enter image description here


Suppose there are two forces, F1 and F2, acting on two ends of the rope, in directions parallel to the rope. Obviously the net force F equals F1 - F2 (in the same direction as F1)

We know F=ma.

Since m=0, we have F = 0. Hence, F1 - F2 = F = 0. F1 = F2.


In other words, as long as there are two forces acting on either end of a weightless rope they will always be equal. Even if it's Superman straining at one end and an infant pulling at the other.

What's wrong with my proof?

  • 2
    $\begingroup$ There is nothing wrong with your proof. It is correct. You have successfully found that the tension in a massless rope is the same everywhere in the rope. $\endgroup$ – John Rennie Sep 18 '20 at 9:38
  • 1
    $\begingroup$ The basic problem is that you apply second Newton law not to these objects. I.E. superman has mass and infant has mass. So superman will be pulling infant towards itself with force $F_{\,sup}-F_{\,inf} = m_{\,inf} ~ a_{\,inf}$. But of course your analysis is somewhat contradictory, cause in your case rope should stay at rest. If rope COM stays at rest, then neither superman, nor infant can pull each other. $\endgroup$ – Agnius Vasiliauskas Sep 18 '20 at 12:17

There is first a physics answer, and then a mathematics answer.

The physics answer is that there is no such thing as a massless rope. As soon as you allow the rope to have some mass, no matter how small, your argument no longer works.

Now, it is quite common to invoke the idea of a "massless" or "weightless" rope in arguments involving force and acceleration. What is really going on in such a case is that one is considering a situation where the mass of the rope is negligible in comparison to some other mass which is also in the system under consideration. In the situation you described there are no other masses with which to make the comparison, so this idea of comparison is not available. Hence the concept is not applicable.

Now for a further comment on the mathematics. There are many situations in mathematics where apparent contradictions can arise, owing to a mistake involving the number zero. In this example you have $$ F = m a $$ and with $m = 0$ you are right that one would normally conclude $F=0$. But in the present example we have another consideration: it is the possibility that $a$ might not be a finite number. If $a$ is infinite then you have the combination $0 \times \infty$ and such a quantity is formally undefined. To fill this out a little, consider not a zero mass but a small mass. Then you have $$ a = F / m $$ so if the mass is small the acceleration is large. For any given amount of force, as $m$ gets smaller, $a$ gets larger, in such a way that $m a$ is constant and equal to the force. So in the limit of $m \rightarrow 0$ then indeed in this example you have $a \rightarrow \infty$. So that is the mathematical solution to your puzzle. It is a case where caution is required regarding the combination of zero and infinity.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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