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Pressure based on Force/Area

  1. Sep 19, 2008 #1


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    Hi i need some help with a problem: A submarine has a 40.0cm diameter window 8.50cm thick. The window can withstand forces up to 1.10x106 N. What is the submarine's maximum safe depth?

    The pressure is maintained at 1.0atm and it's in salt water.

    Salt water density = 1025 kg/m3.

    the equation that i used was p = patmos + [tex]\rho[/tex]gd
    p = 1.013x105 + (1.0253)(9.8)(d)

    What i did to try to find p was p = F/A. <---- If this equation to find pressure is correct, how to I calculate the area from diameter and thickness?

  2. jcsd
  3. Sep 19, 2008 #2


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    you can omit the first term (atmospheric pressure) because the sub inside pressure is maintained at atmospheric pressure, such that it will cancel out this piece. On your second term, you have a typo, it's 1.025(10^3)(9.8)(d), where d is the water depth in meters
    this equation is correct, you need to find the area based on the window diameter; the thickness does not enter into the equation, so you don't need it (that is, the thickness doesn't affect the force on the window, just it's strength, which is not part of the problem.
  4. Sep 19, 2008 #3


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    so if i omit the atmospheric pressure then it'll be p = 1.025(10^3)(9.8)(d) now
    p = F/A = 1.10(10^6) / (pie)(0.04) = 8.75(10^6) Pa.
    8.75(10^6) / 1.025(10^3)(9.8) = d
    d = 871m. answer was 868.

    thanks for the help Phanthom Jay
  5. Sep 19, 2008 #4


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    Your answer looks correct. Did the problem give the density of seawater as 1025 kg/m^3 or did you assume that? Using the same values as you, with the exception of g being 9.81 m/s^2 I get 870.5 m.

  6. Sep 19, 2008 #5


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    The professor gave me the value for seawater as 1025 kg/m^3.
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