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Finding the lift force on an airplane

  1. Sep 14, 2012 #1
    1. The problem statement, all variables and given/known data

    A 2598kg airplane is diving (at constant velocity) making an angle of 10 degrees to the horizontal. Find the lift force on the plane assuming that it is normal to the velocity?

    I think my main problem is the fact that I'm having a bit trouble with understanding the terminology.

    2. The attempt at a solution

    Since it says that the plane makes an angle of 10 degrees to the horizontal whilst diving, I interpreted it as the angle is -10 degrees (i.e. south-east direction). Thus, the lift force would have an angle of 80 degrees.

    The plane is diving at constant velocity, meaning the net force is 0. I tried different methods, but I couldn't come up with the right answer..
     
  2. jcsd
  3. Sep 14, 2012 #2

    lewando

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    What does your free body diagram look like? Always a good place to start.
     
  4. Sep 14, 2012 #3
    Pretty much as described under "attempt at a solution", i.e. one vector directed at a south-east direction (the diving plane) and the lift force at a north-east direction.
     
  5. Sep 14, 2012 #4
    One of the main things which confuses me is that I'm not at all sure what the lift force is. I looked at some pictures found on google and it seemed like the lift force is a vertical force, but that doesn't make sense, considering the fact that the lift force is supposed to be normal to the plane..
     
  6. Sep 14, 2012 #5

    lewando

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    If acceleration is zero, then all the forces must sum to zero. So one of your force vectors is the lifting force normal to the velocity. Check. Your other force vector (south east) needs some more thought. What does it represent? Also don't forget about the force due to gravity (mg).
     
  7. Sep 14, 2012 #6

    lewando

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    To respond to your sidebar question: If the plane is flying horizontally, the lift force would be a vertical force. If the plane was in a vertical dive, the lift force would be a horizontal force--making the plane want to recover. If you wanted to fly a plane straight down, you would actually have to adjust the controls to compensate for the horizontal lift force.
     
  8. Sep 14, 2012 #7
    It represents the velocity of the airplane, doesn't it? And if I've done it correctly so far, mg should be its y-axis component. But it's from here that I get lost..
     
  9. Sep 14, 2012 #8

    lewando

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    Velocity is not a force. mg does have only a y-axis component. Along the axis of the plane's travel are 2 forces: thrust and air resistance. Think of it this way-- when you draw a FBD with mg and the lift force only, does it look like it will balance out to zero? No it will not (to save time--let me know if you disagree). There needs to be another force in opposition to the 2 forces. Since thrust and air resistance are constant and both unknown, consider combining them into an "opposing force" vector operating on the plane's axis of travel.
     
  10. Sep 14, 2012 #9
    Ah! That explains why I'm only running around in circles. I was confused by the phrasing of the question and treated its diving motion as a force.. I agree with the fact that it won't balance out to zero, I'd say that there should be an opposing, left-ward, force.

    If we call the lift force, F, then its vertical component should equal mg, right? If so, F should equal 25980/sin80=26381N. Is this correct?
     
  11. Sep 14, 2012 #10

    lewando

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    Unfortunately, the "leftward" opposing force isn't exactly leftward. It is along the axis of travel, in the "west-west-northwest" (170 degrees) direction. [edit-- by unfortunately, I mean you have a bit more work to do]
     
    Last edited: Sep 14, 2012
  12. Sep 14, 2012 #11
    Thanks for all your help, it's really helping me a lot (I'm studying physics from scratch by myself). A bit of logical thinking can go far. Of course the air resistance in a direction 180 degrees from the motion..

    I think I've solved it now. Splitting up the two forces (Air resistance = O and lift force = R) into their components (excluding the gravity as its along the axis), yields the following:

    Ry=Rsin(80)
    Rx=Rcos(80)

    O(y)=Osin(10)
    Ox=-Ocos(10)

    Now, the two vertical components should cancel out the force of gravity, whereas the horizontal components cancel each other out, giving us a simultaneous equation. Hence:

    Rsin(80)+Osin(10)=25980
    Rcos(80)-Ocos(10)=0

    After eliminating one variable (O, since it'd give us the answer we want directly) and calculating R, we get it to equal 25585N. It'd take too long for me to write out all the steps as I'm not so used to writing stuff like these on a computer. Did you nevertheless get the same answer?
     
  13. Sep 14, 2012 #12

    lewando

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    Your setup looks right, but I get a different answer-- 19425 N (ignoring SDs for now, g=10). I am checking my math.
     
  14. Sep 14, 2012 #13
    I double-checked my calculations and got 25585N again.
     
  15. Sep 14, 2012 #14

    lewando

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    My bad. You are right. Congrats! Don't forget to consider sig-digs in your final answer. [edit-- perhaps use g=9.81m/s^2].
     
    Last edited: Sep 14, 2012
  16. Sep 14, 2012 #15
    Again, thank you very much for your help. You really clarified some things for me!
     
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