PROBLEM 4.62
Solve Problem 4.61, assuming that the 320–lb load is
applied at A.
PROBLEM 4.61 A 48–in. boom is held by a ball–and–
socket joint at C and by two cables BF and DAE; cable
DAE passes around a frictionless pulley at A. For the
loading shown, determine the tension in each cable and
the reaction at C.
SOLUTION
Free–Body Diagram:
Five unknowns and six equations of equilibrium, but equilibrium is
maintained
( 0).
AC
MΣ=
T=
tension in both parts of cable DAE.
30
48
B
A
=
=
rk
rk
20 48 52 in.
20 48 52 in.
16 30 34 in.
AD AD
AE AE
BF BF
=−− =
=−=
=−=
ik
jk
ik
FFF
FFF
FFF
( 20 48 ) ( 5 12 )
52 13
(20 48 ) (5 12 )
52 13
(16 30 ) (8 15 )
34 17
AD
AE
BF BF
BF BF
AD T T
TAD
AE T T
TAE
TT
BF
TBF
= = − − = −−
= = −= −
= = −= −
T ik ik
T jk jk
T ik ik
FFF
FFF
FFF
0: ( 320 lb) 0
C A AD A AE B BF A
MΣ = × +× +× +×− =rT rT rT r j
0 0 48 0 0 48 0 0 30 48 ( 320 ) 0
13 13 17
5 0 12 0 5 12 8 0 15
BF
T
TT
+ + + ×− =
−−−−
i j k ij k ij k
kj
Coefficient of i:
240 15,360 0 832 lb
13 TT
−+ = =
consent of McGraw–Hill Education.
SOLUTION Continued
Coefficient of j:
240 240 0
13 17
BD
TT−+ =
17 17 (832) 1088 lb
13 13
BD BD
TT T= = =
0: 320 0
AD AE BF
Σ= + + − + =
F T T T jC
Coefficient of i:
20 8
(832) (1088) 0
52 17
x
C− + +=
320 512 0 192 lb
xx
CC−+ += =−
Coefficient of j:
20 (832) 320 0
52 y
C
− +=
320 320 0 0
yy
CC− += =
Coefficient of k:
48 48 30
(832) (852) (1088) 0
52 52 34
z
C
− − − +=
768 768 960 0 2496 lb
zz
CC−−−+= =
Answers:
DAE
TT=
832 lb
DAE
T=
1088 lb
BD
T=
(192.0 lb) (2496 lb)=−+C ik
consent of McGraw–Hill Education.
PROBLEM 4.63
The 6–m pole ABC is acted upon by a 455–N force as shown.
The pole is held by a ball-and-socket joint at A and by two
cables BD and BE. For
3 m,a=
determine the tension in
each cable and the reaction at A.
SOLUTION
Free–Body Diagram:
( 0)
AC
M
Σ=
3
6
B
C
=
=
rj
rj
3 6 2 7m
1.5 3 3 4.5 m
1.5 3 3 4.5 m
CF CF
BD BD
BE BE
=−− + =
= −− =
= −+ =
ijk
ijk
ijk
FFFC
FFFC
FFFC
(3 6 2)
7
(1.5 3 3 ) ( 2 2 )
4.5 3
( 2 2)
3
BD BD
BD BD
BD
BE BE
CF P
PCE
TT
BD
TBD
T
BE
TBE
= = −− +
= = −− = −−
= = = −+
P ijk
T ijk i jk
T ijk
FFFC
FFFC
FFFC
0: 0
03 0 030 0 60 0
337
1 2 2 1 22 3 62
A B BD B BE C
BD BE
M
TT P
Σ = × + × + ×=
+ +=
−− − −−
rT rT rP
i j k i jk i jk
Coefficient of i:
12
22 0
7
BD BE
TT P−+ +=
(1)
Coefficient of k:
18 0
7
BD BF
TT P−− + =
(2)
consent of McGraw–Hill Education.
SOLUTION Continued
Eq. (1) + 2 Eq. (2):
48 12
40
77
BD BD
T PT P−+= =
Eq. (2):
12 18 6
0
77 7
BE BE
PT P T P− −+ = =
Since
12
445 N (455)
7
BD
PT= =
780 N
BD
T=
6(455)
7
BE
T=
390 N
BE
T=
0: 0
BD BE
Σ= + + + =F T T PA
Coefficient of i:
780 390 455 (3) 0
337 x
A+ − +=
260 130 195 0 195.0 N
xx
AA+ − += =
Coefficient of j:
780 390 455
(2) (2) (6) 0
337
y
A−−−+=
520 260 390 0 1170 N
yy
AA−− − += =
Coefficient of k:
780 390 455
(2) (2) (2) 0
337
z
A− + + +=
520 260 130 0 130.0 N
zz
AA− + + += =+
(195.0 N) (1170 N) (130.0 N)=−++A ijk
consent of McGraw–Hill Education.
PROBLEM 4.64
A 600–lb crate hangs from a cable that passes over a
pulley B and is attached to a support at H. The 200–lb
boom AB is supported by a ball–and–socket joint at A
and by two cables DE and DF. The center of gravity
of the boom is located at G. Determine (a) the tension
in cables DE and DF, (b) the reaction at A.
SOLUTION
Free–Body Diagram:
600 lb
200 lb
C
G
W
W
=
=
DE DF x y z
the
AB
axis, but equilibrium is maintained, since
0.
AB
MΣ=
We have
(30 ft) (22.5 ft) 37.5 ft
8.8
(13.8 ft) (22.5 ft) (6.6 ft)
12
(13.8 ft) (16.5 ft) (6.6 ft) 22.5 ft
(13.8 ft) (16.5 ft) (6.6 ft) 22.5 ft
BH BH
DE
DE
DF DF
=−=
=−+
=−+ =
=−− =
ij
i jk
i jk
i jk
FFF
FFF
FFF
consent of McGraw–Hill Education.
SOLUTION Continued
Thus:
30 22.5
(600 lb) (480 lb) (360 lb)
37.5
(13.8 16.5 6.6 )
22.5
(13.8 16.5 6.6 )
22.5
BH BH
DE
DE DE
DE
DF DF
BH
BH
T
DE
DE
T
DF
DF
−
= = = −
= = −+
= = −−
ij
TT i j
TT i j k
TT i jk
FFF
FFF
FFF
(a)
0:( )( )( )( )( )0
(12 ) ( 600 ) (6 ) ( 200 ) (18 ) (480 360 )
A J C K G H BH E DE F DF
Σ = × +× +× +× +× =
− ×− − ×− + × −
M rW rW rT rT rT
ijijiij
5 0 6.6 5 0 6.6 0
22.5 22.5
13.8 16.5 6.6 13.8 16.5 6.6
DE DF
TT
+ + −=
− −−
ijk ijk
or
7200 1200 6480 4.84( )
DE DF
TT+−+ −kkk i
58.08 82.5
( ) ( )0
22.5 22.5
DE DF DE DF
TT TT+ −− + =jk
i or j:
0*
DE DF DE DF
TT TT−= =
82.5
: 7200 1200 6480 (2 ) 0
22.5
DE
T+−− =k
261.82 lb
DE
T=
262 lb
DE DF
TT= =
(b)
13.8
0: 480 2 (261.82) 0 801.17 lb
22.5
16.5
0: 600 200 360 2 (261.82) 0 1544.00 lb
22.5
xx x
yy y
FA A
FA A
Σ= + + = =−
Σ= − − − − = =
0: 0
zz
FAΣ= =
(801lb) (1544 lb)=−+Aij
PROBLEM 4.65
The horizontal platform ABCD weighs 60 lb and supports a 240-lb
load at its center. The platform is normally held in position by
hinges at A and B and by braces CE and DE. If brace DE is
removed, determine the reactions at the hinges and the force
exerted by the remaining brace CE. The hinge at A does not exert
any axial thrust.
SOLUTION
Free–Body Diagram:
Express forces, weight in terms of rectangular components:
( ) ( ) ( )
3 ft 4 ft 2 ftEC =++ijk
C
( ) ( ) ( )
222
342
342
CE CE CE
EC
FF
EC
++
= =
++
ijk
F
C
0.55709 0.74278 0.37139
CE CE CE
FFF
=++ijk
( ) (300 lb)mg=−=−Wj j
( ) ( ) ( ) ( )
( ) ( )
0: 4 ft 1.5 ft 2 ft 300 lb
3 ft 4 ft 0
B
CE
Σ = × + + ×−
+ + ×=
M kA i k j
i kF
or
( ) ( ) ( )( ) ( )( )
4 ft 4 ft 1.5 ft 300 lb 2 ft 300 lb
yz
AA−+ − +ij k i
( )
3 0 4 ft 0
0.55709 0.74278 0.37139
CE
F+=
i jk
consent of McGraw–Hill Education.
PROBLEM 4.66
A 100–kg uniform rectangular plate is supported in the
position shown by hinges A and B and by cable DCE that
passes over a frictionless hook at C. Assuming that the
tension is the same in both parts of the cable, determine (a)
the tension in the cable, (b) the reactions at A and B.
Assume that the hinge at B does not exert any axial thrust.
SOLUTION
/
/
/
(960 180) 780
960 450
90
22
390 225
600 450
BA
GA
CA
−=
= −+
= +
= +
r ii
r ik
ik
r ik
Dimensions in mm
(a)
T=
Tension in cable DCE
690 675 450 1065 mm
270 675 450 855 mm
CD CD
CE CE
=−+ − =
=+− =
ijk
ijk
C
C
2
( 690 675 450 )
1065
(270 675 450 )
855
(100 kg)(9.81m/s ) (981 N)
CD
CE
T
T
mg
= −+ −
= +−
=−=− =−
T ijk
T ijk
Wi j j
/// /
0: ( ) 0
A C A CD C A CE G A B A
WΣ = × + × + ×− + × =M rTrTr jrB
600 0 450 600 0 450
1065 855
690 675 450 270 675 450
390 0 225 780 0 0 0
0 981 0 0 yz
TT
BB
+
−− −
+ +=
−
i jk ijk
i j k i jk
consent of McGraw–Hill Education.
SOLUTION Continued
Coefficient of i:
3
(450)(675) (450)(675) 220.73 10 0
1065 855
TT
− − + ×=
344.64 NT=
345 NT=
Coefficient of j:
344.64 344.64
( 690 450 600 450) (270 450 600 450) 780 0
1065 855
z
B−×+× + ×+× − =
185.516 N
z
B=
Coefficient of k:
3
344.64 344.64
(600)(675) (600)(675) 382.59 10 780 0 113.178 N
1065 855 yy
BB+ − ×+ = =
(b)
(113.2 N) (185.5 N)= +B jk
0: 0
CD CE
Σ= + + + + =F ABT T W
Coefficient of i:
690 270
(344.64) (344.64) 0
1065 855
x
A− +=
114.454 N
x
A=
Coefficient of j:
675 675
113.178 (344.64) (344.64) 981 0 377.30 N
1065 855
yy
AA+ + + −= =
Coefficient of k:
450 450
185.516 (344.64) (344.64) 0
1065 855
z
A+− − =
141.496 N
z
A=
(114.4 N) (377 N) (141.5 N)= ++A ij k
consent of McGraw–Hill Education.
PROBLEM 4.62
Solve Problem 4.61, assuming that the 320–lb load is
applied at A.
PROBLEM 4.61 A 48–in. boom is held by a ball–and–
socket joint at C and by two cables BF and DAE; cable
DAE passes around a frictionless pulley at A. For the
loading shown, determine the tension in each cable and
the reaction at C.
SOLUTION
Free–Body Diagram:
Five unknowns and six equations of equilibrium, but equilibrium is
maintained
( 0).
AC
MΣ=
T=
tension in both parts of cable DAE.
30
48
B
A
=
=
rk
rk
20 48 52 in.
20 48 52 in.
16 30 34 in.
AD AD
AE AE
BF BF
=−− =
=−=
=−=
ik
jk
ik
FFF
FFF
FFF
( 20 48 ) ( 5 12 )
52 13
(20 48 ) (5 12 )
52 13
(16 30 ) (8 15 )
34 17
AD
AE
BF BF
BF BF
AD T T
TAD
AE T T
TAE
TT
BF
TBF
= = − − = −−
= = −= −
= = −= −
T ik ik
T jk jk
T ik ik
FFF
FFF
FFF
0: ( 320 lb) 0
C A AD A AE B BF A
MΣ = × +× +× +×− =rT rT rT r j
0 0 48 0 0 48 0 0 30 48 ( 320 ) 0
13 13 17
5 0 12 0 5 12 8 0 15
BF
T
TT
+ + + ×− =
−−−−
i j k ij k ij k
kj
Coefficient of i:
240 15,360 0 832 lb
13 TT
−+ = =
consent of McGraw–Hill Education.
SOLUTION Continued
Coefficient of j:
240 240 0
13 17
BD
TT−+ =
17 17 (832) 1088 lb
13 13
BD BD
TT T= = =
0: 320 0
AD AE BF
Σ= + + − + =
F T T T jC
Coefficient of i:
20 8
(832) (1088) 0
52 17
x
C− + +=
320 512 0 192 lb
xx
CC−+ += =−
Coefficient of j:
20 (832) 320 0
52 y
C
− +=
320 320 0 0
yy
CC− += =
Coefficient of k:
48 48 30
(832) (852) (1088) 0
52 52 34
z
C
− − − +=
768 768 960 0 2496 lb
zz
CC−−−+= =
Answers:
DAE
TT=
832 lb
DAE
T=
1088 lb
BD
T=
(192.0 lb) (2496 lb)=−+C ik
consent of McGraw–Hill Education.
PROBLEM 4.63
The 6–m pole ABC is acted upon by a 455–N force as shown.
The pole is held by a ball-and-socket joint at A and by two
cables BD and BE. For
3 m,a=
determine the tension in
each cable and the reaction at A.
SOLUTION
Free–Body Diagram:
( 0)
AC
M
Σ=
3
6
B
C
=
=
rj
rj
3 6 2 7m
1.5 3 3 4.5 m
1.5 3 3 4.5 m
CF CF
BD BD
BE BE
=−− + =
= −− =
= −+ =
ijk
ijk
ijk
FFFC
FFFC
FFFC
(3 6 2)
7
(1.5 3 3 ) ( 2 2 )
4.5 3
( 2 2)
3
BD BD
BD BD
BD
BE BE
CF P
PCE
TT
BD
TBD
T
BE
TBE
= = −− +
= = −− = −−
= = = −+
P ijk
T ijk i jk
T ijk
FFFC
FFFC
FFFC
0: 0
03 0 030 0 60 0
337
1 2 2 1 22 3 62
A B BD B BE C
BD BE
M
TT P
Σ = × + × + ×=
+ +=
−− − −−
rT rT rP
i j k i jk i jk
Coefficient of i:
12
22 0
7
BD BE
TT P−+ +=
(1)
Coefficient of k:
18 0
7
BD BF
TT P−− + =
(2)
consent of McGraw–Hill Education.
SOLUTION Continued
Eq. (1) + 2 Eq. (2):
48 12
40
77
BD BD
T PT P−+= =
Eq. (2):
12 18 6
0
77 7
BE BE
PT P T P− −+ = =
Since
12
445 N (455)
7
BD
PT= =
780 N
BD
T=
6(455)
7
BE
T=
390 N
BE
T=
0: 0
BD BE
Σ= + + + =F T T PA
Coefficient of i:
780 390 455 (3) 0
337 x
A+ − +=
260 130 195 0 195.0 N
xx
AA+ − += =
Coefficient of j:
780 390 455
(2) (2) (6) 0
337
y
A−−−+=
520 260 390 0 1170 N
yy
AA−− − += =
Coefficient of k:
780 390 455
(2) (2) (2) 0
337
z
A− + + +=
520 260 130 0 130.0 N
zz
AA− + + += =+
(195.0 N) (1170 N) (130.0 N)=−++A ijk
consent of McGraw–Hill Education.
PROBLEM 4.64
A 600–lb crate hangs from a cable that passes over a
pulley B and is attached to a support at H. The 200–lb
boom AB is supported by a ball–and–socket joint at A
and by two cables DE and DF. The center of gravity
of the boom is located at G. Determine (a) the tension
in cables DE and DF, (b) the reaction at A.
SOLUTION
Free–Body Diagram:
600 lb
200 lb
C
G
W
W
=
=
DE DF x y z
the
AB
axis, but equilibrium is maintained, since
0.
AB
MΣ=
We have
(30 ft) (22.5 ft) 37.5 ft
8.8
(13.8 ft) (22.5 ft) (6.6 ft)
12
(13.8 ft) (16.5 ft) (6.6 ft) 22.5 ft
(13.8 ft) (16.5 ft) (6.6 ft) 22.5 ft
BH BH
DE
DE
DF DF
=−=
=−+
=−+ =
=−− =
ij
i jk
i jk
i jk
FFF
FFF
FFF
consent of McGraw–Hill Education.
SOLUTION Continued
Thus:
30 22.5
(600 lb) (480 lb) (360 lb)
37.5
(13.8 16.5 6.6 )
22.5
(13.8 16.5 6.6 )
22.5
BH BH
DE
DE DE
DE
DF DF
BH
BH
T
DE
DE
T
DF
DF
−
= = = −
= = −+
= = −−
ij
TT i j
TT i j k
TT i jk
FFF
FFF
FFF
(a)
0:( )( )( )( )( )0
(12 ) ( 600 ) (6 ) ( 200 ) (18 ) (480 360 )
A J C K G H BH E DE F DF
Σ = × +× +× +× +× =
− ×− − ×− + × −
M rW rW rT rT rT
ijijiij
5 0 6.6 5 0 6.6 0
22.5 22.5
13.8 16.5 6.6 13.8 16.5 6.6
DE DF
TT
+ + −=
− −−
ijk ijk
or
7200 1200 6480 4.84( )
DE DF
TT+−+ −kkk i
58.08 82.5
( ) ( )0
22.5 22.5
DE DF DE DF
TT TT+ −− + =jk
i or j:
0*
DE DF DE DF
TT TT−= =
82.5
: 7200 1200 6480 (2 ) 0
22.5
DE
T+−− =k
261.82 lb
DE
T=
262 lb
DE DF
TT= =
(b)
13.8
0: 480 2 (261.82) 0 801.17 lb
22.5
16.5
0: 600 200 360 2 (261.82) 0 1544.00 lb
22.5
xx x
yy y
FA A
FA A
Σ= + + = =−
Σ= − − − − = =
0: 0
zz
FAΣ= =
(801lb) (1544 lb)=−+Aij
PROBLEM 4.65
The horizontal platform ABCD weighs 60 lb and supports a 240-lb
load at its center. The platform is normally held in position by
hinges at A and B and by braces CE and DE. If brace DE is
removed, determine the reactions at the hinges and the force
exerted by the remaining brace CE. The hinge at A does not exert
any axial thrust.
SOLUTION
Free–Body Diagram:
Express forces, weight in terms of rectangular components:
( ) ( ) ( )
3 ft 4 ft 2 ftEC =++ijk
C
( ) ( ) ( )
222
342
342
CE CE CE
EC
FF
EC
++
= =
++
ijk
F
C
0.55709 0.74278 0.37139
CE CE CE
FFF
=++ijk
( ) (300 lb)mg=−=−Wj j
( ) ( ) ( ) ( )
( ) ( )
0: 4 ft 1.5 ft 2 ft 300 lb
3 ft 4 ft 0
B
CE
Σ = × + + ×−
+ + ×=
M kA i k j
i kF
or
( ) ( ) ( )( ) ( )( )
4 ft 4 ft 1.5 ft 300 lb 2 ft 300 lb
yz
AA−+ − +ij k i
( )
3 0 4 ft 0
0.55709 0.74278 0.37139
CE
F+=
i jk
consent of McGraw–Hill Education.
PROBLEM 4.66
A 100–kg uniform rectangular plate is supported in the
position shown by hinges A and B and by cable DCE that
passes over a frictionless hook at C. Assuming that the
tension is the same in both parts of the cable, determine (a)
the tension in the cable, (b) the reactions at A and B.
Assume that the hinge at B does not exert any axial thrust.
SOLUTION
/
/
/
(960 180) 780
960 450
90
22
390 225
600 450
BA
GA
CA
−=
= −+
= +
= +
r ii
r ik
ik
r ik
Dimensions in mm
(a)
T=
Tension in cable DCE
690 675 450 1065 mm
270 675 450 855 mm
CD CD
CE CE
=−+ − =
=+− =
ijk
ijk
C
C
2
( 690 675 450 )
1065
(270 675 450 )
855
(100 kg)(9.81m/s ) (981 N)
CD
CE
T
T
mg
= −+ −
= +−
=−=− =−
T ijk
T ijk
Wi j j
/// /
0: ( ) 0
A C A CD C A CE G A B A
WΣ = × + × + ×− + × =M rTrTr jrB
600 0 450 600 0 450
1065 855
690 675 450 270 675 450
390 0 225 780 0 0 0
0 981 0 0 yz
TT
BB
+
−− −
+ +=
−
i jk ijk
i j k i jk
consent of McGraw–Hill Education.
SOLUTION Continued
Coefficient of i:
3
(450)(675) (450)(675) 220.73 10 0
1065 855
TT
− − + ×=
344.64 NT=
345 NT=
Coefficient of j:
344.64 344.64
( 690 450 600 450) (270 450 600 450) 780 0
1065 855
z
B−×+× + ×+× − =
185.516 N
z
B=
Coefficient of k:
3
344.64 344.64
(600)(675) (600)(675) 382.59 10 780 0 113.178 N
1065 855 yy
BB+ − ×+ = =
(b)
(113.2 N) (185.5 N)= +B jk
0: 0
CD CE
Σ= + + + + =F ABT T W
Coefficient of i:
690 270
(344.64) (344.64) 0
1065 855
x
A− +=
114.454 N
x
A=
Coefficient of j:
675 675
113.178 (344.64) (344.64) 981 0 377.30 N
1065 855
yy
AA+ + + −= =
Coefficient of k:
450 450
185.516 (344.64) (344.64) 0
1065 855
z
A+− − =
141.496 N
z
A=
(114.4 N) (377 N) (141.5 N)= ++A ij k
consent of McGraw–Hill Education.