This question has an answer to the question of why tension is equal in the string in an atwood machine. The part of that explanation that I don't understand is:

This upward movement would relax the tension in the upper part of the rope (Tt decreases) and increase the tension in the lower part of the rope (Tb increases). This will continue until Tt equals Tb and there is no net force on the rope.

If the string is oscillating with a variable acceleration until it reaches equilibrium, wouldn't the masses also have to accelerate and oscillate, since they are attached to the rope. The answer says this happens instanteously, but that's not possible since the masses on either side of the pulley have mass.

Thank you very much.

P.S. I would have commented on the original post, but I don't have 50 reputation.


3 Answers 3


In part the confusion is because assumptions are made at the start of an analysis and then changed during the analysis.

enter image description here

The simplest system to analyse is an ideal one where the masses are point masses, the string is massless and inextensible, the pulley is massless, there is no friction and the final (steady) state of the system is being considered.
With these assumptions:
For the string the magnitude of the force it exerts on mass $m_1$ at one end, $T_1$, is always exactly equal to the magnitude of the force it exerts on mass $m_2$, $T_2$, at the other end, so $T_1 = T_2 = T$.

Because the string is inextensible the magnitude of the acceleration of one end of the string and hence mass $m_1$ is equal to the magnitude of the acceleration of the string at its other end and hence mass $m_2$, so $a_1=a_2 = a$.

An ideal massless and inextensible string transmits forces between masses and changes the direction of the forces.

Assume that $m_2 > m_1$.
The equations of motion for the two masses can be obtained using Newton's second law:

$m_2 a = m_2g - T$ and $m_1a = T-m_1g$.

With the assumptions that have been made there can be no discussion of how the system reached a state where the acceleration was $a$ from a different set of initial conditions.

If the string had some mass, $m_s$, but was still inextensible then the equations of motion are:

$m_1 a = mg - T_1, m_2a = T_2-m_2g$ and $m_sa = T_1-T_2$.

Note here that as $m_s \rightarrow 0 $ then $T_1 \rightarrow T_2$.

Now consider a real system with the string having a mass, $m_s$ and also capable of being stretched with a "spring" constant $k$.
To make the analysis easier, initially the string has its centre of mass directly over the axle of the pulley, masses $m_1$ and $m_2$ are at rest being held up by upward forces $m_1g$ and $m_2g$ and the string is taut.
The ends of the string exert no forces on the masses.

Remove the upward forces $m_1g$ and $m_2g$ simultaneously and apply Newton's second law.
$m_2 a_2 = m_2 g \Rightarrow a_2 =g, \quad m_1 a_1 = m_1 g \Rightarrow a_1 =g, \quad m_s a_s =0 \Rightarrow a_s = 0$.
The reason for these surprising? accelerations is that it takes time for the information/pulse/wave to travel along the string.
Thus one end of the string does not know what is happening to the other end. That speed depends on, amongst other things, the "spring" constant of the string and the mass per unit length of the string.
As the "spring" constant gets bigger and the mass per unit length gets smaller so the speed of information transfer gets larger.
Linking this with the ideal massless and inextensible string the implication is that for such an ideal string the information travels at an infinite speed.
This would mean that any changes to acceleration etc happen instantaneously.

For a normal string the speed of information transfer is relatively high, so much so that it is unlikely that those initial accelerations would be observed?

So now you have two masses accelerating downwards at $g$ and the centre of mass of the string at rest.
I do not know exactly what happens before the system settles down to some sort of ideal system behaviour but there must be a time when mass $m_1$ actually stops again and then starts moving upwards.
There may well be some oscillations in the string and perhaps even the masses actually oscillate as well as accelerating in a "clockwise" direction.
With friction and other dissipative forces present those initial "oscillations" will be damped down in a short period of time and again I think that it might be difficult to observe them.

A suggestion is that an apparatus is set up, with unequal masses connected by a spring which goes over a pulley, to see if any of the effects that I have described are visible.


If you are analyzing T_t and T_b, those are forces on the rope, not forces on the masses (the action/reaction pair force is on the mass). If the rope had a very tiny mass, and those forces were different, the rope would accelerate very rapidly, but that could not be sustained because the mass won't accelerate so easily. So the forces quickly equilibrate to each other, even if the whole system is oscillating. Think of it as a constraint on the rope, like if you attach one end of a rope to a wall and pull on the other end-- it won't take long for the forces at both ends to be the same. Now attach a massless rope to a block on a frictionless plane, and pull on one end of the rope-- again the force at both ends of the rope quickly becomes the same, even though the whole system will be accelerating. The point is, the acceleration won't be infinite, so the forces can't be different.

  • $\begingroup$ How is the rope able to accelerate at all, since it is attached to the masses? In your second example, there is only one force on the rope: the pull. So the force propogates instantaneously through the rope and it becomes taught and pulls on the block. That I understand. For the exmaple with the wall, this assumes the rope is able to accelerate. So, in the atwood machine, are the two components of tension only not equal when the rope is not taught, and the acceleration it undergoes is the rope becoming taught? $\endgroup$
    – Fazer
    Commented Sep 12, 2016 at 1:39
  • $\begingroup$ That's pretty much the point, the rope cannot accelerate at all, so the forces must immediately become equal to each other. It's an idealized rope, with no mass and no stretchiness, so then the forces must be the same immediately. For a real rope, with some mass and some stretchiness, it could have a different acceleration from the mass, for a short time. The same if it is not taut-- the assumption for the idealized rope is that it is massless, rigid, and taut. $\endgroup$
    – Ken G
    Commented Sep 12, 2016 at 2:37
  • $\begingroup$ What I didn't understand is how the forces become equal. They can't just magicaaly become equal. But I believe it gets summed up well in @Farcher answer when he says 'the masses oscillate while accelerating "clockwise".' $\endgroup$
    – Fazer
    Commented Sep 14, 2016 at 23:22

In these problems we always consider "ideal ropes" that is these ropes possess no mass and they are perfectly inextensible.

If we take a real rope, then yes it would take some time for the tension to reach a constant value.(due to expansion and contraction of rope)

But here, since the rope can't be extended, we can consider that the tension would become constant immediately.

Now consider a small portion of rope just touching the block the block applies a force mg on it, downwards and since this small portion is massless as well as inextensible, therefore it would accelerate with infinite magnitude, which is not the case. Therefore we conclude that a tension must be created in the rope(instantaneouly) which balances the mg from the block so that nett forces on this small portion always remain zero. (As if any forces appear, they would cause infinite acceleration)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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