What is the submarine’s maximum safe depth?

[GreyGus]

Homework Statement

I have tried this problem six times and it's still wrong. The problem is

A research submarine has a 22.8-cm-diameter window 7.9 cm thick. The manufacturer says the window can withstand forces up to 1280000 N. What is the submarine’s maximum safe depth? The pressure inside the submarine is maintained at 1.0 atm.

Homework Equations

p=po+rogh, p=f/a

The Attempt at a Solution

p=f/a, a=2*pi*r^2+ 2*pi*r*h because I think the window is a cylinder since it has thickness. Hence, I find p by pluging in a.

So,

p= 101325 pa + rogh and i solve for h but that is wrong. Can you help me please. Thank you.

 

[minger]

Simply divide the force by the area to get the pressure that the window can withstand. We'll deal with gauge pressures, so the 1 atm interior pressure will cancel with ambient pressure.

So,
$$p = \frac{F}{A} = \frac{1280000 N}{\frac{\pi D^2}{4}} = 31350.9 kPa$$
Since we're dealing with gauge pressurse, this will be equal to hydrostatic pressure, so:
$$p = \rho g h$$
$$31350.0*10^3 Pa = (1000 kg/m^3)(9.81 m/s^2)h;\,\, h = 3.195 km$$

At least that's what I get.

 

[GreyGus]

For some reason it's not working for me. Is (pi*d^2)/4 another equation for the surface area of a cylinder?

 

[dx]

The thickness of the window doesn't matter. The total force on the window is

ρghA - (101325)A

which must equal 1280000 N at the maximum safe depth. In your attempt, you had the wrong sign for 101325A (since it's pointing in the opposite direction to the water pressure force), and the wrong formula for the area of the window. A = πr² or πd²/4.