PROBLEM 2.52
Solve Parts b and d of Problem 2.51, assuming
that the free end of the rope is attached to the
crate.
PROBLEM 2.51 A 600–lb crate is supported
by several rope–and–pulley arrangements as
shown. Determine for each arrangement the
tension in the rope. . (Hint: The tension in the
Ch. 4.)
SOLUTION
Free-Body Diagram of Pulley and Crate
(b)
0: 3 (600 lb) 0
1(600 lb)
3
y
FT
T
Σ= − =
=
200 lbT=
(d)
0: 4 (600 lb) 0
1(600 lb)
4
y
FT
T
Σ= − =
=
150.0 lbT=
consent of McGraw–Hill Education.
PROBLEM 2.53
A 200–kg crate is to be supported by the rope–and–pulley arrangement shown.
Determine the magnitude and direction of the force P that must be exerted on
the free end of the rope to maintain equilibrium. (See the hint for Prob. 2.51.)
SOLUTION
Free–Body Diagram: Pulley A
5
0: 2 cos 0
281
cos 0.59655
53.377
x
FPP
α
α
α

Σ= − + =


=
=±°
For
53.377 :
α
=+°
724 N=P
53.4°
For
53.377 :
α
=−°
16
0: 2 sin( 53.377 ) 1962 N 0
281
y
FP P

Σ = + − °− =


1773=P
53.4°
consent of McGraw–Hill Education.
PROBLEM 2.54
A load Q is applied to the pulley C, which can roll on the
cable ACB. The pulley is held in the position shown by a
second cable CAD, which passes over the pulley A and
supports a load P. Knowing that
750 N,P=
determine
(a) the tension in cable ACB, (b) the magnitude of load Q.
SOLUTION
Free–Body Diagram: Pulley C
(a)
0: (cos25 cos55 ) (750 N)cos55° 0
x ACB
FTΣ = °− ° − =
Hence:
1292.88 N
ACB
T=
1293 N
ACB
T=
(b)
0: (sin 25 sin55 ) (750 N)sin55 0
(1292.88 N)(sin 25 sin55 ) (750 N)sin55 0
y ACB
FT Q
Q
Σ = °+ ° + °− =
°+ ° + °− =
or
2219.8 NQ=
2220 NQ=
PROBLEM 2.55
An 1800–N load Q is applied to the pulley C, which can roll
on the cable ACB. The pulley is held in the position shown by
a second cable CAD, which passes over the pulley A and
supports a load P. Determine (a) the tension in cable ACB,
(b) the magnitude of load P.
SOLUTION
Free–Body Diagram: Pulley C
0: (cos25 cos55 ) cos55 0
x ACB
FT PΣ = °− ° − °=
or
0.58010
ACB
PT
=
(1)
0: (sin 25 sin55 ) sin55 1800 N 0
y ACB
FT PΣ = °+ ° + °− =
or
1.24177 0.81915 1800 N
ACB
TP
+=
(2)
(a) Substitute Equation (1) into Equation (2):
1.24177 0.81915(0.58010 ) 1800 N
ACB ACB
TT+=
Hence:
1048.37 N
ACB
T=
1048 N
ACB
T=
(b) Using (1),
0.58010(1048.37 N) 608.16 NP= =
608 NP=
consent of McGraw–Hill Education.
PROBLEM 2.56
Determine (a) the x, y, and z components of the 900–N force, (b) the
angles
θ
x,
θ
y, and
θ
z that the force forms with the coordinate axes.
SOLUTION
cos 65
(900 N)cos 65
380.36 N
h
h
FF
F
= °
= °
=
(a)
sin 20
(380.36 N)sin20°
xh
FF= °
=
130.091 N,
= −
x
F
130.1 N
x
F= −
sin65
(900 N)sin 65°
815.68 N,
y
y
FF
F
= °
=
= +
816 N
y
F= +
cos20
(380.36 N)cos20
357.42 N
= °
= °
= +
zh
z
FF
F
357 N
z
F= +
(b)
130.091 N
cos 900 N
x
x
F
F
θ
−
= =
98.3
x
θ
= °
815.68 N
cos 900 N
y
y
F
F
θ
+
= =
25.0
y
θ
= °
357.42 N
cos 900 N
z
zF
F
θ
+
= =
66.6
z
θ
= °
consent of McGraw–Hill Education.
PROBLEM 2.57
Determine (a) the x, y, and z components of the 750–N force, (b) the
angles
θ
x,
θ
y, and
θ
z that the force forms with the coordinate axes.
SOLUTION
sin 35
(750 N)sin 35
430.18 N
h
h
FF
F
= °
= °
=
(a)
cos 25
(430.18 N)cos 25°
xh
FF= °
=
389.88 N,
x
F= +
390 N
x
F= +
cos35
(750 N)cos 35°
614.36 N,
y
y
FF
F
= °
=
= +
614 N
y
F= +
sin 25
(430.18 N)sin 25
181.802 N
zh
z
FF
F
= °
= °
= +
181.8 N
z
F= +
(b)
389.88 N
cos 750 N
x
xF
F
θ
+
= =
58.7
x
θ
= °
614.36 N
cos 750 N
y
y
F
F
θ
+
= =
35.0
y
θ
= °
181.802 N
cos 750 N
z
zF
F
θ
+
= =
76.0
z
θ
= °
consent of McGraw–Hill Education.
PROBLEM 2.58
The end of the coaxial cable AE is attached to the pole AB, which is
strengthened by the guy wires AC and AD. Knowing that the tension
in wire AC is 120 lb, determine (a) the components of the force
exerted by this wire on the pole, (b) the angles
θ
x,
θ
y, and
θ
z that
the force forms with the coordinate axes.
SOLUTION
(a)
(120 lb)cos 60 cos 20
x
F= °°
56.382 lb
x
F=
56.4 lb
x
F= +
(120 lb)sin 60
103.923 lb
y
y
F
F
=−°
= −
103.9 lb
y
F= −
(120 lb)cos 60 sin 20
20.521 lb
z
z
F
F
=− °°
= −
20.5 lb
z
F= −
(b)
56.382 lb
cos 120 lb
x
xF
F
θ
= =
62.0
x
θ
= °
103.923 lb
cos 120 lb
y
y
F
F
θ
−
= =
150.0
y
θ
= °
20.52 lb
cos 120 lb
z
zF
F
θ
−
= =
99.8
z
θ
= °
consent of McGraw–Hill Education.
PROBLEM 2.59
the force forms with the coordinate axes.
SOLUTION
(a)
(85 lb)sin 36 sin 48
x
F= °°
37.129 lb=
37.1 lb
x
F=
(85 lb)cos 36
68.766 lb
y
F=−°
= −
68.8 lb
y
F= −
(85 lb)sin 36 cos 48
33.431 lb
z
F= °°
=
33.4 lb
z
F=
(b)
37.129 lb
cos 85 lb
x
x
F
F
θ
= =
64.1
x
θ
= °
68.766 lb
cos 85 lb
y
y
F
F
θ
−
= =
144.0
y
θ
= °
33.431 lb
cos 85 lb
z
z
F
F
θ
= =
66.8
z
θ
= °
consent of McGraw–Hill Education.
PROBLEM 2.60
A gun is aimed at a point A located 35° east of north. Knowing that the barrel of the gun forms an angle
SOLUTION
Recoil force
400 NF=
(400 N)cos40
306.42 N
H
F∴= °
=
(a)
sin35
(306.42 N)sin35
xH
FF=−°
=−°
175.755 N= −
175.8 N
x
F= −
sin 40
(400 N)sin 40
257.12 N
y
FF=−°
=−°
= −
257 N
y
F= −
cos35
(306.42 N)cos35
251.00 N
zH
FF=+°
=+°
= +
251 N
z
F= +
(b)
175.755 N
cos 400 N
x
x
F
F
θ
−
= =
116.1
x
θ
= °
257.12 N
cos 400 N
y
y
F
F
θ
−
= =
130.0
y
θ
= °
251.00 N
cos 400 N
z
zF
F
θ
= =
51.1
z
θ
= °
consent of McGraw–Hill Education.
PROBLEM 2.52
Solve Parts b and d of Problem 2.51, assuming
that the free end of the rope is attached to the
crate.
PROBLEM 2.51 A 600–lb crate is supported
by several rope–and–pulley arrangements as
shown. Determine for each arrangement the
tension in the rope. . (Hint: The tension in the
Ch. 4.)
SOLUTION
Free-Body Diagram of Pulley and Crate
(b)
0: 3 (600 lb) 0
1(600 lb)
3
y
FT
T
Σ= − =
=
200 lbT=
(d)
0: 4 (600 lb) 0
1(600 lb)
4
y
FT
T
Σ= − =
=
150.0 lbT=
consent of McGraw–Hill Education.
PROBLEM 2.53
A 200–kg crate is to be supported by the rope–and–pulley arrangement shown.
Determine the magnitude and direction of the force P that must be exerted on
the free end of the rope to maintain equilibrium. (See the hint for Prob. 2.51.)
SOLUTION
Free–Body Diagram: Pulley A
5
0: 2 cos 0
281
cos 0.59655
53.377
x
FPP
α
α
α

Σ= − + =


=
=±°
For
53.377 :
α
=+°
724 N=P
53.4°
For
53.377 :
α
=−°
16
0: 2 sin( 53.377 ) 1962 N 0
281
y
FP P

Σ = + − °− =


1773=P
53.4°
consent of McGraw–Hill Education.
PROBLEM 2.54
A load Q is applied to the pulley C, which can roll on the
cable ACB. The pulley is held in the position shown by a
second cable CAD, which passes over the pulley A and
supports a load P. Knowing that
750 N,P=
determine
(a) the tension in cable ACB, (b) the magnitude of load Q.
SOLUTION
Free–Body Diagram: Pulley C
(a)
0: (cos25 cos55 ) (750 N)cos55° 0
x ACB
FTΣ = °− ° − =
Hence:
1292.88 N
ACB
T=
1293 N
ACB
T=
(b)
0: (sin 25 sin55 ) (750 N)sin55 0
(1292.88 N)(sin 25 sin55 ) (750 N)sin55 0
y ACB
FT Q
Q
Σ = °+ ° + °− =
°+ ° + °− =
or
2219.8 NQ=
2220 NQ=
PROBLEM 2.55
An 1800–N load Q is applied to the pulley C, which can roll
on the cable ACB. The pulley is held in the position shown by
a second cable CAD, which passes over the pulley A and
supports a load P. Determine (a) the tension in cable ACB,
(b) the magnitude of load P.
SOLUTION
Free–Body Diagram: Pulley C
0: (cos25 cos55 ) cos55 0
x ACB
FT PΣ = °− ° − °=
or
0.58010
ACB
PT
=
(1)
0: (sin 25 sin55 ) sin55 1800 N 0
y ACB
FT PΣ = °+ ° + °− =
or
1.24177 0.81915 1800 N
ACB
TP
+=
(2)
(a) Substitute Equation (1) into Equation (2):
1.24177 0.81915(0.58010 ) 1800 N
ACB ACB
TT+=
Hence:
1048.37 N
ACB
T=
1048 N
ACB
T=
(b) Using (1),
0.58010(1048.37 N) 608.16 NP= =
608 NP=
consent of McGraw–Hill Education.
PROBLEM 2.56
Determine (a) the x, y, and z components of the 900–N force, (b) the
angles
θ
x,
θ
y, and
θ
z that the force forms with the coordinate axes.
SOLUTION
cos 65
(900 N)cos 65
380.36 N
h
h
FF
F
= °
= °
=
(a)
sin 20
(380.36 N)sin20°
xh
FF= °
=
130.091 N,
= −
x
F
130.1 N
x
F= −
sin65
(900 N)sin 65°
815.68 N,
y
y
FF
F
= °
=
= +
816 N
y
F= +
cos20
(380.36 N)cos20
357.42 N
= °
= °
= +
zh
z
FF
F
357 N
z
F= +
(b)
130.091 N
cos 900 N
x
x
F
F
θ
−
= =
98.3
x
θ
= °
815.68 N
cos 900 N
y
y
F
F
θ
+
= =
25.0
y
θ
= °
357.42 N
cos 900 N
z
zF
F
θ
+
= =
66.6
z
θ
= °
consent of McGraw–Hill Education.
PROBLEM 2.57
Determine (a) the x, y, and z components of the 750–N force, (b) the
angles
θ
x,
θ
y, and
θ
z that the force forms with the coordinate axes.
SOLUTION
sin 35
(750 N)sin 35
430.18 N
h
h
FF
F
= °
= °
=
(a)
cos 25
(430.18 N)cos 25°
xh
FF= °
=
389.88 N,
x
F= +
390 N
x
F= +
cos35
(750 N)cos 35°
614.36 N,
y
y
FF
F
= °
=
= +
614 N
y
F= +
sin 25
(430.18 N)sin 25
181.802 N
zh
z
FF
F
= °
= °
= +
181.8 N
z
F= +
(b)
389.88 N
cos 750 N
x
xF
F
θ
+
= =
58.7
x
θ
= °
614.36 N
cos 750 N
y
y
F
F
θ
+
= =
35.0
y
θ
= °
181.802 N
cos 750 N
z
zF
F
θ
+
= =
76.0
z
θ
= °
consent of McGraw–Hill Education.
PROBLEM 2.58
The end of the coaxial cable AE is attached to the pole AB, which is
strengthened by the guy wires AC and AD. Knowing that the tension
in wire AC is 120 lb, determine (a) the components of the force
exerted by this wire on the pole, (b) the angles
θ
x,
θ
y, and
θ
z that
the force forms with the coordinate axes.
SOLUTION
(a)
(120 lb)cos 60 cos 20
x
F= °°
56.382 lb
x
F=
56.4 lb
x
F= +
(120 lb)sin 60
103.923 lb
y
y
F
F
=−°
= −
103.9 lb
y
F= −
(120 lb)cos 60 sin 20
20.521 lb
z
z
F
F
=− °°
= −
20.5 lb
z
F= −
(b)
56.382 lb
cos 120 lb
x
xF
F
θ
= =
62.0
x
θ
= °
103.923 lb
cos 120 lb
y
y
F
F
θ
−
= =
150.0
y
θ
= °
20.52 lb
cos 120 lb
z
zF
F
θ
−
= =
99.8
z
θ
= °
consent of McGraw–Hill Education.
PROBLEM 2.59
the force forms with the coordinate axes.
SOLUTION
(a)
(85 lb)sin 36 sin 48
x
F= °°
37.129 lb=
37.1 lb
x
F=
(85 lb)cos 36
68.766 lb
y
F=−°
= −
68.8 lb
y
F= −
(85 lb)sin 36 cos 48
33.431 lb
z
F= °°
=
33.4 lb
z
F=
(b)
37.129 lb
cos 85 lb
x
x
F
F
θ
= =
64.1
x
θ
= °
68.766 lb
cos 85 lb
y
y
F
F
θ
−
= =
144.0
y
θ
= °
33.431 lb
cos 85 lb
z
z
F
F
θ
= =
66.8
z
θ
= °
consent of McGraw–Hill Education.
PROBLEM 2.60
A gun is aimed at a point A located 35° east of north. Knowing that the barrel of the gun forms an angle
SOLUTION
Recoil force
400 NF=
(400 N)cos40
306.42 N
H
F∴= °
=
(a)
sin35
(306.42 N)sin35
xH
FF=−°
=−°
175.755 N= −
175.8 N
x
F= −
sin 40
(400 N)sin 40
257.12 N
y
FF=−°
=−°
= −
257 N
y
F= −
cos35
(306.42 N)cos35
251.00 N
zH
FF=+°
=+°
= +
251 N
z
F= +
(b)
175.755 N
cos 400 N
x
x
F
F
θ
−
= =
116.1
x
θ
= °
257.12 N
cos 400 N
y
y
F
F
θ
−
= =
130.0
y
θ
= °
251.00 N
cos 400 N
z
zF
F
θ
= =
51.1
z
θ
= °
consent of McGraw–Hill Education.