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Homework Help: Tension in string with ball

  1. Nov 7, 2006 #1
    Hi, Can someone please explain how to do this problem? I really dont understand.

    A ball is held at rest in position A as shown in the figure by 2 light cords. The horizontal cord is cut and the ball swings as a pendulum. What is the ratio of the tensions in the supporting cord, in position A, to that in position B?

    http://img445.imageshack.us/img445/3927/lomprobtensionvx9.png [Broken]

    thanks for your time.
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Nov 7, 2006 #2


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    I assume they want you to compare the tension at A before the horizontal cord is cut to the tension at B. Otherwise, it is not much of a problem. At A, the net force is zero. At B, the ball is accelerating in the direction of the arc. Compare the forces in the two cases.
  4. Nov 7, 2006 #3
    In case-1, T1cosx=mg, while in case-2, T2=mgcosx because in case-2, T cannot be resolved while mg can be resolved as there is resultant acceleration. So T1/T2=secx*secx. Its a situation of temporary rest.
  5. Nov 8, 2006 #4

    I had earlier done this for case 2: Tcosx=mg. why is that wrong? cant tension be resolved along the vertical?
    can you please explain that?

    yup, tension before the cord is cut. Otherwise, the tensions are same right! so T1/T2 = 1..

    but what is causing it to accelerate in the direction of the arc? its own weight?
  6. Nov 8, 2006 #5


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    Yes. At B, the weight has a component in the direction of the arc, and a component perpendicular to the arc (in the direction of the string). The ball is not accelerating perpendicular to the arc, so what does that tell you about the sum of the forces in that direction?

    I'm not sure exactly how the interpret shramana's words, but the idea expressed by the equations is the same. I think he is just saying that in one case you are going to resolve the vectors in one set of directions, and in the other case you are going to resolve them in a different set. In fact you could resolve them in the same set of directions in both cases and get the same result. It is just easier to do the two cases with different directions, keeping one force "unresolved" in both cases.
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