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Chain Falling on a Scale

  1. Sep 16, 2009 #1
    1. The problem statement, all variables and given/known data

    Hi guys, I have a question on problem 4.11 in Kleppner and Kolenkow's mechanics book. A chain of mass M and length L is suspended vertically with its lowest end touching a scale. The chain is released and falls onto the scale. What is the reading of the scale when a length of chain, x, has fallen?

    2. Relevant equations

    I will denote the chain's linear mass density by r = M/L.

    3. The attempt at a solution

    Okay, there are only two forces acting on the chain: the normal force of the scale and the gravity force due to its mass. The normal force N is what we're interested in; we want it as a function of position. I will write Newton's second law for the chain:

    r (d/dt)[(L - x)(-x')] = N - rLg

    The left hand side is the change in vertical momentum of the chain; the right side is the normal force and weight. Note that x' = dx/dt. This rewrites as:

    N/r = Lg - Lx'' + (x')^2 + xx'' (equation 1)

    Now, let's use conservation of energy on the chain. The normal force at the bottom does no work, but the conservative gravitational force does. We get (equating the energy at chain fallen x to the energy at the beginning of the fall):

    (1/2)r(L - x)(x')^2 + r(L - x)g(L - x)/2 = rLg*L/2

    Simplifying,

    (x')^2 + g(L - x) = g(L^2)/(L - x) (equation 2)

    I will now take the time derivative of each side to get:

    x'' = (g/2) + (g/2)(L^2)/[(L - x)^2] (equation 3)

    Now I substitute equations 2 and 3 into equation 1 to come up with the expression for the normal force on the chain. After some algebra, I get:

    N = rg[(-L/2) + (3/2)x + (L^2)/(2L - 2x)]

    which is not the answer. Am I messing up Newton's second law, the energy conservation, or my algebra after I plug things in? Thanks in advance for the help.
     
  2. jcsd
  3. Sep 16, 2009 #2

    turin

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    Homework Helper

    x''=g should be true, shouldn't it?
     
    Last edited: Sep 16, 2009
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