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This should be really easy but

  1. Feb 10, 2008 #1
    [SOLVED] this should be really easy but..

    1. The problem statement, all variables and given/known data
    A chain is held on a frictionless table with one-fourth of its length hanging over the edge, as shown in the figure. If the chain has length L = 1.30m and mass m = 2.6kg, how much work is required to pull the hanging part back onto the table?


    2. Relevant equations



    3. The attempt at a solution
    ΔX=1.3*.25=.325
    mass=2.6
    g=9.8
    F=m*a
    W=FΔX
    W=(2.6)(9.8)(.325)=8.281J


    not the right answer
     
  2. jcsd
  3. Feb 10, 2008 #2
    should i not assume that the mass of the chain is evenly distributed over the length of the chain?
     
  4. Feb 10, 2008 #3

    Doc Al

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    Since the force is not constant, to use this method you'd need to integrate. (The integration is easy though.) Why not use energy methods instead? How does the gravitational PE change?
     
  5. Feb 10, 2008 #4

    Dick

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    Yes, you should. But you can't use F*dx to get the work, since F isn't constant. As more of the chain is pulled onto the table F decreases. So you want to write F as a function of x, the amount of chain left hanging, and then integrate F(x)*dx.
     
  6. Feb 10, 2008 #5
    so, integrate (2.6/1.3)dx from 0 to .325?
     
  7. Feb 10, 2008 #6
    wait no that would be if the mass wasn't distributed evenly, never mind.
     
  8. Feb 10, 2008 #7

    Dick

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    The overhanging mass is the density of the chain times x. What other factor do you need to get a force?
     
  9. Feb 10, 2008 #8
    using energy methods, i have W=mgy1-mgy2
    you're lifting from y1=-.325 to y2=0
    is the mass proportional to the chain or is it constant?
     
    Last edited: Feb 10, 2008
  10. Feb 10, 2008 #9
    The overhanging mass is the density of the chain times x. What other factor do you need to get a force?

    you need acceleration..
     
  11. Feb 10, 2008 #10

    Doc Al

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    Hint: Consider the part initially hanging. Where's the center of mass of that piece? How does it change when the chain is pulled up? (Alternatively, what's the average distance that the hanging mass elements must be raised?)

    The mass of any piece of chain is proportional to its length.
     
  12. Feb 10, 2008 #11

    Dick

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    You need g. Integrate g*density*x*dx from 0.325 to 0. Using energy is also instructive. See if you get the same thing.
     
  13. Feb 10, 2008 #12
    hm, integral of (9.8xdx) from .325 to 0 = .517, which is not what the answer i should get
     
  14. Feb 10, 2008 #13

    Dick

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    Put in the density. (I omitted that when I first posted).
     
  15. Feb 10, 2008 #14
    ah ok, i thought there was something missing, density = mass/volume --> 2.6/....?
     
  16. Feb 10, 2008 #15

    Dick

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    Linear density. (total mass)/(total length).
     
  17. Feb 10, 2008 #16
    never mind ---- got it.
     
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