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Submarine fluid physics problem

  • Thread starter cathliccat
  • Start date
Ok, 'A circular window with a 30 cm diameter in a submarine can withstand a maximum force of 5.2x10^5N. What is the maximum depth (in m) in a lake to which the submarine can go without damaging the window round off to the nearest whole number?'

I tried P=F/A=pAhg/A, so I converted 30 cm to meters, solved for P then tried to find h. p=1000, A=pi*.15^2, h=?, g =9.8. So .070 = 9.8h, h=7.21X10^-6. That can't be right because then the submarine wouldn't move at all before the window busted.

Can someone tell me where I went wrong? Thanks!
 

chroot

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Re: Fluids

Originally posted by cathliccat
I tried P=F/A=pAhg/A
This is correct. It seems that somehow you later derived the equation A = g h, since you said "0.70 = 9.8 h." I'm not sure at all where got this! You have the right equation -- just solve it for h.

The pressure at a depth of h meters in a fluid is given by

P = [rho] g h

The force on the window is indeed equal to

F = P A or P = F / A

where A is the area of the window.

Equating these, we get

P = F / A
[rho] g h = F / A

h = F / (A [rho] g)

Does this make sense?

- Warren
 

HallsofIvy

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I'm trying to figure out where you got the "0.70" in ".070 = 9.8h"!

You are correct that P=F/A=ρAhg/A or, since it is F we are given rather than P, F= ρAhg.

You are given that F= 5.2x10^5N so that hg=9.8h= 5.2x10^5/(ρA)

The Area is π(.15)2= 0.0708 sq.m so F/A= 7356495. Taking &rho= 1000 gives F/(ρA)= 7356.495= 9.8h so h= 751 meters.
 
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