What is M9.1: Moments?

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.

How many marks are M9.1 questions worth?

M9.1 practice drills on this page are worth 2–7 marks each; marks in a real exam vary by question.

Which past exam papers has M9.1 appeared in?

We have not yet indexed a verified past-paper appearance for M9.1 — check the Edexcel A-level Maths 9MA0 past-papers hub directly.

M9.1

Understand and use moments in simple static contexts.

Draft — not yet indexed

Moments

Worked answers and methods for M9.1 on Edexcel A-level Maths 9MA0.

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.

Worked practice

Tier 1 · Easy

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)

Mark scheme

Mark scheme for question 1
Question	Scheme	Marks
1	$4.2\,\text{N m}$	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 marks)

Tier 2 · Standard

A uniform horizontal beam

AB

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 2 is 5 marks)

Mark scheme

Mark scheme for question 2
Question	Scheme	Marks
2	Upward force at $A=95\,\text{N}$ Upward force at $B=145\,\text{N}$	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 $R_A=240-145=95\,\text{N}$ .

(5 marks)

Tier 3 · Hard

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 3 is 7 marks)

Mark scheme

Mark scheme for question 3
Question	Scheme	Marks
3	Wall force $=40\sqrt3\,\text{N}$ Least coefficient of friction $=\dfrac{\sqrt3}{6}$	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 $\mu=40\sqrt3/240=\sqrt3/6$ .

(7 marks)

Tier 1 · Easy

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 4 is 2 marks)

Mark scheme

Mark scheme for question 4
Question	Scheme	Marks
4	$4.0\,\text{N m}$	2
Notes
The perpendicular component of the force is $20\sin30^\circ=10\,\text{N}$ . Hence the moment is $10(0.40)=4.0\,\text{N m}$ .

(2 marks)

Tier 2 · Standard

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 5 is 5 marks)

Mark scheme

Mark scheme for question 5
Question	Scheme	Marks
5	Upward force at $A=107.8\,\text{N}$ Upward force at $B=142.2\,\text{N}$	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 $R_A=107.8\,\text{N}$ .

(5 marks)

Tier 3 · Hard

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 6 is 6 marks)

Mark scheme

Mark scheme for question 6
Question	Scheme	Marks
6	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
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 $C$ .

(6 marks)

Tier 2 · Standard

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 7 is 4 marks)

Mark scheme

Mark scheme for question 7
Question	Scheme	Marks
7	$W=1.8\,\text{N}$	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.

(4 marks)

Tier 3 · Hard

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 8 is 7 marks)

Mark scheme

Mark scheme for question 8
Question	Scheme	Marks
8	Tension $=210\,\text{N}$ Hinge-force magnitude $=30\sqrt{43}\,\text{N}$ Direction $=22.4^\circ$ above the horizontal, towards $B$	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.

(7 marks)

Tier 3 · Hard

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 9 is 6 marks)

Mark scheme

Mark scheme for question 9
Question	Scheme	Marks
9	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
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 $R_A=91\,\text{N}$ .

(6 marks)

Q10

Tier 3 · Hard

10.

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 10 is 6 marks)

Mark scheme

Mark scheme for question 10
Question	Scheme	Marks
10	$P=90\,\text{N}$ Pivot-force magnitude $=30\sqrt{13}\,\text{N}$ Direction $=\tan^{-1}(2/3)$ below the horizontal, towards the left	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.

(6 marks)

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