Question

Three particles of masses 1.0 kg, 2.0 kg and 3.0 kg are placed at the corners A, B and C respectively of an equilateral triangle ABC of edge 1 m. Locate the centre of mass of the system.

Answer 1

Taking BC as the X-axis and point B as the origin, the positions of masses m₁ = 1 kg, m₂ = 2 kg and m₃ = 3 kg are
$\left(0,0\right),\left(1,0\right)\mathrm{and}\left(\frac{1}{2},\frac{\sqrt{3}}{2}\right)$ respectively.


The position of centre of mass is given by:

$X_{\mathrm{cm}},Y_{\mathrm{cm}}=\frac{m_1{x}_1+m_2{x}_2+m_3{x}_3}{m_1+m_2+m_3},\frac{m_1{y}_1+m_2{y}_2+m_3{y}_3}{m_1+m_2+m_3}$
$$\begin{gathered} =\frac{1\times 0+2\times 1+3\times {\displaystyle \frac{1}{2}}}{1+2+3},\frac{1\times 0+2\times 0+3\times {\displaystyle \frac{\sqrt{3}}{2}}}{1+2+3} \\ =\frac{7}{12}\mathrm{m},\frac{\sqrt{3}}{4}\mathrm{m} \end{gathered}$$