M2 June 2014 Q3

Figure 1: lamina ABCDEF, triangle ABC of height 9a and base 12a with triangle FED of height 6a and base 8a removed — Figure 1

The uniform lamina \(ABCDEF\), shown shaded in Figure 1, is symmetrical about the line through \(B\) and \(E\). It is formed by removing the isosceles triangle \(FED\), of height \(6a\) and base \(8a\), from the isosceles triangle \(ABC\) of height \(9a\) and base \(12a\).

(a) Find, in terms of \(a\), the distance of the centre of mass of the lamina from \(AC\). (5)

The lamina is freely suspended from \(A\) and hangs in equilibrium.

(b) Find, to the nearest degree, the size of the angle between \(AB\) and the downward vertical. (4)

Diagram: lamina hanging from A, AF = 2a, FD side 4a, centre of mass 3.8a from AC on the axis through E and B, 6a and 3a along the axis — Mark scheme (a)

Diagram: triangle AGB with AG = 7.1, GB = 5.2, AB = 10.8 and angle theta at A — Mark scheme (b)

Scheme	Marks

Ratio of areas 54 : 24 : 30 (or equivalent)	B1
Distance of c of m from \(AC\) \(3a\), \(2a\)	B1
The two B marks can be implied by a correct moments equation
Moments about \(AC\): \(30d = 3a \times 54 - 2a \times 24\ (= 114a)\)	M1 A1
\(d = \dfrac{114a}{30} = 3.8a\)	A1
	(5)

Scheme	Marks
Correct triangle and use of \(\tan^{-1}\) or equivalent	M1
\(\tan^{-1}\left(\dfrac{3.8a}{6a}\right),\ \tan^{-1}\left(\dfrac{6a}{3.8a}\right),\ 32.3\ldots\) or \(57.65\ldots\)	A1
Required angle \(= \tan^{-1}\left(\dfrac{9a}{6a}\right) - \tan^{-1}\left(\dfrac{3.8a}{6a}\right) = 23.96\ldots = 24^\circ\)	DM1 A1
	(4)
	(9 marks)

	M1 A1
\(\cos\theta = \dfrac{10.8^2 + 7.1^2 - 5.2^2}{2 \times 10.8 \times 7.1}\)	DM1
\(\theta = 24^\circ\)	A1

M2 June 2014 Q3

M2 June 2014 Q3

Notes

Notes

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