Question

masses are distributed in the xy plane as follows: 2kg at (2.0,6.0) m,4.0 kg at (2.0,0.0) m, and 6.0 kg at (0.0,3.0) m. where would a 15 kg mass needto be positioned so that the center of mass of the resulting four mass system would be at (0.-2) m?

Answer 1

here,

m1 = 2 kg is at (x1 , y1) = ( 2 , 6) m

m2 = 4 kg is at (x2 , y2) = ( 2 , 0) m

m3 = 6 kg is at (x3 , y3) = ( 0 , 3) m

m4 = 15 kg is at (x4 , y4)

the x-coordinate of center of mass , x = ( m1 * x1 + m2 * x2 + m3 * x3 + m4 * x4)/(m1 + m2 + m3 + m4)

0 = ( 2 * 2 + 4 * 2 + 6 * 0 + 15 * x4) /(2 + 4 + 6 + 15)

solving for x4

x4 = - 0.8 m

the y-coordinate of center of mass , y = ( m1 * y1 + m2 * y2 + m3 * y3 + m4 * y4)/(m1 + m2 + m3 + m4)

-2 = ( 2 * 6 + 4 * 0 + 6 * 3 + 15 * y4) /(2 + 4 + 6 + 15)

solving for y4

y4 = - 5 m

the corrdinate of 15 kg mass is (- 0.8 m , -5 m)