Three particles of masses 1.0kg, 2.0kg and 3.0kg are placed at the corners A, B and C respectively of an equilateral triangle ABC of edge 1m. Locate the centre of mass of the system.
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$\text{m}_1=1\text{kg},\ \text{m}_2=2\text{kg},\ \text{m}_3=3\text{kg}$ $\text{x}_1=0,\ \text{x}_2=1,\ \text{x}_3=\frac{1}{2}$ $\text{y}_1=0,\ \text{y}_2=0,\ \text{y}_3=\frac{\sqrt3}{2}$ 
The position of centre of mass is, $\text{CM}=\Big(\frac{\text{m}_1\text{x}_1+\text{m}_2\text{x}_2+\text{m}_3\text{x}_3}{\text{m}_1+\text{m}_2+\text{m}_3},\frac{\text{m}_1\text{y}_1+\text{m}_2\text{y}_2+\text{m}_3\text{y}_3}{\text{m}_1+\text{m}_2+\text{m}_3}\Big)$ $=\Bigg(\frac{(1\times0)+(2\times1)\big(3\times\frac{1}{2}\big)}{1+2+3},\frac{(1\times0)+(2\times0)+\big(3\times\frac{\sqrt3}{2}\big)}{1+2+3}\Bigg)$ $=\Big(\frac{7}{12},\frac{3\sqrt3}{12}\Big)$ from the point B.
The position of centre of mass is, $\text{CM}=\Big(\frac{\text{m}_1\text{x}_1+\text{m}_2\text{x}_2+\text{m}_3\text{x}_3}{\text{m}_1+\text{m}_2+\text{m}_3},\frac{\text{m}_1\text{y}_1+\text{m}_2\text{y}_2+\text{m}_3\text{y}_3}{\text{m}_1+\text{m}_2+\text{m}_3}\Big)$ $=\Bigg(\frac{(1\times0)+(2\times1)\big(3\times\frac{1}{2}\big)}{1+2+3},\frac{(1\times0)+(2\times0)+\big(3\times\frac{\sqrt3}{2}\big)}{1+2+3}\Bigg)$ $=\Big(\frac{7}{12},\frac{3\sqrt3}{12}\Big)$ from the point B.
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