Take an arbitrary triangle with vertices $A$, $B$, and $C$ with side lengths opposite to the vertices $a$, $b$, and $c$. Then, assume this triangle has no mass, and hang a series of weights proportional to each side length on their corresponding angles. Let $G$ be the center of mass of this system.
Is $G$ an already-known triangle center? Surely it must be, as the Steiner curvature centroid has a nearly identical definition, but with the measure of the exterior angle rather than the length of the opposite side.
Additionally, I tried to come up with a method of finding this center on the plane. I believe it's right, but while I'm asking about it, I figured I should double check it here as well.
Take an arbitrary triangle $\triangle ABC$ on the Cartesian plane, and translate it such that the point $A$ is at the origin and the side $\overline{AB}$ is aligned with the positive $x$-axis. Then, say that $A=(0,0)$, $B=(c,0)$, and $C=(x_0,y_0)$. Then the center $G$ will be at $$G=\bigg(\frac{c\sqrt{x_0^2+y_0^2}+cx_0}{c+\sqrt{x_0^2+y_0^2}+\sqrt{(x_0-c)^2+y_0^2}},\frac{cy_0}{c+\sqrt{x_0^2+y_0^2}+\sqrt{(x_0-c)^2+y_0^2}}\bigg)$$
Thank you for your time!