\includegraphics{figure_2}
A uniform triangular lamina \(ABC\) of mass \(M\) is such that \(AB = AC\), \(BC = 2a\) and the distance of \(A\) from \(BC\) is \(h\). A line, parallel to \(BC\) and at a distance \(\frac{2h}{3}\) from \(A\), cuts \(AB\) at \(D\) and cuts \(AC\) at \(E\), as shown in Figure 2.
It is given that the mass of the trapezium \(BCED\) is \(\frac{5M}{9}\).
- Show that the centre of mass of the trapezium \(BCED\) is \(\frac{7h}{45}\) from \(BC\). [5]
\includegraphics{figure_3}
The portion \(ADE\) of the lamina is folded through 180° about \(DE\) to form the folded lamina shown in Figure 3.
- Find the distance of the centre of mass of the folded lamina from \(BC\). [4]
The folded lamina is freely suspended from \(D\) and hangs in equilibrium. The angle between \(DE\) and the downward vertical is \(\alpha\).
- Find tan \(\alpha\) in terms of \(a\) and \(h\). [4]