# Force and equilibrium

My book says that if a string was passed through a smooth ring the force of tension at each part of the string will be equal. And as it can be seen here the two parts aren't equal in length. So my question is that I don't understand the relation between the ring being smooth and that the tension in the two parts is equal and I don't understand how are they equal in magnitude but not in the length ? And I want to ask will the ring always stop at the middle between the two parts of the string as long as it is smooth?

• It's one continuous string so isn't it a given that the tension will be the same throughout? Commented Nov 24, 2022 at 21:11

Now, imagine the ring and all the forces acting on it. It has a weight downwards, and two strings with force $$T$$. Here length of the parts of the string doesn't matter. the only thing is, sum of their forces should be zero horizontally, and equal to weight of the ring, vertically. To make net force of zero on the ring.
$$T \cos{\alpha} = T \cos{\beta} \\ \implies \alpha = \beta$$ $$T \sin{\alpha} + T \sin{\beta} = 2 \times T \sin{\alpha} = W \\ \implies T = \frac{W}{2 \sin{\alpha}}$$