PROBLEM 4.20
Two slots have been cut in plate DEF, and the plate has
been placed so that the slots fit two fixed, frictionless
pins A and B. Knowing that P = 15 lb, determine (a) the
force each pin exerts on the plate, (b) the reaction at F.
PROBLEM 4.21
A 6–m telephone pole weighing 1600 N is used to support the ends of two
wires. The wires form the angles shown with the horizontal axis and
the tensions in the wires are, respectively,
1
600 NT=
and
2
375T=
N.
Determine the reaction at the fixed end A.
SOLUTION
Free–Body Diagram:
0: (375 N)cos20 (600 N)cos10 0
xx
FAΣ = + °− °=
238.50 N
x
A= +
0: 1600 N (600 N)sin10 (375 N)sin 20 0
yy
FAΣ = − − °− °=
1832.45 N
y
A= +
22
1
238.50 1832.45
1832.45
tan 238.50
A
θ
−
= +
=
1848 N=A
82.6°
0: (600 N)cos10 (6 m) (375 N)cos20 (6 m) 0
AA
MMΣ= + ° − ° =
1431.00 N m
A
M=−⋅
1431 N m
A
= ⋅M
consent of McGraw–Hill Education.
PROBLEM 4.22
The rig shown consists of a 1200–lb horizontal member ABC
and a vertical member DBE welded together at B. The rig is
being used to raise a 3600–lb crate at a distance x = 12 ft
from the vertical member DBE. If the tension in the cable is
4 kips, determine the reaction at E, assuming that the cable is
(a) anchored at F as shown in the figure, (b) attached to the
vertical member at a point located 1 ft above E.
SOLUTION
Free–Body Diagram:
0: (3600 lb) (1200 lb)(6.5 ft) (3.75 ft) 0
EE
MM x T=+ + −=
3.75 3600 7800
E
M Tx=−−
(1)
(a) For
12 ft and 4000 lbs,xT= =
3.75(4000) 3600(12) 7800
36,000 lb ft
E
M= −−
= ⋅
00
xx
FEΣ= ∴ =
0: 3600 lb 1200 lb 4000 0
yy
FEΣ= − − − =
8800 lb
y
E=
8.80 kips=E
;
36.0 kip ft
E
= ⋅M
consent of McGraw–Hill Education.
SOLUTION Continued
(b)
0: (3600 lb)(12 ft) (1200 lb)(6.5 ft) 0
EE
MMΣ= + + =
51,000 lb ft
E
M=−⋅
00
xx
FEΣ= ∴ =
0: 3600 lb 1200 lb 0
yy
FEΣ= − − =
4800 lb
y
E=
4.80 kips=E
;
51.0 kip ft
E
= ⋅M
consent of McGraw–Hill Education.
PROBLEM 4.23
For the rig and crate of Prob. 4.22, and assuming that cable
is anchored at F as shown, determine (a) the required tension
in cable ADCF if the maximum value of the couple at E as x
varies from 1.5 to 17.5 ft is to be as small as possible, (b) the
corresponding maximum value of the couple.
SOLUTION
Free–Body Diagram:
0: (3600 lb) (1200 lb)(6.5 ft) (3.75 ft) 0
EE
MM x T=+ + −=
3.75 3600 7800
E
M Tx=−−
(1)
For
1.5 ft, Eq. (1) becomesx=
1
( ) 3.75 3600(1.5) 7800
E
MT=−−
(2)
For
17.5 ft, Eq. (1) becomesx=
2
( ) 3.75 3600(17.5) 7800
E
MT=−−
(a) For smallest max value of
| |,
E
M
we set
12
()()
EE
MM
−
=
3.75 13,200 3.75 70,800TT− =−+
11.20 kips=T
(b) From Equation (2), then
3.75(11.20) 13.20
E
M= −
| | 28.8 kip ft
E
M= ⋅
consent of McGraw–Hill Education.
PROBLEM 4.24
A tension of 20 N is maintained in a tape as it passes through the
support system shown. Knowing that the radius of each pulley is
10 mm, determine the reaction at C.
SOLUTION
Free–Body Diagram:
0: (20 N) 0
xx
FCΣ= + =
20 N
x
C= −
0: (20 N) 0
yy
FCΣ= − =
20 N
y
C= +
28.3 N=C
45.0°
0: (20 N)(0.160 m) (20 N)(0.055 m) 0
CC
MMΣ= + + =
4.30 N m
C
M=−⋅
4.30 N m
C
= ⋅M
consent of McGraw–Hill Education.
PROBLEM 4.25
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 and, wherever possible, compute the reactions, assuming
that the magnitude of the force P is 100 lb.
SOLUTION
1. Three non-concurrent, non-parallel reactions:
SOLUTION Continued
5. Four non-concurrent, non-parallel reactions:
PROBLEM 4.26
Eight identical
500 750-mm×
rectangular plates, each of mass
40 kg,m=
are held in a vertical plane as
shown. All connections consist of frictionless pins, rollers, or short links. In each case, determine whether
(a) the plate is completely, partially, or improperly constrained, (b) the reactions are statically determinate
or indeterminate, (c) the equilibrium of the plate is maintained in the position shown. Also, wherever
possible, compute the reactions.
SOLUTION
1. Three non-concurrent, non-parallel reactions:
SOLUTION Continued
5. Two reactions:
PROBLEM 4.21
A 6–m telephone pole weighing 1600 N is used to support the ends of two
wires. The wires form the angles shown with the horizontal axis and
the tensions in the wires are, respectively,
1
600 NT=
and
2
375T=
N.
Determine the reaction at the fixed end A.
SOLUTION
Free–Body Diagram:
0: (375 N)cos20 (600 N)cos10 0
xx
FAΣ = + °− °=
238.50 N
x
A= +
0: 1600 N (600 N)sin10 (375 N)sin 20 0
yy
FAΣ = − − °− °=
1832.45 N
y
A= +
22
1
238.50 1832.45
1832.45
tan 238.50
A
θ
−
= +
=
1848 N=A
82.6°
0: (600 N)cos10 (6 m) (375 N)cos20 (6 m) 0
AA
MMΣ= + ° − ° =
1431.00 N m
A
M=−⋅
1431 N m
A
= ⋅M
consent of McGraw–Hill Education.
PROBLEM 4.22
The rig shown consists of a 1200–lb horizontal member ABC
and a vertical member DBE welded together at B. The rig is
being used to raise a 3600–lb crate at a distance x = 12 ft
from the vertical member DBE. If the tension in the cable is
4 kips, determine the reaction at E, assuming that the cable is
(a) anchored at F as shown in the figure, (b) attached to the
vertical member at a point located 1 ft above E.
SOLUTION
Free–Body Diagram:
0: (3600 lb) (1200 lb)(6.5 ft) (3.75 ft) 0
EE
MM x T=+ + −=
3.75 3600 7800
E
M Tx=−−
(1)
(a) For
12 ft and 4000 lbs,xT= =
3.75(4000) 3600(12) 7800
36,000 lb ft
E
M= −−
= ⋅
00
xx
FEΣ= ∴ =
0: 3600 lb 1200 lb 4000 0
yy
FEΣ= − − − =
8800 lb
y
E=
8.80 kips=E
;
36.0 kip ft
E
= ⋅M
consent of McGraw–Hill Education.
SOLUTION Continued
(b)
0: (3600 lb)(12 ft) (1200 lb)(6.5 ft) 0
EE
MMΣ= + + =
51,000 lb ft
E
M=−⋅
00
xx
FEΣ= ∴ =
0: 3600 lb 1200 lb 0
yy
FEΣ= − − =
4800 lb
y
E=
4.80 kips=E
;
51.0 kip ft
E
= ⋅M
consent of McGraw–Hill Education.
PROBLEM 4.23
For the rig and crate of Prob. 4.22, and assuming that cable
is anchored at F as shown, determine (a) the required tension
in cable ADCF if the maximum value of the couple at E as x
varies from 1.5 to 17.5 ft is to be as small as possible, (b) the
corresponding maximum value of the couple.
SOLUTION
Free–Body Diagram:
0: (3600 lb) (1200 lb)(6.5 ft) (3.75 ft) 0
EE
MM x T=+ + −=
3.75 3600 7800
E
M Tx=−−
(1)
For
1.5 ft, Eq. (1) becomesx=
1
( ) 3.75 3600(1.5) 7800
E
MT=−−
(2)
For
17.5 ft, Eq. (1) becomesx=
2
( ) 3.75 3600(17.5) 7800
E
MT=−−
(a) For smallest max value of
| |,
E
M
we set
12
()()
EE
MM
−
=
3.75 13,200 3.75 70,800TT− =−+
11.20 kips=T
(b) From Equation (2), then
3.75(11.20) 13.20
E
M= −
| | 28.8 kip ft
E
M= ⋅
consent of McGraw–Hill Education.
PROBLEM 4.24
A tension of 20 N is maintained in a tape as it passes through the
support system shown. Knowing that the radius of each pulley is
10 mm, determine the reaction at C.
SOLUTION
Free–Body Diagram:
0: (20 N) 0
xx
FCΣ= + =
20 N
x
C= −
0: (20 N) 0
yy
FCΣ= − =
20 N
y
C= +
28.3 N=C
45.0°
0: (20 N)(0.160 m) (20 N)(0.055 m) 0
CC
MMΣ= + + =
4.30 N m
C
M=−⋅
4.30 N m
C
= ⋅M
consent of McGraw–Hill Education.
PROBLEM 4.25
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 and, wherever possible, compute the reactions, assuming
that the magnitude of the force P is 100 lb.
SOLUTION
1. Three non-concurrent, non-parallel reactions:
SOLUTION Continued
5. Four non-concurrent, non-parallel reactions:
PROBLEM 4.26
Eight identical
500 750-mm×
rectangular plates, each of mass
40 kg,m=
are held in a vertical plane as
shown. All connections consist of frictionless pins, rollers, or short links. In each case, determine whether
(a) the plate is completely, partially, or improperly constrained, (b) the reactions are statically determinate
or indeterminate, (c) the equilibrium of the plate is maintained in the position shown. Also, wherever
possible, compute the reactions.
SOLUTION
1. Three non-concurrent, non-parallel reactions:
SOLUTION Continued
5. Two reactions: