Physics
posted by Jeremy on .
Young acrobats are standing still on a circular horizontal platform suspended at the center. The origin of the twodimensional Cartesian coordinate system is assumed to be at the center of the platform. A 30kg acrobat is located at (4 m, 9 m), and a 36kg acrobat is located at (−2 m, −2 m). Assuming that the acrobats stand still in their positions, where must a 20kg acrobat be located so that the center of mass of the system consisting of the three acrobats is at the origin and the platform is balanced?

x(c.m.) =0,
y(c.m.) =0,
x(c.m.) = m1•x1+m2•x2+m3•x3/(m1+m2+m3) =0,
y(c.m.) = m1•y1+m2•y2+m3•y3/(m1+m2+m3) =0,
30•436•2+20•x3 =0,
30•9+36•2+20•y3 = 0 ,
x3 =  48/20 =  2.4 m,
y3 = 198/20 =  9.9 m.