What is the net force on a miniature sub hatch at 8.5m depth?

[kiwikahuna]

Homework Statement

A miniature sub is lowered to a depth of 8.5 m in sea water of density 1.03 grams/cc. The interior of the research craft is maintained at 1.0 atm. Find the net force that must be withstood by a 1.5 m diameter hatch which leads to the inside.

Homework Equations

Pressure = density * gravity * height
Pressure = P (atmospheric) + density * gravity * height
Mass = density * area* height

The Attempt at a Solution

Pressure = 101300 + 1030 kg/m^3 (9.81m/s^2) * 8.5m
Once I found the pressure, I multipied it by the area (pi r^2)
The answer I got was wrong and I'm wondering what I did wrong. Please help.

 

[CaptainZappo]

You must consider the pressure on BOTH sides of the hatch.

 

[m2003]

There are a few things to consider in this problem. First, the density of sea water is given in grams/cc, so it needs to be converted to kg/m^3. This can be done by dividing by 1000, since 1 cc = 1/1000 m^3. So the density should be 1030 kg/m^3, not 1.03 kg/m^3 as given in the problem.

Next, the pressure at a depth of 8.5 m would be the sum of the atmospheric pressure (1 atm = 101300 Pa) and the pressure due to the depth of the water. So the correct equation to use would be:

P = 101300 Pa + 1030 kg/m^3 * 9.81 m/s^2 * 8.5 m

Once you have the pressure, you can calculate the force using the formula F = PA, where P is the pressure and A is the area of the hatch (pi*r^2). This should give you the correct answer for the net force that must be withstood by the hatch. Keep in mind that the pressure inside the sub will also contribute to the net force, but it is not specified in the problem and therefore cannot be calculated.