Find the centre of mass of three particles at the vertices of an equilateral triangle, The masses of the particles are 100g, 150g and 200g respectively. Each side of the equilateral triangle is 0.5m long.

With the X-and Y-axes chosen as shown in the figure, the coordinates of points O, A and B forming the equilateral triangle are respectively (0,0), (0.5,0), (0.25,0.25$\sqrt{3}$ ). Let the masses 100g, 150g and 200g be located at O, A and B be respectively.
$\begin{array}{l}{\mathrm{m}}_1=100\mathrm{g},{\mathrm{m}}_2=150\mathrm{g},{\mathrm{m}}_3=200\mathrm{g}\\ \left({\mathrm{x}}_1{\mathrm{y}}_1\right)=(0,0),\left({\mathrm{x}}_2{\mathrm{y}}_2\right)=(0,5,0)\\ \left({\mathrm{x}}_3{\mathrm{y}}_3\right)=(0.25,0.25,\sqrt{3})\end{array}$

$\begin{array}{l}\text{Then, }{\mathrm{x}}_{\mathrm{cm}}=\frac{{\mathrm{m}}_1{\mathrm{x}}_1+{\mathrm{m}}_2{\mathrm{x}}_2+{\mathrm{m}}_3{\mathrm{x}}_3}{{\mathrm{m}}_1+{\mathrm{m}}_2+{\mathrm{m}}_3}\\ =\frac{\left[100\right(0)+150(0.5)+200(0.25\left)\right]}{(100+150+200)}=\frac{5}{18}\mathrm{m}\\ {\mathrm{y}}_{\mathrm{cm}}=\frac{{\mathrm{m}}_1{\mathrm{y}}_1+{\mathrm{m}}_2{\mathrm{y}}_2+{\mathrm{m}}_3{\mathrm{y}}_3}{{\mathrm{m}}_1+{\mathrm{m}}_2+{\mathrm{m}}_3}\end{array}$
$=\frac{\left[100\right(0)+150(0)+200(0.25\sqrt{3}\left)\right]}{(100+150+200)}=\frac{1}{3\sqrt{3}}\mathrm{m}$
The coordinates of centre of mass are $(5/18,1/3\sqrt{3})$