PhysicsICSE Class 10 · Moment of a Force and Centre of Gravity

Moment of a Force (Torque) and Couple: Formula, Units and Solved Questions (ICSE Class 10)

Updated 15 October 2026 · Fact-checked

The moment of a force (torque) is its turning effect about a pivot: force × perpendicular distance of its line of action from the pivot, τ = F × d⊥. Its SI unit is the newton metre (N m) and its cgs unit the dyne cm, with 1 N m = 10⁷ dyne cm. A couple gives rotation without translation.

Understand Moment of a Force (Torque) and Couple

A force on a body that is free to move tends to make it move along a line (translational motion). If the body is fixed at a point or axis, called the pivot or fulcrum, the same force tends to turn it (rotational motion). The moment of a force, also called torque, measures this turning effect.

Definition. Moment of a force about a pivot $=$ force $\times$ perpendicular distance of the line of action of the force from the pivot:

$τ = F \times d_{⊥} .$

The distance that matters is the perpendicular distance from the pivot to the line along which the force acts, not the distance to the point where the force is applied. They are equal only when the force is at right angles to the line joining the pivot to that point. If the line of action passes through the pivot, $d_{⊥} = 0$ and the moment is zero however large the force: pushing a door straight towards its hinges does not turn it.

What the moment depends on. For a given turning effect, a larger perpendicular distance needs a smaller force. That is why a door handle is fitted at the edge far from the hinges, a spanner has a long handle, a steering wheel has a large rim, and a bicycle pedal is on a crank rather than at the axle.

Units.

SI unit: newton metre (N m). cgs unit: dyne centimetre (dyne cm).
$1 N m = 1 0^{5} dyne \times 1 0^{2} cm = 1 0^{7} dyne cm$ .
Gravitational units: kgf m and gf cm, where $1 kgf = 9.8$ N (taking $g = 9.8 m s^{- 2}$ ).
Write the unit as N m (a moment), not as J, even though both are newton times metre; the distance in a moment is perpendicular to the force, while in work it is along the force.

Clockwise and anticlockwise moments. A moment that turns a body anticlockwise is taken as positive and one that turns it clockwise as negative. When several forces act, the net moment is the anticlockwise total minus the clockwise total; the body turns the way of the larger total. If the two totals are equal the body is in rotational equilibrium (that is the principle of moments, the next topic).

Couple. Two equal and opposite parallel forces whose lines of action are not the same form a couple. Their resultant force is zero, so the body has no translation, but their moments add, so it rotates. Moment of a couple $=$ either force $\times$ perpendicular distance between the two lines of action (the couple arm): $τ = F \times d$ . A couple can only be balanced by another couple of equal moment in the opposite sense. The moment of a couple is a vector: it has both size and a sense of turning. Examples: turning a steering wheel with both hands, opening a water tap, winding a key, a compass needle turning in a magnetic field.

Where it comes in the paper. The 2024 paper asked for the least force from a given moment and lever length (1 mark), and the 2025 paper had two 1-mark MCQs on a couple: what motion it produces, and that its moment is a vector.

Moment of a Force (Torque) and Couple: formulas and key facts

Moment of a force: $τ = F \times d_{⊥}$; $d_{⊥}$ = perpendicular distance of the line of action from the pivot.
Least force for a given moment: $F = \frac{τ}{d _{⊥}}$; Divide the moment by the distance; the force is least when it is at right angles at the far end.
Unit conversion: $1 N m = 1 0^{7} dyne cm; 1 kgf m = 9.8 N m$; With $g = 9.8 m s^{- 2}$ .
Net moment: $τ_{net} = Σ τ_{anticlockwise} - Σ τ_{clockwise}$; Positive: the body turns anticlockwise; negative: clockwise.
Moment of a couple: $τ = F \times d$; $d$ = couple arm, the perpendicular distance between the two forces.

How to solve Moment of a Force (Torque) and Couple questions

Every moment question follows the same short routine; write each step, because marks go to the formula and the substitution with units.

1Mark the pivot and draw the line of action of each force.
2Find the perpendicular distance from the pivot to each line of action, in metres for SI answers.
3Write $τ = F \times d_{⊥}$ and substitute with units.
4Decide the sense of each moment: anticlockwise (+) or clockwise (−).
5For several forces, subtract the clockwise total from the anticlockwise total to get the net moment and its sense.
6For a couple, multiply one force by the couple arm; state that there is rotation but no translation.
7To find a force or a distance, rearrange: $F = \frac{τ}{d _{⊥}}$ or $d_{⊥} = \frac{τ}{F}$ .

Moment of a Force (Torque) and Couple in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

A larger distance needs a smaller force

Suppose a garden gate 2 m wide turns on hinges at one side. A child pushes it at right angles with 40 N at a point only 0.5 m from the hinges.

What moment does the child produce? What force would an adult need to produce the same moment by pushing at right angles at the free edge?

Try it first, then show the solutionSolution

$τ = F \times d_{⊥} = 40 \times 0.5 = 20$ N m.
At the edge, $d_{⊥} = 2$ m: $F = \frac{20}{2} = 10$ N.

Answer: 20 N m; 10 N at the edge, a quarter of the child's force.

Memory hook: Four times the distance, a quarter of the force.

Choosing the arm of a tool

Imagine tightening a bicycle seat bolt with an L-shaped Allen key. The bolt needs a moment of 6 N m. You can hold the key by its short arm (2 cm long) or its long arm (8 cm long).

Find the force needed at right angles at the end of each arm.

Try it first, then show the solutionSolution

Short arm: $F = \frac{6}{0.02} = 300$ N.
Long arm: $F = \frac{6}{0.08} = 75$ N.

Answer: 300 N on the short arm, 75 N on the long arm.

Memory hook: Hold the long arm: the bolt cannot tell your force from your distance, only their product.

Tricks to remember Moment of a Force (Torque) and Couple

Short memory aids. Say each one aloud once, then use it on a question the same day.

Perpendicular or nothing
Only the perpendicular distance from the pivot to the line of action counts. Drop a perpendicular before you multiply.
Use it when: Forces that are not at right angles to the lever.
Moment over distance
Given the moment, find the force by dividing: F = τ / d. Multiplying a moment by a distance gives nonsense units (N m²).
Use it when: Least-force MCQs like 2024 Q1(ii).
Seven zeros
N to dyne is five zeros, m to cm is two more: 1 N m = 10⁷ dyne cm.
Use it when: Unit conversion questions.
Couple = spin, no shift
Equal, opposite, not in line: the forces cancel for shifting but add for spinning.
Use it when: Couple MCQs like 2025 Q1(i).

Moment of a Force (Torque) and Couple: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Least force from a moment · 1 mark · MCQBoard level

A moment of 12 N m is needed to loosen a wheel nut. Using a wrench 0.4 m long, the least force needed is:
(a) 4.8 N
(b) 30 N
(c) 12.4 N
(d) 3 N

Try it first, then show the solutionSolution

The force is least when applied at right angles at the end of the wrench, so $d_{⊥} = 0.4$ m.
$F = \frac{τ}{d _{⊥}} = \frac{12}{0.4} = 30$ N.

Answer: (b) 30 N

Example 2: Converting units of moment · 2 marks · Short answerBoard level

(a) Express a moment of 0.5 N m in dyne cm.
(b) Express a moment of 2 kgf m in N m. (Take $g = 9.8 m s^{- 2}$ .)

Try it first, then show the solutionSolution

(a) $1 N m = 1 0^{5} dyne \times 100 cm = 1 0^{7}$ dyne cm, so $0.5 N m = 5 \times 1 0^{6}$ dyne cm. (1)
(b) $1 kgf = 9.8$ N, so $2 kgf m = 2 \times 9.8 = 19.6$ N m. (1)

Answer: (a) $5 \times 1 0^{6}$ dyne cm (b) 19.6 N m

Example 3: Net moment of two forces on a wheel · 3 marks · Short answerHarder board level

A wheel of radius 0.5 m can turn about its axle. A force of 20 N acts tangentially at the rim, turning it clockwise. A force of 30 N acts tangentially at 0.2 m from the axle, turning it anticlockwise. Find the net moment and the direction in which the wheel turns.

Try it first, then show the solutionSolution

A tangential force is at right angles to the radius, so $d_{⊥}$ is the distance from the axle.
Clockwise moment $= 20 \times 0.5 = 10$ N m. (1)
Anticlockwise moment $= 30 \times 0.2 = 6$ N m. (1)
Net moment $= 6 - 10 = - 4$ N m: 4 N m clockwise, so the wheel turns clockwise. (1)

Answer: 4 N m, clockwise.

Example 4: Same force, different moments · 3 marks · Short answerHOTS / competency

A force of 50 N is applied to a spanner 0.3 m long in three ways: (i) at right angles at the end; (ii) at right angles at the middle; (iii) at the end, but at $30°$ to the spanner. Find the moment in each case and rank them.

Try it first, then show the solutionSolution

(i) $d_{⊥} = 0.3$ m: $τ = 50 \times 0.3 = 15$ N m. (1)
(ii) $d_{⊥} = 0.15$ m: $τ = 50 \times 0.15 = 7.5$ N m. (1)
(iii) The perpendicular from the nut to the line of action is $0.3 sin 30° = 0.15$ m: $τ = 50 \times 0.15 = 7.5$ N m. Ranking: (i) > (ii) = (iii). (1)

Answer: 15 N m, 7.5 N m, 7.5 N m; (i) is largest, (ii) and (iii) are equal.

Example 5: Couple on a merry-go-round · 3 marks · Short answerHarder board level

Two children stand at opposite ends of a diameter of a merry-go-round of diameter 3 m. Each pushes tangentially with a force of 40 N, in opposite directions.
(a) What do the two forces form?
(b) Find the moment.
(c) Will the merry-go-round move from its place? Give a reason.

Try it first, then show the solutionSolution

(a) Equal and opposite parallel forces not along the same line: a couple. (1)
(b) Couple arm = 3 m: $τ = 40 \times 3 = 120$ N m. (1)
(c) No. The resultant force is zero, so there is no translation; it only rotates. (1)

Answer: (a) A couple (b) 120 N m (c) No; it only rotates, since the net force is zero.

Example 6: Case study: a stuck wheel nut · 4 marks · Case-basedHOTS / competency

A wheel nut on a car needs a moment of 90 N m to loosen it.
(a) What force is needed at right angles at the end of a 0.3 m wrench?
(b) A 0.6 m pipe is slipped over the handle. What force is needed now?
(c) A person of mass 60 kg stands on the end of the 0.6 m handle when it is horizontal. Will the nut turn? ( $g = 9.8 m s^{- 2}$ )

Try it first, then show the solutionSolution

(a) $F = \frac{90}{0.3} = 300$ N. (1)
(b) $F = \frac{90}{0.6} = 150$ N; doubling the arm halves the force. (1)
(c) Weight $= 60 \times 9.8 = 588$ N, acting vertically down, at right angles to the horizontal handle. (1)
Moment $= 588 \times 0.6 = 352.8$ N m, more than 90 N m, so the nut turns. (1)

Answer: (a) 300 N (b) 150 N (c) Yes: 352.8 N m > 90 N m.

Where students lose marks in Moment of a Force (Torque) and Couple

The slips examiners see most often, and what to write instead.

Mistake: Multiplying the moment by the distance
To find a force from a moment, divide by the distance.
Wrong working
$F = 1 \times 0.2 = 0.2$ N for a moment of 1 N m and a handle of 0.2 m.
Right working
$F = \frac{1}{0.2} = 5$ N.
Do this: Write $τ = F \times d$ first, then rearrange; check the unit comes out in N.
Usually costs: The 2024 Q1(ii) MCQ; examiners asked for the formula and substitution to be written.
Mistake: Thinking equal and opposite forces always mean rest
If the two forces are not along the same line they form a couple: no translation, but rotation.
Wrong working
Net force is zero, so the body stays stationary.
Right working
Net force zero, net moment not zero: rotational motion only.
Do this: Check both: the net force (for shifting) and the net moment (for turning).
Usually costs: The 2025 Q1(i) MCQ; examiners noted candidates confused net force with net torque.
Mistake: Calling the moment of a couple a scalar
A moment has a sense of turning (clockwise or anticlockwise), so it is a vector, unlike work, power and energy.
Wrong working
Vector quantity: work.
Right working
Vector quantity: moment of a couple.
Do this: Remember that the product of two vectors is not always a scalar.
Usually costs: The 2025 Q1(ii) MCQ.
Mistake: Using the slant distance instead of the perpendicular distance
When the force is not at right angles to the lever, the distance to the point of application is not $d_{⊥}$ .
Wrong working
50 N at $30°$ to a 0.3 m spanner: $τ = 50 \times 0.3 = 15$ N m.
Right working
$d_{⊥} = 0.3 sin 30° = 0.15$ m; $τ = 7.5$ N m.
Do this: Draw the line of action and drop a perpendicular from the pivot.
Usually costs: The answer mark in any lever or spanner numerical.

Exam tips

Write the formula, the substitution with units, and the answer with its unit; the 2024 examiners' suggestions for Q1(ii) asked for exactly this.
Convert cm to m before using N, so the answer comes out in N m; or stay in gf and cm and answer in gf cm.
In definitions, say 'perpendicular distance of the line of action of the force from the pivot', not just 'distance'.
For any couple question, state both facts: the resultant force is zero (no translation) and the moment is $F \times d$ (rotation).

Practice questions on Moment of a Force (Torque) and Couple

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
A force of 60 N is applied at right angles to the end of a spanner 15 cm long. The moment of the force is:
(a) 9 N m
(b) 900 N m
(c) 0.25 N m
(d) 75 N m
Hint
Convert 15 cm to metres first.
Answer
(a) $60 \times 0.15 = 9$ N m.
Q2 · 2 marks · Short answerBoard level
Convert (a) $3 \times 1 0^{6}$ dyne cm into N m, and (b) 50 gf cm into N m. ( $g = 9.8 m s^{- 2}$ )
Hint
$1 N m = 1 0^{7}$ dyne cm; 50 gf = 0.05 kgf.
Answer
(a) $\frac{3 \times 1 0 ^{6}}{1 0 ^{7}} = 0.3$ N m. (b) $0.05 \times 9.8 \times 0.01 = 0.0049$ N m.
Q3 · 2 marks · Short answerHarder board level
A force $F$ applied at right angles 0.2 m from a pivot has the same moment as a force of 30 N applied at right angles 0.5 m from the pivot. Find $F$ .
Hint
Equal moments: $F \times 0.2 = 30 \times 0.5$ .
Answer
$F = \frac{15}{0.2} = 75$ N.
Q4 · 3 marks · Short answerHarder board level
A pulley wheel of radius 0.3 m turns about its axle. A force of 40 N acts tangentially at the rim, turning it clockwise, and a force of 50 N acts tangentially at 0.2 m from the axle, turning it anticlockwise. Find the net moment and the sense of rotation.
Hint
Work out each moment, then subtract.
Answer
Clockwise $40 \times 0.3 = 12$ N m; anticlockwise $50 \times 0.2 = 10$ N m; net 2 N m clockwise.
Q5 · 2 marks · Short answerHarder board level
A cross-shaped wheel spanner has two arms, each 0.2 m from the centre, in a straight line. A mechanic pushes one arm and pulls the other with equal forces at right angles to the arms, producing a couple of moment 40 N m. Find each force.
Hint
The couple arm is the full distance between the two hands.
Answer
Couple arm $= 0.4$ m; $F = \frac{40}{0.4} = 100$ N.
Q6 · 3 marks · Short answerHarder board level
A door 1 m wide is pushed on opposite faces. Rahul pushes at right angles at the free edge with 15 N; Riya pushes at right angles 0.6 m from the hinges with 20 N, trying to turn it the other way. Which way does the door turn, and what is the net moment?
Hint
The larger force does not always win; compare moments.
Answer
Rahul: $15 \times 1 = 15$ N m; Riya: $20 \times 0.6 = 12$ N m. The door turns Rahul's way with a net moment of 3 N m.
Q7 · 4 marks · Long answerHOTS / competency
A bicycle crank is 0.18 m long and the rider pushes the pedal vertically downwards with 300 N. Find the moment about the axle when the crank is (a) horizontal, (b) $30°$ below the horizontal, (c) vertical. Hence explain why riders push hardest when the crank is horizontal.
Hint
The force is vertical; its perpendicular distance from the axle is the horizontal distance of the pedal from the axle.
Answer
(a) $300 \times 0.18 = 54$ N m. (b) $d_{⊥} = 0.18 cos 30° \approx 0.156$ m, so $τ \approx 46.8$ N m. (c) The force passes through the axle: 0. The same push gives the greatest turning effect when the crank is horizontal.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.A force of 5 N acts at right angles 0.2 m from a pivot. Moment?Show answer
1 N m.
2.How many dyne cm make 1 N m?Show answer
$1 0^{7}$ .
3.When is the moment of a force about a point zero?Show answer
When its line of action passes through that point.
4.What motion does a couple produce in a free body?Show answer
Rotation only, with no translation.
5.Is the moment of a couple a scalar or a vector?Show answer
A vector.

Moment of a Force (Torque) and Couple: frequently asked questions

What is the difference between the moment of a force and torque?

None at ICSE level. Torque is another name for the moment of a force: the force multiplied by the perpendicular distance of its line of action from the pivot.

Why is a door handle fixed far from the hinges?

The moment is force times perpendicular distance. Fixing the handle far from the hinges gives a large distance, so a small force produces enough moment to turn the door.

Is N m the same as joule for a moment?

Both are a newton times a metre, but a moment is written in N m. In a moment the distance is perpendicular to the force; in work it is along the force.