M2 June 2008 Q5

EdexcelOld spec11 marksMoments Resolving Forces

Figure 2: rod AB of length 3a resting on a peg at C, AC = 2a, A on the ground, angle alpha with the ground — Figure 2

A plank rests in equilibrium against a fixed horizontal pole. The plank is modelled as a uniform rod \(AB\) and the pole as a smooth horizontal peg perpendicular to the vertical plane containing \(AB\). The rod has length \(3a\) and weight \(W\) and rests on the peg at \(C\), where \(AC = 2a\). The end \(A\) of the rod rests on rough horizontal ground and \(AB\) makes an angle \(\alpha\) with the ground, as shown in Figure 2.

(a) Show that the normal reaction on the rod at \(A\) is \(\dfrac{1}{4}(4 - 3\cos^2\alpha)W\). (6)

Given that the rod is in limiting equilibrium and that \(\cos\alpha = \dfrac{2}{3}\),

(b) find the coefficient of friction between the rod and the ground. (5)

Force diagram: R up and mu R horizontal at A, W down at 1.5a from A, P at C perpendicular to the rod, 0.5a further on — Mark scheme (a)

Scheme	Marks
R\((\uparrow)\) \(R + P\cos\alpha = W\)	M1 A1
M\((A)\) \(P \times 2a = W \times 1.5a\cos\alpha\)	M1 A1
\(\left(P = \dfrac{3}{4}W\cos\alpha\right)\)
\(R = W - P\cos\alpha = W - \dfrac{3}{4}W\cos^2\alpha\)	DM1
\(= \dfrac{1}{4}(4 - 3\cos^2\alpha)W\) * cso	A1
	(6)

Mark scheme (b)

Scheme	Marks
Using \(\cos\alpha = \dfrac{2}{3}\), \(R = \dfrac{2}{3}W\)	B1
R\((\rightarrow)\) \(\mu R = P\sin\alpha\)	M1 A1
Leading to \(\mu = \dfrac{3}{4}\sin\alpha\) \(\left(\sin\alpha = \sqrt{1 - \tfrac{4}{9}} = \tfrac{\sqrt{5}}{3}\right)\)
\(\mu = \dfrac{\sqrt{5}}{4}\) awrt 0.56	DM1 A1
	(5)
	(11 marks)