It is always best to draw a diagram to convince yourself of things in a case like this.

This is intended to represent a steady state situation: nobody is moving / winning. As you can see, there are two horizontal forces on A: the floor (pushing with 100N) and the rope (pulling with 100N). There will be two vertical forces (gravity pulling down on center of mass, and ground pulling up) to balance the torques - I did not show them because they are not relevant to the answer.

Now I drew a dotted line between A and B. Consider this a curtain. A cannot see whether the rope is attached to B (an opponent) or a wall. A can measure the tension in the rope by looking (for instance) at the speed at which a wave travels along the rope - or by including a spring gage.

Now ask yourself this question: if A feels a tension of 100N in the rope (this is the definition of the force on A), and can confirm (by looking at the gage) that the tension is 100 N, but he cannot see whether the rope is attached to a ring or to an opponent, then how can the tension be 200N? If I pull on a gage with a force of 100N, it *will* read 100N - it cannot read anything else (in a static situation, and where the gage is massless, ... )

I think I understand the source of your confusion based on the earlier q/a that you referenced - so let me draw another diagram:

In this diagram, I have move the point of attachment of the rope with which A pulls B away from B's hands, to his waist. Similarly, the rope with which B pulls on A is moved to A's waist.

What happens? Now there are two distinct points where A experiences a force of 100 N: one, his hands (where he is pulling on the rope attached to B's waits); and another where the rope that B is pulling on is tied around his waist.

The results is that there are two ropes with a tension of 100N each, that together result in a force of 200N on A (two ropes) offset by a force of 200N from the floor, etc.

This is NOT the same thing as the first diagram, where the point on which B's rope is attached is the hands of A - there is only a single line connecting A and B with a tension of 100 N in that case.

As was pointed out in comments, you can put a spring gauge in series with your rope to measure the tension in it; and now the difference between "a single person pulling on a rope attached to a ring at the wall (taken to be the dotted line) and two people pulling across a curtain (so they cannot see what they are doing) is that in one case, a single spring (with spring constant $k$) expands by a length $l$, while in the second case you find a spring that's twice as long, with constant $k/2$), expanding by $2l$.

These are all different ways to look at the same thing.

I have Marion-Thornton 4th ed. around here somewhere. It is an older book and presents some material differently than we are used to in more modern books (for instance they even use the old imaginary time method when discussing some things in special relativity, which I personally dislike). However I agree with DanielSank, different pedagogy does not equal "nonsense".

Newton's laws are presented slightly differently by different books. For instance, it can be argued Newton meant his second law to be $F=dp/dt$ (although he didn't write it in this modern notation), although many books present it as $F=ma$. Some people go even further and try to extract a modern meaning, as I've seen some people say Newton's third law *is* the conservation of momentum. This may be pedagogically useful, but not historically accurate. It is worth reminding that some debate over the exact statements translated to modern language is understandable. Even though Newton invented calculus, some concepts in mechanics still took long after Newton to come into their modern understanding, such as the concept of kinetic energy was put in its modern form much later.

Thus to answer this question requires agreeing on a statement for Newton's third law. I don't have Marion-Thornton handy, so using wikipedia

When one body exerts a force on a second body, the second body simultaneously exerts a force equal in magnitude and opposite in direction on the first body.

The force between two particles in electromagnetism can violate this. For a concrete example consider a positive charged particle A pulled along the x axis at a constant velocity in the positive direction, and another positive charged particle B pulled along the y axis at a constant velocity in the positive direction. If it is arranged such that when A is at (0,0), B is at (0,1), then we can calculate the fields and find:

- the electric forces on the particles will be in opposing directions
- the magnetic force on A is zero
- the magnetic force on B is in the -x direction

Does this mean momentum is not conserved here? No.

If we include the person or device pulling these charges along as part of the system (so there are no external forces), then we should expect the momentum of the system to be conserved.

Where is the missing momentum then? It is in the fields!

I constructed this scenario specially to also help break a bad habit of some descriptions of this phenomena. Because the charges are moving at a constant velocity, there is no radiation. We don't need radiation to provide a force back on the partices or something to solve this. Momentum can be stored in the fields themselves. (While not shown in this example, even static fields can have non-zero momentum.)

## Best Answer

We start by noting that force is the rate of change of momentum. Let's suppose you and I are floating in space (so we are the only two interacting bodies) and you're pushing me so I feel a force $F_{me}$, then:

$$ F_{me} = \frac{dp_{me}}{dt} $$

where $p_{me}$ is my momentum.

But we know that momentum is conserved, so since you are the only thing interacting with me your momentum, $p_{you}$, must be changing in the opposite sense to balance out the changes in my momentum. In other words:

$$ \frac{dp_{you}}{dt} = - \frac{dp_{me}}{dt} $$

And since force is rate of change of momentum that means there is a force on you:

$$ F_{you} = \frac{dp_{you}}{dt} = - \frac{dp_{me}}{dt} = -F_{me} $$

So the two forces are equal and opposite just as Newton's third law tells us.

The details of exactly how the forces are transmitted depend on exactly how the two bodies are interacting, but whatever the interaction the changes in momentum must be equal and opposite, and therefore the forces are equal and opposite.