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Homework Help: Moving Pulley System Tension

  1. Jun 26, 2014 #1
    1. The problem statement, all variables and given/known data
    The diagram shows a system of two blocks suspended by ideal strings and pulleys.
    If the tension of the string attached to block m1 is T, what is the tension of the string attached to block m2?

    2. Relevant equations
    Due to conservation of string, a1/2 = a2
    Force balance on mass 1:
    ∑F1 = m1a1 = m1g - T

    Force balance on mass 2:
    ∑F2 = m2a2 = m2g - T2

    3. The attempt at a solution
    Substituting a1/2 = a2 into the ∑F2 expression, I get

    a1 = 2g - 2T2/m2

    Substituting the above into the ∑F1 expression, I get

    T2 = m2g/2 + m2T/(2m1)

    The correct answer is given as 2T. I'm struggling with this, because I don't see how the pulley could accelerate upwards if this is true, as all forces are balanced.
    Last edited: Jun 26, 2014
  2. jcsd
  3. Jun 26, 2014 #2
    The tension in the string is same since the string is assumed massless .It doesn't depend on the acceleration of the blocks .Since two strings are pulling block 2 upwards the upwards force on block 2 is 2T.
  4. Jun 26, 2014 #3


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    It may also be useful to note the tension across a frictionless pulley is uniform. So if you knew the tension on the left was ##T##, the tension in the two other strings is also ##T##. Since there is two of them, it is ##2T##.
  5. Jun 26, 2014 #4


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    That is true for the magnitude of the accelerations. But they are of opposite directions. If m1 moves downward, m2 moves upward.
    So you chose m1 moving downward.
    You have to change the sign of the right-hand side as m2 accelerates upward: m2a2 = T2-m2g

    The tensions also act on the moving pulley, 2T upward and T2 downward. But the pulley is massless, so ma =0. The net force on the pulley must be zero.

  6. Jun 27, 2014 #5
    Thank you to those that replied. I thought about this over night.... I think I'm getting caught up in the idea that the pulley is a real object, but in the problem we're asked to assume an "ideal pulley".

    How does this explanation sound:
    The moveable pulley (subscript P) must have the same acceleration as m2. The acceleration of the moveable pulley is:
    aP = FNet,P / mP = a2
    FNet,P = 2T - T2

    As long as m1 and m2 are constant, then a2 is constant and aP is constant.

    An ideal pulley is massless, so as mP --> 0 , FNet,P --> 0. The lighter the pulley, the smaller the difference between 2T and T2. Less net force is required to produce the same acceleration of a less massive object.

    This leaves us with

    T2 --> 2T for the idealized case where mP --> 0.
    Last edited: Jun 27, 2014
  7. Jun 27, 2014 #6


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    Yes. More generally, if an object is taken as massless then you can assume there's no net force on it (or its acceleration would be infinite). Similarly torques.
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