M2 January 2011 Q7

EdexcelOld spec10 marksMoments Resolving Forces

Figure 4: plank AB of length 4 m with A on the ground at angle alpha, resting on a cylindrical drum at C, AC = 3 m — Figure 4

A uniform plank \(AB\), of weight 100 N and length 4 m, rests in equilibrium with the end \(A\) on rough horizontal ground. The plank rests on a smooth cylindrical drum. The drum is fixed to the ground and cannot move. The point of contact between the plank and the drum is \(C\), where \(AC = 3\) m, as shown in Figure 4. The plank is resting in a vertical plane which is perpendicular to the axis of the drum, at an angle \(\alpha\) to the horizontal, where \(\sin\alpha = \dfrac{1}{3}\). The coefficient of friction between the plank and the ground is \(\mu\). Modelling the plank as a rod, find the least possible value of \(\mu\). (10)

Force diagram: R up and F horizontal at A, 100 N at 2 m from A, normal reaction S at C perpendicular to the plank at angle alpha to the vertical — Mark scheme

Scheme	Marks
Taking moments about A:
\(3S = 100 \times 2 \times \cos\alpha\)	M1 A1
Resolving vertically:
\(R + S\cos\alpha = 100\)	M1 A1
Resolving horizontally:
\(S\sin\alpha = F\)	M1 A1
(Most alternative methods need 3 independent equations, each one worth M1A1. Can be done in 2 e.g. if they resolve horizontally and take moments about \(X\) then \(\ R \times 2 \times \cos\alpha = S \times (3 - 2 \times \cos^2\alpha)\ \) scores M2A2)
Substitute trig values to obtain correct values for F and R (exact or decimal equivalent).	DM1
\(\left(S = \dfrac{200\sqrt{8}}{9}\right),\ R = 100 - \dfrac{1600}{27} = \dfrac{1100}{27} \approx 40.74,\ \ F = \dfrac{200\sqrt{8}}{27} \approx 20.95\ldots\)	A1
\(F \leqslant \mu R,\ \ 200\sqrt{8} \leqslant \mu \times 1100,\ \ \mu \geqslant \dfrac{200\sqrt{8}}{1100} = \dfrac{2\sqrt{8}}{11}\).	M1
Least possible \(\mu\) is 0.514 (3sf), or exact.	A1
	(10 marks)