Question

A 2.64 kg particle has the xy coordinates (-1.63 m, 0.00697 m), and a 2.31 kg particle has the xy coordinates (0.432 m, -0.944 m). Both lie on a horizontal plane. At what (a) x and (b) y coordinates must you place a 2.63 kg particle such that the center of mass of the three-particle system has the coordinates (-0.251 m, -0.476 m)?

Answer 1

here,

m1 = 2.64 kg is at (x1 , y1) = (- 1.63 m , 0.00697 m)

m2 = 2.31 kg is at (x2 , y2) = (0.432 m , - 0.944 m)

m3 = 2.63 kg is at (x3 , y3)

(x , y) = ( (m1 * x1 + m2 * x2 + m3 * x3 )/(m1 + m2 + m3) , ((m1 * y1 + m2 * y2 + m3 * y3)/(m1 + m2 + m3))

( - 0.251 , - 0.476) = ( (2.64 * (-1.63) + 2.31 * 0.432 + 2.63 * x3)/(2.64 + 2.31 + 2.63) , ( (2.64 * 0.00697 - 2.31 * 0.944 + 2.63 * y3 )/(2.64 + 2.31 + 2.63) )

solving for (x3 , y3)

x3 = 0.53 m

y3 = - 0.55 m

a)

the x-coordinate is 0.53 m

b)

the y-coordinate is -0.55 m