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Homework Help: Airplane wing

  1. Nov 27, 2008 #1
    1. The problem statement, all variables and given/known data

    (a) Find the pressure difference on an airplane wing where air flows over the upper surface with a speed of 117 m/s, and along the bottom surface with a speed of 104 m/s.

    (b) If the area of the wing is 38 m2, what is the net upward force exerted on the wing?

    2. Relevant equations

    i have no clue.

    3. The attempt at a solution

    I think my teacher might have left off a part. I don't know how you could find a change in pressure without maybe the height of the wing. I guess you can assume the density of air is about 1.29 kg/m^3 and g=9.81 m/s^2. Is there a way to find it with just the 2 velocities?
  2. jcsd
  3. Nov 27, 2008 #2
    do you know bernoulli's equation?
  4. Nov 27, 2008 #3
    the height of the wing could be required, but often assumed negligible, especially in such a general question.
  5. Nov 27, 2008 #4

    what would you put as the y's?
    if you ignored it, would the pgy's cancel?

    i guess you could do P1-P2=(1/2)pv2^2-(1/2)pv1^2+pgy2-pgy1
  6. Nov 27, 2008 #5
    exactly! if we assume the height of the wing to be small then y1 = y2 and the pgy's cancel.

    now you have the pressure difference in terms of velocity and density. you have those values so, what is your pressure difference?
  7. Nov 27, 2008 #6
    P1-P2=-1853.09 pa

    well a change would be possitive. that seems kinda large for having only a difference of 13 m/s.
    Last edited: Nov 27, 2008
  8. Nov 27, 2008 #7
    So that's telling us that the air pressure on top of the wing is 1853.09Pa less than the pressure at the bottom of the wing (P1=P2-1853.09), ie. the pressure difference. If it was positive then we would see that pressure at the bottom would be less than the top, which is not the case here. Kinda large is relative, 1853Pa is about 1.85kPa. The air pressure can change due to the weather between 900mbar and 1100mbar (approximately) which is a change of 200mbar or 20 000Pa. And that is just due to weather and you can't even feel it. So, 1853Pa is not really that much.

    So, for the second part of the question, how do we calculate a force from a pressure difference and an area?
  9. Nov 27, 2008 #8
    so 1.85(38)=F=70.014 kN

  10. Nov 28, 2008 #9
    No problem.

    Always remember to check your answer to see if it makes sense. For example, 70kN of lift, how much could I lift with that? using F=mg, I see that I can lift about 7000kg, that sounds about right for a medium sized plane. As long as I didn't get 7grams or 700000000kg, then I know my answer could be reasonable.
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