PROBLEM 2.42
A sailor is being rescued using a boatswain’s chair that is
suspended from a pulley that can roll freely on the support
cable ACB and is pulled at a constant speed by cable CD.
Knowing that α = 25° and
β
= 15° and that the tension in
cable CD is 20 lb, determine (a) the combined weight of
the boatswain’s chair and the sailor, (b) the tension in the
support cable ACB.
SOLUTION
Free–Body Diagram
0: cos 15 cos 25 (20 lb)cos 25 0
x ACB ACB
FT TΣ = °− °− °=
304.04 lb
ACB
T=
0: (304.04 lb)sin 15 (304.04 lb)sin 25
y
FΣ = °+ °
(20 lb)sin 25 0
215.64 lb
W
W
+ °− =
=
(a)
216 lbW=
(b)
304 lb
ACB
T=
PROBLEM 2.43
For the cables of prob. 2.32, find the value of α for which the tension is as small
as possible (a) in cable bc, (b) in both cables simultaneously. In each case
determine the tension in each cable.
SOLUTION
Free-Body Diagram Force Triangle
(a) For a minimum tension in cable BC, set angle between cables to 90 degrees.
PROBLEM 2.44
For the cables of Problem 2.36, it is known that the maximum
allowable tension is 600 N in cable AC and 750 N in cable BC.
Determine (a) the maximum force P that can be applied at C,
(b) the corresponding value of
α
.
SOLUTION
Free–Body Diagram Force Triangle
(a) Law of cosines
222
(600) (750) 2(600)(750)cos(25 45 )P= + − °+ °
784.02 NP=
784 NP=
sin sin(25 45 )
β
°+ °
PROBLEM 2.45
Two cables tied together at C are loaded as shown. Knowing that
the maximum allowable tension in each cable is 800 N, determine
(a) the magnitude of the largest force P that can be applied at C,
(b) the corresponding value of
α
.
SOLUTION
Free–Body Diagram: C Force Triangle
Force triangle is isosceles with
2 180 85
47.5
β
β
= °− °
= °
(a)
2(800 N)cos 47.5° 1081 NP= =
PROBLEM 2.46
Two cables tied together at C are loaded as shown. Knowing
that the maximum allowable tension is 1200 N in cable AC and
600 N in cable BC, determine (a) the magnitude of the largest
force P that can be applied at C, (b) the corresponding value of
α
.
SOLUTION
Free–Body Diagram Force Triangle
(a) Law of cosines:
2 22
(1200 N) (600 N) 2(1200 N)(600 N)cos 85
1294 N
P
P
= +− °
=
Since
1200 N,P.
the solution is correct.
1294 N
P=
(b) Law of sines:
sin sin 85
1200 N 1294 N
67.5
180 50 67.5
β
β
α
°
=
= °
= °− °− °
62.5
α
= °
consent of McGraw–Hill Education.
PROBLEM 2.47
Two cables tied together at C are loaded as shown. Determine the
range of values of Q for which the tension will not exceed 60 lb in
either cable.
SOLUTION
Free-Body Diagram
0: cos60 75 lb 0
x BC
F TQΣ = − − °+ =
75 lb cos60
BC
TQ=−°
(1)
0: sin60 0
y AC
F TQΣ = − °=
sin60
AC
TQ
= °
(2)
Requirement:
60 lb:
AC
T=
From Eq. (2):
sin60 60 lbQ°=
69.3 lbQ=
Requirement:
60 lb:
BC
T=
From Eq. (1):
75 lb cos60 60 lbQ− °=
30.0 lbQ=
30.0 lb 69.3 lbQ≤≤
consent of McGraw–Hill Education.
PROBLEM 2.48
Collar A is connected as shown to a 50–lb load and can
slide on a frictionless horizontal rod. Determine the
magnitude of the force P required to maintain the
equilibrium of the collar when (a)
4.5 in.,x=
(b)
15 in.x=
SOLUTION
(a) Free Body: Collar A Force Triangle
50 lb
4.5 20.5
P=
10.98 lbP=
(b) Free Body: Collar A Force Triangle
50 lb
15 25
P=
30.0 lbP=
consent of McGraw–Hill Education.
PROBLEM 2.49
Collar A is connected as shown to a 50–lb load and can
slide on a frictionless horizontal rod. Determine the
distance x for which the collar is in equilibrium when
P = 48 lb.
SOLUTION
Free Body: Collar A Force Triangle
222
(50) (48) 196
14.00 lb
N
N
=−=
=
Similar Triangles
48 lb
20 in. 14 lb
x=
68.6 in.x=
consent of McGraw–Hill Education.
PROBLEM 2.50
A movable bin and its contents have a combined weight of 2.8 kN. Determine
the shortest chain sling ACB that can be used to lift the loaded bin if the
tension in the chain is not to exceed 5 kN.
SOLUTION
Free–Body Diagram
tan 0.6 m
h
a
=
(1)
Isosceles Force Triangle
1
2
(2.8 kN)
PROBLEM 2.42
A sailor is being rescued using a boatswain’s chair that is
suspended from a pulley that can roll freely on the support
cable ACB and is pulled at a constant speed by cable CD.
Knowing that α = 25° and
β
= 15° and that the tension in
cable CD is 20 lb, determine (a) the combined weight of
the boatswain’s chair and the sailor, (b) the tension in the
support cable ACB.
SOLUTION
Free–Body Diagram
0: cos 15 cos 25 (20 lb)cos 25 0
x ACB ACB
FT TΣ = °− °− °=
304.04 lb
ACB
T=
0: (304.04 lb)sin 15 (304.04 lb)sin 25
y
FΣ = °+ °
(20 lb)sin 25 0
215.64 lb
W
W
+ °− =
=
(a)
216 lbW=
(b)
304 lb
ACB
T=
PROBLEM 2.43
For the cables of prob. 2.32, find the value of α for which the tension is as small
as possible (a) in cable bc, (b) in both cables simultaneously. In each case
determine the tension in each cable.
SOLUTION
Free-Body Diagram Force Triangle
(a) For a minimum tension in cable BC, set angle between cables to 90 degrees.
PROBLEM 2.44
For the cables of Problem 2.36, it is known that the maximum
allowable tension is 600 N in cable AC and 750 N in cable BC.
Determine (a) the maximum force P that can be applied at C,
(b) the corresponding value of
α
.
SOLUTION
Free–Body Diagram Force Triangle
(a) Law of cosines
222
(600) (750) 2(600)(750)cos(25 45 )P= + − °+ °
784.02 NP=
784 NP=
sin sin(25 45 )
β
°+ °
PROBLEM 2.45
Two cables tied together at C are loaded as shown. Knowing that
the maximum allowable tension in each cable is 800 N, determine
(a) the magnitude of the largest force P that can be applied at C,
(b) the corresponding value of
α
.
SOLUTION
Free–Body Diagram: C Force Triangle
Force triangle is isosceles with
2 180 85
47.5
β
β
= °− °
= °
(a)
2(800 N)cos 47.5° 1081 NP= =
PROBLEM 2.46
Two cables tied together at C are loaded as shown. Knowing
that the maximum allowable tension is 1200 N in cable AC and
600 N in cable BC, determine (a) the magnitude of the largest
force P that can be applied at C, (b) the corresponding value of
α
.
SOLUTION
Free–Body Diagram Force Triangle
(a) Law of cosines:
2 22
(1200 N) (600 N) 2(1200 N)(600 N)cos 85
1294 N
P
P
= +− °
=
Since
1200 N,P.
the solution is correct.
1294 N
P=
(b) Law of sines:
sin sin 85
1200 N 1294 N
67.5
180 50 67.5
β
β
α
°
=
= °
= °− °− °
62.5
α
= °
consent of McGraw–Hill Education.
PROBLEM 2.47
Two cables tied together at C are loaded as shown. Determine the
range of values of Q for which the tension will not exceed 60 lb in
either cable.
SOLUTION
Free-Body Diagram
0: cos60 75 lb 0
x BC
F TQΣ = − − °+ =
75 lb cos60
BC
TQ=−°
(1)
0: sin60 0
y AC
F TQΣ = − °=
sin60
AC
TQ
= °
(2)
Requirement:
60 lb:
AC
T=
From Eq. (2):
sin60 60 lbQ°=
69.3 lbQ=
Requirement:
60 lb:
BC
T=
From Eq. (1):
75 lb cos60 60 lbQ− °=
30.0 lbQ=
30.0 lb 69.3 lbQ≤≤
consent of McGraw–Hill Education.
PROBLEM 2.48
Collar A is connected as shown to a 50–lb load and can
slide on a frictionless horizontal rod. Determine the
magnitude of the force P required to maintain the
equilibrium of the collar when (a)
4.5 in.,x=
(b)
15 in.x=
SOLUTION
(a) Free Body: Collar A Force Triangle
50 lb
4.5 20.5
P=
10.98 lbP=
(b) Free Body: Collar A Force Triangle
50 lb
15 25
P=
30.0 lbP=
consent of McGraw–Hill Education.
PROBLEM 2.49
Collar A is connected as shown to a 50–lb load and can
slide on a frictionless horizontal rod. Determine the
distance x for which the collar is in equilibrium when
P = 48 lb.
SOLUTION
Free Body: Collar A Force Triangle
222
(50) (48) 196
14.00 lb
N
N
=−=
=
Similar Triangles
48 lb
20 in. 14 lb
x=
68.6 in.x=
consent of McGraw–Hill Education.
PROBLEM 2.50
A movable bin and its contents have a combined weight of 2.8 kN. Determine
the shortest chain sling ACB that can be used to lift the loaded bin if the
tension in the chain is not to exceed 5 kN.
SOLUTION
Free–Body Diagram
tan 0.6 m
h
a
=
(1)
Isosceles Force Triangle
1
2
(2.8 kN)