Question

The position vectors of three particles of
masses m1 = 1kg, m2 = 2kg and m3 = 3kg
are 1 = (i + 4j + k)m, r2 = (1 + + )m and F
3 = (21 - j - 2x)m respectively. The
position vector of their centre of mass is:
1 (31 + 21 - 4k)
(31 + j-6)​

Answer 1

Answer:

               (1.5,0.5,-0.5)

Explanation: Remark r2 is not given which is (1,1,1)

The center of mass of for three particles given as

  COM = $(M_{1} r_{1} +M_{2} r_{2} +M_{3} r_{3} )/(M_{1} +M_{2} +M_{3})$

Xcom = $(M_{1} x_{1} +M_{2} x_{2} +M_{3} x_{3} )/(M_{1} +M_{2} +M_{3})$

Ycom = $(M_{1} y_{1} +M_{2} y_{2} +M_{3} y_{3} )/(M_{1} +M_{2} +M_{3})$

Zcom = $(M_{1} z_{1} +M_{2} z_{2} +M_{3} z_{3} )/(M_{1} +M_{2} +M_{3})$

Put the value of coordinates and proceed as follow