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Tension in a nonmassless rope

  1. Jul 30, 2014 #1
    When 2 unequal vertical forces act on the 2 ends of a vertical rope, how do I calculate the tension at the midpoint?
    First I calculated the net acceleration of the rope, then I choose any 1 half of the rope for fbd of it and then where and how does the tension act?
  2. jcsd
  3. Jul 31, 2014 #2
    The tension of the rope acts inside the rope. If you cut the rope, you will find the tension, acting outwards in either of the cut ends, as seen in the picture. Then, once you know that you can calculate the tension using any one half of the ropes. In the example, lets say the rope has mass 1 kg. Then, the weight of the role is 1 (10) = 10 N, acting at the center of mass of the rope, that is halfway up or down, i.e. in the middle, not shown on the diagram. Then, for acceleration a, the equation is, Fnet = ma. ∴ 30-10-10(weight of the rope) = 1 (a). This gives a = 10ms-2. Now, taking the upper half, which has half the mass, and again applying the equation, we have 30-T-5 (weight of the upper half now acting downwards at the center of the half) = 0.5 (10). Thus, T = 20 N. You will arrive at the same answer, if you work on the bottom half of the rope. Hope it helped. :smile:

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