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

Principle of Moments and Equilibrium: Metre Rule and Beam Numericals (ICSE Class 10)

Updated 15 October 2026 · Fact-checked

The principle of moments says that when a body is in equilibrium about a pivot, the sum of the anticlockwise moments equals the sum of the clockwise moments. Together with zero resultant force (upward forces equal downward forces), it lets you find an unknown weight, distance or support reaction on a balanced rule or beam.

Each side's moment is force × distance from the pivot; when the two moments match (1200) and the 100 N reaction equals the weights, the beam is balanced.

Illustration drawn for QuestionBuddy.

Understand Principle of Moments and Equilibrium

Equilibrium. A body is in equilibrium when the forces on it cause no change in its state of rest or of uniform motion. It may be static (at rest, like a balanced see-saw) or dynamic (moving or rotating uniformly, like a fan turning steadily). Two conditions must both hold:

1. Translational equilibrium: the resultant force is zero. For forces along one line, total upward force $=$ total downward force.
2. Rotational equilibrium: the resultant moment about any point is zero, so the sum of anticlockwise moments equals the sum of clockwise moments.

Principle of moments. If a body is in equilibrium under several forces, the sum of the anticlockwise moments about any point equals the sum of the clockwise moments about the same point:

$Σ τ_{anticlockwise} = Σ τ_{clockwise} .$

A beam balance and a see-saw work on this principle.

Verification with a metre rule. Hang a metre rule horizontally from two spring balances (or balance it on a knife edge) and suspend slotted weights at known marks with threads. When the rule is horizontal and at rest:

take moments about any mark: the anticlockwise and clockwise totals agree within experimental error;
the sum of the spring-balance readings equals the total weight of the slotted weights and the rule (translational equilibrium).

The rule's own weight. A uniform rule's weight acts at its centre, the 50 cm mark. If the pivot is not at 50 cm, this weight has a moment and must be included. A balanced rule is how you find the weight of the rule itself: the only unknown is $W$ , acting at 50 cm.

Non-uniform beams. If a beam is non-uniform, its weight acts at its centre of gravity, which is not the middle. The pivot (fulcrum) is wherever the beam is supported or hinged; do not confuse it with the C.G.

Reactions. A knife edge, hinge or support pushes up on the beam. Its upward force (the reaction) comes from translational equilibrium: reaction $=$ total downward force minus any other upward forces. Taking moments about the support removes the reaction from the moment equation, because its moment about that point is zero. This is the key trick: take moments about the point where an unknown force acts, and that unknown disappears.

In the paper. The 2024 paper set a 2-mark non-uniform beam pivoted at one end, and the 2025 paper a 3-mark metre-scale question asking for a diagram and the weight of the ruler.

Principle of Moments and Equilibrium: formulas and key facts

Principle of moments: $Σ τ_{anticlockwise} = Σ τ_{clockwise}$; About any point, for a body in equilibrium.
Translational equilibrium: $Σ F_{up} = Σ F_{down}$; Gives a support reaction or the sum of spring-balance readings.
Two weights on a lever: $W_{1} d_{1} = W_{2} d_{2}$; $d_{1}$ , $d_{2}$ measured from the pivot, on opposite sides.
Balance point of several weights on a rule: $x = \frac{W _{1} x _{1} + W _{2} x _{2} + \dots}{W _{1} + W _{2} + \dots}$; Positions measured from the 0 cm end; include the rule's weight at 50 cm if uniform.

How to solve Principle of Moments and Equilibrium questions

Always draw the arrangement first; the 2025 examiners asked for the diagram and gave marks for it.

1Draw the rule or beam with the pivot, every weight as a downward arrow and every support or balance as an upward arrow.
2Put the rule's own weight at 50 cm if it is uniform, or at its C.G. if it is non-uniform.
3Choose the pivot as the point to take moments about; if a reaction is unknown, take moments about the point where it acts.
4Measure each distance from that point (not from the 0 cm end) and convert to one unit.
5Write anticlockwise moments = clockwise moments and solve for the unknown.
6Use upward forces = downward forces for a reaction or a second unknown.
7Check: the heavier weight should be nearer the pivot.

Principle of Moments and Equilibrium 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.

Carrying a load on a pole

Suppose you carry a light pole 1.5 m long over your shoulder, with a 60 N bag hanging from its front end. Your shoulder is 0.5 m from the front end, and your hand holds the back end down.

What force must your hand exert, and what force does your shoulder take? Ignore the pole's weight.

Try it first, then show the solutionSolution

Take moments about the shoulder: $60 \times 0.5 = F \times 1.0$ , so $F = 30$ N downward at the hand.
Upward force = downward forces: shoulder $= 60 + 30 = 90$ N.

Answer: Hand 30 N downward; shoulder 90 N, more than the bag itself.

Memory hook: The hand pushes down too, so the shoulder carries bag plus hand.

Balancing a hanging mobile

Imagine making a hanging mobile from a light stick 30 cm long, with a 20 g paper fish at one end and a 40 g wooden bird at the other.

How far from the fish should you tie the thread so the stick hangs level?

Try it first, then show the solutionSolution

Let the thread be $x$ cm from the fish: $20 \times x = 40 \times (30 - x)$ .
$60 x = 1200$ , so $x = 20$ cm.

Answer: 20 cm from the fish (10 cm from the bird).

Memory hook: Twice the mass, half the distance.

Tricks to remember Principle of Moments and Equilibrium

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

Heavy sits near
On a balanced lever the heavier load is always nearer the pivot. Use it to check every answer.
Use it when: Quick check in metre-rule and see-saw questions.
Fifty is never free
A uniform rule's weight sits at 50 cm. Unless the pivot is at 50, the rule's weight has a moment: write it in.
Use it when: 2025-type 'find the weight of the ruler' questions.
Stand on the unknown
Take moments about the point where the unknown force acts; its moment is zero, so it vanishes from the equation.
Use it when: Support reactions and hinges.
Two checks, two equations
Moments balance (no turning) and forces balance (no shifting). Two equations solve two unknowns.
Use it when: Spring-balance and two-support problems.

Principle of Moments and Equilibrium: solved examples

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

Example 1: Where to hang a weight · 1 mark · MCQBoard level

A uniform metre rule balances at its 50 cm mark. A weight of 36 gf hangs at the 25 cm mark. Where must a weight of 20 gf hang to keep it horizontal?
(a) 95 cm mark
(b) 80 cm mark
(c) 5 cm mark
(d) 70 cm mark

Try it first, then show the solutionSolution

The rule's weight acts at the pivot, so it has no moment.
$36 \times (50 - 25) = 20 \times d \Rightarrow d = 45$ cm on the other side: the 95 cm mark.

Answer: (a) 95 cm mark

Example 2: Weight of a metre rule · 3 marks · Short answerBoard level

A uniform metre rule has a weight of 50 gf hung at its 0 cm end and 10 gf at its 100 cm end. It balances horizontally on a knife edge at the 40 cm mark. Draw a diagram, and find (a) the weight of the rule, (b) the reaction of the knife edge.

Try it first, then show the solutionSolution

Diagram: rule from 0 to 100 cm; 50 gf down at 0; knife edge (up) at 40; rule's weight $W$ down at 50; 10 gf down at 100. (1)
Moments about 40 cm: anticlockwise $50 \times 40 = 2000$ gf cm; clockwise $W \times 10 + 10 \times 60 = 10 W + 600$ . So $W = 140$ gf. (1)
Reaction $= 50 + 140 + 10 = 200$ gf, upward. (1)

Answer: (a) 140 gf (b) 200 gf upward

Example 3: Readings of two spring balances · 3 marks · Short answerHarder board level

A uniform metre rule of weight 1.2 N is held horizontal by two spring balances attached at the 20 cm and 80 cm marks. Weights of 3 N and 1.8 N hang at the 40 cm and 60 cm marks. Find the reading of each balance.

Try it first, then show the solutionSolution

Moments about 20 cm (removes $T_{1}$ ): $T_{2} \times 60 = 3 \times 20 + 1.2 \times 30 + 1.8 \times 40 = 168$ , so $T_{2} = 2.8$ N. (1)
Forces: $T_{1} + T_{2} = 3 + 1.2 + 1.8 = 6$ N, so $T_{1} = 3.2$ N. (1)
Check with moments about 80 cm: $T_{1} \times 60 = 3 \times 40 + 1.2 \times 30 + 1.8 \times 20 = 192$ , $T_{1} = 3.2$ N. (1)

Answer: 3.2 N at 20 cm and 2.8 N at 80 cm.

Example 4: Non-uniform rod held by a rope · 3 marks · Short answerHarder board level

A non-uniform rod $A B$ , 1.5 m long and of weight 90 N, is hinged at $A$ . Its centre of gravity is 0.5 m from $A$ . It is held horizontal by a vertical rope tied 1.2 m from $A$ . Find (a) the tension in the rope, (b) the force at the hinge.

Try it first, then show the solutionSolution

Moments about the hinge $A$ : $T \times 1.2 = 90 \times 0.5 = 45$ , so $T = 37.5$ N. (1½)
Forces: hinge force $= 90 - 37.5 = 52.5$ N upward. (1½)

Answer: (a) 37.5 N (b) 52.5 N upward

Example 5: Finding the weight and C.G. of a log · 4 marks · Long answerHOTS / competency

A non-uniform log $A B$ , 3 m long, lies on level ground. To lift end $A$ just off the ground needs a vertical force of 400 N; to lift end $B$ just off the ground needs 200 N. Find the weight of the log and the position of its centre of gravity.

Try it first, then show the solutionSolution

Let the weight be $W$ at $x$ m from $A$ .
Lifting $A$ (pivot at $B$ ): $400 \times 3 = W (3 - x)$ . (1)
Lifting $B$ (pivot at $A$ ): $200 \times 3 = W x$ . (1)
Add: $1800 = 3 W$ , so $W = 600$ N. (1)
$x = \frac{600}{600} = 1$ m from $A$ , the end that was harder to lift. (1)

Answer: 600 N; C.G. 1 m from $A$ .

Example 6: Case study: a diving board · 4 marks · Case-basedHOTS / competency

A uniform diving board $A B$ is 4 m long and weighs 400 N. It rests on a support 1 m from the back end $A$ and is bolted down at $A$ . A diver of weight 600 N stands at the free end $B$ .
(a) Which way does the bolt pull on the board, and why?
(b) Find the force in the bolt.
(c) Find the force on the support.

Try it first, then show the solutionSolution

(a) The diver and the board's weight (at 2 m from $A$ ) both turn the board clockwise about the support, so the bolt must pull down at $A$ to give an anticlockwise moment. (1)
(b) Moments about the support: clockwise $600 \times 3 + 400 \times 1 = 2200$ N m; anticlockwise $F \times 1$ . So $F = 2200$ N. (2)
(c) Support (up) $= 2200 + 400 + 600 = 3200$ N. (1)

Answer: (a) Downward (b) 2200 N (c) 3200 N

Where students lose marks in Principle of Moments and Equilibrium

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

Mistake: Measuring the distance from the wrong point
Distances in the principle of moments are taken from the pivot, not from the 0 cm end or the centre of the beam.
Wrong working
Beam pivoted at one end, weight 0.2 m from the pivot, force at the far end 0.8 m: $120 \times 0.2 = F \times 0.6$ .
Right working
$120 \times 0.2 = F \times 0.8$ , so $F = 30$ N.
Do this: Mark the pivot and write each distance from it on the diagram.
Usually costs: Reported for 2024 Q2(iii), where candidates also confused the fulcrum with the centre of gravity.
Mistake: Leaving out the rule's own weight
A uniform rule's weight acts at 50 cm and has a moment unless the pivot is at 50 cm.
Wrong working
$2 \times 70 = 5 \times 30$ for a rule balanced at 70 cm.
Right working
$2 \times 70 + W \times 20 = 5 \times 30$ , so $W = 0.5$ N.
Do this: Draw the rule's weight at 50 cm in every metre-rule diagram.
Usually costs: Reported for 2025 Q6(i).
Mistake: Putting the fulcrum on the wrong side
For balance, the pivot must be nearer the larger weight.
Wrong working
Marking the knife edge nearer the 2 N weight than the 5 N weight.
Right working
The knife edge lies nearer the 5 N weight.
Do this: Check your diagram with 'heavy sits near'.
Usually costs: Reported for 2025 Q6(i), along with failing to draw the forces at the ends.
Mistake: Forgetting the support reaction
When asked for the force on a support, translational equilibrium is needed as well as moments.
Wrong working
Stopping after finding the unknown weight.
Right working
Reaction = sum of downward forces − other upward forces.
Do this: Read the question again: count what is asked and write one equation for each.
Usually costs: The last mark of a 3-mark question.

Exam tips

Draw the diagram even if not asked: rule 0 to 100 cm, pivot, each force with its mark, and the rule's weight at 50 cm. The 2025 examiners suggested always drawing it.
State the principle in words before using it: 'sum of anticlockwise moments = sum of clockwise moments'.
Keep units consistent: gf with cm, or N with m. Do not mix them in one equation.
In a definition of equilibrium, give both conditions: resultant force zero and resultant moment zero.

Practice questions on Principle of Moments and Equilibrium

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 body acted on by several forces is in equilibrium. Then:
(a) only the resultant force is zero
(b) only the resultant moment is zero
(c) both the resultant force and the resultant moment are zero
(d) neither need be zero
Hint
Equilibrium means no shifting and no turning.
Answer
(c)
Q2 · 2 marks · Short answerBoard level
A uniform metre rule is pivoted at its 50 cm mark. A weight of 40 gf hangs at the 10 cm mark. What weight hung at the 70 cm mark keeps it horizontal?
Hint
Distances 40 cm and 20 cm from the pivot.
Answer
$40 \times 40 = W \times 20$ , so $W = 80$ gf.
Q3 · 3 marks · Short answerHarder board level
A uniform rod 1 m long and of weight 2 N rests horizontally on two supports at its ends $A$ and $B$ . A load of 6 N hangs 25 cm from $A$ . Find the reaction at each support.
Hint
Take moments about $A$ to find the reaction at $B$ .
Answer
$R_{B} \times 1 = 6 \times 0.25 + 2 \times 0.5 = 2.5$ , so $R_{B} = 2.5$ N; $R_{A} = 8 - 2.5 = 5.5$ N.
Q4 · 3 marks · Short answerHarder board level
A uniform metre rule of weight 60 gf carries 40 gf at the 5 cm mark and 100 gf at the 85 cm mark. At which mark should a knife edge be placed so that it balances?
Hint
Take moments about the 0 cm end; the knife edge's reaction is the total weight.
Answer
$200 x = 40 \times 5 + 60 \times 50 + 100 \times 85 = 11700$ , so $x = 58.5$ cm mark.
Q5 · 2 marks · Short answerHarder board level
A non-uniform pole $A B$ , 2.5 m long and of weight 120 N, is hinged at $A$ and held horizontal by a vertical upward force of 48 N at $B$ . How far from $A$ is its centre of gravity?
Hint
Moments about the hinge.
Answer
$48 \times 2.5 = 120 \times x$ , so $x = 1$ m from $A$ .
Q6 · 3 marks · Short answerHarder board level
A uniform see-saw is pivoted at its centre. A child of weight 300 N sits 1.8 m to the left of the pivot. (a) Where must a child of weight 400 N sit to balance it? (b) A 100 N bag is then placed 0.5 m to the right of the pivot. Where must the 400 N child now sit?
Hint
The bag adds a clockwise moment, so the heavier child moves in.
Answer
(a) $\frac{300 \times 1.8}{400} = 1.35$ m to the right. (b) $400 d + 100 \times 0.5 = 540$ , so $d = 1.225$ m to the right.
Q7 · 4 marks · Long answerHOTS / competency
A uniform metre rule hangs horizontally from two spring balances at its 0 cm and 100 cm marks, which read 3 N and 2 N. A single weight of 4 N hangs from the rule. Find (a) the weight of the rule, (b) the mark at which the 4 N weight hangs.
Hint
Forces first, then moments about the 0 cm end.
Answer
(a) $W = 3 + 2 - 4 = 1$ N. (b) $2 \times 100 = 4 x + 1 \times 50$ , so $x = 37.5$ cm mark.

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.State the principle of moments.Show answer
For a body in equilibrium, the sum of anticlockwise moments about any point equals the sum of clockwise moments about the same point.
2.Where does the weight of a uniform metre rule act?Show answer
At its 50 cm mark.
3.A 2 kgf load is 30 cm from a pivot. What load balances it at 20 cm on the other side?Show answer
3 kgf.
4.Name the two conditions for equilibrium.Show answer
Resultant force zero (translational) and resultant moment zero (rotational).
5.Why take moments about the point where an unknown reaction acts?Show answer
Its moment about that point is zero, so it drops out of the equation.

Principle of Moments and Equilibrium: frequently asked questions

What is the principle of moments in simple words?

If something is balanced, the turning effects trying to turn it one way exactly cancel the turning effects trying to turn it the other way.

How do you find the weight of a metre rule using the principle of moments?

Hang a known weight on the rule and move a knife edge until the rule balances. The rule's weight acts at 50 cm; equate moments about the knife edge and solve for it.

Is a body moving at constant velocity in equilibrium?

Yes. It is in dynamic equilibrium: the resultant force and the resultant moment on it are both zero.