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Increasing Torque, Force, and Work: Increasing crank Length (L)

  1. Jun 23, 2011 #1
    1. The problem statement, all variables and given/known data

    You are pulling water with a constant velocity from a well using a crank of length L. If the length of the crank was doubled, you could ...

    2. Relevant equations
    Work = Torque*angular distance
    Torque = I*angular acceleration
    Torque=F*d*sin(angle)
    It looks like the force is being applied to the lever at 90 degrees


    3. The attempt at a solution
    The answer is "pull up double the amount of water with the same force", however I can't work out why this is the correct answer.

    Maybe working backwards...
    Double the amount of water has double the weight, (force of gravity)
    So The torque from the crank wheel would then need to be double (?)
    Torque = F*D*sin(90 degrees)
    And if D is us doubled, then the same force does twice the Torque.
    Is this right?

    Also, I couldn't convince myself why these other options were wrong:
    Incorrect: pull up the pail with half the number of revolutions
    Incorrect: exert double the torque while pulling up the pail with half the work
    Incorrect: exert four times the torque while pulling up the pail with the same work
    Incorrect: pull up double the amount of water with the same work
    Incorrect: pull up the pail with half the work and half the force

    Thanks!
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jun 23, 2011 #2

    tiny-tim

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    Last edited by a moderator: Apr 26, 2017
  4. Jun 23, 2011 #3
    Oh, sorry, I should have been more clear. The "crank" is a circular pulley sort of crank that winds up a rope with a bucket attached to the end.
    Does this affect your answer?
     
  5. Jun 23, 2011 #4

    tiny-tim

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    It still has a shaft (that the rope winds round before going down the well).
     
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