enter image description hereAs I know, Newton's third law states that "forces occur in pair between two bodies,a pair of equal and opposite forces on each other and Net force on a system of particles is equal to total external force on the system,since internal forces cancel each other." Let's consider a very simple case.

A man pulls a string of mass 'M' on a frictionless surface with a force "F". Let's divide the string into two equal sections,A and B. Only a part of one end of the section B is in contact with the man.

If we take "section B" as our system, there is an external pulling force of "F" at one end by man, and a force of "F" by section A in opposite direction, due to contact action-reaction pair between sections B and A. Making total external force on System " Section B " zero.

I don't get how it can be. Because total external force on section "B" must be "F" if we consider both sections A and B together as a system. So,What is it that I'm missing in my understanding? Please help! As you can see,I'm not a student of physics,just a curious noob.


As the string has mass, tension will not remain same at all points in the string, instead it will vary linearly with distance. This can be proven in the following way :-

Let the length of the string be 'l'. Let tension at a distance 'x' from the point of application of force be 'T'. Divide the string into sections of lengths x and l-x. The net force on the section with length x will be F-T towards the right. Thus acceleration of this section will be $\frac{F-T}{m\frac{x}{l}}$. We know that the acceleration of the string is $\frac{F}{m}$ and so all the points on this string should also have the same acceleration. Therefore:- $\frac{F-T}{m\frac{x}{l}}=\frac{F}{m}$ =>$$F-T=F\frac{x}{l} $$ =>$$T=F(1-\frac{x}{l})$$ Your assumption that the force of tension between sections A and B is equal to F , and that the total external force on section B is also equal to F is thus wrong. The value of tension between sections A and B will be $\frac{F}{2}$ and total external force will also be $\frac{F}{2}$ for both the sections.

  • $\begingroup$ You provided a very important piece of information what I was missing.Thankyou. Can you explain how the force 'F' , man applied at tip of the section B, is exerting and distributing itself on entire system of both sections with varying tension. Like, I can't seem to make an inituitive understanding of it. I understand the mathematical derivation of how the sum of total forces on the system is equal to only total external forces on the system and internal forces whatever occuring cancelling out. $\endgroup$ – Viswanath Reddy Jan 20 '18 at 7:33
  • $\begingroup$ Is it just how intermolecular interactions work and the force you apply on tip travels through the string and all internal force pairs getting cancelled out,remaining only the effect of external force on all over the system? $\endgroup$ – Viswanath Reddy Jan 20 '18 at 7:45
  • $\begingroup$ Imagine the string to be a collection of extremely small discs. When the string is pulled, the disc at the end will move increasing the gap between this disc and the one next to it. This creates an electromagnetic force that tries to restore this gap pulling the disc at the end towards the left and the disc next to it towards right. We call this force tension. The tension increases with the distance between the discs. The gap between the two discs will remain same only if the acceleration and velocity of the part of the string at one side of the gap is equal to that on the other side. $\endgroup$ – Atharva Kulkarni Jan 20 '18 at 8:37
  • $\begingroup$ The tension will keep on increasing till it is able to create an acceleration in the rest of the string which is equal to the acceleration of the disc at the end. If the tension still increases then the gap will become shorter than it was at start. So tension stays at that value. This happens at all the points in the string and is the reason tension varies through it. $\endgroup$ – Atharva Kulkarni Jan 20 '18 at 8:37

Assuming no friction, if the string is “massless” then you do not need a force to accelerate it.
If the string has a mass then there has to be a net force on the string to accelerate it so the forces at each end of the string will not be equal in magnitude.
To accelerate string B the pull on string B by the man will be larger in magnitude than the pull on string B by string A.

  • $\begingroup$ You are saying,the pull on the section B by man must be higher than the pull on string B by string A. We know, total system (both section A and section B) is acted by total external force 'F' .If section B isn't pulling section A with 'F' (which is the reason i thought for why section A will pull Section B with F, how is this external force 'F' acting upon section A? I hope you understand.Sorry,my english is not good. $\endgroup$ – Viswanath Reddy Jan 20 '18 at 6:36
  • $\begingroup$ @ViswanathReddy The tension in the string is only the same at each point if the string is massless. $\endgroup$ – Farcher Jan 20 '18 at 6:41
  • $\begingroup$ Huh?Sorry,I don't understand.Even if it isn't a string, say a block of two connected sections A and B.I uploaded a vague messy drawing ,in case it helps. $\endgroup$ – Viswanath Reddy Jan 20 '18 at 6:59
  • $\begingroup$ See,my question in brief is this. According to newton third law, external force acting on sections A and B is F. I'm asking how is section A acted by external force 'F' ? I been thinking, this is because section B pulls section A with 'F' .But this is contradictory since it would make external force on section B to be zero if we consider section B as system.If we consider both sections as system, then this internal force pair between section B and section A somehow cancels out, and external force "F" acts on both sections A and B. $\endgroup$ – Viswanath Reddy Jan 20 '18 at 7:09

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