978-0133915426 Chapter 5 Part 2

subject Type Homework Help
subject Pages 14
subject Words 1447
subject Authors Russell C. Hibbeler

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© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
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Ans:
Pmin =271 N
SOLUTION
5–30.
Determine the magnitude and direction
u
of the minimum
force P needed to pull the 50-kg roller over the smooth step.
A
B
P
300 mm
50 mm
u
page-pf2
5–31.
The operation of the fuel pump for an automobile depends
on the reciprocating action of the rocker arm ABC, which
is pinned at Band is spring loaded at Aand D.When the
smooth cam Cis in the position shown, determine the
horizontal and vertical components of force at the pin and
the force along the spring DF for equilibrium.The vertical
force acting on the rocker arm at Ais , and at C
it is .FC=125 N
FA=60 N
30°
F
A
= 60 N
C
D
B
A
F
E
F
C
= 125 N
SOLUTION
page-pf3
*5–32.
Determine the magnitude of force at the pin and in the
cable needed to support the 500-lb load. Neglect the
weight of the boom .
SOLUTION
summing moments about point .
Ans.
Thus,Ans.F
A=Ax=2060.9 lb =2.06 kip
Ay=0
Ay+1820.7 sin 13° -500 cos 35° =0a+©F
y=0;
Ax=2060.9 lb
Ax-1820.7 cos 13° -500 sin 35° =0
+
Q©F
x=0;
F
BC =1820.7 lb =1.82 kip
F
BC sin 13°(8) -500 cos 35°(8) =0aMA=0;
A
AB
BC
A
35
22
8 ft
C
B
A
page-pf4
5–33.
SOLUTION
Ans.
Ans.
Ans.Bx=25.4 kN
:
+©Fx=0; Bx-25.4 =0
By=22.8 kN
+c©Fy=0; By-800 (9.81) -15 000 =0
Ax=25.4 kN
The dimensions of a jib crane,which is manufactured by the
Basick Co., are given in the figure. If the crane has a mass of
800 kg and a center of mass at G,and the maximum rated
force at its end is F15 kN,determine the reactions at its
bearings. The bearing at Ais a journal bearing and supports
only a horizontal force,whereas the bearing at Bisathrust
bearing that supports both horizontal and vertical components.
F
G
A
3m
2m
0.75 m
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421
5–34.
The dimensions of a jib crane,which is manufactured by the
Basick Co., are given in the figure. The crane has a mass of
800 kg and a center of mass at G.The bearing at Ais a journal
bearing and can support a horizontal force,whereas the
bearing at Bis a thrust bearing that supports both horizontal
andvertical components.Determine the maximum load Fthat
can be suspended from its end if the selected bearings at A
and Bcan sustain a maximum resultant load of 24 kN and
34 kN,respectively.
SOLUTION
Assume .
Solving,
Ans.
OKFB=(24)2+(21.9)2=32.5 kN 634 kN
F=14.0 kN
By=21.9 kN
Bx=24 kN
Ax=24 000 N
:
+©Fx=0; Bx-Ax=0
+c©Fy=0; By-800 (9.81) -F=0
F
G
A
3m
2m
B
0.75 m
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422
Ans:
NA=173 N
NC=416 N
NB=69.2 N
page-pf7
page-pf8
5–37.
The cantilevered jib crane is used to support the load of
780 lb.If ,determine the reactions at the supports.
Note that the supports are collars that allow the crane to
rotate freely about the vertical axis.The collar at Bsupports a
force in the vertical direction, whereas the one at Adoes not.
x=5 ft
8 ft
4 ft
780 lb
x
T
B
SOLUTION
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5–38.
T
h
e cant
il
evere
d
jib
crane
i
s use
d
to support t
h
e
l
oa
d
of
780 lb.If the trolley Tcan be placed anywhere between
determine the maximum magnitude of
reaction
at the supports Aand B.Note that the supports
are collars that allow the crane to rotate freely about the
vertical axis.The collar at Bsupports a force in the vertical
direction, whereas the one at Adoes not.
1.5 ft x7.5 ft,
SOLUTION
a
Ans.
Ans.
=
1657.5 lb
=
1.66 ki
p
FB=2(1462.5)2+(780)2
By=780 lb
+c©Fy=0; By-780 =0
Ax=1462.5 =1462 lb
:
+©Fx=0; Ax-1462.5 =0
Bx=1462.5 lb
MA=0; -780(7.5) +Bx(4) =0
8ft
4ft
780 lb
x
T
B
A
page-pfa
SOLUTION
5–39.
The bar of negligible weight is supported by two springs,
each having a stiffness k
=
100 N
>
m. If the springs are
originally unstretched, and the force is vertical as shown,
determine the angle
u
the bar makes with the horizontal,
when the 30-N force is applied to the bar. 2 m
1 m
B
C
30 N
k
k
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5–41.
The bulk head AD is subjected to both water and soil-
backfill pressures.Assuming AD is “pinned” to the
ground at A,determine the horizontal and vertical
reactions there and also the required tension in the
ground anchor BC necessary for equilibrium. The bulk
head has a mass of 800 kg.
6m
4m
310 kN/m118 kN/m
0.5 m
CF
A
B
D
SOLUTION
page-pfd
5–42.
SOLUTION
a
Ans.
Ans.
Ans.Ay=681 N
+c©Fy=0; Ay-800 -350 +3
5(781.6) =0
Ax=625 N
:
+©Fx=0; Ax-4
5(781.6) =0
FCB =781.6 =782 N
+4
5FCB (2.5 sin 30°) +3
5FCB(2.5 cos 30°) =0
MA=0;
-800(1.5 cos 30°) -350(2.5 cos 30°)
Theboom supports the two vertical loads.Neglect the size
of
the collars at Dand Band the thickness of the boom,
and
compute the horizontal and vertical components of
force
at the pin Aand the force in cable CB.Set
and F
2=350 N.
F
1=800 N
1.5 m
30
3
C
B
F
1
F
2
D
A
4
5
1m
page-pfe
5–43.
The boom is intended to support two vertical loads,and
If the cable CB can sustain a maximum load of 1500 N before
it fails, determine the critical loads if Also, what is
the magnitude of the maximum reaction at pin A?
F
1=2F
2.
F
2.
F
1
SOLUTION
3
C
4
5
page-pff
page-pf10
432
Ans:
P=660 N
NA=442 N
u=48.0°
b
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5–46.
Three uniform books, each having a weight Wand length a,
are stacked as shown. Determine the maximum distance d
that the top book can extend out from the bottom one so
the stack does not topple over.
SOLUTION
Ans.d=3a
4
ad
page-pf12
5–47.
Determine the reactions at the pin A and the tension in cord
BC. Set F = 40 kN. Neglect the thickness of the beam.
SOLUTION
3
C
A
F
26 kN
13 12
5
53
4
B
4 m
2 m
page-pf13
*5–48.
If rope BC will fail when the tension becomes 50 kN,
determine the greatest vertical load F that can be applied to
the beam at B. What is the magnitude of the reaction at A
for this loading? Neglect the thickness of thebeam.
SOLUTION
3
C
A
F
26 kN
13 12
5
53
4
B
4 m
2 m
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5–49.
The rigid metal strip of negligible weight is used as part of an
electromagnetic switch. If the stiffness of the springs at A
and Bis and the strip is originally horizontal
when the springs are unstretched, determine the smallest
force needed to close the contact gap at C.
k=5N>m,
SOLUTION
Set , then
From Eq. (2),
Ans.F
C=
F
A=
ky
A=
(5)(0.002)
=
10 mN
yB=4mm
yA=2mm
x=16.67
yA
=100 -x
10
x
©Fy=0; FB=2F
©MB=0; FA=FC=F
50 mm 50 mm
10 mm
A
B
C
k
k

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