I was told that when a mass attached to a smooth suspending wire by a ring shaped holder; after reaching the equilibrium position it would be exactly halfway across the wire and, tension on both ends of the wire would be equal. In short, this. But why necessarily so? According to Newton's first law, only that the components add up to zero should be sufficient, no? While the horizontal components of the tension forces must be equal, the vertical ones not necessarily so. Then why are the tension forces equal? Also, why is that achieved only when the mass slides down to the middle of the rope? Are they just assumed for simplification purposes, or are there better reasons to justify them?

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    $\begingroup$ Would you mind adding a sketch of the setup? It is not easy to visualise. $\endgroup$
    – Steeven
    Jan 25, 2021 at 10:40

1 Answer 1


Let's solve this problem experimentally.

Consider that you initially start by placing the mass at any point on the rope. Due to extra length between the joints or elasticity of the rope a certain bending would be observed, forming an angle.

As you can see in this picture

As you can see in this picture, there would be a $\theta_1$ and $\theta_2$ on the two sides of the ring.

Due to this, if the ring could not move freely, both joints would experience different tension.

Now, as the ring can move freely along the rope, the mass would decrease its potential energy by moving to a point on the rope that makes it closest to the ground. This point geometrically would be the center of that rope. At this point both $\theta_1$ and $\theta_2$ would be equal and we would get equal tension.


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