Submarine depth and pressure problem - mastering physics

  • Thread starter Shayna
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  • #1
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Homework Statement


A research submarine has a 40.0 cm-diameter window 8.50 cm thick. The manufacturer says the window can withstand forces up to 1.20×10^6 N. What is the submarine's maximum safe depth?

The Attempt at a Solution


The area which force will be exerted on is

A= Pai * r^2 = 0.5027 m^2

Maximum pressure allowed

p' = 1.20×10^6 N / 0.5027 m^2 = 2387109.608 Pa = 23.559 atm

Maximum pressure allowed from outside

p = 23.559 - 1 = 22.559 atm = 2285784.608 Pa

h =p/rou(density) * g = 2285784.608 / (1030 * 9.8) = 226.449 m

Which is not right.

I have two questions,
1, what is the 0.085m thickness given for
2, Was it correct that I have used density of sea water instead of water since it is a submarine?

Thanks
 

Answers and Replies

  • #2
LowlyPion
Homework Helper
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The area which force will be exerted on is

A= Pai * r^2 = 0.5027 m^2
Are you sure your area is correct? Isn't the radius d/2?
 
  • #3
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First correct your calculation for the window area...You used r=40 cm but that is given as the diameter!
 
  • #4
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What level of education should you want this be done,I think it can done differently by a high school student from a university student!Specify your need...Also i think YES "practically" it makes more sense to use the density of see water,but if it is a text book problem then observe how other relevant questions are being attended in using the densities...Plus remember to correct the window area calculation!
 
  • #5
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What level of approach is it required to approach this problem?Because I think it would be approached by a university student perhaps different from a high school student,so please be specific!..Also I think YES it makes more sense "practically" to use the density of sea water than plain water..But if it is a text book problem then maybe you should observe how other relevant questions are being attended with respect to density specifications!
 
  • #6
mgb_phys
Science Advisor
Homework Helper
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Your method is almost correct.
First correct the area as LowlyPion suggests.
Then there is no need to convert to atmospheres. Although there is presuambly one atmosphere of air pressure inside the sub pushing out, there is also one atmosphere of air pressure on the surface of the water pushing down. Imagine the window was exactly at the surface - you wouldn't have zero atmospheres of water pushing in and one atmosphere inside pushing out, otherwise the windows would fall out!

And yes an extra bonus mark for using seawater!
 
  • #7
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Thanks, the radius was the only thing, thickness was given as a trick condition
 
  • #8
mgb_phys
Science Advisor
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Thanks, the radius was the only thing, thickness was given as a trick condition
Often the question is reprinted from a longer question and the extra parameters are needed for part 2 of a question. Or sometimes it's jsut ot confuse you!
 

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