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Thread: Force times distance... Work problem

  1. #1 Force times distance... Work problem 
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    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 problem 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?


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  3. #2  
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    If you are calculating the work, then distance has to be in there. Wouldn't the dx part of it be the distance? I think you must have misinterpreted something in either the rope problem or the pump problem.
    Edit: I think I know what the problem is. If you are lifting a fixed weight up with a rope, it doesn't matter how high you lift it, the force is the same. The work will still be a function of the height lifted, though. The difference with a pump is that the higher you have to pump the water, the more pressure there will be, so the force applied by the pump is a function of the height.


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  4. #3  
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    In a simple rope problem don't you assume the rope has no mass and therefore no work is expended moving it, but rather in moving the object it is attached to.
    In contrast with a pump problem the fluid that is moving, which you are thinking is equivalent to the rope, does have mass both real and in the problem and it is that which work is being done on.
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  5. #4  
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    I thought the force does change because as you pull the rope up it gets lighter in weight.
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  6. #5  
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    In the rope problem it looks like you have a rope with no weight on it, other than the rope itself. The rope weighs 25 lb (or whatever units) and is 50 ft long. X is the distance.The maximum weight is at the beginning (x=0) where it weighs 25 lb (25-0). When it is all wound up onto the windlass (x=50) it weighs nothing. (25-.5*50=0). So yes there is something that represents the distance.
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