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If there are two people on ice skates named $A$ and $B$, and $A$ pushes on $B$ with a force $F_A$, then both of them will experience a force of magnitude $F_A$ in opposite directions. I'm confused as to what will happen if $B$ is also pushing at the same time with a force $F_B$. If $A$ pushes on $B$, he experiences an opposite force because of his 'crime' (the act of pushing on $B$), but now $B$ is also pushing at the same time with a force $F_B$, so maybe the forces add up on each of them? I'm not sure. I would really appreciate any help!

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@RonMaimon To be honest, it isn't, though I have no problem adding the homework tag. But could you please just explain me what would be the answer? –  Alraxite Oct 31 '12 at 1:42
@RonMaimon So the answer is definitely $F_a$ + $F_b$? I mean if $F_A$ is 10 N and $F_B$ is 7 N, then each will experience 17 N? –  Alraxite Oct 31 '12 at 1:44
@RonMaimon Thank you! If you want you can close this but I just wanted to say something else: I assume then, if both of them push against each other with equal forces then each will experience twice the force? I asked a similar question a few days back (thought it was primarily on tension) and the answer I received was that it wouldn't be twice the force. Here's the link: physics.stackexchange.com/q/41291 –  Alraxite Oct 31 '12 at 2:00
Ok--- I guess I should write an answer. The reason for the confusion is something I didn't say properly (it's correct that the forces add, but I didn't explain the rope issue). –  Ron Maimon Oct 31 '12 at 2:06
@RonMaimon And btw, I've always thought of forces the way you've presented in your answer. I've always found the formula for momentum the most intuitive and natural (unlike the formula for mechanical/kinetic energy; and yes, I have read your brilliant answer on the justification for the formula of kinetic energy), and I do still think of them that way. –  Alraxite Oct 31 '12 at 4:29

3 Answers 3

up vote 8 down vote accepted

General Nonsense

A force is a current, like an electric current. A current is the rate of flow of a conserved quantity from place to place. The force is the current of momentum from one object to another. Like a regular electric current tells you how much charge is going from point to point, a force tells you how momentum goes from point to point.

When you have 20 Coulombs of charge at one point, and 1 second later, you find 20 coulombs at another point attached by a wire, you can say that 20 Amperes of current moved through the wire. It's not 40 coulombs of current, even though one point lost 20 Coulombs, and the other point gained 20 Coulombs. This is what makes currents currents--- you count the flow of the conserved quantity from one place to the other, you don't double count, by saying "oh, 20 Coulombs disappeared here, and 20 appeared there, so it's 40 Coulombs overall". The flow is defined by how much of the conserved quantity crossed an imaginary surface separating the two points, counting once, with one sign for one direction, and the other sign for the other direction. This is why Newton's 3rd law is true--- the force from A to B is opposite to the force from B to A just because it is a current, and the flow in opposite directions is opposite.

The same is true for a hose carrying water from the main to your yard, at a rate of 8 kg/s. In one second, the main loses 8 kg of water, your yard gains 8kg. But the total rate of flow is not 16 kg/s, just 8. The rate of loss of water in the main is equal to the rate of water mass gain in the yard. For water, you don't get confused with double counting.

In the case of 1 dimensional motion, there is a conserved quantity--- the momentum. The momentum associated to a velocity in the positive x direction is positive, and the momentum associated to a velocity in the negative x direction is negative, so unlike Energy (or nonrelativistically, mass), the conserved quantity is signed. It's like charge, not like mass. The rate of momentum flow is in momentums/second, or Newtons. It's annoying that the unit of momentum is not named after anybody, I'll call it "the Galileo". One Galileo is 1 kg times 1 m/s. A Newton is a Galileo per second, it's Galileos of momentum going from one body to another.

When you have two people on two ends of a rope in a tug of war, pulling with equal strength at 10 N, 10 Galileos of momentum are flowing from one to the other every second, through the rope. The tension in the rope is 10N, and the force on opposite sides is opposite sign, and equal to 10N. Just like the hose, if you slice at any point, 8kg/s of water is moving through the hose, when you slice the rope, anyplace you slice it, there is 10N of tension of each end on the other, which is the current of momentum through the rope.

Your thing

Your confusion comes from the issue of reconciling momentum flows with exertion-feeling, that feeling that one is exerting oneself when one is pushing. The exertion-feeling is not directly related to force, rather to energy consumption and production, and without a good model, it is easy to get confused about the forces.

Consider two astronauts in space, pushing on each other. I will model the astronauts as two identical blocks of mass M, and they have a compressed spring attached to a plate. The plate can be attached to the other astronaut, to the other block directly, or you can attach the plate to the other plate. Further, each block can "choose" whether to push--- meaning that there is a removable rigid rod attached to the plate, which prevents it from moving, which you can remove at any time to allow either astronaut's spring to do work.

I will consider four situations, and then the situation should be clear.

  1. The two plates are each touching the other astronaut--- each one is pushing on the other, and both are pushing.
  2. The two plates are each touching the other astronaut, and only one is pushing, the other spring is replaced with is a rigid rod.
  3. The two plates are pushing on each other, and both astronauts are pushing, meaning both springs are free to expand.
  4. The two plates are pushing on each other, and only one astronaut is pushing, meaning only one of the springs is free to expand.

In all situations, each spring starts out compressed an amount x, and has spring constant k. The total spring constant in the first case comes from adding springs in parallel, the total spring constant is 2k, and the total energy is

$$ kx^2 $$

The two forces from the two springs always add up, the total force at the initial time is $2kx$, and you get double the force, as your intuition suggested for two astronauts pushing on each other's chests.

The work done by the force converts all the energy to kinetic energy, so that the final velocity (which is equal by symmetry) is:

$$ Mv^2 = kx^2 $$

So that

$$ v = \sqrt{k\over M} x$$

That's the velocity the two astronauts have when they shove off each other pushing on each other's chest, with their arms modelled as springs. That's situation 1.

In situation 2, only one of the astronauts pushing, the force is halved, the initial energy is reduced by a factor of 2, and the final velocity is reduced by a factor of $\sqrt{2}$, which comes from the energy law:

$$ Mv^2 = {kx^2\over 2} $$

Next, consider situation 3. Here the two plates are attached to each other. The springs are in series, so that the effective spring constant is halved. But when both springs are compressed, the amount of displacement is doubled, so that the total energy in the springs is

$$ {k\over 2} {(2x)^2\over 2} = kx^2 $$

just as for the first case. The final velocity is the same as in case 1. But the force is completely different! The momentum flowing through one spring is the same as the momentum flowing through the second spring, and the force is kx, not 2kx.

The final velocity, though, is unchanged--- when both are pushing on each other, they end up moving equally fast even though the force is exactly the same as if only one were pushing. The reason is that the displacements are larger by a compensating factor, so that the work done by the forces is greater.

The situation 4 gives the same velocity as when only one was pushing before, but the force is still exactly the same--- the force is kx, whether both springs are pushing, or only one is pushing. This is the counterintuitive thing that confused you--- it depends on whether you are pushing at the same spot or at a different spot. When the momentum flow is through both force sources, the force doesn't add up over the two sources, but the work still adds up.

In case 4, the final velocity is still reduced by $\sqrt{2}$. So while the end velocity doesn't depend on where you push, only whether both people are pushing, the force you generate to do the pushing does depend on where exactly you are pushing.

This is what is leading to the intuition paradox.

When the two people are pushing on each other's chest, the force is doubled. When they are pushing on each other's hands, the force isn't doubled, but the displacement is doubled, so that the work is still doubled. The work is what is felt as the exertion, and to produce the same final velocity (although not the same initial force) you need to exert yourself twice as hard when pushing alone.

But if you have a rope you want to break, and you're not strong enoug, and you have a friend, tie it to a wall and ask him to pull on the same end as you. Don't put him on one end and you on the other.

This is the correct answer, I wrote nonsense previously. The gist of the other answers to the other problem was not wrong, but they didn't consider the issue of pushing through the same link, or pushing through parallel links. It makes a difference in the force.

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+1: This answer is more than enough as regards the question. BTW, Good story Ron. Time to bed :-) –  Waffle's Crazy Peanut Oct 31 '12 at 5:40
@CrazyBuddy: I am sorry for going on and on, but this is the first thing that has confused me in Newtonian mechanics for 20 years. Maybe I'm just getting old. –  Ron Maimon Oct 31 '12 at 5:41
@RonMaimon I haven't read your answer completely yet (and will read it and reply on it soon), but I just wanted to know the conclusion: What happens if two astronauts push each other with a force 10 N and 7 N respectively? What will be the overall force exerted on each during the pushing? Just to be clear and rigorous I will bring in the following postulates: next comment –  Alraxite Oct 31 '12 at 6:20
1. The astronauts can extend their hands indefinitely so that there is no problem of not being able to push due to the distance. 2. They apply the force uniformly for exactly the same time. By uniformly I mean that after t seconds they would have given each other 10*t Galileos and 7*t Galileos. 3. And most importantly, they apply the forces hand on hand. I hope it won't be a stupid thing to ask after you writing such a detailed answer! Just want to know the answer. And thanks! –  Alraxite Oct 31 '12 at 6:20
@Alraxite: you aren't asking for a possible situation--- if one astronaut is pushing with 10N of force hand-on-hand, and the other is pushing with 7N hand on hand, the weaker astronaut's hands will bend back until the weaker astronaut is not actively pushing at all, but giving 10N back. You can only ask the hand-on-hand question with equal force, and then it is what I answered. You can model this with unequally compressed springs in the model--- the springs will first come to force equilibrium, then push themselves to expend the rest of the energy. –  Ron Maimon Oct 31 '12 at 6:32

Your Answer is as given below for both force and its reaction-

Let for Push, A applies a force Fa on B and B applies a force of Fb on A, then

  1. Equal forces Neutralize each other.

  2. Unequal forces results in unidirectional motion in the direction of greater force.

This proves for both push or pull.

Hence to move other person, one has to exert double or greater force than which the other applies or it will neutralize if equal and there will be no push or pull.

$Explaination: -$

Here, $F$ force applied on the wall; $R$ reaction from the wall and let S be the strength of the wall.

When a person pushes a wall with a force less than the strength of the wall, there is equal reaction from the wall as per Newtons's third law.

Say $Fw = R$ the reaction from the wall

Therefore $Fa = Fw$ & $Fa - Fw = 0$, Here Fw is the reaction from the wall and force will not result any push.

Now suppose he pushes a person there will be reaction only if other person stands to take the load, else other man will be pushed.

$if Fa \gt Fb$, $Fa$ will move the other person, and reaction will neutralize and other person will be pushed with a force equal to $(Fa - Fb)$.

If the other person applies same force as that of the first person applying force, he will neutralise the incoming force and there won't be any movement.

$if Fa + R = Fb +R $, $Fa - Fb = 0$, No body will move and reactions will neutralize

Now if the other person or the first person applies a force greater than the other, equal portion of the force will get neutralized and extra force will push the other person who applies lesser force.

$\implies$ $if Fa \gt Fb$, $Fa$ will push $Fb$ and R, the reactions will cancel each other or

$\Rightarrow (Fa - Fb)$ will push the second person.

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What a picture ha! +1 for work (pun intended). Hope I'm not forcing the joke, over a distance...I'm out. –  Andres Salas Aug 11 at 22:27
thanks, I really like it. –  Berry P J Aug 12 at 16:45

It appears that the confusion solely arises from the taught picture of 'exerting a force': A exerts a force on B. This kind of 'sender-receiver' understanding of mechanical force is nonsense. A force between two objects should be understood as an interaction between the objects, which has no direction at all. It is meaningless to say "if I give a force and then you give a force and ..., what is the total force ?". In short, to exert a force = to do something so that an interaction comes into being between (the molecules of) the objects. Force is not passed on from ... to ... !!

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