# Mind-spinner 1: When (do/don't) action/reaction force pairs cancel out?

 P: 3 Hello to everyone! This is my first post. I have some basic knowledge in physics back from high-school, but since I decided now to pursue a degree in physics, I am starting from the bottom to get a good base. As I am reading through some of the textbooks on the basics of physics (classical mechanics, the basis of electricity, magnetism, waves, thermodynamics and alike), I come across some "questionable" arguments and examples. I would like to put them forward onto this forum to see if I'm right in posing these, to me at least, very interesting mid-spinners. Here's one of them: Newtons third law of action and reaction: I will first pose some general principles (as I understood them, from how textbooks present them; so please correct me if I make a mistake here) - If we take a point of view looking at the system such that action-reaction force pairs are within the system, they do cancel out to zero and there is no acceleration; while if one of the forces in the pair is outer to the system it can result in an acceleration of the system (unless there is another outer force equal in magnitude but opposite in direction). The examples are such as: • Pushing on the car’s dashboard while inside (within the system) doesn't move (accelerate) the car, while pushing from the outside (of the system), using the force of friction with feet against the road, does accelerate the car; • Trillions of inter-atomic forces within the football do not, while kicking the ball (acting on it outside of the system) does accelerate the ball; etc. The difficulty I have is putting some of the situations into these broad definitions, such as when a canon fires a cannonball and recoils (perspective: cannon and the cannonball are both a part of the observed system), and alike. As there are other forces acting on this system, to simplify, we can imagine the following idealised situation: We are observing a system of two objects A and B acting on each other with an action-reaction force pair. The system is totally isolated from the outer world, i.e. two astronauts in the outer-space, holding hands and then pushing on each other. This would accelerate each astronaut in the opposite direction. Even-though I understand that only if we look at the each astronaut as a separate system, we would say it is accelerating, still our “whole” system is expanding through space. If we take the central point in-between the astronauts as the centre of the system, it is not moving, still if the system itself was to hit an object C on its path of expanding it would act on it with a force. Furthermore if object A has more mass than object B, they would accelerate differently. Would this mean that the central point of the system is moving thus the system itself is moving as well? Let us take an example of inter-atomic forces within the football not accelerating it and replace it with an atomic bomb which explodes. Is there a difference here? Perhaps a better understanding is needed of what is actually meant by saying that “the forces cancel out” or that there is no acceleration of “the system”. If you see a hole in my thinking process, could you please point it out and explain how the above mentioned general rule (that action/reaction force pair within the system cannot accelerate it) holds in this situation. Since it is the law, I presume it should hold, but it does lead me to:... ...another related interesting question that I'm thinking about: if two astronauts hold hands and don’t let go after pushing on each other; then pull and push again but directing the force on an angle such that they start rotating (as a system, one around the other, like two orbiting stars), would it be possible for them to adjust these pushes and pulls and their directions and magnitudes such that they start moving (as a system) through space in a certain direction, or would this be fundamentally impossible (according to the laws of physics) and they would always just be rotating in one central “spot” in space? Happy physics! Vlad
 P: 282 Your textbook should be clearer. The rule is, if there's no external forces on a system, then its centre of mass doesn't accelerate. What's the center of mass? Roughly, it's the average position of the mass of the system. Quantitatively, the center of mass of a system of N particles, each with mass mi and position ri, is $$\vec{r}_{\mathrm{centre \ of \ mass}}=\frac{1}{M}\sum_{i=1}^N m_i \vec{r}_i.$$
 Sci Advisor Thanks PF Gold P: 12,130 Are you confusing the Third law with the First law? Law three has noting to do with equilibrium or acceleration.
 Mentor P: 17,207 Mind-spinner 1: When (do/don't) action/reaction force pairs cancel out? Hi Vlada, welcome to PF! The rule is quite simple. First, you define your free bodies (aka systems). Often a problem will have more than one interacting body. Second, draw a free body diagram for each system. Third, label only the external forces on each free body and neglect all internal forces since they are all canceled out. Fourth, use Newtons 3rd law to define any action-reaction pairs of forces between interacting free bodies. Fifth, use Newtons 2nd law to find the acceleration of the COM for each system.
 P: 3 Hi all @sophiecentaur No confusion here, these are examples that various textbooks use (Hewitt Paul G, 2008, "Conceptual Physics", Addison Wesley, Boston for instance) in the Newton's third law section to illustrate the difference in the result of an action/reaction force pairs on a system depending on whether both are within the system or one of them was outer to the system. @dEdt & DaleSpam Centre of Mass, not the centre in space, of course! That was exactly the concept that I was missing. Thank's guys!
 P: 360 Newtons laws are very redundant to be honest..classical mechanics can be more elegantly put in terms of the functional schemes
 P: 3 Thanks HomogenousCow As a beginner I must say I don't really know how. Can you give an example please
 P: 360 In the more useful Lagrangian scheme, one starts with the Scalar Lagrangian, which in most cases is the kinetic energy minus the potential energy, then one simply feeds this into the euler lagrange equation for x(t). The advantage of this alternate but physically equivalent scheme is that we are only dealing with scalars and never have to worry about which direction things are pointing, the lagrange formalism also makes changes coordinates much easier. http://en.wikipedia.org/wiki/Lagrange_equation If you've taken courses in calculus and differential equations this article will put you up to speed nicely (you might find some of the wording to be confusing, I know I did)