M9 Moments — revision question pack

1 specification points · notes, questions, answers and worked methods

Checked against Edexcel 9MA0 section M9. Review basis: the qualification registry sourced from the Pearson Edexcel Level 3 Advanced GCE in Mathematics (9MA0) specification; registry verification recorded 11 July 2026.

How this checking works

M9.1 · Understand and use moments in simple static contexts.

Explanation

The moment of a force about a point is $F d$ , where $d$ is the perpendicular distance from the point to the force's line of action; state whether it is clockwise or anticlockwise when required.
For a rigid body in equilibrium, choose a pivot that removes unknown forces where possible, set total clockwise moments equal to total anticlockwise moments, then use force equilibrium.
A uniform rod's weight acts at its midpoint, while a non-uniform body's weight acts at its stated centre of mass; contact forces act at their contact points.
A common error is to use the distance along a rod instead of the perpendicular distance to the line of action, or to omit a force whose line of action does not pass through the pivot.

Moment about a pivot is force times perpendicular distance; equilibrium requires clockwise and anticlockwise moments to balance.

Worked example

A uniform horizontal beam $AB$ has length $4\,\text{m}$ and weight $120\,\text{N}$ . It is supported vertically at $A$ and $B$ . Find the upward force at each support.

1.The beam's weight acts at its midpoint, $2\,\text{m}$ from $A$ .
2.Taking moments about $A$ , $R_B(4)=120(2)$ , so $R_B=60\,\text{N}$ .
3.Vertical equilibrium gives $R_A+R_B=120$ , hence $R_A=60\,\text{N}$ .

Answer: Upward force at $A=60\,\text{N}$ ; Upward force at $B=60\,\text{N}$

Common mistakes

Don't use $Fd\cos\theta$ when the stated angle makes $Fd\sin\theta$ the perpendicular moment.
Don't take moments using the sloping distance to a force instead of its perpendicular distance from the pivot.

Exam tip

Choose a pivot that removes an unknown force, assign moment directions consistently and also check vertical equilibrium.

Tier 1 · Easy

1.
A $12\,\text{N}$ force acts perpendicular to a spanner at a distance $0.35\,\text{m}$ from a nut. Find the magnitude of its moment about the nut.
(2)
(Total for Question 1 is 2 marks)
2.
A straight handle extends horizontally to the right from a pivot. A force of $20\,\text{N}$ acts upwards and to the right at $30^\circ$ to the handle, at a point $0.40\,\text{m}$ from the pivot. Find the magnitude of the moment about the pivot.
(2)
(Total for Question 2 is 2 marks)

Tier 2 · Standard

1.
A uniform horizontal beam $AB$ is $6\,\text{m}$ long and weighs $90\,\text{N}$ . It is supported vertically at $A$ and $B$ , and a $150\,\text{N}$ load is placed $4\,\text{m}$ from $A$ . Find the upward force at each support.
(5)
(Total for Question 1 is 5 marks)
2.
A non-uniform horizontal beam $AB$ has length $5\,\text{m}$ and weight $180\,\text{N}$ . Its centre of mass is $2.2\,\text{m}$ from $A$ . The beam is supported vertically at $A$ and $B$ , and a load of $70\,\text{N}$ is placed $4.5\,\text{m}$ from $A$ . Find the upward force at each support.
(5)
(Total for Question 2 is 5 marks)
3.
A uniform metre rule of weight $1.0\,\text{N}$ is horizontal and pivoted at its $40\,\text{cm}$ mark. A weight of $3.0\,\text{N}$ hangs from the $10\,\text{cm}$ mark and a weight of $W\,\text{N}$ hangs from the $85\,\text{cm}$ mark. Find $W$ when the rule is in equilibrium, giving your answer to 2 significant figures.
(4)
(Total for Question 3 is 4 marks)

Tier 3 · Hard

1.
A uniform ladder of length $5\,\text{m}$ and weight $240\,\text{N}$ rests with its lower end on rough horizontal ground and its upper end against a smooth vertical wall. The ladder makes an angle of $60^\circ$ with the ground and is in limiting equilibrium. Find the force exerted by the wall and the least coefficient of friction at the ground.
(7)
(Total for Question 1 is 7 marks)
2.
A uniform horizontal board $AB$ has length $8\,\text{m}$ and weight $200\,\text{N}$ . It rests on small supports at $A$ and $C$ , where $AC=5\,\text{m}$ ; each support can exert only an upward force. A crate of weight $150\,\text{N}$ is placed $6\,\text{m}$ from $A$ . Find the force exerted by each support. The crate is then moved slowly towards $B$ . Find how far it can be from $A$ before the board loses contact with the support at $A$ , and state what begins to happen if it is moved farther.
(6)
(Total for Question 2 is 6 marks)
3.
A uniform horizontal rod $AB$ has length $4\,\text{m}$ and weight $120\,\text{N}$ . It is freely hinged at $A$ and held in equilibrium by a light cable attached at $B$ . The cable makes an angle of $30^\circ$ above the rod and pulls upwards and towards $A$ . A load of weight $60\,\text{N}$ hangs $3\,\text{m}$ from $A$ . Find the tension in the cable. Hence determine the hinge reaction, giving its exact magnitude and its direction to 3 significant figures.
(7)
(Total for Question 3 is 7 marks)
4.
A non-uniform horizontal rod $AB$ has length $5\,\text{m}$ and weight $120\,\text{N}$ . A load of weight $30\,\text{N}$ is attached at $B$ , and the rod balances on a single support $2.3\,\text{m}$ from $A$ . Find the exact distance of the rod's centre of mass from $A$ . The load is then removed, the rod is supported vertically at $A$ and $B$ , and a load of weight $50\,\text{N}$ is attached $4\,\text{m}$ from $A$ . Find the upward force at each support. Give all numerical answers exactly.
(6)
(Total for Question 4 is 6 marks)
5.
A light rigid frame consists of a horizontal section $OA$ of length $3\,\text{m}$ and a vertical section $AB$ of length $2\,\text{m}$ , with $B$ vertically above $A$ . The frame is freely pivoted at $O$ . A force of $40\,\text{N}$ acts vertically downwards at $A$ , a force of $100\,\text{N}$ acts vertically upwards at $B$ , and a horizontal force of magnitude $P\,\text{N}$ acts to the right at $B$ . The frame is in equilibrium. Find $P$ . Hence find the exact magnitude and direction of the force exerted by the pivot on the frame. Give all numerical answers exactly.
(6)
(Total for Question 5 is 6 marks)

Answer key

Answers begin on a new printed page so the question pack can be completed without the solutions alongside it.

M9.1 · Understand and use moments in simple static contexts.

Tier 1 · Easy

Mark scheme for M9.1 Tier 1 · Easy
Question	Scheme	Marks
1	$4.2\,\text{N m}$	2
(2 marks)		2
Notes
The force is perpendicular, so the perpendicular distance is $0.35\,\text{m}$ . The moment is $F d=12(0.35)=4.2\,\text{N m}$ .
2	$4.0\,\text{N m}$	2
(2 marks)		2
Notes
The perpendicular component of the force is $20\sin30^\circ=10\,\text{N}$ . Hence the moment is $10(0.40)=4.0\,\text{N m}$ .

Tier 2 · Standard

Mark scheme for M9.1 Tier 2 · Standard
Question	Scheme	Marks
1	Upward force at $A=95\,\text{N}$ Upward force at $B=145\,\text{N}$	5
(5 marks)		5
Notes
The beam's weight acts $3\,\text{m}$ from $A$ . Taking moments about $A$ , $R_B(6)=90(3)+150(4)=870$ , so $R_B=145\,\text{N}$ . Vertical equilibrium gives $R_A+R_B=90+150=240$ , hence $R_A=240-145=95\,\text{N}$ .
2	Upward force at $A=107.8\,\text{N}$ Upward force at $B=142.2\,\text{N}$	5
(5 marks)		5
Notes
Let the upward forces be $R_A$ and $R_B$ . Taking moments about $A$ , $5R_B=180(2.2)+70(4.5)=711$ , so $R_B=142.2\,\text{N}$ . Vertical equilibrium gives $R_A+R_B=180+70=250$ , hence $R_A=107.8\,\text{N}$ .
3	$W=1.8\,\text{N}$	4
(4 marks)		4
Notes
The rule's weight acts at the $50\,\text{cm}$ mark. Taking moments about the pivot, the $3.0\,\text{N}$ weight has anticlockwise moment $3.0(0.30)$ , while the rule and $W$ have clockwise moments $1.0(0.10)$ and $W(0.45)$ . Hence $3.0(0.30)=1.0(0.10)+0.45W$ , so $0.45W=0.80$ and $W=16/9=1.777\ldots\,\text{N}$ , which is $1.8\,\text{N}$ to 2 significant figures.

Tier 3 · Hard

Mark scheme for M9.1 Tier 3 · Hard
Question	Scheme	Marks
1	Wall force $=40\sqrt3\,\text{N}$ Least coefficient of friction $=\dfrac{\sqrt3}{6}$	7
(7 marks)		7
Notes
Let the horizontal wall force be $S$ . Taking moments about the foot of the ladder, $S(5\sin60^\circ)=240(2.5\cos60^\circ)$ , so $S=40\sqrt3\,\text{N}$ . Horizontal equilibrium gives friction $F=S=40\sqrt3\,\text{N}$ , and vertical equilibrium gives ground reaction $R=240\,\text{N}$ . At limiting equilibrium $F=\mu R$ , so $\mu=40\sqrt3/240=\sqrt3/6$ .
2	Initially, force at $A=10\,\text{N}$ upwards and force at $C=340\,\text{N}$ upwards Contact at $A$ is lost when the crate is $\dfrac{19}{3}\,\text{m}$ from $A$ If moved farther, the board begins to tip clockwise about $C$	6
(6 marks)		6
Notes
Initially, taking moments about $A$ gives $5R_C=200(4)+150(6)=1700$ , so $R_C=340\,\text{N}$ . Vertical equilibrium then gives $R_A=350-340=10\,\text{N}$ . Contact at $A$ is just lost when $R_A=0$ , so $R_C=350\,\text{N}$ . If the crate is then $x$ metres from $A$ , moments about $A$ give $350(5)=200(4)+150x$ , hence $x=950/150=19/3\,\text{m}$ . Farther towards $B$ , equilibrium would require a downward force at $A$ , which the support cannot provide, so the board tips clockwise about $C$ .
3	Tension $=210\,\text{N}$ Hinge-force magnitude $=30\sqrt{43}\,\text{N}$ Direction $=22.4^\circ$ above the horizontal, towards $B$	7
(7 marks)		7
Notes
Let the tension be $T$ . Taking moments about $A$ , only the vertical component of the cable tension contributes, so $4T\sin30^\circ=120(2)+60(3)=420$ . Hence $T=210\,\text{N}$ . The cable pulls left with component $210\cos30^\circ=105\sqrt3\,\text{N}$ and up with component $105\,\text{N}$ . Horizontal and vertical equilibrium therefore give hinge-force components $105\sqrt3\,\text{N}$ towards $B$ and $120+60-105=75\,\text{N}$ upwards. Its magnitude is $\sqrt{(105\sqrt3)^2+75^2}=30\sqrt{43}\,\text{N}$ , and its direction above the horizontal is $\tan^{-1}(75/(105\sqrt3))=22.4^\circ$ to 3 significant figures.
4	Centre of mass is $1.625\,\text{m}$ from $A$ Upward force at $A=91\,\text{N}$ Upward force at $B=79\,\text{N}$	6
(6 marks)		6
Notes
Let the centre of mass be $x\,\text{m}$ from $A$ . Taking moments about the single support gives $120(2.3-x)=30(5-2.3)$ , so $2.3-x=0.675$ and $x=1.625$ . In the second arrangement, taking moments about $A$ gives $5R_B=120(1.625)+50(4)=395$ , hence $R_B=79\,\text{N}$ . Vertical equilibrium gives $R_A+R_B=170$ , so $R_A=91\,\text{N}$ .
5	$P=90\,\text{N}$ Pivot-force magnitude $=30\sqrt{13}\,\text{N}$ Direction $=\tan^{-1}(2/3)$ below the horizontal, towards the left	6
(6 marks)		6
Notes
Taking anticlockwise moments about $O$ as positive gives $100(3)-40(3)-2P=0$ , since the horizontal force at $B$ has perpendicular distance $2\,\text{m}$ . Thus $180-2P=0$ and $P=90\,\text{N}$ . The other forces have resultant $90\,\text{N}$ right and $60\,\text{N}$ up, so the pivot force is $90\,\text{N}$ left and $60\,\text{N}$ down. Its magnitude is $\sqrt{90^2+60^2}=30\sqrt{13}\,\text{N}$ and its direction is $\tan^{-1}(60/90)=\tan^{-1}(2/3)$ below the horizontal towards the left.