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Friction of an object on a moving board

  1. Apr 22, 2012 #1
    1. The problem statement, all variables and given/known data

    A body of mass [itex]m_A=2 kg[/itex] is placed on a long board of mass [itex]m_B=8 kg[/itex] at distance [itex]d=1 m[/itex] from the rear edge of the board. The friction coefficient between the body and the board is [itex]μ=0.2[/itex]. A force of magnitude [itex]30 N[/itex] is applied to the front edge of the board and the body start moving towards the rear edge. How much time will it take to fall off the board?


    2. Relevant equations

    The force of friction is given by [itex]F_f=F_n\cdot μ[/itex] where F_n is the normal force exerted by the object on the surface.


    3. The attempt at a solution

    I tried to write down Newton's equation of motion (on the x-axis) for the body and the board as follows:

    Board: [itex]F=(m_A+m_B)a_1[/itex]

    Body: [itex]F\frac{m_A}{m_A+m_B}-F_f=m_A a_2[/itex] where [itex]F_f=m_Agμ[/itex].

    With these equations the problems doesn't come out right...
     
  2. jcsd
  3. Apr 22, 2012 #2

    Doc Al

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    Staff: Mentor

    What forces act on the body? Apply Newton's 2nd law.
    What forces act on the board? Apply Newton's 2nd law.

    (Don't treat 'board + body' as a single system, since parts are in relative motion.)
     
  4. Apr 22, 2012 #3
    On the board: [itex]F[/itex] (and the weight of [itex]A[/itex] which is equilibrated by the board itself).

    On the body: The force exerted by the board on the body, directed along the direction of [itex]F[/itex] and of a "certain" magnitude and the friction force.
     
    Last edited by a moderator: Apr 22, 2012
  5. Apr 22, 2012 #4

    Doc Al

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    All we care about are the horizontal forces, since vertical forces will cancel.
    You are missing the horizontal force of the body on the board. (Newton's 3rd law.)
    The only horizontal force on the body is the friction force from the board.
     
  6. Apr 22, 2012 #5
    So the body exerts a horizontal force [itex]F_f[/itex] on the board in the opposite direction of [itex]F[/itex]? Then Newton's 2nd law for the board is
    [itex]F-F_f=m_Ba_1[/itex]
    and the Newton's law for the body would be:
    [itex]F_f=m_Aa_2[/itex]?
     
  7. Apr 22, 2012 #6

    Doc Al

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    Staff: Mentor

    Right. And you also know how to calculate the friction force.
     
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