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Airplane wing and forces on it

  1. Feb 6, 2008 #1

    ~christina~

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    1. The problem statement, all variables and given/known data
    An airplane has a mass of 1.60x10^4 kg and each wing has an area of 40.0m^2. During level flight the pressure on the lower wing surface is 7.00x10^4 Pa. Determine the pressure on the upper wing surface.

    [​IMG]
    2. Relevant equations

    [tex]\sum F= F_L -F_u - F_{mg} = 0 [/tex]


    3. The attempt at a solution

    My basic problem is that I keep getting a negative number for the pressure which is odd

    [tex]\sum F= F_L -F_u - F_{mg} = 0 [/tex]

    [tex] F_u = F_{mg}- F_L [/tex]

    P= F/A

    [tex] P_u= Mg - P_L A [/tex]

    [tex]((9.8m/s^2)*(1.60x10^4kg))-((40.0m^2)(7.00x10^4Pa))= P_u[/tex]

    [tex]-2,643,200 = P_u A [/tex]

    [tex] P_u= -66,080 Pa [/tex] =>

    my problem is why is it negative?? is it because it's force is downward??
    Plus..the answer that was given (only the answer) was 6.80x10^4 Pa so I must have done something wrong. However I ccalculated it 3 times and I still get the same answer.

    Thank you very much =D
     
  2. jcsd
  3. Feb 6, 2008 #2

    cepheid

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    First of all, your rearranging of the equation has an error:

    You start with

    FL - Fu - Fmg = 0

    Now add Fmg to both sides of the equation, and subtract FL from both sides:

    - Fu = Fmg - FL

    This is not what you arrived at. The discrepancy explains your errant minus sign
     
  4. Feb 6, 2008 #3

    cepheid

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    Your third equation also makes no sense. You have written:

    Pu = Mg - PLA

    But Pu is a pressure whereas Mg and PLA are forces. You can't equate a pressure to a force. Your equation is not dimensionally consistent. Does this help?
     
  5. Feb 6, 2008 #4

    ~christina~

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    Hm I still get the same answer....just without the negative. 6.60x10^4Pa

    I actually forgot to type that in. On paper I did add that in.
     
  6. Feb 6, 2008 #5
    I get the same answer as you, too.
     
  7. Feb 6, 2008 #6

    ~christina~

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    okay then, I'm suspecting it might be a typo.
     
  8. Feb 6, 2008 #7

    cepheid

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    I don't think you understand. This equation that you have:

    Pu = Mg - PLA

    is WRONG, because the thing on the left-hand side is a pressure, and the quantities on the right-hand side are forces. Therefore your equation is not dimensionally consistent: it is in error and you need to fix it.

    Either have all pressures or all forces.
     
  9. Feb 6, 2008 #8

    ~christina~

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    this is what I have:

    [tex]P_u A= Mg- P_L A [/tex]

    (technically now all forces)
    I used this in my calculations.

    Is its still wrong?
     
  10. Feb 6, 2008 #9

    cepheid

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    Well, it should be

    [tex]-P_u A= Mg- P_L A [/tex]

    because of what we talked about before with the sign error. But other than that, it looks ok.

    When I plug in the numbers, I get

    [tex] P_u = 6.6076 \times 10^4 \ \ \textrm{Pa} [/tex]

    So if the book says 6.8*10^4, then maybe it IS a typo.
     
  11. Feb 6, 2008 #10

    cepheid

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    Sorry I misinterpreted what you were saying before about how you used the correct formula on paper, without the typo.
     
  12. Feb 6, 2008 #11

    ~christina~

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    oh yep and I corrected that..

    Okay, thanks :smile:
     
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