PROBLEM 2.32
Two cables are tied together at C and are loaded as shown. Knowing that α =
30°, determine the tension (a) in cable AC, (b) in cable BC.
SOLUTION
Free–Body Diagram Force Triangle
Law of sines:
6 kN
sin60 sin35 sin85
AC BC
TT
= =
°°°
(a)
6 kN (sin 60 )
sin85
AC
T= °
°
5.22 kN
AC
T=
(b)
6 kN (sin35 )
sin85
BC
T= °
°
3.45 kN
BC
T=
PROBLEM 2.33
Two cables are tied together at C and are loaded as shown.
Determine the tension (a) in cable AC, (b) in cable BC.
SOLUTION
Free-Body Diagram Force Triangle
Law of sines:
400 lb
sin 60 sin 40 sin80
AC BC
TT
= =
°°°
(a)
400 lb (sin 60 )
sin80
AC
T= °
°
352 lb
AC
T=
(b)
400 lb (sin 40 )
sin80
BC
T= °
°
261 lb
BC
T=
PROBLEM 2.34
Two cables are tied together at C and are loaded as shown. Determine the
tension (a) in cable AC, (b) in cable BC.
SOLUTION
Free–Body Diagram Force Triangle
2
mg
(200 kg)(9.81m/s )
1962 N
W=
=
=
Law of sines:
1962 N
sin 15 sin 105 sin 60
AC BC
TT
= =
° °°
(a)
(1962 N) sin 15
sin 60
AC
T°
=°
586 N
AC
T=
(b)
(1962 N)sin 105
sin 60
BC
T°
=°
2190 N
BC
T=
PROBLEM 2.35
Two cables are tied together at C and loaded as shown. Determine
the tension (a) in cable AC, (b) in cable BC.
SOLUTION
Free Body Diagram at C:
12 ft 7.5 ft
0: 0
12.5 ft 8.5 ft
x AC BC
TTΣ= − + =F
1.08800
BC AC
TT
=
3.5 ft 4 ft
0: 396 lb 0
12 ft 8.5 ft
y AC BC
TTΣ= + − =F
(a)
3.5 ft 4 ft (1.08800 ) 396 lb 0
12.5 ft 8.5 ft
(0.28000 0.51200) 396 lb
AC AC
AC
TT
T
+ −=
+=
500.0 lb
AC
T=
500 lb
AC
T=
(b)
(1.08800)(500.0 lb)
BC
T=
544 lb
BC
T=
consent of McGraw–Hill Education.
PROBLEM 2.36
Two cables are tied together at C and are loaded as shown.
Knowing that P = 500 N and
α
= 60°, determine the tension in
(a) in cable AC, (b) in cable BC.
SOLUTION
Free–Body Diagram Force Triangle
Law of sines:
500 N
sin35 sin 75 sin 70°
AC BC
TT
= =
°°
(a)
500 N sin35
sin70
AC
T= °
°
305 N
AC
T=
(b)
500 N sin75
sin70
BC
T= °
°
514 N
BC
T=
consent of McGraw–Hill Education.
PROBLEM 2.37
Two forces of magnitude TA = 8 kips and TB = 15 kips are
applied as shown to a welded connection. Knowing that the
connection is in equilibrium, determine the magnitudes of the
forces TC and TD.
SOLUTION
Free–Body Diagram
0 15 kips 8 kips cos40 0
xD
FTΣ = − − °=
9.1379 kips
D
T=
0
y
FΣ=
sin 40 0
DC
TT°− =
(9.1379 kips)sin 40 0
5.8737 kips
C
C
T
T
°− =
=
5.87 kips
C
T=
9.14 kips
D
T=
consent of McGraw–Hill Education.
PROBLEM 2.38
Two forces of magnitude TA = 6 kips and TC = 9 kips are
applied as shown to a welded connection. Knowing that the
connection is in equilibrium, determine the magnitudes of the
forces TB and TD.
SOLUTION
Free–Body Diagram
0
x
FΣ=
6 kips cos40 0
BD
TT
− − °=
(1)
0
y
FΣ=
sin 40 9 kips 0
9 kips
sin 40
14.0015 kips
D
D
D
T
T
T
°− =
=°
=
Substituting for TD into Eq. (1) gives:
6 kips (14.0015 kips)cos40 0
16.7258 kips
B
B
T
T
− − °=
=
16.73 kips
B
T=
14.00 kips
D
T=
consent of McGraw–Hill Education.
PROBLEM 2.39
Two cables are tied together at C and are loaded as shown. Knowing that P =
300 N, determine the tension in cables AC and BC.
SOLUTION
Free–Body Diagram
0 sin30 sin30 cos45 200N 0
x CA CB
FTT PΣ = − + − °− =
AA
For
200NP=
we have,
0.5 0.5 212.13 200 0
CA CB
TT− + + −=
(1)
0
y
FΣ=
cos30 cos30 sin 45 0
CA CB
TTP°− − =
AA
0.86603 0.86603 212.13 0
CA CB
TT+ −=
(2)
Solving equations (1) and (2) simultaneously gives,
134.6 N
CA
T=
110.4 N
CB
T=
consent of McGraw–Hill Education.
PROBLEM 2.40
Two forces P and Q are applied as shown to an aircraft
connection. Knowing that the connection is in equilibrium
and that
500P=
lb and
650Q=
lb, determine the
magnitudes of the forces exerted on the rods A and B.
SOLUTION
Free–Body Diagram
Resolving the forces into x– and y–directions:
0
AB
=++ + =RPQF F
Substituting components:
(500 lb) [(650 lb)cos50 ]
[(650 lb)sin50 ]
( cos50 ) ( sin50 ) 0
BA A
FF F
=−+ °
−°
+ − °+ °=
Rj i
j
iij
In the y-direction (one unknown force):
500 lb (650 lb)sin50 sin50 0
A
F− − °+ °=
Thus,
500 lb (650 lb)sin50
sin50
A
F+°
=°
1302.70 lb=
1303 lb
A
F=
In the x–direction:
(650 lb)cos50 cos50 0
BA
FF°+ − °=
Thus,
cos50 (650 lb)cos50
(1302.70 lb)cos50 (650 lb)cos50
BA
FF= °− °
= °− °
419.55 lb=
420 lb
B
F=
consent of McGraw–Hill Education.
PROBLEM 2.32
Two cables are tied together at C and are loaded as shown. Knowing that α =
30°, determine the tension (a) in cable AC, (b) in cable BC.
SOLUTION
Free–Body Diagram Force Triangle
Law of sines:
6 kN
sin60 sin35 sin85
AC BC
TT
= =
°°°
(a)
6 kN (sin 60 )
sin85
AC
T= °
°
5.22 kN
AC
T=
(b)
6 kN (sin35 )
sin85
BC
T= °
°
3.45 kN
BC
T=
PROBLEM 2.33
Two cables are tied together at C and are loaded as shown.
Determine the tension (a) in cable AC, (b) in cable BC.
SOLUTION
Free-Body Diagram Force Triangle
Law of sines:
400 lb
sin 60 sin 40 sin80
AC BC
TT
= =
°°°
(a)
400 lb (sin 60 )
sin80
AC
T= °
°
352 lb
AC
T=
(b)
400 lb (sin 40 )
sin80
BC
T= °
°
261 lb
BC
T=
PROBLEM 2.34
Two cables are tied together at C and are loaded as shown. Determine the
tension (a) in cable AC, (b) in cable BC.
SOLUTION
Free–Body Diagram Force Triangle
2
mg
(200 kg)(9.81m/s )
1962 N
W=
=
=
Law of sines:
1962 N
sin 15 sin 105 sin 60
AC BC
TT
= =
° °°
(a)
(1962 N) sin 15
sin 60
AC
T°
=°
586 N
AC
T=
(b)
(1962 N)sin 105
sin 60
BC
T°
=°
2190 N
BC
T=
PROBLEM 2.35
Two cables are tied together at C and loaded as shown. Determine
the tension (a) in cable AC, (b) in cable BC.
SOLUTION
Free Body Diagram at C:
12 ft 7.5 ft
0: 0
12.5 ft 8.5 ft
x AC BC
TTΣ= − + =F
1.08800
BC AC
TT
=
3.5 ft 4 ft
0: 396 lb 0
12 ft 8.5 ft
y AC BC
TTΣ= + − =F
(a)
3.5 ft 4 ft (1.08800 ) 396 lb 0
12.5 ft 8.5 ft
(0.28000 0.51200) 396 lb
AC AC
AC
TT
T
+ −=
+=
500.0 lb
AC
T=
500 lb
AC
T=
(b)
(1.08800)(500.0 lb)
BC
T=
544 lb
BC
T=
consent of McGraw–Hill Education.
PROBLEM 2.36
Two cables are tied together at C and are loaded as shown.
Knowing that P = 500 N and
α
= 60°, determine the tension in
(a) in cable AC, (b) in cable BC.
SOLUTION
Free–Body Diagram Force Triangle
Law of sines:
500 N
sin35 sin 75 sin 70°
AC BC
TT
= =
°°
(a)
500 N sin35
sin70
AC
T= °
°
305 N
AC
T=
(b)
500 N sin75
sin70
BC
T= °
°
514 N
BC
T=
consent of McGraw–Hill Education.
PROBLEM 2.37
Two forces of magnitude TA = 8 kips and TB = 15 kips are
applied as shown to a welded connection. Knowing that the
connection is in equilibrium, determine the magnitudes of the
forces TC and TD.
SOLUTION
Free–Body Diagram
0 15 kips 8 kips cos40 0
xD
FTΣ = − − °=
9.1379 kips
D
T=
0
y
FΣ=
sin 40 0
DC
TT°− =
(9.1379 kips)sin 40 0
5.8737 kips
C
C
T
T
°− =
=
5.87 kips
C
T=
9.14 kips
D
T=
consent of McGraw–Hill Education.
PROBLEM 2.38
Two forces of magnitude TA = 6 kips and TC = 9 kips are
applied as shown to a welded connection. Knowing that the
connection is in equilibrium, determine the magnitudes of the
forces TB and TD.
SOLUTION
Free–Body Diagram
0
x
FΣ=
6 kips cos40 0
BD
TT
− − °=
(1)
0
y
FΣ=
sin 40 9 kips 0
9 kips
sin 40
14.0015 kips
D
D
D
T
T
T
°− =
=°
=
Substituting for TD into Eq. (1) gives:
6 kips (14.0015 kips)cos40 0
16.7258 kips
B
B
T
T
− − °=
=
16.73 kips
B
T=
14.00 kips
D
T=
consent of McGraw–Hill Education.
PROBLEM 2.39
Two cables are tied together at C and are loaded as shown. Knowing that P =
300 N, determine the tension in cables AC and BC.
SOLUTION
Free–Body Diagram
0 sin30 sin30 cos45 200N 0
x CA CB
FTT PΣ = − + − °− =
AA
For
200NP=
we have,
0.5 0.5 212.13 200 0
CA CB
TT− + + −=
(1)
0
y
FΣ=
cos30 cos30 sin 45 0
CA CB
TTP°− − =
AA
0.86603 0.86603 212.13 0
CA CB
TT+ −=
(2)
Solving equations (1) and (2) simultaneously gives,
134.6 N
CA
T=
110.4 N
CB
T=
consent of McGraw–Hill Education.
PROBLEM 2.40
Two forces P and Q are applied as shown to an aircraft
connection. Knowing that the connection is in equilibrium
and that
500P=
lb and
650Q=
lb, determine the
magnitudes of the forces exerted on the rods A and B.
SOLUTION
Free–Body Diagram
Resolving the forces into x– and y–directions:
0
AB
=++ + =RPQF F
Substituting components:
(500 lb) [(650 lb)cos50 ]
[(650 lb)sin50 ]
( cos50 ) ( sin50 ) 0
BA A
FF F
=−+ °
−°
+ − °+ °=
Rj i
j
iij
In the y-direction (one unknown force):
500 lb (650 lb)sin50 sin50 0
A
F− − °+ °=
Thus,
500 lb (650 lb)sin50
sin50
A
F+°
=°
1302.70 lb=
1303 lb
A
F=
In the x–direction:
(650 lb)cos50 cos50 0
BA
FF°+ − °=
Thus,
cos50 (650 lb)cos50
(1302.70 lb)cos50 (650 lb)cos50
BA
FF= °− °
= °− °
419.55 lb=
420 lb
B
F=
consent of McGraw–Hill Education.