Young acrobats are standing still on a circular horizontal platform suspended at the center. The origin of the two-dimensional Cartesian coordinate system is assumed to be at the center of the platform. A 30-kg acrobat is located at (4 m, 9 m), and a 36-kg acrobat is located at (−2 m, −2 m). Assuming that the acrobats stand still in their positions, where must a 20-kg 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?

Answer 1

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•4-36•2+20•x3 =0,
30•9+36•2+20•y3 = 0 ,
x3 = - 48/20 = - 2.4 m,
y3 = -198/20 = - 9.9 m.