M2 January 2007 Q5

EdexcelOld spec12 marksMoments Resolving Forces

Figure 2: horizontal rod AB of length 4a with A against a wall, C on AB with AC = 3a, string from B to D on the wall above A at angle theta to the horizontal — Figure 2

A horizontal uniform rod \(AB\) has mass \(m\) and length \(4a\). The end \(A\) rests against a rough vertical wall. A particle of mass \(2m\) is attached to the rod at the point \(C\), where \(AC = 3a\). One end of a light inextensible string \(BD\) is attached to the rod at \(B\) and the other end is attached to the wall at a point \(D\), where \(D\) is vertically above \(A\). The rod is in equilibrium in a vertical plane perpendicular to the wall. The string is inclined at an angle \(\theta\) to the horizontal, where \(\tan\theta = \tfrac{3}{4}\), as shown in Figure 2.

(a) Find the tension in the string. (5)

(b) Show that the horizontal component of the force exerted by the wall on the rod has magnitude \(\tfrac{8}{3}mg\). (3)

The coefficient of friction between the wall and the rod is \(\mu\). Given that the rod is in limiting equilibrium,

Diagram: rod AB with F and R at A, weights mg and 2mg at 2a and 3a from A, tension T along BD — Mark scheme (a)

Scheme	Marks
M(\(A\)) \(T\sin\theta \times 4a = mg \times 2a + 2mg \times 3a\)	M1* A1=A1
\(T = \dfrac{8mg}{4} \times \dfrac{5}{3} = \dfrac{10}{3}mg\) Accept 32.7m, 33m	DM1* A1
	(5)

Scheme	Marks
\(\rightarrow\) \(R = T\cos\theta = \dfrac{10}{3}mg \times \dfrac{4}{5};\ = \dfrac{8}{3}mg\ \ *\) cso ft their T	M1 A1ft; A1
	(3)

Mark scheme (c)

Scheme	Marks
\(\uparrow\) \(F + T\sin\theta = 3mg\ \ \Rightarrow\ \ F = mg\) ft their T	M1 A1ft
Or: M(B) \(F \times 4a = mg \times 2a + 2mg \times a \Rightarrow F = mg\)
\(F = \mu R\ \ \Rightarrow\ \ \mu = \tfrac{3}{8}\)	M1 A1
	(4)
	(12 marks)

(a) Alternative approach: