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 
\[\left( 0, 0 \right), \left( 1, 0 \right) \text{ and }\left( \frac{1}{2}, \frac{\sqrt{3}}{2} \right)\]  respectively.


The position of centre of mass is given by:
\[X_{cm} , Y_{cm} = \frac{m_1 x_1 + m_2 x_2 + m_3 x_3}{m_1 + m_2 + m_3}, \frac{m_1 y_1 + m_2 y_2 + m_3 y_3}{m_1 + m_2 + m_3}\]

\[= \frac{1 \times 0 + 2 \times 1 + 3 \times \frac{1}{2}}{1 + 2 + 3}, \frac{1 \times 0 + 2 \times 0 + 3 \times \frac{\sqrt{3}}{2}}{1 + 2 + 3}\]
\[ = \frac{7}{12} m, \frac{\sqrt{3}}{4} m\]