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Homework Help: Calc II word problem(work needed to lift a leaky bucket)

  1. Aug 5, 2010 #1
    1. The problem statement, all variables and given/known data
    From Stewart Calculus Concepts and Contexts 4th edition pg.473 section 6.6 #15...

    :A leaky 10-kg bucket is lifted from the ground to a height of 12 meters
    at a constant speed with a rope that weighs 0.8kg/m. Initially the bucket contains 36kg of water but the water leaks at a constant rate and finishes draining just as the bucket reaches the 12 meter level. How much work is done

    2. Relevant equations




    3. The attempt at a solution

    I am able to do these kind of problems, but the only thing different about this one is the weight is constantly changing. The total weight of the water/bucket/rope initially is 55.6 kg, and at the end its just the 10 kg bucket left.

    I dont know how to approach the change in weight of the water leaking out of the bucket. So far I think you integrate from 0-12 meters, and the distance an arbitrary part of the rope has to travel is 12-x. Any hints?
     
  2. jcsd
  3. Aug 5, 2010 #2
    [tex] W = \int\limits_{a}^{b} F\,dr[/tex]
    [tex]F = mg[/tex]
    [tex]m = 10 + 36 + (.8)(12) - .8r - \frac{36r}{12}[/tex]
    [tex]a = 0[/tex]
    [tex]b = 12[/tex]
    [tex]g = 9.8[/tex]
     
  4. Aug 5, 2010 #3
    Ah thats actually quite simple. I thought a new variable would have to be introduced to account for a rate of change. Thanks a lot.
     
  5. Aug 5, 2010 #4

    berkeman

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    Staff: Mentor

    That makes an assumption about how the rope is routed...

    nlsherrill -- can you tell us what assumption he is making about how the rope is routed, and what would be different if it were routed differently?
     
  6. Aug 5, 2010 #5
    Hmm..well there are a bunch of factors that must be taken into consideration I suppose...but as far as the routing of the rope, I imagine it is fixed and stays in a straight line as it is pulled up 12 meters. Otherwise the amount of force required to keep the bucket moving at a constant rate would fluctuate because the bucket would move some in the x or z directions instead of just strictly the y direction. I believe this is the assumption they made, that the bucket travels strictly in a straight line, and the ropes shape does not change.

    I don't know if thats what your looking for, but from the looks of the assumptions they made this problem about as easy as they could(besides the changing of the weight with respect to the distance traveled).
     
  7. Aug 6, 2010 #6

    berkeman

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    Staff: Mentor

    xcv's solution assumes the rope is pulled up to a ledge or something, and dropped there. How would the equation change if the rope were looped over a single pulley, and pulled from the ground in order to hoist up the bucket....? (that is how it normally would be hoisted anyway, right?)
     
  8. Aug 7, 2010 #7
    Ha I'm not really sure! I think it would be a different equation altogether because more rope would be needed in that senario. Otherwise the person on the ground wouldn't have a way to pull the bucket up to 12 meters.

    Other than that...I dont know. Maybe some signs would change because the direction of action on the person pulling the rope up is different.
     
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