AS June 2018 Q3

EdexcelCurrent spec11 marksCentres of Mass

Figure 2: square ABDF of side 2a with A top left, B top right, F bottom left and D bottom right; triangle BDC attached on the right with DC = a along FD extended, and triangle FDE attached below with DE = a along BD extended — Figure 2

The lamina \(L\), shown in Figure 2, consists of a uniform square lamina \(ABDF\) and two uniform triangular laminas \(BDC\) and \(FDE\). The square has sides of length \(2a\). The two triangles are identical.

The straight lines \(BDE\) and \(FDC\) are perpendicular with \(BD = DF = 2a\) and \(DC = DE = a\).
The mass per unit of area of the square is \(M\).
The mass per unit area of each triangle is \(3M\).
The centre of mass of \(L\) is at the point \(G\).

(a) Without doing any calculations, explain why \(G\) lies on \(AD\). (1)

(b) Show that the distance of \(G\) from \(D\) is \(\dfrac{\sqrt{2}}{2}a\) (7)

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

Scheme	Marks	AO
\(L\) is symmetrical about \(AD\)	B1	2.4
	(1)

Mark scheme (b)

Scheme

Marks

	\(ABDF\)	\(BCD\)	\(DEF\)	\(L\)
Mass ratio	\(4a^2 \times M\)	\(a^2 \times 3M\)	\(a^2 \times 3M\)	\(10a^2 \times M\)
C of M from \(BE\)	\(-a\)	\(+\dfrac{a}{3}\)	\(-\dfrac{2a}{3}\)	\(x\)

Mass ratios

1.2

Distances from \(BE\)

1.2

Moments equation

2.1

\(-a \times 4a^2M + \dfrac{a}{3} \times 3a^2M - \dfrac{2a}{3} \times 3a^2M = 10a^2M \times x\)
\((-4a + a - 2a = 10x)\)

1.1b

\(x = -\dfrac{5a}{10} = -\dfrac{a}{2}\)

1.1b

Use symmetry and Pythagoras

1.1a

Distance from \(D = \sqrt{\dfrac{a^2}{4} + \dfrac{a^2}{4}} = \dfrac{\sqrt{2}}{2}a\) *

A1*

2.2a

(7)

Notes

B1: Correct mass ratios

B1: Distance ratios from any horizontal or vertical axis

M1: Moments equation for complete lamina about any horizontal or vertical axis. Must be dimensionally correct

A1: Correct unsimplified equation for their axes

A1: Correct horizontal or vertical distance from \(D\)

M1: Use of Pythagoras with their distance

A1*: Obtain given answer from correct working.

right-angled triangle with vertical side BD, 3a/2 from B to the right angle, horizontal side a/2, and angle theta at B — Mark scheme (c)

Scheme	Marks	AO

Trig ratio of a relevant angle	M1	1.2
\(\tan\theta = \dfrac{1}{3}\) or \(\cos\theta = \dfrac{\dfrac{10}{4}a^2 + 4a^2 - \dfrac{2}{4}a^2}{2 \times \dfrac{\sqrt{10}}{2}a \times 2a} = \dfrac{6}{2\sqrt{10}}\)	A1ft	1.1b
\(\theta = 18.4^\circ\)	A1	1.1b
	(3)
	(11 marks)