How exactly does (the weak form of) Newton's third law hold in the case of a force applied onto a lever?

Question

Suppose we have an ideal massless lever with a fixed pivot point and a mass $m_{1}$ (at radius $L_{1}$) on one end and mass $m_{2}$ (at radius $L_{2}$) on the opposite end, and assume gravity plays no role in the scenario. Suppose also that I apply a constant force of $F_{A}$ onto mass $m_{1}$ perpendicular to the lever. This causes the lever to rotate as indicated in the sketch below.

Of course by Newton's third law, there will be a reaction force $\vec{F}_{R} = -\vec{F}_{A}$ on my hand. Let us calculate the angular acceleration of the lever.

The applied torque on the lever is $\tau = F_{A}\cdot L_{1}$ in the clockwise direction. The moment of inertia of the entire setup is $I = m_{1}L_{1}^{2} + m_{2}L_{2}^{2}$, so the angular acceleration will be $$ \alpha = \frac{F_{A} L_{1}}{m_{1}L_{1}^{2} + m_{2}L_{2}^{2}}. $$

Now I will calculate the net force on mass $m_{2}$. Since the calculation is not as essential as the result, the reader can skip the piece quoted below.

If we take the $xy$-coordinates to be centered at the pivot with mass $m_{2}$ initially at $\vec{x}(0) = (-L_{2}, 0)$, the position of mass $m_{2}$ at later times will be $$ \vec{x}(t) = (-L_{2}\cos(\tfrac{1}{2}\alpha t^{2}), L_{2}\sin(\tfrac{1}{2}\alpha t^{2})). $$ The velocity is $$ \dot{\vec{x}}(t) = (L_{2}\alpha t \cdot \sin(\tfrac{1}{2}\alpha t^{2}), L_{2}\alpha t \cdot \cos(\tfrac{1}{2}\alpha t^{2})), $$ and the acceleration is $$ \ddot{\vec{x}}(t) = (L_{2}\alpha\sin(\tfrac{1}{2}\alpha t^{2}) + L_{2}\alpha^{2}t^{2}\cos(\tfrac{1}{2}\alpha t^{2}), L_{2}\alpha\cos(\tfrac{1}{2}\alpha t^{2}) - L_{2}\alpha^{2}t^{2}\sin(\tfrac{1}{2}\alpha t^{2})). $$ At the initial time $t=0$, we find $\ddot{\vec{x}}(0) = (0, L_{2}\alpha)$, so overall mass $m_{2}$ initially experiences a force $$ F(0) = m_{2}\cdot |\ddot{\vec{x}}(0)| = m_{2}\cdot L_{2}\alpha = \frac{m_{2} L_{1}L_{2}}{m_{1}L_{1}^{2} + m_{2}L_{2}^{2}} F_{A} $$ in the $+y$-direction.

Hence at $t=0$, the net force on $m_{2}$ is $$ \vec{F}(0) = \frac{m_{2} L_{1}L_{2}}{m_{1}L_{1}^{2} + m_{2}L_{2}^{2}} F_{A} \vec{e}_{y}. $$

Now in this scenario, what is the reaction force accompanying $\vec{F}$ and where is it applied? By Newton's third law, there must be an equal and opposite reaction force, but there is no apparent reaction force accompanying the force I calculated above.

In a previous post of mine, I questioned whether the strong form of Newton's third law holds, but now I'm confused as to how even the weak form of the law applies to this situation. Can someone clarify what is going on here?

Answer 1

Hence at $t=0$, the net force on $m_{2}$ is $$ \vec{F}(0) = \frac{m_{2} L_{1}L_{2}}{m_{1}L_{1}^{2} + m_{2}L_{2}^{2}} F_{A} \vec{e}_{y}. $$

Now in this scenario, what is the reaction force accompanying $\vec{F}$ and where is it applied?

Since $\vec F$ is a net force it does not generally have a Newton’s 3rd law pair. Each of the individual forces have their own 3rd law pair forces, but since those forces generally may act on different objects it does not generally make sense to sum them.

However, in this specific case there is only one force acting on the mass. So since the net force is just a single force, in this specific case it does have a 3rd law pair.

That force is the force acting on the left end of the lever. The lever pushes the block in the positive $y$ direction and so the block pushes the lever in the negative $y$ direction. So the end of the lever is loaded in shear.

Answer 2

This problem arises because you are inconsistently choosing to handle the $m_2+$lever$+m_1$ system together as a rigid body, and then suddenly considering $m_2$ separately from that.

Firstly, you have to take $m_2+$lever$+m_1$ system together. The force from your hand onto this combined system is $F_A$, and you drew it originating on the wrong place. It cannot be coming from the centre of $m_1$, but rather must be coming from the junction between your hand and the block. You need to consider them together, in order to get the angular acceleration $\alpha$ as you correctly did. The N3L pair of $F_A$ is the $F_R$ that you drew, but moved to the same junction too.

When you calculated the force felt by $m_2$ separately, using the $\alpha$ found above, the calculation is correct, but you made a swap of viewpoint. This is the force of the lever+$m_1$ system pushing on $m_2$, and its N3L pair is the same magnitude force of $m_2$ pushin on the lever+$m_1$ system.

You can even go into even deeper detail and separate lever and $m_1$ too. In that case, the masslessness of the lever will be a pain. It can be rigorously sorted out, but the masslessness gives an illusion of breakage in the mathematics.

There is just a lot of annoyance in this problem. The pivot is also pushing on the lever. There are a lot of calculations you could do, to show that everything is internally self-consistent.