I'm having a few problems with understanding how to calculate tension in a loop.

If I have a circular loop, and some force is applied uniformly radially outwards in such a way that the force acting on each element of the string is normal to the string at that point, then how will the string develop a tension?

My intuition tells me that some force must act by the string to prevent the string from expanding infinitely. However, how can the string apply such a force, when its only means is tension which acts perpendicular to the force always?

Any answers will be appreciated.

I suppose that this is analogous to asking why pumping air into a soap bubble will make it expand a certain amount (until the external pressure is equivalent to the excess internal pressure).


So, I can simply take the semi-circular half and apply Newton's Laws?


So $2 T = F $

Therefore, tension developed $= \frac F2$

Just want to clarify that by F I mean the total horizontal Force acting on the semicircle, not just the central element.

Is this correct? Just want to verify if I understood the concept.

  • $\begingroup$ If your loop is completely circular, with uniform outward force acting on it, then this is wrong. You need to take each element of the loop individually, not the whole half of the loop. $\endgroup$
    – Sidd
    Jan 11, 2015 at 18:30
  • $\begingroup$ Ah. Just realized how stupid that is. It doesn't include any term signifying the curvature. If the curvature were much smaller, the force should be larger. $\endgroup$ Jan 11, 2015 at 18:34
  • $\begingroup$ Also, if an answers satisfies you, you should mark it as an answer so that it is closed and people don't waste their time answering it. $\endgroup$
    – Sidd
    Jan 11, 2015 at 19:11
  • $\begingroup$ Alright. Done. I'm still not satisfied with your comment though. By symmetry the force acting on any element in the loop must be equal, correct? So tension is equal throughout too. So if I find the tension for a semicircle, it must automatically be the same for any element in the loop, correct? $\endgroup$ Jan 11, 2015 at 19:18
  • $\begingroup$ Sure, you can keep waiting for a better answer. $\endgroup$
    – Sidd
    Jan 11, 2015 at 19:19

3 Answers 3


Alright I'm a bit late here, but I want to give an answer for all who Google this.

Suppose we approximate the loop to n particles joined by n light, inextensible and rigid rods on the vertices of a regular n-gon.

I can't give a diagram here but if you look at the previous answer, you have a radial force of magnitude R/n where R is the total force, and a tension T. If the angle between vertices is θ, then the angle between T and the "horizontal" is θ/2.

θ = 2π/n because 2π in circle and n vertices.

By N2, R/n = 2Tsin(θ/2)

T = R/(2nsin(π/n))

As n tends to infinity and n >> π, sin(π/n) = π/n. So T = R/(2nπ/n) = R/2π.

Surprisingly this seems to be independent of the radius of the circle, but I suppose with dimensional analysis you can't have distances involved in an equation relating two forces.


The loop has a curvature. As a result, when you take a small element of the loop, the tension force applied to it from both the sides is at an angle, which results in a force component to the inside.

enter image description here

Since your question asks for a rationale behind the tension force, I'll leave it here. You can now try yourself to find the tension by equating the outward force with the inward force. Or, if you are given the total outward force, you can integrate the tension force over the loop and then equate the total outward and inward forces.

  • 1
    $\begingroup$ Thank you for the answer. I've edited my question accordingly. Can you please verify my work? $\endgroup$ Jan 11, 2015 at 18:24

I like to challenge the answers above. Instead of looking at local equilibrium of forces, I'm checking the energy balance, meaning work input and output must be equal. Check out the sketch below, assuming the depth of loop is a unit length, not shown here. 1

  • $\begingroup$ There are a number of problems with your soltion - the first problem is this wasn't the OP problem. The second problem is $\vec F$ and $\vec T$ are vectors not scalars. And the third problem is there are 2 solutions, $\vec F=T/2$ and $\vec 0$. Since you integrated over a circle, the vectorial answer (if you had done the vectorial operations correctly, the answer would 0. $\endgroup$ Mar 4, 2022 at 0:56
  • $\begingroup$ Greetings! Please see our guidance about posting photographs of text (short version: please don’t). $\endgroup$
    – rob
    Mar 4, 2022 at 15:17

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