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Work problems

  1. Mar 29, 2009 #1
    This is not really a homework problem but just a question.

    How come while doing pump problems distance is part of the equation but rope problems distance is not...


    For example your rope integral might look like integral from zero to fifty of (25-(.5)x) dx

    Here all that 25-(.5)x is the weight of the rope...


    But a pump integral may look like the integral from zero to three of ((8x)(9.8)(1000)(5-x))dx

    Here (5-x) represents the distance the water travels...


    How come there is nothing that represents the distance the rope travels in the rope problem?
     
  2. jcsd
  3. Mar 30, 2009 #2
    If the actual questions to the examples I gave would make my question more understandable I can write them out, but I figured it was probably not necessary.
     
  4. Mar 30, 2009 #3

    HallsofIvy

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    If you are talking about the work required to lift a rope that is hanging down up to the point from which it is hanging, and x represents the height of a point on the rope then the "piece" of rope is lifted L- x where L is the length of the rope. If, instead, x is the distance from the point on the rope to the point where it is hanging, the distance is x.
     
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