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Work required to pull a hanging chain

  1. Oct 3, 2005 #1
    A chain is held on a frictionless table with one-fourth of its length hanging over the edge. If the chain has length L= 28 cm and mass m=0.012 kg, how much work is required to pull the hanging part back onto the table?

    I have used this model: W horizontal + Work due to gravity = Work

    Wg = (0.012 kg)(9.8 m/s^2)(.07 m)(cos 180)
    = -5.0 * 10^-3 J

    Then I used the work-kinetic energy theorem (change in K = W) because we do not know anything about the force applied to the chain. I get:

    change in K = W due to the horizontal force + Work due to grav. + Work due to normal force

    change in K is zero, because the object is stationary before and after the pull, so...

    0= (W due to horizon. force) - (5.0 * 10^-3) + (5.0 * 10^-3), so...
    Work due to force = 0

    I just get zero work for the horizon. force. Is the only work done here the Work due to gravity? Should I just leave it at -5.0 * 10^-3 J, if that answer is even correct?? Hmm....

    Thank you,
  2. jcsd
  3. Oct 3, 2005 #2

    Doc Al

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    Staff: Mentor

    The only work required is the work done against gravity. Hint: What's the change in gravitational PE of the chain when it's raised from the initial hanging position to be totally on the table?
  4. Oct 3, 2005 #3
    Okay...thanks Doc...I think I am beginning to understand. With that in mind, I used:

    change in potential energy = -W

    change in PE = mg(y-yinitial)

    change in U = (0.012 kg)(9.8 m/s^2)(0.07-0 m)
    change in U = .008232 J, or 8.2 * 10^-3 J

    So Work would be -8.2 * 10^-3 J.

    I appreciate your feedback. Hope I am on the right track now!
  5. Oct 3, 2005 #4

    Doc Al

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    Some comments. The only part of the chain that changes its PE is the part that is hanging off the table. What is the mass of that piece? How high must it be raised to get up to the table? (Note that each segment of the hanging section is a different distance from the table top.)
  6. Oct 3, 2005 #5
    Hmmm..do you suggest that I calculate work for each little piece of chain, and then add them all together? For instance, (.012kg/.28m) = .0429kg/m. Then calculate for this mass at different distances, such as

    change in U = (.0429kg/m)(9.8m/s^2)(.07-0m) = .0294 J, then
    change in U = (.0429kg/m)(9.8m/s^2)(.06-0m) = .0252 J, and so on...

    ...to arrive at 0.1176 J, or W = -0.1176 J.

    You are helping me to think about this in a different way...thank you. I just hope that I get into the right frame of thinking! I sure appreciate it.
  7. Oct 3, 2005 #6

    Doc Al

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    That's the right idea, but you are off by a factor of 100! I assume you are imagining the chain in 0.01m (not 1m) pieces.

    That method will give you an approximate answer. To get the exact answer, you'll have to integrate.

    There's an easier way: Consider the change in the height of the center of mass of the hanging piece.
  8. Oct 3, 2005 #7
    So if use
    m1x1 + m2x2 + m3x3 + etc.../M

    And plug in (.0429kg)(.07m) + (.0429kg)(.06m) +...etc. divided by 0.012 kg,

    = 0.99167m

    This will give me the location of the center of mass...and then I can use this to calculate the work?? I don't know where to go from here.

    What does this mean???:
    Consider the change in the height of the center of mass of the hanging piece.
  9. Oct 4, 2005 #8

    Doc Al

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    Well you could find the center of mass that way (at least approximately) if you do it correctly. If you are taking 1 cm long pieces, what is the mass of each piece? (It's not .0429 kg--that's more mass than the whole chain!) And what's the mass of the hanging piece? Remember, you are trying to find the center of mass of just the hanging piece.

    But there's a much easier way to locate the center of mass of the hanging piece. Think about it. You have a uniform chain that's 7 cm long hanging vertically. Where do you think the center of mass is?

    If you want to find the work done against gravity to raise an object, find the change in its gravitational PE. And that only depends on the location of the center of mass.
  10. Oct 4, 2005 #9
    Alright, thanks Doc! Check this out:

    (1) Center of Mass Method

    m=0.012 kg/.28 m = .0428kg/m*.07m = 0.003 kg. We'll call this m1, as it is the mass of the 0.07 m hanging off the table.

    change in U = (0.003 kg)(9.8 m/s^2)(.07m/2) <----(center of mass)
    change in U = 1.029*10^-3 J, so W is -1.029*10^-3 J.

    (2) Integration Method

    We get the same answer if we integrate:
    change in U = [integral] mgy dy
    change in U = mg (y^2/2)
    density(m) = (0.012 kg/.28 m)
    change in U = (1/2)(0.012 kg/.28m)(9.8 m/s^2)(.07 m)^2
    change in U= 1.029*10^-3 J, therefore W = -1.029*10^-3 J

    BTW, what program are ya'll using to post integrals and other notations?? I'd like to learn how to use it. Thanks a lot -
  11. Oct 5, 2005 #10

    Doc Al

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    Staff: Mentor

    Careful. The center of mass is not at 0.07m below the table, but 0.035m below! (But you got the right answer somehow. :smile: ) Also: Be careful of the sign of the work: The work required to lift the chain is positive, since the force and the displacement are in the same direction (up). (The work that gravity does is negative.)
    It's called Latex and it's easy to learn. (At least the basics are easy.) Learn about it here: https://www.physicsforums.com/showthread.php?t=8997
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