AS June 2022 Q1

EdexcelCurrent spec7 marksCentres of Mass

Figure 1: isosceles trapezium ABCDEF with A, B, C, D along the top edge (AB = a, BC = 3a, CD = a) and F, E along the bottom, with F below B and E below C; the square BCEF is marked with dashed lines — Figure 1

A uniform plane lamina is in the shape of an isosceles trapezium \(ABCDEF\), as shown shaded in Figure 1.

\(BCEF\) is a square
\(AB = CD = a\)
\(BC = 3a\)

(a) Show that the distance of the centre of mass of the lamina from \(AD\) is \(\dfrac{11a}{8}\) (5)

The mass of the lamina is \(M\)

The lamina is suspended by two light vertical strings, one attached to the lamina at \(A\) and the other attached to the lamina at \(F\)

The lamina hangs freely in equilibrium, with \(BF\) horizontal.

(b) Find, in terms of \(M\) and \(g\), the tension in the string attached at \(A\) (2)

Mark scheme (a)

Scheme	Marks	AO
\(ABF\) \(BCEF\) \(CDE\) lamina
\(\dfrac{3}{2}a^2\) \(9a^2\) \(\dfrac{3}{2}a^2\) \(12a^2\)	B1	1.2
\(a\) \(\dfrac{3}{2}a\) \(a\) \(\bar{y}\)	B1	1.2
Moments about \(AD\)	M1	2.1
\(\left(\dfrac{3}{2}a^2 \times a\right) + \left(9a^2 \times \dfrac{3}{2}a\right) + \left(\dfrac{3}{2}a^2 \times a\right) = 12a^2\bar{y}\)	A1	1.1b
\(\bar{y} = \dfrac{11a}{8}\) *	A1*	2.2a
	(5)

Notes

B1: Any equivalent ratios e.g. 3 : 18 : 3 : 24

B1: Or correct distances from a parallel axis

M1: Or moments about a parallel axis

A1: Correct unsimplified equation for their axis

A1*: Correct given answer correctly obtained
If they have centre of mass at \((xa, ya)\) then the \(a\) might not be seen in the working.
Otherwise, with no \(a\) in the working the maximum score is B1B0M1A0A0

Mark scheme (b)

Scheme	Marks	AO
Moments about \(F\), \(Mg \times \left(3a - \dfrac{11a}{8}\right) = 3aT\)	M1	3.1a
\(T = \dfrac{13Mg}{24}\) \((0.54166666\ldots Mg)\)	A1	1.1b
	(2)
	(7 marks)

Notes

M1: A complete method to obtain an equation in \(T\) only

A1: \(0.54Mg\) or better