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Different versions of this question have come up all over the internet. Usually it deals with tension in a rope or two people pushing on each other with the same force. I am trying to understand 2 people pushing each other with different forces.

Say two people of equal mass are standing on a frictionless surface, touching palm to palm, and they push off of each other at the same time with different amounts of force. Say the person on the left pushes with 100 N and the person on the right pushes with 70N. What happens to this system? It seems like there should be a net force of 30 N to the right, but at the same time, each person pushing should feel the same force pushing back as per Newton's 3rd Law.

If I push on you with 100 N then I also feel the 100 N from the force pair and we slide apart each having been accelerated by 100 N. If you push on me with 70 N then I feel the 70 N force and I exert 70 N back on you, we slide apart each having felt 70 N this time. But what happens when we both push with a 100 N and 70 N respectively?

Also, I have seen elsewhere on this site that if 2 people were to push on each other with the same force under conditions like those in the scenario above, they each feel the same force but fly apart at double the final velocity because the same force has been applied for twice the distance. Is this actually the case?

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The human body is terrible laboratory for Gedanken experiments, and this problem proves it. What exactly does pushing with 70N mean? Is it active? Is it passive? Nothing feels correct.

This is why Newton's Laws are quite often imagined with masses and springs, which is the only way to save this problem from comments explaining the answer is insufficient or feels wrong.

So: Imagine each person pushes not with a hand, but with a giant spring--a giant calibrated spring that measures force. If person A pushes with 100 N and person B does nothing, both their springs compress to read 100 N. (This is passive pushing for B). Now B fights back and actively adds 70 N to his spring: both springs compress to read 170 N.

Equal and opposite. Always.

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Say the person on the left pushes with 100 N and the person on the right pushes with 70N. What happens to this system?

This is not possible. They can only push on each other with the same force.

Suppose the person on the left is a weightlifter and capable of pushing very forcefully, and suppose that the person on the right is unconscious and not capable of exerting any force voluntarily. Even in such an asymmetrical circumstance the force will be the same. If the person on the right is rigid then the person on the left can generate their full force and the natural rigidity of the person on the right will return the equal and opposite force without effort. If the person on the right is limp then the person on the left cannot generate much force at all since the limp person will deform under minimal force. Either way the force is automatically guaranteed to be equal and opposite.

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  • $\begingroup$ but what exactly "happens" if we each try to push with the different forces? Perhaps one person begins their 100 N exertion slightly sooner and during the acceleration the other person "pushes" off with their 70 N force. I guess what I am trying to say, how can I visualize or conceptualize the impossibility of this? $\endgroup$ – MattGeo Dec 1 '18 at 2:58
  • $\begingroup$ That is what I explained with the weightlifter and the unconscious person. No matter how asymmetrical the situation the force will always be equal and opposite. $\endgroup$ – Dale Dec 1 '18 at 3:01
  • $\begingroup$ I understand the case of the rigid and limp person and how it makes a difference in being able to generate a force. I do not see how this is the same as someone else actively pushing opposite of me while I am already going to feel the action-reaction force pair of my own push anyway. It feels like they should add, but that also feels incorrect. Like they would each feel 170 N $\endgroup$ – MattGeo Dec 1 '18 at 3:06
  • $\begingroup$ Dale, I think the scenario the OP is thinking is that the two people are using different amounts of "effort". (Maybe one fully conscious person and one semi-conscious person) So I believe the total net force on each would be $170\ \rm N$. I do agree the wording is sloppy though. $\endgroup$ – Aaron Stevens Dec 1 '18 at 6:01
  • $\begingroup$ The amount of effort is not decisive. A weightlifter can put in his maximum effort and still produce less than 1 N force if he is pushing against air. Human musculature is far more complicated than a simple effort = force paradigm. $\endgroup$ – Dale Dec 1 '18 at 11:51
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The problem here stems from understanding closed systems.

The center of mass of a closed system (no external forces) does not move. This comes from the fact that Newton's third law states that any internal force applied will be canceled by another opposite and equal internal force.

The forces you and your friend are applying on each other are internal forces. The argument with 30 N that you made only applies if some external third person pushed your friend 100 N and then you 70 N. Those are forces outside the system where the equal and opposite force was applied to an object outside your system (the third person) causing a net change to your system. Because the forces in your description comes from within what you define as your system, the center of mass doesn't change i.e. no net force. Hope that helps.

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  • $\begingroup$ so what happens to the system then? what is the resultant motion? What I am trying to understand is how can we each feel an action-reaction force pair? $\endgroup$ – MattGeo Dec 1 '18 at 3:01
  • $\begingroup$ Both of you experience 170 N. Now simply calculate the motion based on the masses. $\endgroup$ – Aakash Lakshmanan Dec 1 '18 at 3:03
  • $\begingroup$ I don't think that can be correct. I don't think you can add the applied forces and the contact forces like that. In a lot of textbook examples if two people push on each other with the same force, they each feel the same force, not a double amount. Sort of like tension in a rope. if 2 people pull on a rope at opposite ends with 100 N each, the tension in the rope is 100 N. It is not 200 N $\endgroup$ – MattGeo Dec 1 '18 at 3:14
  • $\begingroup$ The rope is a different type of system which should be discussed separately. Not sure what the textbook exactly said but if you can tell me why you think the forces shouldn't add, I would be happy to help. Consider however, a situation on an ice rink. You push your friend and you both float apart. You push your friend and your friend also pushes on you. You fly apart faster. $\endgroup$ – Aakash Lakshmanan Dec 1 '18 at 3:27
  • $\begingroup$ This makes perfect sense to me. This was my original thought prior to even posting this. It is just that I have been told this is wrong. Please see above in this thread what Dale has given as an answer. Others have told me that the situation I have described is not possible, and yet I don't see how. Your version feels the most correct. $\endgroup$ – MattGeo Dec 1 '18 at 3:43

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