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Kinetic friction and pulleys

  1. Oct 9, 2003 #1
    There are 3 blocks on a table. One of the blocks, block C, is in the middle of the table with mass 2M. There are two pulleys on the edges of the tables. The pulley on the left is attached to block C and is attached to block A (which has a mass of M) such that it is hanging off of the pulley and is along the side of the table. The pulley on the right is also attached to block C and is attached to block B (Which has a mass of 2M) such that it is hanging off of the pulley and is along the right side of the table. They are released from rest and then accelerate with a mag of 0.500 m/s^2. What is the coefficient of kinetic friction (Uk) between the sliding block (block C, I am assuming - since it is attached on both sides by the rope holding the other two blocks that are hanging from the pulley) and the table?

    I had 3 free body diagrams two were vertical (blocks A and B) and one was horizontal (block C)

    Ta
    |
    *
    |
    Mg

    Tb
    |
    *
    |
    2Mg

    Ta--*--Tb (with forces of friction going towards the left since the right side has a block with mass 2M)
    ** also from the block there is a N force up and the Fg force down the block


    Then I had:

    A) Ta - Mg = ma = 0
    Tb = ma + mg

    B) Tb - 2Mg = 2Ma = 0
    Tb = 2Ma + 2Mg

    C) Tb - Ta - f = 2Ma
    Tb - Ta - UkN = 2Ma
    2Ma + 2Mg - Ma - Mg - UkMg = 2Ma
    Uk = a/g + 1
    Uk = 0.5/9.8 + 1 = 1.05102

    Is my process correct?
     
  2. jcsd
  3. Oct 10, 2003 #2
    Well, we have a problem when you get a u (I'm going to use a "u" for the coefficient of friction because it's similar to the greek letter mu) greater than 1. If you think about that it means that friction (remember force-friction Ff=u*Fn force normal for either static or kinetic) is exerting more force than the normal force, in other words, if you put a block on a table like this then stood it on its end the block would stick to the table instead of falling, which doesn't make physical sense.

    Now looking at your equations A and B I see you set them equal to zero. But those blocks are moving, the B-block is falling and the A-block is rising and both have an acceleration a=0.5m/s^2 as does the C-block on the table. The string which is connecting them makes them move as a unit, not that string can push but gravity does the down and the string does the up.

    So the thing here is to draw the picture and pick a block to work with. Since we want the coefficient of friction we chose block C. Now block C has three important forces acting on it, the force of block B falling (as I presume B, the heavier block, is the one causing the systems acceleration) to the right, the force of block A pulling to the left, slowing the fall, and the force of friction from the table which goes to the left because friction always resists motion. So, F=ma, and block C is undergoing acceleration of a=0.5m/s^2 (as are all the blocks) to the right so ma=(2M)*(0.5m/s^2). And the forces sum (with those in the direction of motion positive and those opposed to motion negative) to F=Fb (force by block B) - Fa - Ff (force of friction). then F=ma -> Fb-Fa-Ff=(2M)*(0.5m/s^2) Solve for Ff=Fb-Fa-Mm/s^2

    Then substitute for the forces on the blocks, Fb would be the force due to gravity so Fb=2M*g and similarly Fa=M*g so Ff=2M*g-M*g-Mm/s^2=M*g-Mm/s^2=M(9.8m/s^2 - 1m/s^2)=8.8*Mm/s^2

    Now kinetic friction relates to forces by uk=Ff/Fn so Ff=uk*Fn and we have from above Ff=8.8*Mm/s^2 so uk*Fn=8.8*Mm/s^2 and remember we're talking about block C here, all these forces were acting on block C so Fn=2M*g giving us uk*2M*g=8.8*Mm/s^2 -> uk=(4.4/g)m/s^2=.449 and the units cancelled which is good and the number is less than one which is good and the mass variable M cancelled which is good.
     
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