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.