0
$\begingroup$

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?

enter image description here

$\endgroup$
1
  • $\begingroup$ It's one continuous string so isn't it a given that the tension will be the same throughout? $\endgroup$
    – Rod Bhar
    Commented Nov 24, 2022 at 21:11

1 Answer 1

0
$\begingroup$

First things first. The force tension at each part of the string is equal, only if the string has zero mass (ideal string). The reason is, if the tension is not equal at a point, according to newtons second law the acceleration would be infinite! But if it has weight, upper parts probably have more tension (since they're holding ring + mass of some part of the string).

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.

So the angles shown in the picture should be equal.

Forces on the ring

$$ T \cos{\alpha} = T \cos{\beta} \\ \implies \alpha = \beta $$ $$T \sin{\alpha} + T \sin{\beta} = 2 \times T \sin{\alpha} = W \\ \implies T = \frac{W}{2 \sin{\alpha}}$$

$\endgroup$
3
  • $\begingroup$ Okay I understood the main concept but I have a problem understanding some parts, do you mean that equal tension means this string has no mass ? And what do you mean by "if it has weight upper parts probably have more tension (since they're holding ring and mass of some part of string) ? and the last thing I don't understand the proof under the photo: T cos alfa = T cos b .. etc. could you please explain more ? I'd be really grateful. $\endgroup$
    – Hager
    Commented Nov 25, 2022 at 20:45
  • $\begingroup$ @Hager Sorry for my late answer. I hope it still helps. Yes, the equal tension in all parts of a string means it is mass-less, and vise-versa. To make it more clear, imagine a normal heavy string (like a rope) holding a mass (vertically). Lower parts of the rope only need to hold the mass. But upper parts of the rope need to hold the mass PLUS weight of the lower parts of the string (Sice the string has mass, too). $\endgroup$
    – mokazemi
    Commented Jan 2, 2023 at 10:10
  • $\begingroup$ @Hager And about the proof below, I've just calculated the components of the force vectors in the vertical direction. Since the ring doesn't have acceleration in vertical direction, the vertical net force in vertical direction on it should be zero. $\endgroup$
    – mokazemi
    Commented Jan 2, 2023 at 10:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.