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Air Pressure Question.

  1. Apr 25, 2009 #1

    NPM

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    1. The problem statement, all variables and given/known data
    In a tornado, the pressure can be 15 percent below normal atmospheric pressure. Suppose that a tornado occurred outside a door that is 195 cm high and 91 cm wide. What net force would be exerted on the door by a sudden 15 percent drop in normal atmospheric pressure? In what direction would the force be exerted?


    2. Relevant equations
    Here lies the problem.
    I do no know which formula I should be using.
    I know that atmospheric pressure is about 1.0 x 10^5 N/m squared at sea level.
    Which I'm inferring is what I would use for the problem.
    Could someone post a formula or something?
    I think that's all I would need.
     
  2. jcsd
  3. Apr 25, 2009 #2

    NPM

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    Never mind.
    I just realized that P=F/A would work.
    Whoops xP
     
  4. Apr 25, 2009 #3

    NPM

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    So I tried doing the problem:
    P=F/A
    Regular pressure = 1.0 x 10^5
    15% decrease = 8.5 x 10^3
    85000 = F/(195 x 91)
    = F/177.45<--- Converted to meters
    So I multiplied by the area:
    85000 x 177.45=15,083,250 Pa
    And that, to me, seems like a completely ridiculous answer.
    Would someone care to correct or at least help me?
     
  5. Apr 25, 2009 #4

    Redbelly98

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    Staff Emeritus
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    A couple of comments:

    1. Remember there is a net force, due 1 full atmosphere of pressure, pushing on the other side of the door.

    2. Be careful with the units. It asks for a force, you're answer was in Pa units. Do you see the problem there?
     
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