Find the maximum safe depth of a submarine

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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.
 

Answers and Replies

  • #2
tnich
Homework Helper
1,048
336

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?
 
  • #3
jbriggs444
Science Advisor
Homework Helper
9,597
4,251
p = p0 + ρseawater g d
p is the pressure difference across the window
p0 is the pressure inside the window
ρseawater g d is the pressure outside the window

Does the equation match the physical situation?
 

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