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Homework Help: Force required to keep one block from sliding down another

  1. Sep 18, 2010 #1
    1. The problem statement, all variables and given/known data

    A block (m1) of mass 2.172 kg is placed in front of a larger block (m2) of mass 5.753 kg in such a way that only m2 is in contact with the table (m2 is on the right, and m1 is on the left. there is negligible friction between m2 and table). The static friction between m1 and m2 is fs=.686

    What is the minimum amount of external force applied to m2, to keep m1 off of the table (ie. not sliding down m2)?

    2. Relevant equations

    In my free body diagram for m1, i have mg pointing down, static friction force pointing upwards since the block would slide down if it moves, and a normal force pointing left (perpendicular out from the other block; m2)


    3. The attempt at a solution

    First, I concluded that ignoring m1, to move m2 across the frictionless table i would use Newton's 2nd Law.

    F= 5.753 kg * 9.81m/s^2
    F= 56.437 N

    Next, to also move m2 and keep it from sliding, I would need to add that additional force. To find that value, I concluded that you would need the force that it would take to move m1 by itself but without the friction force since that is doing some of the work holding the block in the air.

    So I calculated F=ma = 2.172 kg * 9.81m/s^2 = 21.307N

    I then subtracted the friction force: 21.307 N - (21.307 N * .686) = 6.69N

    I then added that to 56.437 N to get 62.12 N. This is incorrect, but I'm not sure where I'm making my mistake.

    Any help is appreciated.
  2. jcsd
  3. Sep 19, 2010 #2


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    Science Advisor
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    Gold Member

    Stop here! Force F is in the horizontal direction and the two blocks accelerate together across the table. Why do you say that,
    1. You can ignore the mass of one of the blocks? Both blocks are accelerating.
    2. The acceleration is 9.8 m/s2? Neither block is in free fall.

    It is correct that the static friction is up. That force must be just enough to balance the weight of block m1. Correct application of Newton's Second Law requires you to write
    F = (m1+m2)a, where a is some acceleration yet to be defined.
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