3.
\begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{66c0f4c8-938e-4c05-93a7-99ea26ea0348-08_694_710_382_780}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{figure}
The lamina \(L\), shown in Figure 2, consists of a uniform square lamina \(A B D F\) and two uniform triangular laminas \(B D C\) and \(F D E\). The square has sides of length \(2 a\). The two triangles are identical.
The straight lines \(B D E\) and \(F D C\) are perpendicular with \(B D = D F = 2 a\) and \(D C = D E = a\).
The mass per unit of area of the square is \(M\).
The mass per unit area of each triangle is \(3 M\).
The centre of mass of \(L\) is at the point \(G\).
- Without doing any calculations, explain why \(G\) lies on \(A D\).
- Show that the distance of \(G\) from \(D\) is \(\frac { \sqrt { 2 } } { 2 } a\)
The lamina \(L\) is freely suspended from \(B\) and hangs in equilibrium.
- Find the size of the angle between \(B E\) and the downward vertical.
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