Four particle of masses ${\mathrm{m}}_1 = 2\mathrm{m}, {\mathrm{m}}_2 = 4 \mathrm{m}, {\mathrm{m}}_3 = \mathrm{m} \mathrm{and} {\mathrm{m}}_4$are placed at four corners of a square. What should be the value of ${\mathrm{m}}_4$ so that the centres of mass of all the four particles are exactly at the centre of the square?

Let side of square is a.

${\mathrm{x}}_{\mathrm{cm}} = \frac{\mathrm{a}}{2} = \frac{{\mathrm{m}}_1\times 0+{\mathrm{m}}_2\times \mathrm{a}+{\mathrm{m}}_3\times \mathrm{a}+{\mathrm{m}}_4\times 0}{{\mathrm{m}}_1+{\mathrm{m}}_2+{\mathrm{m}}_3+{\mathrm{m}}_4}$

solve to get ${\mathrm{m}}_4 = 3 \mathrm{m}$

${\mathrm{y}}_{\mathrm{cm}} = \frac{\mathrm{a}}{2} = \frac{{\mathrm{m}}_1\times 0+{\mathrm{m}}_2\times 0+{\mathrm{m}}_3\times \mathrm{a}+{\mathrm{m}}_4\times \mathrm{a}}{{\mathrm{m}}_1+{\mathrm{m}}_2+{\mathrm{m}}_3+{\mathrm{m}}_4}$

solve to get ${\mathrm{m}}_4 = 5\mathrm{m}$

They are contradicting, so it is not possible to have centre of mass at the centre of square.