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Friction, tension, and acceleration combination problem

  1. Aug 7, 2011 #1
    1. An 85-kg man plans to tow a 109000-kg
    airplane along a runway by pulling horizontally on a cable attached
    to it. Suppose that he instead attempts the feat by pulling the cable at
    an angle of 9.00 above the horizontal. The coefficient of static friction
    between his shoes and the runway is 0.77. What is the greatest acceleration
    the man can give the airplane? Assume that the airplane is on
    wheels that turn without any frictional resistance.



    Forces on Plane
    Fn = Normal force
    Weight = (mass of plane)(gravity) = 109000(9.8) = 1068200 N
    Tx = x-component of tension in rope = Tcos(9)
    Ty = y-component of tension in rope = Tsin(9)

    Forces on Man
    Fn = Normal force
    Weight = (mass of man)(gravity) = 85(9.8) = 833 N
    Fk = frictional force = (weight)(coefficient of friction) = (833)(0.77) = 641.41 N
    Tx = x-component of tension in rope = -Tcos(9)
    Ty = y-component of tension in rope = -Tsin(9)


    net equations for the plane
    Fx = Tcos(9) = (Mass of plane)(x-component of acceleration)
    Fy = Tsin(9) = (Mass of plane)(y-component of acceleartion)*********

    net equations for the man
    Fx = Fk - Tcos(9) = (Mass of man)(x-component of acceleration)
    Fy = -Tsin(9) = (Mass of man)(y-component of acceleration)*********

    From here, i think it is plug and chug to find tension. After I find tension, I can find both the x and y components of acceleration. I can then use Pythagorean theorem to find acceleration from the x and y components.

    My question is, have I written my net equations correctly???? I think something is wrong because when I plug in T to the two starred equations above, I get two different answers for the y-component of acceleration...help??? please?
    Last edited: Aug 7, 2011
  2. jcsd
  3. Aug 7, 2011 #2


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    If the plane and man do not leave the ground, is there any acceleration in the y direction?
    And don't forget the Normal force in the y direction in your equations.

    Note also that Fk is not weight times the friction coefficient, it is ????? times the friction coefficient.
  4. Aug 7, 2011 #3
    Fk is the normal force times the frictional coefficient...but I don't see how that changes anything...? :uhh::confused: Hmmm...Does the tension of rope contribute to the normal force??? So that the normal force = mg + Tsin9??? And thus Fk = 0.77[mg + Tsin9]
    Last edited: Aug 7, 2011
  5. Aug 7, 2011 #4


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    Yes, on the man.
    Yes, very good. Now continue... (sorry for delayed response, I fell asleep:zzz:
  6. Aug 7, 2011 #5
    AWESOME!!! THANK YOU SO MUCH, Phanthomjay....ur my physicss superman hahaha
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