June 2025 Paper 3 Mechanics Q6

EdexcelCurrent spec11 marksMoments Resolving Forces

Figure 4: rod AB with end A on horizontal ground and end B against a vertical wall, at angle alpha to the wall, with a particle C of mass 2M on the rod — Figure 4

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

A particle of mass \(2M\) is attached to the rod at the point \(C\), where \(AC = 0.5a\)

The rod rests with end \(A\) on rough horizontal ground and end \(B\) against a vertical wall.

The rod lies in a vertical plane which is perpendicular to the wall.

The rod is in equilibrium at an angle \(\alpha\) to the wall, as shown in Figure 4.

In an initial model

the vertical wall is modelled as being smooth
the magnitude of the normal reaction of the ground on the rod at \(A\) is \(R\)
the magnitude of the force exerted on the rod by the wall at \(B\) is \(S\)

Using the model,

(a) find \(R\) in terms of \(M\) and \(g\) (1)

(b) show that \(S = Mg\tan\alpha\) (3)

In a refined model

the vertical wall is modelled as being rough
the magnitude of the normal reaction of the ground on the rod at \(A\) is \(R_1\)

Figure 5: the rod AB, now with a particle of mass 3M at B as well as the particle C of mass 2M, with end A on the ground and end B against the wall, at angle beta to the wall — Figure 5

A second particle of mass \(3M\) is now attached to the rod at \(B\).

The rod again rests with end \(A\) on rough horizontal ground and end \(B\) against the vertical wall.

The rod lies in a vertical plane which is perpendicular to the wall.

The rod is now in limiting equilibrium at an angle \(\beta\) to the wall, as shown in Figure 5.

The vertical wall is again modelled as being smooth.

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

Given that \(\tan\beta = \dfrac{1}{2}\)

(d) use the model to show that \(\mu = \dfrac{1}{3}\) (6)

force diagram: S horizontal at B, R up and F horizontal at A, 2Mg down at C, Mg down at G, angle alpha between rod and wall — Mark scheme (b)

Scheme	Marks	AO

N.B. Allow consistent use of a different letter for the angle in the working but must be \(\alpha\) in the final answer. BOD if their \(a\)'s and \(\alpha\)'s look the same.
EITHER:
Obtain an equation in \(S\), \(M\), \(a\) and \(\alpha\) only, by taking moments about \(A\)	M1	3.1a
M(\(A\)): \(S \times 2a\cos\alpha = Mga\sin\alpha + 2Mg \times 0.5a\sin\alpha\)	A1	1.1b
OR
Obtain an equation in \(S\), \(M\), \(a\) and \(\alpha\) only, by taking moments about \(B\), \(C\) or \(G\) and resolving once and eliminating \(F\) and \(R\) (N.B. \(R = 3Mg\) from (a))	M1	3.1a
Correct equation in \(S\), \(M\), \(a\) and \(\alpha\) only	A1	1.1b
Possible Moments equations: M(\(B\)): \(R \times 2a\sin\alpha = Mga\sin\alpha + 2Mg \times 1.5a\sin\alpha + F \times 2a\cos\alpha\) or M(\(C\)): \(F \times 0.5a\cos\alpha + S \times 1.5a\cos\alpha = R \times 0.5a\sin\alpha + Mg \times 0.5a\sin\alpha\) or M(\(G\)): \(Sa\cos\alpha + Fa\cos\alpha + 2Mg \times 0.5a\sin\alpha = Ra\sin\alpha\) Possible resolution equations: \((\rightarrow)\): \(F = S\) or \((\nearrow)\): \(R\cos\alpha + F\sin\alpha = S\sin\alpha + (2Mg + Mg)\cos\alpha\) or \((\nwarrow)\): \(R\sin\alpha + S\cos\alpha = (2Mg + Mg)\sin\alpha + F\cos\alpha\)
\(S = Mg\tan\alpha\) *	A1*	2.2a
	(3)

Mark scheme (c)

Scheme	Marks	AO
The friction force at \(B\) will provide an extra upward (not just vertical) force so \(R_1\) will be less i.e. \(R\) will be greater. \(R\) (is greater than \(R_1\)) as the friction act against the weight of the rod. \(R\) (is greater) as the friction acts in the same direction as \(R_1\) \(R\) (is greater) as the friction holds the rod up as well. \(R = 3Mg = R_1 + F\) so \(R\) will be greater N.B. Allow \(R_1\) will be smaller with a correct reason. B0 \(R\) is greater as there is no friction from the wall.	B1	3.5a
	(1)

Notes

B1: Correct explanation
B0 if any incorrect extras

force diagram: P horizontal and 3Mg down at B, N up and F1 horizontal at A, 2Mg down at C, Mg down at G, angle beta between rod and wall — Mark scheme (d)

Scheme	Marks	AO
N.B. Allow consistent use of a different letter for the angle and could score FULL marks.

EITHER: Moments about \(B\) or \(D\), the intersection of the normals	M1	3.4
M(\(B\)): \(N \times 2a\sin\beta = Mga\sin\beta + 2Mg \times 1.5a\sin\beta + F_1 \times 2a\cos\beta\)	A1	1.1b
or M(\(D\)): \(F_1 \times 2a\cos\beta = Mga\sin\beta + 2Mg \times 0.5a\sin\beta + 3Mg \times 2a\sin\beta\)	A1	1.1b
and resolve vertically: \((\uparrow)\ N = Mg + 2Mg + 3Mg\)	B1	3.3
OR: Moments about \(A\) or \(C\) or \(G\)	M1	3.4
A correct moments equation.	A1	1.1b
A correct resolution (N.B. Treat as a B mark so M0A0A1 is a possible score)	A1	1.1b
A correct second resolution or a correct second moments equation	B1	3.3
Possible Moments equations: M(\(A\)): \(2Mg \times 0.5a\sin\beta + Mga\sin\beta + 3Mg \times 2a\sin\beta = P \times 2a\cos\beta\) N.B. (You may see the first 2 terms collected as a single term) or M(\(C\)): \(0.5aF_1\cos\beta + 1.5Pa\cos\beta = 0.5Na\sin\beta + 0.5Mga\sin\beta + 3Mg \times 1.5a\sin\beta\) or M(\(G\)): \(F_1a\cos\beta + Pa\cos\beta + 2Mg \times 0.5a\sin\beta = Na\sin\beta + 3Mga\sin\beta\) Possible resolution equations: \((\uparrow)\): \(N = Mg + 2Mg + 3Mg\) \((\rightarrow)\): \(F_1 = P\) (or \(P = \mu N\)) \((\nearrow)\): \(F_1\sin\beta + N\cos\beta = P\sin\beta + (3Mg + Mg + 2Mg)\cos\beta\) \((\nwarrow)\): \(N\sin\beta + P\cos\beta = (3Mg + Mg + 2Mg)\sin\beta + F_1\cos\beta\)
(N.B. \(F_1 = P = 2Mg\))
Use of \(F_1 = \mu N\), their equations (which must be correct) and \(\tan\beta = \dfrac{1}{2}\), to obtain an equation in \(\mu\) only. N.B. This mark is only available if the previous 4 marks have all been earned.	dM1	3.1a
\(\mu = \dfrac{1}{3}\) *	A1*	2.2a
	(6)
	(11 marks)