What is a moment in the context of Mechanics 1 for Edexcel International A Levels?

In Mechanics 1, a moment refers to the turning effect of a force applied to a point. It is calculated as the product of the force and the perpendicular distance from the pivot or axis of rotation to the line of action of the force. Moments are crucial in understanding how forces cause objects to rotate.

How do you calculate the moment of a force in Mathematics Mechanics 1?

To calculate the moment of a force, you multiply the magnitude of the force by the perpendicular distance from the pivot point to the line of action of the force. The formula is Moment = Force x Distance. The unit of moment is Newton-meters (Nm).

What is the principle of moments and how is it applied in solving problems?

The principle of moments states that for an object in equilibrium, the sum of the clockwise moments about any point is equal to the sum of the anticlockwise moments about that same point. This principle is used to solve problems involving beams, levers, and other systems in equilibrium by setting the total clockwise moments equal to the total anticlockwise moments.

Why is understanding moments important in the Edexcel International A Levels Mechanics 1 syllabus?

Understanding moments is essential in the Mechanics 1 syllabus because it helps students analyze and solve problems involving rotational forces and equilibrium. This knowledge is applicable in real-world scenarios such as engineering and physics, where understanding the balance of forces is crucial for designing stable structures and systems.

What are some common mistakes students make when working with moments in Mechanics 1?

Common mistakes include not identifying the correct pivot point, using incorrect distances (not perpendicular), and neglecting to consider all forces acting on the system. Students may also confuse clockwise and anticlockwise moments, leading to incorrect calculations. Careful analysis and practice can help avoid these errors.

Why is "Using the position of the force, not the perpendicular distance to the line of action" a common mistake in Moments?

Why it happens: Confusing 'distance' with 'lever arm'. How to avoid it: The lever arm is the PERPENDICULAR distance from the pivot to the LINE OF ACTION of the force. For a force inclined at acting at point P, the lever arm is the perpendicular component of OP to the force, i.e. |OP| where is the angle between OP and the force. Source: WME01/01 examiner reports

Why is "Adding clockwise and anticlockwise moments without signing them differently" a common mistake in Moments?

Why it happens: Treating moments as scalars. How to avoid it: Pick a positive sense (typically anticlockwise) and assign signs consistently. Or use the convention: (anticlockwise moments) = (clockwise moments) for equilibrium.

Why is "Assuming centre of mass at midpoint for a non-uniform rod" a common mistake in Moments?

Why it happens: Reflex from uniform-rod problems. How to avoid it: If the question says 'non-uniform', place the centre of mass at an unknown x and solve for it using one moment equation plus vertical equilibrium.

Why is "Choosing the wrong support as the tipping pivot" a common mistake in Moments?

Why it happens: Not visualising which support lifts. How to avoid it: Imagine the load shifting in the question's direction. The support that loses contact first is the one furthest from the load. Tipping is about the OTHER support. Source: WME01/01 examiner reports — flagged annually

Why is "Including the pivot's reaction in the moment equation taken about that pivot" a common mistake in Moments?

Why it happens: Habit of listing all forces. How to avoid it: A force passing through the pivot has zero moment about the pivot (zero lever arm). Strategic pivot choice eliminates one unknown.

Mechanics 1Edexcel International A Levels Mathematics (XMA01-YMA01)

Moments

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Moments Study Notes — Edexcel IAL Mathematics M1 (WME01/01, 2018 spec — 2026 onwards)

The turning effect of a force: $M = F \times d_{\perp}$ . For a rigid body in equilibrium, the resultant force is zero AND the resultant moment about any point is zero. Standard problems: uniform / non-uniform rods on supports, find centre of mass, 'about to tip' threshold, ladder against a wall.

What you’ll learn

Mapped to the Cambridge IGCSE XMA01-YMA01 syllabus (2026 onwards).

M1 8.1 — Define the moment of a force about a point and use $M = F \times d_{\perp}$ .
M1 8.2 — Apply the principle of moments for the equilibrium of a rigid body: $\Sigma F = 0$ and $\Sigma M = 0$ about any point.
M1 8.3 — Solve problems involving uniform and non-uniform rods, supports, loads, tipping and ladder geometry.

Moment of a force

$M = F \times d_{\perp}$ . Sense: clockwise or anticlockwise.

Definition. The moment of a force $F$ about a point $P$ is

$M = F \times d_{\perp},$

where $d_{\perp}$ is the perpendicular distance from $P$ to the LINE OF ACTION of the force.

Units: newton-metre (N m).
Sense: in coplanar problems, signed as anticlockwise positive (mathematical convention) — though clockwise positive is equally fine if used consistently.
In vector form (3-D): $\mathbf{M} = \mathbf{r} \times \mathbf{F}$ . The magnitude $|\mathbf{M}| = |F||r|\sin\theta = F d_{\perp}$ .

Finding $d_{\perp}$ . The perpendicular distance from $P$ to the line of action:

If the force acts at point $Q$ in a direction making angle $\alpha$ with $PQ$ , then $d_{\perp} = |PQ|\sin\alpha$ .
For a force parallel or perpendicular to $PQ$ , the calculation is immediate.

Examples:

Force	Position	Pivot	Moment
$50$ N down	$2$ m horizontally from pivot	pivot	$50 \cdot 2 = 100$ N m (CW or CCW depending on direction)
$30$ N at $60^{\circ}$ to a rod	$1.5$ m along rod from pivot	pivot at end	$30 \cdot 1.5 \sin 60^{\circ} = 38.97$ N m
$20$ N along a rod	any point	any point ON the rod	$0$ (line of action passes through pivot)

Force passing through pivot $\Rightarrow$ zero moment. This is the strategic basis for choosing pivots — eliminate an unknown by placing the pivot ON its line of action.

$M = F \times d_{\perp}$ (force × perpendicular distance).
Sense: clockwise or anticlockwise — be consistent.
Force at angle $\alpha$ to position vector: $M = F r \sin\alpha$ .
Force passing through pivot: zero moment.

Equilibrium of a rigid body

Two conditions: resultant force zero AND resultant moment zero about any point.

For a rigid body, equilibrium requires both:

$\Sigma \mathbf{F} = \mathbf{0}$ (no net translation).
$\Sigma M_P = 0$ about ANY point $P$ (no net rotation).

(The second condition is automatic if the first holds and moments are zero about ANY one point — but in practice you pick a strategic point and verify.)

For coplanar forces:

$\Sigma F_x = 0, \qquad \Sigma F_y = 0, \qquad \Sigma M = 0 \text{ about chosen point}.$

Three equations $\Rightarrow$ up to three unknowns.

Strategic choice of pivot. Take moments about a point through which an unknown force passes — that force then has zero moment and drops out of the equation.

Workflow for a typical rod problem:

Draw the rod with all forces (reactions at supports, weight, applied loads) at their positions.
Choose axes; mark distances.
Vertical equilibrium: $\Sigma F_y = 0 \Rightarrow$ relates reactions.
Moments about a support: eliminate that reaction $\Rightarrow$ solve for the other.
Back-substitute.

Worked example. Uniform rod $AB$ of length $4$ m, mass $8$ kg, on two supports at $C$ ( $1$ m from $A$ ) and $D$ ( $1$ m from $B$ ). Find $R_C$ and $R_D$ .

Vertical: $R_C + R_D = 78.4$ N.
Moments about $C$ (anticlockwise +): $R_D \cdot 2 - 78.4 \cdot 1 = 0 \Rightarrow R_D = 39.2$ N.
So $R_C = 39.2$ N (symmetric — agrees).

$\Sigma F = 0$ AND $\Sigma M = 0$ for equilibrium.
Three coplanar equations, up to three unknowns.
Strategic pivot: through an unknown force, to eliminate it.
Always cross-check by taking moments about a different point.

Uniform vs non-uniform rods

Uniform: weight at midpoint. Non-uniform: weight at unknown $\bar{x}$ .

Uniform rod. Mass evenly distributed; centre of mass at the geometric midpoint. Weight acts at the midpoint for moment calculations.

Non-uniform rod. Mass NOT evenly distributed; centre of mass at some unknown position $\bar{x}$ along the rod (typically measured from one end, say $A$ ).

Standard non-uniform problem:

A non-uniform rod $AB$ of length $L$ and mass $m$ rests on two supports. The reactions $R_A$ and $R_B$ are given (or one is given and the other follows from vertical equilibrium). Find $\bar{x}$ .

Vertical equilibrium gives $R_A + R_B = mg$ .
Take moments about $A$ : $R_B \cdot L - mg \cdot \bar{x} = 0 \Rightarrow \bar{x} = R_B L/(mg)$ .

If $\bar{x} < L/2$ , the rod is heavier on the $A$ side; if $\bar{x} > L/2$ , heavier on the $B$ side.

Worked example. Non-uniform rod $AB$ of length $4$ m, mass $5$ kg, on supports at $A$ and $B$ . $R_A = 20$ N. Find $\bar{x}$ .

$R_A + R_B = 49$ N $\Rightarrow R_B = 29$ N.
Moments about $A$ : $R_B \cdot 4 = 49 \cdot \bar{x} \Rightarrow \bar{x} = 116/49 \approx 2.37$ m.

So the centre of mass is past the midpoint, on the $B$ side.

Uniform: CoM at midpoint; weight acts there.
Non-uniform: CoM at unknown $\bar{x}$ .
Find $\bar{x}$ via: vertical equilibrium + one moment equation.
$\bar{x}$ measured from a stated end.

Tipping problems

At the tipping threshold, the lifting support's reaction is zero.

Setup. A rigid body (e.g. plank) rests on two supports. A load (e.g. a person) is moved or added. At some critical position, the body is on the point of TIPPING — rotating about one support while the other lifts off.

Key insight. At the tipping threshold, the reaction at the lifting support is zero. The body is then in equilibrium with one support reaction, weight, and the applied load.

Identifying which support lifts. The support FURTHEST from the load lifts as the load moves toward the body's end past the closer support.

Workflow:

Identify the support around which tipping occurs (the one closer to the moving load).
Set the reaction at the OTHER (lifting) support to zero.
Take moments about the tipping support.
Solve for the load's critical position.

Worked example. Uniform plank $AB$ of length $5$ m, mass $30$ kg, on supports at $C$ ( $1$ m from $A$ ) and $D$ ( $3$ m from $A$ ). A child of mass $25$ kg walks from $A$ toward $B$ . Find the maximum distance from $A$ the child can reach before the plank tips.

As the child moves past $D$ toward $B$ , the plank tends to tip about $D$ and lift off $C$ . Set $R_C = 0$ .
Take moments about $D$ . Plank weight $294$ N at $2.5$ m from $A$ = $0.5$ m LEFT of $D$ (anticlockwise about $D$ ). Child at $x$ m from $A$ = $(x - 3)$ m RIGHT of $D$ (clockwise about $D$ ).
Balance: $294 \cdot 0.5 = 245 \cdot (x - 3) \Rightarrow x = 3 + 0.6 = 3.6$ m.

Beyond $x = 3.6$ m, the plank tips.

Tipping threshold: $R_{\text{lifting support}} = 0$ .
Take moments about the OTHER support.
Identify which support lifts (furthest from load).
Solve for critical load position or critical load mass.

Ladder against a wall

Combination of moments (about foot) plus horizontal/vertical equilibrium.

Setup. A uniform ladder of length $L$ , mass $m$ , rests with foot $A$ on rough horizontal ground and top $B$ against a smooth vertical wall. The ladder makes angle $\theta$ with the horizontal.

Forces:

Weight $W = mg$ at midpoint (uniform).
Normal reaction $N$ from wall, horizontal (perpendicular to wall, pointing away from wall).
Normal reaction $R$ from ground, vertical (perpendicular to ground).
Friction $F$ from ground, horizontal (opposing slipping — toward the wall).

Equations:

Vertical: $R = mg$ .
Horizontal: $F = N$ .
Moments about $A$ (foot): weight at $L/2$ from foot, horizontal distance $(L/2)\cos\theta$ . Wall reaction at $B$ , height $L\sin\theta$ from foot.

$N \cdot L\sin\theta - mg \cdot (L/2)\cos\theta = 0 \Rightarrow N = \dfrac{mg\cos\theta}{2\sin\theta} = \dfrac{mg}{2\tan\theta}.$

So $F = N = mg/(2\tan\theta)$ .

Does the ladder slip? Compare $F$ with $\mu R = \mu mg$ :

$\dfrac{mg}{2\tan\theta} \le \mu mg \iff \tan\theta \ge \dfrac{1}{2\mu}.$

If $\tan\theta < 1/(2\mu)$ , the ladder slips.

Worked example. $20$ kg ladder, $4$ m long, at $60^{\circ}$ to horizontal, smooth wall, rough ground. Find $N$ and $F$ .

$R = 196$ N.
$N = 196/(2\tan 60^{\circ}) = 196/(2\sqrt{3}) \approx 56.58$ N.
$F = N \approx 56.58$ N.

Forces: weight, $N$ (from wall), $R$ (from ground), $F$ (friction at foot).
Take moments about the FOOT to eliminate $R$ and $F$ .
$N = mg\cos\theta/(2\sin\theta) = mg/(2\tan\theta)$ for uniform ladder.
Slips if $\tan\theta < 1/(2\mu)$ .

How it’s examined

Moments is the anchor question of the second half of WME01/01 — typically Q6 or Q7, worth $10$ - $14$ marks. Common scenarios: uniform plank with loads or supports, non-uniform rod (find $\bar{x}$ ), 'about to tip' problems, ladder against a wall. The 'about to tip' part is a frequent stumbling block — examiners want $R_{\text{lifting}} = 0$ explicitly stated, then a single moment equation about the OTHER support. Mark schemes credit any consistent sign convention, but require distances measured PERPENDICULAR to the line of action. A clear, large, labelled diagram is worth $2$ B-marks reliably.

Sources: Pearson Edexcel International Advanced Level Mathematics specification (2018 — current for 2026 onwards); Edexcel IAL Mathematical Formulae and Statistical Tables (2018); WME01/01 Examiner Reports — January 2023, June 2023, January 2024, June 2024. Last reviewed 2026-05-12.

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Step-by-step worked examples — Moments

Step-by-step solutions to past-paper-style questions on moments, written exactly the way a tutor would explain them at the board.

1Uniform rod on two supports — find reactions (8 marks, M1)
Core• Adapted from WME01/01 January 2024 Q7• moments, uniform rod, reactions
Question
A uniform rod $AB$ of length $4\,\text{m}$ and mass $8\,\text{kg}$ rests horizontally on two supports at $C$ and $D$ , where $AC = 1\,\text{m}$ and $BD = 1\,\text{m}$ . Find the magnitudes of the reactions at $C$ and $D$ .
Step-by-step solution
1. Step 1
  Draw the rod with $A$ at the left end, $B$ at the right. Supports at $C$ (1 m from $A$ ) and $D$ (3 m from $A$ , i.e. $1$ m from $B$ ). The rod is uniform, so weight $W = 8g = 78.4\,\text{N}$ acts at the midpoint, $2\,\text{m}$ from $A$ .
2. Step 2
  Mark normal reactions $R_C$ at $C$ (upward) and $R_D$ at $D$ (upward).
3. Step 3
  Vertical equilibrium ( $\Sigma F_y = 0$ ):
  $R_C + R_D = 8g = 78.4\,\text{N} \quad \text{(1)}$
4. Step 4
  Take moments about $C$ (anticlockwise positive). Distances: weight is $2 - 1 = 1\,\text{m}$ to the RIGHT of $C$ (clockwise moment), $R_D$ is $3 - 1 = 2\,\text{m}$ to the right of $C$ (anticlockwise moment).
  $R_D \cdot 2 - 8g \cdot 1 = 0 \Rightarrow R_D = 4g = 39.2\,\text{N}$
5. Step 5
  From (1): $R_C = 78.4 - 39.2 = 39.2\,\text{N}$ .
6. Step 6
  Symmetry check: the supports are equally placed about the midpoint, so by symmetry the reactions are equal — each is half the weight. ✓
Answer
$R_C = R_D = 39.2\,\text{N}$ .
Examiner tip
B1 for diagram with all forces and distances. M1 for vertical equilibrium. A1. M1 for taking moments about a chosen point. A1 for moment equation. M1 for solving. A1A1 for both reactions. Mark schemes accept moments about ANY point — choosing one of the supports eliminates one unknown.
2Uniform rod with asymmetric supports and load (10 marks, M1)
Extended• Adapted from WME01/01 June 2024 Q7• moments, uniform rod, load
Question
A uniform plank $AB$ of length $6\,\text{m}$ and mass $20\,\text{kg}$ rests on supports at $C$ ( $AC = 1\,\text{m}$ ) and $D$ ( $BD = 1\,\text{m}$ ). A child of mass $30\,\text{kg}$ stands on the plank at the point $E$ where $AE = 2\,\text{m}$ . Find the reactions at $C$ and $D$ .
Step-by-step solution
1. Step 1
  Diagram. $A$ on the left at $0$ m, $C$ at $1$ m, plank midpoint at $3$ m (where the plank's weight $W_p = 20g = 196\,\text{N}$ acts), $E$ at $2$ m (where the child's weight $W_c = 30g = 294\,\text{N}$ acts), $D$ at $5$ m, $B$ at $6$ m.
2. Step 2
  Vertical equilibrium:
  $R_C + R_D = 196 + 294 = 490\,\text{N} \quad \text{(1)}$
3. Step 3
  Take moments about $C$ (anticlockwise positive). Distances from $C$ : child at $2 - 1 = 1$ m right (clockwise), plank weight at $3 - 1 = 2$ m right (clockwise), $R_D$ at $5 - 1 = 4$ m right (anticlockwise).
  $R_D \cdot 4 - 294 \cdot 1 - 196 \cdot 2 = 0$
4. Step 4
  Simplify:
  $4 R_D = 294 + 392 = 686 \Rightarrow R_D = 171.5\,\text{N}$
5. Step 5
  From (1):
  $R_C = 490 - 171.5 = 318.5\,\text{N}$
6. Step 6
  Sanity check by taking moments about $D$ . Distances from $D$ : child at $5 - 2 = 3$ m left (anticlockwise), plank weight at $5 - 3 = 2$ m left (anticlockwise), $R_C$ at $5 - 1 = 4$ m left (clockwise). $-R_C \cdot 4 + 294 \cdot 3 + 196 \cdot 2 = 0 \Rightarrow R_C = (882 + 392)/4 = 318.5$ . ✓
Answer
$R_C = 318.5\,\text{N}$ and $R_D = 171.5\,\text{N}$ .
Examiner tip
B1 for diagram. M1 for vertical equilibrium. M1 for moments about $C$ (clockwise vs anticlockwise consistent). A1 for moment equation. M1 for solving. A1 for $R_D$ . A1 for $R_C$ . B1 for cross-check.
3Non-uniform rod — find centre of mass (8 marks, M1)
Extended• Adapted from WME01/01 January 2023 Q8• moments, non-uniform, centre of mass
Question
A non-uniform rod $AB$ of length $4\,\text{m}$ and mass $5\,\text{kg}$ rests horizontally on supports at $A$ and $B$ . The reaction at $A$ is $20\,\text{N}$ . Find (a) the reaction at $B$ , (b) the distance from $A$ to the centre of mass of the rod.
Step-by-step solution
1. Step 1
  Let the centre of mass be at distance $\bar{x}$ from $A$ . Weight $W = 5g = 49\,\text{N}$ acts at this point.
2. Step 2
  Vertical equilibrium:
  $R_A + R_B = 49 \Rightarrow 20 + R_B = 49 \Rightarrow R_B = 29\,\text{N}$
3. Step 3
  Take moments about $A$ (so that $R_A$ has zero lever arm and drops out):
  $R_B \cdot 4 - 49 \cdot \bar{x} = 0$
4. Step 4
  Solve.
  $\bar{x} = \dfrac{4 R_B}{49} = \dfrac{4 \cdot 29}{49} = \dfrac{116}{49} \approx 2.37\,\text{m}$
5. Step 5
  The centre of mass is $\approx 2.37\,\text{m}$ from $A$ — past the midpoint, so the rod is heavier on the $B$ side (consistent with $R_B > R_A$ ).
Answer
(a) $R_B = 29\,\text{N}$ . (b) Centre of mass $\approx 2.37\,\text{m}$ from $A$ .
Examiner tip
M1 for vertical equilibrium. A1 for $R_B$ . M1 for taking moments about $A$ . A1 for moment equation. M1 for solving. A1 for $\bar{x}$ . The choice of pivot $A$ eliminates $R_A$ — Edexcel mark schemes credit this strategic choice.
4Plank about to tip — find maximum overhang (10 marks, M1)
Extended• Adapted from WME01/01 June 2023 Q7• moments, about to tip, threshold
Question
A uniform plank $AB$ of length $5\,\text{m}$ and mass $30\,\text{kg}$ rests horizontally on two supports at $C$ ( $AC = 1\,\text{m}$ ) and $D$ ( $AD = 3\,\text{m}$ ). A child of mass $25\,\text{kg}$ walks along the plank from $A$ toward $B$ . Find the greatest distance from $A$ that the child can walk before the plank starts to tip.
Step-by-step solution
1. Step 1
  Tipping happens when one of the supports' reactions becomes zero. As the child moves toward $B$ (past $D$ ), the plank tends to rotate about $D$ and lift off $C$ — so $R_C = 0$ at the threshold.
2. Step 2
  Forces: plank weight $W_p = 30g = 294\,\text{N}$ at midpoint ( $2.5$ m from $A$ ), child weight $W_c = 25g = 245\,\text{N}$ at position $x$ from $A$ , $R_D$ upward at $D$ (3 m from $A$ ).
3. Step 3
  At the tipping threshold ( $R_C = 0$ ), take moments about $D$ (so $R_D$ drops out):
  $(W_p) \cdot (3 - 2.5) - W_c \cdot (x - 3) = 0$
4. Step 4
  The plank weight is $0.5$ m to the LEFT of $D$ (anticlockwise moment about $D$ ). The child is to the right of $D$ at $(x - 3)$ m (clockwise moment). For tipping about $D$ , the clockwise moment from the child equals the anticlockwise from the plank weight.
  $294 \cdot 0.5 = 245 \cdot (x - 3)$
5. Step 5
  Solve.
  $x - 3 = \dfrac{147}{245} = 0.6 \Rightarrow x = 3.6\,\text{m}$
6. Step 6
  So the child can walk up to $3.6$ m from $A$ . Beyond that point, $R_C$ would have to be negative (i.e. the support would have to PULL DOWN on the plank, which is impossible), so the plank tips.
Answer
The child can walk at most $3.6\,\text{m}$ from $A$ before the plank tips about $D$ .
Examiner tip
B1 for diagram and identifying the pivot ( $D$ ) and the support that lifts ( $C$ ). M1 for $R_C = 0$ at tipping. M1 for moments about $D$ . A1 for moment equation. M1 for solving. A1 for $x = 3.6$ . B1 for interpretation. Common error: students assume the plank tips about $C$ when the child moves toward $B$ — read the geometry carefully.
5Ladder problem — moments about the foot (10 marks, M1)
Challenge• Adapted from WME01/01 June 2024 Q9• moments, ladder, equilibrium
Question
A uniform ladder of length $4\,\text{m}$ and mass $20\,\text{kg}$ rests with its foot $A$ on rough horizontal ground and its top $B$ against a smooth vertical wall. The ladder makes an angle of $60^{\circ}$ with the horizontal. Find (a) the normal reaction at the wall, (b) the friction force at the ground.
Step-by-step solution
1. Step 1
  Draw the ladder leaning against the wall. Forces: weight $W = 20g = 196\,\text{N}$ vertically down at the midpoint, normal reaction $N$ horizontal (from smooth wall, pushing the ladder AWAY from the wall), normal reaction $R$ vertical (from rough ground, pushing the ladder up), friction $F$ at the ground (horizontal, opposing slipping — toward the wall).
2. Step 2
  Vertical equilibrium:
  $R = 196\,\text{N}$
3. Step 3
  Horizontal equilibrium:
  $F = N$
4. Step 4
  Take moments about $A$ (foot of the ladder, anticlockwise positive). Lever arms: • Weight acts at midpoint of ladder $2\,\text{m}$ along. Horizontal distance from $A$ : $2\cos 60^{\circ} = 1\,\text{m}$ . Moment (clockwise about $A$ ): $196 \cdot 1 = 196\,\text{N m}$ . • $N$ acts horizontally at $B$ , which is $4\sin 60^{\circ} = 2\sqrt{3}\,\text{m}$ above $A$ . Moment (anticlockwise about $A$ ): $N \cdot 2\sqrt{3}$ .
  $N \cdot 2\sqrt{3} - 196 \cdot 1 = 0$
5. Step 5
  Solve:
  $N = \dfrac{196}{2\sqrt{3}} = \dfrac{98}{\sqrt{3}} = \dfrac{98\sqrt{3}}{3} \approx 56.58\,\text{N}$
6. Step 6
  Friction: $F = N \approx 56.58\,\text{N}$ .
Answer
(a) $N = \dfrac{98\sqrt{3}}{3} \approx 56.58\,\text{N}$ . (b) Friction $F \approx 56.58\,\text{N}$ .
Examiner tip
B1 for diagram with all forces and lever arms. M1 for vertical equilibrium. A1 for $R$ . M1 for horizontal equilibrium. M1 for moments about $A$ . A1 for moment equation (with correct lever arms). M1 for solving. A1 for $N$ . A1 for $F$ . Edexcel mark schemes accept moments about $B$ or the centre as alternatives. The geometry of perpendicular distances is the bottleneck — draw the ladder accurately to scale.

Key Formulae — Moments

The formulae you need to memorise for moments on the Pearson Edexcel IAL XMA01 / YMA01 paper, with every variable defined in plain English and a note on when to use it.

Moment of a force about a point
$M = F \times d_{\perp}$
$M$
moment (N m)
$F$
magnitude of force (N)
$d_{\perp}$
perpendicular distance from the point to the line of action of the force
When to use
Whenever a force tends to rotate a rigid body about a point. State the SENSE (clockwise or anticlockwise) along with the magnitude. NOT in the IAL formula booklet — assumed knowledge.
Example
Weight $196\,\text{N}$ acting $1\,\text{m}$ horizontally from a pivot: moment $= 196\,\text{N m}$ (sense depends on orientation).
Equilibrium of a rigid body
$\Sigma F_x = 0, \quad \Sigma F_y = 0, \quad \Sigma M = 0 \text{ about any point}$
$\Sigma F$
sum of forces in two perpendicular directions
$\Sigma M$
sum of moments about any chosen point
When to use
ALL rigid-body equilibrium problems. The moment equation is usually applied about a convenient point (e.g. a support, to eliminate its reaction). NOT in the formula booklet.
Example
Rod on two supports: vertical equilibrium gives $R_C + R_D = W$ ; moments about $C$ gives $R_D$ directly.
Tipping condition
$R_{\text{lifting support}} = 0 \text{ at the point of tipping}$
$R$
reaction at the support that lifts off as the body tips
When to use
When asked for the load position / mass at which a rod or plank is on the point of tipping. Set the lifting support's reaction to zero and take moments about the OTHER support.
Example
Plank tipping about $D$ : set $R_C = 0$ ; take moments about $D$ to find the critical child position.
Centre of mass of a uniform rod
$\bar{x} = \dfrac{L}{2} \text{ (uniform rod of length } L)$
$\bar{x}$
distance of centre of mass from one end
$L$
length of rod
When to use
For a UNIFORM rod, weight acts at the midpoint. For a NON-uniform rod, the centre of mass is at some unknown $\bar{x}$ — find it by combining vertical equilibrium with a moment equation.
Example
Uniform rod $AB$ of length $4$ m: centre of mass at $2$ m from $A$ . Weight acts there.

Key Definitions and Keywords — Moments

Definitions to memorise and the exact keywords mark schemes credit for moments answers — sharpened from recent examiner reports for the 2026 Pearson Edexcel IAL XMA01 / YMA01 sitting.

Moment of a force
Examiner keyword
The turning effect of a force about a point. Magnitude = force × perpendicular distance from the point to the line of action. SI unit: newton-metre (N m). A vector in 3-D, a signed scalar (clockwise/anticlockwise) in coplanar problems.
Rigid body
Examiner keyword
A body that does not deform under applied forces. Distinct from a particle: rigid bodies have size, so moments matter. Equilibrium requires BOTH zero net force AND zero net moment.
Centre of mass
Examiner keyword
The point at which the entire weight of the body can be considered to act for the purposes of taking moments. For a uniform rod, it's at the midpoint; for a non-uniform rod, somewhere else along its length.
Pivot / fulcrum
The point about which a rigid body can rotate. Often a support, hinge or balance point. Reaction forces at the pivot have zero moment about the pivot itself.
Tipping
Examiner keyword
Loss of equilibrium by rotation about an edge or support. At the threshold, the reaction at the support being lifted is zero, and the body is on the point of rotating about the other support.

Common Mistakes and Misconceptions — Moments

The traps other students keep falling into on moments questions — taken from recent Pearson Edexcel IAL XMA01 / YMA01 examiner reports and mark schemes — and how to avoid them.

✕Using the position of the force, not the perpendicular distance to the line of action
WME01/01 examiner reports
Why it happens
Confusing 'distance' with 'lever arm'.
How to avoid it
The lever arm is the PERPENDICULAR distance from the pivot to the LINE OF ACTION of the force. For a force inclined at $\theta$ acting at point $P$ , the lever arm is the perpendicular component of $\vec{OP}$ to the force, i.e. $|OP|\sin\alpha$ where $\alpha$ is the angle between $\vec{OP}$ and the force.
✕Adding clockwise and anticlockwise moments without signing them differently
Why it happens
Treating moments as scalars.
How to avoid it
Pick a positive sense (typically anticlockwise) and assign signs consistently. Or use the convention: $\Sigma$ (anticlockwise moments) = $\Sigma$ (clockwise moments) for equilibrium.
✕Assuming centre of mass at midpoint for a non-uniform rod
Why it happens
Reflex from uniform-rod problems.
How to avoid it
If the question says 'non-uniform', place the centre of mass at an unknown $\bar{x}$ and solve for it using one moment equation plus vertical equilibrium.
✕Choosing the wrong support as the tipping pivot
WME01/01 examiner reports — flagged annually
Why it happens
Not visualising which support lifts.
How to avoid it
Imagine the load shifting in the question's direction. The support that loses contact first is the one furthest from the load. Tipping is about the OTHER support.
✕Including the pivot's reaction in the moment equation taken about that pivot
Why it happens
Habit of listing all forces.
How to avoid it
A force passing through the pivot has zero moment about the pivot (zero lever arm). Strategic pivot choice eliminates one unknown.

Practice questions

Exam-style questions with step-by-step worked solutions. Try one before checking the method.

Past paper style quiz

Get a report showing which sub-topics you've nailed and which ones still need work.

Moments — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Moments Study Notes — Edexcel IAL Mathematics M1 (WME01/01, 2018 spec — 2026 onwards)

The turning effect of a force:

M = F \times d_{\perp}

. For a rigid body in equilibrium, the resultant force is zero AND the resultant moment about any point is zero. Standard problems: uniform / non-uniform rods on supports, find centre of mass, 'about to tip' threshold, ladder against a wall.

Force

Position

Pivot

Moment

50

N down

2

m horizontally from pivot

pivot

50 \cdot 2 = 100

N m (CW or CCW depending on direction)

30

N at

60^{\circ}

to a rod

1.5

m along rod from pivot

pivot at end

30 \cdot 1.5 \sin 60^{\circ} = 38.97

N m

20

N along a rod

any point

any point ON the rod

0

(line of action passes through pivot)

Key insight. At the tipping threshold, the reaction at the lifting support is zero. The body is then in equilibrium with one support reaction, weight, and the applied load.

Identifying which support lifts. The support FURTHEST from the load lifts as the load moves toward the body's end past the closer support.

Workflow:

Identify the support around which tipping occurs (the one closer to the moving load).
Set the reaction at the OTHER (lifting) support to zero.
Take moments about the tipping support.
Solve for the load's critical position.

As the child moves past $D$ toward $B$ , the plank tends to tip about $D$ and lift off $C$ . Set $R_C = 0$ .
Take moments about $D$ . Plank weight $294$ N at $2.5$ m from $A$ = $0.5$ m LEFT of $D$ (anticlockwise about $D$ ). Child at $x$ m from $A$ = $(x - 3)$ m RIGHT of $D$ (clockwise about $D$ ).
Balance: $294 \cdot 0.5 = 245 \cdot (x - 3) \Rightarrow x = 3 + 0.6 = 3.6$ m.

Beyond $x = 3.6$ m, the plank tips.

Moments

Short Study Notes in the form of Slides

Detailed Study Notes

What you’ll learn

How it’s examined

1Uniform rod on two supports — find reactions (8 marks, M1)

2Uniform rod with asymmetric supports and load (10 marks, M1)

3Non-uniform rod — find centre of mass (8 marks, M1)

4Plank about to tip — find maximum overhang (10 marks, M1)

5Ladder problem — moments about the foot (10 marks, M1)

Moment of a force about a point

Equilibrium of a rigid body

Tipping condition

Centre of mass of a uniform rod

Moment of a force

Rigid body

Centre of mass

Pivot / fulcrum

Tipping

✕Using the position of the force, not the perpendicular distance to the line of action

✕Adding clockwise and anticlockwise moments without signing them differently

✕Assuming centre of mass at midpoint for a non-uniform rod

✕Choosing the wrong support as the tipping pivot

✕Including the pivot's reaction in the moment equation taken about that pivot

Practice questions

Past paper style quiz

Assess your understanding

Moments — frequently asked questions

Moments

Short Study Notes in the form of Slides

Detailed Study Notes

What you’ll learn

How it’s examined

1Uniform rod on two supports — find reactions (8 marks, M1)

2Uniform rod with asymmetric supports and load (10 marks, M1)

3Non-uniform rod — find centre of mass (8 marks, M1)

4Plank about to tip — find maximum overhang (10 marks, M1)

5Ladder problem — moments about the foot (10 marks, M1)

Moment of a force about a point

Equilibrium of a rigid body

Tipping condition

Centre of mass of a uniform rod

Moment of a force

Rigid body

Centre of mass

Pivot / fulcrum

Tipping

✕Using the position of the force, not the perpendicular distance to the line of action

✕Adding clockwise and anticlockwise moments without signing them differently

✕Assuming centre of mass at midpoint for a non-uniform rod

✕Choosing the wrong support as the tipping pivot

✕Including the pivot's reaction in the moment equation taken about that pivot

Practice questions

Past paper style quiz

Assess your understanding

Moments — frequently asked questions