# Why I think tension should be twice the force in a tug of war

I'm going to provide my argument for why I think the tension in a rope should be twice the force exerted on either side of it.

First, let's consider a different example. Say, there is a person named `A` and a block in space. `A` pushes on the block with a force of 100 N. Then, the block will also push `A` with a force of 100 N by Newton's third law. Now, consider the case where instead of the block, there is a person `B` who is also pushing on `A` with a force of 100 N while `A` is pushing on him. `A` will experience a force of 100 N because he pushed on `B`, AND another 100 N because he is being pushed by `B`. Hence he will experience a force of 200 N. Similarly, `B` also experiences 200 N of force.

Now, back to the original problem. There are two people `A` and `B` in space with a taut rope (no tension currently) in between them. If only `A` is pulling and `B` is not, then I agree that the tension is equal to the force `A` exerts. This situation (in my opinion) becomes analogous to the above if `B` is also pulling. So, say both of them pull from either side with a force of 100 N. Then the rope at the end of `B` will pull `B` with a force of 100 N (this pull is caused by `A`). By Newton's third law, the rope will experience a pull of 100 N. But `B` is also pulling his end of the rope with 100 N. Therefore, the tension should be 200 N. Similarly, the end of the rope at `A` must pull `A` with 100 N of force (because `B` is pulling from the other side) and hence experience a force of 100 N itself by Newton's third law plus another 100 N because `A` is pulling on the rope.

Apparently, the answer is not this (according to my searching on the web). So, could anyone tell me why this reasoning is wrong? Thanks.

EDIT : So apparently people don't agree with my first example, leave alone the second. This is to the downvoters and the upvoters of the highest-rated answer: You all agree that if only `A` pushes `B` with a force of 100 N, then `A` and `B` both will get pushed by a force of 100 N in opposite directions, right? Then, in the case where `B` is also pushing with a force of 100 N, it doesn't make sense that the answer would be exactly the same. It doesn't seem right that no matter what `B` does, `B` and `A` will always experience the same force as they would have if `B` hadn't applied any force.

EDIT 2 : I'm going to provide here a link to a question that I posted: Two people pushing off each other According to the answer and the comments there, the reason as to why my first example is incorrect is different to the one provided here. So maybe you should all read the answer and the comments provided by the person and reconsider what you think.

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Imagine 1Kg-block hanging on the ceiling. Each hook (the one in the ceiling and the one in the 1Kg-block) pull with 100N still you would not get the Idea that the tension would be 200N. In the rope example the ground takes care of the other 100N. – miceterminator Oct 20 '12 at 14:42
Yes, but that's because only one side is doing the pulling. The block is being pulled by gravity downwards and since it is not moving it must be the case that the rope is pulling upwards with a force equal to its weight which implies by Newton's third law that the block is pulling downwards on the rope with the same force. Similarly, at the other end, the rope pulls the ceiling with a force equal to the weight of the block and hence again by the third law, the ceiling pulls the rope with the same force. So I see no reason to say that the tension would be twice the weight. – Alraxite Oct 20 '12 at 14:58
Sorry I didn't read through your whole text. "If only 'A' is pulling and 'B' is not, then I agree that the tension is equal to the force 'A' exerts." This does not work if you pull a massless rope with 100N there is no tension It would just accelerate infinitely. – miceterminator Oct 20 '12 at 15:09

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@Alaxrite: I suggest you get two spring scales and a rope and perform this experiment yourself. Your example is wrong. The action/reaction pair of A and the block has nothing to do with what B is doing. You can make this really clear by drawing free body diagrams for the three objects. – Jerry Schirmer Oct 20 '12 at 18:27

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By the way, downvoters are of course entitled to their opinion, but I do think this is a good question because it asks about a conceptual problem, and a somewhat subtle one at that. The fact that it's based on a misconception is not a problem IMO, and in fact such questions often turn out to prompt insightful answers. – David Zaslavsky♦ Oct 21 '12 at 10:01
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## 2 Answers

Your first example is facetious. If each is providing 100N then each is feeling 100N, period. In order to feel 200N, each would have to provide 200N. This is what Newton's Laws of Motion are all about; one does not feel their own force, only external forces, or when their own force comes into contact with an external body.

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I absolutely don't follow. Your wording is too vague. Look at this way: You agree that by the third law, forces always come in action-reaction pairs, right? Now, consider the same situation where the two people are pushing on each other. Let's consider these two forces: the force exerted by 'A' on 'B' and the force exerted by 'B' on 'A'. Note here, that I'm talking about the deliberate forces that they exert on each other. These do not form the third law's action-reaction pair. It's entirely B's choice that he is pushing. `continued` – Alraxite Oct 20 '12 at 15:52

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Since these two do not form an action-reaction pair of each other, there must be two other reaction forces for these two. So now I ask, what are the they? Think about it and then reply. – Alraxite Oct 20 '12 at 15:52

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And how do they form a pair? Didn't I mention that it's entirely B's choice that he is pushing? There are four forces in this situation. Two of which comes from each of them which they did deliberately, and the other two are the reaction forces to each of them as a consequence of the third law. And now you've just brought in two redundant equations into this discussion. What are x and y exactly? – Alraxite Oct 20 '12 at 16:12

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I keep trying to explain why, but you say "no, I think I'll ignore the math". You can't ignore the math. You can't trust intuition, or common sense, or the fuzzy-wuzzies when it comes to physics. You must listen to the math; it is the only thing that has proven itself correct time and time again. – Ignacio Vazquez-Abrams Oct 20 '12 at 19:05

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-1: Alraxite makes reasonable arguments (he confused me), and the responses to his questions are flippant and wrong. If you have two astronauts carrying nearly infinitely heavy backpacks, with ropes, one attached from A to B's chest and the other from B to A's chest, you wouldn't get confused that the pulls of A and B add up to determine how fast they approach each other. So why is it confusing when they are both pulling on the same rope? To be honest, I agree that it is confusing, but I haven't sorted out why. There is some failure of intuition here among the trained, not among the untrained. – Ron Maimon Oct 31 '12 at 4:18
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No.

Your mistake is that the block also pushes back on A, with the same 100 N that A is pushing with, by Newton's third law. Assuming both weigh 100 kg, both the block and the person will accelerate at 1 m/s$^2$ for the duration of the push. If there's two people pushing on each other, then clearly they will also accelerate at 1 m/s$^2$ so the forces must be the same. (If this isn't obvious, consider placing a very massive, very thin, very strong wall between them, which can't alter the physics. Then it's just two 100 kg people pushing off a stiff wall with 100 N.)

The difference between the two scenarios is that in the first one the reaction force is provided by the block's structure, while in the second one by a work-performing human. In the first one the block goes away while in the second one A and B's palms stay stationary. Thus A is able to push for twice as long and therefore do ~twice the work, so they will - as intuition says - end up going faster.

This translates directly to the rope-pulling scenario, simply substituting tension for the compression force at the guys' palms.

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I'm not sure about the mistake you're referring here. I have mentioned in my question that the block pushes on A. Your answer then goes on to say that it is obvious in the case where B is a person who is also pushing that he will experience the same force as the block. You have given a reason with a wall which I don't quite understand. Apparently the rest of your answer is about the details of how they push each other and I guess it does get confusing if you think about them pushing directly at each other's hands. So let me alter the situation a little. `continued` – Alraxite Oct 20 '12 at 17:17
Firstly, to keep it simple, both of them are using only one hand of theirs. Secondly, each hand of theirs are resting on each others chest before they push. So, now it should be clear that when A pushes on B's chest, his chest provides an equal and opposite force, during which B's hands which are on A's chest provides another force. Thus twice the force. – Alraxite Oct 20 '12 at 17:18
@Alraxite: Hmm, it seems you settled on the same example I gave. This is very true, and somewhat puzzling still, because when a man is pulling with 100N on a wall, it doesn't matter if it's a wall or another person pulling with 100N back. – Ron Maimon Oct 31 '12 at 4:20
The situation with pushing on each other's hands is different from the situation of pushing on each other's chests, it's serial vs. parallel springs. – Ron Maimon Oct 31 '12 at 5:47
Yes, if they push on each other's chests it's a different game (parallel instead of serial springs) and it cannot be translated to the rope example without introducing a second rope. – Emilio Pisanty Oct 31 '12 at 11:11
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