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Magnitude of Force on a Submarine Hatch

  • Thread starter NeRdHeRd
  • Start date
1. The problem statement, all variables and given/known data

A submarine is operating at 100 m below the surface of the ocean. If the air inside the submarine is maintained at a pressure of 1 atmosphere, what is the magnitude of the force that acts on the rectangular hatch 2.0 x 1.0 m on the deck of the submarine?

2. Relevant equations

Force = Pressure* Area
Pressure = density*gravity*height
?

3. The attempt at a solution

If atmospheric pressure is 1.013 x 10^5 Pa and Area of hatch = width * height Then,

Area = 2*1 = 2 m^2
Force = (1.013 x 10^5 Pa)(2 m^2) = 202600

This doesn't seem right to me and I know I'm missing something. I just can't seem to put my finger on it. Please help.
 
I think you forgot about the 100m of water above the hatch...
 
I think you forgot about the 100m of water above the hatch...
Wouldn't I need to know the density of the water to determine the pressure thai it is exerting on the hatch?
 

HallsofIvy

Science Advisor
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Yes, you would- look it up! The pressure is the weight of water above the hatch divided by the area of the hatch. Since the weight of the water itself is the density of water times its height times the area, the pressure is just the density of water times the height of the column of water above the hatch.
 
Yes, you need the density.

I believe the commonly accepted value is 998 kg / m^3.
 
I guess I could assume that the density of the water is 1.000 * 10^3. If that is the case then:

the pressure of the water is (1.000 * 10^3)(9.8)(100) = 980000

So the force is 980000*2 = 1960000

I guess then I would subtract 202600 from 196000 giving me 1757400...

I think
 
There's a small detail you're missing. If you think about it, on top of the water "column", there is air, which means that the water itself is having the atmospheric pressure added to its own, and exerting the sum of both.

On the sub, you know that there is atmospheric pressure pushing on the hatch from the other side. Think about what this means.
 
On the sub, you know that there is atmospheric pressure pushing on the hatch from the other side. Think about what this means.
The cancel each other because they are pointing in opposite directions. Which means the force acting on the hatch is 1960000 or 1.96 * 10^6
 

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