91Ó°ÊÓ

The position vector of the center of mass of this system is

$\stackrel{\mathbf{→}}{{\mathbf{r}}_{\mathbf{c}\mathbf{m}}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{M}}\mathbf{(}{\mathbf{m}}_{\mathbf{1}}\stackrel{\mathbf{→}}{{\mathbf{r}}_{\mathbf{1}}}\mathbf{+}{\mathbf{m}}_{\mathbf{2}}\stackrel{\mathbf{→}}{{\mathbf{r}}_{\mathbf{2}}}\mathbf{+}\mathbf{.}\mathbf{....}\mathbf{+}{\mathbf{m}}_{\mathbf{n}}\stackrel{\mathbf{→}}{{\mathbf{r}}_{\mathbf{n}}}\mathbf{)}$where $\mathit{M}\mathbf{=}{\mathit{m}}_{\mathbf{1}}\mathbf{+}{\mathit{m}}_{\mathbf{2}}\mathbf{+}\mathbf{.}\mathbf{....}\mathbf{+}{\mathit{m}}_{\mathbf{n}}$.

Insert the required values into the above formula:

$\begin{array}{l}\stackrel{→}{r_{cm}}=\frac{1}{M}(m_1\stackrel{→}{r_1}+m_2\stackrel{→}{r_2}+\mathrm{.....}+m_n\stackrel{→}{r_n})\\ \left[\begin{array}{c}2\\ 2\end{array}\right]=\frac{1}{1}(m_1\left[\begin{array}{c}1\\ 2\end{array}\right]+m_2\left[\begin{array}{c}2\\ 3\end{array}\right]+m_3\left[\begin{array}{c}4\\ 1\end{array}\right])\end{array}$

The masses equal to unity.

$m_1+m_2+m_3=1$

So, the equations will be:

$\begin{array}{c}{m}_1+m_2+m_3=1\mathrm{......}\left(1\right)\\ m_1+2m_2+4m_3=2\mathrm{......}\left(2\right)\\ 2m_1+3m_2+m_3=2\mathrm{......}\left(3\right)\end{array}$

Represent the equations (1), (2) and (3), in the form of matrix.

$\left[\begin{array}{cccc}1& 1& 1& 1\\ 1& 2& 4& 2\\ 2& 3& 1& 2\end{array}\right]$

Now, reduce the obtained matrix as follows:

$\begin{array}{l}\left[\begin{array}{cccc}1& 1& 1& 1\\ 1& 2& 4& 2\\ 2& 3& 1& 2\end{array}\right]\\ R_2→R_2−R_1;R_3→R_3−2R_1\\ \left[\begin{array}{cccc}1& 1& 1& 1\\ 0& 1& 3& 1\\ 0& 1& −1& 0\end{array}\right]\\ R_1→R_1−R_2;R_3→R_3−R_2\\ \left[\begin{array}{cccc}1& 0& −2& 0\\ 0& 1& 3& 1\\ 0& 0& −4& −1\end{array}\right]\end{array}$

$\begin{array}{l}{R}_3→\frac{−1}{4}{R}_3\\ \left[\begin{array}{cccc}1& 0& −2& 0\\ 0& 1& 3& 1\\ 0& 0& 1& \frac{1}{4}\end{array}\right]\\ R_1→R_1+2R_3;R_2→R_2−3R_3\\ \left[\begin{array}{cccc}1& 0& 0& \frac{1}{2}\\ 0& 1& 0& \frac{1}{4}\\ 0& 0& 1& \frac{1}{4}\end{array}\right]\end{array}$

Therefore, the values are, $m_1=\frac{1}{2},m_2=m_3=\frac{1}{4}$.

Hence, a total mass of 1 kg should be distributed among the three vertices such that, are .

$m_1=\frac{1}{2},m_2=m_3=\frac{1}{4}$