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PROBLEM 2.22
Cable AC exerts on beam AB a force P directed along line AC. Knowing
that P must have a 350-lb vertical component, determine (a) the magnitude
of the force P, (b) its horizontal component.
SOLUTION
(a)
cos 55
y
P
P=°
350 lb
cos 55
610.21 lb
=°
=
610 lbP=
(b)
sin 55
x
PP= °
(610.21 lb)sin 55
499.85 lb
= °
=
500 lb
x
P=
PROBLEM 2.23
The hydraulic cylinder BD exerts on member ABC a force P directed
along line BD. Knowing that P must have a 750-N component
perpendicular to member ABC, determine (a) the magnitude of the force
P, (b) its component parallel to ABC.
SOLUTION
(a)
750 N sin 20P= °
2192.9 NP=
2190 NP=
(b)
cos20
ABC
PP= °
(2192.9 N)cos20= °
2060 N
ABC
P=
PROBLEM 2.24
Determine the resultant of the three forces of Problem 2.16.
PROBLEM 2.16 Determine the x and y components of each of
the forces shown.
SOLUTION
Components of the forces were determined in Problem 2.16:
Force x Comp. (N) y Comp. (N)
800 lb +640 +480
424 lb –224 –360
408 lb +192 –360
608
x
R= +
240
y
R= −
(608 lb) ( 240 lb)
tan
240
608
21.541
240 N
sin(21.541°)
653.65 N
xy
y
x
RR
R
R
R
a
a
= +
= +−
=
=
= °
=
=
Rij
ij
654 N=R
21.5°
consent of McGraw-Hill Education.
PROBLEM 2.25
Determine the resultant of the three forces of Problem 2.17.
PROBLEM 2.17 Determine the x and y components of each of the
forces shown.
SOLUTION
29 lb
+21.0 +20.0
50 lb
–14.00 +48.0
51 lb
+24.0 –45.0
31.0
x
R= +
23.0
y
R= +
(31.0 lb) (23.0 lb)
tan
23.0
31.0
36.573
23.0 lb
sin(36.573 )
xy
y
x
RR
R
R
R
a
a
= +
= +
=
=
= °
=°
ij
ij
R
38.601 lb=
38.6 lb=R
36.6°
consent of McGraw-Hill Education.
PROBLEM 2.26
Determine the resultant of the three forces of Problem 2.18.
PROBLEM 2.18 Determine the x and y components of each of
the forces shown.
SOLUTION
Force x Comp. (lb) y Comp. (lb)
40 lb +20.00 –34.64
50 lb –38.30 –32.14
60 lb +54.38 +25.36
36.08
x
R= +
41.42
y
R= −
( 36.08 lb) ( 41.42 lb)
tan
41.42 lb
tan 36.08 lb
tan 1.14800
48.942
41.42 lb
sin 48.942
xy
y
x
RR
R
R
R
a
a
a
a
= +
=+ +−
=
=
=
= °
=°
Rij
ij
54.9 lb=R
48.9°
consent of McGraw-Hill Education.
PROBLEM 2.27
Determine the resultant of the three forces of Problem 2.19.
PROBLEM 2.19 Determine the x and y components of each of the
forces shown.
SOLUTION
Components of the forces were determined in Problem 2.19:
Force x Comp. (N) y Comp. (N)
80 N +61.3 +51.4
120 N +41.0 +112.8
150 N –122.9 +86.0
20.6
x
R= −
250.2
y
R= +
( 20.6 N) (250.2 N)
tan
250.2 N
tan 20.6 N
tan 12.1456
85.293
250.2 N
sin85.293
xy
y
x
RR
R
R
R
a
a
a
a
= +
=−+
=
=
=
= °
=°
Rij
ij
251 N=R
85.3°
consent of McGraw-Hill Education.
PROBLEM 2.28
of
α
if the resultant of the three forces shown is to be vertical,
(b) the corresponding magnitude of the resultant.
SOLUTION
(100 N)cos (150 N)cos( 30 ) (200 N)cos
(100 N)cos (150 N)cos( 30 )
xx
x
RF
R
αα α
αα
= Σ
= + + °−
=− + +°
(1)
(100 N)sin (150 N)sin( 30 ) (200 N)sin
(300 N)sin (150 N)sin( 30 )
yy
y
RF
R
αα α
αα
= Σ
=− − + °−
=− − +°
(2)
x
x
100cos 150cos( 30 ) 0
100cos 150(cos cos 30 sin sin 30 ) 0
29.904cos 75sin
αα
αα α
αα
− + + °=
− + °− ° =
=
29.904
tan 75
0.39872
21.738
α
α
=
=
= °
21.7
α
= °
(b) Substituting for
α
in Eq. (2):
300sin 21.738 150sin 51.738
228.89 N
y
R=− °− °
= −
| | 228.89 N
y
RR= =
229 NR=
consent of McGraw-Hill Education.
PROBLEM 2.29
A hoist trolley is subjected to the three forces shown. Knowing that α = 40°,
determine (a) the required magnitude of the force P if the resultant of
the three forces is to be vertical, (b) the corresponding magnitude of
the resultant.
SOLUTION
x
R=
(200 lb)sin40 (400 lb)cos40
x
FPΣ = + °− °
177.860 lb
x
RP= −
(1)
y
R=
(200 lb)cos40 (400 lb)sin40
y
FΣ = °+ °
410.32 lb
y
R=
(2)
PROBLEM 2.30
A hoist trolley is subjected to the three forces shown. Knowing
that P = 250 lb, determine (a) the required value of
α
if the
resultant of the three forces is to be vertical, (b) the corresponding
magnitude of the resultant.
SOLUTION
x
R=
250 lb (200 lb)sin (400 lb)cos
x
F
αα
Σ= + −
250 lb (200 lb)sin (400 lb)cos
x
R
αα
=+−
(1)
y
R=
(200 lb)cos (400 lb)sin
y
F
αα
Σ= +
x
PROBLEM 2.22
Cable AC exerts on beam AB a force P directed along line AC. Knowing
that P must have a 350-lb vertical component, determine (a) the magnitude
of the force P, (b) its horizontal component.
SOLUTION
(a)
cos 55
y
P
P=°
350 lb
cos 55
610.21 lb
=°
=
610 lbP=
(b)
sin 55
x
PP= °
(610.21 lb)sin 55
499.85 lb
= °
=
500 lb
x
P=
PROBLEM 2.23
The hydraulic cylinder BD exerts on member ABC a force P directed
along line BD. Knowing that P must have a 750-N component
perpendicular to member ABC, determine (a) the magnitude of the force
P, (b) its component parallel to ABC.
SOLUTION
(a)
750 N sin 20P= °
2192.9 NP=
2190 NP=
(b)
cos20
ABC
PP= °
(2192.9 N)cos20= °
2060 N
ABC
P=
PROBLEM 2.24
Determine the resultant of the three forces of Problem 2.16.
PROBLEM 2.16 Determine the x and y components of each of
the forces shown.
SOLUTION
Components of the forces were determined in Problem 2.16:
Force x Comp. (N) y Comp. (N)
800 lb +640 +480
424 lb –224 –360
408 lb +192 –360
608
x
R= +
240
y
R= −
(608 lb) ( 240 lb)
tan
240
608
21.541
240 N
sin(21.541°)
653.65 N
xy
y
x
RR
R
R
R
a
a
= +
= +−
=
=
= °
=
=
Rij
ij
654 N=R
21.5°
consent of McGraw-Hill Education.
PROBLEM 2.25
Determine the resultant of the three forces of Problem 2.17.
PROBLEM 2.17 Determine the x and y components of each of the
forces shown.
SOLUTION
29 lb
+21.0 +20.0
50 lb
–14.00 +48.0
51 lb
+24.0 –45.0
31.0
x
R= +
23.0
y
R= +
(31.0 lb) (23.0 lb)
tan
23.0
31.0
36.573
23.0 lb
sin(36.573 )
xy
y
x
RR
R
R
R
a
a
= +
= +
=
=
= °
=°
ij
ij
R
38.601 lb=
38.6 lb=R
36.6°
consent of McGraw-Hill Education.
PROBLEM 2.26
Determine the resultant of the three forces of Problem 2.18.
PROBLEM 2.18 Determine the x and y components of each of
the forces shown.
SOLUTION
Force x Comp. (lb) y Comp. (lb)
40 lb +20.00 –34.64
50 lb –38.30 –32.14
60 lb +54.38 +25.36
36.08
x
R= +
41.42
y
R= −
( 36.08 lb) ( 41.42 lb)
tan
41.42 lb
tan 36.08 lb
tan 1.14800
48.942
41.42 lb
sin 48.942
xy
y
x
RR
R
R
R
a
a
a
a
= +
=+ +−
=
=
=
= °
=°
Rij
ij
54.9 lb=R
48.9°
consent of McGraw-Hill Education.
PROBLEM 2.27
Determine the resultant of the three forces of Problem 2.19.
PROBLEM 2.19 Determine the x and y components of each of the
forces shown.
SOLUTION
Components of the forces were determined in Problem 2.19:
Force x Comp. (N) y Comp. (N)
80 N +61.3 +51.4
120 N +41.0 +112.8
150 N –122.9 +86.0
20.6
x
R= −
250.2
y
R= +
( 20.6 N) (250.2 N)
tan
250.2 N
tan 20.6 N
tan 12.1456
85.293
250.2 N
sin85.293
xy
y
x
RR
R
R
R
a
a
a
a
= +
=−+
=
=
=
= °
=°
Rij
ij
251 N=R
85.3°
consent of McGraw-Hill Education.
PROBLEM 2.28
of
α
if the resultant of the three forces shown is to be vertical,
(b) the corresponding magnitude of the resultant.
SOLUTION
(100 N)cos (150 N)cos( 30 ) (200 N)cos
(100 N)cos (150 N)cos( 30 )
xx
x
RF
R
αα α
αα
= Σ
= + + °−
=− + +°
(1)
(100 N)sin (150 N)sin( 30 ) (200 N)sin
(300 N)sin (150 N)sin( 30 )
yy
y
RF
R
αα α
αα
= Σ
=− − + °−
=− − +°
(2)
x
x
100cos 150cos( 30 ) 0
100cos 150(cos cos 30 sin sin 30 ) 0
29.904cos 75sin
αα
αα α
αα
− + + °=
− + °− ° =
=
29.904
tan 75
0.39872
21.738
α
α
=
=
= °
21.7
α
= °
(b) Substituting for
α
in Eq. (2):
300sin 21.738 150sin 51.738
228.89 N
y
R=− °− °
= −
| | 228.89 N
y
RR= =
229 NR=
consent of McGraw-Hill Education.
PROBLEM 2.29
A hoist trolley is subjected to the three forces shown. Knowing that α = 40°,
determine (a) the required magnitude of the force P if the resultant of
the three forces is to be vertical, (b) the corresponding magnitude of
the resultant.
SOLUTION
x
R=
(200 lb)sin40 (400 lb)cos40
x
FPΣ = + °− °
177.860 lb
x
RP= −
(1)
y
R=
(200 lb)cos40 (400 lb)sin40
y
FΣ = °+ °
410.32 lb
y
R=
(2)
PROBLEM 2.30
A hoist trolley is subjected to the three forces shown. Knowing
that P = 250 lb, determine (a) the required value of
α
if the
resultant of the three forces is to be vertical, (b) the corresponding
magnitude of the resultant.
SOLUTION
x
R=
250 lb (200 lb)sin (400 lb)cos
x
F
αα
Σ= + −
250 lb (200 lb)sin (400 lb)cos
x
R
αα
=+−
(1)
y
R=
(200 lb)cos (400 lb)sin
y
F
αα
Σ= +
x
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