Position of the Centre of Mass

(15)

[IMAGE 60]

$\Rightarrow m_{1} x_{1} + m_{2} x_{2} = m_{3} x_{3}$

$10 x_{1} + 2 (1.5) = 12 (1.5) \Rightarrow x_{1} = 1.5 cm$

$\Rightarrow m_{1} y_{1} + m_{2} y_{2} = m_{3} y_{3}$

$10 y_{1} + 2 (1.5) = 12 \times 1 \Rightarrow y_{1} = 0.9 cm$

$\frac{x_{1}}{y_{1}} = \frac{1.5}{0.9} = \frac{15}{9} \Rightarrow n = 15$

(3)

[IMAGE 61]

$Position vector {\vec{r}}_{C O M} = \frac{m_{1} {\vec{r}}_{1} + m_{2} {\vec{r}}_{2} + m_{3} {\vec{r}}_{3}}{m_{1} + m_{2} + m_{3}}$

${\vec{r}}_{C O M} = \frac{2 M \times 0 + 2 M \times 4 \hat{i} + 2 M \times 4 \hat{j}}{6 M}$

$\vec{r} = \frac{4}{3} \hat{i} + \frac{4}{3} \hat{j}$

$| \vec{r} |$ $= \sqrt{{(\frac{4}{3})}^{2} + {(\frac{4}{3})}^{2}} = \frac{4}{3} \sqrt{2} = \frac{4 \sqrt{2}}{x}$

$x = 3$

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