M1 June 2013 (R) Q8

EdexcelOld spec13 marksMoments

Figure 5: rod AB of length 2 m on supports at C and D, AC = 0.2 m, DB = x m — Figure 5

A uniform rod \(AB\) has length 2 m and mass 50 kg. The rod is in equilibrium in a horizontal position, resting on two smooth supports at \(C\) and \(D\), where \(AC = 0.2\) metres and \(DB = x\) metres, as shown in Figure 5. Given that the magnitude of the reaction on the rod at \(D\) is twice the magnitude of the reaction on the rod at \(C\),

(a) find the value of \(x\). (6)

The support at \(D\) is now moved to the point \(E\) on the rod, where \(EB = 0.4\) metres. A particle of mass \(m\) kg is placed on the rod at \(B\), and the rod remains in equilibrium in a horizontal position. Given that the magnitude of the reaction on the rod at \(E\) is four times the magnitude of the reaction on the rod at \(C\),

(b) find the value of \(m\). (7)

Force diagram: reactions R at C and 2R at D, weight 50g at the centre; distances 0.2 m, 0.8, 1 - x and x m — Mark scheme (a)

Force diagram: reactions S at C and 4S at E, weight 50g at the centre, mg at B; distances 0.2 m, 0.8, 0.6 and 0.4 m — Mark scheme (b)

Scheme	Marks
Vertical equilibrium: \(R + 2R = 50g\),	M1A1
Moments about \(C\): \(50g \times 0.8 = (1.8 - x) \times 2 \times R\)	M1A1
\(3 \times 0.8 = 3.6 - 2x\), \(x = 0.6\)	DM1A1
	(6)

Scheme	Marks
\(S\), \(4S\)	B1
Vertical equilibrium: \(S + 4S = (50 + m)g = 5S\)	M1A1
Moments about \(B\): \(50g \times 1 = 4S \times 0.4 + S \times 1.8 = 3.4S\)	M1A1
\(50 \times \dfrac{5}{3.4} = (50 + m)\)	DM1
\(m = 400/17\), 24, 23.5 or better	A1
	(7)
	(13 marks)

M1 June 2013 (R) Q8

M1 June 2013 (R) Q8

Notes

Notes