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

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 60

^{o}, so for point A the x-coordinate will be

*a*cos60

^{o}=$\frac{a}{2}$

x-coordinate will be

*a*sin60

^{o}= $\frac{\sqrt{3}a}{2}$

The length of the side of equilateral triangle is taken to be equal to

*a*.

The coordinates of centre of mass are as follows:

${x}_{cm}=\frac{m\times 0+m\times a+m\times \left({\displaystyle \frac{a}{2}}\right)}{3m}=\frac{a}{2}$

${y}_{cm}=\frac{m\times 0+m\times 0+m\times \left({\displaystyle \frac{\sqrt{3}a}{2}}\right)}{3m}=\frac{a}{2\sqrt{3}}$

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