# Find the maximum safe depth of a submarine

## Homework Statement

An ocean-going research submarine has a 40.0 cm -diameter window 9.00 cm thick. The manufacturer says the window can withstand forces up to 1.20×106 N . What is the submarine's maximum safe depth? The pressure inside the submarine is maintained at 1.0 atm.

ρseawater = 1030 kg/m3
r = .2 m
g = 9.8 m/s2

## Homework Equations

p = F/A
p = p0 + ρseawatergd
A = πr2

## The Attempt at a Solution

p0 = 1 atm = 101,300 Pa

p = (1.2*106 N)/(π*.22 m) = 9549296.586 Pa

9549296.586 Pa = 101300 Pa + (1030 kg/m3)(9.8 m/s2)d

d = (9549296.586 Pa - 101300 Pa)/((1030 kg/m3)(9.8 m/s2))

d = 936m

But the actual answer is 947m so I'm pretty sure I did something wrong or I'm missing something at some point because that is a significant difference. I'm pretty sure there's some point where I'm supposed to be using the volume of the window when I'm not but I'm not quite sure where I'm supposed to use it.

tnich
Homework Helper

## Homework Statement

An ocean-going research submarine has a 40.0 cm -diameter window 9.00 cm thick. The manufacturer says the window can withstand forces up to 1.20×106 N . What is the submarine's maximum safe depth? The pressure inside the submarine is maintained at 1.0 atm.

ρseawater = 1030 kg/m3
r = .2 m
g = 9.8 m/s2

## Homework Equations

p = F/A
p = p0 + ρseawatergd
A = πr2

## The Attempt at a Solution

p0 = 1 atm = 101,300 Pa

p = (1.2*106 N)/(π*.22 m) = 9549296.586 Pa

9549296.586 Pa = 101300 Pa + (1030 kg/m3)(9.8 m/s2)d

d = (9549296.586 Pa - 101300 Pa)/((1030 kg/m3)(9.8 m/s2))

d = 936m

But the actual answer is 947m so I'm pretty sure I did something wrong or I'm missing something at some point because that is a significant difference. I'm pretty sure there's some point where I'm supposed to be using the volume of the window when I'm not but I'm not quite sure where I'm supposed to use it.
Wouldn't the force on the window be proportional to the pressure difference between the inside and the outside?

jbriggs444