1.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{0567d068-e23c-446e-9e11-f0c292972093-02_490_824_253_588}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}

Figure 1 shows a uniform rectangular lamina $A B C D$ with $A B = 2 a$ and $A D = a$ The mass of the lamina is $6 m$.

A particle of mass $2 m$ is attached to the lamina at $A$, a particle of mass $m$ is attached to the lamina at $B$ and a particle of mass $3 m$ is attached to the lamina at $D$, to form a loaded lamina $L$ of total mass $12 m$.
\begin{enumerate}[label=(\alph*)]
\item Write down the distance of the centre of mass of $L$ from $A B$. You must give a reason for your answer.
\item Show that the distance of the centre of mass of $L$ from $A D$ is $\frac { 2 a } { 3 }$

A particle of mass $k m$ is now also attached to $L$ at $D$ to form a new loaded lamina $N$.
\item Show that the distance of the centre of mass of $N$ from $A B$ is $\frac { ( k + 6 ) a } { ( k + 12 ) }$

When $N$ is freely suspended from $A$ and is hanging in equilibrium, the side $A B$ makes an angle $\alpha$ with the vertical, where $\tan \alpha = \frac { 3 } { 2 }$
\item Find the value of $k$.
\end{enumerate}

\hfill \mbox{\textit{Edexcel FM2 AS 2020 Q1 [15]}}