M2 June 2013 (R) Q6

EdexcelOld spec13 marksCentres of Mass

6.

Figure 2: isosceles triangle ABC with BC = 2a and height h; line DE parallel to BC at distance 2h/3 from A
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 \(\dfrac{2h}{3}\) 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 \(\dfrac{5M}{9}\).

(a) Show that the centre of mass of the trapezium \(BCED\) is \(\dfrac{7h}{45}\) from \(BC\). (5)
Figure 3: the lamina with triangle ADE folded about DE so that A lies below BC
Figure 3

The portion \(ADE\) of the lamina is folded through 180\(^\circ\) about \(DE\) to form the folded lamina shown in Figure 3.

(b) 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 \(\alpha\).

(c) Find \(\tan\alpha\) in terms of \(a\) and \(h\). (4)
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