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Gragging a board across surfaces with different frictions

  1. Mar 6, 2007 #1
    "A uniform board of length L and mass M lies near a boundary that separates two regions. In region 1, the coefficient of kinetic friction between the board and the surface is (mu1), and in region 2, the coefficient is (mu2). The positive direction is from the region with mu1 to the region with mu2. Find the net work W done by friction in pulling the board directly from region 1 to region 2. Assume that the board moves at constant velocity. Express the net work in terms of M, L, g, mu1 and mu2"

    In my force formula, x is equal to the amount of the board in region 2 (with mu2).

    F = mu1(mg(1-(x/L))) + mu2(mg(x/L)

    I know I'm supposed to integrate this with respect to L (or am I). Unfortunately, I suck at calculus. This is what I got, sorta.

    ((1/2)((mu1*g)/L)L^2) + ((1/2)((mu2*g)/L)L^2

    Again, I suck at calc. Any thoughts?
     
  2. jcsd
  3. Mar 6, 2007 #2

    Dick

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    You should be integrating with respect to x (not L as you said). x going from 0 -> L. And you are doing pretty well. But you forgot to integrate the 1 in the first term of your force expression.
     
  4. Mar 6, 2007 #3
    ((1/2)(x-((mu1*g)/L))L^2) + (1/2)((mu2*g)/L)L^2

    the 1 should just become an x, no?
     
  5. Mar 6, 2007 #4

    Dick

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    Evaluate the x between 0 and L. Just like you did the x^2 terms. And your x should be inside of the m*g*mu. And OUTSIDE of the 1/2. Do it again. Carefully this time.
     
    Last edited: Mar 6, 2007
  6. Mar 6, 2007 #5
    well, justguessing, but:

    [((mu1+mu1)Mg)/L]*(1/2)L^2
     
  7. Mar 6, 2007 #6

    Dick

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    I don't think you're guessing. You can cancel one of the L's too. So you just average the two mu's. Makes sense, right?
     
  8. Mar 6, 2007 #7
    yeah, the board spends the same amount of time in each region so that does make sense.
     
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