Centre of mass of 3 particles 10 kg, 20 kg and 30 kg is at (0, 0, 0). Where should a particle of mass 40 kg be placed so that the combined centre of mass will be at (3, 3, 3)?

Options

A. (0,0,0)
B. (7.5 , 7.5 , 7.5)
C. (1 , 2 , 3)
D. (4 , 4 , 4)

Answer

B

Explanation

Let the position coordinates of 40 kg be (x , y , z)
`therefore X_(CM) = sum_(m_(i) x_(i))/(sum m_(i)) or 3 = (10 xx 0 + 20 xx 0 + 30 xx 0 + 40 xx x)/(10 + 20 + 30 + 40)`
`therefore x = (300)/(40) = 7.5`
`Y_(CM) = (sum m_(i) y_(i))/(sum m_(i))`
`3 = (10 xx 0 + 20 xx 0 + 30 xx 0 + 40 xx y)/(10 + 20 + 30 + 40) = y = (300)/(40) = 7.5`
`Z_(CM) = (sum m_(i) z_(i))/(sum m_(i))`
`3= (10 xx 0 + 20 xx 0 + 30 xx 0 + 40 xx z)/(10 + 20 + 30 + 40) implies z = (300)/(40) = 7.5`