4.
\begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{790546bf-38a4-4eb7-876e-941fe58f4a48-07_671_661_239_635}
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\caption{Figure 2}
\end{figure}
The uniform lamina \(A B C D E\) is made by joining a rectangular lamina \(A B D E\) to a triangular lamina \(B C D\) along the edge \(B D\). The rectangle has length \(6 a\) and width \(3 a\). The triangle is isosceles, with \(B C = C D\), and the distance from \(C\) to \(B D\) is \(3 a\), as shown in Figure 2.
- Find the distance of the centre of mass of the lamina, \(A B C D E\), from \(A E\).
The lamina \(A B C D E\) is freely suspended from \(A\). A horizontal force of magnitude \(F\) newtons is applied to the lamina at \(D\). The line of action of the force lies in the vertical plane containing the lamina. The lamina is in equilibrium with \(A E\) vertical. The mass of the lamina is 4 kg .
- Find the magnitude of the force exerted on the lamina at \(A\).
DO NOT WIRITE IN THIS AREA