Why can internal forces be disregarded considering the motion of the center of mass of the system?

Question

I've been told when considering the net force on a system of particles, only the external forces on all the particles matter. Why don't the internal forces on each particle matter? I've been told it is because of Newton's third law, that:

$$F_{ij}^{int} = -F_{ji}^{int}$$

However it's not clicking with me, mainly because I can't seem to piece together why the above is true. $j$ and $i$ refers to particles, so $F_{ij}$ refers to the force on the $i$th particle due to the $j$th particle. From what I gather, it's saying when some particle $i$ pushes on another particle $j$, that pushed particle $j$ exerts an equal and opposite directed force on the $i$th particle. So the $i$th particle stops accelerating, since it feels an equal and opposite force on impact, and the $j$th particle starts to accelerate. How does this explain why internal forces do not be considered? I feel like I have a fundamental misunderstanding of Newton's Third Law here, or understanding of what's going on here with the particles themselves.

However my logic behind this is that, if all the particles are uniform, and the forces on it act on all particles, and so no particles will accelerate relative to any others. For example, if a clump of particles were dropped from a building (neglecting air resistance) no particles would fall faster than the others, so only external forces can be considered.

EDIT:

I'm going to try to illustrate a situation describing forces on individual particles that amount to an external force:

Suppose a particle in a system of many particles in moving inside the multi-particle mass at some velocity $v$, and suppose it has such a velocity from experiencing an external force that affected it and some particles near the point of impact. Suppose it collides with another particle elastically. The original particle now moves in velocity $-.5v$ while the particle hit goes at $.5v$. Both move until colliding at the opposite ends of the connected mass. This causes a net force of $0$ on the mass as a whole. But the mass as a whole was struck by an external force? It should be experiencing a net force in the direction of the external force? And suppose this scenario happened to all the initial particles near the external force when it made impact, so no bulk motion occurred. Where am I going wrong with this? Sure, some collisions could be inelastic but some wouldn't, surely? And if so this would happen which seems bizarre to me.

Answer 1

I am not sure why you are confused. You need to understand two things. There first is that the total momentum of the system is the sum of each particle's momentum. The second thing is that the rate of change in each particle's momentum is the force that acts on it.

From these two facts it follows that the rate of change of total momentum is the sum of all the forces on each particle. This sum can be decomposed into two pieces. The first piece is the sum of external forces on each particle and the second is the sum of internal forces. The second sum, the one of internal forces, can be grouped as action reaction pairs. Each pair must sum to zero and therefore the total sum just be zero. Since the sum over internal forces us zero, the total rate of momentum change must be the sum of external forces.

As an example, consider two billiards balls (mass $m$), one above the other falling under gravity. Then the external force is $2mg$ pointing down. Now suppose we add a compressed spring between the two balls. This spring exerts a force $F$ on the top ball pointing up and it must exert an opposite force on the bottom ball, namely a force $F$ pointing down. Therefore the total force is $mg -F$ pointing down from the top ball plus $mg+F$ pointing down from the bottom ball. In the sum, the $F$s cancel out so the total force is again $2mg$. Here you can see the internal forces for not need to be considered because the cancel out when doing the sum.

Answer 2

From what I gather, it's saying when some particle i pushes on another particle j, that pushed particle j exerts an equal and opposite directed force on the ith particle. So the ith particle stops accelerating, since it feels an equal and opposite force on impact, and the jth particle starts to accelerate. How does this explain why internal forces do not be considered?

If an abstract or general case doesn't immediately make sense, it may be of some use to consider a specific example of low $N$-count.

As an example, consider a 2 particle system. Particle 1 exerts a force on particle 2, and particle 2 exerts a force equal in magnitude, yet opposite in direction, on particle 1.

$F_{12}$ is the force exerted on particle 2 by particle 1, while $F_{21}$ is the force exerted on particle 1 by particle 2.

Since $F_{12} = -F_{21},$ we can clearly see that the net force of this system of (two) particles is zero (0) via $F^{int}_{net} = F_{12} + F_{21} = F_{12} + -F_{12} = 0.$

Hence, this system's center of mass does not accelerate. Thus, for this system of particles to accelerate, only external forces matter, as such forces are the only ones that can induce any accelerations.

A further assumption must be made for rigid bodies, and that is the condition that the particles within the system remain a fixed distance from one another; if this isn't the case, all the particles will merely clump up as they accelerate towards one another, but this has no effect on the system's acceleration as a whole, which is also another way of saying that the system's center of mass does not accelerate (as we saw, the internal forces sum to 0).

Here is an explicit example of a system of four particles:

$$F_{12} + F_{13} + F_{14} + F_{21} + F_{23} + F_{24} + F_{31} + F_{32} + F_{34} + F_{41} + F_{42} + F_{43} = (F_{12} + F_{21}) + (F_{13} + F_{31}) + (F_{14} + F_{41}) + (F_{23} + F_{32}) + (F_{24} + F_{42}) + (F_{34} + F_{43}) = (F_{12} + -F_{12}) + (F_{13} + -F_{13}) + (F_{14} + -F_{14}) + (F_{23} + -F_{23}) + (F_{24} + -F_{24}) + (F_{34} + -F_{34}) = (0) + (0) + (0) + (0) + (0) + (0) = 0$$

At this point, it should be obvious that the sum of all internal $F_{ij}$'s sums up to zero (0). You can show this mathematically for any system of $N$ particles.

For example, if a clump of particles were dropped from a building (neglecting air resistance) no particles would fall faster than the others, so only external forces can be considered.

This is an entirely separate thing altogether. It seems you are conflating the fact that an object's acceleration due to gravity is independent of its mass with an imposed condition that these various falling objects must be apart of the same system; this is not necessarily true. What is true, however, is the fact that the force of gravity would be an external force to any rigid body, and, as such, the acceleration of this rigid body would only depend on this external force and not on its internal forces.

Regarding your edit:

This causes a net force of 0 on the mass as a whole. But the mass as a whole was struck by an external force? It should be experiencing a net force in the direction of the external force? And suppose this scenario happened to all the initial particles near the external force when it made impact, so no bulk motion occurred.

If the net force is 0, then the net external force must have been 0 as well, for it has already been shown that the net internal forces always sum to 0. Your edit doesn't really make sense, so one cannot give any valid answers regarding it. It is like asking a muffin what its favorite color is... it simply does not make sense to ask such a question.

EDIT: I just saw that this question was asked three years ago, yet it popped up on my screen quite high as if it were a recent question. I suppose I'll leave the answer.

Answer 3

An elastic collision has many different outcomes depending on the mass and velocity of both particles. To find both final velocities of the particles, you will need their masses and their initial velocities and use conservation of energy and momentum equations. In a simple case where the masses are equal and one is stationary, the moving particle will stop and the other particle will have the same velocity as the first particle. If a particle is moving, it will contribute to the total momentum of the system, which is the vector sum of all the momenta of the particles. The force that gave the first particle its momentum contributed to the total momentum of the system. In all collisions, some or all momentum of that particle will be transferred to another particle and the first particle will lose an equal amount of momentum. Newton's third law states that for an action-reaction pair of forces, and all internal forces are action-reaction pairs, the momentum gained by one particle will equal the momentum lost by the other particle over a time interval. The net change in momentum of the system was zero.

Answer 4

The original question is "Why don't the internal forces on each particle matter?" As too much of what is written here, and everywhere, about introductory physics shows, there is almost always a failure to distinguish between the actual physical situation and the abstract mathematical situation. People talk about forces "canceling" and forces "summing to zero," but this is only in the math. In any actual physical situation, no forces cancel and they don't "sum" to zero. All forces exist and act according to Newton's Third Law. If internal forces did cancel in the real world, then nothing would move. But if you had to measure all of the internal forces and do calculations with them, then you would have a lot of work to do. By adding equations, you set up an equivalent, but not identical, physical system in which there are no internal forces. Here is a simple example that can be extended to as many objects as you want. Consider an external force F1 acting on a mass M1, which is connected by a massless string to a second mass M2. M1 pulls on M2 with a force F21(force on 2 of 1), and by Newton's Third Law, M2 pulls back on M1 with a force F12 (force on 1 of 2). F21 accelerates M2, and F12 decelerates M1. These are real forces that continuously act. Sum of the forces on M1 = F1 - F12 = M1 * A1 and sum of the forces on M2 = F21 * A2 How is this usually solved? You combine the equations and create a mathemtical description of an equivalent, but not identical, physical situation. Let's look at the math: F1 - F12 + F21 = M1 * A1 + M2 * A2. As the masses are connected, A1 = A2 = A, and from Newton's Third Law, F12 = F21. Substituting gives: F1 - F12 + F12 = M1 * A + M2 * A = (M1 + M2) * A. This last part could descrbie a single mass M = M1 + M2. The math is the same for describing a single mass M with acceleration A as it is for two connected masses M1 and M2, each with acceleration A, whether they are connected or not. An external force F1 is needed to accelerate M with A, an external force F1 is needed to accelerate M1 with A when M1 is attached to M2 with A. By considering M1 and M2 connected to each other as a "system," the forces "cancel" in the math, and you can get answers more easily than if you had lots of equations with lots of internal forces. You can use the same reasoning for questions about momentum, where particles not actually connected and having different accelerations from different external forces are treated as if they were actually connected and had the same acceleration. The internal forces that would start to act if the particles were connected would result in the same effect as if a single mass were acted on by a single external force equivalent to the vector sum of the actual different external forces. A lot of force and momentum problems are about setting up an equivalent, but not identical, physical situation that can be described by math that is easier to work with because of the "cancellation" that occurs. But no internal forces actually cancel. It's just that two or more masses can often be viewed as a single mass that doesn't have any of the internal forces present in the actual physical situation. You work an equivalent problem to find out something about the actual, more complicated situation.