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Homework Help: Lift Force Generated by a Wing

  1. Nov 26, 2009 #1
    1. The problem statement, all variables and given/known data

    A commercial airliner cruises at a speed of 284 meters per second. Each wing is 28.8 meters long and has an average width of 1.73 meters. The density of air at cruising altitude is 0.47 kilograms per cubic meter. How much lift force is generated by the wings if the air below the wings travels at the plane's cruising speed and the air above the wings travels 9.8 percent faster? Hint: Planes have two wings.

    2. Relevant equations


    3. The attempt at a solution

    No real clear idea where to start...
  2. jcsd
  3. Nov 26, 2009 #2
    What is Bernoulli's equation?
  4. Nov 26, 2009 #3
    p +1/2(rho)v^2 + (rho)gh = p + 1/2(rho)v^2 + (rho)gh

    so the pressures should cancel since it atmospheric....

    and so if the
    air above: 311.832 m/s
    air below: 284 m/s

    1/2(.47)311.832 + im not really sure what to do beyond this....
    so somehow i should relate the force exerted by the air on wing above to the force exerted by the air on the wing below

    so F/A = F/A?

    im full of a lot of random thoughts
  5. Nov 26, 2009 #4
    There's not a lot of pressure difference due to gravity on the top and the bottom of the wing, so we can leave out the rho*gh part.

    p +1/2(rho)v^2 = constant

    What happens to p when the the velocity, v is increased over the top of the wing?
  6. Nov 27, 2009 #5
    I'm not completely sure about this, but i read something about NASA scientists deducing an equation for finding the force of lift on a wing:

    Flift=1/2*density of air*(velocity of air in opposite direction)2*area
    of wing*wing coeficcient

    I barely think this is going to work, unless you know how to find out the wing coeficcient, which I don't know. I just read this on the net. Just trying to help.

    BTW...you didn't get this equation from an engineering book, did you, 'cause if you did, I think you're supposed to have posted it in the engineering section, not in 'introductory physics!
  7. Nov 27, 2009 #6
    This is an AP Physics B question..... so it should be just involving Bournelli's.... and i still can't quite get it...

    so I have P + 1/2(rho)v^2 + (rho)gh = P + 1/2(rho)v^2 + (rho)gh

    so I simplified to P + 1/2(rho)v^2 = P + 1/2(rho)v^2

    so I know (rho),v,and I can get P because P=F/A and I know A and atmospheric pressure is at cruising altitude....

    so then I just compare the two and the difference will be the force?
    Last edited: Nov 27, 2009
  8. Nov 27, 2009 #7


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    This is the flaw in your thinking. What comes out of Bernoulli's equation is the result that in order for the equality to be satisfied, the air pressures above and below the wings should be slightly different due to the different flow rates. That is why there *is* a lift force in the first place. So, obviously you should be getting the answer that p below the wings is slightly higher than p above them.
  9. Nov 27, 2009 #8
    yes i figured that..... so how can i calculate the pressure differential per each wing?

    alright so i got it down to F/A = F/A + 3897.05 somethings....
  10. Nov 27, 2009 #9


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    Er...not really a question that makes any sense. The pressure below the wings has a certain value (period, no matter where exactly you are or what direction you look in). Same with the pressure above the wings. This pressure difference leads to a net force on the wings. The wing geometry comes into play because of the relationship between force and pressure. Pressure is the force per unit area. Does that help?
  11. Nov 27, 2009 #10
    alright so i calculated my 1/2(rho)v^2 values....
    then subtracted over..

    so i got F/A = F/A + 3897.05 somethings....
    and i can calculate the area under and above the wing (they will be equal) so
    F = F/A + 3897.05 * A?
  12. Nov 27, 2009 #11
    hrm still having trouble
  13. Nov 27, 2009 #12
    I got it thank you...
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