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Wrench and nut

  1. Mar 5, 2013 #1
    1. The problem statement, all variables and given/known data
    (see attachment)


    2. Relevant equations



    3. The attempt at a solution
    If a force a F acts on the farther end of the wrench, the torque due to it Fl where l is the length of wrench. The work done by this torque for one full turn is Fl*(2##\pi##)
    For wrench A, l=10 cm, and for wrench B, l=20cm.
    Therefore the ratio of work done is 0.5. But this is wrong. :confused:
    I don't understand why they have given the radius of the nut. I haven't used this information and I have no idea about where to use this.

    Any help is appreciated. Thanks!
     

    Attached Files:

  2. jcsd
  3. Mar 5, 2013 #2

    Doc Al

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    Staff: Mentor

    If a force F is required to tighten the nut using the 10 cm wrench, what force is required with the 20 cm one?
     
  4. Mar 5, 2013 #3
    F/2?
     
  5. Mar 5, 2013 #4

    Doc Al

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    Staff: Mentor

    Right.
     
  6. Mar 5, 2013 #5
    But what next? Does that mean the ratio is 1?
     
  7. Mar 5, 2013 #6

    Doc Al

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    Staff: Mentor

    That's what I would say.
     
  8. Mar 5, 2013 #7
    Thanks a lot Doc Al! :smile:
     
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