M2 June 2013 Q5

EdexcelOld spec13 marksMoments Resolving Forces

Figure 3: rod AB of length 2a at angle theta to the horizontal, hinged at A, particle at B, force F at C perpendicular to the rod, AC = b — Figure 3

A uniform rod \(AB\), of mass \(m\) and length \(2a\), is freely hinged to a fixed point \(A\). A particle of mass \(m\) is attached to the rod at \(B\). The rod is held in equilibrium at an angle \(\theta\) to the horizontal by a force of magnitude \(F\) acting at the point \(C\) on the rod, where \(AC = b\), as shown in Figure 3. The force at \(C\) acts at right angles to \(AB\) and in the vertical plane containing \(AB\).

(a) Show that \(F = \dfrac{3amg\cos\theta}{b}\). (4)

(b) Find, in terms of \(a\), \(b\), \(g\), \(m\) and \(\theta\),

(i) the horizontal component of the force acting on the rod at \(A\),

(ii) the vertical component of the force acting on the rod at \(A\). (5)

Given that the force acting on the rod at \(A\) acts along the rod,

Force diagram: V and H at A, mg at the midpoint of the rod, mg at B, F at C perpendicular to the rod — Mark scheme (a)

Scheme	Marks

Moments about A:	M1
\(bF = a\cos\theta mg + 2a\cos\theta mg\ (= 3a\cos\theta mg)\)	A2
\(F = \dfrac{3amg\cos\theta}{b}\ \ \ \)*Answer given*	A1
	(4)

Mark scheme (b)

Scheme	Marks
\(\rightarrow:\ \ H = F\sin\theta = \dfrac{3amg\cos\theta\sin\theta}{b}\)	M1 A1
\(\uparrow:\ \ 2mg = \pm V + F\cos\theta\)	M1 A1
\(\pm V = 2mg - \dfrac{3amg\cos\theta}{b} \times \cos\theta\ \left(= 2mg - \dfrac{3amg\cos^2\theta}{b}\right)\)	A1
	(5)

Notes

M1 Resolve horizontally. Condone trig confusion

A1 RHS correct. Or equivalent.

M1 Resolve vertically. Condone sign error and trig confusion

A1 Correct equation

A1 RHS correct. Or equivalent

Diagram: three forces on the rod, R at A along the rod, F at C and 2mg — Mark scheme (c)

The centre of mass of the combined rod + particle is \(\dfrac{3}{2}a\) from \(A\)	M1A1

3 forces in equilibrium must be concurrent \(\Rightarrow b = \dfrac{3}{2}a\)	M1
\(\Rightarrow \dfrac{a}{b} = \dfrac{2}{3}\)	A1

\(R\) acts along the rod, so resolve forces perpendicular to the rod. \(F = mg\cos\theta + mg\cos\theta\)	M1
\(2mg\cos\theta = \dfrac{3amg\cos\theta}{b}\)	A1 DM1
\(\Rightarrow \dfrac{a}{b} = \dfrac{2}{3}\)	A1

\(R\) acts along the rod. Take moments about \(C\) \(mg\cos\theta(2a - b) = mg\cos\theta(b - a)\)	M1 A1
\(2a - b = b - a,\ \ \ \Rightarrow \dfrac{a}{b} = \dfrac{2}{3}\)	DM1A1

Resultant parallel to the rod \(\Rightarrow R = 2mg\sin\theta\) And \(V^2 + H^2 = R^2\)	M1
\((2mg\sin\theta)^2 = \left(\dfrac{3amg\cos\theta\sin\theta}{b}\right)^2 + \left(2mg - \dfrac{3amg\cos^2\theta}{b}\right)^2\)	A1
Eliminate \(\theta\)	DM1
\(\Rightarrow \dfrac{a}{b} = \dfrac{2}{3}\)	A1

M2 June 2013 Q5

M2 June 2013 Q5

Notes

Notes

Notes

5c alt 2

alt c 3

alt c 4

c alt 5

Scheme	Marks
\(\dfrac{2mg - \dfrac{3amg\cos^2\theta}{b}}{\dfrac{3amg\cos\theta\sin\theta}{b}} = \tan\theta\)	M1 A1
\(\dfrac{2b - 3a\cos^2\theta}{3a\cos\theta\sin\theta} = \dfrac{\sin\theta}{\cos\theta}\)	DM1
\(\Rightarrow 2b - 3a\cos^2\theta = 3a\sin^2\theta \Rightarrow 2b = 3a,\ \ \dfrac{a}{b} = \dfrac{2}{3}\)	A1
	(4)
	(13 marks)