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Force of friction with multiple coefficient of frictions

  1. Oct 14, 2006 #1
    I'm working on this problem for homework. Basically it's to find the work done by friction. Now for the work it would be the force of friction times the displacement. The problem is the board being pushed is going over 2 different surfaces, with 2 different values for friction, at the same time. Assuming velocity of the board is constant, what would be the force of friction? I thought it was the normal force times the sum of the coefficients (F_friction=N(mu_1+mu_2)) but apparently it's not.
     
  2. jcsd
  3. Oct 14, 2006 #2

    Doc Al

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    Staff: Mentor

    Interesting. Could you describe the problem exactly? (Why would you add the coefficients? Since any board can be thought of as two half-boards, your reasoning leads to a contradiction.)
     
  4. Oct 14, 2006 #3
    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 mu_1, and in region 2, the coefficient is mu_2.

    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, g, L, mu_1, and mu_2.

    EDIT: The board is pulled over the boundary exactly the distance L.
     
  5. Oct 14, 2006 #4

    Doc Al

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    Since the board is uniform, assume that the normal force per unit length is also uniform. Express the force that must be exerted on the board as a function of the position of the leading edge, which moves from x = 0 to x = L.
     
  6. Oct 14, 2006 #5
    So basically because the board is uniform, the normal force is uniform. So if i were to divide the board into 2 seperate boards then the value of the normal force would also be divided into 2? So since there are two different surfaces, it would be dividing the force by two eg. F_friction=Mg(mu_1+mu_2)/2?
     
  7. Oct 14, 2006 #6

    Doc Al

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    Right.
    That will be the average force of friction during its travel across the boundary. The actual force at any point depends on how much surface is on each side. For example, just as it begins to move all of the length (and thus all of the normal force) is on the mu_1 side, so F_friction = Mg(mu_1) at that point.
     
  8. Oct 14, 2006 #7
    Ohhh, so since the board is basically on each surface the same amount of time, the force of friction would be the average of the force of friction for each seperate surface:

    F_f1=Mg(mu_1)
    F_f2=Mg(mu_2)

    => (F_f1+F_f2)/2 = Mg(mu_1+mu_2)/2
     
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