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Force exerted by the floor on the feet of a student doing pushups

  1. Apr 15, 2004 #1
    A physics student is about to do a pushup. Her center of mass lies above a point on the floor which is d1 = 1.01 m from her feet and d2 = 0.65 m from her hands. If her mass is 45.0 kg, what is the magnitude of the force exerted by the floor on her feet? (Neglect friction in this problem)

    I don't think i know where to put the pivot in this problem. I tried a few times at the center of gravity.

    Also I don't know what the "mass" of her feet are is it just (1.01/1.66)*45?

    That's what i did then, once i just tried to find the Normal force with just that.

    The next time I tried , I used the same method with the "mass" of her feet and hands, and then tried to find the torques. Her feet had a larger torque, so I subtracted the torque of her hands, and then divided by 1.01 to try and find the Normal force against her feet.. That didn't work either..

    Anyone have a better approach?
  2. jcsd
  3. Apr 15, 2004 #2

    Doc Al

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    Staff: Mentor

    Just treat her like a plank of wood with two supports: feet and hands. Each support will have a normal force pushing up to counter her weight pulling down (at her center of mass). For equilibrium, the sum of the torques must equal zero. You can solve it in one step if you choose her hands as your pivot point.
  4. Apr 15, 2004 #3
    So, Hands are Pivot.

    Weight of girl*distance from pivot = F* distance of feet from pivot

    45*9.8*.65 = (1.66)F

    F of feet = 172.68

    Only one shot left for right answer.. Does that look good?
  5. Apr 15, 2004 #4
    Yay: )
    I got it Thanks!
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