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A massless string is passed through a smooth ring such that the ring can slide freely. The two ends of the string are attached to two points(C and B) on a vertical surface which are on the same horizontal line. Now if the ring starts to slide downwards from one end of the string and reaches to point A, a horizontal force is applued on it so that it remains in static equilibrium.

My question is that why is the tension same along the AB and AC part of the string? They are attached to two different points so shouldn't their tension be different even though they are part of the same string? Moreover the string has bent at the point A. Won't this bending cause any difference between the tensions of part AB and AC?

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N.B: I questioned about this earlier but forgot to add the image at the beginning so my question seemed to be a different one. And the answers were based on the later one. Afterwards I added the image, however none noticed it I guess

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2 Answers 2

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Look for the words 'smooth' or 'rough' in these types of questions. If it said smooth ring you can assume the tensions in the string on either side are equal.

For the tension to change, even on a bend in the string, the ring would have to exert a force parallel to the string, but if it's smooth then the ring only exerts the normal contact force on the string - that acts at right angles to the string and doesn't change the tension.

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  • $\begingroup$ Yes the ring is smooth. But why are the tensions equal? I want to know about that. It would be very helpful if you could clarify this problem.😅 $\endgroup$
    – MSKB
    Mar 31, 2021 at 11:12
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Think of 3 sections of the string, section AB, section AC, and a small section that is in contact with the ring. If the tensions AB and AC are unequal, there will be a net force on the portion of the string that is in contact with the ring. Since the contact is smooth, the string-ring contact will slide until that net force is zero, which will not occur until the tensions in AB and AC are equal.

For the special case considered here, since the additional force at A is horizontal, the equilibrium position will occur at an elevation midway between the elevations of A and B, so the tensions will be equal by symmetry. This symmetry is not required for the tensions AB and AC to be equal, however. The frictionless contact is sufficient to assure equal tensions.

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