Locate the centre of mass of a system of particles of mass $m_1 = 1kg, m_2 = 2kg$ and $m_3 = 3kg$, situated at the corners of an equilateral triangle of side 1.0 metre.
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Consider an equilateral triangle of side 1m as shown in fig. Take X and Y axes as shown in fig.
By the definition of centre of mass, we have
$\bar{\text{x}}=\frac{\text{m}_1\text{x}_1+\text{m}_2\text{x}_2+\text{m}_3\text{x}_3}{\text{m}_1+\text{m}_2+\text{m}_3}$
$\text{and }\bar{\text{y}}=\frac{\text{m}_1\text{y}_1+\text{m}_2\text{y}_2+\text{m}_3\text{y}_3}{\text{m}_1+\text{m}_2+\text{m}_3}$
Here, $\text{m}_1\text{kg},\text{m}_2=2\text{kg}\text{ and }\text{m}_3=3\text{kg}$
$[\text{x}_1=0,\text{y}_1=0]$
$[\text{x}_2=1,\text{y}_2=0]$
and $\Big[\text{x}_3=0.5,\text{y}_3=\frac{\sqrt{3}}{2}\Big]$
Here, $\bar{\text{y}}=\frac{1\times0+2\times0+3\times\Big(\frac{\sqrt{3}}{2}\Big)}{1+2+3}=\frac{\sqrt{3}}{4}\text{m}$
The co-ordinates of centre of mass are $\Big(\frac{3.5}{6},\frac{\sqrt{3}}{4}\Big)$
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