Question

I cocate the centre of mass of thill particles of most mi = 1 kg,
mes 2kg, m - 3kg at the corners of equilateral triangle
I of each side 1m.​

Answer 1

let 1kg is at orgin so 3kg is at (1,0)

and 2kg is at (1/2,√3/2)

y coordinare of centre of mass will be ((1×0)+(2×√3/2)+(3×0))/(1+2+3)=√3/6=1/2x3

x coordinare of centre of mass will be ((1×0)+(2×1/2)+(3×1))/(1+2+3)=4/6=2/3

so coordinates of centre of mass are (2/3,1/2√3)

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