Question

Two masses of
of 6 and 2 units are at
position of
6i kcap - 7j kcap and 2i kcap + 5j kcap - 8k kcap
Deduce position of their centre of mass.​

Answer 1

Given:

Two masses of 6 and 2 units are at position of (6i cap - 7j cap) and (2i cap + 5j cap - 8k cap).

To find:

Position of centre of mass ?

Calculation:

"x" coordinate of Centre of Mass:

$\therefore \: \bar{x} = \dfrac{ \sum(m_{i}x_{i}) }{ \sum (m_{i})}$

$= > \: \bar{x} = \dfrac{ (6 \times 6) + (2 \times 2)}{6 + 2}$

$= > \: \bar{x} = \dfrac{ 36 + 4}{8}$

$= > \: \bar{x} = \dfrac{40}{8}$

$= > \: \bar{x} = 5$

"y" coordinate of Centre of Mass:

$\therefore \: \bar{y} = \dfrac{ \sum(m_{i}y_{i}) }{ \sum (m_{i})}$

$= > \: \bar{y} = \dfrac{ \{6 \times ( - 7) \} + (2 \times 5)}{6 + 2}$

$= > \: \bar{y} = \dfrac{ - 42 + 10}{8}$

$= > \: \bar{y} = \dfrac{ - 32 }{8}$

$= > \: \bar{y} = - 4$

"z" coordinate of Centre of Mass:

$\therefore \: \bar{z} = \dfrac{ \sum(m_{i}z_{i}) }{ \sum (m_{i})}$

$= > \: \bar{z} = \dfrac{ (6 \times 0) + \{2 \times ( - 8) \}}{6 + 2}$

$= > \: \bar{z} = \dfrac{ - 16}{8}$

$= > \: \bar{z} = - 2$

So, final position of C-O-M will be:

$\boxed{ \bold{ \vec{r} = 5 \hat{i} - 4 \hat{j} - 2 \hat{z}}}$