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Homework Help: Two bloack stack on top, find max force to prevent skidding

  1. Aug 20, 2010 #1
    1. The problem statement, all variables and given/known data

    A block of mass m rests on top of a block of mass M, which in turn rests on a horizontal surface. The coefficient of static friction between m and M is mu-1 and the coefficient of kinetic friction between m and M is mu-2. The coefficient of static friction between M and the surface is mu-3 and the coefficient of kinetic friction between M and the surface is mu-4. What is the maximum horizontal force you can apply to block M such that blocks m and M move at constant velocity and m does not skid off M.

    2. Relevant equations

    Newtons 2nd and 3rd law..

    3. The attempt at a solution

    Well, i solved the problem but i am still not convinced my answer is right.. I am assuming the problem i have is with newtons 3rd law (deciding whether the static friection should point right or left). Here is my answer.

    First, i made a free body diagram from the top block (m). The force down is mg, therefore the upward normal force is the force block M exerts on block m. I put the force of static fricition point left, because without fricition, nothing will keep the block moving to the left. Therefore, the force block M exerts on m (x-direction) is to the right.

    F (M on m in y) = mg UP
    F (M on m in x) = (mu-1)mg RIGHT

    Now going to the FBD for the large box,

    Obviously, F points left, force of kinetic friction points right, Mg points down, and n points up. Now using N3L,

    F (m on M in y) = mg DOWN
    F (m on M in x) = (mu-1)mg LEFT

    Solving this i am left with

    F(max) = (mu-4)(M + m)g - (mu-1)mg

    But conceptually, i would think increasing mu-1 should increase the F max..... so i doubt my answe is right.
  2. jcsd
  3. Aug 20, 2010 #2


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    Homework Helper

    You assumed the force F pointing to the left, did you?

    The boxes interact through the static friction. If the force of static friction on m points to the left why is the force of the big box on m is to the right? The x-force of the big box is the same as the force of static friction.

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