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Work done by a screw

  1. Apr 9, 2013 #1
    1. The problem statement, all variables and given/known data

    It takes 25 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 15 N m when the screw is completely in the wood, what is the total work (in J) required to drive in the screw?

    2. Relevant equations

    Work = ∫torque dθ

    3. The attempt at a solution

    Work = ∫(15/25)x dθ (from 0 to 25)
    Work = 187.5
     
  2. jcsd
  3. Apr 9, 2013 #2

    cepheid

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    Welcome to PF,

    Uh, 25 turns does not correspond to theta = 25. Think about it. What angle do you go through in one turn?
     
  4. Apr 9, 2013 #3
    One revolution would be 2PI, but when I do the same equation with the full 50PI I still get the wrong answer.
     
  5. Apr 9, 2013 #4

    cepheid

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    Well that's no surprise. You've got your integrand being constant. The whole point of this question is that that the torque ISN'T constant with angle. If it was constant, it wouldn't be necessary to integrate at all!

    You have to think more carefully about what the function τ(θ) is. If I were you, I'd find the function τ(x), where x is the depth into the wood, and then translate this into τ(θ).
     
  6. Apr 9, 2013 #5
    Thanks, changed the equation to the integral of (15/(2PI * 25))x from 0 to 50PI
     
  7. Apr 9, 2013 #6

    cepheid

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    I think that's correct, provided x is actually meant to be θ in that equation.
     
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