Question

A porthole on a vertical side of a submarine (submerged in seawater) is 1 square foot. Find the fluid force on the porthole, assuming that the center of the square is 12 feet below the surface.

Can someone help me figure out this problem? I'll give a medal for your help.

Answer 1

pressure = (density) (g) (depth)
force = (pressure)(area)

Answer 2

\[1002.7 *9.8\int\limits_{.5}^{.5}(12-y)(y)dy\] Is that correct?

Answer 3

That looks good, if the lower limit is -0.5.

I think p = rho g h and
F = p A would be close enough.

Answer 4

So I just have to change the lower limit to -.5 and it would be perfect?

Answer 5

Change the lower limit and get rid of the (y) term in the integral.
But check this against the simple approximation that the pressure is virtually uniform at the average depth of 12 so it is equal to 12 rho g and the force is A 12 rho g.

Answer 6

I used \[1,002.7*9.8\int\limits_{.5}^{-.5}(12-y)dy\]

And it didn't match to any of the answers

\[\frac{ 200 }{ 3 }\pi\]
\[\frac{ 275 }{ 6 }\pi\]
\[\frac{ 100 }{ 3 }\pi\]
\[\pi\]
\[\frac{ 275 }{ 3 }\pi\]

Answer 7

I see no reason for pi to enter at all.

Calculate p = rho g h
F = p A

and see which value it matches.