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959
9–81.
SOLUTION
The assembly is made from a steel hemisphere,
. Determine the mass center of the
assembly if the height of the cylinder is .h=200 mm
ral =2.70 Mg>m3
160 mm
h
z
z
G
_
Ans:
z=122
mm
960
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
80 mm
z
G
_
961
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
3.0 ft
G_
y
BG
SOLUTION
In horizontal position
With rear whells elevated
a
+©MB=0; 2576(9.40 cos 12.347°) -4256 cos 12.347°(5.0733)
962
*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
8 Mg>m3.
z=115 mm.
z
Ans:
9–85.
SOLUTION
Determ
i
ne t
h
e
di
stance to t
h
e centro
id
of t
h
e s
h
ape w
hi
c
h
consists of a cone with a hole of height bored into
its base.
h=50 mm
964
9–86.
SOLUTION
Center of mass: The assembly is broken into two composite segments, as shown in
Figs. aand b.
0.6 m
0.4 m
Ans:
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
12 ft
6ft
8ft
5ft
1500 lb
450 lb
280 lb
SOLUTION
Centroid: The floor loadings on the floor and its respective centroid are tabulated
Thus,
Ans:
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
©xW=
E
10p(1)2(20)(150)+7.5p(5)(22-12)(490)
F
1
x
5in. 5in.
10in.
G
2in.
1 in.
x
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
x
z
G
B
A225mm
150 mm
150mm
30 mm
Due to symmetry:
Ans.
y=-15 mm
©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
©m=©rV=c8.74a1
2(0.15)(0.225)(0.03)bd +c7.85a1
2(0.15)(0.225)(0.03)bd
Ans:
9–90.
SOLUTION
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:
969
9–91.
SOLUTION
Determine the outside surface area of the storage tank.
15 ft
4ft
30 ft
970
*9–92.
Determine the volume of the storage tank.
SOLUTION
Volume: Applying the theorem of Pappus and Guldinus, Eq. 9–8 with ,
u=2p
15 ft
4ft
30 ft
9–93.
Determine the surface area of the concrete sea wall,
excluding its bottom.
SOLUTION
Surface Area:Applying Theorem of Pappus and Guldinus,Eq.9–9 with
The surface area of two sides of the wall is
30 ft
8 ft
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
9–95.
SOLUTION
Aring is generated by rotating the quartercircular area about
the xaxis.Determine its volume.
a
2a
974
*9–96.
A ring is generated by rotating the quartercircular area
about the xaxis. Determine its surface area.
SOLUTION
a
2a
975
9–97.
Determine the volume of concrete needed to construct
the curb.
SOLUTION
Ans:
304 m
150 mm
150 mm
100 mm
150 mm
9–98.
Determine the surface area of the curb. Do not include the
area of the ends in the calculation.
SOLUTION
Ans:
304 m
150 mm
150 mm
100 mm
150 mm
9–99.
A ring is formed by rotating the area 360° about the
x
–
x
axes. Determine its surface area.
SOLUTION
50 mm
30 mm 30 mm
80 mm
100 mm
x x
*9–100.
A ring is formed by rotating the area 360° about the
x
–
x
axes. Determine its volume.
SOLUTION
50 mm
30 mm 30 mm
80 mm
100 mm
x x
Ans:
