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Friction force pair between two stacked blocks on a frictionless surface

  1. Jun 5, 2012 #1
    Good afternoon!

    1. The Setup

    One block of Mass M2 is resting on top of a frictionless surface. Another block, M1, is resting on top of the of the first block. There does exist a coefficient between these two blocks, μ.

    2. What happens to the bottom block when a force less than μN is applied to the top block, pushing it to the right?

    3. Attempt at solution

    Free body diagram for M1:

    F to the right
    fs to the left
    M1*g downward
    Normal force, N1, upward

    Free body diagram for M2:

    fs to the right
    M2*g downward
    Normal force, N1, downward
    Normal force, N2, upward

    For M1, F<fsmax, so the force of friction cancels out the applied force. No x direction movement. The weight of the mass is balanced by the normal force from the block below. No y movement.

    For M2, the weight of the mass + the normal force from the block above is being balanced by the normal force from the frictionless surface. No y movement. The frictional force that pushes this block to the right has no pair. Acceleration of the lower block to the right is fs/M2.

    This does not seem to compare favorably with my intuition or my real world experiences. Can someone please care to explain what I am missing here, or would a force less than μN1 applied to the top block cause the bottom block to move to the right?

    The only thing I can think of is that the frictional force on the bottom block essentially becomes a new applied force, and then another frictional force between the two blocks is created that pushes the bottom block to the left and the top block to the right. This pattern repeats itself until the frictional force reaches 0.

    Anyone else got any input? I don't have my intro physics textbook with me currently so I cannot see if there was anything stated in there that applies to this question.

    Thank you all!
     
  2. jcsd
  3. Jun 5, 2012 #2
    I am not sure you can say that with absolute certainty.
    What you can say is that since the two blocks will not slide against one another until fs>fmax, both blocks should stay together and experience the same acceleration a.

    What horizontal forces on your free body diagram are there so that you can write a formula for the acceleration ( ie basic equation is [itex]\sum[/itex]F=ma )
    and see where that gets you.

    You might then want to take another step and try to determine what happens when the force is sufficient enough to cause the two blocks to slide against one another.
     
  4. Jun 5, 2012 #3
    In solving the free body diagram for the top mass, I stated in the introduction that I was assuming an applied force of less than the static friction maximum. In this case, the force of static friction takes on the same magnitude as the applied force. This leaves the top block with no net force and therefore 0 acceleration.

    However, for the bottom block, there is a net force, the reaction friction force to the first block. This means this block should accelerate in the x direction to the right.

    I do not see what is wrong with this setup. Is it possible that the impossibility of a frictionless surface is what creates this supposed paradox?
     
  5. Jun 6, 2012 #4
    Wrong assumption. fs [itex]\neq[/itex] F
    Read the my previous post and do the analysis knowing that there will be movement.
    You cannot analyze a dynamic situation as if it static.

    Maybe it will more clear with the following:
    Two blocks M1 and M2, one behind the other, are acted upon by a force F pushing on block M1. Describe what happens.

    Truly we know that the two blocks accelerate and can be described by the formula,
    F = (M1 + M2) a

    Carrying on your analysis of the two blocks and a frictionless surface the force F on block M1 would be conteracted by a force F in the opposite direction and M1 would not move. M2 on the other hand would have the mating reaction paired force F at the contact between the blocks and fly with acceleration twice as great described by F = M2 a, leaving M1 behind sitting motionless.
    You can see why that would not be correct.

    There is no paradox if you apply F - fs = M1 a ;for M1
     
  6. Jun 9, 2012 #5
    I finally understand the situation. Thanks for the great explanation!
     
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