PROBLEM 2.11
A trolley that moves along a horizontal beam is acted upon by two forces
as shown. Determine by trigonometry the magnitude and direction of the
force P so that the resultant is a vertical force of 2500 N.
PROBLEM 2.12
For the hook support shown, determine by trigonometry the magnitude and
direction of the resultant of the two forces applied to the support.
SOLUTION
Using the law of cosines:
222
(200 lb) (300 lb)
2(200 lb)(300 lb)cos(45 65 )
413.57 lb
R
R
= +
− +°
=
Using the law of sines:
90 25 42.972
β
=+− °

414 lb=R
72.0°
PROBLEM 2.13
The cable stays AB and AD help support pole AC. Knowing
that the tension is 120 lb in AB and 40 lb in AD, determine
graphically the magnitude and direction of the resultant of the
forces exerted by the stays at A using (a) the parallelogram
law, (b) the triangle rule.
SOLUTION
8
tan 10
38.66
6
tan 10
30.96
a
a
β
β
=
= °
=
= °
Using the triangle rule:
180
38.66 30.96 180
110.38
a βψ
ψ
ψ
++= °
°+ °+ = °
= °
Using the law of cosines:
22
2
(120 lb) (40 lb) 2(120 lb)(40 lb)cos110.38
139.08 lb
R
R
= +− °
=
Using the law of sines:
sin sin110.38
40 lb 139.08 lb
γ
°
=
15.64
(90 )
(90 38.66 ) 15.64
66.98
γ
φ aγ
φ
φ
= °
= °− +
= °− ° + °
= °
139.1 lb=R
67.0°
PROBLEM 2.14
Solve Problem 2.4 by trigonometry.
PROBLEM 2.4: Two structural members B and C are bolted to bracket
A. Knowing that both members are in tension and that P = 6 kips and
Q = 4 kips, determine graphically the magnitude and direction of the
resultant force exerted on the bracket using (a) the parallelogram law,
(b) the triangle rule.
SOLUTION
Using the force triangle and the laws of cosines and sines:
We have:
180 (50 25 )
105
γ
= °− °+ °
= °
Then
222
2
(4 kips) (6 kips) 2(4 kips)(6 kips)cos105
64.423 kips
8.0264 kips
R
R
=+− °
=
=
And
4 kips 8.0264 kips
sin(25 ) sin105
sin(25 ) 0.48137
25 28.775
3.775
α
α
α
α
=
°+ °
°+ =
°+ = °
= °
8.03 kips=R
3.8°
PROBLEM 2.15
For the hook support of Prob. 2.9, determine by trigonometry (a) the
magnitude and direction of the smallest force P for which the resultant
R of the two forces applied to the support is horizontal, (b) the
corresponding magnitude of R.
SOLUTION
The smallest force P will be perpendicular to R.
(a)
(50 N)sin 25P= °
21.1 N=P
(b)
(50 N)cos 25R= °
45.3 NR=
consent of McGraw–Hill Education.
PROBLEM 2.16
Determine the x and y components of each of the forces shown.
SOLUTION
Compute the following distances:
22
22
22
(600) (800)
1000 mm
(560) (900)
1060 mm
(480) (900)
1020 mm
OA
OB
OC
= +
=
= +
=
= +
=
800-N Force:
800
(800 N)1000
x
F= +
640 N
x
F= +
600
(800 N)1000
y
F= +
480 N
y
F= +
424-N Force:
560
(424 N)1060
x
F= −
224 N
x
F= −
900
(424 N)1060
y
F= −
360 N
y
F= −
408-N Force:
480
(408 N)1020
x
F= +
192.0 N
x
F= +
900
(408 N)1020
y
F= −
360 N
y
F= −
consent of McGraw–Hill Education.
PROBLEM 2.17
Determine the x and y components of each of the forces shown.
SOLUTION
Compute the following distances:
22
22
22
(84) (80)
116 in.
(28) (96)
100 in.
(48) (90)
102 in.
OA
OB
OC
= +
=
= +
=
= +
=
29-lb Force:
84
(29 lb)116
x
F= +
21.0 lb
x
F= +
80
(29 lb)116
y
F= +
20.0 lb
y
F= +
50-lb Force:
28
(50 lb)100
x
F= −
14.00 lb
x
F= −
96
(50 lb)100
y
F= +
48.0 lb
y
F= +
51-lb Force:
48
(51 lb) 102
x
F= +
24.0 lb
x
F= +
90
(51 lb) 102
y
F= −
45.0 lb
y
F= −
consent of McGraw–Hill Education.
PROBLEM 2.18
Determine the x and y components of each of the forces shown.
SOLUTION
40-lb Force:
(40 lb)cos60
x
F=+°
20.0 lb
x
F=
(40 lb)sin60
y
F=−°
34.6 lb
y
F= −
50-lb Force:
(50 lb)sin50
x
F=−°
38.3 lb
x
F= −
(50 lb)cos50
y
F=−°
32.1 lb
y
F= −
60-lb Force:
(60 lb)cos25
x
F=+°
54.4 lb
x
F=
(60 lb)sin 25
y
F=+°
25.4 lb
y
F=
consent of McGraw–Hill Education.
PROBLEM 2.19
Determine the x and y components of each of the forces shown.
SOLUTION
80-N Force:
(80 N)cos40
x
F=+°
61.3 N
x
F=
(80 N)sin 40
y
F=+°
51.4 N
y
F=
120-N Force:
(120 N)cos70
x
F=+°
41.0 N
x
F=
(120 N)sin70
y
F=+°
112.8 N
y
F=
150-N Force:
(150 N)cos35
x
F=−°
122. 9 N
x
F= −
(150 N)sin35
y
F=+°
86.0 N
y
F=
consent of McGraw–Hill Education.
PROBLEM 2.20
Member BD exerts on member ABC a force P directed along line BD.
Knowing that P must have a 300–lb horizontal component, determine (a) the
magnitude of the force P, (b) its vertical component.
SOLUTION
(a)
sin35 300 lbP°=
300 lb
sin35
P=°
523 lbP=
(b) Vertical component
cos35
v
PP= °
(523 lb)cos35= °
428 lb
v
P=
consent of McGraw–Hill Education.
PROBLEM 2.12
For the hook support shown, determine by trigonometry the magnitude and
direction of the resultant of the two forces applied to the support.
SOLUTION
Using the law of cosines:
222
(200 lb) (300 lb)
2(200 lb)(300 lb)cos(45 65 )
413.57 lb
R
R
= +
− +°
=
Using the law of sines:
90 25 42.972
β
=+− °

414 lb=R
72.0°
PROBLEM 2.13
The cable stays AB and AD help support pole AC. Knowing
that the tension is 120 lb in AB and 40 lb in AD, determine
graphically the magnitude and direction of the resultant of the
forces exerted by the stays at A using (a) the parallelogram
law, (b) the triangle rule.
SOLUTION
8
tan 10
38.66
6
tan 10
30.96
a
a
β
β
=
= °
=
= °
Using the triangle rule:
180
38.66 30.96 180
110.38
a βψ
ψ
ψ
++= °
°+ °+ = °
= °
Using the law of cosines:
22
2
(120 lb) (40 lb) 2(120 lb)(40 lb)cos110.38
139.08 lb
R
R
= +− °
=
Using the law of sines:
sin sin110.38
40 lb 139.08 lb
γ
°
=
15.64
(90 )
(90 38.66 ) 15.64
66.98
γ
φ aγ
φ
φ
= °
= °− +
= °− ° + °
= °
139.1 lb=R
67.0°
PROBLEM 2.14
Solve Problem 2.4 by trigonometry.
PROBLEM 2.4: Two structural members B and C are bolted to bracket
A. Knowing that both members are in tension and that P = 6 kips and
Q = 4 kips, determine graphically the magnitude and direction of the
resultant force exerted on the bracket using (a) the parallelogram law,
(b) the triangle rule.
SOLUTION
Using the force triangle and the laws of cosines and sines:
We have:
180 (50 25 )
105
γ
= °− °+ °
= °
Then
222
2
(4 kips) (6 kips) 2(4 kips)(6 kips)cos105
64.423 kips
8.0264 kips
R
R
=+− °
=
=
And
4 kips 8.0264 kips
sin(25 ) sin105
sin(25 ) 0.48137
25 28.775
3.775
α
α
α
α
=
°+ °
°+ =
°+ = °
= °
8.03 kips=R
3.8°
PROBLEM 2.15
For the hook support of Prob. 2.9, determine by trigonometry (a) the
magnitude and direction of the smallest force P for which the resultant
R of the two forces applied to the support is horizontal, (b) the
corresponding magnitude of R.
SOLUTION
The smallest force P will be perpendicular to R.
(a)
(50 N)sin 25P= °
21.1 N=P
(b)
(50 N)cos 25R= °
45.3 NR=
consent of McGraw–Hill Education.
PROBLEM 2.16
Determine the x and y components of each of the forces shown.
SOLUTION
Compute the following distances:
22
22
22
(600) (800)
1000 mm
(560) (900)
1060 mm
(480) (900)
1020 mm
OA
OB
OC
= +
=
= +
=
= +
=
800-N Force:
800
(800 N)1000
x
F= +
640 N
x
F= +
600
(800 N)1000
y
F= +
480 N
y
F= +
424-N Force:
560
(424 N)1060
x
F= −
224 N
x
F= −
900
(424 N)1060
y
F= −
360 N
y
F= −
408-N Force:
480
(408 N)1020
x
F= +
192.0 N
x
F= +
900
(408 N)1020
y
F= −
360 N
y
F= −
consent of McGraw–Hill Education.
PROBLEM 2.17
Determine the x and y components of each of the forces shown.
SOLUTION
Compute the following distances:
22
22
22
(84) (80)
116 in.
(28) (96)
100 in.
(48) (90)
102 in.
OA
OB
OC
= +
=
= +
=
= +
=
29-lb Force:
84
(29 lb)116
x
F= +
21.0 lb
x
F= +
80
(29 lb)116
y
F= +
20.0 lb
y
F= +
50-lb Force:
28
(50 lb)100
x
F= −
14.00 lb
x
F= −
96
(50 lb)100
y
F= +
48.0 lb
y
F= +
51-lb Force:
48
(51 lb) 102
x
F= +
24.0 lb
x
F= +
90
(51 lb) 102
y
F= −
45.0 lb
y
F= −
consent of McGraw–Hill Education.
PROBLEM 2.18
Determine the x and y components of each of the forces shown.
SOLUTION
40-lb Force:
(40 lb)cos60
x
F=+°
20.0 lb
x
F=
(40 lb)sin60
y
F=−°
34.6 lb
y
F= −
50-lb Force:
(50 lb)sin50
x
F=−°
38.3 lb
x
F= −
(50 lb)cos50
y
F=−°
32.1 lb
y
F= −
60-lb Force:
(60 lb)cos25
x
F=+°
54.4 lb
x
F=
(60 lb)sin 25
y
F=+°
25.4 lb
y
F=
consent of McGraw–Hill Education.
PROBLEM 2.19
Determine the x and y components of each of the forces shown.
SOLUTION
80-N Force:
(80 N)cos40
x
F=+°
61.3 N
x
F=
(80 N)sin 40
y
F=+°
51.4 N
y
F=
120-N Force:
(120 N)cos70
x
F=+°
41.0 N
x
F=
(120 N)sin70
y
F=+°
112.8 N
y
F=
150-N Force:
(150 N)cos35
x
F=−°
122. 9 N
x
F= −
(150 N)sin35
y
F=+°
86.0 N
y
F=
consent of McGraw–Hill Education.
PROBLEM 2.20
Member BD exerts on member ABC a force P directed along line BD.
Knowing that P must have a 300–lb horizontal component, determine (a) the
magnitude of the force P, (b) its vertical component.
SOLUTION
(a)
sin35 300 lbP°=
300 lb
sin35
P=°
523 lbP=
(b) Vertical component
cos35
v
PP= °
(523 lb)cos35= °
428 lb
v
P=
consent of McGraw–Hill Education.