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A uniform lamina \(A B C D E\) consists of a rectangle \(B C D E\) and an isosceles triangle \(A B E\) joined along their common edge \(B E\). For the triangle, \(A B = A E , B E = a \mathrm {~m}\) and the perpendicular height is \(h \mathrm {~m}\). For the rectangle, \(B C = D E = 0.5 \mathrm {~m}\) and \(C D = B E = a \mathrm {~m}\) (see diagram).
- Show that the distance in metres of the centre of mass of the lamina from \(B E\) towards \(C D\) is
$$\frac { 3 - 4 h ^ { 2 } } { 12 + 12 h }$$
The lamina is freely suspended at \(E\) and hangs in equilibrium.
- Given that \(D E\) is horizontal, calculate \(h\).
- Given instead that \(h = 0.5\) and \(A E\) is horizontal, calculate \(a\).