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Finding acceleration with ideal pulley

  1. Oct 2, 2014 #1
    1. The problem statement, all variables and given/known data
    Two blocks are connected by a string as in the figure. (below)
    What is the upper block's acceleration if the coefficient of kinetic friction between the block and the table is 0.18?
    Express your answer to two significant figures and include the appropriate units.

    Bybw6lz.png

    2. Relevant equations
    F = ma
    w = mg
    Since this is an ideal pulley, the tension in each part of the string should be the same.

    3. The attempt at a solution
    Let's call the tension in the string T. T should be equal in both parts of the string since this is an ideal pulley (we have not covered problems with friction in the pulley yet).

    Since the upper block is not moving in the y direction, its net force in the y direction is 0. More specifically, its weight force is being balanced by its normal force, which are both 19.6 N.
    w = mg = 2.0kg * 9.8 m/s2 = 19.6 N.

    Since the weight of the lower block is acting on the string, the weight force of the lower block is the tension in the string because it is moving solely in the y direction.
    Therefore, T = mg of lower block = 1.0 kg * 9.8 = 9.8 N.

    Next, the upper block is moving solely in the x direction because its y forces sum to zero. Therefore:
    Fx = max = T + (-fk).

    We are given the coefficient of kinetic friction as 0.18, and since we know the normal force on the upper block to be 19.6 N, we can algebraically calculate the frictional force.
    0.18 = fk / n --> fk = 3.53 N.

    So, the sum of the forces in the x direction for the upper block = T + (-fk) = 9.8 N - 3.53 N = 6.27 N.

    Using this value, we can algebraically solve for the upper block's acceleration:

    F = max
    max = 6.27 N.
    (2.0 kg) * ax = 6.27 N
    ax = 3.136 m/s2, rounded to 2 sig figs is 3.1.

    The homework system says this is not the correct answer, so I don't know where to go from here.

    Thanks for any guidance.
     
  2. jcsd
  3. Oct 2, 2014 #2

    PhanthomJay

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    you didn't account for the fact that the lower hanging block is also accelerating, so T is not equal to mg. Draw free body diagrams for each block before solving.
     
  4. Oct 3, 2014 #3

    ehild

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    This is not true. Both the weight and the tension act on the lower block and it accelerates, so the tension and the weight does not balance each other.

    ehild
     
  5. Oct 3, 2016 #4
    Try this:

    Only use the external forces acting on the system that are parallel to the motion.
    Only use the friction to the left and the weight of the 1kg mass. These are the only two forces that push/pull the system in its direction of motion.

    (Weight 1kg) - (friction) = (total mass)a
    (1kg*10m/s^2) - (.18*2kg*10 m/s^2) = (3kg)a
    a = 2.133 m/s^2
     
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