Archive for November, 2009

Center of gravity of a trapezium

November 14, 2009

It is worth noting that the center of gravity of a trapezium $ABCD$ does not coincide with the center of gravities of the set $\lbrace A,B,C,D \rbrace$. Indeed, if we consider the special case where the coordinates of $A,B,C,D$ are $(-\alpha,0),(\alpha,0),(\beta,h)$ and $(-\beta,h)$, respectively, the coordinates of the first center of gravity (let’s call it $G_1$) are $(0,\frac{(\alpha+2\beta)h}{3(\alpha+\beta)})$ while the coordinates of the second (let’s call it $G_2$) are $(0,\frac{h}{2})$. If $\alpha>\beta$, then $G_1$ is below $G_2$. Also, $G_1$ satisifies $(2\alpha+beta)(\overrightarrow{GA}+\overrightarrow{GB})+(\alpha+2\beta)(\overrightarrow{GC}+\overrightarrow{GD})=\overrightarrow{0}$ while $G_2$ satisifies $\overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}+\overrightarrow{GD}=\overrightarrow{0}$.