978-0133915426 Chapter 9 Part 5

subject Type Homework Help
subject Pages 14
subject Words 2350
subject Authors Russell C. Hibbeler

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page-pf1
9–81.
SOLUTION
Ans.z=©zm
©m=9.51425(10-3)
77.7706(10-3)
=0.122 m =122 mm
=77.7706(10-3)Mg
©m=
A
2
3
B
p(0.160)3(7.80) +p(0.2)(0.08)2(2.70)
The assembly is made from a steel hemisphere,
, and an aluminum cylinder,
. Determine the mass center of the
assembly if the height of the cylinder is .h=200 mm
ral =2.70 Mg>m3
rst =7.80 Mg>m3
160 mm
h
z
80 mm
z
G
_
page-pf2
9–82.
The assembly is made from a steel hemisphere,
, and an aluminum cylinder,
. Determine the height hof the cylinder
so that the mass center of the assembly is located at
.z=160 mm
ral =2.70 Mg>m3
rst =7.80 Mg>m3
SOLUTION
Solving
Ans.h=0.385 m =385 mm
z=©z m
©m=6.691(10-3)+8.686(10-3)h+27.143(10-3)h2
66.91(10-3)+54.29(10-3)h
=0.160
=66.91(10-3)+54.29(10-3)h
©m=
A
2
3
B
p(0.160)3(7.80) +p(h)(0.08)2(2.70)
=6.691(10-3)+8.686(10-3)h+27.143(10-3)h2
©zm=
C
0.160 -3
8(0.160)
DA
2
3
B
p(0.160)3(7.80) +
A
0.160 +h
2
B
p(h)(0.08)2(2.70)
160 mm
h
z
y
x
80 mm
z
G
_
page-pf3
9–83.
The car rests on four scales and in this position the scale
readings of both the front and rear tires are shown by
and .When the rear wheels are elevated to a height of 3 ft
above the front scales, the new readings of the front wheels
are also recorded. Use this data to compute the location
and to the center of gravity Gof the car.The tires each
have a diameter of 1.98 ft.
y
x
F
B
F
A
F
A
1129 lb 1168 lb 2297 lb
F
B
975 lb 984 lb 1959 lb
A
_
x
B
9.40 ft
G_
y
SOLUTION
page-pf4
*9–84.
Determine the distance hto which a 100-mm diameter hole
must be bored into the base of the cone so that the center of
mass of the resulting shape is located at The
material has a density of
0.4313 -0.2875 h=0.4688 -1.25 h2
1
3 p10.152210.52
A
0.5
4
B
-p10.05221h2
A
h
2
B
1
3 p10.152210.52-p10.05221h2
=0.115
8 Mg>m3.
z=115 mm.
z
C
h
500 mm
_
z
page-pf5
963
9–85.
SOLUTION
Ans.z=©z
'V
©V
=1.463 (10-3)
0.01139 =0.12845 m=128 mm
=0.01139 m3
©V=1
3p(0.15)2(0.5)-p(0.05)2(0.05)
=1.463(10-3)m
4
©z
'V=1
3p(0.15)2
A
0.5
B
a0.5
4b-p(0.05)2(0.05) a0.05
2b
i
h
di
h
id
h
h
hi
h
consists of a cone with a hole of height bored into
its base.
h=50 mm
Ans:
z=128
mm
page-pf6
9–86.
Locate the center of mass of the assembly.The cylinder
and the cone are made from materials having densities of
and , respectively.9Mg>m3
5Mg>m3
z
SOLUTION
z
0.6 m
0.4 m
page-pf7
965
9–87.
Major floor loadings in a shop are caused by the weights of
the objects shown. Each force acts through its respective
center of gravity G. Locate the center of gravity (, ) of all
these components.
y
x
z
y
G
2
G
4
G
3
G
1
600 lb
9ft
7ft
6ft
1500 lb
450 lb
280 lb
SOLUTION
1 450 672700 3150
2 1500 18 16 27000 24000
3 600 26 3 15600 1800
4 280 30 8 8400 2240
2830 53700 31190
Thus,
Ans.
Ans.y=
©yW
©W
=31190
2830 =11.02 ft =11.0 ft
x=©xW
©W
=53700
2830 =18.98 ft =19.0 ft
©
Ans:
x=19.0
ft
y=11.0
ft
page-pf8
966
*9–88.
The assembly consists of a 20-in. wooden dowel rod and a
tight-fitting steel collar. Determine the distance to its
center of gravity if the specific weights of the materials are
and The radii of the dowel
and collar are shown.
gst =490lb>ft3.gw=150 lb>ft3
x
SOLUTION
Ans.x=©xW
©
W
=154.8
18.82 =8.22 in.
=18.82 lb
©W=
E
p(1)2(20)(150)+p(5)(22-12)(490)
F
1
(12)3
=154.8lb #in.
©xW=
E
10p(1)2(20)(150)+7.5p(5)(22-12)(490)
F
1
(12)3
x
5in. 5in.
10in.
G
2in.
1 in.
x
Ans:
x=8.22 in.
page-pf9
967
9–89.
SOLUTION
The composite plate is made from both steel
(
A
)
and brass
(B) segments. Determine the mass and location of
its mass center G.Take and
rbr =8.74 Mg>m3.
rst =7.85 Mg>m3
1x,y,z2
y
z
G
B
A225mm
150 mm
150mm
30 mm
Ans.
Ans.
Due to symmetry:
Ans.
Ans.z=©zm
©
m=1.8221(10-3)
16.347(10
-3
)
=0.1115 m =111 mm
y=-15 mm
x=©xm
©m=2.4971(10-3)
16.347(10-3)
=0.153 m =153 mm
=1.8221
A
10-3
B
kg #m
©zm=a1
3(0.225)b(4.4246)
A
10-3
B
+a2
3(0.225)b(3.9471)
A
10-3
B
+a0.225
2b(7.9481)
A
10-3
B
+1
2(0.150)(7.9481)(10-3)=2.4971(10-3)kg#m
©xm=a0.150 +2
3(0.150)b(4.4246)
A
10-3
B
+a0.150 +1
3(0.150)b(3.9741)
A
10-3
B
=16.347
A
10-3
B
=16.4 kg
=
C
4.4246
A
10-3
BD
+
C
3.9741
A
10-3
BD
+
C
7.9481
A
10-3
BD
+[7.85(0.15)(0.225)(0.03)]
©mrV=c8.74a1
2(0.15)(0.225)(0.03)bd +c7.85a1
2(0.15)(0.225)(0.03)bd
Ans:
Σm=16.4 kg
x=153
mm
y=-15
mm
z=111
mm
page-pfa
968
9–90.
SOLUTION
VurA=2pc4(10)
3pa1
4bp(10)2+5(80)(10) d
Determ
i
ne t
h
e vo
l
ume of t
h
e s
il
o w
hi
c
h
cons
i
sts of a
cylinder and hemispherical cap.Neglect the thickness of
the plates.
10 ft 10 ft
10 ft
80 ft
Ans:
V
=
27.2
(
10
3
)
ft
3
page-pfb
969
9–91.
SOLUTION
,, and , we have
Ans.A=u©r
~L=2p
C
7.5
A
2241
B
+15(30)
D
=3.56
A
103
B
ft2
r2=15 ftr1=7.5 ftL2=30 ftL1=2152+42=2241 ft
Determine the outside surface area of the storage tank.
15 ft
4ft
30 ft
Ans:
A
=
3.56
(
10
3
)
ft
2
page-pfc
970
*9–92.
Determine the volume of the storage tank.
SOLUTION
,, and , we have
Ans.V=u©rA=2p[5(30.0) +7.5(450)] =22.1
A
103
B
ft3
A2=30(15) =450 ft2
A1=1
2(15)(4) =30.0 ft2
r2=7.5 ftr1=5ft
15 ft
4ft
30 ft
Ans:
V
=
22.1
(
10
3
)
ft
3
page-pfd
971
9–93.
Determine the surface area of the concrete sea wall,
excluding its bottom.
SOLUTION
,and as indicated in Fig. a,
The surface area of two sides of the wall is
Thus the total surface area is
Ans.=4856 ft2
=4856.25 ft2
A=A1+A2=4166.25 +690
A2=2c1
2(8+15)(30)d=690 ft2
=4166.25 ft2
A1=u©NL=5
18 p [75(30)+71(8)+63.5(2949)]
N
3=63.5 ftN
2=71 ftN
1=75 ft
30 ft
8 ft
Ans:
A=4856
ft2
page-pfe
972
9–94.
SOLUTION
A circular sea wall is made of concrete. Determine the total
weight of the wall if the concrete has a specific weight of
gc=150 lb>ft3.
30 ft
8ft
60 ft
page-pff
973
9–95.
SOLUTION
and , we have
Ans.V=urA=2p6p+4
3p
ap
4a2=
p(6p+4)
6a3
A=
p
4a2
r=2a+4a
3p
=6p+4
3p
a
Aring is generated by rotating the quartercircular area about
the xaxis.Determine its volume.
a
x
2a
Ans:
V=
p
(6
p
+4)
6
a3
page-pf10
*9–96.
A ring is generated by rotating the quartercircular area
about the xaxis. Determine its surface area.
SOLUTION
,,,and ,we have
r3=5
2ar2=2(p+1)
p
ar1=2aL2=pa
2
L1=L3=a
a
2a
page-pf11
9–97.
Determine the volume of concrete needed to construct
the curb.
SOLUTION
p
p
304 m
150 mm
150 mm
100 mm
150 mm
page-pf12
9–98.
Determine the surface area of the curb. Do not include the
area of the ends in the calculation.
SOLUTION
p
2
304 m
150 mm
150 mm
100 mm
150 mm
page-pf13
977
9–99.
A ring is formed by rotating the area 360° about the
x
x
axes. Determine its surface area.
SOLUTION
2
2
50 mm
30 mm 30 mm
80 mm
100 mm
x x
Ans:
A
=
276(10
3
) mm
2
page-pf14
*9–100.
A ring is formed by rotating the area 360° about the
x
x
axes. Determine its volume.
SOLUTION
1
50 mm
30 mm 30 mm
80 mm
100 mm
x x

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