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Net Force on airplance cabin surface

  1. Nov 10, 2008 #1
    1. The problem statement, all variables and given/known data

    5. The air pressure at sea level on a cold day is P0 = 1.013 x 105 Pa. At 10.0 km above sea level the air pressure is 0.30 P0 ; however a commercial jet plane keeps the pressure in the passenger and crew section at 0.75 P0 . If the surface area of the cabin structure is 1200 m2 , find the net force on the cabin surface due to the differential of pressure.

    2. Relevant equations

    3. The attempt at a solution

    P net pressure = 0.75 - 0.30 = 0.45 Pa
  2. jcsd
  3. Nov 10, 2008 #2


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    Welcome to PF!

    Hi tbstar ! Welcome to PF! :smile:

    Pressure is defined as force per area.

    So force = … ? :smile:
  4. Nov 10, 2008 #3
    I wanted to understand this solution:

    F = P x A = 0.45 x 1.013x10e5 x 1200
    F= 5.47 x 10e7

    Why we are multiplying the difference pressure between the air and the cabin by the pressure at sea level?

    Thanks for the warm welcoming
  5. Nov 10, 2008 #4


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    You can do it in steps if you want.
    What is the outward force (caused by the pressure inside the cabin?)
    What is the inward force (caused by the pressure outside the cabin?)
    What is the net force (mind the direction of the forces, e.g. relative minus signs!)
    Do you see why you can take the difference of the pressures to begin with?
  6. Nov 10, 2008 #5


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    Hi tbstar! :smile:

    (btw, you can use the X2 tag above the reply box to write e7 :wink:)
    Because the question gave you the pressure as multiples of P0, the sea-level pressure.

    This is quite usual … pressure in many situations (particularly where people have to breathe) is measured as multiples of "atmospheric pressure", just as acceleration (of an airplane, for example) is often measured as multiples of g. :smile:
  7. Nov 10, 2008 #6


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    If you do the steps I told you, you will see that the formula for the force is
    F = (p2 - p1) x A
    where p1 and p2 are the outside and inside pressure, respectively. If you plug in the numbers,
    F = (0.75 p0 - 0.30 p0) x A
    Now, just as you can write x2 + x = x(x + 1), you can rewrite
    (0.75 p0 - 0.30 p0) = (0.75 - 0.30) p0
    F = (0.75 - 0.30) p0 A
  8. Nov 10, 2008 #7
    I am so stupid, I didn't even pay attention to the details that's all. But I have been working on physics problems from 5pm to 2pm straight up because I have an exam tomorrow. So thank you very much guys for helping me to understand how stupid I was asking that stupid question.

  9. Nov 11, 2008 #8


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    Actually it's a very useful question.
    IMO it is far more stupid to learn the formulas by heart without understanding them, real physics is also about grasping the physical concepts (and usually, when you have those right, it's also easier to remember the formulas... or derive them when you forget them).

    Basically here, all you have to remember is what pressure is (force per area), then by logical steps you can solve the question (it must be F/A then, so we can write F = p A... what is the outward force, what is the inward force and what is the net force - hey, it's the pressure difference that causes it! That makes sense, otherwise we would also have a force when both the cabin and the ouside air were at sea level pressure p0, for example.). Also, it makes it easier to check your answer (for example: the inside pressure is higher than the outside pressure, so you expect an outward force... does your calculation give that as well?)
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