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PROBLEM 2.2
Two forces are applied as shown to a bracket support. Determine
graphically the magnitude and direction of their resultant using (a) the
parallelogram law, (b) the triangle rule.
SOLUTION
PROBLEM 2.3
Two structural members B and C are bolted to bracket A. Knowing that
both members are in tension and that P = 10 kN and Q = 15 kN,
determine graphically the magnitude and direction of the resultant force
exerted on the bracket using (a) the parallelogram law, (b) the triangle
rule.
SOLUTION
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
PROBLEM 2.5
The 300-lb force is to be resolved into components along lines a-a′ and b-b′.
(a) Determine the angle
α
by trigonometry knowing that the component along
line a-a′ is to be 240 lb. (b) What is the corresponding value of the
component along b-b′?
SOLUTION
(a) Using the triangle rule and law of sines:
sin sin60
240 lb 300 lb
sin 0.69282
43.854
60 180
180 60 43.854
76.146
b
b
b
αb
α
°
=
=
= °
+ + °= °
= °− °− °
= °
76.1
α
= °
(b) Law of sines:
300 lb
sin76.146 sin60
bb
F′=
°°
336 lb
bb
F
′
=
PROBLEM 2.6
The 300-lb force is to be resolved into components along lines a-a′ and b-b′.
(a) Determine the angle
α
by trigonometry knowing that the component along
line b-b′ is to be 120 lb. (b) What is the corresponding value of the component
along a-a′?
SOLUTION
Using the triangle rule and law of sines:
(a)
sin sin60
120 lb 300 lb
α
°
=
sin 0.34641
20.268
α
α
=
= °
20.3
α
= °
(b)
60 180
180 60 20.268
99.732
αβ
β
+ + °= °
= °− °− °
= °
300 lb
sin99.732 sin60
aa
F′=
°°
341 lb
aa
F
′
=
PROBLEM 2.7
A trolley that moves along a horizontal beam is acted upon by two
forces as shown. (a) Knowing that
α
= 25°, determine by trigonometry
the magnitude of the force P so that the resultant force exerted on the
trolley is vertical. (b) What is the corresponding magnitude of the
resultant?
SOLUTION
Using the triangle rule and the law of sines:
(a)
1600 N
sin 25° sin 75
P
=°
3660 NP=
(b)
25 75 180
180 25 75
80
β
β
°+ + °= °
= °− °− °
= °
1600 N
sin 25° sin80
R
=°
3730 NR=
PROBLEM 2.8
A disabled automobile is pulled by means of two ropes as
shown. The tension in rope AB is 2.2 kN, and the angle α
is 25°. Knowing that the resultant of the two forces
applied at A is directed along the axis of the automobile,
determine by trigonometry (a) the tension in rope AC, (b)
the magnitude of the resultant of the two forces applied at
A.
SOLUTION
Using the law of sines:
2.2 kN
sin30° sin125 sin 25
AC
TR
= =
°
A
2.603 kN
4.264 kN
AC
T
R
=
=
(a)
2.60 kN
AC
T=
(b)
4.26 kNR=
PROBLEM 2.9
Two forces are applied as shown to a hook support. Knowing that the
magnitude of P is 35 N, determine by trigonometry (a) the required
angle
α
if the resultant R of the two forces applied to the support is to
be horizontal, (b) the corresponding magnitude of R.
SOLUTION
Using the triangle rule and law of sines:
(a)
sin sin 25
50 N 35 N
sin 0.60374
α
α
°
=
=
37.138
α
= °
37.1
α
= °
(b)
25 180
180 25 37.138
117.862
αβ
β
+ + °= °
= °− °− °
= °
35 N
sin117.862 sin 25
R=
°°
73.2 NR=
PROBLEM 2.10
A disabled automobile is pulled by means of two ropes as
shown. Knowing that the tension in rope AB is 3 kN,
determine by trigonometry the tension in rope AC and the
value of α so that the resultant force exerted at A is a 4.8-
kN force directed along the axis of the automobile.
SOLUTION
Using the law of cosines:
22 2
(3 kN) (4.8 kN) 2(3 kN)(4.8 kN)cos 30°
2.6643 kN
AC
AC
T
T
=+−
=
Using the law of sines:
sin sin30
3 kN 2.6643 kN
34.3
α
α
°
=
= °
2.66 kN
AC
=T
34.3°
PROBLEM 2.2
Two forces are applied as shown to a bracket support. Determine
graphically the magnitude and direction of their resultant using (a) the
parallelogram law, (b) the triangle rule.
SOLUTION
PROBLEM 2.3
Two structural members B and C are bolted to bracket A. Knowing that
both members are in tension and that P = 10 kN and Q = 15 kN,
determine graphically the magnitude and direction of the resultant force
exerted on the bracket using (a) the parallelogram law, (b) the triangle
rule.
SOLUTION
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
PROBLEM 2.5
The 300-lb force is to be resolved into components along lines a-a′ and b-b′.
(a) Determine the angle
α
by trigonometry knowing that the component along
line a-a′ is to be 240 lb. (b) What is the corresponding value of the
component along b-b′?
SOLUTION
(a) Using the triangle rule and law of sines:
sin sin60
240 lb 300 lb
sin 0.69282
43.854
60 180
180 60 43.854
76.146
b
b
b
αb
α
°
=
=
= °
+ + °= °
= °− °− °
= °
76.1
α
= °
(b) Law of sines:
300 lb
sin76.146 sin60
bb
F′=
°°
336 lb
bb
F
′
=
PROBLEM 2.6
The 300-lb force is to be resolved into components along lines a-a′ and b-b′.
(a) Determine the angle
α
by trigonometry knowing that the component along
line b-b′ is to be 120 lb. (b) What is the corresponding value of the component
along a-a′?
SOLUTION
Using the triangle rule and law of sines:
(a)
sin sin60
120 lb 300 lb
α
°
=
sin 0.34641
20.268
α
α
=
= °
20.3
α
= °
(b)
60 180
180 60 20.268
99.732
αβ
β
+ + °= °
= °− °− °
= °
300 lb
sin99.732 sin60
aa
F′=
°°
341 lb
aa
F
′
=
PROBLEM 2.7
A trolley that moves along a horizontal beam is acted upon by two
forces as shown. (a) Knowing that
α
= 25°, determine by trigonometry
the magnitude of the force P so that the resultant force exerted on the
trolley is vertical. (b) What is the corresponding magnitude of the
resultant?
SOLUTION
Using the triangle rule and the law of sines:
(a)
1600 N
sin 25° sin 75
P
=°
3660 NP=
(b)
25 75 180
180 25 75
80
β
β
°+ + °= °
= °− °− °
= °
1600 N
sin 25° sin80
R
=°
3730 NR=
PROBLEM 2.8
A disabled automobile is pulled by means of two ropes as
shown. The tension in rope AB is 2.2 kN, and the angle α
is 25°. Knowing that the resultant of the two forces
applied at A is directed along the axis of the automobile,
determine by trigonometry (a) the tension in rope AC, (b)
the magnitude of the resultant of the two forces applied at
A.
SOLUTION
Using the law of sines:
2.2 kN
sin30° sin125 sin 25
AC
TR
= =
°
A
2.603 kN
4.264 kN
AC
T
R
=
=
(a)
2.60 kN
AC
T=
(b)
4.26 kNR=
PROBLEM 2.9
Two forces are applied as shown to a hook support. Knowing that the
magnitude of P is 35 N, determine by trigonometry (a) the required
angle
α
if the resultant R of the two forces applied to the support is to
be horizontal, (b) the corresponding magnitude of R.
SOLUTION
Using the triangle rule and law of sines:
(a)
sin sin 25
50 N 35 N
sin 0.60374
α
α
°
=
=
37.138
α
= °
37.1
α
= °
(b)
25 180
180 25 37.138
117.862
αβ
β
+ + °= °
= °− °− °
= °
35 N
sin117.862 sin 25
R=
°°
73.2 NR=
PROBLEM 2.10
A disabled automobile is pulled by means of two ropes as
shown. Knowing that the tension in rope AB is 3 kN,
determine by trigonometry the tension in rope AC and the
value of α so that the resultant force exerted at A is a 4.8-
kN force directed along the axis of the automobile.
SOLUTION
Using the law of cosines:
22 2
(3 kN) (4.8 kN) 2(3 kN)(4.8 kN)cos 30°
2.6643 kN
AC
AC
T
T
=+−
=
Using the law of sines:
sin sin30
3 kN 2.6643 kN
34.3
α
α
°
=
= °
2.66 kN
AC
=T
34.3°
