4.
\begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{d5f06fe7-4d9c-4009-8931-3ecbc31fa5e5-10_552_680_255_447}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{figure}
\begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{d5f06fe7-4d9c-4009-8931-3ecbc31fa5e5-10_547_494_255_1165}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{figure}
The uniform rectangular lamina \(A B C D\), shown in Figure 2, has \(D C = 4 a\) and \(A D = 5 a\)
The points \(S\) on \(A B\) and \(T\) on \(B C\) are such that \(S B = B T = 3 a\)
The lamina is folded along \(S T\) to form the folded lamina \(L\), shown in Figure 3.
The distance of the centre of mass of \(L\) from \(A D\) is \(d\).
- Show that \(d = \frac { 71 } { 40 } a\)
The weight of \(L\) is \(4 W\). A particle of weight \(W\) is attached to \(L\) at \(C\).
The folded lamina \(L\) is freely suspended from \(S\).
A force of magnitude \(F\), acting parallel to \(D C\), is applied to \(L\) at \(D\) so that \(A D\) is vertical. - Find \(F\) in terms of \(W\)