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Friction in vertical directions

  1. Feb 16, 2012 #1
    1. The problem statement, all variables and given/known data
    Two blocks (m1 and m2) are being pushed to the right along a frictionless table with such a force that the left block (m1), which is smaller, is above the table and not falling. It is pushed up against m2 (which is on the table), and they have a coefficient of static friction of 0.4 between them. The mass of m1 is 16 kg and m2 is 80 kg. What is the minimum applied force from the left required to keep m1 from sliding down m2?

    2. Relevant equations
    fs≤usFnormal
    Fnet=ma

    3. The attempt at a solution
    I drew a force body diagram and found that since there is no vertical movement for m1, the net force is zero so the downward weight force must cancel with the upward frictional force of m2 on m1. Since the weight = 156.96 N, the frictional force must also. This is set equal to 0.4Fnormal to find that the minimum normal force required is equal to 392.4 N. Since this force acts opposite to the applied force, the applied force must be greater than or equal to this number. However, the system is not accepting my answer of 392.4 N. Can anyone point out to me what I did wrong?
     
  2. jcsd
  3. Feb 16, 2012 #2

    SammyS

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    Let's see if I understand this. An applied force pushes to the right on m1 which in turn pushes m2 to the right.

    If the normal force the blocks exert on each other is 392.4 N, then you can find the acceleration of m2. But m1 also has that acceleration.

    The applied force must be sufficient to give m1 + m2 that acceleration.
     
    Last edited: Feb 17, 2012
  4. Feb 17, 2012 #3
    Ahh I see, I ignored m2's opposing force. Thanks!
     
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