# three particles each of mass m are placed at three corner of an equilateral triangle of length l. find the position of center of mass in terms of coordinates

Taking origin at vertex B and taking axis along the side BC .
so the coordinates of the B are (0,0) and that of C are (a,0)
As the angle between the sides of an equilateral triangle is 60o , so for point A the x-coordinate will be a cos60o =$\frac{a}{2}$
x-coordinate will be a sin60o= $\frac{\sqrt{3}a}{2}$

The length of the side of equilateral triangle is taken to be equal to a .
The coordinates of centre of mass are as follows:
${x}_{cm}=\frac{m×0+m×a+m×\left(\frac{a}{2}\right)}{3m}=\frac{a}{2}$
${y}_{cm}=\frac{m×0+m×0+m×\left(\frac{\sqrt{3}a}{2}\right)}{3m}=\frac{a}{2\sqrt{3}}$

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