# Rigorously proving equal tension on both ends

There is always one point in introductory mechanics that has been continuously bothering me ever since I had taken my freshman physics course - Why does a massless rope have its tension equal on both ends?

I have been searching for a good explanation on various textbooks at different levels, but none of them provided me a satisfactory explanation. They either

1. State directly "we assume the string is massless and the pulley has no friction, therefore the tension must be the same throughout the string" without further any elaboration and continue the calculation,

2. Claim without any reasons that "whenever the rope and the pulley are massless and frictionless, the tension will be the same on both sides", or

3. Explain using a lot of hand-waving arguments regarding infinite acceleration, non-zero moment of inertia, etc. , in one line or two.

Although those hand-waving arguments of type $(3)$ do sound quite reasonable to me and will convince me for a while every time I review them. But after a while, I will start feeling not well about this whole tension concept again.

Therefore, I am looking for a rigorous (in the physicist's sense) proof of exactly why and when do the tension in rope will be uniform, and hopefully from that I can clean up all of my conceptual difficulties.

• Take a straight segment AB of the rope. If this segment is moving rectilinearly along the line AB with acceleration $\:\mathbf{a}\:$ and $\:\mathbf{F}_{A},\mathbf{F}_{B}\:$ the tensions at the ends A,B respectively then $$\mathbf{F}_{A}+\mathbf{F}_{B}=m\cdot\mathbf{a}=0\cdot\mathbf{a}=\boldsymbol{0}$$ so $\:\mathbf{F}_{A}=-\mathbf{F}_{B}\:$. Commented Jan 27, 2017 at 19:58
• Looks like an interesting question, but how can a rope have no mass? Even the scholar case of a string with zero rigidity has a linear mass. What is the physics behind? If the situation is physically impossible, then it may explain why it is difficult or impossible to justify unless you recreate an ad hoc model (I wonder if it make sense to write $F=ma$ when there is no mass).
– user130529
Commented Jan 27, 2017 at 20:31
• @dmckee So you are telling that Newton's 2d law $F=ma$ still makes sense (can be applied) when the mass is zero?
– user130529
Commented Jan 28, 2017 at 7:54
• @claudechuber No. I said that such a situation isn't physical. It's a shorthand for an approximation in which the mass of the rope is small enough to to meaningfully affect the things you are trying to calculate. As you see below explicitly worrying about the mass of the rope can sometime make a calculation hard enough to obscure the lesson. Students in the first class are not well served by throwing a pile of nitpicky math at them. They can pick that up the second time 'round. Commented Jan 28, 2017 at 17:59
• Related: physics.stackexchange.com/q/156413/2451 and links therein. Commented Feb 18, 2017 at 23:22

I will only consider taut, inextensible ropes. This represents an approximation regime where tension is much less than the Young's modulus of the rope material times the cross-sectional area of the rope. (As we will see this condition is enough to guarantee the conclusion for straight segments of the rope as long as they are light compared to what ever they are attached to.)

1. Let's start with the easiest case: there is a frame of reference in which the entire rope is still.

In this case we can reason thus: the rope is subject to net zero external force (by Newton's 2nd Law). Further any internal element of the rope must be subject to equal and opposite forces from neighboring elements. Those forces are the tension in the rope so the tension is the same throughout.

[Note that the rope need not be massless.]

2. Next easiest case is the rope is in motion with constant speed along it's own length, but it may pass over circular pulleys with no friction at bearing (but with enough friction at the grove that the rope does not slip) and the like so any given part of the rope might change directions at times.

The argument in part (1) applies to segments between the pulleys.

The pulleys themselves have no angular acceleration and are therefore subject to zero net torque. So, the segments on either side of the pulley have the same tension because they have to exert equal and opposite torques.

[Note that the neither the rope nor the pulleys need to be massless.]

3. Now we come to cases where the rope is accelerating along it's own length.

If we assume a massless rope then any change of tension along the length would cause arbitrary acceleration. That's a nasty case because it s non-physical, but it represents an approximation of a regime where the mass of the rope is much smaller than the mass of anything it is connected to. Demonstration Atwood's machines are roughly like that.

4. A straight, massive rope (segment) accelerating along it's own length. The segment is subject to a net external force $F_n = F_1 - F_2$ where $F_1$ and $F_2$ are the net forces at either end, so that the acceleration is $a = F_n/m$. The rope segment is assumed to have uniform mass density $\lambda = m/l$. At any fraction $f$ along the length of the rope the tension must be such that the segment on each side of conceptual divide has the same acceleration (to prevent extension without allowing slackness). So tension $\tau$ (which is the same in each direction from Newton's 3rd Law) at position $fl$ must meet the requirements $$a_1 = \frac{F_1 - \tau}{\lambda fl} \;,$$ and $$a_2 = \frac{\tau - F_2}{\lambda (1-f)l} \;,$$ where $a_i$ is the acceleration of the segment closer to force $F_i$, and $a_1 = a_2$ is required. Thus \begin{align*} \frac{F_1 - \tau}{\lambda fl} &= \frac{\tau - F_2}{\lambda (1-f)l} \\ \frac{F_1 - \tau}{f} &= \frac{\tau - F_2}{1-f} \\ (1 - f)(F_1 - \tau) &= f(\tau - F_2) \\ -\tau &= (f - 1)F_1 - fF_2 \\ \tau &= F_1 - fF_n \;. \end{align*} Now, this depends on $f$, which means that it is not constant, right?

OK, but the term is $fF_n$, and $F_n = ma$ where $m$ is the mass of the rope segment. So, if the rope is much less massive than the things attached to it's ends, this terms could be negligible compared to $F1$ (or, indeed $F_2$ since the labeling is arbitrary). Then we get $$\tau \approx F_1 \,,$$ which justifies the claim in item (3) that the hypothetical "massless" condition is an approximation for "very light".

5. Next we could let these things run over low friction pulleys. We can recover the 'tension the same throughout' conclusion only if the pulleys are also "very light" so that we can use an argument like that in item (2) because the torque on them will arise from force small compared to $F_1$ or $F_2$.

6. Once the pulleys have non-trivial mass and the rope is accelerating than the segments must be assumed to have different tensions even if the rope is light.

• Thank you very much, this is exactly the kind of "proof" I am looking for. There is just one point I don't understand (or I somehow missed) - in (4), what make the force $F_1$ so special that it is appeared in the final expression for the tension (and is roughly the tension for the massless case) but not $F_2$? While they are arbitrarily labeled, they are unequal in general so what make the equation favor one of them over the other? Commented Jan 28, 2017 at 6:30
• @AstroK this is only due to the choice of the fraction: in (4) you can also arrive to $\tau = (1-f)F_n+F_2$ and conclude next that $\tau \approx F_2$. This is not contradictory with the above claim because if you suppose that $m$ is small while $a$ remains constant, then $F_1-F_2=F_n=am \approx 0$.
– user130529
Commented Jan 28, 2017 at 13:34
• Nevertheless, in 1., I still worry about the use of Newton's 2d law $F=ma$ when the mass is zero.
– user130529
Commented Jan 28, 2017 at 13:39
• @AstroK As claude says, it is an artifact of the meaning of $f$ (which I didn't make explicit, but is measured from the side of the rope where $F_1$ is applied. I have been wondering if there is a way to clarify that section, but it hasn't srprung to mind. Commented Jan 28, 2017 at 18:01

Let mass M be on the left of a pulley ,a string connects it to mass m on the right of the pulley .1. The pull of M on string is down (T tension ) 2 this pull is transmitted to mass m T up 3 reaction of string on M is T up 4 reaction of m on string T down . So forces on string T down on left and T down on right ! Cheers Eddie It’s Newtons 3 rd Law in action !

This isn't a rigorous proof, but moreso the intuitive answer that I use. Say you're pulling an object with a heavy rope. The end where you pull the rope will be pulling the object's mass and the rope's mass. However, the end attached to the object will only be responsible for pulling the object's mass.

If F = ma and the acceleration is the same at all points on the rope, then since the total mass being pulled is greater at the end of the rope being pulled, the force will also be greater at that end than the end of the object. This would mean that in a rope with mass, the tension force at all ends of the rope is a sort of gradient, getting larger as you move away from the object.

This is inconvenient for most physics textbooks, which is why they use the term "massless rope." Of course a rope actually can't be massless-- they should really say "negligible mass" to avoid confusion.

As for why it's the same at all points along the rope, and not just the start and the end, you have to understand how tension works. In something like a rope, the inside of a rope is wound with fibers. If you apply a pull to one end, then the fibers at that end pull on the fibers right next to it, and those fibers pull on the fibers next to it, etc. until you reach the end of the rope. So that pull sort of travels through the medium of the rope.

Indeed, let us assume that the two ends of the rope A and B have co-ordinates $-d$ and $d$ along an x-axis. The rope is supposed to be homogeneous, in static equilibrium. It does not matter whether it has a positive mass or not, whether it is extensible or not, rigid or not... The only thing that matters is that the problem is perfectly symmetric with respect to the origin. Hence the tensions ${\bf T}_A$ and ${\bf T}_B$ at the two ends have same magnitude and opposite sign: $${\bf T}_A=-{\bf T}_B.$$