M2 June 2014 (R) Q3

EdexcelOld spec10 marksMoments Resolving Forces

Figure 1: rod AB with A on horizontal floor and B against vertical wall, angle 60 degrees at A — Figure 1

A non-uniform rod, \(AB\), of mass \(m\) and length \(2l\), rests in equilibrium with one end \(A\) on a rough horizontal floor and the other end \(B\) against a rough vertical wall. The rod is in a vertical plane perpendicular to the wall and makes an angle of 60\(^\circ\) with the floor as shown in Figure 1. The coefficient of friction between the rod and the floor is \(\dfrac{1}{4}\) and the coefficient of friction between the rod and the wall is \(\dfrac{2}{3}\). The rod is on the point of slipping at both ends.

(a) Find the magnitude of the vertical component of the force exerted on the rod by the floor. (5)

The centre of mass of the rod is at \(G\).

(b) Find the distance \(AG\). (5)

Force diagram: S up and F to the right at A, mg down at G, R to the left and Q up at B — Mark scheme (a)

Scheme	Marks
\(R = F\)	B1
\(S + Q = mg\)	B1
\(Q = \dfrac{2}{3}R,\ \ F = \dfrac{1}{4}S\)	B1
\(Q = \dfrac{2}{3}R = \dfrac{2}{3} \times \dfrac{1}{4}S,\ \ \ S + \dfrac{1}{6}S = mg,\ \ S = \dfrac{6}{7}mg\)	M1 A1
	(5)

Mark scheme (b)

Scheme	Marks
M\((A)\) \(\ mg \times x\cos 60 = Q \times 2l\cos 60 + R \times 2l\sin 60\)	M1
M \((B)\) \(\ mg(2l - x)\cos 60 + F \times 2l\sin 60 = S \times 2l\cos 60\)
M (c of m) \(\ Sx\cos 60 = Fx\sin 60 + R(2l - x)\sin 60 + Q(2l - x)\cos 60\)	A2
\(mgx\cos 60 = \dfrac{1}{6} \times \dfrac{6}{7}mg \times 2l\cos 60 + \dfrac{1}{4} \times \dfrac{6}{7}mg \times 2l\sin 60\)	DM1
\(\dfrac{1}{2}x = \dfrac{1}{7} \times 2l \times \dfrac{1}{2} + \dfrac{3}{14} \times l\sqrt{3}\)
\(AG = x = 1.028\ldots l\ \ \ x = 1.03l\)	A1
	(5)
	(10 marks)

Notes

M1 Moments equation – must include all terms. Condone sign errors and sin/cos confusion

A2 Correct unsimplified equation (for their \(S\).) -1 each error

DM1 Form an equation in \(x\). Depends on the preceding M

A1 \(1.03l\) or better \(\ \dfrac{l\left(2 + 3\sqrt{3}\right)}{7}\)