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Dynamics problem: Rod on a rough surface

  1. Dec 30, 2011 #1
    Not exactly homework, but this is a problem that a guy posed me.. and we seem to be getting different answers. I'm not entirely sure where either of us is going wrong.

    The problem statement, all variables and given/known data
    A uniform rod of mass M and length L lies on the x-axis, with its ends at x = 0 and x = L. A varying force of magnitude F is pulling the rod in the right direction, such that the rod is moving with constant velocity. The floor beneath the rod is rough only in the region from x = 0 to x = L, with the coefficient of friction between the rod and floor being 'kx', where k is a constant and x is the x-coordinate of the point. Find the magnitude of the work done by the frictional force in moving the entire rod out of the rough region.

    The attempt at a solution
    Here's what I came up with:
    I take a length 'dx' of the rod at a distance 'x' from the origin. Its mass is (M/L)dx, force on it [(M/L)dx * gkx], and work done by friction on it [(M/L)dx * gkx * (L-x)]. I integrate x from 0 to L to get [itex] \frac{kMgL^{2}}{6} [/itex].

    A person who gave me the question has this solution:
    At an instant, let the left end of the rod have an x-coordinate "y". Consider a part of the rod with coordinate "x" of length dx. Force on it:[itex] \int \frac{Mgkx(dx)}{L} [/itex] from "y" to "L". Then dW = (That integral) * dy, and he then integrates from 0 to L to get [itex] \frac{kMgL^{2}}{3} [/itex].

    I think his solution is wrong because he's assumed a constant force to be displacing the entire rod by "dy" to get "dW", which won't make sense because different forces are acting on each part of the rod. I'm not fully confident about my own solution either, so it'd be helpful if someone could clarify.
  2. jcsd
  3. Dec 30, 2011 #2


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    Homework Helper
    Gold Member

    The work you got is wrong. The force changes while a tiny piece of the rod moves from x to L. You can not use the same "dx" for the length of a piece and for the displacement of the same piece.
    You have a piece of length δl situated at x=l initially. The work δW done by friction when the piece moves from x=l to x=L is
    [tex]δ W= \int _l^L{(kMg\delta l/L) xdx}=(kMg\delta l/L)\frac{L^2-l^2}{2}[/tex]

    The total work is obtained by integrating for all pieces from l=0 to L.
    [tex]W=(kMg/L)\int _0^L{\frac{L^2-l^2}{2}dl}[/tex]

  4. Jan 1, 2012 #3
    Great, thanks a lot, I knew I was going wrong somewhere.

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