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Equal tension on rope in a pulley system?

  1. Apr 2, 2014 #1
    Suppose the rope involved in the pulley is massless and experience no friction.

    Why is the force on the right hand side equal to the tension experienced by the rope on the left hand side?
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  3. Apr 2, 2014 #2

    Meir Achuz

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    Apply Newton's second law, F=ma, with m=0.
  4. Apr 2, 2014 #3


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    Newton's second law.

    If the rope is massless then each tiny section of the rope must be massless. The net force on that tiny incremental section [in the tangential direction] is equal to the difference in tension, if any, between the two ends of the section.

    But, by Newton's second law: f=ma. We know that the mass of the section is zero. So the net force must be zero. So the tensions at each end must be identical.

    Apply this logic incrementally from one end of the rope to the other. The tension must be the same throughout the length of the rope.

    [Edit: I see that Meir Achuz beat me to it]
    Last edited: Apr 2, 2014
  5. Apr 2, 2014 #4
    That explanation is for a massless rope. You also can consider the same section of massless rope described above but with friction from contact with the pulley acting on it. In this case there would be a difference of tension in the rope when the pulley is accelerating.

    Since the rope is frictionless as well as massless then this is also the reason that forces are the same on both sides of the pulley.
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