M2 June 2014 (R) Q4

EdexcelOld spec10 marksCentres of Mass

Figure 2: isosceles triangle ABC with AB = 12 cm vertical, DC = 8 cm, square PQRS removed with P and Q on AB and centre O — Figure 2

Figure 2 shows a lamina \(L\). It is formed by removing a square \(PQRS\) from a uniform triangle \(ABC\). The triangle \(ABC\) is isosceles with \(AC = BC\) and \(AB = 12\) cm. The midpoint of \(AB\) is \(D\) and \(DC = 8\) cm. The vertices \(P\) and \(Q\) of the square lie on \(AB\) and \(PQ = 4\) cm. The centre of the square is \(O\). The centre of mass of \(L\) is at \(G\).

(a) Find the distance of \(G\) from \(AB\). (4)

When \(L\) is freely suspended from \(A\) and hangs in equilibrium, the line \(AB\) is inclined at 25\(^\circ\) to the vertical.

(b) Find the distance of \(O\) from \(DC\). (6)

Mark scheme (a)

Scheme

Marks

	ABC	PQRS	Result
Mass ratio	48	16	32
Dist of c of m from \(AB\)	\(\dfrac{8}{3}\)	2	\(\bar{x}\)

\(48 \times \dfrac{8}{3} - 16 \times 2 = 32\bar{x}\)

M1 A2

\(\bar{x} = (8 - 2) \div 2 = 3\ \ \text{(cm)}\)

(4)

Notes

M1 Take moments about \(AB\). Must be subtracting the square from the triangle for mass & moments.

A2 Correct unsimplified equation. -1 each error. Could have a common factor of \(g\).

Diagram: AB hanging at 25 degrees to the vertical AG, with N on AD and NG = 3 cm — Mark scheme (b)

Scheme	Marks
\(\tan 25 = \dfrac{\text{their } 3}{AN}\) or \(\dfrac{AN}{\text{their } 3} = \tan 65\)	M1 A1
Dist of \(G\) from \(DC = \dfrac{3}{\tan 25} - 6\)	M1
\((-)\bar{y} = 2\left(\dfrac{\text{their } 3}{\tan 25} - 6\right)\)	M1 A1ft
Distance \(= 0.867\ldots\) cm	A1
	(6)
	(10 marks)