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Hard Friction/Energy Problem

  1. Nov 6, 2006 #1
    A uniform chain of 8.00 m lies stretched out on a horizontal table. 3m of it is hanging of the end of the table. Determine the speed of the chain as all of it leaves the table, assuming a coefficient of kinetic friction of 0.400.

    Here's what I have done so far:

    Let x represent the amount the chain has slipped off the table, so (5 - x) is the amount left on the table. U is the mass per meter. So the normal force is

    N = (5-x)U*g

    and the friction force is

    Fk = 0.4(5-x)U*g

    Conservation gives:

    U0 + K0 - Fk = U1 + K1

    K0 = 0 so

    m0*g*y0 + m1*g*y1 - integral_0_5(Fk) = mT*v^2/2 + mT*g*y

    And substituting the mass values and friction function:

    (5*U*g)y0 + (3*U*g)y1 - integral_0_5(0.4(5-x)U*g) = (8*U*g)v^2/2 + (8*U*g)y

    I think this is right, (please tell me if it isn't) but I can't figure out what to do about the heights, y0, y1 and y? :confused:

    y0 should just be the height of the table, but I don't know that, it's not given in the problem.

    And y1 should be the height of the dangling part of the chain from the floor. But at what point? The beginning, the end? Do I just take the average height (the middle of the chain? And what is the final y???

    I think most tables are about 4m high. And when I think about adding this to the problem, I just get that tired feeling :rolleyes:

    Thanks for any advice or help.

    Last edited: Nov 6, 2006
  2. jcsd
  3. Nov 6, 2006 #2


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    If you say the chain has zero potential energy on the table, then the potential energy as it slides off is increasingly negative. It would be the mass of the part of the chain hanging over times the negative distance of the center of that mass from the top of the table, or half the length of the overhang.
  4. Nov 6, 2006 #3

    Doc Al

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    Is the chain 8.00 m long? Or 5 m long? Where did that 5 come from?


    Don't know what happened here. What are all those heights? y0 and y1 make sense, but what is y? And all those masses? And what does T stand for?
    The height of the table doesn't matter. All that matters is the change in potential energy of the chain. Hint: How does the position of the center of mass change?

    Your overall idea of using energy methods is good:
    What's the change in PE of the chain? (Clean that up.)
    What's the work done against friction? (Your method looks OK.)
    What's the final KE of the chain? (That's what you'll determine.)
  5. Nov 6, 2006 #4
    So you are saying I don't need the mgy0 term at all?

    and y1 = -1.5, y = -4? I will try this, thank you very much.

  6. Nov 6, 2006 #5
    Last edited: Nov 6, 2006
  7. Nov 6, 2006 #6

    Doc Al

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    My bad. I didn't read carefully.

    I see your point about the chain piling up on the floor, but my guess is that you are to assume that at no point does the chain touch the floor. (If it does, things are different--energy will be lost as the chain hits the floor.) As long as the chain slides freely, all you need to worry about is the change in PE and the work done by friction.

    Assume it doesn't touch the floor.
  8. Nov 6, 2006 #7
    Ok, I think I have it, but I am still not sure about this center of mass thing, and the chain on the floor part, if I should do anything about that.

    E0 = 3Ug(-1.5) - Integral_0_5(0.4(5-x)Ug) dx

    E1 = 8Ugv^2/2 + 8Ug(-4)

    Am I right that this assumes a table that is at least as high as the length of the chain?

    When I solve this, I get v = 7.42 m/s, which seems reasonable.

    Does this seem right?

  9. Nov 6, 2006 #8
    Thanks Doc Al. I see you answered my question about the center of mass and the floor while I was typing.

  10. Nov 6, 2006 #9

    Doc Al

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    Looks good, except for that typo. (No "g" in the KE term.)


    I didn't check your arithmetic carefully, but it seems OK.
  11. Nov 6, 2006 #10
    Thanks again, to both you and OlderDan.

    Yes, that was a typo, sorry. Not my day (smile).

  12. Nov 7, 2006 #11


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    An interesting feature with this type of problem is to determine the least possible length that must hang down in order for the chain to begin sliding.

    Let [itex]\mu_{s}[/itex] is the coefficient of maximal static friction, L the total length of the chain, and [itex]x_{min}[/itex] the length of the piece hanging over the corner, [itex]\rho[/itex] the density of the chain and g the acceleration due to gravity.

    We must require that the tension in the chain is continuous; if we follow the vertical piece upwards, the tension must support the weight of the hanging part. At the corner then, the tension is: [itex]T=\rho{g}x_{min}[/itex].

    Following the horizontal part of the chain to the corner, the tension must balance the frictional force; thus, at the corner, we must have: [itex]T=\mu_{s}g(L-x_{min})[/itex]

    Thus, equating these expressions, we get:

    If we are to calculate the final speed v the chain will have when leaving the table in this case, we get:
    where [itex]\mu_{k}[/itex] is the coefficient of kinetic friction.
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