I can't understand the direction of tension. Why is the direction of tension at the ends of a string away from the object or block of mass? Can someone tell me what happens internally in a string? ps: string is massless

  • $\begingroup$ Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. $\endgroup$
    – ZaellixA
    Mar 23, 2022 at 4:58
  • $\begingroup$ Duplicate/related Conceptual doubt in Tension force $\endgroup$
    – Farcher
    Mar 23, 2022 at 8:51

4 Answers 4


I can't understand the direction of tension. Why is the direction of tension at the ends of a string away from the object or block of mass?

I too had many misconceptions about tension and used to struggle with it. Now, this is how I treat it: Tension is no 'extra force' that you need to learn about. Given that it is massless and inextensible, it is simply the means of transfer of forces between 2 objects tied to the 2 ends.

Imagine 2 objects A and B tied to each other using a massless, inextensible string. Start pulling A. As you do so, the string becomes taut, A pulls B with a certain force that is equal to the force with which B pulls A, by Newton's third law. This would answer your question about the direction of tension in the string. It pulls A towards B because B is pulling A and pulls B towards A because A is pulling B ( the pair of 3rd law forces between A and B).

Now, in the same case of pulling, if there had been a massless rod, the effect would have been the same.( only in case of pulling. Note that it doesn't work in the case where you push A towards B because a rod would stay stiff between them and a string would become slack) Thus, an inextensible, massless string is just like a 'connector' between 2 objects.

Observe that I stress on the string being massless and inextensible. If it is not, the string will be an extra object (say C) between the 2 masses and there will be 2 more pairs of 3rd law forces: That between A and C and another (and a different force) between B and C.

Your second question has already been answered by others. In summary, a string is an assembly of many particles in a line, which tend to remain at a constant distance from its neighbours (if it is an inextensible string) due to electromagnetic forces, and hence simply transmit the force by object A till the other end at B.( apply Newton's third law (edit : and also second law) on every pair of particles in the string for better understanding)

  • $\begingroup$ I used only 3ed law on each and every particle on a string just as you told and it works. But here I will be linking a video from youtube that uses 2nd law as well as the 3ed law... KInda different approach.... Is this approach also correct? youtu.be/VXu2gatnMWE $\endgroup$ Mar 24, 2022 at 8:15
  • $\begingroup$ Yes, the second law should also be used, and you would have done so implicitly if the string is not moving, because net force on the fragments on the string will be zero (and that's when it becomes easier and you simply move on with the third law ). Otherwise, you will have to use the 2nd law as done in the video. Also, it is not a 'different' approach. Rather, my writing was less precise by not including the 2nd law. Remember that both the second and the third law are always applicable where ever there is force involved. $\endgroup$ Mar 24, 2022 at 11:24
  • $\begingroup$ To understand why the application of second law is necessary, consider a string with some mass ( let it be inextensible to make it simple) and divide it into several fragments $m_1,m_2, etc.$. Object A pulls object B, both connected by the string, and the whole system accelerates. When A pulls $m_1$, $m_1$ pulls A with the same force. However, the force exerted by $m_1$ on $m_2$ is a little lesser. Why? $m_1$ is itself accelerating and hence net force on it is now non-zero. So, as you move towards end B, the tension is lesser than the force A exerts on that end of the string attached to it. $\endgroup$ Mar 24, 2022 at 11:39
  • $\begingroup$ Now, construct and solve a problem to understand better. A scenario similar to the one mentioned previously. Consider the string with mass m. Apply a force F1 on A (Take your favorite numbers for forces/masses). If the string is inextensible, the whole system moves with the same acceleration a (implying that A, the string and B have the same 'a'=$F_1/(m_A+m_B+m)$). Use the second (you know the acceleration and mass of each object) and the third laws to calculate force of string on B and that by B on string. (tension in the string at the pint of attachment of string to B). Similarly at A. $\endgroup$ Mar 24, 2022 at 11:47

Tension force lies in electromagnetic category of the four fundamental forces.

When you try to stretch a string more than its length, you are actually trying to increase the intermolecular spaces between the atoms of the string. Atoms contain charged particles (initially in equilibrium) and when you disturb them with external force, they show reluctance (opposing force). This opposing force which is electromagnetic in nature, is the fundamental cause of tension in the string.

For direction, it is totally justified that tension (opposing force) will be applied in opposite direction of stretch tendency.


At the ends, the molecules of the object tries to stretch the string by pulling it's end molecule, so the string wants to regain it's shape and pulls back the molecule away from the object.

Microscopic interactions can be approximated as spring force at very small scale.


Tie one end of a string to a wall and pull on it. Pull so it's taut and pull some more. The string stretches just as a spring would. The string is then in tension. It is stretched, and just like a spring, exerts a force to bring it back to an unstretched position. At each end, it exerts a force towards its center.

Unlike a spring, a string has no strength in compression. Push on the ends of a string and it just folds, giving no resistance. Push on a spring or even a solid bar, however, will put into compression, causing it to push outward from its center. Compression is negative tension.


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