Question

three equal masses are positioned on a Cartesian plane at positions (-2,0) (0,1) and (1,-1) where is the center mass of this system particles located

Answer 1

Given

   three equal masses positioned at

   let mass 1 at (x1,y1) = (-2,0)   

       mass 1 at (x2,y2) = (0,1)

       mass 1 at (x3,y3) = (1,-1)
we know that the centre of mass of the three body system is

along x axis

       Xcm = m1x1+m2x2+m3x3/(m1+m2+m3)

and along y -axis

       Y cm = m1y1+m2y2+m3y3/(m1+m2+m3)

substituting the values

       Xcm = m(x1+x2+x3)/3m = (x1+x2+x3)/3 = (-2+0+1)/3 = -1/3 = -0.333 m

       Ycm = m(y1+y2+y3)/3m = ((y1+y2+y3)/3 = (0+1-1)/3 = 0

so the centre of mass of the given three equal mass body system is (-0.333,0)