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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.
Solution
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:
\[X_{cm} , Y_{cm} = \frac{m_1 x_1 + m_2 x_2 + m_3 x_3}{m_1 + m_2 + m_3}, \frac{m_1 y_1 + m_2 y_2 + m_3 y_3}{m_1 + m_2 + m_3}\]
\[= \frac{1 \times 0 + 2 \times 1 + 3 \times \frac{1}{2}}{1 + 2 + 3}, \frac{1 \times 0 + 2 \times 0 + 3 \times \frac{\sqrt{3}}{2}}{1 + 2 + 3}\]
\[ = \frac{7}{12} m, \frac{\sqrt{3}}{4} m\]
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