Question

If the centre of mass of three particles of masses $10$, $20$, $30$ units be at a point $(1,-1,3)$ where should a fourth particle of mass 40 units be placed so that the combined centre of mass may be at the point $(1,1,1)$?

Answer 1

Let $4^{th}$ mass position is $(x,y,z)$ then,

$1=\frac{40x+(10+20+30)1}{40+10+20+30}$

$1=\frac{40x+60}{100}$

$10=4x+6$

$x=1$

$1=\frac{40y+\left(60\right)(-1)}{100}$

$y=\frac{160}{40}=4$

$1=\frac{40z+60\times 3}{100}$

$z=\frac{-80}{40}=-2$