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Work, force, and distance problem

  1. Nov 17, 2007 #1
    Hi everyone,

    Could someone please tell me if this is correct?

    1. The problem statement, all variables and given/known data
    A chain lying on the ground is 10m long and its mass is 80kg. How much work is required to raise one end of the chain to a height of 6m?

    2. Relevant equations
    w=int. from a to b f(x)dx
    w=(force)(distance)

    3. The attempt at a solution
    force=(80kg)(9.8 m/s^2)
    =784N
    distance=10m
    w=784(10)
    w=7840 J

    Thank you very much
     
  2. jcsd
  3. Nov 17, 2007 #2
    No. Thats not right. You're only raising the chain by 6 m. And even so, you cant do it that way. The linear mass density of the chain is constant. So, say you raise the chain by a distance x, as x increases or decreases, the mass raised or lowered also varies. As the linear mass density is constant, you may easily find the mass raised or lowered and the work required to do so.

    Alternatively, you may find the position of the center of mass of the chain and find the height by which it has been raised. You will get the same answer. Be careful in what you take the length of the chain to be in this case.
     
  4. Nov 17, 2007 #3

    mjsd

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    what does this mean? are u actually lifting the entire chain or only part of it?
     
  5. Nov 18, 2007 #4
    Thank you very much

    Could you show me how to set it up, please?
     
  6. Nov 18, 2007 #5
    I am not following this very clearly. Would you do something like this:

    [tex]\frac{80kg}{10m} = 8 \frac{kg}{m}[/tex]

    and then:

    W = [tex]\int 8y dy 9.8 J[/tex]

    AAAAHHHHHHHHH!!!!!!!!!!!!!! THIS FORMATTING IS MAKING ME LOSE MY MIND!!!! IT IS RUINING MY HOUSEHOLD!!!!!!!!!!!!!!!!!!!!!!!

    I'm not understanding how you would do this one...
     
    Last edited: Nov 18, 2007
  7. Nov 18, 2007 #6

    nrqed

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    the starting point is correct [itex] W = \int f_y dy [/itex]
    The key point is that as you lift the chain, the force you apply varies (it increases as you pull up more and more of the chain). The force is [itex] f_y = m(y) g [/itex] where y is the height at which the extremity of the cahin is.. We have a variable m(y) here which represents the mass of the part of the chain that is above the floor. What you need to figure out is how m varies with y. That should be easy if we assume that the mass is uniformly distributed along the chain.
     
  8. Nov 18, 2007 #7
    If [tex]\lambda =\frac{m}{l}[/tex] is the linear mass density of the chain (8kg/m), then the weight of the chain a distance x above the ground is [tex]\lambda xg[/tex]

    To raise the chain a distance dx above the ground, the work done is [tex]dw=\lambda xgdx[/tex]. Integrating this expression from 0 to 6 gives you the required answer.
     
  9. Nov 18, 2007 #8

    nrqed

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    I had given all the information for the OP to figure out the last steps. I think that it is always better not to provide a full solution (and is actually against the policy of the website).
     
  10. Nov 18, 2007 #9
    Thank you very much

    Does this look right?

    F=80kg/10m=
    F=8kg/m

    d=(change in x)(xi*)

    w=int. o to 10 8xdx
    =400kg/m

    Thank you very much
     
  11. Nov 18, 2007 #10
    Isn't the distance 6m ?

     
  12. Nov 18, 2007 #11

    nrqed

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    You should not call this F. It's a linear mass density so we usually call this [itex] \lambda [/itex]
    I am not sure what this means
    No. First, your limits of integration are wrong. secondly, your result should have the units of energy which it has not. You di dnot use the correct formula for the integrand.
     
  13. Nov 18, 2007 #12
    Thank you very much
     
  14. Nov 19, 2007 #13
    Does 144kg/m look right?

    int. 0 to 6 (8x)dx=
    144

    Thank you very much
     
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