Sign up ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Suppose two cases where masses are attached to massless, inextensible strings.

In the first case a body which has a weight of 20 Newtons(downwards) is attached to a string which is acted upon by a force of 100 Newtons upwards.

In the second case, a body of weight 10 Newtons is attached to a string and again a force of 100 Newtons upwards is applied on the string.

These two cases are identical except for the magnitudes of the weights attached to the strings. However, I think that the tension in both strings will equal 100 Newtons(the reaction force for the upward force).

If this is correct, how is it that the string has the same tension when different forces are applied on it? Intuitively, shouldn't the string stretch more and have more tension when a larger force acts on it on one side.

And if the tension in both cases is not equal then how can the tension in the strings be calculated?

share|cite|improve this question
It depends on what is meant by a string which is acted upon by a force of 100 Newtons upwards. Assume for the moment the top end of the string is tied to something - obviously the tension in the string is 20N in the first case and 10N in the second. If acted upon by a force of 100N means 100N extra then the tension will be 120N and 110N respectively. If it means 100N total, i.e. 80N extra in the first case and 90N extra in the second case, then the tensions will both be 100N. – John Rennie Jun 14 '14 at 7:14
Tension of a fixed-length, massless string is transitive because of Newton's third law, therefore the tension exerted on one end will transfer to the end connected to the mass as is. The gist of your question is: how is it possible that a force other than weight can be exerted on a mass, which is trivial to answer in this form. There's a difference between a force exerted on a mass and a force exerted on a massless object, which is simply a force carrier. – auxsvr Jun 14 '14 at 8:08

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.