PROBLEM 2.72
Find the magnitude and direction of the resultant of the two forces
shown knowing that P = 300 N and Q = 400 N.
SOLUTION
(300 N)[ cos30 sin15 sin30 cos30 cos15 ]
(67.243 N) (150 N) (250.95 N)
(400 N)[cos50 cos20 sin 50 cos50 sin 20 ]
(400 N)[0.60402 0.76604 0.21985]
(241.61 N) (306.42 N) (87.939 N)
(174.
= − ° °+ °+ ° °
=− ++
= ° °+ °− ° °
= +−
=+−
= +
=
P ij k
ij k
Q ij k
ij
i jk
RPQ
22 2
367 N) (456.42 N) (163.011 N)
(174.367 N) (456.42 N) (163.011 N)
515.07 N
R
++
= ++
=
ij k
515 NR=
174.367 N
cos 0.33853
515.07 N
x
xR
R
θ
= = =
70.2
x
θ
= °
456.42 N
cos 0.88613
515.07 N
y
y
R
R
θ
= = =
27.6
y
θ
= °
163.011 N
cos 0.31648
515.07 N
z
zR
R
θ
= = =
71.5
z
θ
= °
consent of McGraw–Hill Education.
PROBLEM 2.73
Find the magnitude and direction of the resultant of the two forces
shown knowing that P = 400 N and Q = 300 N.
SOLUTION
(400 N)[ cos30 sin15 sin30 cos30 cos15 ]
(89.678 N) (200 N) (334.61 N)
(300 N)[cos50 cos 20 sin50 cos50 sin 20 ]
(181.21 N) (229.81 N) (65.954 N)
(91.532 N) (429.81 N) (268.66 N)
(91.5
R
= − ° °+ °+ ° °
=− ++
= ° °+ °− ° °
=+−
= +
=++
=
P ij k
ij k
Q ij k
ijk
RPQ
ijk
222
32 N) (429.81 N) (268.66 N)
515.07 N
++
=
515 NR=
91.532 N
cos 0.177708
515.07 N
x
x
R
R
θ
= = =
79.8
x
θ
= °
429.81 N
cos 0.83447
515.07 N
y
y
R
R
θ
= = =
33.4
y
θ
= °
268.66 N
cos 0.52160
515.07 N
z
z
R
R
θ
= = =
58.6
z
θ
= °
PROBLEM 2.74
Knowing that the tension is 425 lb in cable AB and 510 lb
in cable AC, determine the magnitude and direction of the
resultant of the forces exerted at A by the two cables.
SOLUTION
222
222
(40 in.) (45 in.) (60 in.)
(40 in.) (45 in.) (60 in.) 85 in.
(100 in.) (45 in.) (60 in.)
(100 in.) (45 in.) (60 in.) 125 in.
(40 in.) (45 in.) (60 in.)
(425 lb) 85 in.
AB AB AB AB
AB
AB
AC
AC
AB
TT
AB
=−+
= ++=
= −+
= ++=
−+
= = =
ijk
ijk
ijk
Tλ
C
C
C
(200 lb) (225 lb) (300 lb)
(100 in.) (45 in.) (60 in.)
(510 lb) 125 in.
(408 lb) (183.6 lb) (244.8 lb)
(608) (408.6 lb) (544.8 lb)
AB
AC AC AC AC
AC
AB AC
AC
TT
AC
=−+
−+
= = =
=−+
=+= − +
T ijk
ijk
Tλ
T ijk
RT T i j k
C
Then:
912.92 lbR=
913 lbR=
and
608 lb
cos 0.66599
912.92 lb
x
θ
= =
48.2
x
θ
= °
408.6 lb
cos 0.44757
912.92 lb
y
θ
= = −
116.6
y
θ
= °
544.8 lb
cos 0.59677
912.92 lb
z
θ
= =
53.4
z
θ
= °
consent of McGraw–Hill Education.
PROBLEM 2.75
Knowing that the tension is 510 lb in cable AB and 425 lb
in cable AC, determine the magnitude and direction of the
resultant of the forces exerted at A by the two cables.
SOLUTION
222
222
(40 in.) (45 in.) (60 in.)
(40 in.) (45 in.) (60 in.) 85 in.
(100 in.) (45 in.) (60 in.)
(100 in.) (45 in.) (60 in.) 125 in.
(40 in.) (45 in.) (60 in.)
(510 lb) 85 in.
AB AB AB AB
AB
AB
AC
AC
AB
TT
AB
=−+
= ++=
= −+
= ++=
−+
= = =
ijk
ijk
ijk
Tλ
C
C
C
(240 lb) (270 lb) (360 lb)
(100 in.) (45 in.) (60 in.)
(425 lb) 125 in.
(340 lb) (153 lb) (204 lb)
(580 lb) (423 lb) (564 lb)
AB
AC AC AC AC
AC
AB AC
AC
TT
AC
=−+
−+
= = =
=−+
=+= − +
T ijk
ijk
Tλ
T ijk
RT T i j k
C
Then:
912.92 lbR=
913 lbR=
and
580 lb
cos 0.63532
912.92 lb
x
θ
= =
50.6
x
θ
= °
423 lb
−
z
θ
= °
z
θ
= °
PROBLEM 2.76
A frame ABC is supported in part by cable DBE that passes
through a frictionless ring at B. Knowing that the tension in
the cable is 385 N, determine the components of the force
exerted by the cable on the support at D.
SOLUTION
222
(480 mm) (510 mm) (320 mm)
(480 mm) (510 mm) (320 mm) 770 mm
BD
BD
=−+ −
= ++=
ijk
(385 N) [ (480 mm) (510 mm) (320 mm) ]
(770 mm)
(240 N) (255 N) (160 N)
BD BD BD BD BD
TT
BD
= =
= −+ −
=−+ −
Fλ
ijk
i jk
222
(270 mm) (400 mm) (600 mm)
(270 mm) (400 mm) (600 mm) 770 mm
BE
BE
=−+ −
= ++=
i jk
(385 N) [ (270 mm) (400 mm) (600 mm) ]
(770 mm)
(135 N) (200 N) (300 N)
BE BE BE BE BE
TT
BE
= =
= −+ −
=−+ −
Fλ
i jk
i jk
(375 N) (455 N) (460 N)
BD BE
=+=− + −RF F i j k
222
(375 N) (455 N) (460 N) 747.83 N
R= ++ =
748 NR=
375 N
cos 747.83 N
x
θ
−
=
120.1
x
θ
= °
455 N
cos 747.83 N
y
θ
=
52.5
y
θ
= °
460 N
cos 747.83 N
z
θ
−
=
128.0
z
θ
= °
consent of McGraw–Hill Education.
PROBLEM 2.77
For the plate of Prob. 2.68, determine the tensions in cables AB
and AD knowing that the tension in cable AC is 54 N and that
the resultant of the forces exerted by the three cables at A must
be vertical.
SOLUTION
We have:
( )
(320 mm) (480 mm) (360 mm) 680 mm
(450 mm) (480 mm) (360 mm) 750 mm
(250 mm) (480 mm) 360 mm 650 mm
AB AB
AC AC
AD AD
=−−+ =
=−+ =
=−− =
ijk
ijk
ijk
C
C
C
Thus:
( )
( )
( )
320 480 360
680
54 450 480 360
750
250 480 360
650
AB
AB AB AB AB
AC AC AC AC
AD
AD AD AD AD
T
AB
TT
AB
AC
TT
AC
T
AD
TT
AD
= = = −− +
= = = −+
= = = −−
Tλ i jk
Tλ i jk
Tλ i jk
C
C
C
Substituting into the Eq.
= ΣRF
and factoring
, , :i jk
320 250
32.40
680 650
480 480
34.560
680 650
360 360
25.920
680 650
AB AD
AB AD
AB AD
TT
TT
TT
=− ++
+− − −
+ +−
Ri
j
k
consent of McGraw–Hill Education.
SOLUTION (Continued)
:i
320 250
32.40 0
680 650
AB AD
TT
− ++ =
(1)
:k
360 360
25.920 0
680 650
AB AD
TT+− =
(2)
Multiply (1) by 3.6 and (2) by 2.5 then add:
252 181.440 0
680 AB
T−+=
489.60 N
AB
T=
490 N
AB
T=
Substitute into (2) and solve for
:
AD
T
360 360
(489.60 N) 25.920 0
680 650
AD
T+− =
514.80 N
AD
T=
515 N
AD
T=
consent of McGraw–Hill Education.
PROBLEM 2.78
The boom OA carries a load P and is supported by
two cables as shown. Knowing that the tension in
cable AB is 183 lb and that the resultant of the load P
and of the forces exerted at A by the two cables must
be directed along OA, determine the tension in cable
AC.
SOLUTION
Cable AB:
183 lb
AB
T=
( 48 in.) (29 in.) (24 in.)
(183 lb) 61in.
(144 lb) (87 lb) (72 lb)
AB AB AB AB
AB
AB
TT
AB
−+ +
= = =
=− ++
ijk
T
T ijk
l
Cable AC:
( 48 in.) (25 in.) ( 36 in.)
65 in.
48 25 36
65 65 65
AC AC AC AC AC
AC AC AC AC
AC
TTT
AC
TTT
− + +−
= = =
=−+−
ij k
T
T i jk
l
Load P:
P=Pj
36
0: (72 lb) 0
65
z z AC
RF T
′
= Σ= − =
130.0 lb
AC
T=
PROBLEM 2.79
For the boom and loading of Problem. 2.78, determine
the magnitude of the load P.
PROBLEM 2.78 The boom OA carries a load P and is
supported by two cables as shown. Knowing that the
tension in cable AB is 183 lb and that the resultant of
the load P and of the forces exerted at A by the two
cables must be directed along OA, determine the
tension in cable AC.
SOLUTION
See Problem 2.78. Since resultant must be directed along OA, i.e., the x–axis, we write
25
0: (87 lb) 0
65
y y AC
R F TP= Σ = + −=
130.0 lb
AC
T=
from Problem 2.97.
Then
25
(87 lb) (130.0 lb) 0
65 P+ −=
137.0 lbP=
consent of McGraw–Hill Education.
PROBLEM 2.72
Find the magnitude and direction of the resultant of the two forces
shown knowing that P = 300 N and Q = 400 N.
SOLUTION
(300 N)[ cos30 sin15 sin30 cos30 cos15 ]
(67.243 N) (150 N) (250.95 N)
(400 N)[cos50 cos20 sin 50 cos50 sin 20 ]
(400 N)[0.60402 0.76604 0.21985]
(241.61 N) (306.42 N) (87.939 N)
(174.
= − ° °+ °+ ° °
=− ++
= ° °+ °− ° °
= +−
=+−
= +
=
P ij k
ij k
Q ij k
ij
i jk
RPQ
22 2
367 N) (456.42 N) (163.011 N)
(174.367 N) (456.42 N) (163.011 N)
515.07 N
R
++
= ++
=
ij k
515 NR=
174.367 N
cos 0.33853
515.07 N
x
xR
R
θ
= = =
70.2
x
θ
= °
456.42 N
cos 0.88613
515.07 N
y
y
R
R
θ
= = =
27.6
y
θ
= °
163.011 N
cos 0.31648
515.07 N
z
zR
R
θ
= = =
71.5
z
θ
= °
consent of McGraw–Hill Education.
PROBLEM 2.73
Find the magnitude and direction of the resultant of the two forces
shown knowing that P = 400 N and Q = 300 N.
SOLUTION
(400 N)[ cos30 sin15 sin30 cos30 cos15 ]
(89.678 N) (200 N) (334.61 N)
(300 N)[cos50 cos 20 sin50 cos50 sin 20 ]
(181.21 N) (229.81 N) (65.954 N)
(91.532 N) (429.81 N) (268.66 N)
(91.5
R
= − ° °+ °+ ° °
=− ++
= ° °+ °− ° °
=+−
= +
=++
=
P ij k
ij k
Q ij k
ijk
RPQ
ijk
222
32 N) (429.81 N) (268.66 N)
515.07 N
++
=
515 NR=
91.532 N
cos 0.177708
515.07 N
x
x
R
R
θ
= = =
79.8
x
θ
= °
429.81 N
cos 0.83447
515.07 N
y
y
R
R
θ
= = =
33.4
y
θ
= °
268.66 N
cos 0.52160
515.07 N
z
z
R
R
θ
= = =
58.6
z
θ
= °
PROBLEM 2.74
Knowing that the tension is 425 lb in cable AB and 510 lb
in cable AC, determine the magnitude and direction of the
resultant of the forces exerted at A by the two cables.
SOLUTION
222
222
(40 in.) (45 in.) (60 in.)
(40 in.) (45 in.) (60 in.) 85 in.
(100 in.) (45 in.) (60 in.)
(100 in.) (45 in.) (60 in.) 125 in.
(40 in.) (45 in.) (60 in.)
(425 lb) 85 in.
AB AB AB AB
AB
AB
AC
AC
AB
TT
AB
=−+
= ++=
= −+
= ++=
−+
= = =
ijk
ijk
ijk
Tλ
C
C
C
(200 lb) (225 lb) (300 lb)
(100 in.) (45 in.) (60 in.)
(510 lb) 125 in.
(408 lb) (183.6 lb) (244.8 lb)
(608) (408.6 lb) (544.8 lb)
AB
AC AC AC AC
AC
AB AC
AC
TT
AC
=−+
−+
= = =
=−+
=+= − +
T ijk
ijk
Tλ
T ijk
RT T i j k
C
Then:
912.92 lbR=
913 lbR=
and
608 lb
cos 0.66599
912.92 lb
x
θ
= =
48.2
x
θ
= °
408.6 lb
cos 0.44757
912.92 lb
y
θ
= = −
116.6
y
θ
= °
544.8 lb
cos 0.59677
912.92 lb
z
θ
= =
53.4
z
θ
= °
consent of McGraw–Hill Education.
PROBLEM 2.75
Knowing that the tension is 510 lb in cable AB and 425 lb
in cable AC, determine the magnitude and direction of the
resultant of the forces exerted at A by the two cables.
SOLUTION
222
222
(40 in.) (45 in.) (60 in.)
(40 in.) (45 in.) (60 in.) 85 in.
(100 in.) (45 in.) (60 in.)
(100 in.) (45 in.) (60 in.) 125 in.
(40 in.) (45 in.) (60 in.)
(510 lb) 85 in.
AB AB AB AB
AB
AB
AC
AC
AB
TT
AB
=−+
= ++=
= −+
= ++=
−+
= = =
ijk
ijk
ijk
Tλ
C
C
C
(240 lb) (270 lb) (360 lb)
(100 in.) (45 in.) (60 in.)
(425 lb) 125 in.
(340 lb) (153 lb) (204 lb)
(580 lb) (423 lb) (564 lb)
AB
AC AC AC AC
AC
AB AC
AC
TT
AC
=−+
−+
= = =
=−+
=+= − +
T ijk
ijk
Tλ
T ijk
RT T i j k
C
Then:
912.92 lbR=
913 lbR=
and
580 lb
cos 0.63532
912.92 lb
x
θ
= =
50.6
x
θ
= °
423 lb
−
PROBLEM 2.76
A frame ABC is supported in part by cable DBE that passes
through a frictionless ring at B. Knowing that the tension in
the cable is 385 N, determine the components of the force
exerted by the cable on the support at D.
SOLUTION
222
(480 mm) (510 mm) (320 mm)
(480 mm) (510 mm) (320 mm) 770 mm
BD
BD
=−+ −
= ++=
ijk
(385 N) [ (480 mm) (510 mm) (320 mm) ]
(770 mm)
(240 N) (255 N) (160 N)
BD BD BD BD BD
TT
BD
= =
= −+ −
=−+ −
Fλ
ijk
i jk
222
(270 mm) (400 mm) (600 mm)
(270 mm) (400 mm) (600 mm) 770 mm
BE
BE
=−+ −
= ++=
i jk
(385 N) [ (270 mm) (400 mm) (600 mm) ]
(770 mm)
(135 N) (200 N) (300 N)
BE BE BE BE BE
TT
BE
= =
= −+ −
=−+ −
Fλ
i jk
i jk
(375 N) (455 N) (460 N)
BD BE
=+=− + −RF F i j k
222
(375 N) (455 N) (460 N) 747.83 N
R= ++ =
748 NR=
375 N
cos 747.83 N
x
θ
−
=
120.1
x
θ
= °
455 N
cos 747.83 N
y
θ
=
52.5
y
θ
= °
460 N
cos 747.83 N
z
θ
−
=
128.0
z
θ
= °
consent of McGraw–Hill Education.
PROBLEM 2.77
For the plate of Prob. 2.68, determine the tensions in cables AB
and AD knowing that the tension in cable AC is 54 N and that
the resultant of the forces exerted by the three cables at A must
be vertical.
SOLUTION
We have:
( )
(320 mm) (480 mm) (360 mm) 680 mm
(450 mm) (480 mm) (360 mm) 750 mm
(250 mm) (480 mm) 360 mm 650 mm
AB AB
AC AC
AD AD
=−−+ =
=−+ =
=−− =
ijk
ijk
ijk
C
C
C
Thus:
( )
( )
( )
320 480 360
680
54 450 480 360
750
250 480 360
650
AB
AB AB AB AB
AC AC AC AC
AD
AD AD AD AD
T
AB
TT
AB
AC
TT
AC
T
AD
TT
AD
= = = −− +
= = = −+
= = = −−
Tλ i jk
Tλ i jk
Tλ i jk
C
C
C
Substituting into the Eq.
= ΣRF
and factoring
, , :i jk
320 250
32.40
680 650
480 480
34.560
680 650
360 360
25.920
680 650
AB AD
AB AD
AB AD
TT
TT
TT
=− ++
+− − −
+ +−
Ri
j
k
consent of McGraw–Hill Education.
SOLUTION (Continued)
:i
320 250
32.40 0
680 650
AB AD
TT
− ++ =
(1)
:k
360 360
25.920 0
680 650
AB AD
TT+− =
(2)
Multiply (1) by 3.6 and (2) by 2.5 then add:
252 181.440 0
680 AB
T−+=
489.60 N
AB
T=
490 N
AB
T=
Substitute into (2) and solve for
:
AD
T
360 360
(489.60 N) 25.920 0
680 650
AD
T+− =
514.80 N
AD
T=
515 N
AD
T=
consent of McGraw–Hill Education.
PROBLEM 2.78
The boom OA carries a load P and is supported by
two cables as shown. Knowing that the tension in
cable AB is 183 lb and that the resultant of the load P
and of the forces exerted at A by the two cables must
be directed along OA, determine the tension in cable
AC.
SOLUTION
Cable AB:
183 lb
AB
T=
( 48 in.) (29 in.) (24 in.)
(183 lb) 61in.
(144 lb) (87 lb) (72 lb)
AB AB AB AB
AB
AB
TT
AB
−+ +
= = =
=− ++
ijk
T
T ijk
l
Cable AC:
( 48 in.) (25 in.) ( 36 in.)
65 in.
48 25 36
65 65 65
AC AC AC AC AC
AC AC AC AC
AC
TTT
AC
TTT
− + +−
= = =
=−+−
ij k
T
T i jk
l
Load P:
P=Pj
36
0: (72 lb) 0
65
z z AC
RF T
′
= Σ= − =
130.0 lb
AC
T=
PROBLEM 2.79
For the boom and loading of Problem. 2.78, determine
the magnitude of the load P.
PROBLEM 2.78 The boom OA carries a load P and is
supported by two cables as shown. Knowing that the
tension in cable AB is 183 lb and that the resultant of
the load P and of the forces exerted at A by the two
cables must be directed along OA, determine the
tension in cable AC.
SOLUTION
See Problem 2.78. Since resultant must be directed along OA, i.e., the x–axis, we write
25
0: (87 lb) 0
65
y y AC
R F TP= Σ = + −=
130.0 lb
AC
T=
from Problem 2.97.
Then
25
(87 lb) (130.0 lb) 0
65 P+ −=
137.0 lbP=
consent of McGraw–Hill Education.