June 2024 Paper 3 Mechanics Q6

EdexcelCurrent spec9 marksMoments Resolving Forces

Figure 5 shows a uniform rod \(AB\) of mass \(M\) and length \(2a\).

the rod has its end \(A\) on rough horizontal ground
the rod rests in equilibrium against a small smooth fixed horizontal peg \(P\)
the point \(C\) on the rod, where \(AC = 1.5a\), is the point of contact between the rod and the peg
the rod is at an angle \(\theta\) to the ground, where \(\tan\theta = \dfrac{4}{3}\)

The rod lies in a vertical plane perpendicular to the peg.

The magnitude of the normal reaction of the peg on the rod at \(C\) is \(S\).

(a) Show that \(S = \dfrac{2}{5}Mg\) (3)

The coefficient of friction between the rod and the ground is \(\mu\).

Given that the rod is in limiting equilibrium,

(b) find the value of \(\mu\). (6)

Scheme	Marks	AO

Take moments about \(A\)	M1	3.1a
\(S \times 1.5a = Mga\cos\theta = \left(Mga \times \dfrac{3}{5}\right)\)	A1	1.1b
\(S = \dfrac{2}{5}Mg\) *	A1*	2.2a
	(3)

Scheme	Marks	AO

N.B. Marks for the equations should be awarded in the order in which they appear on the script.
Resolve horizontally:	M1	3.4
\(F = S\sin\theta\)	A1	1.1b
Resolve vertically:	M1	3.3
\(R = Mg - S\cos\theta\)	A1	1.1b
Other possible equations: (any of which is worth max M1A1) (parallel to the rod): \(F\cos\theta + R\sin\theta = Mg\sin\theta\) (perp to the rod): \(F\sin\theta + Mg\cos\theta = S + R\cos\theta\) M(\(B\)): \((S \times 0.5a) + (R \times 2a\cos\theta) = (Mg \times a\cos\theta) + (F \times 2a\sin\theta)\) M(\(C\)): \((R \times 1.5a\cos\theta) = (Mg \times 0.5a\cos\theta) + (F \times 1.5a\sin\theta)\) M(\(G\)): \((R \times a\cos\theta) = (S \times 0.5a) + (F \times a\sin\theta)\) N.B. If they have more than two equations, mark only those that they use to try to find \(\mu\)
\(F = \mu R\) and two of their equations used to solve for \(\mu\)	DM1	3.1a
\(\mu = \dfrac{8}{19} = 0.42105\ldots\)	A1	2.2a
	(6)
	(9 marks)