M2 June 2009 Q5

Figure 2: rectangle ABCD, BC 120 cm, DC 90 cm, with isosceles triangle EFG of base 60 cm and height 60 cm removed from the edge AD — Figure 2

A shop sign \(ABCDEFG\) is modelled as a uniform lamina, as illustrated in Figure 2. \(ABCD\) is a rectangle with \(BC = 120\) cm and \(DC = 90\) cm. The shape \(EFG\) is an isosceles triangle with \(EG = 60\) cm and height 60 cm. The mid-point of \(AD\) and the mid-point of \(EG\) coincide.

(a) Find the distance of the centre of mass of the sign from the side \(AD\). (5)

The sign is freely suspended from \(A\) and hangs at rest.

(b) Find the size of the angle between \(AB\) and the vertical. (4)

Mark scheme (a)

Scheme	Marks
Ratio of areas triangle:sign:rectangle = 1 : 5 : 6 (1800:9000:10800)	B1
Centre of mass of the triangle is 20cm down from \(AD\) (seen or implied)	B1
\(\Rightarrow 6 \times 45 - 1 \times 20 = 5 \times \bar{y}\)	M1A1
\(\bar{y} = 50\text{cm}\)	A1
	(5)

Scheme	Marks
Distance of centre of mass from \(AB\) is 60cm	B1
Required angle is \(\tan^{-1}\dfrac{60}{50}\) (their values)	M1A1ft
\(= 50.2^\circ\) (0.876 rads)	A1
	(4)
	(9 marks)