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Homework Help: Pulley question

  1. Nov 1, 2007 #1
    1. The problem statement, all variables and given/known data

    http://img214.imageshack.us/img214/6531/pulleysystemph6.th.jpg [Broken]

    2. Relevant equations

    sum of forces

    3. The attempt at a solution

    So I tried summing the forces in the x-direction where

    m*a (block a) = -Ff + T = m(block a)*g*u + m*a(block b) + m*g(block b)

    where u is the coefficient of friction, T is the tension in the rope and Ff is the frictional force.

    I'm not, however, getting an answer on the multi choice list. Is there something I'm missing?
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Nov 1, 2007 #2
  4. Nov 1, 2007 #3


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    You've got to look at each block separately when doing these problems. This is essential. Draw a free body diagram for each. Identify the forces acting on each block and use newton 2. You'll have 2 equations with 2 unknowns that you can solve. Note that the magnitudes of the tension and acceleration for each block are the same.
  5. Nov 1, 2007 #4
    well i've been doing that but i cant seem to get the right answer.... i was wondering if maybe i was screwing something up with the directions...

    so far ive come to the following;

    for block A;

    Fx = T - Uk*N = mA*ax
    Fy = N - mA*g = 0

    So essentially, Fx = T - Uk*mA*g = mA*ax

    For block C;

    Fy = T - mC*g = mC*ay

    I understand ay = ax but i cant seem to get the right answer... i'm getting something around 1.6 :(
  6. Nov 2, 2007 #5


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    Oh, good, you are on track; only error you made was for block C, since it accelerates down, its weight must be greater than T, so your plus/minus signs are incorrect. It should be mCg - T = mC*ay. Do you see why?
  7. Nov 2, 2007 #6
    ahhh yeah of course. and since the tension is obviously greater in block A the signs are opposite to block C. thanks :)
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