0
$\begingroup$

I'm having a little trouble understanding why conceptually the two oppose each other with equal magnitudes. Let's say you have two people playing tug of war, with person A on the left much stronger than person B on the right. Person B is getting pulled by the rope because they are not strong enough. Person A is pulling with greater force, so shouldn't the tension force pulling person B to the left be greater than the tension pulling tension A to the right? Or, from what I understand about Atwood Machines, they reach some sort of "equilibrium", where the two forces oppose each other just enough so they accelerate with F/m: m being the combined mass and F being the net force.

$\endgroup$
1
$\begingroup$

I gave a detailed answer to this in here.

A quick answer, though, is if the rope has mass and is accelerating because A is stronger than B, then, yes the tension at A will be greater. If the rope is moving at a constant speed (or zero) then the net force on the rope is zero, and the tension on either end is the same.

However if the rope is massless, which is an approximation frequently made, then Newton's second law tells us that the net force on the rope is zero regardless of acceleration. In the massless rope approximation, the tension on either end is the same.

$\endgroup$
  • $\begingroup$ OK, that makes a little more sense. Thank you! But because F=ma, if mass = 0 for the rope, then acceleration is what? Nothing? Infinity? Because I keep thinking that the rope has an acceleration of the system because it's part of the system. $\endgroup$ – rb612 Oct 23 '15 at 2:15
  • 1
    $\begingroup$ The acceleration of the rope can be anything. So if it's being held by A and B, then A, B, and the rope all have the same acceleration. The acceleration is determined by the masses of A and B, and the difference of the force of friction between the ground and their respective feet. The rope plays no role other than tying A and B together. $\endgroup$ – garyp Oct 23 '15 at 2:28

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.