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Q.

Three particles of masses 1 kg, 2kg and 3kg are placed at the vertices A, B and C of an equilateral triangle ABC If A and B lie at (0,0) and (1, 0) m, the Co-ordinates of their centre of mass are

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a

$(\frac{\sqrt{3}}{2} m and \frac{7}{6} m)$

b

$(\frac{7}{12} m and \frac{7}{12} m)$

c

$(\frac{7}{6} m and \frac{\sqrt{3}}{4} m)$

d

$(\frac{7}{12} m and \frac{\sqrt{3}}{4} m)$

answer is C.

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Detailed Solution

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$x_{c} = \frac{m_{1} x_{1} + m_{2} x_{2} + m_{3} x_{3}}{m_{1} + m_{2} + m_{3}} = \frac{0 + 2 + \frac{3}{2}}{6} = \frac{7}{12}$

$y_{c} \frac{m_{1} y_{1} + m_{2} y_{2} + m_{3} y_{3}}{m_{1} + m_{2} + m_{3}} = \frac{0 + 0 + \frac{3 \sqrt{3}}{2}}{6} = \frac{\sqrt{3}}{4}$

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Three particles of masses 1 kg, 2kg and 3kg are placed at the vertices A, B and C of an equilateral triangle ABC If A and B lie at (0,0) and (1, 0) m, the Co-ordinates of their centre of mass are