In all books and texts I've seen so far, Newton's Second Law is used to prove that the net torque acting upon a system of particles with respect to its center of mass is equal to the rate of change of this system's angular momentum with respect to its center of mass. At this point, a crucial assumption is introduced: the internal force pairs obey the strong form of Newton's Third Law. It is then shown that the rate of change of the system's angular momentum is in fact equal to the external net torque acting on it. This result is then applied to rigid bodies without any caveats, and, surprisingly?, it works.
The issue with this is: for a rigid body to remain rigid, Newton's Third Law cannot hold in its strong form. A simple example shows this: picture a rigid body system made up of three particles, initially at rest and positioned along a vertical line, with the middle particle separated by some distance $d$ from the other two particles. A horizontal force directed to the right is applied to the upper-most particle, and another horizontal force of the same magnitude directed to the left is applied to the middle one (it's as if the middle one was a pivot). The linear momentum and angular momentum theorems tell us that both the upper and bottom particles will experience acceleration in the horizontal direction. This must be the case for the rigid-body condition to hold. So the bottom particle experiences net force. Since there is no external force acting on it, that force must originate from within the system, i.e., the bottom particle is acted upon by either the top particle, the middle particle, or both particles. Since the line of action from the bottom particle to those other particles is vertical and the net force it experiences lies in the horizontal direction, Newton's Third Law does not hold in its strong form.
All this means that internal force pairs in a rigid body produce torque. So this must mean that it is not true that the net torque acting upon a system of particles (both torque from the surroundings and from within the system) with respect to its center of mass is equal to the rate of change of this system's angular momentum with respect to its center of mass, if the system is a rigid body. But Newton's Second Law implies such. So Newton's Second Law, in turn, is also violated for rigid bodies.
The real, physical, world, deals with this easily: rigid bodies don't in fact exist. Any body undergoes stresses and strains when under the action of non-body forces.
What is bewildering to me is how the mathematical model of a rigid body seemingly doesn't make any sense in the context of Newtonian mechanics, and yet it still produces precisely the results we would expect. It's as if we start with a set of laws, deduce some properties of systems after applying some hypotheses, then naively translate this to rigid bodies, which, apparently, should work under a different set of laws, and yet it all works out in the end mathematically. How does this not break apart? What is going on? Have I made a gross mistake somewhere?
Best Answer
Your analysis is correct, you cannot have interactions satisfying the strong form of Newton's third law. As Dale pointed out, you cannot achieve rigidity by central forces between the consecutive particles. Either you need to make the middle articulation rigid (which does not fall easily in your discussion), or you can add a third interaction between the uppermost and lower most particle (view it as a degenerate rigid triangle).
To make things quantitative, I'll assume that the rod is in the $y$ direction and the force applied in the $x$ direction. I'll set mass of each point to $m$, space the linked ones with distance $L$ and an applied force of $F$, and number the points $1,2,3$ from top to bottom. We know that the system's CoM (at point $2$) has acceleration: $$ a = \frac{F}{3m} $$ and the body has angular acceleration: $$ \alpha = \frac{F}{2mL} $$ while being initially at rest. This gives the accelerations of the particles: $$ \begin{align} \ddot x_1 &= a+\alpha L & \ddot x_2 &= a & \ddot x_3 &= a-\alpha L \\ \ddot y_1 &= 0 & \ddot y_2 &= 0 & \ddot y_3 &= 0 \\ \end{align} $$ From Newton's second law: $$ \begin{align} m\ddot r_1 &= Fe_x+F_{2\to1}+F_{3\to1} & m\ddot r_2 &= F_{3\to2}+F_{1\to2} & m\ddot r_3 &= F_{1\to3}+F_{2\to3} \end{align} $$
We know the accelerations and we want to know the forces. Your approach was to set $F_{3\to1}=F_{1\to3}=0$. This gives: $$ F_{2\to1} = F_{2\to3} =-\frac{F}{6}e_x $$ and the equations are consistent with Newton's third law in weak form, setting: $$ \begin{align} F_{1\to2}&=-F_{2\to1} & F_{2\to3}&=-F_{3\to2} \end{align} $$ will give you the second equation of acceleration.
The forces have lateral components to ensure the rigidity of the link as explained before, and therefore each interaction pari generates torque. However, since the two extremal forces are equal, the net torque of the internal forces is $0$ (thankfully). Unlike the usually presented argument, this cancellation is not because each term is individually zero. This careful cancelling out needs to be explained.
Btw, because the triangle is degenerate, you cannot have interactions satisfying the strong version of Newton's 3rd law, even if you add the interaction between $1$ and $3$. If there was a slight angle, it would be possible though.
Note that in your case, we were "lucky" to be able to determine the constraint forces. The fact that the constraint forces are not determined is a general feature of the Newtonian approach to solid mechanics. Note that it is not because we are not "smart enough" to determine them, these forces are fundamentally undeterminable (the problem is incomplete logically). You can already see the issue that if you have $N$ particles. You'll need $N(N-1)/2$ scalar forces (assuming the strong version of Newton's third law), but you'll only have $N$ 3-dimensional accelerations (obtained by applying formulas of rigid body dynamics), so you can already anticipate an issue. Note that $3N$ scalar equations are constrained, since they must give the correct acceleration of CoM and angular acceleration, so you only really have $3N-6$ equations. This means that the problem is complete only when $N=3$ or $N=4$, and this assumes that the points are in general positions (affinely free, which was not the case in your problem).
To link this back to your elastic analogy, you should recover the rigid body dynamics in the limit of infinite rigidity. The advantage of the elastic case is that the distribution of interaction forces is well defined. However, even when the limit of infinite elasticity is taken, the distribution of stresses will typically depend on the ratio of elasticities which remain finite. This is another way of seeing that the original rigid problem is incomplete, since these elasticity ratios are not specified. The classic example is to take three rigid vertical pillars of same height with a heavy plank laid on top of them. Note that in your example, the elastic limit won't be of interest, unless if you allow the articulation to be elastic or again add a small angle.
When dealing with rigid bodies, it is therefore more convenient to invoke the principle of virtual work, where the fact that the forces are collinear is irrelevant (in fact, their entire existence is irrelevant thankfully). This allows you to disregard the internal constraining forces and only focus on the allowed rotational/translational motions. in particular, this is the general argument to verify that the torque of the constraint forces to not generate net torque. Check out an older answer of mine for more on this approach: Why is torque defined as $\vec r\times F$?.
Hope this helps.