Question

A 2.63 kg particle has the xy coordinates (-1.92 m, 0.422 m), and a 4.44 kg particle has the xy coordinates (0.115 m, -0.745 m). Both lie on a horizontal plane. At what (a)x and (b)y coordinates must you place a 4.89 kg particle such that the center of mass of the three-particle system has the coordinates (-0.477 m, -0.829 m)?

Answer 1

let

m1 = 2.63 kg, x1 = -1.92 m, y1 = 0.422 m
m2 = 4.44 kg, x2 = 0.115 m, y2 = -0.745 m
Xcm = -0.477 m, Ycm = -0.829 m
m3 = 4.89 kg, x3 = ?, y3 = ?

use, Xcm = (m1*x1 + m2*x2 + m3*x3)/(m1+m2+m3)

-0.477 = (2.63*(-1.92) + 4.44*0.155 + 4.89*x3)/(2.63 + 4.44 + 4.89)

x3 = -0.275 m   <<<<<<-------------Answer

use, Ycm = (m1*y1 + m2*y2 + m3*y3)/(m1+m2+m3)

-0.829 = (2.63*0.422 + 4.44*(-0.745) + 4.89*y3)/(2.63 + 4.44 + 4.89)

y3 = -1.58 m <<<<<<-------------Answer