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B Tension in a basic pulley system

  1. Jun 8, 2017 #1
    I've got a problem with understanding the pulley system.

    In a basic pulley, if the force pulling M upwards is T, and the force pulling m upwards is also T, why is it that T is equal for both m & M? For some reason I feel inclined to believe that the T pulling m upwards is equal to the weight of M, and the T pulling M upwards is equal to the weight of m. Is that right or wrong? Why?
  2. jcsd
  3. Jun 8, 2017 #2
    I believe this relates to Newton's third law. If mass m is pulling on mass M with a certain force, M must be pulling on m with the same magnitude since the tension in the rope must be constant (ignoring friction).

    Imagine an object hanging from the ceiling. The mass of the object itself would influence the magnitude of the upward force which is being exerted on the mass by the rope (the tension). This is because the magnitude of the force which the mass applies on the ceiling is the same as that which the ceiling applies on the mass. The same should hold true for the pulley (the force which m applies on M is the same magnitude as that which M applies on m).

    I hope all of this was both correct and clear.
    Last edited: Jun 8, 2017
  4. Jun 8, 2017 #3


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    If you fixed the top of the string, then you would have equilibrium. The tension on the right would support mass M and the tension on the left would suppose mass m.

    However, if the string is not attached to anything (and there is no friction on the pulley), then the tension must be constant throughout. In this case, assuming mass M is larger, it will fall and pull the lower mass m upwards.
  5. Jun 8, 2017 #4
    This is correct, provided that the pulley is massless. If there is mass, and more specifically a mass moment of inertia for the pulley, the tensions will not be equal on both sides.
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