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Pressure in airplane

  1. Nov 18, 2007 #1
    1. The problem statement, all variables and given/known data
    I know this is probably a very easy question, but i'm new to this and never worked with it before.

    [​IMG]






    3. The attempt at a solution

    so you have 0.25atm pressure going into the window from outside, and 0.75atm from inside pushing out.

    Am i assuming that with 0.50atm more pressure coming inward from a person would equal the force of pressure pushing out and therefore any more than that would theoretically pull the window inward?

    P = F/A
     
  2. jcsd
  3. Nov 18, 2007 #2

    Doc Al

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    Staff: Mentor

    You're on the right track. Figure out the force required to pull the window open. Is it likely that a passenger can exert such a force?
     
  4. Nov 18, 2007 #3
    Well, if i'm doing this right, it seems pathetically small, but I guess this is assuming there are no latches of some sort. Maybe i need to convert my dimensions?

    so P = F/A ---> F = PA

    so for F to be able to pull the window in, it must be [tex]\geq[/tex] (0.75 atm/(0.50 * 0.90).

    If the outside pressure is 0.25atm, then only 0.50 more is required..(?)

    so one must apply a force of (0.50atm) * (0.50 *0.90) = 0.225 N

    I thinking the dimensions need to be converted?
     
  5. Nov 18, 2007 #4

    Doc Al

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    Staff: Mentor

    What is atmospheric pressure?
     
  6. Nov 18, 2007 #5
    at sea level 1 atm = 14.7 psi = 760mm mercury = 101.3 kPa......a ha, i need to use Pascals...?
     
  7. Nov 18, 2007 #6
    0.50 atm x 101.3kPa = 50.65 kPa

    50.65 kPa * (0.50 * 0.90) = 22.8 kN which is a lot for a human being. Looks good?
     
  8. Nov 18, 2007 #7

    Doc Al

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    Staff: Mentor

    Looks good.
     
  9. Nov 18, 2007 #8
    to me, that's a dumb questions.

    Thanks again Al
     
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