Explanation

(c) We can assume that three particles of equal mass $m$ are placed at the corners of triangle
${\overrightarrow{r}}_{1} = 0\widehat{i} + 0\widehat{j},\ {\overrightarrow{r}}_{2} = b\widehat{i} + 0\widehat{j}$
and ${\overrightarrow{r}}_{3} = 0\widehat{i} + h\widehat{j}$
$\therefore\overrightarrow{r_{cm}} = \frac{m_{1}{\overrightarrow{r}}_{1} + m_{2}{\overrightarrow{r}}_{2} + m_{3}{\overrightarrow{r}}_{3}}{m_{1} + m_{2} + m_{3}}$
[image:7_image145.png]
$= \frac{b}{3}\widehat{i} + \frac{h}{3}\widehat{j}$
$i.e.$ coordinates of centre of mass is $\left( \frac{b}{3},\frac{h}{3} \right)$