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9–101.
The water-supply tank has a hemispherical bottom and
cylindrical sides. Determine the weight of water in the tank
when it is filled to the top at C.Take gw=62.4 lb>ft3.
SOLUTION
6ft
8ft
C
6ft
9–102.
SOLUTION
Determine the number of gallons of paint needed to paint
the outside surface of the water-supply tank, which consists
of a hemispherical bottom, cylindrical sides, and conical top.
Each gallon of paint can cover 250 ft2.
6ft
8ft
C
6ft
9–103.
SOLUTION
Determine the surface area and the volume of the ring
formed by rotating the square about the vertical axis
.
b
a
a
45
*9–104.
Determine the surface area of the ring.The cross section is
circular as shown.
SOLUTION
A=ur
'L=2p(3) 2p(1)
8in.
4in.
983
9–105.
The heat exchanger radiates thermal energy at the rate of
2500 kJ h for each square meter of its surface area.
Determine how many joules (J) are radiated within a
5-hour period.
SOLUTION
Ans.Q=2500
A
103
B
aJ
h#m2b
A
16.416 m2
B
(5 h) =205 MJ
=16.419 m2
A=©urL=(2p)
B
2a0.75 +0.5
2b2(0.75)2+(0.25)2+(0.75)(1.5) +(0.5)(1)
R
0.75 m
1.5 m
0.75 m
0.5 m
0.75 m
1m
0.5 m
9–106.
Determine the interior surface area of the brake piston. It
consists of a full circular part. Its cross section is shown in
the figure.
40 mm
60 mm
80 mm
9–107.
SOLUTION
Ans.W
=
0.8 gV
=
0.8(50)(3142)
=
125 664 lb
=
126 kip
V=urA=2p(3.75) (133.3) =3142 ft3
x=
L
A
xdA
LA
dA
=
500
133.3 =3.75 ft
L
A
xdA =L20
0
y
0.4 dy =
y2
0.8 220
0
=500 ft3
LA
dA =L20
0
C
y
0.2 dy =
2
320.2 y3
2220
0
=133.3 ft2
dA =xdy
y
'
=y
x=
x
2
The suspension bunker is made from plates which are
curved to the natural shape which a completely flexible
membrane would take if subjected to a full load of coal.This
curve may be approximated by a parabola,
Determine the weight of coal which the bunker would
contain when completely filled. Coal has a specific weight of
and assume there is a 20% loss in volume due
to air voids. Solve the problem by integration to determine
the cross-sectional area of ABC; then use the second
theorem of Pappus–Guldinus to find the volume.
g=50 lb>ft3,
y=0.2x2.
y
x
10 ft
20 ft
A
y0.2x
2
C
B
*9–108.
Determine the height h to which liquid should be poured
into the cup so that it contacts three-fourths the surface
area on the inside of the cup. Neglect the cup’s thickness for
the calculation.
SOLUTION
160 mm
h
40 mm
9–109.
Determine the surface area of the roof of the structure if it
is formed by rotating the parabola about the yaxis.
SOLUTION
and its centroid is . Here,.Evaluating the
integrals, we nave
Applying Eq. 9–5, we have
Surface Area: Applying the theorem of Pappus and Guldinus, Eq. 9–7, with ,
,,we have
Ans.A=urL=2p(9.178) (23.663) =1365 m2
r=x=9.178L=23.663 m
u=2p
x=LL
x
'dL
LL
dL
=217.181
23.663 =9.178 m
L
L
x
''
dL =L16 m
0
x
¢
C
1+x2
64
≤
dx =217.181 m2
L=LdL =L16 m
0
¢
C
1+x2
64
≤
dx =23.663 m
dy
dx =-
x
8
x=x=
¢
C
1+ady
dx b2
≤
dx
16 m
y
x
y16 (x
2
/16)
988
9–110.
A steel wheel has a diameter of 840 mm and a cross section
as shown in the figure. Determine the total mass of the
wheel if r=5Mg>m3.
30 mm
30 mm
80 mm
100 mm
250 mm
420 mm
840 mm
60 mm
A
SOLUTION
evah ew dna
The mass of the wheel is
Ans.=138 kg
m=rV=51103238.775110-32p4
=8.775p110-32m3
V=u©rA=2p30.09510.0032+0.23510.00752+0.3910.00624
A3=10.1210.062=0.006 m2,A2=0.2510.032=0.0075 m2
A1=0.110.032=0.003 m2,r3=0.39 m,r2=0.235 m,r1=0.095 m,
Ans:
m=138
kg
989
9–111.
Half the cross section of the steel housing is shown in the
figure.There are six 10-mm-diameter bolt holes around its
rim. Determine its mass.The density of steel is 7.85 Mg m3.
The housing is a full circular part.
20 mm
40 mm
10 mm
10 mm
10 mm
10 mm
30 mm
30 mm
SOLUTION
Ans.=2.68 kg
m=rV=a7850 kg
m3b(340.9)
A
103
BA
10-9
B
m3
V=2p[ (40)(40)(10) +(55)(30)(10) +(75)(30)(10)] -6
C
p(5)2(10)
D
=340.9
A
103
B
mm3
Ans:
m=
2.68 kg
*9–112.
The water tank has a paraboloid-shaped roof. If one liter of
paint can cover 3 m2of the tank, determine the number of
liters required to coat the roof.
SOLUTION
where
Integrating,
The centroid of the line can be obtained by applying Eq.9–5 with
Surface Area:Applying the first theorem of Pappus and Guldinus and using the
results obtained above with
Thus,the amount of paint required is
Ans.# of liters =
459.39
3
=153 liters
A=2prL=2p(6.031)(12.124)=459.39 m2
r=x=6.031 m, we have
x=L
L
x
~ dL
LL
dL
=L12 m
0
xc1
48 2482+x2 dx d
12.124
=73.114
12.124
=6.031 m
xc=x.x
L=LL
dL =L12 m
0
1
48 2482+x2 dx =12.124 m
dL =B1+a-
1
48xb2
dx =B1+x2
482
dx =1
48 2482+x2 dx
dy
dx
=-
1
48x. Thus,
dx b2
x
y
2.5 m
y (144 x2)
1
––
96
991
9–113.
Determ
i
ne t
h
e vo
l
ume of mater
i
a
l
nee
d
e
d
to ma
k
e t
h
e
casting.
SOLUTION
Ans.V=1.40(103)in3
=1402.8 in3
=2pc2a1
4pb(6)2a4(6)
3pb+2(6)(4) (3) -2a1
2pb(2)2a6-4(2)
3pbd
V=© uAy
6in. 4in.
6in.
2in.
Side View Front View
Ans:
V
=
1.40
(
10
3
)
in
3
992
9–114.
SOLUTION
Ans.h
=
29.9 mm
10.77h+0.2154h2=513.5
2p
b
5(10) +a10 +
h
5bBa2h
5b2
+h2
r
=1
2(2p)(1127.03)
x=20h
50 =2h
5
=2p(1127.03) mm2
A=uzrL=2p{202(20)2+(50)2+5(10)}
Determine the height hto which liquid should be poured into
the cup so that it contacts half the surface area on the inside
of the cup.Neglect the cup’s thickness for the calculation.
50 m
m
h
30 mm
Ans:
h=29.9
mm
993
9–115.
The pressure loading on the plate varies uniformly along
each of its edges. Determine the magnitude of the resultant
force and the coordinates (,) of the point where the line of
action of the force intersects the plate. Hint: The equation
defining the boundary of the load has the form pax
by c,where the constants a,b,and chave to be determined.
y
x
At
At
At
Thus,
Ans.
Ans.
Ans.y=L
A
yp(x,y)dA
p(x,y)dA
=5416.67
1250 =4.33 ft
=5416.67 lb #ft
=-2
A
1
2(5)2
BA
1
2(10)2
B
-2
A
1
3(10)3
B
(5) +40(5)
A
1
2(10)2
B
LA
yp(x,y)dA =L5
0L10
0
(-2xy -2y2+40y)dy dx
x=L
A
xp(x,y)dA
LA
p(x,y)dA
=2916.67
1250 =2.33 ft
=2916.67 lb #ft
=-2
A
1
3(5)2
B
(10) -2
A
1
2(10)2
BA
1
2(5)2
B
+40
A
1
2(5)2
B
(10)
L
A
xp(x,y)dA =L5
0L10
0
(-2x2-2yx +40 x)dy dx
=1250 lb
=-2
A
1
2(5)2
B
(10) -2
A
1
2(10)2
B
5+40(5)(10)
FR=LA
p(x,y)dA =L5
0L10
0
(-2x-2y+40) dy dx
p=-2x-2y+40
20 =0+b(10) +40;
b=-2
x=0; y=10, p=20
30 =a(5) +0+40;
a=-2
x=5, y=0, p=30
40 =0+0+c;c=40
x=0, y=0; p=40
p=ax +by +c
p
y
5ft
40 lb/ft
20 lb/ft30 lb/ft
10 lb/ft
FR=
1250 lb
x=2.33
ft
y=4.33
ft
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*9–116.
The load over the plate varies linearly along the sides of the
plate such that
p= (12 -6x+4y) kPa.
Determine the
magnitude of the resultant force and the coordinates
(
x, y
)
of the point where the line of action of the force intersects
the plate.
SOLUTION
p
1.5 m
2 m
y
x
18 kPa
12 kPa
6 kPa
