# Why does Action-Reaction pair cancel out in a system?

When I first studied Newton's third law, I heard that the action-reaction pair of forces do not cancel out as they are applicable on two different bodies. And that makes sense.

However, people say in a system of particles, the internal forces of the system cancel out by Newton's third law. How can the forces on individual particles in the system cancel out if the forces are applicable on different constituent particles?

For e.g. let's say $$F_{ij}$$ is the force on the i-th particle due to the j-th particle. By Newton's third law, another force, $$F_{ji}$$ will be applicable on the j-th particle due to the i-th particle. Good till now. But how can these two forces cancel out if they are applicable on each other?

I have seen forces cancelling out only if they are on the same body with equal magnitude and opposite direction.

• If they didn't cancel out, the two sides would accelerate away from each other. Mar 29 at 20:19

tldr:

It's a matter of switching your viewpoint; that is, it's a matter of changing what we view as the "system" and calculating appropriately. In addition, thinking about this as the "forces canceling out" is not really the best way to think about things. Instead, you should be thinking about the outcomes of those forces acting by determining whether or not the the impulses or internal works done on the objects inside the system by each other cancel.

When we think of two objects 1 and 2 exerting forces on each other, and we care only about the motion of one of the objects (object 1, say), then we consider only the forces acting on that object. To figure out what happens to object 1, we can compute the impulse imparted to object 1 by object 2 as $$\vec{J}_{\textrm{2 on 1}}=\vec{F}_{\textrm{2 on 1}}\Delta t\,,$$ where $$\Delta t$$ is the amount of time that the force acts (I am assuming for simplicity that the force is constant; we would need an integral otherwise).

In the case where this is the only force acting on object 1, this is equal to its change in momentum, $$\Delta \vec{p}_1 = \vec{J}_{\textrm{net on 1}} = \vec{J}_{\textrm{2 on 1}}\,.$$ Similarly, $$\Delta \vec{p}_2 = \vec{J}_{\textrm{net on 2}} = \vec{J}_{\textrm{1 on 2}} = \vec{F}_{\textrm{1 on 2}}\Delta t\,,$$ again assuming that the only force acting on object 2 is $$\vec{F}_{\textrm{1 on 2}}$$. That's the end of the story, as far as it goes. We can now compute what happens to the two objects individually as a result of their collision with each other.

However, it is often the case that we know very little about the details (which is to say, the forces) of the collision between the two objects. If we can find a way to eliminate these forces from the calculation, then we don't need to know these details, and that can help us. So we note the following interesting phenomenon, which is a consequence of the third law: $$\Delta \vec{p}_1 + \Delta \vec{p}_2 = \vec{F}_{\textrm{2 on 1}}\Delta t + \vec{F}_{\textrm{1 on 2}}\Delta t = \vec{F}_{\textrm{2 on 1}}\Delta t +(-\vec{F}_{\textrm{2 on 1}})\Delta t =0\,.$$ Again assuming that no other forces act during the collision (or, more true-to-life, they're small enough that we can neglect them), the sum of the changes in momenta of the objects is zero!

For this reason, we redefine our system to be composed of both objects together, and we define the total momentum $$\vec{P}$$ to be $$\vec{p}_1+\vec{p}_2$$. Our calculations show that even though the objects are exerting forces on each other during the collision$$-$$and thereby changing each others' momentum$$-$$the total momentum of the system is left unchanged. This is what is really meant by "the internal-to-the-system forces cancel each other out during the collision".

So, just to comment, then: the point is that we need to be careful and specific about what we mean when we say the forces "cancel each other out", because they "can" in one context and not in another, but that's because we are computing different things. (Really, we shouldn't talk about the forces canceling each other out; instead we should be talking about the impulses canceling, or, in an energy context, the internal works canceling (or not).)

To add fruit to the fire, to coin a mixed metaphor, things are different again when we are thinking about the work done during collisions, i.e., when we are taking an energy perspective on the system rather than a momentum perspective.

If two objects are colliding and exerting contact forces on each other (like two billiard balls), then one of the objects does positive work and the other does negative work, and these works are equal and opposite, so there is no change in kinetic energy (assuming the collision is elastic).

However, if two objects inside the system are interacting via conservative long-range forces like the gravitational force, then the two objects can both do positive work on each other, increasing the kinetic energy of the system! So in this latter case, the forces don't "cancel out", even though they are equal and opposite forces acting on two parts of a single system. But, we still need to be careful! The forces do "cancel out" in the sense that the total momentum is still conserved, because the changes in momentum are equal and opposite.

• Thanks,another question:Two boys playing tug of war exert equal forces on each other.Is it true that they both are at same position?If yes,then why is it that in other cases particles exert equal forces on each other,and get displaced in the system,while keeping the net momentum of the system conserved?If particles do not change their positions,then also momentum is conserved.But why is that sometimes particles exert forces on each other but remain at the same position,as in tug of war.In both cases,net momentum is conserved but why in some cases initial position of particles remain unchanged? Mar 30 at 7:30
• In the tug of war case, there are other important forces acting: the friction forces exerted on the kids by the ground. Mar 30 at 13:51
• Thanks again , for replying . So, you are basically saying that if the tug of war was played on a frictionless ground , the individual components of the system (the boys) would come closer ? In that case the rope would become slack , and then we can conclude tug of war cannot be played on smooth surfaces . Is that true enough ? Apr 1 at 12:23