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Force on an object being dragged

  1. Feb 14, 2008 #1
    1. The problem statement, all variables and given/known data
    You drag a suitcase of mass 9.2 kg with a force of F at an angle 52.1 degrees with respect to the horizontal along a surface with kinetic coefficient of friction 0.54. The acceleration of gravity is 9.8 m/s^2.
    If the suitcase is moving with constant velocity 1.65 m/s, what is F ?

    2. Relevant equations

    3. The attempt at a solution

    a = 0

    FcosΘ - Friction = (9.2)(0)
    F - µR = 0

    F - (0.54)((9.2)(9.8)-mgsinΘ)) = 0
    F- (0.54)(90.16 - 9.2(9.8)sin52.1) = 0
    F - 0.54(19.016) = 0
    F = 10.268 N

    but this is not right..I think that I am doing something wrong....Any help would be great. Thanks
  2. jcsd
  3. Feb 15, 2008 #2
    this should be enough...

    If the object moves with constant velocity then NET force on it is zero.

    Fcos theta - friction= zero

    >fcos theta = frictional force.
    >f cos theta=48.68
    >f=48.68/cos 52.1
    >f=79 N
    Last edited: Feb 15, 2008
  4. Feb 15, 2008 #3
    The frictional force depends on F and theta. If you pull upwards the suitcase, the force that the suitcase exerts on the ground will be smaller, and therefore the friction will decrease.
    with theta = 60 your answer would have F > mg and the suitcase flying up in the air.:smile:

    The force with which you pull upward on the suitcase is F sin(theta), so the frictional force is 0.54 (mg - F sin(theta))
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