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Rope integral

  1. Oct 16, 2007 #1
    1. The problem statement, all variables and given/known data
    A freight elevator weighing 3000 pounds is supported by a 12 foot long cable that weighs 14 pounds per linear foot. Approximate the work required to lift the elevator 9 feet by winding the cable onto a winch

    2. Relevant equations
    W = int(f)dy

    3. The attempt at a solution
    The work lifting the elevator is 3000lb(9ft)=27000ft*lb

    The winch is lifting a piece of rope dy a distance 12-y from 0 to 9
    Each section dy weighs 14dy

    (sorry for the lack of latex)
    W = int0-9[14(12-y)]dy = 14int0-9[(12-y)]dy
    W = 14[12y-.5y2]0-9
    W = 945 ft*lb

    Total work = 27000+945 = 27,945ft*lb
    But the book has an answer of 36,945. The 945 leads me to think I set up the integral correctly, but where did the extra 900 come from?
  2. jcsd
  3. Oct 16, 2007 #2


    User Avatar
    Science Advisor
    Homework Helper

    The book's answer is clearly wrong. You knew that, right?
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