the collar
45o. determine (a) the
at B can move freely on the rod and that
tension in cord OB, (b) the reactions at A and D.
SOLUTION
(a) From f.b.d. of rod ABCD
ME
0:
25 N cos 60 dOE
T
T cos 45
dOE
0
17.6777 N
or T
17.68 N
(b) From f.b.d. of rod ABCD
Fx
0:
17.6777 N cos 45
25 N cos 60
N D cos 45
NA
or
0: N A sin 45
NA
NA
(1)
N D sin 45
25 N sin 60
ND
0
0
ND
ND
Fy
N A cos 45
17.6777 N sin 45
0
48.296 N
(2)
Substituting Equation (1) into Equation (2),
2N A
48.296 N
NA
24.148 N
or N A
24.1 N
45.0
and N D
24.1 N
45.0
PROBLEM 4.39
Rod ABCD is bent in the shape of a circular arc of radius 80 mm and
rests against frictionless surfaces at A and D. Knowing that the collar
at B can move freely on the rod, determine (a) the value of
for
which the tension in cord OB is as small as possible, (b) the
corresponding value of the tension, (c) the reactions at A and D.
SOLUTION
(a) From f.b.d. of rod ABCD
ME
0:
25 N cos 60 dOE
dOE
0
12.5 N
cos
T
or
T cos
(1)
T is minimum when cos is maximum,
or
0
(b) From Equation (1)
12.5 N
cos 0
T
12.5 N
or Tmin
Fx
(c)
0:
N A cos 45
N D cos 45
25 N cos 60
ND
or
ND
Fy
0: N A sin 45
ND
12.5 N
0
0
NA
NA
(2)
N D sin 45
NA
12.50 N
25 N sin 60
0
30.619 N
(3)
Substituting Equation (2) into Equation (3),
2N A
NA
30.619
15.3095 N
or N A
15.31 N
45.0
and N D
15.31 N
45.0
PROBLEM 4.40
Bar AC supports two 100-lb loads as shown. Rollers 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
First note that from similar triangles
yDB
6
10
20
BD
and
yDB
3
2
14
2
3 in.
in.
14.3178 in.
Tx
14
T
14.3178
0.97780T
Ty
3
T
14.3178
0.20953T
(a) From f.b.d. of bar AC
ME
0:
0.97780T 7 in.
100 lb 16 in.
T
0.20953T 6 in.
100 lb 4 in.
0
357.95 lb
or T
358 lb
(b) From f.b.d. of bar AC
Fy
0: A
100
A
0.20953 357.95
100
0
275.00 lb
or A
275 lb
(c) From f.b.d of bar AC
Fx
0: 0.97780 357.95
C
C
0
350.00 lb
or C
350 lb
PROBLEM 4.41
A parabolic slot has been cut in plate AD, and the plate has been
placed so that the slot fits two fixed, frictionless pins B and C. The
equation of the slot is y x 2 /100, where x and y are expressed in mm.
Knowing that the input force P 4 N, determine (a) the force each
pin exerts on the plate, (b) the output force Q.
SOLUTION
The equation of the slot is
x2
100
y
dy
dx
Now
slope of the slot at C
C
2x
100
tan
1
90
and
1.200
x 60 mm
1.200
50.194
90
50.194
39.806
Coordinates of C are
xC
60 mm,
Also, the coordinates of D are
yD
60
100
xD
60 mm
46 mm
where
tan
yD
1
2
yC
36 mm
40 mm sin
120 66
240
46 mm
12.6804
40 mm tan12.6804
55.000 mm
PROBLEM 4.41 CONTINUED
60 mm
tan
yED
Also,
60 mm
tan12.6804
266.67 mm
From f.b.d. of plate AD
ME
N C cos 39.806
0:
NC cos
266.67
55.0
yED
36.0
yD
yC
mm
NC sin
N C sin 39.806
NC
NC
or
Fx
0:
4N
NC cos
4N
3.70 N
Q sin
Q
0: N B
NB
NC sin
(a)
(b)
yD
4 N 266.67
0
55.0 mm
39.8
Q sin12.6804
0
5.2649 N
5.26 N
Q cos
77.3
0
3.7025 N sin 39.806
NB
or
60 mm
yED
0
Q
Fy
4N
3.7025 N
3.7025 N cos 39.806
or
xC
NB
5.2649 N cos12.6804
0
2.7662 N
2.77 N
NB
2.77 N , NC
Q
5.26 N
3.70 N
39.8
77.3 output
0
PROBLEM 4.42
A parabolic slot has been cut in plate AD, and the plate has been placed
so that the slot fits two fixed, frictionless pins B and C. The equation of
the slot is y x 2 /100, where x and y are expressed in mm. Knowing that
the maximum allowable force exerted on the roller at D is 8.5 N,
determine (a) the corresponding magnitude of the input force P, (b) the
force each pin exerts on the plate.
SOLUTION
x2
100
y
The equation of the slot is,
dy
dx
Now
slope of slot at C
C
2x
100
1
tan
and
90
1.200
x 60 mm
1.200
90
50.194
50.194
39.806
Coordinates of C are
2
xC
60
100
60 mm, yC
36 mm
Also, the coordinates of D are
xD
yD
where
46 mm
tan
yD
60 mm
46 mm
1
40 mm sin
120 66
240
40 mm tan12.6804
xE
Note:
yE
12.6804
yC
55.000 mm
0
60 mm tan
36 mm
60 mm tan 39.806
86.001 mm
(a) From f.b.d. of plate AD
ME
0: P yE
8.5 N sin
8.5 N cos
60 mm
yE
yD
0
PROBLEM 4.42 CONITNIUED
P 86.001 mm
8.5 N sin12.6804
8.5 N cos12.6804
P
31.001 mm
60 mm
0
6.4581 N
or P
(b)
Fx
6.458 N
0: P
NC cos
8.5 N sin
NB
0: N B
5.9778 N sin 39.806
NB
0
5.9778 N
or NC
Fy
0
NC cos 39.806
8.5 N sin12.6804
NC
6.46 N
NC sin
39.8
5.98 N
8.5 N cos
8.5 N cos12.6804
0
0
4.4657 N
or N B
4.47 N
PROBLEM 4.43
A movable bracket is held at rest by a cable attached at E and by frictionless
rollers. Knowing that the width of post FG is slightly less than the distance
between the rollers, determine the force exerted on the post by each roller
when
20o.
SOLUTION
From f.b.d. of bracket
Fy
0: T sin 20
60 lb
T
Tx
0
175.428 lb
175.428 lb cos 20
Ty
164.849 lb
175.428 lb sin 20
60 lb
Note: Ty and 60 lb force form a couple of
60 lb 10 in.
MB
600 lb in.
0: 164.849 lb 5 in.
FCD
or
0: FCD
FAB
28.030 lb
FAB
0
28.030 lb
Tx
FAB
FAB
or
FCD 8 in.
28.0 lb
FCD
Fx
600 lb in.
0
164.849 lb
0
192.879 lb
192.9 lb
Rollers A and C can only apply a horizontal force to the right onto the
vertical post corresponding to the equal and opposite force to the left on
the bracket. Since FAB is directed to the right onto the bracket, roller B
will react FAB. Also, since FCD is acting to the left on the bracket, it will
act to the right on the post at roller C.
PROBLEM 4.43 CONTINUED
A
D
B
192.9 lb
C
28.0 lb
0
Forces exerted on the post are
A
B
C
D
192.9 lb
28.0 lb
0
PROBLEM 4.44
30o.
Solve Problem 4.43 when
P4.43 A movable bracket is held at rest by a cable attached at E and by
frictionless rollers. Knowing that the width of post FG is slightly less
than the distance between the rollers, determine the force exerted on the
20o.
post by each roller when
SOLUTION
From f.b.d. of bracket
Fy
0: T sin 30
Tx
60 lb
0
T
120 lb
120 lb cos 30
Ty
103.923 lb
120 lb sin 30
60 lb
Note: Ty and 60 lb force form a couple of
60 lb 10 in.
MB
0:
103.923 lb 5 in.
FCD
0: FCD
FAB
Tx
FAB
or
0
10.05 lb
FAB
10.0481 lb
FCD 8 in.
600 lb in.
10.0481 lb
FCD
or
Fx
600 lb in.
FAB
0
103.923 lb
0
93.875 lb
93.9 lb
Rollers A and C can only apply a horizontal force to the right on the
vertical post corresponding to the equal and opposite force to the left on
the bracket. The opposite direction apply to roller B and D. Since both
FAB and FCD act to the right on the bracket, rollers B and D will react
these forces.
A
C
B
93.9 lb
D
10.05 lb
0
Forces exerted on the post are
A
B
D
C
93.9 lb
10.05 lb
0
PROBLEM 4.45
A 20-lb weight can be supported in the three different ways shown.
Knowing that the pulleys have a 4-in. radius, determine the reaction at A
in each case.
SOLUTION
(a) From f.b.d. of AB
Fx
0: Ax
0
Fy
0: Ay
20 lb
Ay
or
0
20.0 lb
and A
MA
0: M A
MA
20 lb 1.5 ft
0
30.0 lb ft
or M A
4 in.
(b) Note:
20.0 lb
1 ft
12 in.
30.0 lb ft
0.33333 ft
From f.b.d. of AB
Fx
0: Ax
Ax
or
Fy
20 lb
Ay
A
Ax2
0
20.0 lb
0: Ay
or
Then
20 lb
0
20.0 lb
Ay2
20.0
2
20.0
2
A
MA
0: M A
28.3 lb
45
20 lb 0.33333 ft
20 lb 1.5 ft
MA
28.284 lb
0.33333 ft
0
30.0 lb ft
or M A
30.0 lb ft
PROBLEM 4.45 CONTINUED
(c) From f.b.d. of AB
Fx
0: Ax
0
Fy
0: Ay
20 lb
Ay
or
20 lb
0
40.0 lb
and A
MA
0: M A
20 lb 1.5 ft
20 lb 1.5 ft
MA
40.0 lb
0.33333 ft
0.33333 ft
0
60.0 lb ft
or M A
60.0 lb ft
PROBLEM 4.46
A belt passes over two 50-mm-diameter pulleys which are mounted on a
bracket as shown. Knowing that M 0 and Ti TO 24 N, determine
the reaction at C.
SOLUTION
From f.b.d. of bracket
Fx
Fy
0: Cx
24 N
0: C y
0
Cx
24 N
24 N
0
Cy
Then
C
C x2
C y2
24 N
24
2
24
2
C
MC
0: M C
24 N
24 N
MC
25
33.941 N
33.9 N
45
25 mm
50
60 mm
45.0
0
120 N mm
or M C
0.120 N m
PROBLEM 4.47
A belt passes over two 50-mm-diameter pulleys which are mounted on a
bracket as shown. Knowing that M 0.40 N m m and that Ti and TO
are equal to 32 N and 16 N, respectively, determine the reaction at C.
SOLUTION
From f.b.d. of bracket
Fx
0: C x
32 N
0
Cx
Fy
0: C y
16 N
32 N
0
Cy
Then
C x2
C
and
tan
C y2
1
Cy
Cx
16 N
32
tan
2
1
16
16
32
2
35.777 N
26.565
or C
MC
0: M C
32 N 45 mm
16 N 25 mm
MC
35.8 N
26.6
25 mm
50 mm
60 mm
400 N mm
800 N mm
or M C
0.800 N m
0
PROBLEM 4.48
A 350-lb utility pole is used to support at C the end of an electric wire.
The tension in the wire is 120 lb, 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 200 lb ft.
SOLUTION
First note
LBD
Tmax :
4.5
2
10
2
From f.b.d. of utility pole with M A
MA
0:
200 lb ft
10.9659 ft
200 lb ft
120 lb cos15
4.5
Tmax 10 ft
10.9659
Tmax
14 ft
0
444.19 lb
or Tmax
Tmin :
From f.b.d. of utility pole with M A
MA
0: 200 lb ft
200 lb ft
120 lb cos15
4.5
Tmin 10 ft
10.9659
Tmin
444 lb
14 ft
0
346.71 lb
or Tmin
347 lb
PROBLEM 4.49
In a laboratory experiment, students hang the masses shown from a beam
of negligible mass. (a) Determine the reaction at the fixed support A
knowing that end D of the beam does not touch support E. (b) Determine
the reaction at the fixed support A knowing that the adjustable support E
exerts an upward force of 6 N on the beam.
SOLUTION
WB
mB g
1 kg 9.81 m/s 2
WC
mC g
0.5 kg 9.81 m/s 2
9.81 N
4.905 N
(a) From f.b.d. of beam ABCD
Fx
0: Ax
0
Fy
0: Ay
WB
Ay
WC
9.81 N
0
4.905 N
Ay
0
14.715 N
or A
MA
0: M A
WB 0.2 m
MA
WC 0.3 m
9.81 N 0.2 m
MA
14.72 N
0
4.905 N 0.3 m
0
3.4335 N m
or M A
3.43 N m
(b) From f.b.d. of beam ABCD
Fx
0: Ax
0
Fy
0: Ay
WB
WC
Ay
9.81 N
Ay
8.715 N
MA
0: M A
MA
9.81 N 0.2 m
6N
4.905 N
6N
0
or A
WB 0.2 m
MA
0
8.72 N
WC 0.3 m
6 N 0.4 m
0
4.905 N 0.3 m
6 N 0.4 m
0
1.03350 N m
or M A
1.034 N m
PROBLEM 4.50
In a laboratory experiment, students hang the masses shown from a beam
of negligible mass. Determine the range of values of the force exerted on
the beam by the adjustable support E for which the magnitude of the
couple at A does not exceed 2.5 N m.
SOLUTION
WB
mB g
1 kg 9.81 m/s 2
WC
mC g
0.5 kg 9.81 m/s 2
9.81 N
4.905 N
Maximum M A value is 2.5 N m
Fmin :
From f.b.d. of beam ABCD with M A
MA
0: 2.5 N m
WB 0.2 m
Fmin 0.4 m
2.5 N m
9.81 N 0.2 m
Fmax :
From f.b.d. of beam ABCD with M A
0:
2.5 N m
2.5 N m
9.81 N 0.2 m
Fmin 0.4 m
0
2.33 N
2.5 N m
WB 0.2 m
Fmax 0.4 m
WC 0.3 m
0
4.905 N 0.3 m
Fmax
or
0
2.3338 N
Fmin
MA
WC 0.3 m
4.905 N 0.3 m
Fmin
or
2.5 N m
Fmax
Fmax 0.4 m
0
14.8338 N
14.83 N
or 2.33 N
FE
14.83 N
PROBLEM 4.51
Knowing that the tension in wire BD is 300 lb, determine the reaction at
fixed support C for the frame shown.
SOLUTION
From f.b.d. of frame with T
Fx
300 lb
0: C x
Cx
Fy
15.3846 lb
0: C y
C
C x2
456.92 lb
C y2
and
15.3846
tan
1
0
or
Cx
12
300 lb
13
180 lb
Cy
Then
5
300 lb
13
100 lb
Cy
Cx
0
or
2
Cy
456.92
tan
15.3846 lb
1
456.92 lb
2
457.18 lb
456.92
15.3846
88.072
or C
MC
0: M C
180 lb 20 in.
MC
769.23 lb in.
100 lb 16 in.
457 lb
12
300 lb 16 in.
13
or M C
88.1
0
769 lb in.
PROBLEM 4.52
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 75 lb ft.
SOLUTION
Tmax
From f.b.d. of frame with M C
MC
0: 900 lb in.
75 lb ft
900 lb in.
180 lb 20 in.
100 lb 16 in.
Tmax
Tmin
From f.b.d. of frame with
MC
0:
900 lb in.
MC
180 lb 20 in.
Tmin
12
Tmax 16 in.
13
0
413.02 lb
75 lb ft
100 lb 16 in.
900 lb in.
12
Tmin 16 in.
13
0
291.15 lb
291 lb
T
413 lb
PROBLEM 4.53
Uniform rod AB of length l and weight W lies in a vertical plane and is
acted upon by a couple M. The ends of the rod are connected to small
rollers which rest against frictionless surfaces. (a) Express the angle
corresponding to equilibrium in terms of M, W, and l. (b) Determine the
value of
corresponding to equilibrium when M 1.5 lb ft,
W 4 lb, and l 2 ft.
SOLUTION
(a) From f.b.d. of uniform rod AB
Fx
0:
A cos 45
A
Fy
B
0: A sin 45
A
B
B cos 45
0
or
0
B sin 45
B
W
A
(1)
0
2W
(2)
From Equations (1) and (2)
2A
2W
1
W
2
A
From f.b.d. of uniform rod AB
MB
0: W
l
cos
2
1
W
2
M
l cos 45
0
From trigonometric identity
cos
cos cos
sin sin
Equation (3) becomes
Wl
cos
2
M
Wl
2
cos
sin
0
(3)
PROBLEM 4.53 CONTINUED
Wl
cos
2
or
M
sin
Wl
cos
2
2M
Wl
or
(b)
sin
1
2 1.5 lb ft
4 lb 2 ft
Wl
sin
2
sin
1
2M
Wl
22.024
or
22.0
0
PROBLEM 4.54
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 which 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
MC
0: T l sin
T
l
cos
2
W
T l cos
0
W cos
2 cos
sin
Dividing both numerator and denominator by cos ,
T
W
2 1
1
tan
W
2
1 tan
or T
(b) For T
3W ,
W
2
1 tan
3W
1
or
tan
tan
1
5
6
1
6
39.806
or
39.8
PROBLEM 4.55
A thin, uniform ring of mass m and radius R is attached by a frictionless
pin to a collar at A and rests against a small roller at B. The ring lies in a
vertical plane, and the collar can move freely on a horizontal rod and is
acted upon by a horizontal force P. (a) Express the angle
corresponding to equilibrium in terms of m and P. (b) Determine the
value of corresponding to equilibrium when m 500 g and P 5 N.
SOLUTION
(a) From f.b.d. of ring
MC
0: P R cos
2P
W tan
R cos
where W
tan
W R sin
0
mg
2P
mg
or
(b) Have
m
500 g
tan
1
0.500 kg and P
tan
1
2P
mg
5N
2 5N
0.500 kg 9.81 m/s 2
63.872
or
63.9
PROBLEM 4.56
Rod AB is acted upon by a couple M and two forces, each of magnitude
P. (a) Derive an equation in , P, M, and l which must be satisfied when
corresponding to
the rod is in equilibrium. (b) Determine the value of
equilibrium when M 150 lb in., P 20 lb, and l 6 in.
SOLUTION
(a) From f.b.d. of rod AB
MC
0: P l cos
P l sin
M
0
or sin
(b) For
M
150 lb in., P
sin
20 lb, and l
150 lb in.
20 lb 6 in.
cos
sin 2
Using identity
sin
sin 2
1
1
sin 2
2sin 2
1
2
1.25
1.5625
2.5sin
1.25
1
1
2
sin 2
1
M
Pl
6 in.
5
4
cos 2
cos
1.25
sin
sin 2
2.5sin
0.5625
0
Using quadratic formula
2.5
sin
6.25
4 2 0.5625
2 2
2.5
1.75
4
or
sin
0.95572
72.886
and
and
sin
0.29428
17.1144
or
17.11 and
72.9
PROBLEM 4.57
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
90o. (a) Neglecting the
corresponding to equilibrium in
weight of the rod, express the angle
corresponding to
terms of P, k, and l. (b) Determine the value of
1
kl.
equilibrium when P
4
SOLUTION
First note
T
where
tension in spring
s
ks
elongation of spring
AB
AB
2l sin
T
2l sin
2
2l sin
90
90
2
1
2
2
2kl sin
2
1
2
(1)
(a) From f.b.d. of rod BC
MC
0: T l cos
P l sin
2
0
Substituting T From Equation (1)
2kl sin
2kl 2 sin
2
Factoring out
1
2
2
1
2
cos
l cos
2
2l cos
2
P l sin
Pl 2sin
2
2
, leaves
cos
0
2
0
PROBLEM 4.57 CONTINUED
kl sin
1
2
2
P sin
0
2
or
sin
1
kl
2 kl P
2
2sin
(b) P
kl
2 kl
1
P
kl
4
2sin
1
2sin
1
kl
2 kl
kl
4
2sin
1
kl
4
2 3 kl
2sin
1
4
3 2
0.94281
141.058
or
141.1
PROBLEM 4.58
Solve Sample Problem 4.5 assuming that the spring is unstretched when
90o.
SOLUTION
First note
T
where
tension in spring
s
ks
deformation of spring
r
F
kr
From f.b.d. of assembly
M0
0: W l cos
k
250 lb/in., r
cos
or
cos
0
kr 2
Wl
cos
For
kr 2
Wl cos
or
0
F r
3 in., l
8 in., W
250 lb/in. 3 in.
400 lb 8 in.
400 lb
2
0.703125
Solving numerically,
0.89245 rad
or
Then
51.134
90
51.134
141.134
or
141.1
PROBLEM 4.59
A collar B of weight W can move freely along the vertical rod shown. The
0.
constant of the spring is k, and the spring is unstretched when
(a) Derive an equation in , W, k, and l which must be satisfied when the
collar is in equilibrium. (b) Knowing that W 3 lb, l 6 in., and
k 8 lb/ft, determine the value of corresponding to equilibrium.
SOLUTION
T
First note
k
where
s
ks
spring constant
elongation of spring
l
cos
l
l
cos
T
kl
cos
1
cos
W
0
1
cos
(a) From f.b.d. of collar B
Fy
0: T sin
kl
cos
or
1
cos
sin
W
or tan
(b) For W
3 lb, l
6 in., k
l
tan
sin
0
sin
W
kl
8 lb/ft
6 in.
12 in./ft
0.5 ft
3 lb
8 lb/ft 0.5 ft
0.75
Solving Numerically,
57.957
or
58.0
PROBLEM 4.60
A slender rod AB, of mass m, is attached to blocks A and B which move
freely in the guides shown. The constant of the spring is k, and the spring
is unstretched when
0 . (a) Neglecting the mass of the blocks, derive
an equation in m, g, k, l, and which must be satisfied when the rod is in
when m 2 kg, l 750
equilibrium. (b) Determine the value of
mm, and k 30 N/m.
SOLUTION
First note
Fs
where
ks
spring force
k
spring constant
s
spring deformation
l
l cos
l 1
cos
Fs
kl 1
cos
0: Fs l sin
W
l
cos
2
(a) From f.b.d. of assembly
MD
kl 1
l sin
cos
kl sin
0
W
cos sin
l
cos
2
0
W
cos
2
0
Dividing by cos
kl tan
tan
sin
sin
W
2
W
2kl
or tan
(b) For m
2 kg, l
750 mm, k
l
750 mm
30 N/m
0.750 m
sin
mg
2kl
PROBLEM 4.60 CONTINUED
Then
tan
sin
2 kg 9.81 m/s 2
2 30 N/m 0.750 m
0.436
Solving Numerically,
50.328
or
50.3
PROBLEM 4.61
The bracket ABC can be supported in the eight different ways shown. All
connections consist of smooth 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 assuming that the magnitude of
the force P is 100 N.
SOLUTION
1. Three non-concurrent, non-parallel reactions
(a)
Completely constrained
(b)
Determinate
(c)
Equilibrium
From f.b.d. of bracket:
MA
0: B 1 m
100 N 0.6 m
0
B
Fx
0: Ax
Ax
Fy
0: Ay
Ay
Then
and
A
60.0
2
tan
1
60 N
60.0 N
0
60.0 N
100 N
0
100 N
100
100
60.0
2
116.619 N
59.036
A
116.6 N
59.0
2. Four concurrent reactions through A
(a)
Improperly constrained
(b)
Indeterminate
(c)
No equilibrium
3. Two reactions
(a)
Partially constrained
(b)
Determinate
(c)
Equilibrium
PROBLEM 4.61 CONTINUED
From f.b.d. of bracket
MA
Fy
0: C 1.2 m
0: A
100 N 0.6 m
100 N
50 N
0
C
50.0 N
A
50.0 N
0
4. Three non-concurrent, non-parallel reactions
(a)
Completely constrained
(b)
Determinate
(c)
Equilibrium
From f.b.d. of bracket
tan
BC
MA
1.2
2
1
1.0
1.2
1.0
39.8
2
1.56205 m
1.2
B 1m
1.56205
0:
100 N 0.6 m
B
Fx
Fy
0: C
0: A
78.102 N cos 39.806
78.102 N sin 39.806
0
39.8
78.1 N
0
C
60.0 N
100 N
0
A
50.0 N
5. Four non-concurrent, non-parallel reactions
(a)
Completely constrained
(b)
Indeterminate
(c)
Equilibrium
From f.b.d. of bracket
MC
0:
100 N 0.6 m
Ay
Ay 1.2 m
50 N
0
or A y
50.0 N
6. Four non-concurrent non-parallel reactions
(a)
Completely constrained
(b)
Indeterminate
(c)
Equilibrium
PROBLEM 4.61 CONTINUED
From f.b.d. of bracket
MA
0:
Bx 1 m
Bx
Fx
0:
60
Ax
100 N 0.6 m
0
60.0 N
Ax
or B x
60.0 N
or A x
60.0 N
0
60.0 N
7. Three non-concurrent, non-parallel reactions
(a)
Completely constrained
(b)
Determinate
(c)
Equilibrium
From f.b.d. of bracket
Fx
MA
0: Ax
0
0: C 1.2 m
C
100 N 0.6 m
0
50.0 N
or C
Fy
0: Ay
Ay
100 N
50.0 N
50.0 N
0
50.0 N
A
50.0 N
8. Three concurrent, non-parallel reactions
(a)
Improperly constrained
(b)
Indeterminate
(c)
No equilibrium
PROBLEM 4.62
Eight identical 20 30-in. rectangular plates, each weighing 50 lb, are
held in a vertical plane as shown. All connections consist of frictionless
pins, rollers, or short links. For each case, answer the questions listed in
Problem 4.61, and, wherever possible, compute the reactions.
P6.1 The bracket ABC can be supported in the eight different ways
shown. All connections consist of smooth 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 assuming
that the magnitude of the force P is 100 N.
SOLUTION
1. Three non-concurrent, non-parallel reactions
(a)
Completely constrained
(b)
Determinate
(c)
Equilibrium
From f.b.d. of plate
MA
0: C 30 in.
Fx
0: Ax
0
Fy
0: Ay
50 lb
Ay
50 lb 15 in.
25 lb
0
C
25.0 lb
A
25.0 lb
0
25 lb
2. Three non-current, non-parallel reactions
(a)
Completely constrained
(b)
Determinate
(c)
Equilibrium
From f.b.d. of plate
Fx
0:
MB
0:
Fy
B
50 lb 15 in.
0: 25.0 lb
50 lb
D 30 in.
C
0
D
25.0 lb
C
25.0 lb
0
0
PROBLEM 4.62 CONTINUED
3. Four non-concurrent, non-parallel reactions
(a)
Completely constrained
(b)
Indeterminate
(c)
Equilibrium
From f.b.d. of plate
MD
Fx
0: Ax 20 in.
0: Dx
50 lb 15 in.
37.5 lb
Ax
37.5 lb
Dx
37.5 lb
0
4. Three concurrent reactions
(a)
Improperly constrained
(b)
Indeterminate
(c)
No equilibrium
5. Two parallel reactions
(a)
Partial constraint
(b)
Determinate
(c)
Equilibrium
From f.b.d. of plate
MD
Fy
0: C 30 in.
0: D
50 lb
50 lb 15 in.
25 lb
0
C
25.0 lb
D
25.0 lb
0
6. Three non-concurrent, non-parallel reactions
(a)
Completely constrained
(b)
Determinate
(c)
Equilibrium
From f.b.d. of plate
MD
0: B 20 in.
50 lb 15 in.
0
B
Fx
0: Dx
37.5 lb
Fy
0: Dy
50 lb
0
0
Dx
37.5 lb
37.5 lb
Dy
50.0 lb
or D
62.5 lb
53.1
PROBLEM 4.62 CONTINUED
7. Two parallel reactions
(a)
Improperly constrained
(b)
Reactions determined by dynamics
(c)
No equilibrium
8. Four non-concurrent, non-parallel reactions
(a)
Completely constrained
(b)
Indeterminate
(c)
Equilibrium
From f.b.d. of plate
MD
Fy
Fx
0: B 30 in.
0: Dy
0: Dx
50 lb 15 in.
50 lb
C
0
25.0 lb
0
B
25.0 lb
Dy
25.0 lb
0
PROBLEM 4.63
Horizontal and vertical links are hinged to a wheel, and forces are applied
to the links as shown. Knowing that a 3.0 in., determine the value of P
and the reaction at A.
SOLUTION
As shown on the f.b.d., the wheel is a three-force body. Let point D be
the intersection of the three forces.
From force triangle
A
5
P
and
A
tan
P
4
21 lb
3
4
21 lb
3
5
21 lb
3
1
3
4
28 lb
or P
28.0 lb
35.0 lb
36.9
35 lb
36.870
A
PROBLEM 4.64
Horizontal and vertical links are hinged to a wheel, and forces are applied
to the links as shown. Determine the range of values of the distance a for
which the magnitude of the reaction at A does not exceed 42 lb.
SOLUTION
Let D be the intersection of the three forces acting on the wheel.
From the force triangle
21 lb
a
or
A
For
1
42 lb
A
21 lb
a
or
16
a2
21
A
42 lb
a2
16
4
16
3
a
a2
16
a2
or
a2
16
2.3094 in.
or a
Since
as a increases, A decreases
A
21
16
a2
1
2.31 in.
PROBLEM 4.65
Using the method of Section 4.7, solve Problem 4.21.
P4.21 The required tension in cable AB is 800 N. Determine (a) the
vertical force P which must be applied to the pedal, (b) the corresponding
reaction at C.
SOLUTION
Let E be the intersection of the three forces acting on the pedal device.
First note
tan
1
180 mm sin 60
21.291
400 mm
From force triangle
(a)
800 N tan 21.291
P
311.76 N
or P
(b)
C
312 N
800 N
cos 21.291
858.60 N
or C
859 N
21.3
PROBLEM 4.66
Using the method of Section 4.7, solve Problem 4.22.
P4.22 Determine the maximum tension which can be developed in cable
AB if the maximum allowable value of the reaction at C is 1000 N.
SOLUTION
Let E be the intersection of the three forces acting on the pedal device.
First note
tan
1
180 mm sin 60
400 mm
21.291
From force triangle
Tmax
1000 N cos 21.291
931.75 N
or Tmax
932 N
PROBLEM 4.67
To remove a nail, a small block of wood is placed under a crowbar, and a
horizontal force P is applied as shown. Knowing that l 3.5 in. and
P 30 lb, determine the vertical force exerted on the nail and the
reaction at B.
SOLUTION
Let D be the intersection of the three forces acting on the crowbar.
First note
tan
1
36 in. sin 50
82.767
3.5 in.
From force triangle
FN
P tan
30 lb tan 82.767
236.381 lb
on nail FN
RB
P
cos
30 lb
cos82.767
236 lb
238.28 lb
or R B
238 lb
82.8
PROBLEM 4.68
To remove a nail, a small block of wood is placed under a crowbar, and a
horizontal force P is applied as shown. Knowing that the maximum
vertical force needed to extract the nail is 600 lb and that the horizontal
force P is not to exceed 65 lb, determine the largest acceptable value of
distance l.
SOLUTION
Let D be the intersection of the three forces acting on the crowbar.
From force diagram
FN
P
tan
600 lb
65 lb
9.2308
83.817
From f.b.d.
tan
l
36 in. sin 50
l
36 in. sin 50
tan 83.817
2.9876 in.
or l
2.99 in.
PROBLEM 4.69
For the frame and loading shown, determine the reactions at C and D.
SOLUTION
Since member BD is acted upon by two forces, B and D, they must be colinear, have the same magnitude, and
be opposite in direction for BD to be in equilibrium. The force B acting at B of member ABC will be equal in
magnitude but opposite in direction to force B acting on member BD. Member ABC is a three-force body with
member forces intersecting at E. The f.b.d.’s of members ABC and BD illustrate the above conditions. The
force triangle for member ABC is also shown. The angles and are found from the member dimensions:
tan
1
tan
1
0.5 m
1.0 m
26.565
1.5 m
1.0 m
56.310
Applying the law of sines to the force triangle for member ABC,
150 N
sin
or
C
sin 90
150 N
sin 29.745
B
sin 90
C
sin116.565
150 N sin116.565
sin 29.745
C
B
sin 33.690
270.42 N
or C
and
D
B
150 N sin 33.690
sin 29.745
270 N
56.3
167.7 N
26.6
167.704 N
or D
PROBLEM 4.70
For the frame and loading shown, determine the reactions at A and C.
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
is found from the member dimensions:
tan
1
60 m
100 m
30.964
Applying of the law of sines to the force triangle for member BCD,
130 N
sin 45
130 N
sin14.036
or
A
and
B
C
B
sin
B
sin 30.964
130 N sin 30.964
sin14.036
130 N sin135
sin14.036
C
sin135
C
sin135
275.78 N
or A
276 N
45.0
or C
379 N
59.0
379.02 N
PROBLEM 4.71
To remove the lid from a 5-gallon pail, the tool shown is used to apply an
upward and radially outward force to the bottom inside rim of the lid.
Assuming that the rim rests against the tool at A and that a 100-N force is
applied as indicated to the handle, determine the force acting on the rim.
SOLUTION
The three-force member ABC has forces that intersect at D, where
tan
90 mm
yBC 45 mm
1
yDC
and
yDC
360 mm cos 35
tan 20
xBC
tan 20
810.22 mm
yBC
360 mm sin 35
206.49 mm
tan
1
90
558.73
9.1506
Based on the force triangle, the law of sines gives
100 N
sin
A
or
A
sin 20
100 N sin 20
sin 9.1506
A
215 N
and A
215.07 N
80.8 on tool
215 N
80.8 on rim of can
PROBLEM 4.72
To remove the lid from a 5-gallon pail, the tool shown is used to apply
an upward and radially outward force to the bottom inside rim of the
lid. Assuming that the top and the rim of the lid rest against the tool at
A and B, respectively, and that a 60-N force is applied as indicated to
the handle, determine the force acting on the rim.
SOLUTION
The three-force member ABC has forces that intersect at point D, where,
from the law of sines CDE
150 mm
L
sin 95
L
19 mm tan 35
sin 30
325.37 mm
Then
45 mm
yBD
tan
1
L
y AE
where
yBD
22 mm
325.37 mm
19 mm
cos 35
22 mm
280.18 mm
tan
45 mm
280.18 mm
1
9.1246
Applying the law of sines to the force triangle,
B
sin150
Or, on member
and, on lid
60 N
sin 9.1246
B
189.177 N
B
189.2 N
80.9
B
189.2 N
80.9
PROBLEM 4.73
A 200-lb crate is attached to the trolley-beam system shown. Knowing
that a 1.5 ft, determine (a) the tension in cable CD, (b) the reaction
at B.
SOLUTION
From geometry of forces
tan
1
yBE
1.5 ft
where
yBE
2.0
yDE
2.0
1.5 tan 35
0.94969 ft
tan
and
1
0.94969
1.5
90
32.339
90
35
32.339
32.339
57.661
35
67.339
Applying the law of sines to the force triangle,
200 lb
sin
or
200 lb
sin 67.339
(a)
T
T
sin
B
sin 55
T
sin 57.661
B
sin 55
200 lb sin 57.661
sin 67.339
183.116 lb
or T
(b)
B
200 lb sin 55
sin 67.339
183.1 lb
177.536 lb
or B
177.5 lb
32.3
PROBLEM 4.74
Solve Problem 4.73 assuming that a
3 ft.
P4.73 A 200-lb crate is attached to the trolley-beam system shown.
Knowing that a 1.5 ft, determine (a) the tension in cable CD, (b) the
reaction
at B.
SOLUTION
From geometry of forces
tan
1
yBE
3 ft
where
yBE
yDE
2.0 ft
3tan 35
2.0
0.100623 ft
tan
and
1
0.100623
3
1.92103
90
90
1.92103
91.921
35
35
1.92103
33.079
Applying the law of sines to the force triangle,
200 lb
sin
or
200 lb
sin 33.079
(a)
T
T
sin
B
sin 55
B
sin 55
T
sin 91.921
200 lb sin 91.921
sin 33.079
366.23 lb
or T
(b)
B
200 lb sin 55
sin 33.079
366 lb
300.17 lb
or B
300 lb
1.921
PROBLEM 4.75
A 20-kg roller, of diameter 200 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 8 mm, determine the force P required to move the roller
onto the tiles if the roller is pushed to the left.
SOLUTION
Based on the roller having impending motion to the left, the only contact
between the roller and floor will be at the edge of the tile.
First note
W
20 kg 9.81 m/s 2
mg
196.2 N
From the geometry of the three forces acting on the roller
cos
1
92 mm
100 mm
and
23.074
90
30
60
23.074
36.926
Applying the law of sines to the force triangle,
W
sin
or
P
sin
196.2 N
sin 36.926
P
P
sin 23.074
127.991 N
or P
128.0 N
30
PROBLEM 4.76
A 20-kg roller, of diameter 200 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 8 mm, determine the force P required to move the roller
onto the tiles if the roller is pulled to the right.
SOLUTION
Based on the roller having impending motion to the right, the only
contact between the roller and floor will be at the edge of the tile.
First note
W
20 kg 9.81 m/s 2
mg
196.2 N
From the geometry of the three forces acting on the roller
cos
1
92 mm
100 mm
and
90
23.074
30
120
23.074
96.926
Applying the law of sines to the force triangle,
W
sin
or
P
sin
196.2 N
sin 96.926
P
P
sin 23.074
77.460 N
or P
77.5 N
30
PROBLEM 4.77
A small hoist is mounted on the back of a pickup truck and is used to lift
a 120-kg crate. Determine (a) the force exerted on the hoist by the
hydraulic cylinder BC, (b) the reaction at A.
SOLUTION
First note
W
120 kg 9.81 m/s 2
mg
1177.2 N
From the geometry of the three forces acting on the small hoist
x AD
1.2 m cos 30
1.03923 m
y AD
1.2 m sin 30
0.6 m
and
yBE
Then
tan
x AD tan 75
1
yBE
0.4 m
tan
x AD
75
180
1.03923 m tan 75
75
73.366
15
165
1
3.4785
1.03923
3.8785 m
73.366
1.63412
1.63412
163.366
Applying the law of sines to the force triangle,
W
sin
or
(a)
B
sin
1177.2 N
sin1.63412
B
A
sin15
B
sin163.366
11 816.9 N
or B
(b)
A
A
sin15
11.82 kN
75.0
10.68 kN
73.4
10 684.2 N
or A
PROBLEM 4.78
The clamp shown is used to hold the rough workpiece C. Knowing that
the maximum allowable compressive force on the workpiece is 200 N
and neglecting the effect of friction at A, determine the corresponding
(a) reaction at B, (b) reaction at A, (c) tension in the bolt.
SOLUTION
From the geometry of the three forces acting on the clamp
Then
y AD
105 mm tan 78
yBD
y AD
tan
1
493.99 mm
70 mm
493.99
yBD
195 mm
90
tan
12
78
70 mm
423.99
195
1
423.99 mm
65.301
65.301
12.6987
(a) Based on the maximum allowable compressive force on the
workpiece of 200 N,
RB
200 N
y
RB sin
or
RB
200 N
200 N
sin 65.301
220.14 N
or R B
220 N
65.3
Applying the law of sines to the force triangle,
RB
sin12
or
(b)
NA
sin
220.14 N
sin12
NA
T
sin 90
NA
sin12.6987
T
sin155.301
232.75 N
or N A
(c)
T
233 N
442.43 N
or T
442 N
PROBLEM 4.79
A modified peavey is used to lift a 0.2-m-diameter log of mass 36 kg.
Knowing that
45 and that the force exerted at C by the worker is
perpendicular to the handle of the peavey, determine (a) the force exerted
at C, (b) the reaction at A.
SOLUTION
First note
W
36 kg 9.81 m/s 2
mg
353.16 N
From the geometry of the three forces acting on the modified peavey
tan
1
1.1 m
1.1 m 0.2 m
45
45
40.236
40.236
4.7636
Applying the law of sines to the force triangle,
W
sin
or
(a)
(b)
353.16 N
sin 40.236
C
A
C
sin
A
sin135
C
sin 4.7636
A
sin135
45.404 N
or C
45.4 N
45.0
or A
387 N
85.2
386.60 N
PROBLEM 4.80
A modified peavey is used to lift a 0.2-m-diameter log of mass 36 kg.
Knowing that
60 and that the force exerted at C by the worker is
perpendicular to the handle of the peavey, determine (a) the force exerted
at C, (b) the reaction at A.
SOLUTION
First note
W
36 kg 9.81 m/s 2
mg
353.16 N
From the geometry of the three forces acting on the modified peavey
1
tan
DC
where
1.1 m
DC 0.2 m
1.1 m
a
a tan 30
R
tan 30
R
0.1 m
tan 30
0.1 m
0.073205 m
DC
1.173205 tan 30
0.67735 m
and
tan
60
1.1
0.87735
1
60
51.424
51.424
8.5756
Applying the law of sines to the force triangle,
W
sin
or
(a)
C
sin
353.16 N
sin 51.424
C
A
sin120
C
sin 8.5756
A
sin120
67.360 N
or C
(b)
A
67.4 N
30
391.22 N
or A
391 N
81.4
PROBLEM 4.81
Member ABC is supported by a pin and bracket at B and by an
inextensible cord at A and C and passing over a frictionless pulley at D.
The tension may be assumed to be the same in portion 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.
SOLUTION
From the f.b.d. of member ABC, it is seen that the member can be treated
as a three-force body.
From the force triangle
T
300
T
3T
4T
3
4
1200
T
B
T
Also,
B
5
T
4
tan
1
5
4
5
1200 lb
4
3
4
1200 lb
1500 lb
36.870
and B
1500 lb
36.9
PROBLEM 4.82
Member ABCD is supported by a pin and bracket at C and by an
inextensible cord attached at A and D and passing over frictionless
pulleys at B and E. Neglecting the size of the pulleys, determine the
tension in the cord and the reaction at C.
SOLUTION
From the geometry of the forces acting on member ABCD
tan
1
tan
1
200
300
33.690
375
200
61.928
61.928
180
33.690
180
28.237
61.928
118.072
Applying the law of sines to the force triangle,
T
sin
or
Then
80 N
T 80 N
sin 28.237
T
T
sin
C
sin 180
T
sin 33.690
80 N sin 33.690
T
C
sin118.072
T sin 28.237
543.96 N
or T
and
543.96 N sin118.072
C
544 N
C sin 33.690
865.27 N
or C
865 N
33.7
PROBLEM 4.83
Using the method of Section 4.7, solve Problem 4.18.
P4.18 Determine the reactions at A and B when (a) h
0 , (b) h
8 in.
or A
40.0 lb
30
and B
40.0 lb
30
SOLUTION
(a) Based on symmetry
30
From force triangle
A
B
40 lb
(b) From geometry of forces
tan
1
8 in.
10 in. tan 30
12.5521
10 in.
Also,
30
30
12.5521
17.4479
90
90
12.5521
102.5521
Applying law of sines to the force triangle,
40 lb
sin 30
or
A
sin 60
40 lb
sin17.4479
A
B
sin 90
A
sin 60
B
sin102.5521
115.533 lb
or A
B
115.5 lb
12.55
130.217 lb
or B
130.2 lb
30.0
PROBLEM 4.84
Using the method of Section 4.7, solve Problem 4.28.
P4.28 A lever is hinged at C and is attached to a control cable at A. If the
lever is subjected to a 300-N vertical force at B, determine
(a) the tension in the cable, (b) the reaction at C.
SOLUTION
From geometry of forces acting on lever
yDA
xDA
1
tan
where
yDA
0.24 m
y AC
0.24 m
0.2 m sin 20
0.171596 m
xDA
0.2 m cos 20
0.187939 m
tan
0.171596
0.187939
1
90
where
tan
42.397
y AC yEA
xCE
1
xCE
0.3 m cos 20
0.28191 m
y AC
0.2 m sin 20
0.068404 m
yEA
xDA
xCE tan
0.187939
0.28191 tan 42.397
0.42898 m
90
Also,
90
90
tan
1
90
90
0.49739
0.28191
29.544
71.941
18.0593
42.397
132.397
PROBLEM 4.84 CONTINUED
Applying the law of sines to the force triangle,
300 N
sin 90
or
300 N
sin18.0593
(a) T
477.18 N
(b) C
714.67 N
T
sin
T
sin 29.544
C
sin 90
C
sin132.397
or T
or C
715 N
477 N
60.5
PROBLEM 4.85
Knowing that
35o , determine the reaction (a) at B, (b) at C.
SOLUTION
From the geometry of the three forces applied to the member ABC
tan
1
yCD
R
where
yCD
R tan 55
tan
Then
1
R
0.42815R
0.42815
23.178
55
55
23.178
31.822
90
90
23.178
113.178
Applying the law of sines to the force triangle,
P
sin 55
or
(a)
B
sin 90
P
sin 31.822
C
sin 35
B
sin113.178
B
1.74344P
or B
(b)
C
C
sin 35
1.743P
55.0
1.088P
23.2
1.08780P
or C
PROBLEM 4.86
Knowing that
50o , determine the reaction (a) at B, (b) at C.
SOLUTION
From the geometry of the three forces acting on member ABC
tan
1
yDC
R
where
yDC
R
y AD
R 1
tan 90
50
0.160900 R
1
tan
Then
0.160900
9.1406
90
90
9.1406
80.859
40
40
9.1406
49.141
Applying the law of sines to the force triangle,
P
sin 40
or
(a)
(b)
B
sin 90
P
sin 49.141
B
C
C
sin 50
B
sin 80.859
C
sin 50
1.30540P
or B
1.305P
40.0
or C
1.013P
9.14
1.01286P
PROBLEM 4.87
A slender rod of length L and weight W is held in equilibrium as shown,
with one end against a frictionless wall and the other end attached to a
cord of length S. Derive an expression for the distance h in terms of L and
S. Show that this position of equilibrium does not exist if S 2L.
SOLUTION
From the f.b.d of the three-force member AB, forces must intersect at D.
Since the force T intersects point D, directly above G,
yBE
h
For triangle ACE:
S2
2
AE
2h
2
(1)
For triangle ABE:
L2
AE
2
2
h
(2)
Subtracting Equation (2) from Equation (1)
S2
L2
3h 2
(3)
S2
or h
As length S increases relative to length L, angle
is vertical. At this vertical position:
h
L
S
h
or
S2
or
or
3
S2
L2
3 S
L
2
For
3 S2
0
S2
S
3SL
L
6SL
2 L2
0
L
h
S
S
L
L2
2SL
2S 2
0
or
and
L2
L
3S 2
(4)
6SL
3L2
4 L2
S
S
3
increases until rod AB
S
Therefore, for all positions of AB
L2
L S
2L
L
Minimum value of S is L
For
S
2L
0
S
2L
Maximum value of S is 2L
Therefore, equilibrium does not exist if S
2L
PROBLEM 4.88
A slender rod of length L 200 mm is held in equilibrium as shown,
with one end against a frictionless wall and the other end attached to a
cord of length S 300 mm. Knowing that the mass of the rod is 1.5 kg,
determine (a) the distance h, (b) the tension in the cord, (c) the reaction
at B.
SOLUTION
From the f.b.d of the three-force member AB, forces must intersect at D.
Since the force T intersects point D, directly above G,
yBE
h
For triangle ACE:
S2
2
AE
2h
2
(1)
For triangle ABE:
L2
AE
2
h
2
(2)
Subtracting Equation (2) from Equation (1)
S2
L2
3h 2
S2
or h
(a) For L
S
200 mm and
300
h
L2
3
300 mm
2
200
2
3
129.099 mm
or h
(b) Have
W
1.5 kg 9.81 m/s 2
mg
and
sin
1
2h
s
129.1 mm
14.715 N
sin
1
2 129.099
300
59.391
From the force triangle
T
W
sin
14.715 N
sin 59.391
17.0973 N
or T
(c)
B
W
tan
14.715 N
tan 59.391
17.10 N
8.7055 N
or B
8.71 N
PROBLEM 4.89
A slender rod of length L and weight W is attached to collars which 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 f.b.d of the slender rod AB, the three forces intersect at
C. From the force geometry
xGB
y AB
tan
where
and
tan
y AB
L cos
xGB
1
L sin
2
1
2
L sin
L cos
1
tan
2
or tan
2 tan
PROBLEM 4.90
A 10-kg slender rod of length L is attached to collars which can slide
freely along the guides shown. Knowing that the rod is in equilibrium and
that
25 , determine (a) the angle
that the rod forms with the
vertical, (b) the reactions at A and B.
SOLUTION
(a) As shown in the f.b.d. of the slender rod AB, the three forces
intersect at C. From the geometry of the forces
xCB
yBC
tan
where
and
xCB
1
L sin
2
yBC
L cos
1
tan
2
tan
or
tan
For
2 tan
25
tan
2 tan 25
0.93262
43.003
or
(b)
W
10 kg 9.81 m/s 2
mg
43.0
98.1 N
From force triangle
A
W tan
98.1 N tan 25
45.745 N
or A
and
B
W
cos
98.1 N
cos 25
45.7 N
108.241 N
or B
108.2 N
65.0
PROBLEM 4.91
A uniform slender rod of mass 5 g and length 250 mm is balanced on a
glass of inner diameter 70 mm. Neglecting friction, determine the angle
corresponding to equilibrium.
SOLUTION
From the geometry of the forces acting on the three-force member AB
Triangle ACF
Triangle CEF
yCF
d tan
xFE
yCF tan
d tan 2
Triangle AGE
d
cos
xFE
L
2
2d
1
L
Now
1
tan 2
Then
d
tan 2
sec 2
cos
and
2d
sec 2
L
d
70 mm and
2d
L
cos
and
L
2 70
250
cos3
1
cos
sec
2d
1
L cos 2
cos3
For
d tan 2
L
2
250 mm
0.56
0.82426
34.487
or
34.5
PROBLEM 4.92
Rod AB is bent into the shape of a circular arc 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 1 in. and R 5 in.
SOLUTION
Since
Slope of ED is
yED
a,
xED
45
45
slope of HC is
Also
DE
DH
and
2a
1
DE
2
HE
a
2
For triangles DHC and EHC
a/ 2
R
sin
c
Now
For
a
2R
R sin 45
a
1 in. and
sin
1 in.
2 5 in.
R
5 in.
0.141421
or
8.1301
and
c
5 in. sin 45
8.1301
8.13
3.00 in.
or c
3.00 in.
PROBLEM 4.93
A uniform rod AB of weight W and 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
or
x AG
xA
AE cos 2
AG cos
and
2 R cos 2
Now
cos 2
then
4cos 2
4cos 2
or
R cos
2cos 2
1
2
cos
cos
2
0
Applying the quadratic equation
cos
0.84307
32.534
and
and
cos
0.59307
126.375 (Discard)
or
32.5
PROBLEM 4.94
A uniform slender rod of mass m and length 4r rests on the surface shown
and is held in the given equilibrium position by the force P. Neglecting
the effect of friction at A and C, (a) determine the angle (b) derive an
expression for P in terms of m.
SOLUTION
The forces acting on the three-force member intersect at D.
(a) From triangle ACO
tan
1
r
3r
tan
1
3
1
18.4349
(b) From triangle DCG
r
tan
DO
and
r
tan18.4349
DC
r
tan
where
yDO
18.43
r
DC
tan
DC
or
3r
r
3r
4r
yDO
x AG
1
DO cos
4r cos18.4349
3.4947r
and
x AG
2r cos
2r cos18.4349
1.89737r
1
tan
where
90
3.4947r
1.89737r
63.435
90
45
135.00
Applying the law of sines to the force triangle,
mg
RA
sin
sin 90
RA
Finally,
P
0.44721 mg
RA cos
0.44721mg cos 63.435
0.20000mg
or P
mg
5
PROBLEM 4.95
A uniform slender rod of length 2L and mass m rests against a roller at D
and is held in the equilibrium position shown by a cord of length a.
Knowing that L 200 mm, determine (a) the angle , (b) the length a.
SOLUTION
(a) The forces acting on the three-force member AB intersect at E. Since
triangle DBC is isosceles, DB a.
From triangle BDE
ED
DB tan 2
a tan 2
From triangle GED
L a
tan
ED
L a
tan
a tan 2
or
a
From triangle BCD
1
2
a tan tan 2
1.25L
cos
L
a
or
1
L
(1)
1.6cos
(2)
Substituting Equation (2) into Equation (1) yields
1.6cos
Now
tan tan 2
1
tan tan 2
sin sin 2
cos cos 2
sin
cos
2sin cos
2 cos 2
1
2 1 cos 2
2cos 2
1
2 1 cos 2
2cos 2
1
Then
1.6cos
1
or
3.2cos3
1.6cos
Solving numerically
0
23.515
(b) From Equation (2) for L
a
1
200 mm and
5 200 mm
8 cos 23.515
or
23.5
23.5
136.321 mm
or a
136.3 mm
PROBLEM 4.96
Gears A and B are attached to a shaft supported by bearings at C and D.
The diameters of gears A and B are 150 mm and 75 mm, respectively, and
the tangential and radial forces acting on the gears are as shown.
Knowing that the system rotates at a constant rate, determine the
reactions at C and D. Assume that the bearing at C does not exert any
axial force, and neglect the weights of the gears and the shaft.
SOLUTION
Assume moment reactions at the bearing supports are zero. From f.b.d. of
shaft
0:
Fx
MD
z -axis
0:
0
Dx
C y 175 mm
482 N 75 mm
2650 N 50 mm
Cy
or
963.71 N
Cy
MD
y -axis
0
964 N j
0: Cz 175 mm
1325 N 75 mm
964 N 50 mm
Cz
843.29 N
Cz
or
843 N k
and C
MC
z -axis
0:
0
482 N 100 mm
964 N j
Dy 175 mm
2650 N 225 mm
Dy
or
y -axis
0:
3130 N j
1325 N 100 mm
Dz 175 mm
964 N 225 mm
Dz
or
0
3131.7 N
Dy
MC
843 N k
0
482.29 N
Dz
482 N k
and D
3130 N j
482 N k
PROBLEM 4.97
Solve Problem 4.96 assuming that for gear A the tangential and radial
forces are acting at E, so that FA (1325 N)j (482 N)k.
P4.96 Gears A and B are attached to a shaft supported by bearings at C and
D. The diameters of gears A and B are 150 mm and 75 mm, respectively,
and the tangential and radial forces acting on the gears are as shown.
Knowing that the system rotates at a constant rate, determine the reactions at
C and D. Assume that the bearing at C does not exert any axial force, and
neglect the weights of the gears and the shaft.
SOLUTION
Assume moment reactions at the bearing supports are zero. From f.b.d. of
shaft
Fx
MD
z -axis
0:
0:
Dx
C y 175 mm
0
1325 N 75 mm
2650 N 50 mm
Cy
or
189.286 N
Cy
MD
y -axis
0
189.3 N j
0: C z 175 mm
482 N 75 mm
964 N 50 mm
Cz
482.00 N
Cz
or
482 N k
and C
MC
z -axis
0:
0
189.3 N j
D y 175 mm
1325 N 100 mm
2650 N 225 mm
Dy
or
y -axis
0:
4160 N j
482 N 100 mm
964 N 225 mm
Dz
or
Dz
0
4164.3 N
Dy
MC
482 N k
Dz 175 mm
0
964.00 N
964 N k
and D
4160 N j
964 N k
PROBLEM 4.98
Two transmission belts pass over sheaves welded to an axle supported by
bearings at B and D. The sheave at A has a radius of 50 mm, and the
sheave at C has a radius of 40 mm. Knowing that the system rotates with
a constant rate, determine (a) the tension T, (b) the reactions at B and D.
Assume that the bearing at D does not exert any axial thrust and neglect
the weights of the sheaves and the axle.
SOLUTION
Assume moment reactions at the bearing supports are zero. From f.b.d. of shaft
(a)
M x-axis
0:
240 N
180 N 50 mm
300 N
T
40 mm
0
T
(b)
Fx
MD
z -axis
0: Bx
0:
0
300 N
375 N 120 mm
By
MD
y -axis
0:
240 N
By 240 mm
0
Bz 240 mm
0
337.5 N
180 N 400 mm
Bz
700 N
or B
MB
z -axis
0:
300 N
y -axis
0:
240 N
700 N k
0
337.5 N
180 N 160 mm
Dz
338 N j
D y 240 mm
375 N 120 mm
Dy
MB
375 N
Dz 240 mm
0
280 N
or D
338 N j
280 N k
PROBLEM 4.99
For the portion of a machine shown, the 4-in.-diameter pulley A and
wheel B are fixed to a shaft supported by bearings at C and D. The spring
0, and the bearing at C does
of constant 2 lb/in. is unstretched when
180° and that the machine is
not exert any axial force. Knowing that
at rest and in equilibrium, determine (a) the tension T, (b) the reactions at
C and D. Neglect the weights of the shaft, pulley, and wheel.
SOLUTION
First, determine the spring force, FE , at
180 .
FE
ks x
ks
where
x
yE
final
yE
2 lb/in.
12 in.
initial
FE
3.5 in.
12 in.
2 lb/in. 7.0 in.
3.5 in.
7.0 in.
14.0 lb
(a) From f.b.d. of machine part
Mx
0:
34 lb 2 in.
T
(b)
MD
z -axis
T 2 in.
0
34 lb
0:
or T
C y 10 in.
FE 2 in.
C y 10 in.
Cy
MD
y -axis
4.2 lb
0: C z 10 in.
Cz
or
34 lb 4 in.
27.2 lb
or
1 in.
0
14.0 lb 3 in.
Cy
0
4.20 lb j
34 lb 4 in.
Cz
34.0 lb
0
27.2 lb k
and C
4.20 lb j
27.2 lb k
PROBLEM 4.99 CONTINUED
Fx
0: Dx
MC
z -axis
0
0: Dy 10 in.
or
Dy 10 in.
Dy
MC
y -axis
Dz
18.2 lb
0:
40.8 lb
or
FE 12 in.
14.0 13 in.
Dy
2 34 lb 6 in.
or
Dz
1 in.
0
0
18.20 lb j
Dz 10 in.
0
40.8 lb k
and D
18.20 lb j
40.8 lb k
PROBLEM 4.100
Solve Problem 4.99 for
90°.
P4.99 For the portion of a machine shown, the 4-in.-diameter pulley A and
wheel B are fixed to a shaft supported by bearings at C and D. The spring of
0, and the bearing at C does not
constant 2 lb/in. is unstretched when
180° and that the machine is at rest
exert any axial force. Knowing that
and in equilibrium, determine (a) the tension T, (b) the reactions at C and D.
Neglect the weights of the shaft, pulley, and wheel.
SOLUTION
First, determine the spring force, FE , at
90 .
FE
where
2 lb/in.
ks
and
x
Lfinal
Linitial
3.5
FE
Then
FE
2
ks x
12
2
12
2 lb/in. 4.0 in.
12.0
8.0 lb j
12.5
3.5
8.0 lb k
12.5
3.5
12.5
8.5
4.0 in.
8.0 lb
7.68 lb j
2.24 lb k
(a) From f.b.d. of machine part
Mx
0:
T
(b)
MD
z -axis
34 lb 2 in.
y -axis
7.68 lb 3.5 in.
0
20.56 lb
0:
or T
C y 10 in.
Cy
MD
T 2 in.
2.304 lb
0: Cz 10 in.
Cz
7.68 lb 3.0 in.
or
34 lb 4.0 in.
21.152 lb
or
20.6 lb
0
Cy
2.30 lb j
20.56 lb 4.0 in.
Cz
2.24 lb 3 in.
0
21.2 lb k
and C
2.30 lb j
21.2 lb k
PROBLEM 4.100 CONTINUED
Fx
MC
0: Dx
z -axis
0
0: Dy 10 in.
Dy
MC
y -axis
0:
7.68 lb 13 in.
9.984 lb
34 lb 6 in.
Dz
0
or
Dy
20.56 lb 6 in.
35.648 lb
or
9.98 lb j
Dz 10 in.
Dz
2.24 lb 13 in.
0
35.6 lb k
and D
9.98 lb j
35.6 lb k
PROBLEM 4.101
A 1.2 2.4-m sheet of plywood having a mass of 17 kg has been
temporarily placed among three pipe supports. The lower edge of the
sheet rests on small collars A and B and its upper edge leans against pipe
C. Neglecting friction at all surfaces, determine the reactions at A, B,
and C.
SOLUTION
W
First note
17 kg 9.81 m/s 2
mg
h
1.2
2
2
1.125
166.77 N
0.41758 m
From f.b.d. of plywood sheet
Mz
0: C h
W
1.125 m
C 0.41758 m
C
MB
y -axis
0:
x -axis
0:
y -axis
0:
224.65 N
112.324 N
166.77 N 0.3 m
Ay
MA
166.77 N 0.5625 m
or
41.693 N
224.65 N 0.6 m
Bx
112.325 N
Ax 1.2 m
or
Ax
Ay 1.2 m
or
0
225 N i
C
224.65 N 0.6 m
Ax
MB
0
2
Ay
0
112.3 N i
0
41.7 N j
Bx 1.2 m
0
or
112.3 N i
Bx
PROBLEM 4.101 CONTINUED
MA
x -axis
0: By 1.2 m
By
166.77 N 0.9 m
125.078 N
or
By
0
125.1 N j
A
B
112.3 N i
41.7 N j
112.3 N i
125.1 N j
C
225 N i
PROBLEM 4.102
The 200 200-mm square plate shown has a mass of 25 kg and is
supported by three vertical wires. Determine the tension in each wire.
SOLUTION
W
First note
25 kg 9.81 m/s 2
mg
245.25 N
From f.b.d. of plate
Mx
0:
TA
Mz
0: TB 160 mm
2TC
0: TA
TB
TC
TC
TC
0
(1)
245.25 N 100 mm
0
153.281 N
245.25 N
TB
TC 200 mm
245.25 N
TC 160 mm
TB
Fy
TA 100 mm
245.25 N 100 mm
(2)
0
245.25
TA
(3)
91.969 N
(4)
Equating Equations (2) and (3) yields
TA
245.25 N
TA
or
153.281 N
92.0 N
Substituting the value of TA into Equation (1)
TC
245.25 N
91.969 N
2
TC
or
76.641 N
(5)
76.6 N
Substituting the value of TC into Equation (2)
TB
153.281 N
76.641 N
76.639 N
or
TB
76.6 N
TA
92.0 N
TB
76.6 N
TC
76.6 N
PROBLEM 4.103
The 200 200-mm square plate shown has a mass of 25 kg and is
supported by three vertical wires. Determine the mass and location of the
lightest block which should be placed on the plate if the tensions in the
three cables are to be equal.
SOLUTION
WG
First note
25 kg 9.81 m/s 2
m p1g
W1
m 9.81 m/s 2
mg
245.25 N
9.81m N
From f.b.d. of plate
Fy
Mx
0: 3T
W1
0 (1)
0: WG 100 mm
300T
or
Mz
WG
W1 z
100WG
0: 2T 160 mm
W1z
100WG
Eq. 1
Eq. 2
W1x
100W1
z
Now, 3
Eq. 3
320
Eq. 1
3 320T
100 mm
0
T 200 mm
0
0
(2)
WG 100 mm
or 320T
Eliminate T by forming 100
T 100 mm
W1 x
0
0
(3)
W1z
z
0
200 mm,
okay
yields
3 100 WG
3W1x
320 3T
320WG
320W1
0
PROBLEM 4.103 CONTINUED
20WG
or
W1
WG
or
The smallest value of
320
0
20
320
W1
will result in the smallest value of W1 since WG is given.
WG
Use x
W1
WG
and then
WG
14
W1
and
3x
3x W1
m
W1
g
xmax
200 mm
20
3 200 320
245.25 N
14
17.5179 N
9.81 m/s 2
1
14
17.5179 N minimum
1.78571 kg
at x
or m
1.786 kg
200 mm, z
100 mm
PROBLEM 4.104
A camera of mass 240 g is mounted on a small tripod of mass 200 g.
Assuming that the mass of the camera is uniformly distributed and that
the line of action of the weight of the tripod passes through D, determine
0,
(a) the vertical components of the reactions at A, B, and C when
(b) the maximum value of if the tripod is not to tip over.
SOLUTION
WC
mC g
0.24 kg 9.81 m/s 2
2.3544 N
Wtp
mtp g
0.20 kg 9.81 m/s 2
1.9620 N
First note
For
0
xC
zC
60 mm
24 mm
36 mm
0
(a) From f.b.d. of camera and tripod as projected onto plane ABCD
Fy
0: Ay
By
Ay
Cy
By
WC
Cy
Wtp
2.3544 N
Mx
0: C y 38 mm
By 38 mm
Mz
0: By 35 mm
C y 35 mm
9 Ay
0
7By
7C y
1.9620 N
Cy
0
4.3164 N
(1)
By
(2)
2.3544 N 36 mm
Ay 45 mm
0
16.9517
Substitute C y with By from Equation (2) into Equations (1) and (3), and solve by elimination
7 Ay
2By
4.3164
9 Ay
14By
16.9517
16 Ay
47.166
(3)
PROBLEM 4.104 CONTINUED
Ay
2.9479 N
or A y
Substituting Ay
2.9479 N into Equation (1)
2.9479 N
2By
4.3164
By
0.68425 N
Cy
0.68425 N
or B y
(b) By
2.95 N
Cy
0.684 N
0 for impending tipping
From f.b.d. of camera and tripod as projected onto plane ABCD
Fy
0: Ay
Cy
WC
Wtp
Ay
Mx
0: C y 38 mm
0: C y 35 mm
Eq. 1
Eq. 3
7C y
(1)
4.3164 N
36 mm sin
0
(2)
2.2305sin
Ay 45 mm
9 Ay
Forming 7
Cy
2.3544 N
Cy
Mz
0
2.3544 N
36 mm cos
16.9517 N cos
0
(3)
yields
16 Ay
30.215 N
16.9517 N cos
(4)
PROBLEM 4.104 CONTINUED
Substituting Equation (2) into Equation (3)
9 Ay
Forming 9
Eq. 4
16
15.6134 N sin
Eq. 5
(5)
yields
249.81 N sin
or
16.9517 N cos
cos 2
271.93 N
2.2916 N
cos 2
Now
5.4323sin 2
118.662 N cos
2
2.1053 N sin
1
sin 2
9.6490sin
4.2514
0
Using quadratic formula to solve,
sin
0.80981 and sin
54.078
and
0.96641
75.108
or
max
54.1 before tipping
PROBLEM 4.105
Two steel pipes AB and BC, each having a weight per unit length of
5 lb/ft, are welded together at B and are supported by three wires.
Knowing that a 1.25 ft, determine the tension in each wire.
SOLUTION
First note
WAB
5 lb/ft 2 ft
10 lb
WBC
5 lb/ft 4 ft
20 lb
W
WAB
WBC
30 lb
To locate the equivalent force of the pipe assembly weight
rG/B
xG i
or
W
ri
zGk
Wi
rG
30 lb j
30 lb xGk
From i-coefficient
k-coefficient
WAB
AB
1 ft k
rG
BC
10 lb j
30 lb zG i
10 lb ft i
zG
10 lb ft
30 lb
1
ft
3
xG
40 lb ft
30 lb
1
1 ft
3
WBC
2 ft i
20 lb j
40 lb ft k
From f.b.d. of piping
Mx
TA
0: W zG
1
1
ft 30 lb ft
2
3
Fy
0: 5 lb
TD
TA 2 ft
5 lb
TD
TC
TC
0
or
TA
30 lb
0
25 lb
5.00 lb
(1)
PROBLEM 4.105 CONTINUED
Mz
0: TD 1.25 ft
1.25TD
4TC
Equation (3)
TD
From Equation (1)
Results:
21.818 lb
TC
25
30 lb
4
ft
3
0
40 lb ft
4TD
4 Equation 1
Equation (2)
TC 4 ft
or
21.818
(2)
4TC
100
2.75TD
60
TD
3.1818 lb
(3)
21.8 lb
or
TC
3.18 lb
TA
5.00 lb
TC
3.18 lb
TD
21.8 lb
PROBLEM 4.106
For the pile assembly of Problem 4.105, determine (a) the largest
permissible value of a if the assembly is not to tip, (b) the corresponding
tension in each wire.
P4.105 Two steel pipes AB and BC, each having a weight per unit length
of 5 lb/ft, are welded together at B and are supported by three wires.
Knowing that a 1.25 ft, determine the tension in each wire.
SOLUTION
First note
WAB
5 lb/ft 2 ft
10 lb
WBC
5 lb/ft 4 ft
20 lb
From f.b.d. of pipe assembly
Fy
Mx
0: TA
TC
TD
10 lb
TA
TC
TD
30 lb
0:
or
0
5.00 lb
(2)
TC
From Equations (1) and (2)
Mz
0
(1)
TA 2 ft
10 lb 1 ft
TA
or
20 lb
0: TC 4 ft
4 ft TC
TD amax
TD amax
TD
25 lb
20 lb 2 ft
40 lb ft
(3)
0
(4)
PROBLEM 4.106 CONTINUED
Using Equation (3) to eliminate TC
4 25
TD
amax
or
TD amax
40
60
TD
4
By observation, a is maximum when TD is maximum. From Equation (3), TD
Therefore, TD
max
max
occurs when TC
0.
25 lb and
amax
4
60
25
1.600 ft
Results: (a)
amax
1.600 ft
(b)
TA
5.00 lb
TC
TD
0
25.0 lb
PROBLEM 4.107
A uniform aluminum rod of weight W is bent into a circular ring of radius
R and is supported by three wires as shown. Determine the tension in
each wire.
SOLUTION
From f.b.d. of ring
Fy
0: TA
TB
TA
Mx
0: TA R
TA
Mz
TC
TB
TC
0
W
TC R sin 30
(1)
0
0.5TC
0: TC R cos 30
TB
W
(2)
TB R
0
0.86603TC
(3)
Substituting TA and TB from Equations (2) and (3) into Equation (1)
0.5TC
0.86603TC
TC
TC
W
0.42265W
From Equation (2)
TA
0.5 0.42265W
0.21132W
From Equation (3)
TB
0.86603 0.42265W
0.36603W
or TA
0.211W
TB
0.366W
TC
0.423W
PROBLEM 4.108
A uniform aluminum rod of weight W is bent into a circular ring of radius
R and is supported by three wires as shown. A small collar of weight W
is then placed on the ring and positioned so that the tensions in the three
wires are equal. Determine (a) the position of the collar, (b) the value of
W , (c) the tension in the wires.
SOLUTION
Let
angle from x-axis to small collar of weight W
From f.b.d. of ring
Fy
Mx
0: 3T
W
0: T R
W
Mz
W R sin
0
1
T
2
0: T R cos 30
or
(1)
T R sin 30
W sin
or
0
W cos
(2)
W R cos
T R
0
3
T
2
1
(3)
Dividing Equation (2) by Equation (3)
1
2
tan
1
75.000
1
3.7321
and
Based on Equations (2) and (3),
for W , which is not acceptable.
(a)
3
2
W is located at
255.00
75.000 will give a negative value
255 from the x-axis or 15 from A
towards B.
(b) From Equation (1) and Equation (2)
W
3
2W
W
sin 255
W
0.20853W
or W
0.209W
or T
0.403W
(c) From Equation (1)
T
2 0.20853W sin 255
0.40285W
PROBLEM 4.109
An opening in a floor is covered by a 3 4-ft sheet of plywood weighing
12 lb. 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
From f.b.d. of plywood sheet
Mx
0:
12 lb 2 ft
Cy
MB
z -axis
0:
MA
z -axis
0:
6.8571 lb
12 lb 1 ft
Ay
7.7143 lb
12 lb 1 ft
By
C y 3.5 ft
2.5714 lb
or
0
Cy
6.8571 lb 0.5 ft
or
By 2 ft
or
Ay
6.86 lb
Ay 2 ft
7.71 lb
6.8571 lb 2.5 ft
By
0
0
2.57 lb
(a)
Ay
7.71 lb
(b)
By
2.57 lb
(c)
Cy
6.86 lb
PROBLEM 4.110
Solve Problem 4.109 assuming that the small block C is moved and
placed under edge DE at a point 0.5 ft from corner E.
SOLUTION
First,
rB/ A
2 ft i
rC/ A
2 ft i
4 ft k
rG/ A
1 ft i
2 ft k
From f.b.d. of plywood sheet
MA
0: rB/ A
2 ft i
By j
Bzk
2 ft i
By j
1 ft i
2B yk
2 Bz j
i-coeff.
4C y
24
j-coeff.
2 Bz
0
rC/ A
Cy j
Bz k
2 ft k
2C yk
0
k-coeff.
2B y
2C y
12
0
or
2By
2 6
12
0
4C y i
rG/ A
2 ft i
12 lb j
12k
Wj
0
4 ft k
0
24i
0
Cy
6.00 lb
Bz
0
By
0
Cy j
PROBLEM 4.110 CONTINUED
F
0: Ay j
Azk
By j
Ay j
Az k
0j
12
0
j-coeff.
Ay
6
k-coeff.
Az
0
Bzk
0k
Cy j
6j
12 j
Wj
0
0
Ay
6.00 lb
Az
0
a) Ay
6.00 lb
b) By
0
c) C y
6.00 lb
PROBLEM 4.111
The 10-kg square plate shown is supported by three vertical wires.
Determine (a) the tension in each wire when a 100 mm, (b) the value
of a for which tensions in the three wires are equal.
SOLUTION
First note
(a)
W
10 kg 9.81 m/s 2
mg
98.1 N
(a) From f.b.d. of plate
Fy
Mx
0: TA
TB
TC
W
0
TA
TB
TC
0: W 150 mm
TB 300 mm
6TB
Mz
0: TB 100 mm
2TC
(1)
TC 100 mm
0
294.3
TC 300 mm
6TB
Equation (2)
98.1 N
18TC
(2)
98.1 N 150 mm
0
882.9
(3)
Equation (3)
16TC
588.6
TC
36.788 N
or
TC
36.8 N
TB
36.8 N
TA
24.5 N
Substitution into Equation (2)
6TB
2 36.788 N
TB
294.3 N
36.788 N
or
From Equation (1)
TA
36.788
TA
36.788
24.525 N
98.1 N
or
PROBLEM 4.111 CONTINUED
(b)
(b) From f.b.d. of plate
Fy
0: 3T
W
0
1
W
3
T
Mx
0: W 150 mm
(1)
T a
T 300 mm
150W
a 300
T
0
(2)
Equating Equation (1) to Equation (2)
1
W
3
or
150W
a 300
a
300
3 150
or a
150.0 mm
PROBLEM 4.112
The 3-m flagpole AC forms an angle of 30o 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 0.9 m, determine the tension in each brace and the
reaction at C.
SOLUTION
TBE can be found from M about line CE
From f.b.d. of flagpole
M CE
where
0.9 m i
CE
rB/C
0.9
0.9
0.9 m sin 30
m
j
rA/C
CE
0
FA
j
0.9 m cos 30 k
0.9 m i
0.9 m
0.9
0.9 m i
0.70711i
FA
1
i
2
TBD
0.77942 m k
BDTBD
rA/C
2
rB/C
CE
0.9 m j
2
0.45 m j
TBD
0:
0.45 m j
0.35355 j
3 m sin 30 j
0.9 m sin 30
2
0.45
0.77942 m k
2
j
0.77942
0.9 m cos 30 k
2
m
TBD
1.62
0.61237k TBD
3 m cos30 k
1.5 m j
2.5981 m k
300 N j
1
1
0
TBD
0
0.45
0.77942
2
0.70711 0.35355 0.61237
1
0
0
1
0
1.5 2.5981
300
0
1
2
0
TBD
PROBLEM 4.112 CONTINUED
or
1.10227TBD
TBD
TBE
The reaction forces at C are found from F
0:
Fy
0:
TBD
TBD
Cy
0: Cz
Cz
TBD
707.12 N
or TBD
707 N
or TBE
707 N
0
TBE
x
y
TBE
300 N
TBD
Cx
x
Cy
y
0
300 N
or
Cx
0
0
2 0.35355 707.12 N
Cy
Fz
0
707.12 N
Based on symmetry with yz-plane,
Fx
779.43
200.00 N
TBE
z
z
0
2 0.61237 707.12 N
Cz
866.04 N
or C
200 N j
866 N k
PROBLEM 4.113
A 3-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.
SOLUTION
From f.b.d. of boom
M AE
where
0:
AE
2.1 m j
AE
2.1
2
0.27451j
rB/ A
1.8 m i
TBD
BDTBD
rB/ A
TBD
FC
3.0 m i
4 kN j
rC/ A
FC
0
1.8 m k
1.8
2
m
0.23529k
1.8 m i
1.8
2
0.54545i
rC/ A
AE
2.1 m j
2.1
2
1.8 m k
1.8
0.63636 j
2
m
TBD
0.54545k TBD
PROBLEM 4.113 CONTINUED
0
0.27451 0.23529
1.8
0
0
TBD
0.54545 0.63636 0.54545
0.149731
TBD
0.149729 1.8TBD
TBD
Fz
0: Az
TBD
z
TBE
Fy
0: Ay
TBD
y
TBD
Ay
0: Ax
Ax
2.82348
z
y
0
4 kN
2 0.63636 5.2381 kN
TBD
Az
or TBD
5.24 kN
or TBE
5.24 kN
0
0
4 kN
0
2.6666 kN
x
TBE
x
0
2 0.54545 5.2381 kN
Ax
0
0
5.2381 kN
Ay
Fx
0.23529
0
0
5.2381 kN
TBE
Based on symmetry,
0 0.27451
3
0
0
4
0
5.7142 kN
and A
5.71 N i
2.67 N j
PROBLEM 4.114
An 8-ft-long boom is held by a ball-and-socket joint at C and by two
cables AD and BE. Determine the tension in each cable and the reaction
at C.
SOLUTION
From f.b.d. of boom
M CE
where
0:
CE
2 ft j
CE
rA/C
TAD
2
2
rA/C
TAD
3 ft k
3
2
1
2j
13
ft
rA/C
FA
0
3k
8 ft i
ADTAD
8 ft i
8
2
1
TAD
9
FA
CE
1 ft j
1
8i
2
4 ft k
4
j
2
ft
TAD
4k
198 lb j
0 2 3
TAD
8 0 0
9 13
8 1 4
0 2 3
198
8 0 0
13
0 1 0
0
PROBLEM 4.114 CONTINUED
64
TAD
9 13
24
TAD
24
198
13
0
486.00 lb
or TAD
M CD
where
1 ft j
CD
rB/C
TBE
0:
CD
4 ft k
rB/C
TBE
1
1j
17
17 ft
rA/C
CD
486 lb
FA
4k
6 ft i
BETBE
6 ft i
6
2
2 ft j
2
2
3 ft k
3
0 1 4
T
6 0 0 BE
7 17
6 2 3
18
TBE
7
48
TBE
2
ft
1
TBE
7
TBE
0 1 4
198
8 0 0
17
0 1 0
32 198
6i
2j
3k
0
0
672.00 lb
or TBE
Fx
0: C x
Cx
Fy
0: C y
Cy
TAD
x
0: Cz
Cz
0
x
8
486
9
6
672
7
Cx
1008 lb
TAD
y
1
486
9
Cy
Fz
TBE
TAD
z
4
486
9
Cz
TBE
672 lb
0
198 lb
y
2
672
7
0
198 lb
0
48.0 lb
TBE
3
7
0
z
672
0
72.0 lb
or C
1008 lb i
48.0 lb j
72.0 lb k
PROBLEM 4.115
Solve Problem 4.114 assuming that the given 198-lb load is replaced
with two 99-lb loads applied at A and B.
P4.114 An 8-ft-long boom is held by a ball-and-socket joint at C and by
two cables AD and BE. Determine the tension in each cable and the
reaction at C.
SOLUTION
From f.b.d. of boom
M CE
where
0:
CE
rA/C
TAD
2 ft j
CE
2
2
rA/C
8 ft i
rB/C
6 ft i
TAD
ADTAD
rA/C
CE
3 ft k
3
2
1
2j
13
ft
8 ft i
8
2
1
TAD
9
FA
99 lb j
FB
99 lb j
0 2 3
T
8 0 0 AD
9 13
8 1 4
FA
CE
8i
0 2 3
99
8 0 0
13
0 1 0
2
4 ft k
4
j
FB
3k
1 ft j
1
rB/C
2
ft
TAD
4k
0 2 3
99
6 0 0
13
0 1 0
0
0
PROBLEM 4.115 CONTINUED
64
TAD
9 13
24
24
TAD
or
99
13
18
0
425.25 lb
or TAD
M CD
where
0:
CD
1 ft j
TBE
4 ft k
CD
6 ft i
rA/C
8 ft j
BETBE
6 ft i
0 1 4
TBE
6 0 0
7 17
6 2 3
18
48
2
2
3 ft k
3
2
ft
0 1 4
8 0 0
0 1 0
99
17
32
24
TBE
7 17
TBE
or
FA
CD
rB/C
FB
0
4k
2 ft j
2
6
rA/C
CD
1
j
17
17
rB/C
TBE
rB/C
TBE
TBE
7
0 1 4
6 0 0
0 1 0
99
17
6i
99
17
2j
3k
0
0
588.00 lb
or TBE
Fx
0: C x
Cx
TAD
x
TBE
x
8
425.25
9
0: C y
Cy
TAD
y
TBE
1
425.25
9
y
Fz
0: Cz
Cz
TAD
z
TBE
4
425.25
9
0
882 lb
99
99
2
588.00
7
Cy
588 lb
0
6
588.00
7
Cx
Fy
425 lb
0
198
0
17.25 lb
z
0
3
588.00
7
Cz
0
63.0 lb
or C
882 lb i
17.25 lb j
63.0 lb k
PROBLEM 4.116
The 18-ft pole ABC is acted upon by a 210-lb force as shown. The pole is
held by a ball-and-socket joint at A and by two cables BD and BE. For
a 9 ft, determine the tension in each cable and the reaction at A.
SOLUTION
From f.b.d. of pole ABC
M AE
where
4.5 ft i
AE
4.5
9 ft j
rC/ A
18 ft j
TBD
BDTBD
AE
9 ft k
2
rB/ A
0:
9
2
4.5 ft i
TBD
13.5
FC
CF
TBD
AE
1
4.5i
101.25
ft
4.5
rB/ A
9 ft j
2
9
4.5i
9i
210 lb
9
2
9
9j
9k
18 j
18
2
ft
6k
2
6
4.5 0 9
TBD
0 9 0
13.5 101.25
4.5 9 9
0
FC
9k
9 ft k
2
rC/ A
2
TBD
210 lb
10 lb
4.5 0 9
0 18 0
9 18 6
9i
18 j
10 lb
101.25
6k
0
PROBLEM 4.116 CONTINUED
364.5
364.5
486
TBD
13.5 101.25
10 lb
101.25
TBD
and
1458
0
360.00 lb
or TBD
M AD
where
4.5 ft i
AD
4.5
2
rB/ A
9 ft j
rC/ A
18 ft j
TBE
BETBE
0:
rB/ A
AD
9 ft k
9
2
4.5 ft i
9 ft j
2
9
2
9
4.5 0
9
TBE
0 9 0
13.5 101.25
4.5 9 9
364.5
364.5
13.5 101.25
or
ft
TBE
4.5i
13.5
TBE
4.5 0
9
0 18 0
9 18 6
486
TBE
9j
10 lb
101.25
1458 10 lb
101.25
TBE
0
FC
9k
9 ft k
2
rC/ A
AD
1
4.5i
101.25
ft
4.5
TBE
9k
0
0
180.0 lb
or TBE
Fx
0: Ax
Ax
TBD
TBE
x
4.5
360
13.5
0: Ay
Ay
TBD
y
Fz
0: Az
Az
TBD
TBE
9
360
13.5
Az
0
9
210
21
FC
y
y
9
180
13.5
0
0
18
210
21
0
540 lb
TBE
z
x
180.0 lb
90.0 lb
9
360
13.5
Ay
FC
x
4.5
180
13.5
Ax
Fy
360 lb
FC
z
z
9
180
13.5
0
6
210
21
0
60.0 lb
or A
90.0 lb i
540 lb j
60.0 lb k
PROBLEM 4.117
Solve Problem 4.116 for a
4.5 ft.
P4.116 The 18-ft pole ABC is acted upon by a 210-lb force as shown. The
pole is held by a ball-and-socket joint at A and by two cables BD and BE.
For a 9 ft, determine the tension in each cable and the reaction at A.
SOLUTION
From f.b.d. of pole ABC
M AE
where
0:
AE
4.5 ft i
AE
4.5
2
rB/ A
9 ft j
rC/ A
18 ft j
TBD
BDTBD
rB/ A
9 ft k
9
2
CF
ft
4.5
210 lb
9 ft j
2
9
4.5i
4.5i
4.5
2
210 lb
19.5
4.5 0 9
TBD
0 9 0
13.5 101.25
4.5 9 9
rC/ A
AE
1
4.5i
101.25
4.5 ft i
TBD
13.5
FC
TBD
9
9j
9k
18 j
18
2
ft
6k
2
4.5i
6
18 j
0
9k
9 ft k
2
FC
2
TBD
210 lb
6k
4.5 0 9
210 lb
0
18 0
19.5 101.25
4.5 18 6
0
PROBLEM 4.117 CONTINUED
364.5
364.5
13.5 101.25
or
where
0:
4.5 ft i
AD
4.5
rB/ A
9 ft j
rC/ A
18 ft j
TBE
BETBE
9
2
TBE
4.5 ft i
9 ft j
2
9
364.5
or
9
Ax
TBD
0: Ay
TBE
x
Ay
y
0: Az
Az
x
z
or TBE
48.5 lb
0
FC
ft
TBE
4.5i
13.5
TBE
729 210 lb
FC
x
9j
9k
0
0
0
4.5
48.462
13.5
TBE
y
9
242.31
13.5
TBD
242 lb
4.5
210
19.5
0
48.459 lb
Ay
Fz
or TBD
48.462 lb
4.5
242.31
13.5
TBD
rC/ A
19.5 101.25
Ax
Fy
2
486
TBE
0: Ax
0
4.5 0
9
210 lb
0
18 0
19.5 101.25
4.5 18 6
TBE
13.5 101.25
Fx
210 lb
9k ,
9 ft k
2
4.5 0
9
TBE
0 9 0
13.5 101.25
4.5 9 9
364.5
AD
1
4.5i
101.25
ft
4.5
19.5 101.25
rB/ A
AD
9 ft k
2
729
242.31 lb
TBD
M AD
486
TBD
TBE
9
242.31
13.5
Az
FC
y
0
9
48.462
13.5
18
210
19.5
387.69 lb
z
FC
z
0
9
48.462
13.5
6
2
19.5
64.591 lb
or A
48.5 lb i
388 lb j
64.6 lb k
PROBLEM 4.118
Two steel pipes ABCD and EBF are welded together at B to form the
boom shown. The boom is held by a ball-and-socket joint at D and by
two cables EG and ICFH; cable ICFH passes around frictionless pulleys
at C and F. For the loading shown, determine the tension in each cable
and the reaction at D.
SOLUTION
From f.b.d. of boom
Mz
where
rC/D
TCI
0: k
rC/D
TCI
FA
Mz
or
rA/D
FA
0
1.8 m i
1.8 m i
CI TCI
1.8
TCI
2.12
rA/D
k
2
1.12 m j
1.12
1.8i
2
m
TCI
1.12 j
3.5 m i
560 N j
0
0 1
TCI
1.8
0 0
2.12
1.8 1.12 0
0 0 1
3.5 0 0 560 N
0
1 0
2.016
TCI
2.12
3.5 560
TCI
TFH
2061.1 N
0
0
TICFH
2.06 kN
PROBLEM 4.118 CONTINUED
My
where
rG/D
rH /D
TEG
TFH
0: j rG/D
TEG
j rH /D
0
TFH
3.4 m k
2.5 m k
3.0 m i
3
2
3.15 m k
3.15
2
m
3.0 m i
FH TFH
3
2
TEG
4.35
TEG
2.25 m k
2.25
2
2061.1 N
3.75
2061.1 N
0 1
0 0
3 0
TEG
4.35
7.5
TEG
or
3.15k
m
0 1 0
TEG
0 0 3.4
4.35
3 0 3.15
10.2
3i
0
2061.1 N
2.5
3.75
2.25
2061.1 N
3.75
3i
2.25k
0
0
1758.00 N
1.758 kN
TEG
Fx
0: Dx
Dx
TCI
x
1.8
2.12
TFH
TEG
x
3.0
3.75
2061.1 N
0: Dy
Dy
TCI
y
1.12
2.12
560 N
0: Dz
Dz
TEG
z
2061.1 N
TFH
3
1758 N
4.35
2061.1 N
0
0
560 N
Dy
Fz
0
4611.3 N
Dx
Fy
x
0
528.88 N
0
z
3.15
1758 N
4.35
2.25
3.75
Dz
2061.1 N
0
36.374 N
and D
4610 N i
529 N j
36.4 N k
PROBLEM 4.119
Solve Problem 4.118 assuming that the 560-N load is applied at B.
P4.118 Two steel pipes ABCD and EBF are welded together at B to form
the boom shown. The boom is held by a ball-and-socket joint at D and by
two cables EG and ICFH; cable ICFH passes around frictionless pulleys
at C and F. For the loading shown, determine the tension in each cable
and the reaction at D.
SOLUTION
From f.b.d. of boom
Mz
where
rC/D
TCI
0: k
rC/D
TCI
FB
1.8 m i
CI TCI
1.8
FB
0
2
1.12 m j
1.12
1.8i
2
m
TCI
1.12 j
3.0 m i
560 N j
0
0 1
TCI
1.8
0 0
2.12
1.8 1.12 0
or
rB/D
1.8 m i
TCI
2.12
rB/D
k
2.016
TCI
2.12
TCI
TFH
0
3
0
0 1
0 0 560 N
1 0
3 560
0
0
1766.67 N
TICFH
1.767 kN
PROBLEM 4.119 CONTINUED
My
where
0: j rG/D
TEG
j rH /D
0
TFH
3.4 m k
rG/D
2.5 m k
rH /D
TEG
EGTEG
TFH
FH TFH
3.0 m i
3
3.15 m k
2
3.15
3.0 m i
3
2
2
2.25 m k
2.25
2
0 1 0
TEG
0 0 3.4
4.35
3 0 3.15
10.2
m
TEG
4.35
TFH
1766.67 N
3.75
0 1
0 0
3 0
TEG
4.35
7.5
TEG
or
m
TEG
3i
0
1766.67
2.5
3.75
2.25
1766.67
3.75
3.15k
3i
2.25k
0
0
1506.86 N
1.507 kN
TEG
Fx
0: Dx
Dx
TCI
x
TFH
x
TEG
1.8
1766.67 N
2.12
0: Dy
Dy
TCI
y
560 N
0: Dz
1.12
1766.67 N
2.12
Dz
TEG
z
TFH
z
3
1506.86 N
4.35
0
3952.5 N
0
560 N
Dy
Fz
0
3
1766.67 N
3.75
Dx
Fy
x
0
373.34 N
0
3.15
1506.86 N
4.35
2.25
1766.67 N
3.75
Dz
0
31.172 N
D
3950 N i
373 N j
31.2 N k
PROBLEM 4.120
The lever AB is welded to the bent rod BCD which is supported by
bearings at E and F and by cable DG. Knowing that the bearing at E does
not exert any axial thrust, determine (a) the tension in cable DG, (b) the
reactions at E and F.
SOLUTION
(a) From f.b.d. of assembly
TDG
0.12 m j
DGTDG
My
0.12
0:
0.225 m k
2
0.225
2
m
0.225
0.255
TDG
220 N 0.24 m
TDG
TDG
0.255
0.12 j
0.16 m
0.225 k
0
374.00 N
or TDG
(b) From f.b.d. of assembly
MF
z -axis
0:
220 N 0.19 m
Ex 0.13 m
Ex
Fx
0: Fx
104.923 N
MF
x -axis
0: Ez 0.13 m
374 N
Ez
0.120
0.255
104.923 N
220 N
Fx
374 N
0
115.077 N
0.225
0.255
0.06 m
152.308 N
0
0.16 m
0
374 N
PROBLEM 4.120 CONTINUED
Fz
Fy
0: Fz
0: Fy
152.308 N
0.225
0.255
374 N
Fz
482.31 N
374 N
0.12
0.255
Fy
176.0 N
0
0
E
F
115.1 N i
104.9 N i
176.0 N j
152.3 N k
482 N k
PROBLEM 4.121
A 30-kg cover for a roof opening is hinged at corners A and B. The roof
forms an angle of 30o with the horizontal, and the cover is maintained in
a horizontal position by the brace CE. Determine (a) the magnitude of the
force exerted by the brace, (b) the reactions at the hinges. Assume that the
hinge at A does not exert any axial thrust.
SOLUTION
First note
W
30 kg 9.81 m/s2
mg
EC FEC
FEC
sin15 i
294.3 N
cos15 j FEC
From f.b.d. of cover
(a)
Mz
0:
FEC cos15
W 0.5 m
FEC cos15 1.0 m
or
FEC
(b)
1.0 m
Mx
or
0: W 0.4 m
294.3 N 0.5 m
Ay 0.8 m
or
0.8 m
0.8 m
0
0.8 m
39.429 N
Bx
FEC sin15
39.429 N
Bx
Bx
0
152.341 N sin15
0
0
0
152.3 N
0
152.341 N cos15
0
152.341 N sin15
Ax
0: Ax
FEC cos15
FEC sin15
Ax 0.8 m
Fx
or FEC
Ay 0.8 m
Ay
0: Ax 0.8 m
0
152.341 N
294.3 N 0.4 m
My
0
0.8 m
0
PROBLEM 4.121 CONTINUED
Fy
or
0: FEC cos15
W
152.341 N cos15
By
By
0
294.3 N
By
0
147.180 N
or A
39.4 N i
B
147.2 N j
PROBLEM 4.122
The rectangular plate shown has a mass of 15 kg and is held in the
position shown by hinges A and B and 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.
SOLUTION
W
First note
TEF
EF TEF
0.08 m i
0.08
15 kg 9.81 m/s 2
mg
0.25 m j
2
0.25
2
0.2 m k
0.2
2
m
147.15 N
TEF
TEF
0.08i
0.33
0.25j
0.2k
From f.b.d. of rectangular plate
Mx
or
0:
147.15 N 0.1 m
y
0.25
TEF
0.33
14.715 N m
or
TEF
0.2 m
0
0.2 m
0
97.119 N
TEF
or TEF
Fx
0: Ax
Ax
TEF
x
0.08
0.33
Ax
0
97.119 N
23.544 N
0
97.1 N
PROBLEM 4.122 CONTINUED
MB
0:
z -axis
Ay 0.3 m
or
TEF
y
Ay
y -axis
0: Az 0.3 m
W 0.15 m
0.25
97.119 N 0.04 m
0.33
Ay 0.3 m
MB
0.04 m
TEF
0.2 m
x
TEF
z
0.04 m
W
TEF
y
63.765 N
By
0: Az
TEF
z
Bz
7.848 N
0.04 m
0
23.5 N i
63.8 N j
7.85 N k
9.81 N j
66.7 N k
0
147.15 N
By
Fz
0
0.2
TEF
0.33
0.2 m
and A
0: Ay
0
7.848 N
Az
Fy
147.15 N 0.15 m
63.765 N
0.08
TEF
0.33
Az 0.3 m
0
0.25
0.33
97.119 N
By
0
9.81 N
0
0.2
0.33
Bz
97.119 N
Bz
0
66.708 N
and B
PROBLEM 4.123
Solve Problem 4.122 assuming that cable EF is replaced by a cable
attached at points E and H.
P4.122 The rectangular plate shown has a mass of 15 kg and is held in
the position shown by hinges A and B and 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.
SOLUTION
W
First note
TEH
0.3 m i
EH TEH
0.3
2
15 kg 9.81 m/s 2
mg
0.12 m j
0.2 m k
2
0.12
2
0.2
m
147.15 N
TEH
TEH
0.38
0.2 m
0
0.3 i
0.12 j
0.2 k
From f.b.d. of rectangular plate
Mx
or
0:
147.15 N 0.1 m
TEH
147.15 N 0.1 m
TEH
or
y
0.12
TEH
0.38
0.2 m
0
232.99 N
or TEH
Fx
0: Ax
Ax
TEH
0.3
0.38
x
0
232.99 N
Ax
0
183.938 N
233 N
PROBLEM 4.123 CONTINUED
MB
z -axis
0:
Ay 0.3 m
or
TEH
y -axis
0: Az 0.3 m
TEH
W 0.15 m
x
Ay
63.765 N
0.2 m
TEH
0.3
0.38
Az 0.3 m
or
0.04 m
0.12
232.99 N
0.38
Ay 0.3 m
MB
y
232.99 N
z
0.04 m
147.15 N 0.15 m
0.04 m
0
and A
0: Ay
By
W
TEH
By
63.765 N
y
0: Az
Bz
TEH
z
138.976 N
232.99
0.04 m
0
183.9 N i
63.8 N j
139.0 N k
9.81 N j
16.35 N k
0
147.15 N
By
Fz
0.2
0.38
0.2 m
0
138.976 N
Az
Fy
0
0.12
0.38
232.99 N
0
9.8092 N
0
Bz
0.2
0.38
Bz
232.99 N
0
16.3497 N
and B
PROBLEM 4.124
A small door weighing 16 lb is attached by hinges A and B to a wall and
is held in the horizontal position shown by rope EFH. The rope passes
around a small, frictionless pulley at F and is tied to a fixed cleat at H.
Assuming that the hinge at A does not exert any axial thrust, determine
(a) the tension in the rope, (b) the reactions at A and B.
SOLUTION
First note
12 in. i
EF T
T
54 in. j
2
12
T
12i
62
W
2
54
54 j
28 in. k
2
28
T
6i
31
28k
T
in.
27 j
14k
16 lb j at G
From f.b.d. of door ABCD
Mx
(a)
0: Ty 28 in.
T
27
31
W 14 in.
28 in.
0
16 lb 14 in.
T
0
9.1852 lb
or T
(b)
MB
z -axis
0:
Ay 30 in.
Ay 30 in.
W 15 in.
Ty 4 in.
16 lb 15 in.
Ay
0
9.1852 lb
6.9333 lb
27
31
4 in.
0
9.19 lb
PROBLEM 4.124 CONTINUED
MB
y -axis
0: Az 30 in.
Tx 28 in.
Az 30 in.
Tz 4 in.
6
31
9.1852 lb
Az
0
28 in.
9.1852 lb
0: Bx
Tx
9.1852 lb
Bx
Bx
Fy
0: By
By
Ty
W
Ay
0: Az
Tz
Bz
1.10617 lb
6
31
6.93 lb j
1.106 lb k
0
1.77778 lb
16 lb
By
Fz
0
0
27
31
9.1852 lb
4 in.
1.10617 lb
or A
Fx
14
31
6.9333 lb
0
1.06666 lb
0
9.1852 lb
Bz
14
31
Bz
0
5.2543 lb
or B
1.778 lb i
1.067 lb j
5.25 lb k
PROBLEM 4.125
Solve Problem 4.124 assuming that the rope is attached to the door at I.
P4.124 A small door weighing 16 lb is attached by hinges A and B to a
wall and is held in the horizontal position shown by rope EFH. The rope
passes around a small, frictionless pulley at F and is tied to a fixed cleat
at H. Assuming that the hinge at A does not exert any axial thrust,
determine (a) the tension in the rope, (b) the reactions at A and B.
SOLUTION
First note
T
3 in. i
IF T
3
T
3i
55
W
54 in. j
2
54
54 j
2
10 in. k
10
2
T
in.
10k
16 lb j
From f.b.d. of door ABCD
(a)
Mx
0: W 14 in.
Ty 10 in.
0
54
T 10 in.
55
16 lb 14 in.
T
0
22.815 lb
or T
(b)
MB
z -axis
0:
Ay 30 in.
W 15 in.
Ay 30 in.
Ty 5 in.
16 lb 15 in.
Ay
11.7334 lb
0
22.815 lb
54
55
5 in.
0
22.8 lb
PROBLEM 4.125 CONTINUED
MB
y -axis
0: Az 30 in.
Tx 10 in.
Az 30 in.
Tz 5 in.
3
55
22.815 lb
Az
0
10 in.
22.815 lb
0: Bx
Bx
0: Ay
3
55
W
22.815 lb
Ty
11.7334 lb
0: Az
Tz
11.73 lb j
1.106 lb k
Bz
1.10618 lb
Bz
0
1.24444 lb
By
16 lb
By
Fz
0
0
Tx
Bx
Fy
5 in.
1.10618 lb
or A
Fx
10
55
0
22.815 lb
54
55
By
0
18.1336 lb
0
22.815 lb
10
55
Bz
0
5.2544 lb
or B
1.244 lb i
18.13 lb j
5.25 lb k
PROBLEM 4.126
A 285-lb uniform rectangular plate is supported in the position shown by
hinges A and B and by cable DCE, which 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
23 in. i
First note
CD
1
35.5
22.5 in. j
35.5 in.
15 in. k
23i 22.5 j 15k
9 in. i
22.5 in. j
28.5 in.
CE
15 in. k
1
9i 22.5 j 15k
28.5
W
285 lb j
From f.b.d. of plate
(a)
Mx
0:
22.5
T 15 in.
35.5
285 lb 7.5 in.
T
22.5
T 15 in.
28.5
0
100.121 lb
or T
100.1 lb
PROBLEM 4.126 CONTINUED
(b)
Fx
0: Ax
T
Ax
23
35.5
T
9
28.5
23
35.5
100.121 lb
z -axis
or
Ay 26 in.
0:
22.5
28.5
y -axis
0: Az 26 in.
T
15
35.5
6 in.
1
90
35.5
Az 26 in.
6 in.
Az
0: By
By
W
T
22.5
35.5
285 lb
T
100.121 lb
By
Fz
0: Bz
Bz
Az
T
15
35.5
41.106 lb
22.5
35.5
6 in.
0
6 in.
0
T
T
345
23
35.5
T
15 in.
1
90
28.5
135
0
100.121 lb
0
41.106 lb
22.5
28.5
33.3 lb i
Ay
22.5
35.5
109.6 lb j
41.1 lb k
0
22.5
28.5
109.615 lb
0
32.885 lb
15
28.5
100.121 lb
Bz
15 in.
9
28.5
or A
Fy
22.5
28.5
T
6 in.
100.121 lb
6 in.
15
28.5
T
or
22.5
35.5
109.615 lb
Ay
MB
T
285 lb 13 in.
100.121 lb
0
33.250 lb
W 13 in.
Ay 26 in.
9
28.5
100.121 lb
Ax
MB
0
0
15
35.5
15
28.5
0
53.894 lb
or B
32.9 lb j
53.9 lb k
PROBLEM 4.127
Solve Problem 4.126 assuming that cable DCE is replaced by a cable
attached to point E and hook C.
P4.126 A 285-lb uniform rectangular plate is supported in the position
shown by hinges A and B and by cable DCE, which 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
9 in. i
First note
22.5 in. j
28.5 in.
CE
15 in. k
1
9i 22.5 j 15k
28.5
W
285 lb j
From f.b.d. of plate
(a)
Mx
0:
22.5
T 15 in.
28.5
285 lb 7.5 in.
T
0
180.500 lb
or T
(b)
Fx
0: Ax
Ax
9
28.5
0
180.5 lb
9
28.5
T
Ax
57.000 lb
0
180.5 lb
PROBLEM 4.127 CONTINUED
MB
z -axis
0:
Ay 26 in.
W 13 in.
Ay 26 in.
y -axis
0: Az 26 in.
6 in.
180.5 lb
45
28.5
Az
0: By
By
W
T
285 lb
22.5
28.5
Ay
180.5 lb
By
Fz
0: Bz
Bz
Az
T
180.5 lb
0
22.5
28.5
6 in.
0
T
9
28.5
15 in.
0
0
10.9615 lb
or A
Fy
6 in.
109.615 lb
15
28.5
T
Az 26 in.
22.5
28.5
285 lb 13 in.
Ay
MB
T
Bz
109.6 lb j
10.96 lb k
0
22.5
28.5
109.615 lb
0
32.885 lb
15
28.5
10.9615 lb
57.0 lb i
0
180.5 lb
15
28.5
0
105.962 lb
or B
32.9 lb j
106.0 lb k
PROBLEM 4.128
The tensioning mechanism of a belt drive consists of frictionless pulley
A, mounting plate B, and spring C. Attached below the mounting plate is
slider block D which is free to move in the frictionless slot of bracket E.
Knowing that the pulley and the belt lie in a horizontal plane, with
portion F of the belt parallel to the x axis and portion G forming an angle
of 30° with the x axis, determine (a) the force in the spring, (b) the
reaction at D.
SOLUTION
From f.b.d. of plate B
(a)
Fx
0: 12 N
12 N cos 30
T
T
(b)
Fy
0: Dy
0
Fz
0: Dz
12 N sin 30
0
22.392 N
0: M Dx
6N
12 N sin 30
22 mm
M Dx
MD
y -axis
0: M Dy
z -axis
0: M Dz
or D
6.00 N k
0
132.0 N mm
22.392 N 30 mm
M Dy
MD
22.4 N
0
Dz
Mx
or T
12 N 75 mm
12 N cos 30
75 mm
0
1007.66 N mm
22.392 N 18 mm
M Dz
or M D
12 N 22 mm
12 N cos 30
22 mm
0
89.575 N mm
0.1320 N m i
1.008 N m j
0.0896 N m k
PROBLEM 4.129
The assembly shown is welded to collar A which 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 y axis. For the loading shown,
determine the tension in each cable and the reaction at A.
SOLUTION
First note
TCF
CF TCF
0.16 m i
TDE
DETDE
2
0.16
TCF
0.12 m j
0.8i
0.12
0.24
TDE 0.8j
m
TCF
0.6 j
0.24 m j
2
2
0.18 m k
0.18
2
m
TDE
0.6k
(a) From f.b.d. of assembly
Fy
0: 0.6TCF
0.8TDE
800 N
0.6TCF
0.8TDE
800 N
or
My
or
0:
0.8TCF
0.27 m
TDE
2.25TCF
0
0.6TDE 0.16 m
(1)
0
(2)
PROBLEM 4.129 CONTINUED
Substituting Equation (2) into Equation (1)
0.6TCF
0.8 2.25 TCF
333.33 N
TCF
and from Equation (2)
TDE
800 N
2.25 333.33 N
750.00 N
or TCF
333 N
or TDE
750 N
(b) From f.b.d. of assembly
Fz
0: Az
0.6 750.00 N
0
Az
450.00 N
Fx
0: Ax
0.8 333.33 N
0
Ax
266.67 N
or A
Mx
0: M Ax
800 N 0.27 m
333.33 N 0.6
M Ax
Mz
0: M Az
800 N 0.16 m
0.27 m
267 N i
450 N k
750 N 0.8
0.18 m
0
750 N 0.8
0.16 m
0
54.001 N m
333.33 N 0.6
M Az
0.16 m
0
or M A
54.0 N m i
PROBLEM 4.130
The lever AB is welded to the bent rod BCD which is supported by
bearing E and by cable DG. Assuming that the bearing can exert an axial
thrust and couples about axes parallel to the x and z axes, determine
(a) the tension in cable DG, (b) the reaction at E.
SOLUTION
First note
TDG
0.12 m j
DGTDG
2
0.12
TDG
0.255
0.12 j
0.225 m k
0.225
2
m
TDG
0.225k
(a) From f.b.d. of weldment
My
0:
0.225
TDG
0.255
0.16 m
TDG
220 N 0.24 m
374.00 N
(b) From f.b.d. of weldment
Fx
0: Ex
220 N
0
Ex
Fy
0: E y
220.00 N
374.00 N
0.12
0.255
Ey
176.000 N
0
0
or TDG
374 N
PROBLEM 4.130 CONTINUED
Fz
0: Ez
0.225
0.255
374.00 N
Ez
0
330.00 N
or E
Mx
0: M Ex
330.00 N 0.19 m
M Ex
Mz
0:
220 N 0.06 m
M Ez
M Ez
220 N i
176.0 N j
330 N k
0
62.700 N m
374.00 N
0.12
0.255
0.16 m
0
14.9600 N m
or M E
62.7 N m i
14.96 N m k
PROBLEM 4.131
Solve Problem 4.124 assuming that the hinge at A is removed and that the
hinge at B can exert couples about the y and z axes.
P4.124 A small door weighing 16 lb is attached by hinges A and B to a
wall and is held in the horizontal position shown by rope EFH. The rope
passes around a small, frictionless pulley at F and is tied to a fixed cleat
at H. Assuming that the hinge at A does not exert any axial thrust,
determine (a) the tension in the rope, (b) the reactions at A and B.
SOLUTION
From f.b.d. of door
MB
(a)
0: rG/B
W
rE/B
TEF
MB
0
where
W
MB
TEF
16 lb j
M By j
M Bz k
12 in. i
EF TEF
12
54 in. j
2
TEF
6i
31
27 j
rG/B
15 in. i
14 in. k
rE/B
4 in. i
28 in. k
i
j k
15 0 14 16 lb
0 1 0
or
54
224
i j
4 0
6 27
224
TEF
TEF
14k
M By j
3.6129TEF
3.4839TEF
From i-coefficient
2
28 in.
k
T
28 EF
31
14
24.387TEF i
240
2
28 in. k
M Bz k
0
24.387TEF
0
9.1852 lb
3.6129 9.1852
M By
0
M By j
or TEF
(b) From j-coefficient
M Bz k
33.185 lb in.
M By
0
9.19 lb
PROBLEM 4.131 CONTINUED
From k-coefficient
240
3.4839 9.1852
M Bz
208.00 lb in.
or M B
Fx
0: Bx
0: By
0: Bz
16 lb
208 lb in. k
0
27
9.1852 lb
31
0
8.0000 lb
14
9.1852 lb
31
Bz
or B
0
1.77778 lb
By
Fz
33.2 lb in. j
6
9.1852 lb
31
Bx
Fy
M Bz
0
4.1482 lb
1.778 lb i
8.00 lb j
4.15 lb k
PROBLEM 4.132
The frame shown is supported by three cables and a ball-and-socket joint
at A. For P 0, determine the tension in each cable and the reaction at A.
SOLUTION
First note
TDI
0.65 m i
DI TDI
TEH
0.65i
0.2 j
0.45 m i
EH TEH
0.45
TEH
0.51
TFG
0.2
0.65
TDI
0.81
0.2 m j
2
FGTFG
2
0.44
0.24
2
m
0.45i
0.24 j
0.45 m i
0.2 m j
2
0.2
0.45i
0.2 j
0.45
TFG
0.61
2
m
TDI
0.44k
0.24 m j
2
0.44 m k
2
TEH
0.36 m k
0.36
2
m
TFG
0.36k
From f.b.d. of frame
MA
or
0: rD/ A
i
j
0.65 0.2
0.65 0.2
k
TDI
0
0.81
0.44
i
0.45
0
or
0.088i
TDI
rC/ A
i
0.65
0
j
k
0 0.06 360 N
1 0
0.286 j
0.012i
0.26k
0.189 j
TDI
0.81
0.09k
280 N j
rH / A
j k
0 0 280 N
1 0
TEH
rF / A
TFG
i
j k
T
0
0.32 0 EH
0.51
0.45 0.24 0
0
0.65k 280 N
TFG
0.61
0.06i
0.144k
TEH
0.51
0.45k 360 N
0
rF / A
360 N j
0
i
j
k
T
0.45 0 0.06 FG
0.61
0.45 0.2 0.36
PROBLEM 4.132 CONTINUED
From i-coefficient
0.088
TDI
0.81
0.108642TDI
From j-coefficient
TFG
0.61
0.012
0.0196721TFG
TDI
0.81
0.286
TFG
0.189
0.06 360 N
0
(1)
21.6
TFG
0.61
0
1.13959TDI
(2)
From k-coefficient
0.26
TDI
0.81
0.65 280 N
0.32099TDI
TEH
0.51
0.144
0.45 360 N
0
0.28235TEH
0.147541TFG
TFG
0.61
0.09
344 N
(3)
Substitution of Equation (2) into Equation (1)
0.108642TDI
0.0196721 1.13959TDI
TDI
21.6
164.810 N
or
164.8 N
TDI
Then from Equation (2)
TFG
1.13959 164.810 N
187.816 N
or
TFG
187.8 N
And from Equation (3)
0.32099 164.810 N
0.28235TEH
TEH
0.147541 187.816 N
344 N
932.84 N
or
TEH
933 N
The vector forms of the cable forces are:
TDI
164.810 N
0.81
0.65i
132.25 N i
TEH
932.84 N
0.51
TFG
187.816 N
0.61
0.2 j
0.44k
40.694 N j
0.45i
0.24 j
0.45i
138.553 N i
0.2 j
89.526 N k
823.09 N i
0.36k
61.579 N j
110.842 N k
438.98 N j
PROBLEM 4.132 CONTINUED
Then, from f.b.d. of frame
Fx
0: Ax
132.25
823.09
0: Ay
40.694
438.98
Ay
Fz
0: Az
89.526
or
A
61.579
360
280
0
98.747 N
110.842
Az
0
1093.89 N
Ax
Fy
138.553
0
21.316 N
1094 N i
98.7 N j
21.3 N k
PROBLEM 4.133
The frame shown is supported by three cables and a ball-and-socket joint
at A. For P 50 N, determine the tension in each cable and the reaction
at A.
SOLUTION
First note
0.65 m i
DI TDI
TDI
0.65
TDI
81
65i
20 j
15i
0.45
TFG
61
0.44 m k
0.44
2
m
TDI
44k
2
0.24
m
TEH
8j
0.45 m i
FGTFG
2
0.24 m j
2
0.45
TEH
17
TFG
0.2
0.45 m i
EH TEH
TEH
0.2 m j
2
0.2 m j
2
0.2
45i
20 j
2
0.36 m k
0.36
2
m
TFG
36k
From f.b.d. of frame
MA
0: rD/ A
TDI
rH / A
or
i
j
0.65 0.2
65 20
k
TDI
0
81
44
rC/ A
TEH
i
0.65
0
i
j
k
T
0.45 0 0.06 FG
61
45 20 36
and
8.8i
28.6 j
1.2i
26k
18.9 j
280 N j
rF / A
j
k
0
0
280 50
i
0.45
0
TDI
81
9.0k
TFG
rF / A
360 N j
i
j k
T
0 0.32 0 EH
17
15 8 0
j
k
0 0.06 360 N
1 0
32.5 j 182k
TFG
61
50 N k
0.06i
0
4.8k
0.45k 360
TEH
17
0
PROBLEM 4.133 CONTINUED
From i-coefficient
8.8
TDI
81
0.108642TDI
From j-coefficient 28.6
TDI
81
32.5
0.32099TDI
3.25
Equation 1
Add Equation (2)
18.9
9.0
0.28235TEH
TFG
61
0
(1)
21.6
TFG
61
0.30984TFG
TEH
17
4.8
0.06 360
0.0196721TFG
0.35309TDI
From k-coefficient
T
26 DI
182
81
TFG
61
1.2
0
(2)
32.5
0.45 360
0.147541TFG
0
(3)
344
0.35309TDI
0.063935TFG
70.201
0.35309TDI
0.30984TFG
32.5
0.37378TFG
37.701
100.864 N
TFG
or
TFG
100.9 N
Then from Equation (1)
0.108642TDI
0.0196721 100.864
TDI
21.6
180.554 N
or
TDI
180.6 N
and from Equation (3)
0.32099 180.554
0.28235TEH
TEH
0.147541 100.864
344
960.38 N
or
TEH
960 N
The vector forms of the cable forces are:
TDI
180.554 N
81
65i
144.889 N i
TEH
960.38 N
17
TFG
100.864 N
61
15i
45i
74.409 N i
20 j
44k
44.581 N j
8j
20 j
98.079 N k
847.39 N i
451.94 N j
36k
33.070 N j
59.527 N k
PROBLEM 4.133 CONTINUED
Then from f.b.d. of frame
Fx
0: Ax
144.889
847.39
Ax
Fy
0: Ay
44.581
Fz
0: Az
98.079
Therefore,
A
33.070
360
280
0
110.409 N
59.527
Az
0
1066.69 N
451.94
Ay
74.409
50
0
11.448 N
1067 N i
110.4 N j
11.45 N k
PROBLEM 4.134
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.
SOLUTION
First note
BGTBG
TBG
18 in. i
18
TBG
DH TDH
TDH
0.8i
18 in. i
18
TDH
Since
2
2
0.6i
13.5 in. k
13.5
2
in.
TBG
0.6k
24 in. j
24
2
in.
TDH
0.8 j
DH ,
FJ
TFJ
TFJ
0.6i
0.8 j
From f.b.d. of member ABF
MA
x -axis
0:
0.8TFJ
48 in.
3.2TFJ
MA
z -axis
0:
1.6TDH
0.8TFJ 18 in.
3.2TFJ
Equation (1)
0.8TDH
24 in.
120 lb 12 in.
(1)
120 lb 18 in.
120 lb 18 in.
(2)
Equation (2)
y -axis
0:
0.6TFJ
0
960
TDH
Substituting in Equation (1)
MA
0
480
0.8TDH 18 in.
3.2TDH
120 lb 36 in.
300 lb
TFJ
48 in.
0.6 300 lb
24 in.
0.6TBG 18 in.
0
0
TBG
400 lb
PROBLEM 4.134 CONTINUED
Fx
0:
0.6TFJ
0.6TDH
0.6 300 lb
Fy
0: 0.8TFJ
0: 0.6TBG
0.8 400 lb
Ax
500 lb
0.8TDH
240 lb
0.8 300 lb
Fz
240
Ay
Ay
0
Az
0.6 400 lb
Ax
0
Ax
0
Ay
0
0
0
Az
Az
Therefore,
0.8TBG
0
240 lb
A
500 lb i
240 lb k
PROBLEM 4.135
Solve Problem 4.134 assuming that the load at C has been removed.
P4.134 The rigid L-shaped member ABF is supported by a ball-andsocket joint at A and by three cables. For the loading shown, determine
the tension in each cable and the reaction at A.
SOLUTION
First
TBG
18 in. i
BGTBG
18
TBG
TDH
18 in. i
DH TDH
18
Since
TFJ
0.6i
2
in.
TBG
0.6k
24 in. j
24
0.6i
FJ
2
13.5
0.8i
TDH
TFJ
2
13.5 in. k
2
in.
TDH
0.8 j
DH
0.8 j
From f.b.d. of member ABF
MA
x -axis
0:
0.8TFJ
48 in.
3.2TFJ
MA
z -axis
0:
0.8TFJ 18 in.
3.2TFJ
0.8TDH
24 in.
1.6TDH
0
(1)
360
0.8TDH 18 in.
3.2TDH
120 lb 36 in.
120 lb 18 in.
0
(2)
480
Equation (2)
TDH
75.0 lb
Substituting into Equation (2)
TFJ
75.0 lb
TBG
300 lb
Equation (1)
MA
or
y -axis
0:
0.6TFJ
48 in.
75.0 lb 48 in.
0.6TDH
24 in.
75.0 lb 24 in.
0.6TBG 18 in.
0
TBG 18 in.
PROBLEM 4.135 CONTINUED
Fx
0:
0.6TFJ
0.6TDH
0.6 75.0
75.0
0: 0.8TFJ
Az
0.6 300 lb
Az
Therefore
120 lb
120 lb
Ay
0: 0.6TBG
0.8 300
0.8TDH
0.8 150 lb
Fz
Ax
0
Ax
0
330 lb
Ax
Fy
0.8TBG
Ay
Ay
0
0
0
0
Az
0
180 lb
A
330 lb i
180 lb k
PROBLEM 4.136
In order to clean the clogged drainpipe AE, a plumber has disconnected
both ends of the pipe and inserted a power snake through the opening at
A. The cutting head of the snake is connected by a heavy cable to an
electric motor which rotates at a constant speed as the plumber forces the
cable into the pipe. The forces exerted by the plumber and the motor on
the end of the cable can be represented by the wrench F
60 N k ,
M
108 N m k. Determine the additional reactions at B, C, and D
caused by the cleaning operation. Assume that the reaction at each
support consists of two force components perpendicular to the pipe.
SOLUTION
From f.b.d. of pipe assembly ABCD
Fx
MD
x -axis
0: Bx
0:
0
60 N 2.5 m
Bz
Bz 2 m
0
75.0 N
and B
MD
z -axis
0: C y 3 m
108 N m
Cy
MD
y -axis
0:
Cz 3 m
75 N 4 m
36.0
Dy
Fz
0: Dz
60 N 4 m
0
20.0 N
and C
0: Dy
0
36.0 N
Cz
Fy
75.0 N k
36.0 N j
20.0 N k
0
36.0 N
20.0 N
75.0 N
Dz
5.00 N
and D
60 N
0
36.0 N j
5.00 N k
PROBLEM 4.137
Solve Problem 4.136 assuming that the plumber exerts a force
F
60 N k and that the motor is turned off M 0 .
P4.136 In order to clean the clogged drainpipe AE, a plumber has
disconnected both ends of the pipe and inserted a power snake through
the opening at A. The cutting head of the snake is connected by a heavy
cable to an electric motor which rotates at a constant speed as the
plumber forces the cable into the pipe. The forces exerted by the plumber
and the motor on the end of the cable can be represented by the wrench
F
60 N k ,
M
108 N m k. Determine the additional
reactions at B, C, and D caused by the cleaning operation. Assume that
the reaction at each support consists of two force components
perpendicular to the pipe.
SOLUTION
From f.b.d. of pipe assembly ABCD
Fx
MD
x -axis
0: Bx
0:
0
60 N 2.5 m
Bz
Bz 2 m
0
75.0 N
and B
MD
z -axis
0: C y 3 m
Bx 2 m
Cy
MD
y -axis
0: Cz 3 m
0
0
75.0 N 4 m
Cz
60 N 4 m
Fz
0: Dy
0: Dz
Dz
0
20 N
and C
Fy
75.0 N k
20.0 N k
0
Cy
Dy
0
Cz
F
Bz
75 N
20 N
Dz
5.00 N
0
60 N
0
and D
5.00 N k
PROBLEM 4.138
Three rods are welded together to form a “corner” which is supported by
three eyebolts. Neglecting friction, determine the reactions at A, B, and C
when P 240 N, a 120 mm, b 80 mm, and c 100 mm.
SOLUTION
From f.b.d. of weldment
MO
0: rA/O
i
j k
120 0 0
0 Ay Az
120 Az j
120 Ayk
A
rB/O
B
i j k
0 80 0
Bx 0 Bz
80 Bz i
From i-coefficient
80Bz
j-coefficient
k-coefficient
100Cx
Cx
1.2 Az
120 Ay
0: A B C P
Bx
Cx i
or
Ay
80Bx
0
Cy
240 N j
Cx
(3)
Cy
Bz k
0
0
(4)
240 N
240 N
Az
Bz
Az
Bx
Cy
k-coefficient
Az
(2)
0
Ay
Ay
0
0
1.5 Ay
Cx
j-coefficient
100Cx j
(1)
Bx
Bx
or
0
0
120 Az
From i-coefficient
or
100C y
1.25C y
or
or
100C y i
Bz
or
0
C
i
j k
0 0 100
Cx Cy 0
80Bxk
or
F
rC/O
Bz
0
(5)
0
(6)
PROBLEM 4.138 CONTINUED
Substituting Cx from Equation (4) into Equation (2)
1.2 Az
Bz
(7)
Using Equations (1), (6), and (7)
Cy
Bz
1.25
Az
1.25
1 Bx
1.25 1.2
Bx
1.5
(8)
From Equations (3) and (8)
1.5 Ay
Cy
1.5
or
Cy
Ay
and substituting into Equation (5)
Ay
2 Ay
240 N
Cy
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)
Cx
180.0 N
From Equation (6)
Az
150.0 N
Therefore
A
120.0 N j
150.0 N k
B
180.0 N i
150.0 N k
C
180.0 N i
120.0 N j
PROBLEM 4.139
Solve Problem 4.138 assuming that the force P is removed and is
replaced by a couple M
6 N m j acting at B.
P4.138 Three rods are welded together to form a “corner” which is
supported by three eyebolts. Neglecting friction, determine the reactions
at A, B, and C when P 240 N, a 120 mm, b 80 mm, and
c 100 mm.
SOLUTION
From f.b.d. of weldment
MO
0: rA/O
i
j k
0.12 0 0
0 Ay Az
A
rB/O
B
i
j
k
0 0.08 0
Bx 0 Bz
0.12 Az j
0.08Bz j
0.1Cx j
From i-coefficient
0.1C y
Cy
0.8Bz
0.12 Az
or
F
6N m j
0
0
0
(1)
6
0
60
0.12 Ay
0.08Bx
Bx
1.5 Ay
or
0
0.08Bxk
0.1Cx
1.2 Az
Cx
k-coefficient
M
6N m j
0.08Bz
or
j-coefficient
C
i
j k
0 0 0.1
Cx Cy 0
0.12 Ayk
0.1C y i
rC/O
(2)
0
(3)
0: A B C 0
Bx
Cx i
Ay
Cy j
Az
Bz k
0
From i-coefficient
Cx
Bx
(4)
j-coefficient
Cy
Ay
(5)
k-coefficient
Az
Bz
(6)
Substituting Cx from Equation (4) into Equation (2)
Az
50
Bx
1.2
(7)
PROBLEM 4.139 CONTINUED
Using Equations (1), (6), and (7)
Cy
0.8Bz
2
Bx
3
0.8 Az
(8)
40
From Equations (3) and (8)
Cy
40
Ay
Substituting into Equation (5)
2 Ay
Ay
20.0 N
From Equation (5)
Cy
20.0 N
Equation (1)
Bz
25.0 N
Equation (3)
Bx
Equation (4)
Cx
Equation (6)
Az
Therefore
40
30.0 N
30.0 N
25.0 N
A
20.0 N j
25.0 N k
B
30.0 N i
25.0 N k
30.0 N i
20.0 N j
C
PROBLEM 4.140
The uniform 10-lb rod AB is supported by a ball-and-socket joint at A and
leans against both the rod CD and the vertical wall. Neglecting the effects
of friction, determine (a) the force which rod CD exerts on AB, (b) the
reactions at A and B. (Hint: The force exerted by CD on AB must be
perpendicular to both rods.)
SOLUTION
(a) The force acting at E on the f.b.d. of rod AB is perpendicular to AB
and CD. Letting E direction cosines for force E,
rB/ A k
E
rB/ A k
32 in. i
24 in. j
32
2
24
40 in. k
2
k
in.
0.6i 0.8 j
Also,
W
10 lb j
B
Bk
E
E 0.6i 0.8 j
From f.b.d. of rod AB
MA
i
j
16 12
0
1
0: rG/ A
k
20 10 lb
0
20i 16k 10 lb
W
i
j
24 18
0.6 0.8
E
k
30 E
0
0
i
j
32 24
0
0
k
40 B
1
0
0
0
5.3333 lb
E 5.3333 lb 0.6i 0.8 j
or
E
18 5.3333 lb
B
or
B
24i 32 j B
160 30 E
E
(b) From j-coefficient
rB/ A
24i 18 j 30k E
From k-coefficient
and
rE/ A
3.20 lb i
32 B
4.27 lb j
0
3.00 lb
B
3.00 lb k
PROBLEM 4.140 CONTINUED
From f.b.d. of rod AB
F
Ax i
Ay j Az k
0: A W E B
10 lb j
From i-coefficient
3.20 lb i
Ax
k-coefficient
Therefore
3.20 lb
10 lb
Ay
4.27 lb j
3.00 lb k
0
0
3.20 lb
Ax
j-coefficient
0
4.27 lb
Ay
5.73 lb
Az
3.00 lb
Az
3.00 lb
A
3.20 lb i
0
0
5.73 lb j
3.00 lb k
PROBLEM 4.141
A 21-in.-long uniform rod AB weighs 6.4 lb and is attached to a ball-andsocket joint at A. The rod rests against an inclined frictionless surface and
is held in the position shown by cord BC. Knowing that the cord is 21 in.
long, determine (a) the tension in the cord, (b) the reactions at A and B.
SOLUTION
First note
W
6.4 lb j
NB
N B 0.8j
LAB
21 in.
xB
2
0.6k
13
3
2
xB
BCTBC
TBC
4
2
xB2
16
2
4
2
13 in.
13 in. i
TBC
13i
21
16 in. j
21 in.
16 j
4 in. k
TBC
4k
From f.b.d. of rod AB
MA
i
6.5
0
12.8i
0: rG/ A
j k
8 2
6.4 0
41.6k
W
rB/ A
i
j
k
13 16 4 N B
0 0.8 0.6
12.8i
7.8 j
NB
rC/ A
TBC
0
i j k
26TBC
1 0 0
21
13 16 4
10.4k N B
4j
16k
0
26TBC
21
0
PROBLEM 4.141 CONTINUED
From i-coeff.
12.8
12.8N B
or
NB
From j-coeff.
7.8 N B
0
NB
0.800 lb j
4
26
TBC
21
1.00 lb
0.600 lb k
TBC
0
1.575 lb
From f.b.d. of rod AB
F
0: A
Axi
W
NB
Ay j
TBC
Azk
0
6.4 lb j
0.800 lb j
From i-coefficient
Ax
j-coefficient
Ay
4.40 lb
k-coefficient
Az
0.3 lb
0.600 lb k
1.575
13i
21
4k
0
0.975 lb
(a)
(b)
16 j
TBC
A
0.975 lb i
NB
1.575 lb
4.40 lb j
0.300 lb k
0.800 lb j
0.600 lb k
PROBLEM 4.142
While being installed, the 56-lb chute ABCD is attached to a wall with
brackets E and F and is braced with props GH and IJ. Assuming that the
weight of the chute is uniformly distributed, determine the magnitude of
the force exerted on the chute by prop GH if prop IJ is removed.
SOLUTION
First note
tan
xG
yG
1
42 in.
144 in.
16.2602
50 in. cos16.2602
78 in.
50 in. sin16.2602
144 in. i
BA
2
144
42 in. j
42
2
72 in. i
21 in. j
rG/ A
48 in. i
78 in.
PHG
64 in.
1
25
in.
rK / A
W
48 in.
24i
9 in. k
64 in. j
18 in. k
56 lb j
HG PHG
2 in. i
2
PHG
33
i
2
64 in. j
64
32 j
2
8k
7j
16 in. k
16
2
in.
PHG
48 in. i
14 in. j
18 in. k
PROBLEM 4.142 CONTINUED
From the f.b.d. of the chute
M BA
0:
BA
24
72
0
rK / A
W
BA
7 0
56
21 9
25
1 0
216 56
25
24
rG/ A
24
48
1
14
PHG
0
7 0
PHG
14 18
33 25
32 8
8
24 18 32
PHG
0
7 18
1
7 48
8
PHG
33 25
0
29.141 lb
or PHG
29.1 lb
PROBLEM 4.143
While being installed, the 56-lb chute ABCD is attached to a wall with
brackets E and F and is braced with props GH and IJ. Assuming that the
weight of the chute is uniformly distributed, determine the magnitude of
the force exerted on the chute by prop IJ if prop GH is removed.
SOLUTION
First note
1
tan
42 in.
144 in.
16.2602
xI
100 in. cos16.2602
yI
78 in.
100 in. sin16.2602
144 in. i
BA
144
42 in. j
2
42
2
72 in. i
21 in. j
rI / A
96 in. i
78 in.
PJI
1
25
in.
rK / A
W
96 in.
50 in.
24i
9 in. k
50 in. j
18 in. k
56 lb j
JI PJI
1 in. i
1
PJI
51
i
2
50 in. j
50
50 j
2
10k
7j
10 in. k
10
2
in.
PJI
96 in. i
28 in. j
18 in. k
PROBLEM 4.143 CONTINUED
From the f.b.d. of the chute
M BA
0:
BA
24
72
0
rK / A
W
BA
7 0
56
21 9
25
1 0
216 56
25
24
rI / A
24 7
96 28
1 50
28
PJI
0
0
PJI
18
51 25
10
0
24 18 50
7 18
10
PJI
1
7 96
10
PJI
51 25
0
28.728 lb
or PJI
28.7 lb
PROBLEM 4.144
To water seedlings, a gardener joins three lengths of pipe, AB, BC,
and CD, fitted with spray nozzles and suspends the assembly using
hinged supports at A and D and cable EF. Knowing that the pipe
weighs 0.85 lb/ft, determine the tension in the cable.
SOLUTION
First note rG/ A
WAB
rF / A
T
1.5 ft i
0.85 lb/ft 3 ft j
2 ft i
2 ft i
FET
2
3 ft j
2
3
T
33.25
rB/ A
WBC
rH / A
WCD
2
4.5 ft k
4.5
2i
3j
2
T
ft
4.5k
3 ft i
0.85 lb/ft 1 ft j
3 ft i
0.85 lb j
2.25 ft k
0.85 lb/ft 4.5 ft j
3 ft i
AD
2.55 lb j
3
1 ft j
2
1
2
3.825 lb j
4.5 ft k
4.5
2
ft
1
3i
5.5
j
4.5k
PROBLEM 4.144 CONTINUED
From f.b.d. of the pipe assembly
M AD
0:
AD
rG/ A
AD
3
1.5
0
1
0
2.55
3
3
0
17.2125
1
0
0.85
WAB
rB/ A
4.5
1
0
5.5
0
36
WBC
3 1
2 0
2 3
4.5
1
0
5.5
0
T
33.25
T
rF / A
AD
3
3
0
AD
T
rH / A
WCD
0
4.5
T
0
5.5 33.25
4.5
1
0
3.825
11.475
4.5
1
2.25
5.5
0
25.819
0
0
8.7306 lb
or T
8.73 lb
PROBLEM 4.145
Solve Problem 4.144 assuming that cable EF is replaced by a cable
connecting E and C.
P4.144 To water seedlings, a gardener joins three lengths of pipe, AB,
BC, and CD, fitted with spray nozzles and suspends the assembly using
hinged supports at A and D and cable EF. Knowing that the pipe weighs
0.85 lb/ft, determine the tension in the cable.
SOLUTION
First note
rG/ A
WAB
rC/ A
T
1.5 ft i
0.85 lb/ft 3 ft j
3 ft i
2.55 lb j
1 ft j
3 ft i
CET
3
4 ft j
2
4
T
45.25
rB/ A
WBC
rH / A
WCD
2
4.5
3i
4j
2
T
ft
4.5k
3 ft i
0.85 lb/ft 1 ft j
3 ft i
0.85 lb j
2.25 ft k
0.85 lb/ft 1 ft j
3 ft i
AD
4.5 ft k
3
1 ft j
2
1
2
3.825 lb j
4.5 ft k
4.5
2
ft
1
3i
5.5
j
4.5k
PROBLEM 4.145 CONTINUED
From f.b.d. of the pipe assembly
M AD
0:
rG/ A
AD
AD
3
1.5
0
1
0
2.55
3
3
0
17.2125
WAB
rB/ A
WBC
4.5
1
0
5.5
0
1
0
0.85
40.5
3
3
3
4.5
1
0
5.5
0
3
3
0
T
45.25
T
rC/ A
AD
AD
1
1
4
T
rH / A
0
WCD
4.5
T
0
5.5 45.25
4.5
1
0
3.825
11.475
4.5
1
2.25
5.5
0
25.819
0
0
9.0536 lb
or T
9.05 lb
PROBLEM 4.146
The bent rod ABDE is supported by ball-and-socket joints at A and E and
by the cable DF. If a 600-N load is applied at C as shown, determine the
tension in the cable.
SOLUTION
First note
70 mm i
AE
70
rC/ A
2
240
90 mm i
FC
240 mm k
2
1
25
mm
7i
24k
100 mm k
600 N j
rD/ A
90 mm i
T
240 mm k
160 mm i
DF T
160
T
21
110 mm j
2
16 i
110
11j
2
80 mm k
80
2
T
mm
8k
From the f.b.d. of the bend rod
M AE
0:
AE
rC/ A
7 0 24
600
90 0 100
25
0
1 0
700
2160
600
25
FC
T
0
7 0 24
T
90 0 240
25 21
16 11 8
0
18 480
T
AE
rD/ A
23 760
T
25 21
0
853.13 N
or T
853 N
PROBLEM 4.147
Solve Problem 4.146 assuming that cable DF is replaced by a cable
connecting B and F.
P4.146 The bent rod ABDE is supported by ball-and-socket joints at A
and E and by the cable DF. If a 600-N load is applied at C as shown,
determine the tension in the cable.
SOLUTION
First note
70 mm i
AE
2
70
90 mm i
rC/ A
240 mm k
2
240
mm
1
25
7i
24k
100 mm k
600 N j
FC
90 mm i
rB/ A
160 mm i
BF T
T
160
1
251.59
2
110 mm j
110
2
160 mm k
2
mm
rB/ A
T
160
T
160 i 110 j 160k
From the f.b.d. of the bend rod
M AE
7
90
0
0:
AE
rC/ A
0 24
600
0 100
25
1 0
700 2160
FC
AE
0
7
0
24
T
90
0
0
25 251.59
160 110 160
600
25
237 600
T
T
25 251.59
0
0
1817.04 N
or T
1817 N
PROBLEM 4.148
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.
SOLUTION
80 mm i
First note
BD
80
1
17
rA/B
8i
2
90
9j
12k
2
120 mm k
120
2
mm
60 mm i
P
200 N k
rC/D
80 mm i
C
90 mm j
C j
From the f.b.d. of the plates
M BD
0:
BD
rA/B
P
8 9 12
60 200
1 0 0
17
0 0 1
BD
rC/D
C
8 9 12
C 80
1 0 0
17
0 1 0
9 60 200
12 80 C
C
112.5 N
0
0
0
or C
112.5 N j
PROBLEM 4.149
Two 1 2-m plywood panels, each of mass 15 kg, 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.
SOLUTION
Let
W1
W2
15 kg 9.81 m/s 2 j
mg j
147.15 N j
From the f.b.d. of the panels
M AF
where
0:
AF
rG/ A
W1
2m i
AF
2
rG/ A
1m i
rB/ A
2m i
rI / A
2m i
AF
rB/ A
1m j
2
1
2
1m k
T
2m k
2
2
m
AF
1
2i
3
rT / A
j
W2
2k
0
PROBLEM 4.149 CONTINUED
x
BH
x
T
2 1 2
147.15
1 0 0
3
0 1 0
2
2
2
4y
x
2
147.15
1 y
x
Tmin
dT
dy
Setting the numerator equal to zero,
1
(b)
2
y2
2
y2
2 1 2
147.15
2 0
1
3
0 1 0
y2
y y
2
2
1 2
y
2
4
147.15
3
0
y2
4
0
2
147.15 2
y
1 y
y
2
2
2
y2
0:
147.15
1 4
2
2
y2
2
2
2
1
y
(a)
2
Tmin
T min
2
2 k
T
x
The y-value for Tmin is found from
Then
2
y j
T
3
T
2
2
2 i
x
4
2 k
y2
x
BH T
x
or
2 m, T
y j
2
1 2
0 0
2 y 2 3
2 147.15
3
For x
2 i
4
1
2
1
2
4
2y
1
2
1
0
2
1
y
2
131.615 N
4
4m
2
2
2
4
2
2
x
2.00 m, y
Tmin
4.00 m
131.6 N
PROBLEM 4.150
Solve Problem 4.149 subject to the restriction that H must lie on the
y axis.
P4.149 Two 1 2-m plywood panels, each of mass 15 kg, 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.
SOLUTION
Let
W1
W2
AF
rG/ A
mg j
15 kg 9.81 m/s 2 j
147.15 N j
rB/ A
W2
From the f.b.d. of the panels
M AF
where
0:
W1
2m i
AF
2
1m j
2
rG/ A
1m i
rB/ A
2m i
rI / A
2m i
T
BH T
AF
1
2
T
2m k
2
m
1
2i
3
y j
2m k
2
rI / A
AF
j
1m k
2m i
2
2
T
8
y2
y
2i
2
2
yj
2
m
2k
T
2k
0
PROBLEM 4.150 CONTINUED
2
1
0
1
0
1
2
147.15
0
3
0
2
1 2
T
2 0 0
2
2 y 2 3 8 y
2 147.15
4
For Tmin ,
dT
dy
y2
4y T 8
T
0
y
1
2
1
0
1
2
147.15
1
3
0
2 147.15
147.15 8
1 y
1
2
2
0
8
0
0
y2
1
2
y2
1
2y
y
8
y2
1
2
2
1
0
Setting the numerator equal to zero,
1
y y
y
and
Tmin
(a)
(b)
y2
8
8.00 m
147.15
8
1
8
8
2
138.734 N
x
0, y
Tmin
8.00 m
138.7 N
PROBLEM 4.151
A uniform 20 30-in. steel plate ABCD weighs 85 lb and is attached to
ball-and-socket joints at A and B. Knowing that the plate leans against a
frictionless vertical wall at D, determine (a) the location of D, (b) the
reaction at D.
SOLUTION
(a) Since rD/ A is perpendicular to rB/ A ,
rD/ A rB/ A
0
where coordinates of D are 0, y, z , and
rD/ A
4 in. i
rB/ A
12 in. i
y j
48
or
z
Since
LAD
900
4
16
y
or
28 in. k
16 in. k
rD/ A rB/ A
30
z
16 z
448
0
25 in.
30 in.
2
y
y2
9
875 in.
2
25
28
2
29.580 in.
Coordinates of D :
x
0, y
ND
AB
29.6 in., z
25.0 in.
(b) From f.b.d. of steel plate ABCD
M AB
where
0:
rD/ A
AB
12 in. i
AB
rD/ A
ND
12
2
4 in. i
N Di
16 in. k
16
2
in.
1
3i
5
29.580 in. j
rG/B
W
4k
3 in. k
0
PROBLEM 4.151 CONTINUED
rG/B
W
1
rD/B
2
1
2
16 in. i
29.580 in. j
25 in. 12 in. k
85 lb j
3
0
4 29.580
1
0
4
ND
3
5
0
3
0
4
85
16 29.580 13
2 5
0
1
0
118.32 N D
39
ND
64 42.5
0
8.9799 lb
or N D
8.98 lb i
0
PROBLEM 4.152
Beam AD carries the two 40-lb loads shown. The beam is held by a fixed
support at D and by the cable BE which is attached to the counter-weight
W. Determine the reaction at D when (a) W 100 lb, (b) W 90 lb.
SOLUTION
(a) W
100 lb
From f.b.d. of beam AD
Fx
0: Dx
0
Fy
0: Dy
40 lb
Dy
40 lb
100 lb
0
20.0 lb
or D
MD
0:
MD
100 lb 5 ft
40 lb 8 ft
40 lb 4 ft
MD
0
20.0 lb ft
or M D
(b) W
20.0 lb
20.0 lb ft
90 lb
From f.b.d. of beam AD
Fx
0: Dx
0
Fy
0: Dy
90 lb
Dy
40 lb
40 lb
0
10.00 lb
or D
MD
0:
MD
90 lb 5 ft
40 lb 4 ft
MD
10.00 lb
40 lb 8 ft
0
30.0 lb ft
or M D
30.0 lb ft
PROBLEM 4.153
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 40 lb ft.
SOLUTION
For Wmin ,
MD
40 lb ft
From f.b.d. of beam AD
MD
0:
40 lb 8 ft
Wmin 5 ft
Wmin
For Wmax ,
MD
40 lb 4 ft
40 lb ft
0
40 lb 4 ft
40 lb ft
0
88.0 lb
40 lb ft
From f.b.d. of beam AD
MD
0:
40 lb 8 ft
Wmax 5 ft
Wmax
104.0 lb
or 88.0 lb
W
104.0 lb
PROBLEM 4.154
Determine the reactions at A and D when
30 .
SOLUTION
From f.b.d. of frame ABCD
MD
0:
A 0.18 m
150 N sin 30
150 N cos 30
A
0.10 m
0.28 m
0
243.74 N
or A
Fx
0:
243.74 N
150 N sin 30
Dx
Fy
0: Dy
and
D
Dx
tan
2
1
Dx2
Dy
Dx
Dx
0
318.74 N
150 N cos 30
0
129.904 N
Dy
Then
244 N
318.74
tan
1
2
129.904
129.904
318.74
or D
2
344.19 N
22.174
344 N
22.2
PROBLEM 4.155
Determine the reactions at A and D when
60 .
SOLUTION
From f.b.d. of frame ABCD
MD
0:
A 0.18 m
150 N sin 60
150 N cos 60
A
0.10 m
0.28 m
0
188.835 N
or A
Fx
0:
188.835 N
150 N sin 60
0: Dy
150 N cos 60
and
D
Dx
tan
2
1
Dy
Dy
Dx
2
tan
0
75.0 N
Dy
Then
0
Dx
318.74 N
Dx
Fy
188.8 N
318.74
1
2
75.0
318.74
or D
75.0
2
327.44 N
13.2409
327 N
13.24
PROBLEM 4.156
A 2100-lb tractor is used to lift 900 lb of gravel. Determine the reaction at
each of the two (a) rear wheels A, (b) front wheels B.
SOLUTION
(a) From f.b.d. of tractor
MB
0:
2100 lb 40 in.
2 A 60 in.
900 lb 50 in.
0
or A
325 lb
A
325 lb
(b) From f.b.d. of tractor
MA
0:
2 B 60 in.
B
2100 lb 20 in.
1175 lb
900 lb 110 in.
0
or B
1175 lb
PROBLEM 4.157
A tension of 5 lb is maintained in a tape as it passes the support system
shown. Knowing that the radius of each pulley is 0.4 in., determine the
reaction at C.
SOLUTION
From f.b.d. of system
Fx
0: C x
5 lb
0
5 lb
Cx
Fy
0: C y
5 lb
0
5 lb
Cy
Then
C
Cx
2
and
2
Cy
tan
1
5
2
5
5
5
2
7.0711 lb
45
or C
MC
0: M C
MC
5 lb 6.4 in.
43.0 lb in
5 lb 2.2 in.
7.07 lb
45.0
0
or M C
43.0 lb in.
PROBLEM 4.158
Solve Problem 4.157 assuming that 0.6-in.-radius pulleys are used.
P4.157 A tension of 5 lb is maintained in a tape as it passes the support
system shown. Knowing that the radius of each pulley is 0.4 in., determine
the reaction at C.
SOLUTION
From f.b.d of system
Fx
0: C x
5 lb
0
Cx
Fy
0: C y
5 lb
5 lb
0
Cy
Then
C
Cx
2
and
Cy
2
tan
5 lb
5
1
2
5
5
5
0: M C
5 lb 6.6 in.
MC
7.0711 lb
45.0
or C
MC
2
7.07 lb
5 lb 2.4 in.
45.0
0
45.0 lb in.
or M C
45.0 lb in.
PROBLEM 4.159
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
(a) From f.b.d. of bent rod
M CD
where
0:
CD
CD
rH /B
rH /B
T
F
0
i
0.25 m j
T
AH T
y AH j
y AH
y AH
rF /E
CD
z AH k
2
0.25 m
z AH
2
T
0.25 m sin 30
0.125 m
z AH
0.25 m cos 30
0.21651 m
T
rF /E
F
1
0
0
1
0 0.125
T
0.125 j
0.25
0.21651k
0.25 m k
400 N j
0
0
0.25
0.21651
T
0.25
0.21651T
T
1 0 0
0 0 1 0.25 400 N
0 1 0
0.25 400 N
0
0
461.88 N
or T
462 N
PROBLEM 4.159 CONTINUED
(b) From f.b.d. of bent rod
Fx
MD
z -axis
0: C x
0:
0
461.88 N sin 30
0.35 m
400 N 0.05 m
y -axis
0: Cz 0.3 m
Cz
461.88 N cos 30
0: Dy
336.10 N
Dy
Fz
0: Dz
0
336 N j
461.88 N sin 30
467 N k
400 N
0
505.16 N
466.67 N
Dz
0.35 m
466.67 N
or C
Fy
0
336.10 N
Cy
MD
C y 0.3 m
461.88 N cos30
0
66.670 N
or D
505 N j
66.7 N k
PROBLEM 4.160
For the beam and loading shown, determine (a) the reaction at A, (b) the
tension in cable BC.
SOLUTION
(a) From f.b.d of beam
Fx
MB
0: Ax
0:
0
15 lb 28 in.
20 lb 22 in.
20 lb 6 in.
Ay
35 lb 14 in.
Ay 6 in.
0
245 lb
or A
245 lb
(b) From f.b.d of beam
MA
0:
15 lb 22 in.
20 lb 16 in.
15 lb 6 in.
TB
35 lb 8 in.
TB 6 in.
0
140.0 lb
or TB
Check:
Fy
0:
15 lb
20 lb
35 lb
20 lb
15 lb
140 lb
245 lb
245 lb
245 lb
0 ok
0?
140.0 lb
PROBLEM 4.161
Frame ABCD is supported by a ball-and-socket 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
0.48 m i
DGTDG
0.48
0.14 m j
2
2
0.14
m
TDG
0.48i 0.14 j
TDG
0.50
TDG
24i
25
0.48 m i
BETBE
TBE
7j
0.48
2
0.2 m k
2
0.2
m
TBE
0.48i 0.2k
TBE
0.52
TBE
13
12 j
5k
From f.b.d. of frame ABCD
Mx
0:
7
TDG
25
0.3 m
350 N 0.15 m
0
or TDG
My
Mz
0:
24
25
5
TBE
13
625 N 0.3 m
0: TCF 0.14 m
7
25
625 N
0.48 m
0
or TBE
975 N
625 N 0.48 m
350 N 0.48 m
0
or TCF
600 N
PROBLEM 4.161 CONTINUED
Fx
0: Ax
Ax
TCF
TBE
12
13
600 N
0: Ay
Ay
TDG
7
25
625 N
Ay
Fz
Therefore
0: Az
TBE
Az
5
13
975 N
350 N
y
x
0
24
25
625 N
0
2100 N
Ax
Fy
TDG
x
0
350 N
0
175.0 N
0
z
975 N
0
Az
375 N
A
2100 N i
175.0 N j
375 N k
PROBLEM 4.162
Frame ABCD is supported by a ball-and-socket 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
TDG
First note
0.48 m i
DGTDG
0.48
0.14 m j
2
2
0.14
m
TDG
0.48i 0.14 j
TDG
0.50
TDG
24i
25
TBE
7j
0.48 m i
BETBE
0.48
2
0.2 m k
2
0.2
m
TBE
0.48i 0.2k
TBE
0.52
TBE
13
12i
5k
From f.b.d of frame ABCD
Mx
0:
7
TDG
25
0.3 m
350 N 0.3 m
0
or TDG
My
0:
24
25
1250 N 0.3 m
5
TBE
13
0.48 m
or TBE
Mz
0: TCF 0.14 m
7
25
1250 N
0
1950 N
1250 N 0.48 m
350 N 0.48 m
0
or TCF
0
PROBLEM 4.162 CONTINUED
Fx
0: Ax
Ax
TCF
TBE
12
13
0
TDG
x
Fy
0: Ay
TDG
Ay
7
25
y
0: Az
TBE
Az
5
13
350 N
0
0
1250 N
350 N
0
0
0
z
1950 N
Az
Therefore
1250 N
3000 N
Ay
Fz
24
25
1950 N
Ax
0
x
0
750 N
A
3000 N i
750 N k
PROBLEM 4.163
In the problems listed below, the rigid bodies considered were completely
constrained and the reactions were statically determinate. For each of
these rigid bodies it is possible to create an improper set of constraints by
changing a dimension of the body. In each of the following problems
determine the value of a which results in improper constraints.
(a) Problem 4.81, (b) Problem 4.82.
SOLUTION
(a)
MB
(a)
0:
300 lb 16 in.
T a
0
300 lb 16 in.
16 a in.
T
or
T 16 in.
T becomes infinite when
a
16
0
or a
(b)
MC
(b)
0:
T
15
T
17
0.2T
or
16.0
8
T
17
80 N 0.2 m
0.4 m
0.82353T
T
a
0.35294T
16.00 in.
0.175 m
0
0.88235Ta
0
16.0
0.88235a 0.23529
T becomes infinite when
0.88235a
a
0.23529
0
0.26666 m
or a
267 mm