I started mechanics this year. There was a question about a smooth ring threaded on a light string at rest. The string's ends were each connected to a rod forming an isosceles triangle with the ring at one vertex. We were asked to find the tension in the string. My teacher said that since the string is light, there would be no normal contact force pair between the ring and the string. I don't understand why for 2 reasons : 1. The rope, eventhough m=0 for ease, is still rope. It is a surface on which the ring rests. Why won't contact forces exist. 2. If there are no normal contact forces, how is the ring at rest. The tensions aren't "pulling" on the ring. What am I missing? enter image description here

  • $\begingroup$ Your explanation of the situation isn't clear, can you explain in greater details. $\endgroup$
    – TechDroid
    Mar 10, 2019 at 10:26
  • $\begingroup$ My view is that the ring itself has it's weight and the normal contact force. The bit of rope there has the tension from two sides and the other pair of the normal contact. $\endgroup$
    – Sal_99
    Mar 10, 2019 at 10:57
  • $\begingroup$ Can you provide an image of the setup, maybe hand drawn, that will better picture the situation than just words $\endgroup$
    – TechDroid
    Mar 10, 2019 at 11:21
  • $\begingroup$ I have put a picture of my interpretation. $\endgroup$
    – Sal_99
    Mar 10, 2019 at 11:49
  • $\begingroup$ Can you edit the main question so it explains the image because it isn't clear how the "rope view" and "ring view" merge. $\endgroup$
    – TechDroid
    Mar 10, 2019 at 11:55

1 Answer 1


There is indeed a normal contact force.

It sounds like what happened is your teacher wanted to model the free body diagram of the ring with gravity and two tension forces. This is the conventional way to calculate things like the tension in the strings. Adding a pair of normal contact forces to this free body diagram complicates your diagram without changing the answer. It's simpler to think the ring and the tiny bit of the string it touches as your "free body". In this model, the normal contact force is an internal part of your free body.

If this is true then you and your teacher are both correct, from a certain point of view. You are correct in that there is a normal contact force. Your teacher is correct in that it shouldn't be drawn on the free body diagram.

There is an alternative explanation for which your teacher is incorrect. In almost every physics practice problem not involving pulleys, strings are secured to the object they're attached to, thereby exerting tension instead of a normal contact force. A string through a ring is a rare exception. Your teacher could have overlooked this oddity.

  • $\begingroup$ What I don't understand about the teacher's model is how are the tension forces pulling on the ring? With the contact forces, it makes sense to say that bit of rope has 6N ( action reaction contact force pair) down and two tensions pulling on it. Doesn't the ring itself just have it's weight and the normal contact upwards? $\endgroup$
    – Sal_99
    Mar 10, 2019 at 10:50
  • $\begingroup$ By 6N, I'm just referring to the actual number in the question. $\endgroup$
    – Sal_99
    Mar 10, 2019 at 10:58
  • 1
    $\begingroup$ Yes, the ring has just those two forces. Sometimes you're right and a teacher is wrong and it's not worth arguing with them about it. The important thing is you know you're right. If it makes you feel better you can tie the ring to two half-length strings or ropes in your mind. Then there really will be two separate tensions pulling on ring and you'll still get the right answer. $\endgroup$
    – lsusr
    Mar 10, 2019 at 11:01

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