18.
(a) The force on face A of area A
A
is
F
A
=
p
A
A
A
= ρ
w
gh
A
A
A
= 2ρ
w
gd
3
=
2
1.0
× 10
3
kg/m
3
9.8 m/s
2
(5.0 m)
3
= 2.5
× 10
6
N .
(b) The force on face B is
F
B
=
p
avgB
A
B
= ρ
w
g
5d
2
d
2
=
5
2
ρ
w
gd
3
=
5
2
1.0
× 10
3
kg/m
3
9.8 m/s
2
(5.0 m)
3
= 3.1
× 10
6
N .
Note that these figures are due to the water pressure only. If you add the contribution from the
atmospheric pressure, then you need to add F
= (1.0
× 10
5
Pa)(5.0 m)
2
= 2.5
× 10
6
N to each of
the figures above. The results would then be 5.0
× 10
6
N and 5.6
× 10
6
N, respectively.