M2 January 2012 Q5

EdexcelOld spec11 marksMoments Resolving Forces

Figure 2: rod AB of length 1.4 m with A on the ground, inclined at 20 degrees, string BC perpendicular to the rod at B — Figure 2

A uniform rod \(AB\) has mass 4 kg and length 1.4 m. The end \(A\) is resting on rough horizontal ground. A light string \(BC\) has one end attached to \(B\) and the other end attached to a fixed point \(C\). The string is perpendicular to the rod and lies in the same vertical plane as the rod. The rod is in equilibrium, inclined at 20\(^\circ\) to the ground, as shown in Figure 2.

(a) Find the tension in the string. (4)

Given that the rod is about to slip,

(b) find the coefficient of friction between the rod and the ground. (7)

Scheme	Marks
Taking moments about A:	M1
\(4g \times 0.7 \times \cos 20^\circ = 1.4T\)	A1 A1
\(T = 18.4\) N	A1
	(4)

Mark scheme (b)

Scheme	Marks
\(\uparrow\quad R + T\cos 20 = 4g\)
\(R = 4g - T\cos 20^\circ\)	M1 A1
\(\rightarrow\quad F = T\sin 20\)	M1 A1
\(F = \mu R \Rightarrow T\sin 20^\circ = \mu\left(4g - T\cos 20^\circ\right)\)	DM1 A1
\(\mu = \dfrac{T\sin 20^\circ}{4g - T\cos 20^\circ} = 0.29\)	A1
	(7)
	(11 marks)