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Chapter 14
431
P14.61
The torque is
ττ
==
zz
dr
d
F
From the figure
τρ
ρ
=−
=
z
ygHyw
d
y
g
w
H
H
bg
0
3
1
6
The total force is given as
1
2
2
gwH
If this were applied at a height
y
eff
such that the torque remains
unchanged, we have
1
6
1
2
32
ρρ
gwH
y
gwH
eff
=
L
N
M
O
Q
P
and
yH
eff
=
1
3
.
FIG. P14.61
P14.62
(a)
The pressure on the surface of the two hemispheres is constant
at all points, and the force on each element of surface area is
directed along the radius of the hemispheres. The applied force
along the axis must balance the force on the “effective” area,
which is the projection of the actual surface onto a plane
perpendicular to the
x
axis,
AR
=
π
2
Therefore,
FP
P
R
0
2
FIG. P14.62
(b)
For the values given
P
P
=
=
×
00
2
0
4
0 100
0 300
0 254
2 58 10
..
af
m
N
P14.63
Looking first at the top scale and the iron block, we have:
TBF
g
1
+=
, iron
where
T
1
is the tension in the spring scale,
B
is the buoyant force, and
F
g
, iron
is the weight of the iron
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This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .
 Fall '11
 Staff
 Physics, Force

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