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Total work done on a screw.

  1. Apr 19, 2010 #1
    1. The problem statement, all variables and given/known data
    It takes 20 turns to drive a screw completely into a block of wood. Because the
    friction force between the wood and the screw is proportional to the contact area between
    the wood and the screw, the torque required for turning the screw increases linearly with
    the depth that the screw has penetrated into the wood. If the maximum torque is 12 N m
    when the screw is completely in the wood, what is the total work (in Joules) required to
    drive in the screw?

    3. The attempt at a solution
    So I tried to do an integral by finding out how much the torque changed per rotation. Then using that as the equation and the total distance turned to be plugged in.
    [tex]\int.6x[/tex] from 0 to 7200. I got a very large number and I don't think I did it right....
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Apr 19, 2010 #2


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    Rather than use calculus, which is bound to throw you off, what is the average torque if it linearly increases? Then what is the definition of Work for a torque acting through one rotation?
  4. Apr 20, 2010 #3
    Well, the equation for how it increases is just .6*(rotation number). The average torque would be the maximum and minimum divided by two...The maximum is obviously 12 so would that make the minimum be 0? That doesn't really make sense though. In that case the average would be 6. The equation for work is torque times the rotation distance. This gives 754 J which is the correct answer now that I check it. Thanks Phantom...guess I just needed another person to point me in the right direction!
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