1.
\begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{0567d068-e23c-446e-9e11-f0c292972093-02_490_824_253_588}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\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 \(12 m\).
- Write down the distance of the centre of mass of \(L\) from \(A B\). You must give a reason for your answer.
- 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 \(N\).
- 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 \(\tan \alpha = \frac { 3 } { 2 }\)
- Find the value of \(k\).