M2 June 2010 Q6

EdexcelOld spec8 marksMoments

Figure 2: horizontal rod AB hinged to a wall at A with particle (m) at B; string from C on the rod to D on the wall, AD = 2a, AC = 3a, CB = a — Figure 2

Figure 2 shows a uniform rod \(AB\) of mass \(m\) and length \(4a\). The end \(A\) of the rod is freely hinged to a point on a vertical wall. A particle of mass \(m\) is attached to the rod at \(B\). One end of a light inextensible string is attached to the rod at \(C\), where \(AC = 3a\). The other end of the string is attached to the wall at \(D\), where \(AD = 2a\) and \(D\) is vertically above \(A\). The rod rests horizontally in equilibrium in a vertical plane perpendicular to the wall and the tension in the string is \(T\).

(a) Show that \(T = mg\sqrt{13}\). (5)

The particle of mass \(m\) at \(B\) is removed from the rod and replaced by a particle of mass \(M\) which is attached to the rod at \(B\). The string breaks if the tension exceeds \(2mg\sqrt{13}\). Given that the string does not break,

(b) show that \(M \leqslant \dfrac{5}{2}m\). (3)

Force diagram: rod with forces F up and R horizontal at A, mg at 2a from A, tension T at 3a along the string to D (angle theta at D), and mg at B — Mark scheme (a)

Scheme	Marks
\(\text{M}(A)\quad 3a \times T\cos\theta = 2amg + 4amg\)	M1 A1 A1
\(\cos\theta = \left(\dfrac{2}{\sqrt{9 + 4}} =\right)\dfrac{2}{\sqrt{13}}\)	B1
\(\dfrac{6}{\sqrt{13}}T = 6mg\)
\(T = mg\sqrt{13}\ *\)	A1
	(5)

Mark scheme (b)

Scheme	Marks
\(3a \times T \times \cos\theta = 2amg + 4aMg\)	M1
\(T = \dfrac{(2mg + 4Mg)}{6}\sqrt{13} \leqslant 2mg\sqrt{13}\)	A1
\(mg + 2Mg < 6mg\)
\(M \leqslant \dfrac{5}{2}m\ *\) cso	A1
	(3)
	(8 marks)