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Push ups

  1. Nov 10, 2004 #1
    Not quite sure what to do about this one:

    -----A person whose weight is W = 601N is doing push-ups. Assume L1 = 0.859m and L2 = 0.353m. Calculate the normal force exerted by the floor on each hand, assuming that the person holds this position. ------

    I can't get the picture on here, but basically it shows a person in the postion of a push up. L1 is from his feet to the center of gravity (in his stomach in the picture). L2 is from the center of gravity to his hands. I know the equation for center of gravity is Xcg = W1X1 + W2X2 +... / W1 + W2 +...
    I'm not really sure what to do with this. I thought maybe I would calculate for the center of gravity and just multiply by gravity to get the normal force...? Any guidance would be appreciated. Thanks :)
     
  2. jcsd
  3. Nov 10, 2004 #2
    Do you know about torques? Consider the torque done by gravity about the pivot point (your feet) and compare to how much torque at the hands you must do to counteract this.
     
  4. Nov 10, 2004 #3
    hmmm....i do know about torque...that it equals the product of the force and the lever arm...but i didn't consider using it for this problem...hmmm
     
  5. Nov 10, 2004 #4
    It's definitely the way to go. The sum of the two torques must be 0 if he's not moving around the pivot.
     
  6. Nov 10, 2004 #5
    So the sum of the two torques is 0... so Fd + Fd = 0....(601)(0.859) + (601)(0.353) =0...i need another equation to put on the other side to solve for the force...so Fd=0...i'm looking for force...but i need a distance...i thought i could use 0.353...but i just tried this and i got the wrong answer. hmmm
     
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