June 2023 Paper 3 Mechanics Q6

EdexcelCurrent spec13 marksMoments Resolving Forces

Figure 3: rod AB with end A on horizontal ground and end B against a vertical wall, making angle theta with the ground — Figure 3

A rod \(AB\) has mass \(M\) and length \(2a\).

The rod has its end \(A\) on rough horizontal ground and its end \(B\) against a smooth vertical wall.

The rod makes an angle \(\theta\) with the ground, as shown in Figure 3.

The rod is at rest in limiting equilibrium.

(a) State the direction (left or right on Figure 3 above) of the frictional force acting on the rod at \(A\). Give a reason for your answer. (1)

The magnitude of the normal reaction of the wall on the rod at \(B\) is \(S\).

In an initial model, the rod is modelled as being uniform.

Use this initial model to answer parts (b), (c) and (d).

(b) By taking moments about \(A\), show that\[S = \frac{1}{2}Mg\cot\theta\] (3)

The coefficient of friction between the rod and the ground is \(\mu\)

Given that \(\tan\theta = \dfrac{3}{4}\)

(d) find, in terms of \(M\) and \(g\), the magnitude of the resultant force acting on the rod at \(A\). (3)

In a new model, the rod is modelled as being non-uniform, with its centre of mass closer to \(B\) than it is to \(A\).

A new value for \(S\) is calculated using this new model, with \(\tan\theta = \dfrac{3}{4}\)

(e) State whether this new value for \(S\) is larger, smaller or equal to the value that \(S\) would take using the initial model. Give a reason for your answer. (1)

Mark scheme (a)

Scheme	Marks	AO
The normal reaction at \(B\) is acting to the left so it must act to the right, right as it needs to balance (oppose, counter) the force at \(B\), right as it prevents the rod from sliding (slipping, falling), right as the weight (mass) of the rod will mean the rod tends to slip left, mass or weight will be pushing the rod to the left so friction will oppose that. N.B. You may see an arrow on the diagram at \(A\), instead of ‘right’. B0 if they say the rod is moving oe Accept towards the wall instead of to the right.	B1	2.4
	(1)

Notes

B1: Any equivalent appropriate statement.

Mark scheme (b)

Scheme	Marks	AO
Take moments about \(A\)	M1	3.4
\(S \times 2a\sin\theta = Mga\cos\theta\)	A1	1.1b
\(S = \dfrac{1}{2}Mg\cot\theta\) *	A1*	2.2a
	(3)

Notes

M1: Correct no. of terms, dimensionally correct, condone sin/cos confusion and sign errors.
N.B. If \(a\)’s never appear, M0

A1: Correct equation

A1*: Correct given answer correctly obtained, with no wrong working seen.
Allow \(\dfrac{1}{2}Mg\cot\theta = S\) or \(S = \dfrac{Mg\cot\theta}{2}\) or \(\dfrac{Mg\cot\theta}{2} = S\) or \(S = \dfrac{Mg}{2}\cot\theta\) or similar
but NOT \(S = \dfrac{1}{2}\cot\theta\,\text{Mg}\) or similar
N.B. Allow \(m\) instead of \(M\)
Must be \(\theta\) in final answer but allow a different angle in the working.

Mark scheme (c)

Scheme	Marks	AO
Resolve vertically, \(R = Mg\)	B1	3.3
Resolve horizontally, \(F = S\)	B1	3.3
Other possible equations: Resolve along the rod, \(F\cos\theta + R\sin\theta = S\cos\theta + Mg\sin\theta\) Resolve perp to the rod, \(R\cos\theta + S\sin\theta = F\sin\theta + Mg\cos\theta\) M(\(B\)), \(R \times 2a\cos\theta = F \times 2a\sin\theta + Mga\cos\theta\) M(\(G\)), \(Ra\cos\theta = Fa\sin\theta + Sa\sin\theta\) N.B. When entering these two B marks on ePEN, First B1 is for a vertical resolution, second B1 is for a horizontal resolution, and if either is replaced by a different equation, enter appropriately. If both are replaced by other equations, enter in the order in which they appear in their working.
\(F = \mu R\)	B1	1.2
\(\dfrac{1}{2}Mg \times \dfrac{4}{3} = \mu Mg\)	dM1	2.1
\(\mu = \dfrac{2}{3}\) oe Accept 0.67 or better	A1	2.2a
S.C. For \(F \leqslant \mu R\), B0 \(\dfrac{1}{2}Mg \times \dfrac{4}{3} \leqslant \mu Mg\) M1 \(\dfrac{2}{3} \leqslant \mu\) A0 N.B. If \(\mu = \dfrac{2}{3}\) follows this, they could score all the marks.
	(5)

Notes

B1: cao

B1: Seen anywhere, e.g. on the diagram

dM1: Using \(F = \mu R\), their two equations and substitute for trig (not necessarily correctly) to produce an equation in \(\mu\) only.
This mark is dependent on the 3 previous B marks.

A1: Accept 0.67 or better

triangle of forces with sides X, Mg (vertical, downwards) and S (horizontal, to the left), with angle alpha between X and S — Mark scheme (d)

Scheme	Marks	AO
\(\sqrt{F^2 + R^2}\)	M1	3.1a
\(\sqrt{\left(\dfrac{2}{3}Mg\right)^2 + (Mg)^2}\)	M1	1.1b
\(\dfrac{1}{3}Mg\sqrt{13}\) or \(1.2Mg\) or better	A1	2.2a
	(3)

Notes

Notes

Notes

Notes

Alternative

Notes