SOLUTION (Continued)
Equilibrium condition:
0: 0
AB AC AD
FΣ= ∴ + + + =
TTTW
From i:
45 50 0
75 86
AB AD
TT−+ =
(1)
From j:
60 60 60 0
75 68 86
AB AC AD
T T TW+ + −=
(2)
From k:
32 36 0
68 86
AC AD
TT
−+=
(3)
AB
6.1920 kN
5.5080 kN
AC
AC
T
T
=
=
13.98 kNW=
consent of McGraw–Hill Education.
PROBLEM 2.81
A container is supported by three cables that are attached to
a ceiling as shown. Determine the weight W of the container,
knowing that the tension in cable AD is 4.3 kN.
SOLUTION
Free-Body Diagram at A:
PROBLEM 2.81 (Continued)
Equilibrium condition:
0: 0
AB AC AD
F
Σ= ∴ + + + =TTTW
From i:
45 50 0
75 86
AB AD
TT−+ =
From j:
60 60 60 0
75 68 86
AB AC AD
T T TW+ + −=
From k:
32 36 0
68 86
AC AD
TT
−+=
Setting
4.3 kN
AD
T=
into the above equations gives
4.1667 kN
3.8250 kN
AB
AC
T
T
=
=
9.71kNW=
PROBLEM 2.82
Three cables are used to tether a balloon as shown. Knowing that the
balloon exerts an 800–N vertical force at A, determine the tension in
each cable.
SOLUTION
The forces applied at A are:
, , , and
AB AC AD
TTT P
(4.20 m) (5.60 m) 7.00 m
(2.40 m) (5.60 m) (4.20 m) 7.40 m
(5.60 m) (3.30 m) 6.50 m
AB AB
AC AC
AD AD
=−− =
=−+ =
=−− =
ij
ijk
jk
C
C
C
and
( 0.6 0.8 )
(0.32432 0.75676 0.56757 )
( 0.86154 0.50769 )
AB AB AB AB AB
AC AC AC AC AC
AD AD AD AD AD
AB
TT T
AB
AC
TT T
AC
AD
TT T
AD
= = =−−
===−+
===−−
Tλ ij
Tλ jk
Tλ jk
C
C
C
PROBLEM 2.82 (Continued)
Equilibrium condition
0: 0
AB AC AD
FP
Σ= + + + =TTT j
Substituting the expressions obtained for
, , and
AB AC AD
TT T
and factoring i, j, and k:
( 0.6 0.32432 ) ( 0.8 0.75676 0.86154 )
(0.56757 0.50769 ) 0
AB AC AB AC AD
AC AD
T T T T TP
TT
− + +− − − +
+− =
ij
k
Equating to zero the coefficients of i, j, k:
0.6 0.32432 0
AB AC
TT−+ =
(1)
0.8 0.75676 0.86154 0
AB AC AD
T T TP− − − +=
(2)
0.56757 0.50769 0
AC AD
TT−=
(3)
From Eq. (1)
0.54053
AB AC
TT=
From Eq. (3)
1.11795
AD AC
TT=
Substituting for
AB
T
and
AD
T
in terms of
AC
T
into Eq. (2) gives:
0.8(0.54053 ) 0.75676 0.86154(1.11795 ) 0
AC AC AC
T T TP− − − +=
2.1523 ; 800 N
800 N
2.1523
371.69 N
AC
AC
T PP
T
= =
=
=
Substituting into expressions for
AB
T
and
AD
T
gives:
0.54053(371.69 N)
1.11795(371.69 N)
AB
AD
T
T
=
=
201 N, 372 N, 416 N
AB AC AD
TTT
= = =
consent of McGraw–Hill Education.
PROBLEM 2.83
A crate is supported by three cables as shown. Determine the
weight of the crate knowing that the tension in cable AD is
616 lb.
SOLUTION
The forces applied at A are:
, , and
AB AC AD
TTT W
where
.P
=Pj
To express the other forces in terms of the unit vectors i, j, k, we write
(36 in.) (60 in.) (27 in.)
75 in.
(60 in.) (32 in.)
68 in.
(40 in.) (60 in.) (27 in.)
77 in.
AB
AB
AC
AC
AD
AD
=−+ −
=
= +
=
=+−
=
ijk
jk
ijk
C
C
C
and
( 0.48 0.8 0.36 )
(0.88235 0.47059 )
(0.51948 0.77922 0.35065 )
AB AB AB AB
AB
AC AC AC AC
AC
AD AD AD AD
AD
AB
TT
AB T
AC
TT
AC T
AD
TT
AD T
= =
=− +−
= =
= +
= =
= +−
Tλ
ij k
Tλ
jk
Tλ
i jk
C
C
C
Equilibrium Condition with
W= −Wj
0: 0
AB AC AD
FWΣ= + + − =TTT j
PROBLEM 2.83 (Continued)
Substituting the expressions obtained for
, , and
AB AC AD
TT T
and factoring i, j, and k:
( 0.48 0.51948 ) (0.8 0.88235 0.77922 )
( 0.36 0.47059 0.35065 ) 0
AB AD AB AC AD
AB AC AD
TTTTTW
TTT
−+ ++ + −
+− + − =
ij
k
0.48 0.51948 0
AB AD
TT−+ =
0.8 0.88235 0.77922 0
AB AC AD
T T TW+ + −=
0.36 0.47059 0.35065 0
AB AC AD
TTT
−+ − =
AD
equations using conventional algorithms, gives:
667.67 lb
969.00 lb
AB
AC
T
T
=
=
1868 lbW=
consent of McGraw–Hill Education.
PROBLEM 2.84
A crate is supported by three cables as shown. Determine the weight
of the crate knowing that the tension in cable AC is 544 lb.
SOLUTION
See Problem 2.83 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and (3)
below:
0.48 0.51948 0
AB AD
TT−+ =
(1)
0.8 0.88235 0.77922 0
AB AC AD
T T TW+ + −=
(2)
0.36 0.47059 0.35065 0
AB AC AD
TTT−+ − =
(3)
AC
conventional algorithms, gives:
374.27 lb
345.82 lb
AB
AD
T
T
=
=
1049 lbW=
consent of McGraw–Hill Education.
PROBLEM 2.85
A 1600–lb crate is supported by three cables as shown. Determine
the tension in each cable.
SOLUTION
See Problem 2.83 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and (3)
below:
0.48 0.51948 0
AB AD
TT−+ =
(1)
0.8 0.88235 0.77922 0
AB AC AD
T T TW+ + −=
(2)
0.36 0.47059 0.35065 0
AB AC AD
TTT−+ − =
(3)
conventional algorithms, gives
571 lb
AB
T=
830 lb
AC
T=
528 lb
AD
T=
consent of McGraw–Hill Education.
SOLUTION (Continued)
Equilibrium condition:
0: 0
AB AC AD
FΣ= ∴ + + + =
TTTW
From i:
45 50 0
75 86
AB AD
TT−+ =
(1)
From j:
60 60 60 0
75 68 86
AB AC AD
T T TW+ + −=
(2)
From k:
32 36 0
68 86
AC AD
TT
−+=
(3)
AB
6.1920 kN
5.5080 kN
AC
AC
T
T
=
=
13.98 kNW=
consent of McGraw–Hill Education.
PROBLEM 2.81
A container is supported by three cables that are attached to
a ceiling as shown. Determine the weight W of the container,
knowing that the tension in cable AD is 4.3 kN.
SOLUTION
Free-Body Diagram at A:
PROBLEM 2.81 (Continued)
Equilibrium condition:
0: 0
AB AC AD
F
Σ= ∴ + + + =TTTW
From i:
45 50 0
75 86
AB AD
TT−+ =
From j:
60 60 60 0
75 68 86
AB AC AD
T T TW+ + −=
From k:
32 36 0
68 86
AC AD
TT
−+=
Setting
4.3 kN
AD
T=
into the above equations gives
4.1667 kN
3.8250 kN
AB
AC
T
T
=
=
9.71kNW=
PROBLEM 2.82
Three cables are used to tether a balloon as shown. Knowing that the
balloon exerts an 800–N vertical force at A, determine the tension in
each cable.
SOLUTION
The forces applied at A are:
, , , and
AB AC AD
TTT P
(4.20 m) (5.60 m) 7.00 m
(2.40 m) (5.60 m) (4.20 m) 7.40 m
(5.60 m) (3.30 m) 6.50 m
AB AB
AC AC
AD AD
=−− =
=−+ =
=−− =
ij
ijk
jk
C
C
C
and
( 0.6 0.8 )
(0.32432 0.75676 0.56757 )
( 0.86154 0.50769 )
AB AB AB AB AB
AC AC AC AC AC
AD AD AD AD AD
AB
TT T
AB
AC
TT T
AC
AD
TT T
AD
= = =−−
===−+
===−−
Tλ ij
Tλ jk
Tλ jk
C
C
C
PROBLEM 2.82 (Continued)
Equilibrium condition
0: 0
AB AC AD
FP
Σ= + + + =TTT j
Substituting the expressions obtained for
, , and
AB AC AD
TT T
and factoring i, j, and k:
( 0.6 0.32432 ) ( 0.8 0.75676 0.86154 )
(0.56757 0.50769 ) 0
AB AC AB AC AD
AC AD
T T T T TP
TT
− + +− − − +
+− =
ij
k
Equating to zero the coefficients of i, j, k:
0.6 0.32432 0
AB AC
TT−+ =
(1)
0.8 0.75676 0.86154 0
AB AC AD
T T TP− − − +=
(2)
0.56757 0.50769 0
AC AD
TT−=
(3)
From Eq. (1)
0.54053
AB AC
TT=
From Eq. (3)
1.11795
AD AC
TT=
Substituting for
AB
T
and
AD
T
in terms of
AC
T
into Eq. (2) gives:
0.8(0.54053 ) 0.75676 0.86154(1.11795 ) 0
AC AC AC
T T TP− − − +=
2.1523 ; 800 N
800 N
2.1523
371.69 N
AC
AC
T PP
T
= =
=
=
Substituting into expressions for
AB
T
and
AD
T
gives:
0.54053(371.69 N)
1.11795(371.69 N)
AB
AD
T
T
=
=
201 N, 372 N, 416 N
AB AC AD
TTT
= = =
consent of McGraw–Hill Education.
PROBLEM 2.83
A crate is supported by three cables as shown. Determine the
weight of the crate knowing that the tension in cable AD is
616 lb.
SOLUTION
The forces applied at A are:
, , and
AB AC AD
TTT W
where
.P
=Pj
To express the other forces in terms of the unit vectors i, j, k, we write
(36 in.) (60 in.) (27 in.)
75 in.
(60 in.) (32 in.)
68 in.
(40 in.) (60 in.) (27 in.)
77 in.
AB
AB
AC
AC
AD
AD
=−+ −
=
= +
=
=+−
=
ijk
jk
ijk
C
C
C
and
( 0.48 0.8 0.36 )
(0.88235 0.47059 )
(0.51948 0.77922 0.35065 )
AB AB AB AB
AB
AC AC AC AC
AC
AD AD AD AD
AD
AB
TT
AB T
AC
TT
AC T
AD
TT
AD T
= =
=− +−
= =
= +
= =
= +−
Tλ
ij k
Tλ
jk
Tλ
i jk
C
C
C
Equilibrium Condition with
W= −Wj
0: 0
AB AC AD
FWΣ= + + − =TTT j
PROBLEM 2.83 (Continued)
Substituting the expressions obtained for
, , and
AB AC AD
TT T
and factoring i, j, and k:
( 0.48 0.51948 ) (0.8 0.88235 0.77922 )
( 0.36 0.47059 0.35065 ) 0
AB AD AB AC AD
AB AC AD
TTTTTW
TTT
−+ ++ + −
+− + − =
ij
k
0.48 0.51948 0
AB AD
TT−+ =
0.8 0.88235 0.77922 0
AB AC AD
T T TW+ + −=
0.36 0.47059 0.35065 0
AB AC AD
TTT
−+ − =
AD
equations using conventional algorithms, gives:
667.67 lb
969.00 lb
AB
AC
T
T
=
=
1868 lbW=
consent of McGraw–Hill Education.
PROBLEM 2.84
A crate is supported by three cables as shown. Determine the weight
of the crate knowing that the tension in cable AC is 544 lb.
SOLUTION
See Problem 2.83 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and (3)
below:
0.48 0.51948 0
AB AD
TT−+ =
(1)
0.8 0.88235 0.77922 0
AB AC AD
T T TW+ + −=
(2)
0.36 0.47059 0.35065 0
AB AC AD
TTT−+ − =
(3)
AC
conventional algorithms, gives:
374.27 lb
345.82 lb
AB
AD
T
T
=
=
1049 lbW=
consent of McGraw–Hill Education.
PROBLEM 2.85
A 1600–lb crate is supported by three cables as shown. Determine
the tension in each cable.
SOLUTION
See Problem 2.83 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and (3)
below:
0.48 0.51948 0
AB AD
TT−+ =
(1)
0.8 0.88235 0.77922 0
AB AC AD
T T TW+ + −=
(2)
0.36 0.47059 0.35065 0
AB AC AD
TTT−+ − =
(3)
conventional algorithms, gives
571 lb
AB
T=
830 lb
AC
T=
528 lb
AD
T=
consent of McGraw–Hill Education.