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Work done lifting a chain

  1. Oct 25, 2009 #1
    1. The problem statement, all variables and given/known data

    A chain 40 feet in length and weighing 3 pounds per foot is hanging fully extended from a winch. Find the work done by the winch in winding up 30 feet of the chain.

    2. Relevant equations

    Typical segment of the chain: [tex]\Delta y_{i}[/tex]

    Weight of typical segment of the chain: [tex]3\Delta y_{i}[/tex]

    Distance of typical segment being lifted: [tex]30 - c_{i}[/tex], where [tex]c_{i}[/tex] is a point in the interval [tex]\Delta y_{i}[/tex]

    Work done in lifting typical segment: [tex]W = Fd = 3(30 - c_{i})\Delta y_{i}[/tex]

    3. The attempt at a solution

    Total work done in lifting chain: [tex]3\int_0^{30}(30-y)dy[/tex]

    I know this is wrong, I just don't know why.
     
  2. jcsd
  3. Oct 25, 2009 #2
    You need to write an equation for the total hanging mass as a function of how much mass has already been pulled in. Then the force of gravity on this hanging mass is the force that you're doing work against. You're close, but not quite there.
     
  4. Oct 25, 2009 #3
    Also, you're given pounds and feet, make sure you're not expected to use Newtons and Meters.
     
  5. Oct 25, 2009 #4

    Delphi51

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    Homework Helper

    If you take y as the distance down from the winch,
    dW = Fd = 3*dy*y
    Integrate that and you get the right answer, 1350.
    Oh, you get the same answer with your integral!
    It is right because in effect you are lifting the center of mass up 15 feet and W = Fd = 3*30*15 = 1350.
     
  6. Oct 25, 2009 #5
    There's something missing here. You didn't account for the extra mass on the remaining 10 feet that wern't winched up. Immagine the chain were 100 feet long and you winched up 30 feet. This is still mass that would need to have been lifted.

    This means the origional integral is correct, excecpt that inside the integral should be 40-y rather than 30-y
     
  7. Oct 25, 2009 #6

    Delphi51

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    Homework Helper

    Thanks flatmaster! I completely missed that 40 in the question.
     
  8. Oct 25, 2009 #7
    Upon revising, I ended up with

    [tex]900 + 3\int_{10}^{40}(30 - y)dy[/tex]

    Does this look a little better?

    (Also note that the assignment is to find the integral, but not to work it out.)
     
  9. Oct 25, 2009 #8
    Your first integral was better. Your entire integrand is the total leingth of chain remaining after a leingth of chain y has been pulled in. You simply needed to replace the initial leingth with 40 m instead of 30 m.
     
  10. Oct 25, 2009 #9
    OK, so

    [tex]\int_0^{30}(40 - y)dy[/tex]
     
  11. Oct 25, 2009 #10
    That result really makes no sense to me. I can figure out the math, I guess, but I can't see the physics in it.
     
  12. Oct 26, 2009 #11
    You lost the 3 you were multiplying by for density. Immagine the first bit of chain you pull up. y=0, so the integrand contains the entire 40 feet of cable. This is multiplied by the density of 3plb/ft to give you the total hanging mass of cable. Integrating from 0 to 30 corresponds to winding in the 30 feet. The leingth of cable left hanging changes continuiously as 40-y where y is the amount of cable that's been drawn in. Thus you integrate y from 0 to 30.
     
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