Moment of a Force and Centre of Gravity ICSE Class 10: Torque, Equilibrium and Principle of Moments

What Moment of a Force and Centre of Gravity is about

Weightage: CISCE publishes no chapter or unit weightage for ICSE Physics. Moment of a Force and Centre of Gravity is in the 1. Force, Work, Power and Energy (i) part of the 2027 syllabus and can be asked in both Section A and Section B of the 80-mark paper.Source: CISCE ICSE Physics (Science Paper 1) syllabus, examination year 2027

Key concepts

Translational and rotational motion

• Translational motion: every particle of the body moves the same distance in the same direction (a box sliding on the floor).
• Rotational motion: the body turns about a fixed point or axis, called the pivot or fulcrum (a door on its hinges, a wheel on its axle).

A force on a body that is free to move tends to produce translation. If the body is pivoted, the force tends to turn it, unless its line of action passes through the pivot.

Worked example

State the type of motion produced in each case: (a) a ball pushed along a table, (b) a ceiling fan when switched on, (c) a rolling wheel of a bicycle.

Try it first, then show the solutionSolution

1. (a) The ball's centre moves in a line: translational (it may also roll).
2. (b) The blades turn about the fixed axle: rotational.
3. (c) The wheel turns about its axle and the axle moves forward: both.

Answer: (a) translational (b) rotational (c) both translational and rotational

Moment of a force (torque)

Moment of a force about a pivot $=$ force $\times$ perpendicular distance of the line of action of the force from the pivot:
$\tau =F\times d_{\perp }.$

• SI unit: newton metre (N m). cgs unit: dyne cm. $1\text{ N m}=10^5\text{ dyne}\times 10^2\text{ cm}=10^7$ dyne cm.
• Gravitational units: kgf m, gf cm.
• The moment is zero if the force passes through the pivot ($d_{\perp }=0$).

The same turning effect needs a smaller force at a larger distance: a door handle is fitted far from the hinges, and a spanner has a long handle.

Worked example

A force of 40 N is applied at right angles to a spanner at 0.25 m from the nut. Find the moment, in N m and in dyne cm.

Try it first, then show the solutionSolution

1. $\tau =40\times 0.25=10$ N m.
2. $10\times 10^7=10^8$ dyne cm.

Answer: 10 N m = 10⁸ dyne cm

Clockwise and anticlockwise moments; examples

A moment that turns a body anticlockwise is taken as positive and one that turns it clockwise as negative (the usual ICSE convention).

Everyday examples:

• Door: pushed at the handle, far from the hinges, for a large moment.
• Steering wheel: a large rim radius gives a large moment for a small effort.
• Bicycle pedal: the crank gives a moment about the axle; the moment is greatest when the crank is horizontal and the foot pushes straight down, and zero when the crank is vertical.

Worked example

A cyclist pushes vertically down with 300 N on a pedal whose crank is 0.17 m long. Find the moment when the crank is (a) horizontal, (b) vertical.

Try it first, then show the solutionSolution

1. (a) Perpendicular distance $=0.17$ m: $\tau =300\times 0.17=51$ N m.
2. (b) The force passes through the axle: $\tau =0$.

Answer: (a) 51 N m (b) 0

Couple

Two equal and opposite parallel forces whose lines of action are not the same form a couple. The net force is zero, so there is no translation, but the body rotates.

Moment of a couple $=$ either force $\times$ perpendicular distance between them (the couple arm): $\tau =F\times d$.

Moment of a couple is a vector quantity (it has a direction of turning). Examples: turning a steering wheel with both hands, opening a water tap, winding a clock key.

Worked example

A driver turns a steering wheel of diameter 0.4 m by applying 15 N tangentially with each hand at opposite ends of a diameter. Find the moment of the couple.

Try it first, then show the solutionSolution

1. Couple arm $=0.4$ m.
2. $\tau =15\times 0.4=6$ N m.

Answer: 6 N m

Equilibrium and its conditions

A body is in equilibrium if the forces on it produce no change in its state of rest or motion.

• Static equilibrium: the body stays at rest (a book on a table, a balanced see-saw).
• Dynamic equilibrium: the body keeps moving uniformly or rotating uniformly (a raindrop falling at constant speed, a fan rotating steadily).

Conditions:
1. Translational: the resultant force is zero (upward forces = downward forces, and so on).
2. Rotational: the sum of anticlockwise moments about any point equals the sum of clockwise moments.

Worked example

Two equal and opposite forces act on a body but not along the same line. Is the body in equilibrium?

Try it first, then show the solutionSolution

1. The resultant force is zero: no translation.
2. They form a couple, so there is a net moment: the body rotates.
3. So it is not in rotational equilibrium.

Answer: No: it rotates (only rotational motion).

Principle of moments and its verification

Principle of moments: for a body in equilibrium about a pivot, the sum of anticlockwise moments equals the sum of clockwise moments.

Verification: suspend a metre rule horizontally from two spring balances (or balance it on a knife edge). Hang slotted weights at different marks. When the rule is horizontal and at rest, take moments about any point: anticlockwise and clockwise moments are equal, and the sum of the spring-balance readings equals the total weight.

For a uniform metre rule, its own weight acts at the 50 cm mark; include its moment unless the pivot is at 50 cm.

Worked example

A uniform metre rule balances horizontally on a knife edge at the 40 cm mark when a 20 gf weight hangs at the 10 cm mark. Find the weight of the rule.

Try it first, then show the solutionSolution

1. Anticlockwise moment (20 gf at 30 cm from the pivot) $=20\times 30=600$ gf cm.
2. Clockwise moment (rule's weight $W$ at 50 cm, i.e. 10 cm from the pivot) $=10W$.
3. $10W=600$, so $W=60$ gf.

Answer: 60 gf

Centre of gravity

The centre of gravity (C.G.) of a body is the point through which the whole weight of the body acts, i.e. the point about which the sum of the moments of the weights of all its particles is zero.

It depends on the shape and on how the mass is distributed; it need not lie in the material (a ring, a hollow sphere).

• Uniform rod: its mid-point. Circular disc or ring: its centre. Triangular lamina: the centroid (meeting point of the medians). Rectangle or square lamina: where the diagonals meet.
• Hollow cone of height $h$: on the axis at $\frac{h}{3}$ from the base. Solid cone: $\frac{h}{4}$ from the base.

Irregular lamina (plumb-line method): suspend it from a hole, hang a plumb line from the same point and draw the vertical line; repeat from another hole. The C.G. is where the lines cross, because a freely suspended body rests with its C.G. vertically below the point of suspension.

Worked example

Where is the centre of gravity of a hollow cone of height 9 cm, and of a solid cone of the same height?

Try it first, then show the solutionSolution

1. Hollow cone: $\frac{9}{3}=3$ cm from the base, on the axis.
2. Solid cone: $\frac{9}{4}=2.25$ cm from the base, on the axis.

Answer: 3 cm and 2.25 cm from the base, on the axis

Physics in real life: Moment of a Force and Centre of Gravity

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.

Topics in this chapter

• Moment of a Force (Torque) and Couple
• Principle of Moments and Equilibrium
• Centre of Gravity

Moment of a Force and Centre of Gravity formula sheet

Moment of a force

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

d⊥ = perpendicular distance of the line of action from the pivot.

Units of moment

$1\text{N m}=10^7\text{dyne cm}$

Also kgf m and gf cm.

Moment of a couple

$\tau =F\times d$

d = couple arm (distance between the two forces).

Principle of moments

$\Sigma {\tau }_{\text{anticlockwise}}=\Sigma {\tau }_{\text{clockwise}}$

For a body in rotational equilibrium.

Translational equilibrium

$\Sigma F_{\text{up}}=\Sigma F_{\text{down}}$

Gives the reaction at a pivot or the reading of a balance.

C.G. of cones

$\text{hollow: }\frac{h}{3},\text{solid: }\frac{h}{4}\text{ from the base}$

On the axis.

Moment of a Force and Centre of Gravity: diagrams to practise

Practise each one on paper: draw it from the checklist, then tick off every label.

How the board asks Moment of a Force and Centre of Gravity

The question patterns to expect, and the method that works for each.

Tricks to remember Moment of a Force and Centre of Gravity

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

• Far is easy

  The farther from the pivot you push, the less force you need (τ = F × d).

  Use it when: Explaining doors, spanners and steering wheels.

• Couple: spin, no shift

  Equal, opposite, not in line: the body spins but does not move off.

  Use it when: Couple MCQs.

• Rule weighs at fifty

  A uniform metre rule's weight always acts at 50 cm.

  Use it when: Metre-rule numericals.

• Hollow three, solid four

  Cone C.G.: h/3 from the base if hollow, h/4 if solid.

  Use it when: C.G. of cones.

Moment of a Force and Centre of Gravity: board-level solved questions

Ordered from board level to HOTS. Cover the solution, write your own, then compare step by step.

Case-study practice (Section B style)

Read the situation, then answer every part. In the board paper this is a 10-mark question split into 3 + 3 + 4 marks.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Where students lose marks in Moment of a Force and Centre of Gravity

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

• ✗Mistake: Multiplying instead of dividing

  CISCE examiners' comments, 2024: to find the minimum force from a moment of 1 N m and a lever of 0.2 m, candidates multiplied 1 by 0.2 instead of dividing.

  Wrong working

  F = 1 × 0.2 = 0.2 N

  Right working

  F = τ ÷ d = 1 ÷ 0.2 = 5 N

  ✓Do this: Write τ = F × d first, then rearrange.

  Usually costs: 1 mark.

• ✗Mistake: Wrong distance or wrong pivot

  CISCE examiners' comments, 2024: for a non-uniform beam, candidates took the distance as 0.6 m instead of 0.8 m and confused the fulcrum with the centre of gravity.

  ✓Do this: Sketch the beam; mark the pivot, the C.G. and every force; measure each distance from the pivot.

  Usually costs: Most of a 2-mark part.

• ✗Mistake: Forgetting the rule's own weight

  CISCE examiners' comments, 2025: in a metre-rule balance problem, candidates omitted the moment due to the weight of the ruler at the 50 cm mark and marked the fulcrum wrongly.

  ✓Do this: A uniform rule's weight acts at 50 cm; include it unless the pivot is at 50 cm.

  Usually costs: 2 marks in a 3-mark numerical.

• ✗Mistake: Incomplete definition of C.G.

  CISCE examiners' comments, 2024: definitions missed 'point', 'moment of weights' or 'sum of moments is zero', and answers for a cone's C.G. wrote h/3 without the value, gave a wrong value, or said only 'at the centre'.

  ✓Do this: Learn the full definition; for a cone give a number and say 'from the base, on the axis'.

  Usually costs: 1 mark each.

• ✗Mistake: Equal and opposite forces always mean rest

  CISCE examiners' comments, 2025: for equal and opposite forces not in the same line, candidates thought the body stays stationary or moves in a line, confusing net force with net moment.

  ✓Do this: Zero net force means no translation; a couple still rotates the body.

  Usually costs: 1 mark.

• ✗Mistake: Moment of a couple taken as scalar

  CISCE examiners' comments, 2025: asked which quantity is a vector, candidates chose work, power or energy instead of the moment of a couple.

  ✓Do this: Moment (torque) has a turning direction: it is a vector. Work, energy and power are scalars.

  Usually costs: 1 mark.

Practice questions on Moment of a Force and Centre of Gravity

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

1. Q1 · 1 mark · MCQBoard level

   The SI unit of moment of a force is:
   (a) N
   (b) N m
   (c) N m⁻¹
   (d) J s

   Hint

   Force × distance.

   Answer

   (b)

2. Q2 · 1 mark · MCQBoard level

   A couple acting on a body that is free to move produces:
   (a) translational motion only
   (b) rotational motion only
   (c) both translational and rotational motion
   (d) no motion

   Hint

   Net force is zero.

   Answer

   (b)

3. Q3 · 1 mark · MCQHarder board level

   A force of 10 N acts on a body pivoted at $O$. The perpendicular distance of its line of action from $O$ is 0.3 m, though the point of application is 0.5 m from $O$. The moment is:
   (a) 2 N m
   (b) 3 N m
   (c) 5 N m
   (d) 8 N m

   Hint

   Use the perpendicular distance.

   Answer

   (b)

4. Q4 · 1 mark · Assertion-reasonBoard level

   Assertion (A): The centre of gravity of a uniform ring lies at its centre, where there is no material.
   Reason (R): The centre of gravity is the point through which the whole weight of a body acts; it need not lie within the material.
   (a) (A) is true and (R) is false.
   (b) (A) is false and (R) is true.
   (c) Both (A) and (R) are true, and (R) is the correct explanation for (A).
   (d) Both (A) and (R) are true, but (R) is not the correct explanation for (A).

   Hint

   Symmetry of the ring.

   Answer

   (c)

5. Q5 · 2 marks · Short answerHarder board level

   A hollow cone and a solid cone each have height 12 cm. State the position of the centre of gravity of each.

   Hint

   h/3 and h/4.

   Answer

   Hollow: 4 cm from the base on the axis. Solid: 3 cm from the base on the axis.

6. Q6 · 2 marks · Very short answerHarder board level

   Fill in the blanks: (a) The moment of a force is zero when its line of action passes through the __________. (b) 1 N m = __________ dyne cm.

   Hint

   Units: 10⁵ dyne × 10² cm.

   Answer

   (a) pivot (axis of rotation) (b) 10⁷

7. Q7 · 3 marks · Long answerBoard level

   (a) Define moment of a force. (b) State the principle of moments. (c) State the condition for translational equilibrium.

   Hint

   Three definitions.

   Answer

   (a) Product of the force and the perpendicular distance of its line of action from the axis of rotation. (b) In equilibrium, the sum of anticlockwise moments about the pivot equals the sum of clockwise moments. (c) The resultant force on the body is zero.

8. Q8 · 3 marks · Long answerHarder board level

   A uniform metre rule of weight 50 gf balances horizontally on a knife edge at the 60 cm mark when a weight $X$ hangs at the 100 cm mark. Find $X$.

   Hint

   Rule's weight acts 10 cm from the knife edge.

   Answer

   $50\times 10=X\times 40$, so $X=12.5$ gf.

9. Q9 · 4 marks · Long answerHOTS / competency

   A uniform beam 4 m long and of weight 200 N rests horizontally on supports at its ends $A$ and $B$. A load of 300 N is placed 1 m from $A$. Find the reactions at $A$ and $B$.

   Hint

   Take moments about A first.

   Answer

   About $A$: $R_B\times 4=200\times 2+300\times 1=700$, so $R_B=175$ N. $R_A=500-175=325$ N.

10. Q10 · 3 marks · Long answerHarder board level

    A nut needs a moment of 30 N m to turn. (a) What is the shortest spanner that lets a person exert 120 N do it? (b) What force is needed with a 0.5 m extension? (c) Why is it better to push at right angles to the spanner?

    Hint

    d = τ ÷ F.

    Answer

    (a) $\frac{30}{120}=0.25$ m. (b) $\frac{30}{0.5}=60$ N. (c) At right angles the perpendicular distance is largest, so the moment is greatest for a given force.

11. Q11 · 1 mark · MCQHOTS / competency

    An irregular cardboard lamina is suspended freely from a pin. At rest, its centre of gravity:
    (a) is at the pin
    (b) lies vertically above the pin
    (c) lies vertically below the pin
    (d) can be anywhere

    Hint

    Weight must have no moment about the pin.

    Answer

    (c)

12. Q12 · 4 marks · Long answerHOTS / competency

    Two identical uniform rods, each 40 cm long, are joined at one end to form an L shape. Taking the corner as the origin with the rods along the axes, find the position of the centre of gravity. Does it lie on the rods? Explain.

    Hint

    Each rod's weight acts at its mid-point.

    Answer

    The rods' weights act at (20, 0) and (0, 20) cm and are equal, so the C.G. is midway: (10, 10) cm from the corner. It does not lie on either rod; the C.G. depends on mass distribution and can be outside the material.

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. 1.Define moment of a force.Show answer

   Force × perpendicular distance of its line of action from the pivot.

2. 2.SI and cgs units of moment?Show answer

   N m and dyne cm (1 N m = 10⁷ dyne cm).

3. 3.What is a couple?Show answer

   Two equal, opposite, parallel forces not along the same line.

4. 4.Conditions of equilibrium?Show answer

   Resultant force zero; sum of anticlockwise moments = sum of clockwise moments.

5. 5.State the principle of moments.Show answer

   In rotational equilibrium, total anticlockwise moment = total clockwise moment about the pivot.

6. 6.C.G. of a solid cone of height h?Show answer

   h/4 from the base, on the axis.

7. 7.Where is the C.G. of a freely suspended body?Show answer

   Vertically below the point of suspension.

Quick revision checklist

The night before the exam: can you do each of these without looking?

• ☐Can you write τ = F × d⊥ and rearrange it?
• ☐Can you convert N m to dyne cm?
• ☐Do you include a uniform rule's weight at 50 cm?
• ☐Can you find a pivot reaction with ΣF = 0?
• ☐Can you define C.G. with all key words?
• ☐Do you know the C.G. of hollow and solid cones?

Moment of a Force and Centre of Gravity: frequently asked questions