# Is there any case in classical mechanics where Newton's (strong) third law doesn't hold?

Is there any case in classical (non relativistic) mechanics where the strong form of Newton's third law does not hold (that is, reaction forces are not collinear)? For example, if we consider a system of two point particles in equilibrium with each other upon which a constraint acts so that the reaction forces are directed in a direction that is not collinear. Is such a situation possible?

-
Do you mean (weak form) of Newtons third law? There are examples in EM. – hwlau Jan 21 at 5:19
I'm specifically asking about Newtonian mechanics. – Sumukh Atreya Jan 21 at 7:27
– Qmechanic Jan 21 at 9:54

 Fair enough. Another question though; The (Weak) Form of Newton's Third Law says that, if we consider two particles, 1 and 2, the force of 1 acting on 2 is equal in magnitude and opposite in direction to the force exerted by 2 on 1. So that would mean that the reaction forces are in vectorial equilibrium. But for this to be the case, don't the reaction forces have to act in the collinear direction? If so, what is even the point of explicitly stating the strong form of this law? – Sumukh Atreya Jan 21 at 4:21 Okay, I had never seen "weak" form applied to Newton's 3rd Law, and a quick googling doesn't give me the answer in the top 2 hits, but I assume you mean the case where the net force is zero, but the net torque is not. For example, suppose we have a 2-dimensional system, with particles at positions $\textbf{p}_1=1m \textbf{i}$ and $\textbf{p}_2=-1m \textbf{i}$, but forces $\textbf{F}_1=1N \textbf{j}$ and $\textbf{F}_2=-1N \textbf{j}$. This isn't made explicit, but it's not seen in nature. If seen, it would violate conservation of angular momentum. – thatnerd Jan 21 at 13:50 oops, used p rather than r, sorry. I was learning the LaTex to make it look good, and went over the 5 minute deadline for re-editing. For $\textbf{p}$, please pretend it's $\textbf{r}$ – thatnerd Jan 21 at 13:57 Look at this. Look at the first two lines of the answer where the strong form is mentioned (I didn't know about the 'strong' or 'weak' form until a couple of days ago either, hence the confusion.) – Sumukh Atreya Jan 21 at 14:38 @SumukhAtreya Ah, yes, I was right in my guess. The weak form simply implies conservation of linear momentum, the strong form conservation of angular momentum. And for conservative, spherically symmetric fields centered on the point particles, you're right. You can't have a torque between two point particles if they can only push & pull each other along the line. Let them "shove" each other "sideways" (by which I mean perpendicular to the line joining them), and you still have an action/reaction pair, but it violates conservation of angular momentum. – thatnerd Jan 22 at 16:22