This is what I comprehend as the tension of a string:
Suppose we take a string suspended from a rigid support. If a weight $w$ is suspended on it then, it will exert force downward due to the force of gravity. This causes the string to increase in length as a result of which it will exert an equal and opposite force of tension so as to regain its original length by virtue of its elasticity.

Now, this is a passage from my book:

Suppose we take a simple fixed pulley like this:

enter image description here

Suppose the string is inextensible, massless and strong. One end of the string is connected to load $L$ and the effort $E$ is applied at other end of the string. The load $L$ and effort $E$ both act downwards. Here, the tension throughout the string is $T$ upwards(same).

I don't understand right here why the tension must be the same. In the footnote, it was given,

If tension is not same throughout the string, then the string moves even when the pulley is not rotating.

I could not comprehend how can an inextensible string have tension. So, I want to know what my comprehension regarding tension of strings in the beginning is lacking. Further, I have difficulty understanding the footnote. I am a tenth grader. Thus, I would like a simple explanation.


3 Answers 3


If you're comfortable with tension in extensible strings, you can think of the inextensible string as the limit of an extensible string. That is, think of extensibility as a variable rather than a constant and take the limit as this extensibility vanishes. You will see that tension itself does not vanish in this limit. This is, in a sense, the most physical explanation, since no physical strings are truly inextensible (or frictionless, massless, etc.).

However, tension in a perfectly inextensible string is certainly well-posed mathematically. Consider such a string stretched to its maximum length horizontally. If I grab both ends and raise the upward, the string will accelerate like a particle, with its inextensibility having no effect. If I grab both ends and pull the left end left and the right end right, the string must not move anywhere, due to inextensibility. Since Newton's laws must be obeyed, the only possible explanation is that there is a force arising on the left end acting to the right to balance the force to the left, and similarly a force to the left on the right end. We call these forces tension forces. They can act only along the length of the taut string (and only in one direction) and their action is always to prevent the string from extending.

You may already have encountered a similar situation in considering a mass under gravity lying on a slanted hill. Gravity alone would cause the mass to accelerate into the hill, but the hill is treated as incompressible (just as the string is inextensible) and so there must arise a normal force to prevent this. The net result is that the mass slides down the surface of the hill rather than through it.

  • $\begingroup$ I could not upvote you.But thanks for a nice explanation. $\endgroup$ Commented Jul 11, 2021 at 14:57
  • $\begingroup$ I had been reflecting upon the case of pulleys. I have noticed that no matter what, whenever the string moves it must rotate the pulley. Is the footnote regarding the non-uniform tension of pulley is not true? It seemed a bit unnatural to me. $\endgroup$ Commented Jul 11, 2021 at 15:13
  • $\begingroup$ @EdmundBlackadder thanks, glad it was helpful! It is true that if the tension is not uniform throughout the string it will move, though I'm not sure what scenario the footnote is contemplating. If a string is under positive (extensile) tension, a bit of the string somewhere in the middle will experience a force proportional to the tension pulling to the left and a force proportional to the tension pulling it to the right. If the tension is uniform throughout, these forces will balance, but if the tension is varying then these forces will be different and that portion of the string must move. $\endgroup$
    – David
    Commented Jul 11, 2021 at 20:31

Your understanding is correct for real strings - a real string will always stretch slightly under tension. An inextensible string is an idealisation of a real string in which the stiffness of the string (the force required to stretch it by a given proportion) is assumed to be infinite. An inextensible string can produce any required amount of tension without changing its length. Replacing a real string with an ideal inextensible string simplifies the analysis in string and pulley problems.


I just want to add a little in addition to other answers.

Do not think of the tension as being "up" or "down." Tension is a force carried inside of a string and if you were to dissect a piece of the string as it is in tension, then the tension force would act away from the string material. In contrast, think of popsicle stick. A stick can withstand loads that both try to extend it and compress it. If you try to compress the stick, an internal compressive load develops inside of the stick, and if you were to dissect the stick as it is in compression, you would see an internal load that acts towards the material of the stick. A string cannot carry such a load. It can only withstand extension which results in an internal tension.

For the inextensible massless string, the tension force is whatever it needs to be in order to keep the string in equilibrium. Think of a string in your hands. If you pull on each end of it with 10 Newtons, then the string carries a tension force of 10 Newtons. If you pull on it with your two hands with 40 Newtons, then the tension force inside the string is 40 Newtons.

Another subtlety is in strings that may start moving, as in the case of pulley problems. Often the assumption is that strings are massless and the pulley drum is frictionless. From $F = ma$, we see that, even if $a$ is non-zero and the string is moving, if $m$ is zero, then the string must be in force equilibrium. This is why the tension force is constant throughout. If the pulley drum has friction, then it is a different story, and you should start reading about belt friction and the Capstan equation (http://mechanicsmap.psu.edu/websites/6_friction/belt_friction/beltfriction.html).

  • $\begingroup$ I could not follow the last paragraph. Please elaborate it a bit. $\endgroup$ Commented Jul 11, 2021 at 17:17
  • $\begingroup$ Consider a string suspended from the ceiling. At the bottom end, you have a large mass $M$. Kind of like a chandelier. Now imagine that the string is a sturdy, heavy cable with large cross-section, like you might see on the golden gate bridge. In this case we cannot neglect the mass of the cable itself. The tension at the bottom of the cable is $Mg$. But at the top of the cable, the tension is $(M+m)g$ where $m$ is the mass of the cable. In this example, the tension of the cable varies linearly along its length, being minimum at $M$ and maximum at its connection with the ceiling. $\endgroup$
    – Evan
    Commented Jul 11, 2021 at 17:48
  • $\begingroup$ Ah!I got it.In case when the string is massless m=0. $\endgroup$ Commented Jul 11, 2021 at 17:53
  • $\begingroup$ Now if you have that massive cable resting on a table, and you apply two tensions at its ends, then the tensions must be unequal to effect a movement to the cable, because $m$ is large: $T_1 - T_2 = m a$ tells us that $T_1 - T_2$ must be nonzero for $a$ to be nonzero, that is, $T_1$ and $T_2$ must be disparate. However, if the massive cable becomes a virtually massless string in the limit, then the requirement that $T_1 -T_2$ be nonzero for $a$ to be nonzero goes away. Since $m$ goes to zero, that requirement vanishes, and $T_1$ can be equal to $T_2$ for an accelerating massless string. $\endgroup$
    – Evan
    Commented Jul 11, 2021 at 17:54
  • $\begingroup$ I got it. Thank you very much for explaining patiently. $\endgroup$ Commented Jul 11, 2021 at 18:02

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