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Finding acceleration of two objects in a pulley system

  1. May 27, 2009 #1
    1. The problem statement, all variables and given/known data
    "In the diagram shown below, the lower block is acted on by a force, F, which has a magnitude of 74.4 N. The coefficient of kinetic friction between the lower block and the surface is 0.269. The coefficient of kinetic friction between the lower block and the upper block is also 0.269. What is the acceleration of the lower block, if the mass of the lower block is 3.27 kg and the mass of the upper block is 2.02 kg?"

    http://i4.photobucket.com/albums/y111/kathy_felldown/kn-pic0828.png

    I think I understand the system and the forces involved, and I was able to come up with the equations and substitute them, can someone just verify that my methods are correct?

    2. Relevant equations
    Fnet = ma


    3. The attempt at a solution
    For object m2, on the bottom:
    Fnet(x) = n2 - m2*g = 0 --> n2 = m2*g
    Fnet(y) = Fa - u*n2 - T = ma --> T = Fa - u*n2 - m2*a [EQUATION #1]

    For object m1, on the top:
    Fnet(x) = n1 - m1*g = 0 --> n1 = m1*g
    Fnet(y) = T - u*n1 = m1*a --> T = m1*a + u*n1 [EQUATION #2]

    Substitute m1*g for n1 and m2*g for n2, then....
    Equate equation #1 to equation #2 and isolate for acceleration, "a", so that:
    a = Fa - u*g(m2 - m1) / (m1 + m2)
    Since the two objects are connected in the pulley system, their accelerations will be the same.
     
  2. jcsd
  3. May 27, 2009 #2
    I agree with your Fnet(x)'s (where it seems you have x-axis running vertically).

    However, check your freebody diagram's again. There are two friction forces working an the bottom block (top and bottom). It looks to me like you left one out of the equation.
     
  4. May 27, 2009 #3
    Why would both friction forces be affecting the bottom block? Isn't it just the friction on the surface that the bottom block is sliding on (as in the ground) that affects its movement?
     
  5. May 27, 2009 #4
    It goes something like this: Newton said for every action there is an equal and opposite reaction (or something along those lines, too lazy to look it up).

    If a friction force is being generated on the top block by the motion of the bottom block, then the bottom block feels an equal but opposite reaction from the top block.

    Another way to think of it is to imagine sitting on the top block. You see yourself moving across the bottom block and friction force is being generated to oppose your motion. Now imagine sitting on the bottom block. Not only do you see yourself moving across the floor with a friction force opposing your motion, but you also see a top block moving on top of you also with friction opposing your motion.

    Hmm...now that I think of it, I think we both forgot that in the vertical direction on the bottom block you need to have the floor normal force upwards, the gravity force downwards, and, another normal force pointing upwards (EDIT: Ack downwards I mean - need sleep) that is equal and opposite to the normal force of the top block. The floor normal force should come out to be equal to the weight of both blocks. But right now in your equation, it only equals the weight of the bottom block.
     
    Last edited: May 27, 2009
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