Question

Three point masse $m_1=2kg,m_2=4kg$ and $m_3=6kg$ are kept at the three corners of an equilateral triangle of side 1 m. Find the location of their centre of mass.

Answer 1

Consider the masses be kept in a coordinate system with $(m_1=2kg)$ at the origin as follows:- [$C$ is the centre of mass]

from geometry location of $m_3=(\frac{1}{2},\frac{\sqrt{3}}{2})$

from centre of mass formula

$Xem=\frac{m_1{x}_1+m_2{x}_2+m_3{x}_3}{m_1+m_2+m_3}Yem=\frac{m_1{y}_1+m_2{y}_2+m_3{y}_3}{m_1+m_2+m_3}$

$=\frac{2\times 0+4\times 1+6\times 1}{2+4+6}=\frac{2\times 0+4\times 0+6\frac{\sqrt{3}}{2}}{2+4+6}$

$=\frac{3\sqrt{3}}{12}$

$Xem=\frac{7}{12}\Rightarrow \frac{\sqrt{3}}{4}$

$\therefore$ Location of centre of man $=(\frac{7}{12},\frac{\sqrt{3}}{4})$