What is a moment in the context of Physics, specifically in Mechanics and Materials?

In Physics, particularly in the study of Mechanics and Materials, a moment refers to the turning effect produced by a force acting on an object at a distance from a pivot point. It is calculated as the product of the force and the perpendicular distance from the pivot to the line of action of the force. Moments are crucial in understanding how forces cause objects to rotate, which is a key concept in the Edexcel International A Levels curriculum.

How do you calculate the moment of a force in Physics?

To calculate the moment of a force, you use the formula: Moment = Force x Distance. Here, the 'Force' is the magnitude of the force applied, and 'Distance' is the perpendicular distance from the pivot point to the line of action of the force. This calculation helps determine the rotational effect of the force on an object, which is an essential concept in Mechanics and Materials for Edexcel International A Levels.

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

The principle of moments states that for an object to be in equilibrium, the sum of the clockwise moments about any pivot point must equal the sum of the anticlockwise moments. This principle is applied in Physics to solve problems involving balanced forces and torques, ensuring that objects remain stationary or rotate at a constant rate. Understanding this principle is vital for students studying Mechanics and Materials in the Edexcel International A Levels.

What are some real-world applications of moments in Mechanics and Materials?

Moments have numerous real-world applications, such as in the design of levers, gears, and bridges. For example, levers use moments to amplify force, making it easier to lift heavy objects. In bridge engineering, moments help in analyzing the forces acting on different parts of the structure to ensure stability and safety. These applications illustrate the practical importance of moments in the field of Mechanics and Materials, which is covered in the Edexcel International A Levels Physics curriculum.

How do moments relate to torque in Physics?

In Physics, torque is essentially another term for moment, specifically referring to the rotational effect of a force. Both concepts involve the same calculation: Torque (or Moment) = Force x Distance. However, torque is often used in contexts involving rotational dynamics, such as in engines or machinery. Understanding the relationship between moments and torque is crucial for students studying Mechanics and Materials in the Edexcel International A Levels, as it helps in analyzing and solving problems involving rotational motion.

Why is "Using the length of the lever instead of the perpendicular distance when the force acts at an angle" a common mistake in Moments?

Why it happens: Students apply moment = F × d mechanically and forget the 'd' must be perpendicular to the force's line of action. How to avoid it: Always resolve: the moment of an angled force is F × d (or take the perpendicular component F). **Use sin of the angle between the force and the lever, never cos.** Sketch the perpendicular from the pivot to the line of action. Source: Edexcel IAL Physics Unit 1 examiner reports

Why is "Pairing a force with the wrong perpendicular distance in ladder/beam problems" a common mistake in Moments?

Why it happens: In a leaning-ladder problem there are both horizontal and vertical forces, and students use the along-the-ladder distance for both. How to avoid it: The moment of a **vertical** force depends on its **horizontal** distance from the pivot; the moment of a **horizontal** force depends on its **vertical** distance. Draw the perpendicular distance for each force separately before multiplying. Source: Edexcel IAL Physics Unit 1 examiner reports

Why is "Forgetting that a beam, plank or shelf has its own weight acting at its centre of gravity" a common mistake in Moments?

Why it happens: Students focus on the added loads and overlook the weight of the structure itself. How to avoid it: For any uniform beam, **include its weight acting at the midpoint** (centre of gravity). Mark it on the free-body diagram before taking moments. For a non-uniform beam, its weight acts at the stated/calculated centre of gravity, not the midpoint. Source: Edexcel IAL Physics Unit 1 examiner reports

Why is "Choosing a pivot that leaves two unknown forces in the moment equation" a common mistake in Moments?

Why it happens: Students take moments about an arbitrary point rather than strategically eliminating an unknown. How to avoid it: **Take moments about the point where an unknown force acts** (e.g. about a support, hinge or ladder foot). That force then has zero perpendicular distance and drops out, leaving a single unknown to solve for. Source: Edexcel IAL Physics Unit 1 examiner reports

Why is "Using the radius instead of the full separation when finding the moment of a couple" a common mistake in Moments?

Why it happens: Students treat the couple like a single force about the centre and use the radius of a wheel. How to avoid it: The moment of a couple uses the **perpendicular distance between the two forces** — the full diameter for a wheel, since both forces contribute equally. Couple moment = F × s where s is the whole separation. Source: Edexcel IAL Physics Unit 1 examiner reports

Why is "Writing the moment unit as N/m, or rounding intermediate values too early" a common mistake in Moments?

Why it happens: Confusion with other quantities' units, and a habit of rounding each line of working. How to avoid it: The unit of a moment is **N m (newton metre)**, never N/m. Carry full precision through the working and **round only the final answer to 3 significant figures**, with the correct unit, to secure the accuracy marks. Source: Edexcel IAL Physics Unit 1 examiner reports

Edexcel International A Levels Physics (XPH11-YPH11)

Moments

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Short Notes - Moments

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Detailed Study Notes

Detailed notes on Mechanics and Materials for Edexcel International A Levels Physics, covering key concepts, explanations, examples, and exam-focused revision points.

Moments — Pearson Edexcel International A Level Physics (XPH11/YPH11) Study Notes (Unit 1)

The moment of a force (turning effect), the principle of moments, centre of gravity and the equilibrium of extended bodies. Learn to use the perpendicular distance correctly, choose a clever pivot, and solve beam, seesaw, ladder and bracket problems with full mark-scheme discipline.

At a glance

Moment of a force = force × perpendicular distance from the pivot: $M = F \times d_{\perp}$ . Unit: N m (newton metre), never N/m.
When a force acts at an angle $\theta$ to a lever, use the perpendicular part: $M = F d\sin\theta$ — use sin, not cos.
Principle of moments: for a body in equilibrium, total clockwise moment = total anticlockwise moment, about any point.
Choose the pivot through an unknown force to eliminate it from the equation — a huge time-saver in beam and ladder problems.
Centre of gravity = the single point where the whole weight acts. Uniform body → geometric centre; non-uniform → offset towards the heavier part.
An extended body in equilibrium needs two conditions: resultant force $= 0$ and resultant moment $= 0$ .
The weight of a beam/shelf/ladder always acts at its centre of gravity — don't forget it.
A couple is two equal, opposite, parallel forces; its moment $= F \times s$ (the full separation) with zero resultant force.

What you’ll learn

Mapped to the Pearson Edexcel International A Levels XPH11-YPH11 syllabus (2018-onwards).

Unit 1 (WPH11) — Define and calculate the moment of a force as force × perpendicular distance from the pivot ( $M = Fd_{\perp}$ ), in N m.
Unit 1 — State and apply the principle of moments to a body in equilibrium (sum of clockwise moments = sum of anticlockwise moments about any point).
Unit 1 — Resolve a force acting at an angle to find the perpendicular distance / perpendicular component when calculating a moment.
Unit 1 — Understand centre of gravity as the point at which the whole weight of a body acts, and locate it for uniform and non-uniform bodies.
Unit 1 — Apply the two conditions for equilibrium of an extended body (zero resultant force and zero resultant moment) to beams, seesaws, shelf brackets and ladders, finding unknown reaction forces and tensions.
Unit 1 (A* extension) — Recognise a couple and calculate its moment as one force × the perpendicular distance between the two forces.

The moment of a force — the turning effect

Force × perpendicular distance from the pivot. Get the distance right.

The moment of a force is its turning effect about a point (the pivot). It is defined as:

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

where $F$ is the force in newtons and $d_{\perp}$ is the perpendicular distance from the pivot to the line of action of the force, in metres. The unit of a moment is the newton metre (N m) — write it as N m, never N/m.

The word perpendicular is everything. It is not the length of the lever; it is the shortest (90°) distance from the pivot to the line along which the force acts. When the force is at right angles to the lever, these are the same — but when the force is at an angle they are not, and that is where marks are lost.

Two equivalent ways to handle an angled force of magnitude $F$ applied at a distance $d$ from the pivot, at angle $\theta$ to the lever:

Use the perpendicular distance $d_{\perp} = d\sin\theta$ , giving $M = F d\sin\theta$ .
Or take the perpendicular component of the force, $F\sin\theta$ , acting at the full distance $d$ — the same result.

Either way the factor is $\sin\theta$ (where $\theta$ is the angle between the force and the lever). This is why a force applied at 90° gives the maximum moment for a given distance, and why a longer spanner (larger $d$ ) undoes a stiffer nut for the same effort.

Left: a force at 90° to the lever gives the full moment $M = Fd$. Right: a force at angle θ to the lever gives $M = Fd\sin\theta$ — only the component perpendicular to the lever turns it. The maximum moment occurs at θ = 90°.

$M = F \times d_{\perp}$ ; unit N m (not N/m).
$d_{\perp}$ is the perpendicular distance from the pivot to the line of action.
Angled force: $M = Fd\sin\theta$ — use sin of the angle between force and lever.
Maximum moment when the force acts at 90° to the lever.

See the full worked example for moments →

The principle of moments and equilibrium

Clockwise moments = anticlockwise moments — and the resultant force is zero too.

The principle of moments states that, for a body in equilibrium, the sum of the clockwise moments about any point equals the sum of the anticlockwise moments about that same point.

For an extended body (a beam, not a point), equilibrium needs two conditions, and both must be checked:

Resultant force = 0 — the forces balance (e.g. total upward force = total downward force).
Resultant moment = 0 — the turning effects balance (the principle of moments).

A point object only needs condition 1; an extended body that can rotate needs both.

The smart move: choosing the pivot. You may take moments about any point. The trick is to choose the point through which an unknown force acts. That force then has zero perpendicular distance, so its moment is zero and it disappears from the equation — leaving a single unknown to solve for. For a beam on two supports, taking moments about one support eliminates that support's reaction; for a ladder, taking moments about the foot eliminates the (messy) forces there.

A uniform seesaw is in equilibrium because the anticlockwise moment (320 × 1.5) equals the clockwise moment (400 × 1.2). The uniform beam's own weight acts at the central pivot, so it produces no moment about the pivot.

Principle of moments: $\sum M_{cw} = \sum M_{acw}$ about any point.
Extended body in equilibrium: resultant force = 0 AND resultant moment = 0.
Take moments about an unknown force to eliminate it.
Always sense-check: heavier object sits closer to the pivot to balance.

See the full worked example for moments →

Centre of gravity

The single point where the whole weight acts — central if uniform, offset if not.

The centre of gravity of an object is the single point at which the entire weight of the object may be considered to act. Whenever you draw the weight of a beam, shelf, plank or ladder, it acts as one downward arrow ( $W = mg$ ) at the centre of gravity — not spread along the object.

For a uniform body of regular shape, the centre of gravity is at its geometric centre (e.g. the midpoint of a uniform beam).
For a non-uniform body, the mass is unevenly distributed, so the centre of gravity is offset towards the heavier part. You can locate it by taking moments (as in the non-uniform plank and bridge worked examples).

In a uniform gravitational field the centre of mass and the centre of gravity coincide, so the terms are often used interchangeably at this level.

Finding the centre of gravity of an irregular lamina (flat sheet): hang it freely from a point and let it settle — the centre of gravity hangs directly below the suspension point, so a plumb line marks a vertical through it. Repeat from a second point; the centre of gravity is where the two lines cross.

Stability. An object topples when the line of action of its weight (the vertical line through the centre of gravity) falls outside its base of support. So an object is more stable when it has a low centre of gravity and a wide base — which is why racing cars are low and wide, and a tall narrow box tips easily.

An object is stable while the vertical line through its centre of gravity stays inside its base of support (left). Tilt it until that line falls outside the base and it topples (right). A low centre of gravity and a wide base make toppling harder.

Centre of gravity = single point where the whole weight acts.
Uniform body → geometric centre; non-uniform → offset to the heavier side.
Locate an irregular lamina's CofG by suspending it from two points (plumb lines cross).
Stability: low CofG + wide base; topples when the weight line leaves the base.

See the full worked example for moments →

Equilibrium of extended bodies — beams, brackets and ladders

A reliable method for reaction-force and tension problems.

Most exam moment problems are about an extended body held still by several forces, and you must find an unknown reaction or tension. Use this method every time:

Draw a free-body diagram. Mark every force with magnitude and direction: loads, the weight of the body at its centre of gravity, and the reactions/tensions at supports or hinges.
Take moments about a clever point — usually where an unknown force acts, so it drops out. State your pivot.
Write moments = moments. For each force use its perpendicular distance from the pivot. Remember: a vertical force pairs with a horizontal distance, and a horizontal force pairs with a vertical distance.
Resolve forces (e.g. vertically: total up = total down) to find any remaining unknown.
Answer to 3 s.f. with units, and sense-check (the support nearer a load/CofG carries more).

For an angled support (a cable at angle $\theta$ to a beam), only the perpendicular component $T\sin\theta$ produces a moment about the hinge — never use the full $T$ .

Free-body diagram of a hinged beam held by a cable. Taking moments about the hinge eliminates the hinge reaction. The beam's weight (at the midpoint) and the load (at the end) give clockwise moments; the cable's perpendicular component $T\sin\theta$ gives the balancing anticlockwise moment.

For a ladder against a smooth wall, the wall (smooth → no friction) pushes horizontally only; the ground gives an upward normal reaction and a horizontal friction force. Taking moments about the foot, the weight (vertical) uses the horizontal distance to the centre of gravity, while the wall's push (horizontal) uses the vertical height of the top of the ladder.

Method: free-body diagram → moments about an unknown → resolve forces → 3 s.f.
Vertical force × horizontal distance; horizontal force × vertical distance.
Angled cable/support: only $T\sin\theta$ gives a moment about the hinge.
Smooth wall → horizontal force only; rough ground → normal + friction.

See the full worked example for moments →

Couples (A* extension)

Two equal, opposite forces — pure rotation, zero resultant force.

A couple is a pair of equal, opposite, parallel forces whose lines of action do not coincide. Think of two hands turning a steering wheel, or two fingers undoing a tap.

A couple has a special property: because the two forces are equal and opposite, the resultant force is zero — a couple produces pure rotation with no tendency to move the body bodily. Its turning effect is its moment (torque):

$M = F \times s$

where $F$ is the magnitude of one of the forces and $s$ is the perpendicular distance between the two lines of action. For a wheel, $s$ is the full diameter (both forces contribute), not the radius — a common slip.

A couple is the reason a single force can never, by itself, produce a moment without also exerting a resultant force, whereas two balanced forces can. In equilibrium of an extended body, any couple acting on it must be balanced by an equal and opposite couple.

Couple = two equal, opposite, parallel forces (lines of action don't coincide).
Resultant force of a couple = 0 → pure rotation.
Moment of a couple $M = F \times s$ , with $s$ the full separation of the forces.
For a wheel, use the diameter, not the radius.

See the full worked example for moments →

Quick recap

Moment of a force $M = F \times d_{\perp}$ (N m); $d_{\perp}$ is the perpendicular distance to the line of action.
Angled force: $M = Fd\sin\theta$ — use sin of the angle between force and lever.
Principle of moments: for equilibrium, $\sum M_{cw} = \sum M_{acw}$ about any point.
Extended-body equilibrium needs BOTH zero resultant force and zero resultant moment.
Centre of gravity is the single point where the whole weight acts — central if uniform, offset if not; always include a beam's own weight.
Take moments about an unknown force to eliminate it; pair vertical forces with horizontal distances and vice versa.
A couple's moment is $F \times s$ (full separation) with zero resultant force.

Memorise this

Verbatim phrases and definitions Edexcel mark schemes credit.

Moment of a force = force × perpendicular distance from the pivot ( $M = F \times d_{\perp}$ ); unit N m.
For an angled force, $M = Fd\sin\theta$ (θ between force and lever) — sin, not cos.
Principle of moments: for a body in equilibrium, sum of clockwise moments = sum of anticlockwise moments about any point.
An extended body in equilibrium has zero resultant force AND zero resultant moment.
Centre of gravity = the single point at which the entire weight of the body acts.
Weight = mg acts vertically downwards through the centre of gravity (uniform body → geometric centre).
A couple = two equal, opposite, parallel forces; moment of a couple = F × perpendicular distance between them, with zero resultant force.

How it’s examined

Moments is core Unit 1 (Mechanics and Materials, WPH11) content. Common formats: define the moment of a force or state the principle of moments (1–2 marks, AO1, where 'perpendicular' is the credit-bearing word); single-force moment calculations including angled forces needing sin θ (2–3 marks); seesaw/beam balance and reaction-force problems where you take moments about a support and then resolve forces (4–5 marks, AO2); locate the centre of gravity of a non-uniform body by moments; and a levels-marked extended question on the equilibrium of a ladder or bracket (6 marks) requiring correct free-body reasoning and perpendicular distances. Show full M/A working: equation, substitution with units, answer to 3 s.f. with the unit N m. In 'show that' parts, work through to the value and give one extra significant figure. The most penalised errors are using cos instead of sin for angled forces, forgetting the body's own weight, and pairing a force with the wrong perpendicular distance.

Sources: Pearson Edexcel International Advanced Level Physics (YPH11/XPH11) specification (Issue 3, July 2021); Edexcel IAL Physics Unit 1 examiner reports; Edexcel IAL Physics Unit 1 sample assessment materials and mark schemes. Last reviewed 2026-06-26.

Take this whole topic with you

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.

Question type:

1Calculating the moment of a force (2 marks)
Getting startedDirect calculation• moment, AO2, Unit 1
Question
A spanner is used to undo a nut. A force of $48\ \text{N}$ is applied at right angles to the spanner, $0.25\ \text{m}$ from the centre of the nut. Calculate the moment of the force about the nut. (2 marks)
Step-by-step solution
1. Step 1
  The moment of a force is the force multiplied by the perpendicular distance from the pivot to the line of action of the force. Here the force is already perpendicular to the spanner.
  $\text{moment} = F \times d$
2. Step 2
  Substitute the values with units and evaluate.
  $\text{moment} = 48 \times 0.25 = 12\ \text{N m}$
Answer
Moment $= 12\ \text{N m}$ .
Examiner tip
Mark scheme: (1) correct substitution into moment = F × d; (2) answer 12 N m with the unit. The unit is N m (newton metre), NOT N/m. Omitting or mis-writing the unit loses the accuracy mark.
2Finding the force from a required moment (2 marks)
Getting startedDirect calculation• moment, rearrange, Unit 1
Question
A door handle is $0.80\ \text{m}$ from the hinge. A moment of $6.0\ \text{N m}$ is needed to push the door open. Calculate the minimum force that must be applied at right angles to the door at the handle. (2 marks)
Step-by-step solution
1. Step 1
  Rearrange moment = F × d to make the force the subject.
  $F = \frac{\text{moment}}{d}$
2. Step 2
  Substitute and evaluate. Applying the force perpendicular at the greatest distance (the handle) needs the smallest force.
  $F = \frac{6.0}{0.80} = 7.5\ \text{N}$
Answer
Minimum force $= 7.5\ \text{N}$ .
Examiner tip
Mark scheme: (1) correct rearrangement/substitution; (2) answer 7.5 N. The 'minimum' force is obtained at the maximum perpendicular distance and acting at 90° — a common 'explain why the handle is far from the hinge' follow-up.
3Balancing a seesaw using the principle of moments (3 marks)
Getting startedWord problem• principle of moments, AO2, Unit 1
Question
A uniform seesaw is balanced on a central pivot. A child of weight $320\ \text{N}$ sits $1.5\ \text{m}$ to the left of the pivot. How far to the right of the pivot must a child of weight $400\ \text{N}$ sit so that the seesaw balances? (3 marks)
Step-by-step solution
1. Step 1
  For equilibrium, the principle of moments says the total clockwise moment equals the total anticlockwise moment about the pivot. The seesaw is uniform and pivoted at its centre, so its weight produces no moment about the pivot.
  $\text{clockwise moment} = \text{anticlockwise moment}$
2. Step 2
  Take moments about the pivot. The left child gives an anticlockwise moment; the right child a clockwise moment.
  $400 \times d = 320 \times 1.5$
3. Step 3
  Rearrange for the distance d of the heavier child from the pivot.
  $d = \frac{320 \times 1.5}{400} = 1.2\ \text{m}$
Answer
The $400\ \text{N}$ child must sit $1.2\ \text{m}$ from the pivot.
Examiner tip
Mark scheme: (1) principle of moments stated/used (sum CW = sum ACW); (2) correct moments 320 × 1.5 and 400 × d; (3) d = 1.2 m. Note the heavier child sits closer to the pivot — a good sense-check.
4Moment of a force applied at an angle (3 marks)
Building confidenceMulti-step problem• moment, perpendicular distance, AO2, Unit 1
Question
A force of $60\ \text{N}$ is applied to the end of a $0.40\ \text{m}$ lever. The force acts at $35°$ to the lever. Calculate the moment of the force about the pivot at the other end of the lever. (3 marks)
Step-by-step solution
1. Step 1
  The moment uses the perpendicular distance from the pivot to the line of action of the force — not the length of the lever. Because the force is at an angle, resolve the geometry: the perpendicular distance is $d\sin\theta$ , or equivalently take the perpendicular component of the force $F\sin\theta$ acting at distance $d$ .
  $\text{moment} = F \times d\sin\theta$
2. Step 2
  Substitute $F = 60\ \text{N}$ , $d = 0.40\ \text{m}$ , $\theta = 35°$ . Keep full precision; $\sin 35° = 0.5736$ .
  $\text{moment} = 60 \times 0.40 \times \sin 35°$
3. Step 3
  Evaluate to 3 significant figures.
  $\text{moment} = 60 \times 0.40 \times 0.5736 = 13.8\ \text{N m}$
Answer
Moment $= 13.8\ \text{N m}$ (3 s.f.).
Examiner tip
Mark scheme: (1) realising the perpendicular distance/component is needed (use of sin θ); (2) correct substitution 60 × 0.40 × sin 35°; (3) 13.8 N m. Using cos 35° (a frequent error) gives 19.7 N m and scores zero for that step. The angle is measured between the force and the lever, so the perpendicular component is F sin θ.
5Reaction forces on a loaded shelf (4 marks)
Building confidenceMulti-step problem• Adapted from Edexcel IAL Physics Unit 1• equilibrium, principle of moments, AO2, Unit 1
Question
A uniform horizontal shelf of weight $24\ \text{N}$ and length $1.2\ \text{m}$ rests on two supports, A at the left-hand end and B at $0.90\ \text{m}$ from A. A book of weight $9.0\ \text{N}$ is placed $0.30\ \text{m}$ from A. Calculate the upward force the support at B exerts on the shelf. Take $g = 9.81\ \text{m s}^{-2}$ (weights are already given). (4 marks)
Step-by-step solution
1. Step 1
  The shelf is in equilibrium, so the resultant moment about any point is zero. Take moments about A to eliminate the unknown reaction at A (its line of action passes through A, so it has no moment about A).
  $\sum \text{clockwise moment} = \sum \text{anticlockwise moment about A}$
2. Step 2
  The shelf is uniform, so its weight $24\ \text{N}$ acts at its centre of gravity, $0.60\ \text{m}$ from A. The book $9.0\ \text{N}$ acts at $0.30\ \text{m}$ . These give clockwise moments; the reaction $R_B$ at $0.90\ \text{m}$ gives the anticlockwise moment.
  $R_B \times 0.90 = (24 \times 0.60) + (9.0 \times 0.30)$
3. Step 3
  Evaluate the right-hand side.
  $R_B \times 0.90 = 14.4 + 2.7 = 17.1$
4. Step 4
  Solve for the reaction at B.
  $R_B = \frac{17.1}{0.90} = 19.0\ \text{N}$
Answer
The support at B exerts an upward force of $19.0\ \text{N}$ (3 s.f.).
Examiner tip
Mark scheme: (1) take moments about A (so R_A is eliminated); (2) weight of shelf acts at the centre, 0.60 m from A; (3) correct moment equation R_B × 0.90 = 24 × 0.60 + 9.0 × 0.30; (4) R_B = 19.0 N. Forgetting the shelf's own weight (the most common error) gives 3.0 N and loses 3 marks.
6Show that — second reaction by resolving forces (4 marks)
Building confidenceShow that / prove• equilibrium, resultant force, AO2, Unit 1
Question
Using the shelf in the previous example (shelf weight $24\ \text{N}$ , book $9.0\ \text{N}$ , reaction at B $= 19.0\ \text{N}$ ), show that the upward force exerted by the support at A is approximately $14\ \text{N}$ . (4 marks)
Step-by-step solution
1. Step 1
  For an extended body in equilibrium, as well as the moments balancing, the resultant force is zero. Resolve vertically: total upward forces equal total downward forces.
  $R_A + R_B = W_{shelf} + W_{book}$
2. Step 2
  Substitute the known weights and the reaction at B.
  $R_A + 19.0 = 24 + 9.0$
3. Step 3
  Rearrange to make the reaction at A the subject.
  $R_A = 33 - 19.0$
4. Step 4
  Evaluate. Giving the result to one more significant figure than the target ( $14.0\ \text{N}$ ) confirms it rounds to the stated $\approx 14\ \text{N}$ .
  $R_A = 14.0\ \text{N} \approx 14\ \text{N}$
Answer
$R_A = 14.0\ \text{N} \approx 14\ \text{N}$ , as required.
Examiner tip
Mark scheme: (1) state/use resultant force = 0 vertically; (2) total weight 24 + 9.0 = 33 N; (3) subtract R_B; (4) reach 14.0 N and comment that it agrees with the given value. In a 'show that' you must work THROUGH to the value and give one extra significant figure — never just quote 14 N.
7Centre of gravity of a non-uniform plank (4 marks)
Building confidenceMulti-step problem• centre of gravity, principle of moments, AO2, Unit 1
Question
A non-uniform plank of length $2.0\ \text{m}$ and weight $50\ \text{N}$ is pivoted at its midpoint. It balances horizontally when a $15\ \text{N}$ weight is hung from one end. Calculate the distance of the plank's centre of gravity from the pivot. (4 marks)
Step-by-step solution
1. Step 1
  Because the plank is non-uniform, its weight does NOT act at the geometric centre (the pivot). Let its centre of gravity be a distance $x$ from the pivot, on the opposite side to the hung weight.
  $\text{clockwise moment} = \text{anticlockwise moment}$
2. Step 2
  The $15\ \text{N}$ weight hangs at the end, $1.0\ \text{m}$ from the central pivot, giving one moment. The plank's weight $50\ \text{N}$ acts at distance $x$ on the other side, balancing it.
  $50 \times x = 15 \times 1.0$
3. Step 3
  Solve for x.
  $x = \frac{15 \times 1.0}{50} = 0.30\ \text{m}$
4. Step 4
  The centre of gravity is therefore $0.30\ \text{m}$ from the midpoint, towards the end OPPOSITE the hung weight (the heavier end of the plank).
Answer
The centre of gravity is $0.30\ \text{m}$ from the pivot, towards the end away from the hung weight.
Examiner tip
Mark scheme: (1) recognise weight of a non-uniform plank does not act at the centre; (2) take moments about the pivot, weight acts at distance x; (3) 50x = 15 × 1.0 → x = 0.30 m; (4) state the direction (towards the heavier end). The key insight is that the plank's own weight provides a balancing moment because its CofG is offset from the pivot.
8Ladder against a wall — taking moments about the foot (5 marks)
StretchMulti-step problem• Adapted from Edexcel IAL Physics Unit 1• equilibrium, ladder, moments, AO2, Unit 1
Question
A uniform ladder of weight $180\ \text{N}$ and length $4.0\ \text{m}$ leans against a smooth (frictionless) vertical wall. The foot of the ladder is on rough ground, $2.4\ \text{m}$ horizontally from the base of the wall. The ladder's centre of gravity is at its midpoint. Calculate the horizontal force $N$ the wall exerts on the top of the ladder. (5 marks)
Step-by-step solution
1. Step 1
  The wall is smooth, so it exerts only a horizontal (normal) force $N$ on the top of the ladder. Take moments about the foot of the ladder to eliminate the (unknown) forces acting there.
  $\sum \text{clockwise} = \sum \text{anticlockwise about the foot}$
2. Step 2
  Find the vertical height of the top of the ladder using Pythagoras (so we have the perpendicular distance for the horizontal wall force).
  $h = \sqrt{4.0^2 - 2.4^2} = \sqrt{16 - 5.76} = 3.2\ \text{m}$
3. Step 3
  The weight (vertical, $180\ \text{N}$ ) acts at the midpoint; its horizontal (perpendicular) distance from the foot is half the base $= 1.2\ \text{m}$ , giving the clockwise moment. The wall force $N$ (horizontal) acts at the top; its perpendicular distance from the foot is the height $h = 3.2\ \text{m}$ , giving the anticlockwise moment.
  $N \times 3.2 = 180 \times 1.2$
4. Step 4
  Evaluate the right-hand side.
  $N \times 3.2 = 216$
5. Step 5
  Solve for the horizontal force from the wall.
  $N = \frac{216}{3.2} = 67.5\ \text{N}$
Answer
The wall pushes horizontally on the ladder with $N = 67.5\ \text{N}$ .
Examiner tip
Mark scheme: (1) moments taken about the foot; (2) height 3.2 m by Pythagoras; (3) perpendicular distances correct — weight uses horizontal distance 1.2 m, wall force uses vertical distance 3.2 m; (4)/(5) N = 67.5 N. The classic trap is pairing each force with the wrong perpendicular distance: the moment of a horizontal force depends on the VERTICAL distance, and of a vertical force on the HORIZONTAL distance.
9Hinged beam supported by a cable (5 marks)
StretchMulti-step problem• equilibrium, tension, moments, AO2, Unit 1
Question
A uniform horizontal beam of weight $120\ \text{N}$ and length $3.0\ \text{m}$ is hinged to a wall at one end. The far end is held up by a cable attached to the wall above the hinge; the cable makes an angle of $30°$ with the beam. A load of $200\ \text{N}$ hangs from the far end of the beam. Calculate the tension $T$ in the cable. (5 marks)
Step-by-step solution
1. Step 1
  Take moments about the hinge to eliminate the unknown hinge reaction. The beam is in equilibrium, so clockwise moments balance anticlockwise moments.
  $\sum \text{clockwise} = \sum \text{anticlockwise about the hinge}$
2. Step 2
  Clockwise (about the hinge): the beam's weight $120\ \text{N}$ acts at its midpoint $1.5\ \text{m}$ from the hinge; the load $200\ \text{N}$ acts at the far end $3.0\ \text{m}$ from the hinge.
  $\text{clockwise moment} = (120 \times 1.5) + (200 \times 3.0) = 180 + 600 = 780\ \text{N m}$
3. Step 3
  Anticlockwise: only the cable tension lifts the beam. The cable acts at the far end ( $3.0\ \text{m}$ from the hinge), but at $30°$ to the beam, so use the perpendicular component $T\sin 30°$ .
  $\text{anticlockwise moment} = T\sin 30° \times 3.0$
4. Step 4
  Set the moments equal.
  $T\sin 30° \times 3.0 = 780$
5. Step 5
  Solve for T, using $\sin 30° = 0.500$ .
  $T = \frac{780}{3.0 \times 0.500} = \frac{780}{1.5} = 520\ \text{N}$
Answer
The tension in the cable is $T = 520\ \text{N}$ .
Examiner tip
Mark scheme: (1) moments about the hinge (eliminates hinge force); (2) clockwise moment 120 × 1.5 + 200 × 3.0 = 780 N m; (3) use the perpendicular component T sin 30° at 3.0 m; (4)/(5) T = 520 N. Using T (not T sin 30°), or cos 30°, is the dominant error here — only the component perpendicular to the beam produces a moment about the hinge.
10Moment of a couple on a steering wheel (3 marks)
StretchDirect calculation• couple, torque, AO2, Unit 1
Question
A driver turns a steering wheel of diameter $0.36\ \text{m}$ by applying two equal and opposite forces of $8.0\ \text{N}$ at opposite ends of a diameter (a couple). Calculate the moment (torque) of the couple. (3 marks)
Step-by-step solution
1. Step 1
  A couple is a pair of equal, opposite, parallel forces whose lines of action do not coincide. Its moment (torque) is one force multiplied by the perpendicular distance between the two forces — here the diameter of the wheel.
  $\text{moment of couple} = F \times s$
2. Step 2
  The perpendicular separation of the forces is the full diameter, $s = 0.36\ \text{m}$ (not the radius — both forces contribute).
  $s = 0.36\ \text{m}$
3. Step 3
  Substitute and evaluate.
  $\text{moment} = 8.0 \times 0.36 = 2.88\ \text{N m}$
Answer
Moment of the couple $= 2.88\ \text{N m}$ .
Examiner tip
Mark scheme: (1) moment of a couple = one force × perpendicular distance between the forces; (2) use the diameter 0.36 m (NOT the radius); (3) 2.88 N m. A couple produces a turning effect but no resultant force. Halving by using the radius is the standard slip and gives 1.44 N m.

Model Answers — Moments

High-scoring sample answers for moments on the Cambridge IGCSE paper, with examiner-style notes mapping each response to the mark scheme and assessment objectives.

Question 1
2 marks
Q (2 marks). Define the moment of a force about a point.
Model answer
The moment of a force about a point is the force multiplied by the perpendicular distance from the point (pivot) to the line of action of the force. In symbols, $M = F \times d_{\perp}$ , and it is measured in newton metres (N m). It is a measure of the turning effect of the force about that point.
Why this scores
Why this scores 2/2. (1) force × distance; (2) the distance must be the perpendicular distance from the pivot to the line of action of the force. Omitting 'perpendicular' is the single most common reason this 2-mark definition scores only 1.
Question 2
3 marks
Q (3 marks). State the principle of moments and explain what is meant by the centre of gravity of an object.
Model answer
The principle of moments states that, for a body in equilibrium, the sum of the clockwise moments about any point equals the sum of the anticlockwise moments about that same point.

The centre of gravity of an object is the single point at which the entire weight of the object may be considered to act. For a uniform object of regular shape, the centre of gravity is at its geometric centre.
Why this scores
Why this scores 3/3. (1) sum of clockwise moments = sum of anticlockwise moments; (2) 'for a body in equilibrium' / 'about any point'; (3) centre of gravity = single point where the whole weight acts. Saying 'where the object balances' alone is too weak for the CofG mark.
Question 3
2 marks
Q (2 marks). A mechanic cannot undo a tight bolt using a short spanner, but succeeds after fitting a longer spanner. Using ideas about moments, explain why the longer spanner works.
Model answer
The turning effect (moment) of a force equals the force multiplied by the perpendicular distance from the pivot, $M = F \times d$ . The longer spanner increases the perpendicular distance $d$ from the bolt at which the force is applied. Therefore, for the same applied force, a larger moment is produced, which is enough to overcome the resistance and turn the bolt.
Why this scores
Why this scores 2/2. (1) moment = force × (perpendicular) distance; (2) greater distance → greater moment for the same force. A frequent weak answer says 'more leverage' without referencing distance and moment — that does not score.
Question 4
4 marks
Q (4 marks). A uniform plank is balanced on a pivot at its centre. A $25\ \text{N}$ weight is placed $0.80\ \text{m}$ to the left of the pivot, and a $40\ \text{N}$ weight is placed $0.30\ \text{m}$ to the right of the pivot. (a) Show that the plank is not balanced. (b) State and calculate where a single $20\ \text{N}$ weight could be placed to restore balance.
Model answer
(a) Take moments about the central pivot. The plank is uniform and pivoted at its centre, so its own weight gives no moment. Anticlockwise moment (left weight) $= 25 \times 0.80 = 20\ \text{N m}$ . Clockwise moment (right weight) $= 40 \times 0.30 = 12\ \text{N m}$ . Since $20\ \text{N m} \neq 12\ \text{N m}$ , the moments are unequal, so the plank is not balanced — there is a net anticlockwise moment of $8\ \text{N m}$ .

(b) The net moment is anticlockwise, so the extra $20\ \text{N}$ weight must be placed on the right (clockwise) side to supply the missing $8\ \text{N m}$ . Its distance $d$ from the pivot satisfies $20 \times d = 8 \Rightarrow d = \dfrac{8}{20} = 0.40\ \text{m}$ . So the $20\ \text{N}$ weight is placed $0.40\ \text{m}$ to the right of the pivot.
Why this scores
Why this scores 4/4. (a) 2 marks: both moments calculated (20 N m and 12 N m) and a clear statement they are unequal. (b) 2 marks: identify the correct side (right) AND distance 0.40 m. Stating the side without justification, or placing the weight on the wrong side, loses a mark.
Question 5
5 marks
Q (5 marks). A uniform beam of weight $300\ \text{N}$ and length $4.0\ \text{m}$ rests horizontally on two supports, P at the left end and Q at the right end. A load of $600\ \text{N}$ is placed $1.0\ \text{m}$ from P. Calculate the upward force exerted by each support.
Model answer
Take moments about P to eliminate the reaction $R_P$ at that end (it has no moment about P). The beam is uniform, so its weight $300\ \text{N}$ acts at its centre, $2.0\ \text{m}$ from P. The load $600\ \text{N}$ acts $1.0\ \text{m}$ from P, and the reaction $R_Q$ acts at the right end, $4.0\ \text{m}$ from P.

$R_Q \times 4.0 = (300 \times 2.0) + (600 \times 1.0)$ $R_Q \times 4.0 = 600 + 600 = 1200$ $R_Q = \dfrac{1200}{4.0} = 300\ \text{N}$ .

Now use the condition that the resultant vertical force is zero: $R_P + R_Q = 300 + 600 = 900\ \text{N}$ $R_P = 900 - 300 = 600\ \text{N}$ .

So the support at P exerts $600\ \text{N}$ and the support at Q exerts $300\ \text{N}$ , both upwards.
Why this scores
Why this scores 5/5. (1) take moments about P (R_P eliminated); (2) beam weight acts at the centre 2.0 m from P; (3) correct moment equation giving R_Q = 300 N; (4) total upward = total downward (900 N); (5) R_P = 600 N. The load is nearer P, so P carries more — a useful sense-check. Forgetting the beam's own weight is the dominant lost-marks error.
Question 6
4 marks
Q (4 marks). A force of $50\ \text{N}$ is applied at the end of a rigid rod of length $0.60\ \text{m}$ . The rod is pivoted at the other end. The force acts at $40°$ to the rod. (a) Calculate the moment of the force about the pivot. (b) State, with a reason, the angle at which the force should act to give the maximum possible moment.
Model answer
(a) Only the component of the force perpendicular to the rod produces a moment, so use $M = F \times d \times \sin\theta$ , where $\theta$ is the angle between the force and the rod. $M = 50 \times 0.60 \times \sin 40° = 50 \times 0.60 \times 0.6428 = 19.3\ \text{N m}$ (3 s.f.).

(b) The moment is greatest when $\sin\theta$ is greatest, i.e. when $\theta = 90°$ (the force acts perpendicular to the rod). Then the full force acts at the full distance and $M = 50 \times 0.60 = 30\ \text{N m}$ , the maximum possible.
Why this scores
Why this scores 4/4. (a) 2 marks: use of the perpendicular component (sin 40°) and the value 19.3 N m. (b) 2 marks: angle of 90° AND the reason that the perpendicular distance / component is then maximum (sin 90° = 1). Using cos 40° in part (a) gives 23.0 N m and loses both marks there.
Question 7
6 marks
Q (6 marks). A non-uniform footbridge of length $6.0\ \text{m}$ and weight $4800\ \text{N}$ rests on two pillars, X at the left end and Y at the right end. The reaction at X is measured to be $3000\ \text{N}$ . (a) Calculate the reaction at Y. (b) Determine the distance of the bridge's centre of gravity from pillar X. (c) Explain why a non-uniform bridge can have unequal reactions at its two supports even when no external load is on it.
Model answer
(a) With no external load, the two reactions support the bridge's weight. Resultant vertical force is zero: $R_X + R_Y = 4800\ \text{N}$ $R_Y = 4800 - 3000 = 1800\ \text{N}$ .

(b) Take moments about X. The weight $4800\ \text{N}$ acts at the centre of gravity, a distance $x$ from X; the reaction $R_Y = 1800\ \text{N}$ acts at Y, $6.0\ \text{m}$ from X (the reaction $R_X$ has no moment about X). $4800 \times x = R_Y \times 6.0 = 1800 \times 6.0 = 10\,800$ $x = \dfrac{10\,800}{4800} = 2.25\ \text{m}$ . The centre of gravity is $2.25\ \text{m}$ from X.

(c) In a non-uniform bridge the mass is not evenly distributed, so the centre of gravity is not at the midpoint. The weight therefore acts closer to one pillar than the other. Taking moments shows the nearer pillar carries the larger share of the weight, so the support nearer the centre of gravity exerts the greater reaction — here X (at $2.25\ \text{m}$ , nearer the CofG than Y at $3.75\ \text{m}$ ) carries $3000\ \text{N}$ against Y's $1800\ \text{N}$ .
Why this scores
Why this scores 6/6. (a) 2 marks: resultant force = 0 → R_Y = 1800 N. (b) 2 marks: moments about X with weight at distance x → x = 2.25 m. (c) 2 marks: non-uniform → CofG off-centre → support nearer the CofG takes more weight, consistent with the numbers. Levels-style credit rewards linking the calculation back to the physics rather than restating it.
Question 8
6 marks
Q (6 marks). A uniform ladder leans against a smooth vertical wall with its foot on rough horizontal ground. By considering the forces and moments acting on the ladder, explain how the ladder is kept in equilibrium and why friction at the foot is essential. Refer to the perpendicular distances involved. (This question is marked using levels.)
Model answer
For the ladder to be in equilibrium, two conditions must both be satisfied: the resultant force must be zero and the resultant moment about any point must be zero.

Four forces act on the ladder. Its weight $W$ acts vertically downwards through the centre of gravity at the midpoint. The wall is smooth, so it exerts only a horizontal normal force $N$ on the top of the ladder (there is no friction at the wall). At the foot, the ground exerts a vertical normal reaction $R$ upwards and a horizontal friction force $F$ acting towards the wall.

Resolving forces: vertically the reaction balances the weight, $R = W$ . Horizontally, the only forces are the wall's push $N$ (away from the wall) and friction $F$ (towards the wall), so for zero resultant force $F = N$ . Without friction at the foot, there would be nothing to balance the horizontal push of the wall, and the foot would slide outwards — so friction is essential.

Taking moments about the foot confirms the magnitude of $N$ . The weight (vertical) has a perpendicular distance equal to the horizontal distance from the foot to the centre of gravity, while the wall force (horizontal) has a perpendicular distance equal to the vertical height of the top of the ladder. Setting these moments equal, $N \times (\text{height}) = W \times (\text{horizontal distance to CofG})$ , fixes $N$ , and hence the friction $F = N$ needed at the foot. Because the height is larger than the horizontal distance to the midpoint, $N$ (and the required friction) is comfortably less than the weight, which is why a steeper ladder needs less friction and is less likely to slip.
Why this scores
Why this reaches the top level (5–6/6). Level 3 needs both equilibrium conditions stated AND applied: all four forces correctly identified (smooth wall → horizontal force only; rough ground → normal + friction); horizontal balance F = N showing friction is essential; and moments about the foot using the correct perpendicular distances (horizontal distance for the vertical weight, vertical height for the horizontal wall force). A sustained, logically linked argument referring to perpendicular distances scores the top band; merely listing forces without the moment reasoning stays in Level 1–2.
Question 9
Adapted from Edexcel IAL Physics Unit 16 marks
Q (6 marks). A uniform horizontal beam of weight $160\ \text{N}$ and length $2.0\ \text{m}$ is hinged to a wall at one end. A load of $240\ \text{N}$ hangs from the far end. The far end is held up by a cable that runs from the far end up to a point on the wall, making an angle of $30°$ with the beam. (a) By taking moments about the hinge, calculate the tension $T$ in the cable. (b) Hence determine the magnitude and direction of the force the hinge exerts on the beam.
Model answer
(a) Take moments about the hinge so the (unknown) hinge force is eliminated. The beam's weight $160\ \text{N}$ acts at its midpoint $1.0\ \text{m}$ from the hinge, and the load $240\ \text{N}$ acts at the far end $2.0\ \text{m}$ from the hinge — both clockwise. The cable acts at the far end ( $2.0\ \text{m}$ from the hinge) at $30°$ to the beam, so only its perpendicular component $T\sin 30°$ produces a moment. $T\sin 30° \times 2.0 = (160 \times 1.0) + (240 \times 2.0)$ $T \times 0.500 \times 2.0 = 160 + 480 = 640$ $T = \dfrac{640}{1.0} = 640\ \text{N}$ .

(b) Now resolve the forces on the beam (resultant force = 0), treating the hinge force as a horizontal component $H$ and a vertical component $V$ . The cable tension $640\ \text{N}$ has components $T\cos 30° = 640 \times 0.866 = 554\ \text{N}$ horizontal (towards the wall) and $T\sin 30° = 640 \times 0.500 = 320\ \text{N}$ vertical (upward). Vertically: $V + T\sin 30° = W_{beam} + W_{load}$ , so $V = (160 + 240) - 320 = 80\ \text{N}$ (upward). Horizontally: $H = T\cos 30° = 554\ \text{N}$ (away from the wall, balancing the cable's pull). The magnitude of the hinge force is $\sqrt{H^2 + V^2} = \sqrt{554^2 + 80^2} = \sqrt{306\,916 + 6400} = \sqrt{313\,316} = 560\ \text{N}$ (3 s.f.), at an angle $\theta = \tan^{-1}\!\left(\dfrac{V}{H}\right) = \tan^{-1}\!\left(\dfrac{80}{554}\right) = 8.2°$ above the horizontal.
Why this scores
Why this scores 6/6. (a) 3 marks: moments about the hinge (hinge force eliminated); weight at midpoint and load at the far end; correct use of the perpendicular component $T\sin 30°$ at 2.0 m, giving T = 640 N. (b) 3 marks: resolve the tension into components; vertical balance V = 400 − 320 = 80 N; horizontal balance H = T cos 30° = 554 N; resultant $\sqrt{H^2+V^2} = 560\ \text{N}$ at 8.2° above horizontal. Using T instead of T sin 30° in (a), or forgetting that the hinge also has a vertical component in (b), are the dominant errors.

Key Formulae — Moments

The formulae you need to memorise for moments on the Cambridge IGCSE paper, with every variable defined in plain English and a note on when to use it.

Moment of a force
$M = F \times d_{\perp}$
$M$
moment of the force, in N m
$F$
applied force, in N
$d_{\perp}$
perpendicular distance from the pivot to the line of action of the force, in m
When to use
To find the turning effect of a single force about a point. If the force is at an angle θ to a lever of length d, use $d_{\perp} = d\sin\theta$ (or the perpendicular component $F\sin\theta$ at distance d).
Example
$48\ \text{N}$ at right angles, $0.25\ \text{m}$ from the pivot: $M = 48 \times 0.25 = 12\ \text{N m}$ .
Principle of moments (equilibrium condition)
$\sum M_{clockwise} = \sum M_{anticlockwise}$
$\sum M_{clockwise}$
total clockwise moment about a chosen point, in N m
$\sum M_{anticlockwise}$
total anticlockwise moment about the same point, in N m
When to use
For any body in equilibrium, about ANY point. Choose the pivot through an unknown force to eliminate it from the equation.
Example
Seesaw: $400 \times d = 320 \times 1.5 \Rightarrow d = 1.2\ \text{m}$ .
Translational equilibrium (resultant force = 0)
$\sum F_{up} = \sum F_{down}$
$\sum F_{up}$
sum of upward forces, in N
$\sum F_{down}$
sum of downward forces, in N
When to use
An extended body in equilibrium needs BOTH zero resultant force AND zero resultant moment. Resolve forces (e.g. vertically) to find a second reaction once moments give the first.
Example
$R_A + R_B = W_{shelf} + W_{book} \Rightarrow R_A = 33 - 19.0 = 14.0\ \text{N}$ .
Moment of a couple
$M = F \times s$
$M$
moment (torque) of the couple, in N m
$F$
magnitude of one of the two equal forces, in N
$s$
perpendicular distance between the lines of action of the two forces, in m
When to use
When two equal, opposite, parallel forces act on a body (e.g. turning a tap or steering wheel). A couple gives a pure turning effect with zero resultant force.
Example
$8.0\ \text{N}$ at the ends of a $0.36\ \text{m}$ diameter: $M = 8.0 \times 0.36 = 2.88\ \text{N m}$ .

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 Cambridge IGCSE sitting.

Moment of a force
Examiner keyword
The turning effect of a force about a point, equal to the force multiplied by the perpendicular distance from the point to the line of action of the force: $M = F \times d_{\perp}$ . Measured in newton metres (N m).
Perpendicular distance
Examiner keyword
The shortest (90°) distance from the pivot to the line of action of the force. It is this distance, not the length of the lever, that must be used in moment = F × d. For a force at angle θ to a lever of length d, the perpendicular distance is $d\sin\theta$ .
Line of action
The straight line through the point of application of a force, extended in the direction the force acts. The perpendicular distance for a moment is measured from the pivot to this line.
Pivot (fulcrum)
The fixed point about which a body turns or is in equilibrium. When taking moments you may choose any point as the pivot; a wise choice is one that eliminates an unknown force.
Principle of moments
Examiner keyword
For a body in equilibrium, the sum of the clockwise moments about any point equals the sum of the anticlockwise moments about that same point.
Equilibrium (of an extended body)
Examiner keyword
The state in which a body has both zero resultant force AND zero resultant moment about any point. The body remains at rest (or moves with constant velocity) and does not rotate.
Resultant force
Examiner keyword
The single force equivalent to all the forces acting on a body (their vector sum). For an object in equilibrium the resultant force is zero, so the upward forces balance the downward forces.
Centre of gravity
Examiner keyword
The single point at which the entire weight of an object may be considered to act. For a uniform, regularly shaped object it lies at the geometric centre.
Centre of mass
The point at which the whole mass of a body may be considered to be concentrated. In a uniform gravitational field the centre of mass and the centre of gravity coincide.
Weight
Examiner keyword
The gravitational force on an object, $W = mg$ , acting vertically downwards through its centre of gravity. Measured in newtons (N). With $g = 9.81\ \text{m s}^{-2}$ , a $1\ \text{kg}$ mass weighs $9.81\ \text{N}$ .
Uniform body
Examiner keyword
A body whose mass is evenly distributed, so its centre of gravity lies at its geometric centre (e.g. the midpoint of a uniform beam). For a non-uniform body the centre of gravity is offset towards the heavier part.
Couple
Examiner keyword
A pair of equal, opposite, parallel forces whose lines of action do not coincide. A couple produces a turning effect (its moment is one force × the perpendicular distance between them) but no resultant force.
Torque of a couple
The moment of a couple, equal to one of the forces multiplied by the perpendicular distance between the two lines of action: $M = F \times s$ . Measured in N m.
Normal reaction (support force)
The force a support or surface exerts on a body, perpendicular to the surface. At a smooth (frictionless) wall the reaction is purely normal; on rough ground there is also a friction force parallel to the surface.
Stability
How resistant an object is to toppling. An object is more stable when it has a low centre of gravity and a wide base; it topples once its line of weight (the vertical through the centre of gravity) falls outside its base of support.

Common Mistakes and Misconceptions — Moments

The traps other students keep falling into on moments questions — taken from recent Cambridge IGCSE examiner reports and mark schemes — and how to avoid them.

✕Using the length of the lever instead of the perpendicular distance when the force acts at an angle
Edexcel IAL Physics Unit 1 examiner reports
Why it happens
Students apply moment = F × d mechanically and forget the 'd' must be perpendicular to the force's line of action.
How to avoid it
Always resolve: the moment of an angled force is $F \times d\sin\theta$ (or take the perpendicular component $F\sin\theta$ ). Use sin of the angle between the force and the lever, never cos. Sketch the perpendicular from the pivot to the line of action.
✕Pairing a force with the wrong perpendicular distance in ladder/beam problems
Edexcel IAL Physics Unit 1 examiner reports
Why it happens
In a leaning-ladder problem there are both horizontal and vertical forces, and students use the along-the-ladder distance for both.
How to avoid it
The moment of a vertical force depends on its horizontal distance from the pivot; the moment of a horizontal force depends on its vertical distance. Draw the perpendicular distance for each force separately before multiplying.
✕Forgetting that a beam, plank or shelf has its own weight acting at its centre of gravity
Edexcel IAL Physics Unit 1 examiner reports
Why it happens
Students focus on the added loads and overlook the weight of the structure itself.
How to avoid it
For any uniform beam, include its weight acting at the midpoint (centre of gravity). Mark it on the free-body diagram before taking moments. For a non-uniform beam, its weight acts at the stated/calculated centre of gravity, not the midpoint.
✕Choosing a pivot that leaves two unknown forces in the moment equation
Edexcel IAL Physics Unit 1 examiner reports
Why it happens
Students take moments about an arbitrary point rather than strategically eliminating an unknown.
How to avoid it
Take moments about the point where an unknown force acts (e.g. about a support, hinge or ladder foot). That force then has zero perpendicular distance and drops out, leaving a single unknown to solve for.
✕Using the radius instead of the full separation when finding the moment of a couple
Edexcel IAL Physics Unit 1 examiner reports
Why it happens
Students treat the couple like a single force about the centre and use the radius of a wheel.
How to avoid it
The moment of a couple uses the perpendicular distance between the two forces — the full diameter for a wheel, since both forces contribute equally. Couple moment $= F \times s$ where $s$ is the whole separation.
✕Writing the moment unit as N/m, or rounding intermediate values too early
Edexcel IAL Physics Unit 1 examiner reports
Why it happens
Confusion with other quantities' units, and a habit of rounding each line of working.
How to avoid it
The unit of a moment is N m (newton metre), never N/m. Carry full precision through the working and round only the final answer to 3 significant figures, with the correct unit, to secure the accuracy marks.

Moments — frequently asked questions

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

Moments

Short Notes - Moments

Detailed Study Notes

At a glance

What you’ll learn

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

1Calculating the moment of a force (2 marks)

2Finding the force from a required moment (2 marks)

3Balancing a seesaw using the principle of moments (3 marks)

4Moment of a force applied at an angle (3 marks)

5Reaction forces on a loaded shelf (4 marks)

6Show that — second reaction by resolving forces (4 marks)

7Centre of gravity of a non-uniform plank (4 marks)

8Ladder against a wall — taking moments about the foot (5 marks)

9Hinged beam supported by a cable (5 marks)

10Moment of a couple on a steering wheel (3 marks)

Moment of a force

Principle of moments (equilibrium condition)

Translational equilibrium (resultant force = 0)

Moment of a couple

Moment of a force

Perpendicular distance

Line of action

Pivot (fulcrum)

Principle of moments

Equilibrium (of an extended body)

Resultant force

Centre of gravity

Centre of mass

Weight

Uniform body

Couple

Torque of a couple

Normal reaction (support force)

Stability

✕Using the length of the lever instead of the perpendicular distance when the force acts at an angle

✕Pairing a force with the wrong perpendicular distance in ladder/beam problems

✕Forgetting that a beam, plank or shelf has its own weight acting at its centre of gravity

✕Choosing a pivot that leaves two unknown forces in the moment equation

✕Using the radius instead of the full separation when finding the moment of a couple

✕Writing the moment unit as N/m, or rounding intermediate values too early

Moments — frequently asked questions