I have multiple ways of showing that the lines of action of 3 co-planar forces in static equilibrium meet at a point. But they combine physical arguments and math. It seems like there should be some purely mathematical way of proving the claim, treating it as a system of equations having a unique solution.

Is there a way to express this as a uniquely determined system of linear equations?

Here are the equations of equilibrium corresponding to the illustration, where $\vec{r}_1=\mathfrak{p}_1-\mathscr{O},$ etc. The heavy gray line is a rigid rod in equilibrium under the applied forces.

$$\begin{aligned} \vec{0}&=\vec{f}_{1}+\vec{f}_{2}+\vec{f}_{3}\\ \vec{0}&=\vec{r}_{1}\times\vec{f}_{1}+\vec{r}_{2}\times\vec{f}_{2}+\vec{r}_{3}\times\vec{f}_{3}\\ \vec{0}&=\left(\vec{r}_{1}-\vec{p}\right)\times\vec{f}_{1}+\left(\vec{r}_{2}-\vec{p}\right)\times\vec{f}_{2}+\left(\vec{r}_{3}-\vec{p}\right)\times\vec{f}_{3} \end{aligned}$$

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The equation of a line through $r_i$ in the direction $f_i$ is $(r-r_i)\times f_i=0$ which we can rewrite as $r\times f_i=r_i\times f_i$. Now assuming $f_1$ and $f_2$ are not parallel, they intersect at some point $p$, so $$p\times f_1=r_1\times f_1$$ and $$p\times f_2=r_2\times f_2$$ which gives that

$$p\times(f_1+f_2)=r_1\times f_1+r_2\times f_2.$$ Now add $p\times f_3$ to both sides, to give $$0=r_1\times f_1+r_2\times f_2+p\times f_3$$ by the fact that the forces are in equilibrium (your first equation). But using the fact that the moments are in equilibrium (your second equation) gives that $$r_1\times f_1+r_2\times f_2+r_3\times f_3=r_1\times f_1+r_2\times f_2+p\times f_3$$ so $$r_3\times f_3=p\times f_3$$ which exactly says that $p$ (the intersection of the first two lines) also lies on the third line, thus the lines all intersect at a common point.

  • $\begingroup$ When I asked a physicist about the case of parallel forces, he used the assumption of projective geometry that parallel lines meet at a point at infinity. $\endgroup$ – Steven Thomas Hatton Apr 8 at 14:27
  • $\begingroup$ @StevenThomasHatton Yes, if you allow the point at infinity, then your claim is also true in the cases with two or three of the forces being parallel. $\endgroup$ – acernine Apr 8 at 14:45
  • $\begingroup$ A moment's reflection will convince you that if any two of the forces are parallel, all three must be. $\endgroup$ – Steven Thomas Hatton Apr 8 at 15:18
  • $\begingroup$ The hook I was missing was the equation of a line in the form $(r-r_i)\times f_i=0$. I'm certain I've encountered it before, but its practical utility was lost on me until now. $\endgroup$ – Steven Thomas Hatton Apr 8 at 17:48

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