Is there a proof that a constraint force between two particles must be parallel to the vector joining them?

Question

I'm in the process of self-studying (classical) analytical mechanics, and in various textbooks I have come across the assumption that if two particles are constrained to remain at a fixed distance from each other (for instance, by a massless rod), then the constraint force that each one exerts on the other is parallel to the vector joining them. I am not comfortable with just assuming this, and thus I am trying to prove it from first principles (Newton's laws), as follows:

Suppose we have to particles A and B, with masses $m_1$ and $m_2$, respectively, moving in a plane, and constrained to remain at a fixed distance $d$ from each other. We consider two external forces, $\boldsymbol{F_A}$ and $\boldsymbol{F_B}$ acting on the particles, which account for the total external force exerted on each of them, respectively. Since $\boldsymbol{F_A}$ and $\boldsymbol{F_B}$ are arbitrary, there must also be a pair of internal forces acting on the particles, so that the imposed constraint is satisfied. Now, given that the internal force acting on particle A is exerted by particle B and vice-versa, from Newton's third law we know that these two forces must be of equal magnitude and opposite direction, so we label them $\boldsymbol{F}$ and $\boldsymbol{-F}$. However, we do not want to assume that they act along the line of the relative position vector $\boldsymbol{r_{B/A}} = \boldsymbol{r_B} - \boldsymbol{r_A}$, and thus we leave both their magnitude and direction as unknowns.

The equations of motion of the system thus become:

$$ m_1 \boldsymbol{\ddot{r}_A} = \boldsymbol{F_A} + \boldsymbol{F}$$ $$ m_2 \boldsymbol{\ddot{r}_B} = \boldsymbol{F_B} - \boldsymbol{F}$$

By considering the constraint, through a cinematic analysis we can also conclude that:

$$ |\boldsymbol{r_{B/A}}| = d \Rightarrow \frac d {dt}|\boldsymbol{r_{B/A}}| = 0 \Rightarrow \boldsymbol{\dot{r}_{B/A}} \cdot \boldsymbol{r_{B/A}} = 0 $$

I have been trying to prove, from these three equations alone, that $\boldsymbol{F}$ must in fact be parallel to $\boldsymbol{r_{B/A}}$ (for instance, by showing that $\boldsymbol{F} \cdot \boldsymbol{\dot{r}_{B/A}} = 0 $ or some equivalent statement), but I am struggling to arrive at this result. Am I missing something here? Is there maybe another hidden assumption or equation that I am failing to take into account? Or else, if it cannot be proven in this way, what is the justification for assuming that this is is true?

Answer 1

1. The assumption of collinarity of forces in Newton's third law is known as the strong Newton's third law, cf. e.g. this related Phys.SE post.

2. For examples that violate the strong Newton's third law, see e.g. this related Phys.SE post.

Answer 2

No, there is no such constraint. An easy way to see this is by using a counterexample. Imagine you had a force that always act perpendicular to the line joining the two particles (chose any dependence of the force magnitude as a function of the particle's distance as you want). There is no inconsistency between this law and newton's laws. The force on each particle is still equal to the change in momentum of the particle, and the two forces are of equal magnitude but opposite directions.

The problem though is that if the forces are not collinear, then angular momentum is not conserved, you can see that the two non collinear forces would create a torque if the two particles were at the two ends of a rod. Angular momentum is related to the invariance of the laws of physics under rotations, so the original Newton's laws do not guarantee the conservation of angular momentum.

Answer 3

The EOM's \begin{align*} &m_1\,\ddot{r}_A=F_A+F\\ &m_2\,\ddot{r}_B=F_B-F \end{align*} and the holonomic constraint equation (a rod between A and B ) \begin{align*} &(r_B-r_A)\cdot (r_B-r_A)-L^2=0\tag 1 \end{align*} $~\quad\Rightarrow~$ \begin{align*} &( r_B-r_A)\cdot (\dot r_B-\dot r_A)=0\\ & \underbrace{\begin{bmatrix} r_A& -r_B \\ \end{bmatrix}}_{C}\,\begin{bmatrix} \dot{r}_A \\ \dot{r}_B \\ \end{bmatrix}=0 \end{align*} \begin{align*} &\dot{C}\,\begin{bmatrix} \dot{r}_A \\ \dot{r}_B \\ \end{bmatrix}+ C\,\begin{bmatrix} \ddot{r}_A \\ \ddot{r}_B \\ \end{bmatrix}=0\tag 2 \end{align*} where $~L~$ is the rod length

from here with $~F=C^T\,F_q~$ , $~F_q~$ the generalized force ,the EOM's \begin{align*} &m_1\,\ddot{r}_A=F_A+r_A\,F_q\\ &m_2\,\ddot{r}_B=F_B-r_B\,F_q \end{align*}

to solve this problem , you obtained the 6 scalar EOM's and 1 constraint equation (Equation (2)) , for the 7 unknowns $~\ddot{r}_A~,\ddot{r}_B~,F_q~$

for the solution the initial conditions must fulfilled equation (1)

---

How to eliminate the constraint force $~F_q~$ from the EOM's

with

\begin{align*} &C\,\begin{bmatrix} \dot{r}_A \\ \dot{r}_B \\ \end{bmatrix}=0\quad\Rightarrow \underbrace{\begin{bmatrix} \dot{r}_A \\ \dot{r}_B \\ \end{bmatrix}}_{\dot{w}}=J\,\dot{q} \quad\text{where } ,J^T\,C^T=0 \end{align*} where $~J~$ is the jacobian matrix $~6\times 5~$ and $~\dot{q}~$ are the 5 generalized velocity coordinates

from here: \begin{align*} &\dot{w}=J\,\dot{q}\quad,\ddot{w}=J\,\ddot{q}+\dot{J}\dot{q}\\ & M\,\ddot{w}=M\,(J\,\ddot{q}+\dot{J}\dot{q})=F_q+C^T\,F_q\\\\ &\boxed{\,J^T\,[M\,(J\,\ddot{q}+\dot{J}\dot{q})]=J^T\,F_q~} \end{align*}