Homework solution. Chapter 14.
P14.6
(a)
(
)
(
)
(
)
5
3
2
0
1.013
10
Pa
1024 kg
m
9.80 m
s
1000 m
P
P
gh
ρ
=
+
=
×
+
7
1.01
10
Pa
P
=
×
(b)
The gauge pressure is the difference in pressure between the water outside and the air
inside the submarine, which we suppose is at 1.00 atmosphere.
7
gauge
0
1.00
10
Pa
P
P
P
gh
ρ
=
−
=
=
×
The resultant inward force on the porthole is then
(
)
2
7
5
gauge
1.00
10
Pa
0.150 m
7.09
10
N
F
P
A
π
⎡
⎤
=
=
×
=
×
⎣
⎦
.
2.00 m
1.00 m
2.00 m
P14.14
The air outside and water inside both exert atmospheric pressure,
so only the excess water pressure
gh
ρ
counts for the net force.
Take a strip of hatch between depth
h
and
h
dh
+
. It feels force
(
)
2.00 m
dF
PdA
gh
dh
ρ
=
=
.
(a)
The total force is
(
)
2.00 m
1.00 m
2.00 m
h
F
dF
gh
d
ρ
=
=
=
∫
∫
h
FIG. P14.14
(
)
(
)
(
)
(
)
(
)
(
)
(
)
2.00 m
2
2
2
3
2
1.00 m
2.00 m
2.00 m
1000 kg
m
9.80 m
s
2.00 m
1.00 m
2
2
29.4 kN to the right
h
F
g
F
ρ
⎡
⎤
=
=
−
⎣
⎦
=
(b)
The lever arm of
dF
is the distance
(
)
1.00 m
h
−
from hinge to strip:
(
)
(
)
(
)
(
)
(
)
(
)
(
)
2.00 m
1.00 m
2.00 m
3
2
1.00 m
3
3
3
2
2.00 m
1.00 m
2.00 m
1.00 m
3
2
7.00 m
3.00 m
1000 kg
m
9.80 m
s
2.00 m
3
2
16.3 kN
m counterclockwise
h
d
gh
h
dh
h
h
g
τ
τ
ρ
τ
ρ
τ
τ
=
=
=
−
⎡
⎤
=
−
⎢
⎥
⎣
⎦
⎛
⎞
=
−
⎜
⎟
⎝
⎠
=
⋅
∫
∫
411

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