Question

Three particles of mass $1kg,2kg$ and $3kg$ are placed at the corners $A,B$ and $C$ respectively of an equilateral triangle $ABC$ of edge $1m$. Find the distance of their centre of mass from $A$.

Answer 1

Assume , Point $A$ is origin.

From the figure, $CP=1\times sin\left({60}^o\right)=\frac{\sqrt{3}}{2}$

Centre of mass from Point $A$ in direction of $x-axis$, $x_{cm}=\frac{m_A{x}_A+m_B{x}_B+m_C{x}_C}{m_A+m_B+m_C}$

$\Rightarrow x_{cm}=\frac{1\times 0+2\times 1+3\times \frac{1}{2}}{1+2+3}=\frac{7}{12}$

Centre of mass from Point $A$ in direction of $y-axis$, $y_{cm}=\frac{m_A{y}_A+m_B{y}_B+m_C{y}_C}{m_A+m_B+m_C}$

$\Rightarrow y_{cm}=\frac{1\times 0+2\times 0+3\times \frac{\sqrt{3}}{2}}{1+2+3}=\frac{\sqrt{3}}{4}$

So coordinate of centre of mass $(\frac{7}{12},\frac{\sqrt{3}}{4})$, Distance from $A$, $=\sqrt{{\left(\frac{7}{12}\right)}^2+{\left(\frac{\sqrt{3}}{4}\right)}^2}=\frac{\sqrt{19}}{6}$