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Really hard intergration problem\ take a look

  1. May 10, 2009 #1
    Really hard intergration problem\!! take a look

    A 20-lb. monkey is attached to a 50-ft. chain that weighs 0.5 lb.
    per (linear) foot. The other end of the chain is attached to the
    40-ft.-high ceiling of the monkey's cage. Find the amount of work the
    monkey does in climbing up her chain to the ceiling.

    anwser is 1087.5 ft-lb

    can you figure it out??
     
  2. jcsd
  3. May 10, 2009 #2

    Cyosis

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    Re: Really hard intergration problem\!! take a look

    As the monkey climbs up the mass he has to pull up becomes greater and greater, because he now also needs to lift a piece of the chain. Lets call the height z. The chain is 10 ft longer than the height of the roof. So for the first 10 meters the monkey has to climb every foot he needs to lift an additional 0.5 lb. So what is the mass as a function of z for the first 10 ft?

    After those first 10 ft the entire chain is suspended in the air. From hat point on for every ft he climbs the length of the chain between the ceiling and the "bend" of the chain becomes shorter. Whereas the amount of chain between the the monkey and the "bend" becomes longer. Again you can express m in terms of z, but this time for every ft he has to climb there won't be 0.5 lb added but less. Think about it this way if the monkey reaches the ceiling both pieces of the chain will be 25 ft long. Now try to find both mass functions, after that its a simple integration over z.
     
  4. May 10, 2009 #3

    HallsofIvy

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    Re: Really hard intergration problem\!! take a look

    I certainly don't see anything difficult about it, except perhaps determining when there is part of the chain lying on the floor and when there is not. When the monkey is at height a ft., at which the chain just touches the floor, the total length of the chain will be 40+ a= 50 so a= 10. When the monkey is at height x, with x< 10, there will be x ft of chain hanging below the monkey so going up by distance dx require work of (20+ .5x)dx foot-pounds. The work done climbing to that point is
    [tex]\int_0^{10} (20+ .5x)dx[/tex]

    Once x> 10, there will be 40- x feet of chain above the monkey leaving 50-(40-x)= 10+ x feet below the monkey. Half of that hangs from the ceiling and half from the monkey so there will be 5+ (1/2)x feet of chain hanging from the monkey and so he will be supporting (5+ (1/2)x)(0.5) pounds of chain. Raising himself and that chain dx feet will require work of (20+ 0.5(5+ 1/2)x))dx= (22.5+ (1/4)x)dx foot-pounds. The work done climbing from 10 feet above the ground to the ceiling is
    [tex]\int_{10}^{40} (22.5+ (1/4)x)dx[/tex]
     
    Last edited: May 10, 2009
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