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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)

June 2024 Paper 3 Mechanics Q6

EdexcelCurrent spec9 marksMoments Resolving Forces

Figure 5: rod AB with end A on horizontal ground at angle theta, resting on a peg P at point C where AC = 1.5a — Figure 5

Figure 5 shows a uniform rod \(AB\) of mass \(M\) and length \(2a\).

the rod has its end \(A\) on rough horizontal ground
the rod rests in equilibrium against a small smooth fixed horizontal peg \(P\)
the point \(C\) on the rod, where \(AC = 1.5a\), is the point of contact between the rod and the peg
the rod is at an angle \(\theta\) to the ground, where \(\tan\theta = \dfrac{4}{3}\)

The rod lies in a vertical plane perpendicular to the peg.

The magnitude of the normal reaction of the peg on the rod at \(C\) is \(S\).

(a) Show that \(S = \dfrac{2}{5}Mg\) (3)

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

Given that the rod is in limiting equilibrium,

(b) find the value of \(\mu\). (6)

force diagram: rod AB at angle theta to the ground, R up and F horizontal at A, Mg down at G with AG = a, S perpendicular to the rod at C with AC = 1.5a — Mark scheme (a)

Scheme	Marks	AO

Take moments about \(A\)	M1	3.1a
\(S \times 1.5a = Mga\cos\theta = \left(Mga \times \dfrac{3}{5}\right)\)	A1	1.1b
\(S = \dfrac{2}{5}Mg\) *	A1*	2.2a
	(3)

Scheme	Marks	AO

N.B. Marks for the equations should be awarded in the order in which they appear on the script.
Resolve horizontally:	M1	3.4
\(F = S\sin\theta\)	A1	1.1b
Resolve vertically:	M1	3.3
\(R = Mg - S\cos\theta\)	A1	1.1b
Other possible equations: (any of which is worth max M1A1) (parallel to the rod): \(F\cos\theta + R\sin\theta = Mg\sin\theta\) (perp to the rod): \(F\sin\theta + Mg\cos\theta = S + R\cos\theta\) M(\(B\)): \((S \times 0.5a) + (R \times 2a\cos\theta) = (Mg \times a\cos\theta) + (F \times 2a\sin\theta)\) M(\(C\)): \((R \times 1.5a\cos\theta) = (Mg \times 0.5a\cos\theta) + (F \times 1.5a\sin\theta)\) M(\(G\)): \((R \times a\cos\theta) = (S \times 0.5a) + (F \times a\sin\theta)\) N.B. If they have more than two equations, mark only those that they use to try to find \(\mu\)
\(F = \mu R\) and two of their equations used to solve for \(\mu\)	DM1	3.1a
\(\mu = \dfrac{8}{19} = 0.42105\ldots\)	A1	2.2a
	(6)
	(9 marks)

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)

Scheme	Marks	AO
New value of \(S\) would be larger as the moment of the weight about A would be larger	B1	3.5a
	(1)
	(13 marks)

June 2022 Paper 3 Mechanics Q4

EdexcelCurrent spec11 marksMoments Resolving Forces

Figure 2: rod AB with end A on horizontal ground at angle theta, particle at C on the rod, and a string attached at B perpendicular to the rod — Figure 2

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 \(AC = 1.5a\)

The rod rests with its end \(A\) on rough horizontal ground.

The rod is held in equilibrium at an angle \(\theta\) to the ground by a light string that is attached to the end \(B\) of the rod.

The string is perpendicular to the rod, as shown in Figure 2.

(a) Explain why the frictional force acting on the rod at \(A\) acts horizontally to the right on the diagram. (1)

The tension in the string is \(T\)

(b) Show that \(T = 2Mg\cos\theta\) (3)

Given that \(\cos\theta = \dfrac{3}{5}\)

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

Given that the rod is in limiting equilibrium,

(d) show that \(\mu = \dfrac{8}{19}\) (4)

Mark scheme (a)

Scheme	Marks	AO
The horizontal component of \(T\) acts to the left and since the only other horizontal force is friction, it must act to the right oe	B1	2.4
	(1)

Notes

B1: Any equivalent explanation

Mark scheme (b)

Scheme	Marks	AO
Take moments about \(A\) or any other complete method to obtain an equation in T, M and \(\theta\) only. (see possible equations below that they may use)	M1	3.1b
\(T.2a = Mga\cos\theta + 2Mg \times 1.5a\cos\theta\) (A0 if \(a\)’s missing)	A1	1.1b
Other possible equations but \(F\) and \(R\) would need to be eliminated. \((\nwarrow)\), \(R\cos\theta + T = F\sin\theta + Mg\cos\theta + 2Mg\cos\theta\) \((\nearrow)\), \(R\sin\theta + F\cos\theta = Mg\sin\theta + 2Mg\sin\theta\) \((\rightarrow)\), \(F = T\sin\theta\) M(\(B\)), \(R.2a\cos\theta = Mga\cos\theta + 2Mg \times 0.5a\cos\theta + F.2a\sin\theta\) M(\(G\)), \(Fa\sin\theta + Ta = Ra\cos\theta + 2Mg \times 0.5a\cos\theta\) M(\(C\)), \(R \times 1.5a\cos\theta = T \times 0.5a + Mg \times 0.5a\cos\theta + F \times 1.5a\sin\theta\)
\(T = 2Mg\cos\theta\) *	A1*	1.1b
	(3)

Notes

M1: Correct no. of terms, dimensionally correct, condone sin/cos confusion and sign errors

A1: Correct equation, trig does not need to be substituted
(Allow: \(T.2a = Mga\cos\theta + 3Mga\cos\theta\))

A1*: Given answer correctly obtained with no wrong working seen.
Allow \(2Mg\cos\theta = T\)
But not \(T = 2\cos\theta Mg\)

Mark scheme (c)

Scheme	Marks	AO
e.g. Resolve vertically	M1	3.4
\((\uparrow)\), \(R + T\cos\theta = Mg + 2Mg\)	A1	1.1b
\(R = \dfrac{57Mg}{25}\) *	A1*	1.1b
Other possible equations but \(F\) would need to be eliminated. \((\nwarrow)\), \(R\cos\theta + T = F\sin\theta + Mg\cos\theta + 2Mg\cos\theta\) \((\nearrow)\), \(R\sin\theta + F\cos\theta = Mg\sin\theta + 2Mg\sin\theta\) \((\rightarrow)\), \(F = T\sin\theta\) M(\(B\)), \(R.2a\cos\theta = Mga\cos\theta + 2Mg \times 0.5a\cos\theta + F.2a\sin\theta\) M(\(G\)), \(Fa\sin\theta + Ta = Ra\cos\theta + 2Mg \times 0.5a\cos\theta\) M(\(C\)), \(R \times 1.5a\cos\theta = T \times 0.5a + Mg \times 0.5a\cos\theta + F \times 1.5a\sin\theta\)
	(3)

Notes

M1: For an equation in \(R\), \(M\), \(T\) and \(\theta\) only
Correct no. of terms, dimensionally correct, condone sin/cos confusion and sign errors, each term that needs to be resolved must be resolved

A1: Correct equation, \(T\) and trig do not need to be substituted

A1*: Given answer correctly obtained with no wrong working seen

Mark scheme (d)

Scheme	Marks	AO
Find an equation containing \(F\) e.g. Resolve horizontally	M1	3.4
\((\rightarrow)\), \(F = T\sin\theta\)	A1	1.1b
Other possible equations \((\nwarrow)\), \(R\cos\theta + T = F\sin\theta + Mg\cos\theta + 2Mg\cos\theta\) \((\nearrow)\), \(R\sin\theta + F\cos\theta = Mg\sin\theta + 2Mg\sin\theta\) \((\rightarrow)\), \(F = T\sin\theta\) M(\(B\)), \(R.2a\cos\theta = Mga\cos\theta + 2Mg \times 0.5a\cos\theta + F.2a\sin\theta\) M(\(G\)), \(Fa\sin\theta + Ta = Ra\cos\theta + 2Mg \times 0.5a\cos\theta\) M(\(C\)), \(R \times 1.5a\cos\theta = T \times 0.5a + Mg \times 0.5a\cos\theta + F \times 1.5a\sin\theta\)
\(F = \mu R\) used i.e. both \(F\) and \(R\) are substituted.	M1	3.1b
\(\mu = \dfrac{8}{19}\) *	A1*	2.2a
	(4)
	(11 marks)

Notes

M1: For any equation with \(F\) in it
Correct no. of terms, dimensionally correct, condone sin/cos confusion and sign errors, each term that needs to be resolved must be resolved

A1: Correct equation, trig does not need to be substituted

M1: Must be used i.e M0 if merely quoting it.

A1*: Given answer correctly obtained with no wrong working seen

October 2021 Paper 3 Mechanics Q3

EdexcelCurrent spec10 marksMoments Resolving Forces

Figure 2: beam AB of length 2a with A on horizontal ground and B against a vertical wall, inclined at angle theta to the horizontal — Figure 2

A beam \(AB\) has mass \(m\) and length \(2a\).

The beam rests in equilibrium with \(A\) on rough horizontal ground and with \(B\) against a smooth vertical wall.

The beam is inclined to the horizontal at an angle \(\theta\), as shown in Figure 2.

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

The beam is modelled as a uniform rod resting in a vertical plane that is perpendicular to the wall.

Using the model,

(a) show that \(\mu \geqslant \dfrac{1}{2}\cot\theta\) (5)

A horizontal force of magnitude \(kmg\), where \(k\) is a constant, is now applied to the beam at \(A\).

This force acts in a direction that is perpendicular to the wall and towards the wall.

Given that \(\tan\theta = \dfrac{5}{4}\), \(\mu = \dfrac{1}{2}\) and the beam is now in limiting equilibrium,

(b) use the model to find the value of \(k\). (5)

force diagram: beam from A on the ground to B on the wall, R up and F towards the wall at A, mg down at G, S horizontal away from the wall at B; C is level with B above A, D is the foot of the wall — Mark scheme (a)

Scheme	Marks	AO
Part (a) is a ‘Show that..’ so equations need to be given in full to earn A marks

Moments equation: (M1A0 for a moments inequality)	M1	3.3
M\((A)\), \(mga\cos\theta = 2Sa\sin\theta\) M\((B)\), \(mga\cos\theta + 2Fa\sin\theta = 2Ra\cos\theta\) M\((C)\), \(F \times 2a\sin\theta = mga\cos\theta\) M\((D)\), \(2Ra\cos\theta = mga\cos\theta + 2Sa\sin\theta\) M\((G)\), \(Ra\cos\theta = Fa\sin\theta + Sa\sin\theta\).	A1	1.1b
\((\updownarrow)\ R = mg\) OR \((\leftrightarrow)\ F = S\)	B1	3.4
Use their equations (they must have enough) and \(F \leqslant \mu R\) to give an inequality in \(\mu\) and \(\theta\) only (allow DM1 for use of \(F = \mu R\) to give an equation in \(\mu\) and \(\theta\) only)	DM1	2.1
\(\mu \geqslant \dfrac{1}{2}\cot\theta\) *	A1*	2.2a
	(5)

force diagram: beam from A on the ground to B on the wall, R up at A, kmg towards the wall and one half mg away from the wall at A, mg down at G, N horizontal away from the wall at B; C is level with B above A, D is the foot of the wall — Mark scheme (b)

Scheme	Marks	AO

Moments equation:	M1	3.4
M\((A)\), \(mga\cos\theta = 2Na\sin\theta\) M\((B)\), \(mga\cos\theta + 2kmga\sin\theta = 2Ra\cos\theta + \dfrac{1}{2}mg2a\sin\theta\) M\((D)\), \(2Ra\cos\theta = mga\cos\theta + N2a\sin\theta\) M\((G)\), \(kmga\sin\theta + Na\sin\theta = \dfrac{1}{2}mga\sin\theta + Ra\cos\theta\)	A1	1.1b
S.C. M\((C)\), \(mga\cos\theta + \dfrac{1}{2}mg2a\sin\theta = kmg2a\sin\theta\) M1A1B1 \(1 + \dfrac{5}{4} = \dfrac{5k}{2}\) M1 \(k = 0.9\) A1
\(N = kmg - F\) OR \(R = mg\)	B1	3.3
Use their equations (they must have enough) to solve for \(k\) (numerical)	DM1	3.1b
\(k = 0.9\) oe	A1	1.1b
	(5)
	(10 marks)

October 2020 Paper 3 Mechanics Q4

EdexcelCurrent spec10 marksMoments Resolving Forces

Figure 1: ladder AB of length 6a with A on horizontal ground, resting on a rail at C which is at height 4a above the ground, ladder inclined at angle alpha to the horizontal — Figure 1

A ladder \(AB\) has mass \(M\) and length \(6a\).

The end \(A\) of the ladder is on rough horizontal ground.

The ladder rests against a fixed smooth horizontal rail at the point \(C\).

The point \(C\) is at a vertical height \(4a\) above the ground.

The vertical plane containing \(AB\) is perpendicular to the rail.

The ladder is inclined to the horizontal at an angle \(\alpha\), where \(\sin\alpha = \dfrac{4}{5}\), as shown in Figure 1.

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

The ladder rests in limiting equilibrium.

The ladder is modelled as a uniform rod.

Using the model,

(a) show that the magnitude of the force exerted on the ladder by the rail at \(C\) is \(\dfrac{9Mg}{25}\) (3)

(b) Hence, or otherwise, find the value of \(\mu\). (7)

Mark scheme (a)

Scheme	Marks	AO
Take moments about \(A\)	M1	3.3
\(N \times \dfrac{4a}{\sin\alpha} = Mg \times 3a\cos\alpha\)	A1	1.1b
\(\dfrac{9Mg}{25}\) *	A1*	1.1b
	(3)

Notes

M1: Correct no. of terms, dim correct, condone sin/cos confusion and sign errors for an equation in \(N\) and \(Mg\) only.
For perp distance allow any of: \(\dfrac{4a}{\sin\alpha}, \dfrac{4a}{\cos\alpha}, 5a\) but
use of any of: \(6a, 5a\sin\alpha, 4a\cos\alpha, \ldots\) or anything involving \(\tan\alpha\) is M0
Also M0 if no \(a\)’s in their first equation.

A1: Correct equation, trig does not need to be substituted

A1*: Given answer correctly obtained.

Mark scheme (b)

Scheme	Marks	AO
Resolve horizontally	M1	3.4
\((\rightarrow)\ F = \dfrac{9Mg}{25}\sin\alpha\)	A1	1.1b
Resolve vertically	M1	3.4
\((\uparrow)\ R + \dfrac{9Mg}{25}\cos\alpha = Mg\)	A1	1.1b
Other possible equations: \((\nwarrow),\ R\cos\alpha + \dfrac{9Mg}{25} = Mg\cos\alpha + F\sin\alpha\) \((\nearrow),\ Mg\sin\alpha = F\cos\alpha + R\sin\alpha\) M\((C)\), \(Mg.2a\cos\alpha + F.5a\sin\alpha = R.5a\cos\alpha\) M\((G)\), \(\dfrac{9Mg}{25}.2a + F.3a\sin\alpha = R.3a\cos\alpha\) M\((B)\), \(Mg.3a\cos\alpha + F.6a\sin\alpha = R.6a\cos\alpha + \dfrac{9Mg}{25}a\) \(\left(F = \dfrac{36Mg}{125},\ R = \dfrac{98Mg}{125}\right)\)
\(F = \mu R\) used	M1	3.4
Eliminate \(R\) and \(F\) and solve for \(\mu\)	M1	3.1b
Alternative equations if they have at \(A\): \(X\) horizontally and \(Y\) perpendicular to the rod. \((\nwarrow),\ Y + \dfrac{9Mg}{25} = Mg\cos\alpha + X\sin\alpha\) \((\nearrow),\ Mg\sin\alpha = X\cos\alpha\) \((\uparrow),\ \dfrac{9Mg}{25}\cos\alpha + Y\cos\alpha = Mg\) \((\rightarrow),\ Y\sin\alpha + \dfrac{9Mg}{25}\sin\alpha = X\) M\((C)\), \(Mg.2a\cos\alpha + X.5a\sin\alpha = Y.5a\) M\((G)\), \(\dfrac{9Mg}{25}.2a + X.3a\sin\alpha = Y.3a\) M1A1 M1A1 M\((B)\), \(Mg.3a\cos\alpha + X.6a\sin\alpha = Y.6a + \dfrac{9Mg}{25}a\) \(\left(X = \dfrac{4Mg}{3},\ Y = \dfrac{98Mg}{75}\right)\) Then \(F = \mu R\) becomes: \(X - Y\sin\alpha = \mu Y\cos\alpha\) M1 Eliminate \(X\) and \(Y\) and solve for \(\mu\) M1
\(\mu = \dfrac{18}{49}\) (0.3673…..accept 0.37 or better)	A1	2.2a
	(7)
	(10 marks)

Notes

M1: Correct no. of terms, dim correct, condone sin/cos confusion and sign errors

A1: Correct equation, trig does not need to be substituted but \(N\) does.

M1: Correct no. of terms, dim correct, condone sin/cos confusion and sign errors

A1: Correct equation, trig does not need to be substituted but \(N\) does.

N.B. The above 4 marks are for any two equations, either resolutions or moments or one of each. Mark best two equations.
Equations may appear in part (a) but must be used in (b) to earn marks.

M1: Must be used, e.g. seen on the diagram. i.e. M0 if merely quoting it.
(M0 if \(F = \mu \times \dfrac{9Mg}{25}\) used)

M1: Must have 3 equations (and all 3 previous M marks)

A1: Accept 0.37 or better

June 2019 Paper 3 Mechanics Q4

EdexcelCurrent spec11 marksMoments Resolving Forces

Figure 2: ramp AB with end A on horizontal ground at angle theta, resting on a cylindrical drum partly under the ground at C, where AC = 5 m and CB = 3 m — Figure 2

A ramp, \(AB\), of length 8 m and mass 20 kg, rests in equilibrium with the end \(A\) on rough horizontal ground.

The ramp rests on a smooth solid cylindrical drum which is partly under the ground. The drum is fixed with its axis at the same horizontal level as \(A\).

The point of contact between the ramp and the drum is \(C\), where \(AC = 5\) m, as shown in Figure 2.

The ramp is resting in a vertical plane which is perpendicular to the axis of the drum, at an angle \(\theta\) to the horizontal, where \(\tan\theta = \dfrac{7}{24}\)

The ramp is modelled as a uniform rod.

(a) Explain why the reaction from the drum on the ramp at point \(C\) acts in a direction which is perpendicular to the ramp. (1)

(b) Find the magnitude of the resultant force acting on the ramp at \(A\). (9)

The ramp is still in equilibrium in the position shown in Figure 2 but the ramp is not now modelled as being uniform.

Given that the centre of mass of the ramp is assumed to be closer to \(A\) than to \(B\),

Scheme	Marks	AO
Drum smooth, or no friction, (therefore reaction is perpendicular to the ramp)	B1	2.4
	(1)

force diagram: ramp AB at angle theta, R vertically up and F horizontal (towards the drum) at A, 20g vertically down, N perpendicular to the ramp at C on the drum — Mark scheme (b)

Scheme	Marks	AO
N.B. In (b), for a moments equation, if there is an extra \(\sin\theta\) or \(\cos\theta\) on a length, give M0 for the equation e.g. M(\(A\)): \(20g \times 4\cos\theta = 5N\sin\theta\) would be given M0A0

Possible equns \((\nearrow)\): \(F\cos\theta + R\sin\theta = 20g\sin\theta\) \((\nwarrow)\): \(N + R\cos\theta = 20g\cos\theta + F\sin\theta\) \((\uparrow)\): \(R + N\cos\theta = 20g\) \((\rightarrow)\): \(F = N\sin\theta\) M(\(A\)): \(20g \times 4\cos\theta = 5N\) M(\(B\)): \(3N + R \times 8\cos\theta = F \times 8\sin\theta + 20g \times 4\cos\theta\) M(\(C\)): \(R \times 5\cos\theta = F \times 5\sin\theta + 20g \times \cos\theta\) M(\(G\)): \(R \times 4\cos\theta = F \times 4\sin\theta + N\)	M1 A1 M1 A1 M1 A1	3.3 1.1b 3.4 1.1b 3.4 1.1b
(The values of the 3 unknowns are: \(N = 150.528\); \(F = 42.14784\); \(R = 51.49312\))
Solve their 3 equations for \(F\) and \(R\) OR \(X\) and \(Y\) OR \(H\) and \(S\)	M1	1.1b
\(\|\text{Force}\| = \sqrt{R^2 + F^2}\) Main scheme OR \(= \sqrt{X^2 + Y^2}\) Alternative 1 OR \(= \sqrt{(H^2 + S^2 - 2HS\cos(90^\circ - \theta))}\) Alternative 2	M1	3.1b
Magnitude = 67 or 66.5 (N)	A1	2.2a
	(9)

Scheme	Marks	AO
Possible equations: \((\nearrow)\): \(X = 20g\sin\theta\) \((\nwarrow)\): \(Y + N = 20g\cos\theta\) \((\uparrow)\): \(X\sin\theta + Y\cos\theta + N\cos\theta = 20g\) \((\rightarrow)\): \(X\cos\theta = Y\sin\theta + N\sin\theta\) M(\(A\)): \(20g \times 4\cos\theta = 5N\) M(\(B\)): \(20g \times 4\cos\theta = 8Y + 3N\) M(\(C\)): \(20g \times \cos\theta = 5Y\) M(\(G\)): \(4Y = N \times 1\)	M1 A1 M1 A1 M1 A1	3.3 1.1b 3.4 1.1b 3.4 1.1b
(The values of the 3 unknowns are: \(N = 150.528\); \(X = 54.88\); \(Y = 37.632\))

Scheme	Marks	AO
\((\nearrow)\): \(H\cos\theta = 20g\sin\theta\) \((\nwarrow)\): \(S + N = H\sin\theta + 20g\cos\theta\) \((\uparrow)\): \(S\cos\theta + N\cos\theta = 20g\) \((\rightarrow)\): \(H = S\sin\theta + N\sin\theta\) M(\(A\)): \(20g \times 4\cos\theta = 5N\) M(\(B\)): \(20g \times 4\cos\theta + H \times 8\sin\theta = 8S + 3N\) M(\(C\)): \(20g \times \cos\theta + H \times 5\sin\theta = 5S\) M(\(G\)): \(4S = N \times 1 + H \times 4\sin\theta\)	M1 A1 M1 A1 M1 A1	3.3 1.1b 3.4 1.1b 3.4 1.1b
(The values of the 3 unknowns are: \(N = 150.528\); \(H = 57.1666…\); \(S = 53.638666…\))

Scheme	Marks	AO
Magnitude of the normal reaction (at \(C\)) will decrease.	B1	3.5a
	(1)
	(11 marks)

June 2025 Paper 2 Q18

AQACurrent spec6 marksMoments

18 In this question use \(g = 9.81\ \text{m s}^{-2}\)

A uniform platform \(AD\) has length 2.5 metres and weight 400 newtons.

The platform is attached to a chain at \(A\). The other end of the chain is fixed to the floor.

The platform rests on a support at a point \(B\) which is 0.6 metres from \(A\)

The support exerts an upwards reaction force on the platform.

A child of mass 30 kilograms stands at a point \(C\)

The point \(C\) is 0.2 metres from \(D\), as shown in the diagram.

A horizontal platform AD; a chain at A goes down to the floor, a support under B, and point C is 0.2 m from the end D

The system is in equilibrium with the platform resting horizontally.

(a) By taking moments about \(A\), find the reaction force at \(B\) [4 marks]

(b)

(i) The child moves from \(C\) to \(B\)

State what happens to the reaction force at \(B\) during this movement.

[1 mark]

(ii) Having reached \(B\), the child steps off the platform.

State what happens to the tension in the chain during this movement.

[1 mark]

Mark scheme (a)

Scheme	Marks	AO
Obtains one of the following moments about A. \(\pm 2.3 \times 30g\) OE \(\pm 1.25 \times 400\) OE PI AWRT 1960 or AWRT 1983	B1	3.3
Forms a three term moments equation about A with at least one correct term. Must be dimensionally correct. PI 0.6R = AWFW [1176, 1177] or 0.6R = 1190	M1	3.3
Forms a fully correct moments equation about A. PI AWRT 1960 or AWRT 1983	A1	1.1b
Obtains \(1960\ \text{N}\) Must include units	A1	3.2a
	(4)

Typical solution

clockwise: \(2.3 \times 30 \times 9.81 + 1.25 \times 400\)

anticlockwise: \(0.6\mathbf{R}\)

\[0.6\mathbf{R} = 2.3 \times 30 \times 9.81 + 1.25 \times 400\]\[\mathbf{R} = \frac{2.3 \times 30 \times 9.81 + 1.25 \times 400}{0.6}\]\[\mathbf{R} = 1960\ \text{N}\]

Mark scheme (b)

Scheme	Marks	AO
(i) States that the reaction force decreases with no other incorrect reasoning.	E1	3.5a
	(1)
(ii) States that the tension in the chain does not change.	E1	3.5a
	(1)
	(6 marks)

Typical solution

(b)(i)

As the child walks towards \(B\) the reaction force at \(B\) decreases.

(b)(ii)

The tension in the chain does not change.

June 2024 Paper 2 Q17

AQACurrent spec4 marksMoments

17 A uniform rod is resting on two fixed supports at points \(A\) and \(B\).

\(A\) lies at a distance \(x\) metres from one end of the rod.

\(B\) lies at a distance \((x + 0.1)\) metres from the other end of the rod.

The rod has length \(2L\) metres and mass \(m\) kilograms.

The rod lies horizontally in equilibrium as shown in the diagram below.

Horizontal rod of length 2L on two supports: A at distance x from the left end, B at distance (x + 0.1) from the right end

The reaction force of the support on the rod at \(B\) is twice the reaction force of the support on the rod at \(A\).

Show that

\[L - x = k\]

where \(k\) is a constant to be found. [4 marks]

Mark scheme

Scheme	Marks	AO
Forms a dimensionally correct moment in \(L\) and \(x\) May see : \((L - x)R\) \((L - (x + 0.1))2R\) \((L - x)mg\) \((2L - (2x + 0.1))2R\) OE	B1	3.3
Forms dimensionally correct moments equation with at least one term correct May also see: \((L - x)mg = (2L - (2x + 0.1))2R\) Or \((L - (x + 0.1))mg = (2L - (2x + 0.1))R\) Condone missing brackets	M1	1.1a
Obtains a fully correct equation in \(L\) and \(x\) only	A1	1.1b
Completes reasoned argument to obtain \(L - x\) = 0.2 Must show expansion of all brackets before the final answer	R1	2.1
	(4 marks)

Typical solution

The reaction at A is \(R\)

\[(L - x)R = (L - (x + 0.1))2R\]\[L - x = 2(L - (x + 0.1))\]\[L - x = 2L - 2x - 0.2\]\[L - x = 0.2\]

June 2023 Paper 2 Q17

AQACurrent spec6 marksMoments

17 A uniform plank \(PQ\), of length 7 metres, lies horizontally at rest, in equilibrium, on two fixed supports at points \(X\) and \(Y\)

The distance \(PX\) is 1.4 metres and the distance \(QY\) is 2 metres as shown in the diagram below.

Horizontal plank PQ of length 7 m resting on supports at X and Y, with PX = 1.4 m and QY = 2 m

(a) The reaction force on the plank at \(X\) is \(4g\) newtons.

(i) Show that the mass of the plank is 9.6 kilograms. [2 marks]

(ii) Find the reaction force, in terms of \(g\), on the plank at \(Y\) [2 marks]

(b) The support at \(Y\) is moved so that the distance \(QY = 1.4\) metres.

The plank remains horizontally at rest in equilibrium.

It is claimed that the reaction force at \(Y\) remains unchanged.

Explain, with a reason, whether this claim is correct. [2 marks]

Mark scheme (a)

Scheme	Marks	AO
(i) Forms a moments equation by taking moments about \(Y\) with one term correct. or Forms two correct moments equations by taking moments about \(P\) and \(Q\) Must clearly use force \(\times\) distance for every term. For example: Moments about \(P\) \(3.5mg = (1.4)(4g) + 5R\) Moments about \(Q\) \(3.5mg = (5.6)(4g) + 2R\) Moments about the centre \(1.5(Yg) = (2.1)(4g)\)	M1	3.1b
Completes reasoned argument with at least one intermediate step to obtain 9.6 kilograms AG Condone 9.6	R1	2.1
	(2)
(ii) Resolves vertically to find \(R\) or Forms a moments equation about any point other than about \(Y\) with the correct number of terms and at least one term correct. For example: Moments about \(X\) \((2.1)(9.6g) = 3.6R\) Moments about midpoint of plank \((2.1)(4g) = 1.5R\) PI by \(5.6g\)	M1	3.3
Obtains \(5.6g\) N Condone missing units.	A1	1.1b
	(2)

Typical solution

(i)

Taking moments about \(Y\)

\[1.5mg = (3.6)(4g)\]\[m = \frac{(3.6)(4g)}{1.5g}\]

\(m = 9.6\) kilograms

(ii)

Resolving vertical forces:

\[4g + R = 9.6g\]\[R = 5.6g\text{ N}\]

Mark scheme (b)

Scheme	Marks	AO
Obtains \(4.8g\) or States that the two supports are equidistant from the centre/the ends of the plank or Refers to symmetry	B1	3.1b
States one of the following: The reaction force at \(Y\) changes. The reaction force at \(Y\) decreases. \(4.8g \neq 5.6g\) \(4.8g \lt 5.6g\) and concludes that the claim is incorrect.	E1	2.4
	(2)
	(6 marks)

Typical solution

\(X\) and \(Y\) are the same distance from the midpoint, so their reaction forces are equal.

The reaction force at \(Y\) decreases.

Therefore, the claim is wrong.

June 2022 Paper 2 Q14

AQACurrent spec4 marksMoments

14 A £2 coin has a diameter of 28 mm and a mass of 12 grams.

A uniform rod \(AB\) of length 160 mm and a fixed load of mass \(m\) grams are used to check that a £2 coin has the correct mass.

The rod rests with its midpoint on a support.

A £2 coin is placed face down on the rod with part of its curved edge directly above \(A\).

The fixed load is hung by a light inextensible string from a point directly below the other end of the rod at \(B\), as shown in the diagram.

Horizontal rod AB resting at its midpoint on a triangular support; a flat coin lies on the rod at end A; a circular load of mass m hangs from a string below end B

(a) Given that the rod is horizontal and rests in equilibrium, find \(m\). [3 marks]

(b) State an assumption you have made about the £2 coin to answer part (a). [1 mark]

Mark scheme (a)

Scheme	Marks	AO
Forms a dimensionally correct non-zero moment For example \(0.012g \times 0.066\) \(0.012g \times 66\) \(12g \times 66\) \(160mg\)	B1	3.3
Obtains a correct single equation where the only unknown is the mass This might come from correct moments about a point other than the centre with forces correctly resolved in the vertical direction, but these equations must be combined to eliminate the force at the centre PI by 9.9	M1	1.1a
Obtains 9.9 Condone \(m\) = 0.0099 kg OE provided units are included if \(m\) is not in grams	A1	1.1b
	(3)

Typical solution

\[66 \times 12g = 80mg\]\[m = 9.9\]

Mark scheme (b)

Scheme	Marks	AO
States a valid assumption about the £2 coin with no contradictions or incorrect statements For example Models the coin as uniform, or weight acts through the centre or centre of mass is at the centre Do not accept ‘mass acts’ or ‘centre of mass acts’ at the centre or ‘centre of weight’ or ‘weight is at’	E1	3.5a
	(1)
	(4 marks)

Typical solution

The coin is uniform

June 2025 Paper 3 Q10

OCR ACurrent spec8 marksMoments Resolving Forces

Horizontal rod AB with end A against a vertical wall, right angle at A; G marked on AB; a string from B to a point on the wall above A makes angle theta degrees with the rod at B

The diagram shows a non-uniform rod \(AB\) of mass 4 kg and length 2 m. The end \(A\) of the rod rests against a rough vertical wall. The rod is held in a horizontal position, perpendicular to the wall, by a light inextensible string attached to the rod at \(B\). The other end of the string is attached to the wall at a point vertically above \(A\). The string is inclined at an angle of \(\theta^\circ\) to the horizontal, where \(\sin\theta = \frac{4}{5}\).

The rod rests in equilibrium in a vertical plane perpendicular to the wall. The rod’s weight acts at the point \(G\) on \(AB\).

You are given that the magnitude of the moment of the rod’s weight about \(A\) is 47.04 N m.

(a) Show that the distance \(AG\) is 1.2 m. [1]

(b) Find the tension in the string. [2]

Mark scheme (a)

Scheme	Marks	AO
\(AG \times 4g = 47.04\) or \((AG =)\ \dfrac{47.04}{4g}\) or \(4g \times 1.2 = 47.04\)	B1	3.1b
	[1]

Notes

B1: Taking moments about \(A\) for the rod’s weight – allow \(\dfrac{47.04}{4 \times 9.8}\) oe (so using 9.8 for \(g\)) but B0 for only \(\dfrac{47.04}{39.2}\) oe
B0 for \(AB \times 4g = 47.04\) or similar

AG – www but need not see 1.2 explicitly stated

Mark scheme (b)

Scheme	Marks	AO
\(2 \times \frac{4}{5}T = 47.04\) or \(2 \times \frac{4}{5}T = 1.2 \times 4g\)	M1	3.3
\(T = 29.4\) (N)	A1	1.1
	[2]

Notes

M1: Taking moments about \(A\) for the rod – M1 for \(2 \times \frac{4}{5}T = 47.04\) or \(2 \times T\sin 53 = 47.04\) or \(2 \times \frac{3}{5}T = 47.04\) or \(2 \times T\cos 53 = 47.04\) only (or \(1.2 \times 4g\) for 47.04)

Allow use of 53 or better (53.1301…) for \(\theta\) - note that e.g. \(2 \times T\sin 37 = 47.04\) or \(2 \times T\cos 37 = 47.04\) (therefore using the complementary angle) is M1

A1: cao exact e.g. \(\frac{147}{5}\) or awrt 29.4

Mark scheme (c)

Scheme	Marks	AO
\(X = \pm\text{‘}29.4\text{’} \times \frac{3}{5}\) or \(\pm\text{‘}29.4\text{’} \times \cos 53 \quad (= \pm 17.64)\)	B1FT	3.3
Resolve vertically or take moments about \(B\) or \(G\) for the forces acting on the rod to form an equation or an expression for \(Y\)	M1*	3.3
\(\mathrm{R}(\uparrow)\): \(\text{‘}29.4\text{’} \times \frac{4}{5} + Y = 4g\) or \(\text{‘}29.4\text{’} \times \frac{4}{5} = 4g + Y\) or \(\mathfrak{M}(B)\): \(0.8 \times 4g = 2Y\) or \(\mathfrak{M}(G)\): \(0.8 \times \text{‘}29.4\text{’} \times \frac{4}{5} = Y \times 1.2\)	A1FT	1.1
\(\sqrt{17.64^2 + 15.68^2}\)	M1dep*	3.4
Magnitude is 23.6 (N)	A1	1.1
	[5]

Notes

B1FT: Follow through their value of \(T\) only. Allow use of 53 or better (53.1301…) for \(\theta\)

\(X\) is the horizontal component of the contact force at \(A\)

M1*: Must be the correct number of dimensionally correct terms but allow sign errors and sin/cos confusion only – all values must be substituted. If taking moments about \(B\) or \(G\) allow sin/cos confusion only

\(Y\) is the vertical contact force at \(A\)

A1FT: Correct equation for \(Y\) follow through their value of \(T\) only – if correct then \(Y = \pm 15.68\)

Allow \(\sin 53\) or \(\cos 37\) or better for \(\frac{4}{5}\)

M1dep*: Correct use of Pythagoras with their values of \(X\) and \(Y\) which must come from equations with the correct number of dimensionally correct relevant terms

Relevant for \(X\) requires \(\text{‘}29.4\text{’} \times \frac{3}{5}\) or \(\text{‘}29.4\text{’} \times \frac{4}{5}\) oe only

A1: awrt 23.6 or exact e.g. \(\dfrac{49\sqrt{145}}{25}\) oe

23.60152…

June 2024 Paper 3 Q11

OCR ACurrent spec7 marksMoments Resolving Forces

Rectangle ABCD leaning with corner A on horizontal ground and corner D against a vertical wall; AD makes an angle of 50 degrees with the ground, AB = 2 m and BC = 6 m

A uniform rectangular lamina \(ABCD\) has a mass of 0.5 kg. The length of \(AB\) is 2 m, and the length of \(BC\) is 6 m. The lamina is in limiting equilibrium with corner \(A\) in contact with rough horizontal ground and corner \(D\) in contact with a smooth vertical wall. The lamina rests in a vertical plane that is perpendicular to the wall, with \(AD\) inclined at 50° to the horizontal (see diagram).

(a) By taking moments, show that the magnitude of the normal contact force between the lamina and the wall is 1.24 N, correct to 3 significant figures. [4]

(b) Determine the coefficient of friction between the lamina and the ground. [3]

Mark scheme (a)

Scheme	Marks	AO
\(R_D \times k_1\sin 50 = 0.5g \times k_2\) or \(R_D \times k_1\cos 50 = 0.5g \times k_2\) (see notes)	M1	3.3
\(6(R_D\sin 50) = \ldots\)	A1	1.1
\(\ldots 3(0.5g\cos 50) - 1(0.5g\sin 50)\)	A1	3.1b
\(R_D = \dfrac{1.5g\cos 50 - 0.5g\sin 50}{6\sin 50} = 1.24\) (N)	A1	2.2a
	[4]

Notes

M1: Attempt at moments about \(A\) – at least two terms (one term for the weight and one term with a component of \(R_D\)) – must be written as an equation to score any marks in this part (see Answer column for the only acceptable forms)
with \(k_1, k_2 \gt 0\) but not \(= \pm 1\)

A1: Correct moment for the contact force at \(D\)

A1: Correct moment for the weight of the lamina (See Appendix) (e.g. \(0.5g \times (3 - \tan 50)\cos 50\) or \(0.5g \times \cos 50 \times 1.808\ldots\) etc. e.g. \(0.5g \times \sqrt{10}\sin\left(40 - \arctan\left(\tfrac{1}{3}\right)\right)\) or \(0.5g \times \sqrt{10}\sin(21.565\ldots)\) etc.) – if value stated with no or unclear working, then must see at least 5.6953… to award this mark - if moment stated as 5.6993… then A0 (and A0 for the next mark)
For reference 5.695360…

A1: AG – as this answer is given then working must be checked carefully – all previous marks must have been awarded
1.23912743…
A fully correct equation followed by stating 1.24 can score full marks

Additional guidance for 11(a) (Appendix)

There are many ways that candidates are correctly approaching this part:
For example,
\(6(R_D\sin 50) = 3(0.5g\cos 50) - (0.5g\sin 50)\)
\(6(R_D\sin 50) = 0.5g \times (3 - \tan 50)\cos 50\)
\(6(R_D\sin 50) = 0.5g \times \sqrt{10}\sin\left(40 - \arctan\left(\tfrac{1}{3}\right)\right)\)
\(6(R_D\sin 50) = 0.5g \times \sqrt{10}\cos(140 - \arctan(3))\)
\(6(R_D\sin 50) = 0.5g \times \sqrt{10}\cos\left(50 + \arctan\left(\tfrac{1}{3}\right)\right)\)

Any of these followed by the correct answer of 1.24 would score full marks (we do not need to see any intermediate working). If these exact expressions are not seen then sufficient working with values that are correct to at least 3 significant figures should be awarded full marks (and those that have 2 significant figures with sufficient working should get partial credit), e.g.

• \(6(R_D\sin 50) = 9.448977\ldots - 3.7536177\ldots\) so seeing \(6(R_D\sin 50) = 9.45 - 3.75\) followed by 1.24 is M1 A1 A1 A1
• \(6(R_D\sin 50) = 9.4 - 3.8\) followed by 1.24 is M1 A1 A1 A0 (leads to 1.22 not 1.24)
• \(6(R_D\sin 50) = 5.7(0)\) is no marks (assume working backwards from the AG and no indication that the RHS contains a weight component)
• \(6(R_D\sin 50) = 0.5g \times \cos 50 \times 1.808246\ldots\) so seeing \(6(R_D\sin 50) = 0.5g \times \cos 50 \times 1.81\) followed by 1.24 is M1 A1 A1 A1
• \(6(R_D\sin 50) = 0.5g \times \cos 50 \times 1.80\) followed by 1.24 is M1 A1 A1 A0 (leads to 1.23 not 1.24)
• \(6(R_D\sin 50) = 0.5g \times 0.643 \times 1.81\) followed by 1.24 is M1 A1 A1 A1
• \(6(R_D\sin 50) = 0.5g \times 0.64 \times 1.8\) followed by 1.24 is M1 A1 A1 A0 (leads to 1.23 not 1.24)
• \(6(R_D\sin 50) = 0.5g \times 1.16\) followed by 1.24 is M1 A1 A0 A0 (assume working backwards from the given answer on the RHS)
• \(6(R_D\sin 50) = 0.5g \times \sqrt{10}\sin(40 - 18.434948\ldots)\) so seeing \(6(R_D\sin 50) = 0.5g \times 3.16 \times \sin(21.6)\) followed by 1.24 is M1 A1 A1 A1
• \(6(R_D\sin 50) = 0.5g \times 3.16 \times \sin(22)\) followed by 1.24 is M1 A1 A1 A0 (leads to 1.26 not 1.24)
• \(6(R_D\sin 50) = 0.5g \times 3.2 \times \sin(22)\) followed by 1.24 is M1 A1 A1 A0 (leads to 1.28 not 1.24)
• \(6(R_D\sin 50) = 0.5g \times \sqrt{10}\cos(68.4349\ldots)\) so seeing \(6(R_D\sin 50) = 0.5g \times \sqrt{10} \times \cos(68)\) followed by 1.24 is M1 A1 A1 A0 (leads to 1.26 not 1.24)

Mark scheme (b)

Scheme	Marks	AO
\(R(\uparrow): R_A = 0.5g\) \(R(\rightarrow): 1.24 = F_A\)	B1*	3.3
\(1.24 = 0.5g\mu\)	M1dep*	3.4
\(\mu = 0.253\)	A1	2.2a
	[3]

Notes

B1*: Resolving horizontally and vertically (possibly implied by later working) – must be using given 1.24 (or a more accurate correct value) only (so not their incorrect value from part (a))
If taking moments about another point e.g. \(D\), then the corresponding equation(s) must be correct

M1dep*: Use of \(F = \mu R\) or \(F \leqslant \mu R\) with \(0.5g\) and 1.24 (or better) for \(F\)

A1: awrt 0.253
If final answer is \(\mu \geqslant 0.253\) then A0
Using the exact value or 1.24 for \(R_D\) leads to the same answer to 3 significant figures

June 2023 Paper 3 Q11

OCR ACurrent spec8 marksMoments Resolving Forces

A rod AB with end A on horizontal ground, inclined at 30 degrees to the horizontal; end B rests against a wall that makes an angle of 55 degrees with the ground

A uniform rod \(AB\), of weight \(20\,\mathrm{N}\) and length \(2.8\,\mathrm{m}\), rests in equilibrium with the end \(A\) in contact with rough horizontal ground and the end \(B\) resting against a smooth wall inclined at \(55^\circ\) to the horizontal. The rod, which rests in a vertical plane that is perpendicular to the wall, is inclined at \(30^\circ\) to the horizontal (see diagram).

(a) Show that the magnitude of the force acting on the rod at \(B\) is \(9.56\,\mathrm{N}\), correct to 3 significant figures. [3]

(b) Determine the magnitude of the contact force between the rod and the ground. Give your answer correct to 3 significant figures. [5]

Mark scheme (a)

Scheme	Marks	AO
	M1	3.1b
\(1.4(20\cos 30) = R_B(2.8\cos 25)\)	M1	1.1
\(R_B = \dfrac{28\cos 30}{2.8\cos 25} = 9.5555330\ldots\) \(= 9.56\) (N) (correct to 3 significant figures)	A1	1.1
	[3]

Notes

M1: Moments about \(A\) to form an equation with the correct number of terms and both the weight and contact force at \(B\) resolved - condone \(20g\) for the weight of the rod. All relevant values must have been substituted. May take moments about another point (and resolve) but must end up after elimination with an equation in \(R_B\) only
Resolved for the weight means either sin or cos with an angle of either 30 or 60 only, and for the contact force at \(B\) either sin or cos with an angle of 25 or 35 or 55 or 65 only

M1: One of the two moment terms correct
Must be as part of a two-term moment equation (so dimensionally correct so M0 if \(20g\) for the weight) but the other term need not be resolved (or resolved correctly) e.g. \(1.4(20\cos 30) = 2.8R_B\) scores M0 M1

A1: AG - correct equation followed by 9.56 scores all 3 marks
Allow awrt 9.56

For reference in parts (a) and (b):
Moments about \(B\): \(R_A(2.8\cos 30) = F_A(2.8\sin 30) + 20(1.4\cos 30)\)
Moments about the mid-point: \(R_A(1.4\cos 30) = F_A(1.4\sin 30) + R_B(1.4\cos 25)\)

Mark scheme (b)

Scheme	Marks	AO
	M1*	3.3
\(R_A + 9.56\sin 35 = 20\) \(F_A = 9.56\cos 35\)	A1 A1	1.1 1.1
\(\sqrt{(9.56\cos 35)^2 + (20 - 9.56\sin 35)^2}\)	M1dep*	1.1
16.5 (N)	A1	2.2a
	[5]

Notes

M1*: Attempt at resolving vertically or horizontally – correct number of terms with the contact force at \(B\) resolved (angle must be one of 25, 35, 55 or 65 only) – or taking moments again (see below) (same conditions for taking moments as in part (a)). M0 if using \(W\) or \(mg\) only for the weight of the rod (must substitute in the given value before awarding this mark – see second guidance column if using \(20g\))
Allow sign errors, sin/cos confusion and \(20g\) used as the weight for the M mark. Allow \(R_B\) instead of the given value of 9.56 for the M mark

A1: Correct expression/equation for \(R_A\)
\(R_A = 20 - 5.48\ldots = 14.5\ldots\)

A1: Correct expression/equation for \(F_A\)
\(F_A = 7.83\ldots\)

M1dep*: Correct method for calculating the magnitude of the contact force at \(A\) where \(R_A\) and \(F_A\) are a linear combination of the correct number of terms with 9.56 resolved in both equations (angle must be one of 25, 35, 55 or 65 only)

A1: www awrt 16.5. For reference if using 9.56 for \(R_B\) then \(16.494179\ldots\)
If using more accurate value for \(R_B\) then \(16.494698\ldots\)

June 2022 Paper 3 Q11

OCR ACurrent spec7 marksMoments Resolving Forces

Rod AB, 3 m long, with A on horizontal ground, making 60° with the ground, with a particle at B; a string from point C on the rod, x m from A, goes to point D vertically above A, with angle 60° between DC and the vertical

A uniform rod \(AB\) of mass 4 kg and length 3 m rests in a vertical plane with \(A\) on rough horizontal ground.

A particle of mass 1 kg is attached to the rod at \(B\). The rod makes an angle of 60° with the horizontal and is held in limiting equilibrium by a light inextensible string \(CD\). \(D\) is a fixed point vertically above \(A\) and \(CD\) makes an angle of 60° with the vertical. The distance \(AC\) is \(x\) m (see diagram).

(a) Find, in terms of \(g\) and \(x\), the tension in the string. [3]

The coefficient of friction between the rod and the ground is \(\dfrac{9\sqrt{3}}{35}\).

(b) Determine the value of \(x\). [4]

Mark scheme (a)

Scheme	Marks	AO
Taking moments about \(A\) – correct number of terms, allow sin/cos confusion and sign errors – may take moments about another point (and resolve) but must end up after elimination with an equation in \(T\) and \(x\) only	M1	3.1b
\(4g\left(\frac{3}{2}\cos 60\right) + g(3\cos 60) = xT\)	A1	1.1
\(T = \dfrac{9g}{2x}\) (N)	A1	2.2a
	[3]

Notes

M1: Dimensionally correct.
Must be \(xT\) and not \(xT\cos(\ldots)\) or \(xT\sin(\ldots)\)

A1: Correct equation in \(g\), \(x\) and \(T\) only – condone \(g\) replaced by 9.8
\(T\) is the tension in the string

A1: An answer of \(\frac{44.1}{x}\) (or with trigonometric terms) is A0 unless correct answer in terms of \(g\) seen
oe exact answers in terms of \(g\) and \(x\) (condone correct triple decker fractions)

For reference for parts (a) and (b):
Moments about \(C\): \(R_A(x\cos 60) + g(3 - x)\cos 60 = 4g(x - 1.5)\cos 60 + F_A(x\sin 60)\)
Moments about \(B\): \(T(3 - x) + R_A(3\cos 60) = 4g(1.5\cos 60) + F_A(3\sin 60)\)
Moments about midpoint of \(AB\): \(R_A(1.5\cos 60) + g(1.5\cos 60) = T(x - 1.5) + F_A(1.5\sin 60)\)
Resolving perpendicular to \(AB\): \(T + R_A\cos 60 = 4g\cos 60 + g\cos 60 + F_A\sin 60\)
Resolving parallel to \(AB\): \(R_A\sin 60 + F_A\cos 60 = 4g\sin 60 + g\sin 60\)

Mark scheme (b)

Scheme	Marks	AO
Resolve vertically or horizontally – correct number of terms with the tension at \(C\) in terms of cos/sin, condone sign errors, allow sin/cos confusion but forces that require resolving must be (and correspondingly those that don’t require resolving e.g. the weights if resolving vertically, should not be resolved)	M1	3.3
\(T\cos 60 + R_A = 4g + g\) \(\left(\Rightarrow R_A = 5g - \dfrac{9g}{4x}\right)\) \(F_A = T\sin 60\) \(\left(\Rightarrow F_A = \dfrac{9\sqrt{3}g}{4x}\right)\)	A1	1.1
\(\dfrac{9\sqrt{3}g}{4x} = \dfrac{9\sqrt{3}}{35}\left(5g - \dfrac{9g}{4x}\right)\)	M1dep*	3.4
\(x = 2.2\)	A1	2.2a
	[4]

Notes

M1: Or obtain an equation in \(F_A\) and/or \(R_A\) in terms of \(T\) (or their \(T\)) only (see list of equations below)

A1: Both correct (unsimplified) – allow with \(T\) or their (possibly incorrect) \(T\) (oe eg two valid equations in \(R_A\) and \(F_A\))
\(R_A\) is the normal contact force at \(A\)
\(F_A\) is the frictional contact force at \(A\)

M1dep*: Use of \(F = \mu R\) with correct \(\mu\) to form an equation in \(x\) only – no forces missing from their \(R_A\) and \(F_A\) and all required forces resolved accordingly or not e.g. if resolving vertically the two weights should not contain sin/cos

A1: awrt 2.2
www

October 2021 Paper 3 Q12

OCR ACurrent spec7 marksMoments

Horizontal beam AB of length 4 m hanging from two vertical ropes attached at C and D; AC = 0.5 m and DB = 0.7 m

A beam, \(AB\), has length \(4\,\mathrm{m}\) and mass \(20\,\mathrm{kg}\). The beam is suspended horizontally by two vertical ropes. One rope is attached to the beam at \(C\), where \(AC = 0.5\,\mathrm{m}\). The other rope is attached to the beam at \(D\), where \(DB = 0.7\,\mathrm{m}\) (see diagram).

The beam is modelled as a non-uniform rod and the ropes as light inextensible strings.

It is given that the tension in the rope at \(C\) is three times the tension in the rope at \(D\).

(a) Determine the distance of the centre of mass of the beam from \(A\). [5]

A particle of mass \(m\,\mathrm{kg}\) is now placed on the beam at a point where the magnitude of the moment of the particle's weight about \(C\) is \(3.5mg\,\mathrm{N\,m}\). The beam remains horizontal and in equilibrium.

(b) Determine the largest possible value of \(m\). [2]

Mark scheme (a)

Scheme	Marks	AO
\(\begin{array}{ll}0.5T_C + (4 - 0.7)T_D = 20g\bar{x} & \text{(moments about } A\text{)}\\ (\bar{x} - 0.5)(20g) = (4 - 1.2)T_D & \text{(moments about } C\text{)}\\ (4 - 1.2)T_C = 20g(4 - 0.7 - \bar{x}) & \text{(moments about } D\text{)}\\ 0.7T_D + (4 - 0.5)T_C = 20g(4 - \bar{x}) & \text{(moments about } B\text{)}\\ (\bar{x} - 0.5)T_C = (4 - 0.7 - \bar{x})T_D & \text{(moments about com)}\end{array}\)	M1* A1FT	3.3 1.1
\(T_C = 3T_D\) or \(T_C = 3T\) and \(T_D = T\)	B1	3.3
\(T_C = 147\,(= 15g)\) and/or \(T_D = 49\,(= 5g)\) (from \(T_C + T_D = 20g\)) used in relevant moment equation(s)	M1dep*	1.1
\(\bar{x} = 1.2\ (\mathrm{m})\)	A1	2.2a
	[5]

Notes

M1*: Moments (with correct number of terms) about \(A\), \(C\), \(D\), \(B\) or com
\(\bar{x}\) is the centre of mass of the rod from \(A\)
Allow \(g\) to be absent

A1FT: Follow though their \(T_C\) and \(T_D\) (allow just \(T\)) and allow \(W\) for \(20g\)

B1: Correct relationship(s) for the tensions at \(C\) and \(D\) (soi)

M1dep*: Equation in \(\bar{x}\) only, e.g.,
• \(W = 4T\) and any one mom. eq.
• \(4T = 20g\) and any one mom. eq.
• \(W = 20g\) and any two mom. eq.
M0 if tension used are the same in both ropes

A1: Must be 1.2 as question asks for com from \(A\)
Alternative: B1 as main scheme (soi), then M1 for splitting 2.8 in the ratio 1: 3 or 3:1 then A1 for correct 1:3 then B2 for \(0.5 + 0.25(2.8) = 1.2\,(\mathrm{m})\)

Mark scheme (b)

Scheme	Marks	AO
\(0.7mg = 20g(4 - \bar{x} - 0.7)\) (moments about \(D\)) \(20g(\bar{x} - 0.5) + 3.5mg = (4 - 0.5 - 0.7)(20g + mg)\) (moments about \(C\))	M1	3.1b
\(m = 60\)	A1	2.2a
	[2]

Notes

M1: Moments about \(D\) (oe) – correct number of terms – if taking moments about another point then must set \(T_C = 0\) and \(T_D = 20g + mg\)
Allow \(g\) to be absent

June 2025 Paper 1 Q5

OCR MEICurrent spec3 marksMoments

5 A uniform rectangular lamina ABCD has a mass of 0.12 kg. Length AB is 11 cm and length AD is 20 cm.

The lamina ABCD is held by a smooth hinge at A.

A horizontal force of magnitude \(P\) N is applied at B so that the lamina is in equilibrium in a vertical plane with AD horizontal as shown in the diagram.

Rectangle ABCD with A top left (hinge), D top right, AD = 20 cm, AB = 11 cm vertical, horizontal force P N acting at B

Show that \(P\) is 1.07 to 3 significant figures. [3]

Mark scheme

Scheme	Marks	AO
Moments about A \(0.1\times 0.12g = 0.11P\)	M1 A1	3.1a 2.1
\(P = 1.07\)	A1	2.1
	[3]

Notes

M1: Attempt to calculate a moment of at least one force about any point. Do not award for oblique distance(s) used without resolving the forces

A1: Correct moment of weight about A

A1: AG 1.07 obtained from correct total moment equation seen. 1.06909…
Award if full equation or \(10\times 0.12g\) and \(\frac{11.76}{11} = 1.07\) seen

Allow full credit for taking moments about another point where the correct force(s) at the hinge is seen.

June 2024 Paper 1 Q7

OCR MEICurrent spec7 marksMoments

7 A rectangular book ABCD rests on a smooth horizontal table. The length of AB is 28 cm and the length of AD is 18 cm. The following five forces act on the book, as shown in the diagram.

4 N at A in the direction AD
5 N at B in the direction BC
3 N at B in the direction BA
9 N at D in the direction DA
3 N at D in the direction DC

Rectangle ABCD with A top left, B top right, C bottom right, D bottom left; 4 N down at A, 5 N down at B, 3 N leftwards at B, 3 N rightwards at D and 9 N up at D

(a) Show that the resultant of the forces acting on the book has zero magnitude. [2]

(b) Find the total moment of the forces about the centre of the book. Give your answer in N m. [3]

Mark scheme (a)

Scheme	Marks	AO
Total in the AB direction is \(3 - 3 = 0\) Total in the direction AD is \(4 + 5 - 9 = 0\)	M1	2.1
So zero resultant force	A1	2.1
	[2]

Notes

M1: Considers forces in each direction

A1: AG Allow if clear statement of zero force in two perpendicular directions, Do not allow for “equilibrium” on its own eg 3=3 and 4+5=9 is not sufficient without a comment

Mark scheme (b)

Scheme	Marks	AO
Take moments anticlockwise (N cm)	B1	1.1
\(4 \times 14 - 5 \times 14 + 3 \times 9 - 9 \times 14 + 3 \times 9\)	M1	1.1
\(= -86\) N cm
\(= -0.86\) [N m]	A1	1.1
	[3]

Notes

B1: At least one moment about the centre or any corner

M1: Combines moments of 5 forces about the centre. Allow only sign errors

\(-86\) N cm: Do not allow as final answer

A1: Allow for 0.86 N m clockwise but not \(-0.86\) N m clockwise
Allow B1 SC1 for \(-0.86\) Nm as the total moment about a different point.

Mark scheme (c)

Scheme	Marks	AO
The book will rotate	B1	2.2a
clockwise [about the centre of mass]	B1	1.2
	[2]

Notes

B1: Allow turn or spin but not move, tilt or flip

B1: Allow “there is a clockwise moment” for the second mark

June 2023 Paper 1 Q4

OCR MEICurrent spec4 marksMoments

4 A ruler PQRS is a uniform rectangular lamina with mass 20 grams. The length of PQ is 30 cm and the length of PS is 4 cm. The ruler is attached at P to a smooth hinge and held with S vertically below P by a horizontal force of magnitude \(F\) N as shown in the diagram.

Rectangle PQRS with PQ horizontal at the top, hinge at P, S vertically below P, horizontal force F N acting at S

(a) Calculate the value of \(F\). [3]

(b) Explain what would happen to the lamina if the force at S were removed. [1]

Mark scheme (a)

Scheme	Marks	AO
Weight acts through the centre of the ruler 15cm from P	B1	1.1a
Take moments about P \(0.02g \times 15 = F \times 4\)	M1	3.1b
\(F = \dfrac{0.02g \times 15}{4}\) \(= \frac{3g}{40} = 0.735\)	A1	1.1b
	[3]

Notes

B1: Uses uniformity of the ruler, by distance shown on diagram or 15 used in a calculation or equation

M1: Taking moments about any suitable point with all relevant forces and distances attempted. Must include a weight (not mass only) term. Allow one incorrect distance used. Allow 20g for weight or error in converting mass to kilograms or lengths to another unit.

A1: Allow \(0.075g\)

Mark scheme (b)

Scheme	Marks	AO
Without \(F\) the ruler would rotate clockwise [about P]	E1	2.4
	[1]

Notes

E1: Must indicate the turning effect of the weight and the direction. “Clockwise” implies rotation.

June 2022 Paper 1 Q6

OCR MEICurrent spec9 marksMoments

6 A shelf consists of a horizontal uniform plank AB of length 0.8 m and mass 5 kg with light inextensible vertical strings attached at each end. A stack of bricks each of mass 2.3 kg is placed on the plank as shown in the diagram.

Horizontal plank AB hanging from vertical strings at A and B, with a stack of bricks standing on the plank to the right of its centre

(a) Explain the meaning of each of the following modelling assumptions.

The stack of bricks is modelled as a particle.
The plank is modelled as uniform.

[2]

Either of the strings will break if the tension exceeds 75 N.

(b) Find the greatest number of bricks that can be placed at the centre of the plank without breaking the strings. [2]

(c) Find an expression for the moment about A of the weight of a stack of \(n\) bricks when the stack is at a distance of \(x\) m from A. State the units for your answer. [2]

(d) Calculate the greatest distance from A that the largest stack of bricks can be placed without a string breaking. [3]

Mark scheme (a)

Scheme	Marks	AO
the bricks have negligible size so contact force with the plank acts at a point The mass of plank is evenly spread across its length the weight of plank acts at centre of plank.	B1 B1	3.3 3.3
	[2]

Notes

B1: Allow “no size” or “size doesn’t matter” or “shape is not relevant” etc
Allow “weight of bricks acts on the plank at point”

B1: Allow for either statement
Allow the plank is the same throughout or centre of mass at centre
Do not allow “mass acts” at the centre

Mark scheme (b)

Scheme	Marks	AO
when placed at the centre tensions are equal \(2 \times 75 = 2.3ng + 5g\)	M1	3.1b
\(n = \left[\dfrac{101}{2.3g}\right] = 4.48\ldots\) so 4 bricks	A1	1.1b
	[2]

Notes

M1: Using symmetry to establish equation for \(n\) soi. Allow if the weight of the plank or one of the tensions is missing
Trial and improvement may be used

A1: Final answer must be the integer
Allow if 4 seen www

Alternative using moments

Scheme	Marks
\(5g \times 0.4 + 2.3gn \times 0.4 = 75 \times 0.8\)	M1
\(n = \left[\dfrac{40.4}{9.016}\right] = 4.48\ldots\) so 4 bricks	A1

M1: Allow for missing moment of weight or one of the tensions
Every term must be a moment

A1: Final answer must be the integer
Allow if 4 seen www

Mark scheme (d)

Scheme	Marks	AO
4 bricks on the point of breaking \(x\) m from A Taking moments about A \(5g \times 0.4 + 4 \times 2.3gx = 75 \times 0.8\)	M1	3.1b
\(9.2gx = 60 - 2g\)	A1	1.1b
\(x = 0.448\) [so 44.8 cm from A]	A1	1.1b
	[3]

Notes

M1: Taking moments about any point to form an equation. FT their \(n\)
All forces used in a moment. Allow sign errors. Allow an incorrect distance used. Could be an inequality

A1: Fully correct equation FT their \(n\). Allow corresponding inequality
Need not be simplified

A1: cao

October 2021 Paper 1 Q5

OCR MEICurrent spec5 marksMoments

5 ABCD is a rectangular lamina in which AB is 30 cm and AD is 10 cm.

Three forces of 20 N and one force of 30 N act along the sides of the lamina as shown in the diagram.

Rectangle ABCD with AB along the top: 20 N acts to the left at A along BA extended, 20 N acts upwards at B, 20 N acts to the right at C along DC extended, 30 N acts upwards at D; a force F N acts downwards on AB at a point x cm from A

An additional force \(F\) N is also applied at right angles to AB to a point on the edge AB \(x\) cm from A.

(a) Given that the lamina is in equilibrium, calculate the values of \(F\) and \(x\). [3]

The point of application of the force \(F\) N is now moved to B, but the magnitude and direction of the force remain the same.

(b) Explain the effect of the new system of forces on the lamina. [2]

Mark scheme (a)

Scheme	Marks	AO
\(F = 50\)	B1	1.1b
Take moments about A (or D) \(Fx = 20 \times 30 + 20 \times 10\) OR Take moments about B (or C) \(F(30 - x) + 20 \times 10 = 30 \times 30\)	M1	1.1a
\(x = \dfrac{800}{50} = 16\)	A1	1.1b
	[3]

Notes

B1: cao

M1: Forming an equation for moments about any point – allow their value of \(F\) used. Allow one incorrect distance or a missing term
Do not allow for moment = force

A1: Allow 0.16 m if unit given

Mark scheme (b)

Scheme	Marks	AO
Forces have a resultant moment	E1	2.4
which have a clockwise turning effect on the lamina	E1	2.4
	[2]

Notes

E1: Allow ‘resultant moment’ or ‘not in equilibrium’ soi
Do not allow for ‘a resultant force’ on its own

E1: Clockwise must be indicated

October 2020 Paper 1 Q6

OCR MEICurrent spec4 marksMoments

6 A uniform ruler AB has mass 28 g and length 30 cm. As shown in Fig. 6, the ruler is placed on a horizontal table so that it overhangs a point C at the edge of the table by 25 cm.

A downward force of \(F\) N is applied at A. This force just holds the ruler in equilibrium so that the contact force between the table and the ruler acts through C.

Fig. 6: ruler AB on a table, with A on the table and the edge of the table at C; CB = 25 cm overhangs the edge — Fig. 6

(a) Complete the force diagram in the Printed Answer Booklet, labelling the forces and all relevant distances.

Force diagram from the Printed Answer Booklet: ruler AB on the table edge C, with CB = 25 cm and the upward force R N drawn at C

[2]

(b) Calculate the value of \(F\). [2]

Completed force diagram: F N downwards at A, R N upwards at C, W N downwards 10 cm from C towards B; AC = 5 cm — Mark scheme (a)

Mark scheme (b)

Scheme	Marks	AO
Take moments about C \(10 \times 0.028g = 5F\)	M1	1.1a
\(F = 0.549\) N	A1	1.1
	[2]

Notes

M1: Moments about any point with all relevant forces seen. Allow their weight and one incorrect distance
[Value of R is 0.8232 N]

A1: allow \(\dfrac{7}{125}g\) oe