978-0133915426 Chapter 2 Part 1

subject Type Homework Help
subject Pages 14
subject Words 2422
subject Authors Russell C. Hibbeler

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page-pf1
2–1.
Ans.
This yields
Thus,the direction of angle of measured counterclockwise from the
positive axis,is
Ans.
60° 95.19° 60° 155°
F
sin
700
sin 45°
497.01 95.19°
497.01 N 497 N
If and ,determine the magnitude of the
resultant force and its direction, measured counterclockwise
from the positive xaxis.
4
5
0 N60°
x
y
700 N
F
15
page-pf2
2–2.
Ans.
Applying the law of sines to Fig. b, and using this result, yields
Ans.u=45.2°
sin (90° +u)
700 =sin 105°
959.78
=959.78 N =960 N
If the magnitude of the resultant force is to be 500 N,
directed along the positive yaxis, determine the magnitude
of force Fand its direction .u
x
y
700 N
F
u
15
page-pf3
2–3.
SOLUTION
Ans.f=360° -45° +37.89° =353°
u=37.89°
393.2
sin 75° =250
sin u
Determine the magnitude of the resultant force
and its direction, measured counterclockwise from the positive
xaxis.
F
R
=F
1
+F
2
y
F
2
375 lb
x
F
1
250 lb
30
page-pf4
2 5
*2–4.
SOLUTION
Ans.
Ans.
366 N
sin 45°
500
sin 75°
448 N
The vertical force acts downward at on the two-membered
frame. Determine the magnitudes of the two components of
directed along the axes of and . Set .
500 NF
F
B
A
page-pf5
2–5.
Solve Prob. 2-4 with .F=350 lb
SOLUTION
Ans.
Ans.FAC =256 lb
FAC
sin 45° =350
sin 75°
FAB =314 lb
sin 60° =350
sin 75°
B
A
45
page-pf6
page-pf7
2 8
Ans:
(F
1)v=2.93 kN
(F
1)u=2.07 kN
sin 30°
sin 105°
page-pf8
page-pf9
page-pfa
page-pfb
2–11.
SOLUTION
Ans.
The angle can be determined using law of sines (Fig. b).
Thus, the direction of FRmeasured from the xaxis is
Ans.f=33.16° -30° =3.16°
f
u=33.16°
sin u=0.5470
sin u
6=sin 100°
10.80
u
=10.80 kN =10.8 kN
The plate is subjected to the two forces at Aand Bas
shown. If , determine the magnitude of the resultant
of these two forces and its direction measured clockwise
from the horizontal.
u=60°
A
F
A
8kN
F
B
6kN
40
u
page-pfc
*2–12.
Determ
i
ne t
h
e ang
l
e of for connect
i
ng mem
b
er Ato t
h
e
plate so that the resultant force of FAand FBis directed
horizontally to the right. Also,what is the magnitude of the
resultant force?
u
SOLUTION
Ans.
From the triangle,.Thus, using law of
cosines, the magnitude of FRis
Ans.=10.4 kN
F
R=282+62-2(8)(6) cos 94.93°
f=180° -(90° -54.93°) -50° =94.93°
u=54.93° =54.9°
sin (90° -u)=0.5745
A
F
A
8kN
F
B
6kN
u
page-pfd
2–13.
SOLUTION
Ans.
20
sin 40° =
F
b
sin 60°;F
b=26.9 lb
20
sin 40° =
F
a
sin 80°;F
T
h
eforce act
i
ng on t
h
egeartoot
hi
sReso
l
ve
this force into two components acting alongthe linesaa
and bb.
F
=20 lb.
80
60
a
a
b
b
F
page-pfe
2–14.
T
h
e component of force Fact
i
ng a
l
ong
li
ne aa
i
s requ
i
re
d
to
be 30 lb.Determine the magnitude of Fand its component
along line bb.
SOLUTION
Ans.
30
sin 80° =
Fb
sin 60°;Fb=26.4 lb
30
sin 80° =F
sin 40°;F=19.6 lb
80
60
a
a
b
b
F
page-pff
2–15.
SOLUTION
Ans.
Using this result and applying the law of sines to Fig. b, yields
Ans.u=31.8°
sin u
500 =sin 105°
916.91
=916.91 lb =917 lb
Force Facts on the frame such that its component acting
along member is 650 lb, directed from towards , and
the component acting along member is 500 lb, directed
from towards . Determine the magnitude of Fand its
direction . Set .f=60°u
CB
BC
ABAB
A
B
C
F
45
u
f
page-pf10
*2–16.
Force Facts on the frame such that its component acting
along member AB is 650 lb, directed from Btowards A.
Determine the required angle and the
component acting along member BC. Set and
.u=30°
F=850 lb
f (0° f45°)
Ans.
Using this result and applying the sine law to Fig. b, yields
Ans.
sin (45° +f)
850 =sin 30°
433.64 f=33.5°
=433.64 lb =434 lb
A
B
C
F
45
u
f
page-pf11
3 8
2–17.
SOLUTION
Ans.
Ans.
19.18
sin 1.47° =30.85
sin u;u=2.37°
FR=2(30.85)2+(50)2-2(30.85)(50) cos 1.47° =19.18 =19.2 N
30.85
sin 73.13° =30
sin (70° -u¿);u¿=1.47°
F¿=2(20)2+(30)2-2(20)(30) cos 73.13° =30.85 N
Determinethe magnitudeand directionofthe resultant
of the three forces by first finding the
resultant and then forming FR=F¿+F3.F¿=F1+F2
FR=F1+F2+F3
y
x
F
1
30 N
20
35
4
F
3
50 N
page-pf12
3 9
2–18.
Determ
i
ne t
h
emagn
i
tu
d
ean
ddi
rect
i
on of t
h
eresu
l
tant
of the three forces by first finding the
resultant and then forming FR=F¿+F1.F¿=F2+F3
FR=F1+F2+F3
SOLUTION
Ans.
Ans.u=23.53° -21.15° =2.37°
19.18
sin 13.34° =30
sin f;f=21.15°
F
R=2(47.07)2+(30)2-2(47.07)(30) cos 13.34° =19.18 =19.2 N
20
sin u¿=47.07
sin 70°;u¿=23.53°
F¿=2(20)2+(50)2-2(20)(50) cos 70° =47.07 N
y
x
F
1
30 N
20
3
5
4
F
3
50 N
page-pf13
2–19.
Determine the design angle for strut AB
so that the 400-lb horizontal force has a component of 500 lb
directed from Atowards C.What is the component of force
acting along member AB?Take .f=40°
u
(0°
u
90°)
SOLUTION
Ans.
Thus,
Using law of sines (Fig. b)
Ans.F
AB =621 lb
F
AB
sin 86.54° =400
sin 40°
c=180° -40° -53.46° =86.54°
u=53.46° =53.5°
sin u=0.8035
A
B
400 lb
u
f
page-pf14
*2–20.
The angle can be determined using law of sines (Fig. b).
Ans.f=38.3°
sin f=0.6193
sin f
400 =sin 30°
322.97
f
Determine the design angle between
struts AB and AC so that the 400-lb horizontal force has a
component of 600 lb which acts up to the left, in the same
direction as from Btowards A.Take .u=30°
f
(0°
f
90°)
A
B
400 lb
u
f

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