M2 June 2011 Q7

EdexcelOld spec13 marksMoments Resolving Forces

Figure 3: horizontal rod AB of length 4a with A against a vertical wall, string BD with D 3a above A, particle at C where AC = x — Figure 3

A uniform rod \(AB\), of mass \(3m\) and length \(4a\), is held in a horizontal position with the end \(A\) against a rough vertical wall. One end of a light inextensible string \(BD\) is attached to the rod at \(B\) and the other end of the string is attached to the wall at the point \(D\) vertically above \(A\), where \(AD = 3a\). A particle of mass \(3m\) is attached to the rod at \(C\), where \(AC = x\). The rod is in equilibrium in a vertical plane perpendicular to the wall as shown in Figure 3. The tension in the string is \(\dfrac{25}{4}mg\).

Show that

(a) \(x = 3a\), (5)

(b) the horizontal component of the force exerted by the wall on the rod has magnitude \(5mg\). (3)

The coefficient of friction between the wall and the rod is \(\mu\). Given that the rod is about to slip,

Force diagram: F up and R horizontal at A, 3mg at 2a from A, 3mg at C (AC = x), tension T along BD of length 5a, angle theta at D — Mark scheme (a)

Scheme	Marks
\(\text{M(A)}\quad 3mg \times 2a + 3mgx = T\cos\theta \times 4a\)	M1 A2,1,0
\(= \dfrac{12}{5}aT\)
\(\dfrac{12}{5}aT = 6mga + 3mgx\)
\(T = \dfrac{25}{4}mg\qquad \dfrac{12}{5}a \times \dfrac{25}{4}mg = 6mga + 3mgx\)	M1
\(15a = 6a + 3x\)
\(x = 3a\quad **\)	A1
	(5)

Scheme	Marks
\(\text{R}(\rightarrow)\quad R = T\sin\theta\)	M1
\(= \dfrac{25}{4}mg \times \dfrac{4}{5}\)	A1
\(= 5mg\quad **\)	A1
	(3)

Mark scheme (c)

Scheme	Marks
\(\text{R}(\uparrow)\quad F + \dfrac{25}{4}mg \times \dfrac{3}{5} = 3mg + 3mg\)	M1 A2,1,0
\(F = 6mg - \dfrac{15}{4}mg = \dfrac{9}{4}mg\)
\(\mu = \dfrac{F}{R} = \dfrac{\frac{9}{4}mg}{5mg} = \dfrac{9}{20}\)	DM1 A1
	(5)
	(13 marks)