AS June 2023 Q4

EdexcelCurrent spec14 marksCentres of Mass

4.

Figure 1: isosceles triangle ABC with AB = 18a horizontal at the top, M the midpoint of AB and C vertically below M with CM = 18a; triangle CDE, with D on MC, DE = 6a parallel to AB and CD = 12a, is removed from the left side, leaving the shaded lamina L
Figure 1

A uniform triangular lamina \(ABC\) is isosceles, with \(AC = BC\). The midpoint of \(AB\) is \(M\).
The length of \(AB\) is \(18a\) and the length of \(CM\) is \(18a\).

The triangular lamina \(CDE\), with \(DE = 6a\) and \(CD = 12a\), has \(ED\) parallel to \(AB\) and \(MDC\) is a straight line.

Triangle \(CDE\) is removed from triangle \(ABC\) to form the lamina \(L\), shown shaded in Figure 1.

The distance of the centre of mass of \(L\) from \(MC\) is \(d\).

(a) Show that \(d = \dfrac{4}{7}a\) (4)

The lamina \(L\) is suspended by two light inextensible strings. One string is attached to \(L\) at \(A\) and the other string is attached to \(L\) at \(B\).
The lamina hangs in equilibrium in a vertical plane with the strings vertical and \(AB\) horizontal.
The weight of \(L\) is \(W\)

(b) Find, in terms of \(W\), the tension in the string attached to \(L\) at \(B\) (3)

The string attached to \(L\) at \(B\) breaks, so that \(L\) is now suspended from \(A\).
When \(L\) is hanging in equilibrium in a vertical plane, the angle between \(AB\) and the downward vertical through \(A\) is \(\theta^\circ\)

(c) Find the value of \(\theta\) (7)
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