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Forces on a Pilot im stuck!

  1. Oct 1, 2007 #1
    To appreciate the forces that a fighter pilot must endure, consider the magnitude of the normal force that the pilot's seat exerts on him at the bottom of a dive. The plane is traveling at 252 m/s on a vertical circle of radius 616 m. Determine the ratio of the normal force to the magnitude of the pilot's weight. For comparison, note that black-out can occur for ratios as small as 2 if the pilot is not wearing an anti-G suit.
     
  2. jcsd
  3. Oct 1, 2007 #2

    Doc Al

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    centripetal acceleration

    Identify the forces acting on the pilot (normal force is one) and apply Newton's 2nd law.
     
  4. Oct 1, 2007 #3
    gravity and Fn and ???

    F= ma but there is no m

    a = 9.8 being gravity?


    Before solving this problem, answer the following two questions:
    (1) What is the direction of the acceleration of the pilot at the bottom of the dive?
    (2) What forces are acting on the pilot in the vertical direction at the bottom of the dive?
    Apply Newton's second law in the vertical direction, .


    thats the clue it gave me...

    1. the acceleration is down at the bottom of the dive..right?
    2. (from above)
     
  5. Oct 1, 2007 #4

    Doc Al

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    The two forces acting are gravity and the normal force. But no, the acceleration is not equal to 9.8 m/s^2. (That would be the acceleration of a falling object.)

    The acceleration is not downward. Hint: Think circular motion and centripetal acceleration.
     
  6. Oct 1, 2007 #5
    centripetal acceleration = v^2/r or 252^2/616 or 103.0909091 m/s^2

    and if its not down its up then, but how much up?

    what about uniform circular velocity?

    v=2pir/T
     
  7. Oct 1, 2007 #6

    Doc Al

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    Since the pilot (and plane) is executing circular motion, his acceleration is centripetal (which means "toward the center"). At the bottom of the loop the center is up, so the acceleration is up. Now apply Newton's 2nd law. (You don't need the actual mass of the pilot; just call it "m".)
     
  8. Oct 1, 2007 #7
    F = mv^2/r = m252^2/616 = 103.090909 thats excluding m...

    f=ma = 103.090909??
     
  9. Oct 1, 2007 #8

    Doc Al

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    Newton's second law says this (for circular motion):
    [tex]F_{net} = ma_c = mv^2/r[/tex]

    But you have to figure out the net force. There are only two forces: the pilot's weight (mg) which acts down and the normal force (Fn) which acts up, so:
    [tex]F_{net} = F_n - mg[/tex]

    Combine those equations and solve for Fn. Remember that they only want the ratio of Fn to mg. (The mass will cancel out.)
     
  10. Oct 1, 2007 #9
    Fnet = mv2/r = 252^2/616 = 103.090909

    Fn = Fnet + mg = 103.090909 + 9.8 = 112.890909?

    Fn/mg? 112.890909/9.8 = 11.5 = the ratio?
     
  11. Oct 1, 2007 #10

    Doc Al

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    Good!

    Just be careful when you show your work, for example, when you wrote this:
    You just dropped off the mass. If you do that, your equation is no longer true. Instead write it as:
    Fnet = mv2/r = m(252^2/616) = m(103.090909)

    Then: Fn = m(103.1) + mg
    And: Fn/mg = [m(103.1) + mg]/mg = [(103.1) + g]/g = 11.5

    Note that this means that at the bottom of that dive, the pilot feels a force pushing up on him that is 11.5 times his weight. (He feels 11.5 times heavier.)
     
  12. Oct 1, 2007 #11
    so my ratio answer is no units then right?
     
  13. Oct 1, 2007 #12
    sweet thanks for your help
     
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