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Forces on a slab of mass Question

  1. Feb 27, 2008 #1
    1. The problem statement, all variables and given/known data
    A slab of mass [tex]m_{1} = 40 \mbox{kg}[/tex] rests on a frictionless floor and a block [tex]m_{2} = 10 \mbox{kg}[/tex] rests on top of the mass. Between the block and the slab, the coefficient of static friction is 0.60 and the coefficient of kinetic friction is 0.40. The block ontop of the mass is pulled by a horizontal force of 100 N. What are the resulting accelerations of the block and the slab?


    2. Relevant equations



    3. The attempt at a solution
    What I'm having trouble is identify the forces. My friend says that with the force being applied on [tex]m_{2}[/tex], there would be friction resisting the motion. Consequently, a force of the same magnitude of the mentioned friction force would occur on [tex]m_{1}in the direction of the applied force[/tex]. How does that work?
     
  2. jcsd
  3. Feb 28, 2008 #2
    The max value of static friction possible between the two is [tex]\mu_sm_2g[/tex] =0.60*10*10=60 N.
    Since the applied force is 100 N, definitely there will be slipping between the block and the slab and ultimately, they will get separated.

    Until they get seperated, Kinetic friction will act and both of them will have different accelerations.

    [tex]f_k=\mu_km_2g[/tex]=40 N acting on [tex]m_2[/tex] in the backward direction and on [tex]m_1[/tex] in forward direction.

    so [tex]a_1= [/tex]<< solution deleted by berkeman >>



    and [tex] a_2=[/tex]<< solution deleted by berkeman >>
    _________
    I may be wrong.
     
    Last edited by a moderator: Feb 29, 2008
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