PROBLEM 2.93
Three cables are connected at A, where the forces P and Q
are applied as shown. Knowing that
1200 N,P=
determine
the values of Q for which cable AD is taut.
SOLUTION
PROBLEM 2.94
A container of weight W is suspended from ring A.
Cable BAC passes through the ring and is attached to
fixed supports at B and C. Two forces
P=Pi
and Q
Q=k
are applied to the ring to maintain the container
in the position shown. Knowing that W
376 N,=
determine P and Q. (Hint: The tension is the same in
both portions of cable BAC.)
SOLUTION
( 130 mm) (400 mm) (160 mm)
450 mm
13 40 16
45 45 45
AB AB
T
AB
TAB
T
T
=
=
−+ +
=
=−+ +
Tλ
i jk
ijk
Free-Body A:
( 150 mm) (400 mm) ( 240 mm)
490 mm
15 40 24
49 49 49
0: 0
AC AC
AB AC
T
AC
TAC
T
T
F
=
=
− + +−
=
=−+−
Σ= + + ++ =
Tλ
ij k
ijk
T T QPW
13 15
: 0 0.59501
45 49
T TP TP− − += =i
(1)
40 40
: 0 1.70521
45 49
T TW TW++−= =j
(2)
16 24
: 0 0.134240
45 49
T TQ TQ+−+= =k
(3)
PROBLEM 2.94 (Continued)
Data:
376 N 1.70521 376 N 220.50 NW TT= = =
0.59501(220.50 N) P=
131.2 NP=
0.134240(220.50 N) Q=
29.6 NQ=
PROBLEM 2.95
knowing that
164 N.P=
PROBLEM 2.94 A container of weight W is suspended
from ring A. Cable BAC passes through the ring and is
attached to fixed supports at B and C. Two forces
P
=Pi
and
Q=Qk
are applied to the ring to maintain the container
in the position shown. Knowing that W
376 N,=
determine
P and Q. (Hint: The tension is the same in both portions of
cable BAC.)
SOLUTION
terms of T below. Setting
164 NP=
we have:
Eq. (1):
0.59501 164 NT=
275.63 NT=
Eq. (2):
1.70521(275.63 N) W=
470 NW=
Eq. (3):
0.134240(275.63 N) Q=
37.0 NQ=
consent of McGraw–Hill Education.
PROBLEM 2.96
Cable BAC passes through a frictionless ring A and is
attached to fixed supports at B and C, while cables AD and
AE are both tied to the ring and are attached, respectively, to
supports at D and E. Knowing that a 200–lb vertical load P is
applied to ring A, determine the tension in each of the three
cables.
SOLUTION
Free Body Diagram at A:
Since
tension in
BAC
T=
cable BAC, it follows that
AB AC BAC
TTT
= =
( 17.5 in.) (60 in.) 17.5 60
62.5 in. 62.5 62.5
(60 in.) (25 in.) 60 25
65 in. 65 65
(80 in.) (60 in.) 4 3
100 in. 5 5
AB BAC AB BAC BAC
AC BAC AC BAC BAC
AD AD AD AD AD
AE AE AE AE
TT T
TT T
TT T
TT
−+ −
= = = +
+
= = = +
+
= = = +
= =
ij
Tλ ij
ik
Tλ jk
ij
Tλ ij
Tλ(60 in.) (45 in.) 4 3
75 in. 5 5
AE
T
−
= −
jk jk
consent of McGraw–Hill Education.
SOLUTION Continued
17.5 4
PROBLEM 2.97
Knowing that the tension in cable AE of Prob. 2.96 is 75 lb,
determine (a) the magnitude of the load P, (b) the tension in
cables BAC and AD.
PROBLEM 2.96 Cable BAC passes through a frictionless
ring A and is attached to fixed supports at B and C, while
cables AD and AE are both tied to the ring and are attached,
respectively, to supports at D and E. Knowing that a 200–lb
vertical load P is applied to ring A, determine the tension in
each of the three cables.
SOLUTION
Refer to the solution to Problem 2.96 for the figure and analysis leading to the following set of
equilibrium equations, Equation (2) being modified to include Pj as an unknown quantity:
17.5 4 0
62.5 5
BAC AD
TT− +=
(1)
60 60 3 4 0
62.5 65 5 5
BAC AD AE
T T TP
+ + + −=
(2)
25 3 0
65 5
BAC AE
TT−=
(3)
Substituting for
75 lb
AE
T=
and solving simultaneously gives:
(a)
305 lbP=
(b)
117.0 lb; 40.9 lb
BAC AD
TT= =
consent of McGraw–Hill Education.
SOLUTION Continued
Then from the specifications of the problem,
155 mm 0.155 my= =
2 22
0.23563 m (0.155 m)
0.46 m
z
z
= −
=
PROBLEM 2.98
A container of weight W is suspended from ring
A, to which cables AC and AE are attached. A
force P is applied to the end F of a third cable
that passes over a pulley at B and through ring A
and that is attached to a support at D. Knowing
that
1000 N,W=
determine the magnitude of P.
(Hint: The tension is the same in all portions of
cable FBAD.)
SOLUTION
along the cable. That is, with
2 22
(0.78 m) (1.6 m) (0 m)
( 0.78 m) (1.6 m) (0)
1.78 m
[ (0.78 m) (1.6 m) (0 m) ]
1.78 m
( 0.4382 0.8989 0 )
AB AB AB
AB
AB AB
AB
AB
AB
TT
AB
T
T
=− ++
=−++
=
= =
=− ++
=−+ +
i jk
Tλ
i jk
T i jk
and
222
(0) (1.6 m) (1.2 m)
(0 m) (1.6 m) (1.2 m) 2 m
[(0) (1.6 m) (1.2 m) ]
2 m
(0.8 0.6 )
AC
AC AC AC
AC AC
AC
AC
T
AC
TT
AC
T
=++
=++=
=== ++
= +
ijk
Tλ i jk
T jk
and
222
(1.3 m) (1.6 m) (0.4 m)
(1.3 m) (1.6 m) (0.4 m) 2.1 m
[(1.3 m) (1.6 m) (0.4 m) ]
2.1 m
(0.6190 0.7619 0.1905 )
AD
AD AD AD
AD AD
AD
AD
T
AD
TT
AD
T
=++
= ++ =
=== ++
= ++
ijk
Tλ ijk
T ijk
consent of McGraw–Hill Education.
PROBLEM 2.93
Three cables are connected at A, where the forces P and Q
are applied as shown. Knowing that
1200 N,P=
determine
the values of Q for which cable AD is taut.
SOLUTION
PROBLEM 2.94
A container of weight W is suspended from ring A.
Cable BAC passes through the ring and is attached to
fixed supports at B and C. Two forces
P=Pi
and Q
Q=k
are applied to the ring to maintain the container
in the position shown. Knowing that W
376 N,=
determine P and Q. (Hint: The tension is the same in
both portions of cable BAC.)
SOLUTION
( 130 mm) (400 mm) (160 mm)
450 mm
13 40 16
45 45 45
AB AB
T
AB
TAB
T
T
=
=
−+ +
=
=−+ +
Tλ
i jk
ijk
Free-Body A:
( 150 mm) (400 mm) ( 240 mm)
490 mm
15 40 24
49 49 49
0: 0
AC AC
AB AC
T
AC
TAC
T
T
F
=
=
− + +−
=
=−+−
Σ= + + ++ =
Tλ
ij k
ijk
T T QPW
13 15
: 0 0.59501
45 49
T TP TP− − += =i
(1)
40 40
: 0 1.70521
45 49
T TW TW++−= =j
(2)
16 24
: 0 0.134240
45 49
T TQ TQ+−+= =k
(3)
PROBLEM 2.94 (Continued)
Data:
376 N 1.70521 376 N 220.50 NW TT= = =
0.59501(220.50 N) P=
131.2 NP=
0.134240(220.50 N) Q=
29.6 NQ=
PROBLEM 2.95
knowing that
164 N.P=
PROBLEM 2.94 A container of weight W is suspended
from ring A. Cable BAC passes through the ring and is
attached to fixed supports at B and C. Two forces
P
=Pi
and
Q=Qk
are applied to the ring to maintain the container
in the position shown. Knowing that W
376 N,=
determine
P and Q. (Hint: The tension is the same in both portions of
cable BAC.)
SOLUTION
terms of T below. Setting
164 NP=
we have:
Eq. (1):
0.59501 164 NT=
275.63 NT=
Eq. (2):
1.70521(275.63 N) W=
470 NW=
Eq. (3):
0.134240(275.63 N) Q=
37.0 NQ=
consent of McGraw–Hill Education.
PROBLEM 2.96
Cable BAC passes through a frictionless ring A and is
attached to fixed supports at B and C, while cables AD and
AE are both tied to the ring and are attached, respectively, to
supports at D and E. Knowing that a 200–lb vertical load P is
applied to ring A, determine the tension in each of the three
cables.
SOLUTION
Free Body Diagram at A:
Since
tension in
BAC
T=
cable BAC, it follows that
AB AC BAC
TTT
= =
( 17.5 in.) (60 in.) 17.5 60
62.5 in. 62.5 62.5
(60 in.) (25 in.) 60 25
65 in. 65 65
(80 in.) (60 in.) 4 3
100 in. 5 5
AB BAC AB BAC BAC
AC BAC AC BAC BAC
AD AD AD AD AD
AE AE AE AE
TT T
TT T
TT T
TT
−+ −
= = = +
+
= = = +
+
= = = +
= =
ij
Tλ ij
ik
Tλ jk
ij
Tλ ij
Tλ(60 in.) (45 in.) 4 3
75 in. 5 5
AE
T
−
= −
jk jk
consent of McGraw–Hill Education.
SOLUTION Continued
17.5 4
PROBLEM 2.97
Knowing that the tension in cable AE of Prob. 2.96 is 75 lb,
determine (a) the magnitude of the load P, (b) the tension in
cables BAC and AD.
PROBLEM 2.96 Cable BAC passes through a frictionless
ring A and is attached to fixed supports at B and C, while
cables AD and AE are both tied to the ring and are attached,
respectively, to supports at D and E. Knowing that a 200–lb
vertical load P is applied to ring A, determine the tension in
each of the three cables.
SOLUTION
Refer to the solution to Problem 2.96 for the figure and analysis leading to the following set of
equilibrium equations, Equation (2) being modified to include Pj as an unknown quantity:
17.5 4 0
62.5 5
BAC AD
TT− +=
(1)
60 60 3 4 0
62.5 65 5 5
BAC AD AE
T T TP
+ + + −=
(2)
25 3 0
65 5
BAC AE
TT−=
(3)
Substituting for
75 lb
AE
T=
and solving simultaneously gives:
(a)
305 lbP=
(b)
117.0 lb; 40.9 lb
BAC AD
TT= =
consent of McGraw–Hill Education.
SOLUTION Continued
Then from the specifications of the problem,
155 mm 0.155 my= =
2 22
0.23563 m (0.155 m)
0.46 m
z
z
= −
=
PROBLEM 2.98
A container of weight W is suspended from ring
A, to which cables AC and AE are attached. A
force P is applied to the end F of a third cable
that passes over a pulley at B and through ring A
and that is attached to a support at D. Knowing
that
1000 N,W=
determine the magnitude of P.
(Hint: The tension is the same in all portions of
cable FBAD.)
SOLUTION
along the cable. That is, with
2 22
(0.78 m) (1.6 m) (0 m)
( 0.78 m) (1.6 m) (0)
1.78 m
[ (0.78 m) (1.6 m) (0 m) ]
1.78 m
( 0.4382 0.8989 0 )
AB AB AB
AB
AB AB
AB
AB
AB
TT
AB
T
T
=− ++
=−++
=
= =
=− ++
=−+ +
i jk
Tλ
i jk
T i jk
and
222
(0) (1.6 m) (1.2 m)
(0 m) (1.6 m) (1.2 m) 2 m
[(0) (1.6 m) (1.2 m) ]
2 m
(0.8 0.6 )
AC
AC AC AC
AC AC
AC
AC
T
AC
TT
AC
T
=++
=++=
=== ++
= +
ijk
Tλ i jk
T jk
and
222
(1.3 m) (1.6 m) (0.4 m)
(1.3 m) (1.6 m) (0.4 m) 2.1 m
[(1.3 m) (1.6 m) (0.4 m) ]
2.1 m
(0.6190 0.7619 0.1905 )
AD
AD AD AD
AD AD
AD
AD
T
AD
TT
AD
T
=++
= ++ =
=== ++
= ++
ijk
Tλ ijk
T ijk
consent of McGraw–Hill Education.