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Physics ICSE Class 10 CBSE CUET

Force – ICSE Class 10 Physics

📅 Published: June 2026

Complete chapter notes on Force for ICSE Class 10 Physics. Covers translational and rotational motion, torque, principle of moments, equilibrium, centre of gravity, centripetal and centrifugal force, uniform circular motion, solved numericals, and exam tips.

1) What is Force?

Force is an external agent that changes or tries to change the state of rest or of uniform motion of a body, or changes the direction of motion, or changes the shape and size of a body.

In simple terms, a force is a push or a pull. It is a vector quantity — it has both magnitude and direction. The SI unit of force is newton (N).

Effects of Force

Force (F) = mass (m) × acceleration (a) F = ma SI unit: newton (N) — 1 N = 1 kg m/s²

2) Translational and Rotational Motion

When a force is applied on a body, it can produce two types of motion depending on where and how the force is applied:

Translational Motion

If the line of action of the applied force passes through the centre of gravity of the body, the body moves as a whole from one point to another without rotating. This is called translational motion.

Rotational Motion

If the line of action of the applied force does NOT pass through the centre of gravity, the body rotates about a fixed axis. This is called rotational motion.

Translational Motion Rotational Motion
Body moves as a whole from one point to another Body turns about a fixed axis
Force passes through the centre of gravity Force does not pass through the centre of gravity
All particles have same displacement Particles have different displacements
Linear acceleration is produced Angular acceleration is produced
Example: a sliding block Example: a spinning top, a door opening
In ICSE, the most important idea is: a force passing through the centre of gravity produces only translational motion, while a force not passing through the centre of gravity produces both translational and rotational motion.

3) Moment of a Force — Torque

The turning effect of a force about a fixed point or axis is called the moment of force or torque. It is the rotational equivalent of force.

Definition

The moment of force about a point is defined as the product of the force and the perpendicular distance of the line of action of the force from that point.

Moment of Force (τ) = Force (F) × Perpendicular distance (d) τ = F × d

SI Unit

The SI unit of moment of force is newton-metre (N m). Its CGS unit is dyne-cm.

Clockwise and Anticlockwise Moments

Moment of force = F × d, where d is the perpendicular distance from the pivot to the line of action of the force.

Factors Affecting the Moment of Force

Examples of Torque in Daily Life

Pro Tip: To maximise the turning effect, apply the force as far as possible from the pivot, and at right angles (90°) to the line joining the pivot and the point of application.

4) Principle of Moments

The principle of moments states that when a body is in equilibrium under the action of a number of coplanar forces, the sum of the anticlockwise moments about any point is equal to the sum of the clockwise moments about the same point.

Sum of anticlockwise moments = Sum of clockwise moments Σ Anticlockwise moment = Σ Clockwise moment

Conditions for a Body to Be in Equilibrium Under Parallel Forces

Applications of the Principle of Moments

Numerical Example — Principle of Moments

A uniform metre scale of weight 50 gf is balanced at the 40 cm mark when a weight of 100 gf is suspended at the 10 cm mark. Find the position where a weight of 200 gf must be suspended to balance it.

Given: Weight of scale = 50 gf acts at centre (50 cm mark) Pivot at 40 cm mark. Scale weight moment arm = 50 - 40 = 10 cm Clockwise moment of scale = 50 × 10 = 500 gf cm 100 gf at 10 cm → moment arm = 40 - 10 = 30 cm Anticlockwise moment of 100 gf = 100 × 30 = 3000 gf cm Net anticlockwise = 3000 - 500 = 2500 gf cm Let 200 gf be at distance x from pivot on right side Then 200 × x = 2500 → x = 12.5 cm ∴ 200 gf is suspended at 40 + 12.5 = 52.5 cm mark

5) Equilibrium

A body is said to be in equilibrium when the net force and the net moment (torque) acting on it are both zero. In other words, the body is at rest or in a state of uniform motion with no rotation.

Types of Equilibrium

There are three types of equilibrium:

Stable Equilibrium

When a body is slightly displaced from its equilibrium position and it tends to return back to its original position, it is said to be in stable equilibrium.

Unstable Equilibrium

When a body is slightly displaced from its equilibrium position and it tends to move further away from its original position, it is said to be in unstable equilibrium.

Neutral Equilibrium

When a body is slightly displaced from its equilibrium position and it remains in its new position without any tendency to return or move further, it is said to be in neutral equilibrium.

Property Stable Unstable Neutral
On displacement Returns to original position Moves further away Stays in new position
Centre of gravity Rises Falls Stays at same height
Example Cone on its base Cone on its tip Cone on its side

6) Centre of Gravity (COG)

The centre of gravity of a body is the point through which the entire weight of the body acts, irrespective of the position or orientation of the body. It is the point where the whole mass of the body is assumed to be concentrated.

Characteristics of Centre of Gravity

Examples of COG

Relation Between Centre of Gravity and Stability

A body is in stable equilibrium as long as the vertical line through its centre of gravity falls within its base of support. If the COG falls outside the base, the body topples.

7) Centripetal Force

When a body moves along a circular path with uniform speed, its direction changes continuously at every point. This change in direction requires a force, which acts towards the centre of the circle. This force is called the centripetal force.

Centripetal Force (Fc) = mv² / r Where: m = mass of the body (kg) v = linear speed (m/s) r = radius of the circular path (m)

Characteristics of Centripetal Force

Examples of Centripetal Force

8) Centrifugal Force

Centrifugal force is the apparent outward force experienced by a body moving in a circular path, when observed from a rotating reference frame. It is equal in magnitude to the centripetal force but acts in the opposite direction — away from the centre.

Centrifugal Force = mv² / r (outward direction)

Important Note

Centrifugal force is a pseudo force or fictitious force. It does not exist in an inertial (non-rotating) frame of reference. It is felt only when the observer is in the rotating frame.

Examples of Centrifugal Force

Centripetal Force Centrifugal Force
Acts towards the centre of the circle Acts away from the centre of the circle
Real force — exists in inertial frames Pseudo force — exists only in rotating frames
Causes circular motion Is a reaction felt by the moving body
Required for circular motion Not required for circular motion

9) Uniform Circular Motion

When a body moves along a circular path with constant speed, its motion is called uniform circular motion. Although the speed is constant, the velocity changes because the direction changes continuously. Hence, it is an accelerated motion.

Important Terms in Circular Motion

Quantity Symbol Definition Unit
Radius r Radius of the circular path m
Linear speed v Distance travelled along the circle per unit time m/s
Angular speed ω Angle swept per unit time rad/s
Time period T Time taken for one complete revolution s
Frequency f Number of revolutions per second Hz
Centripetal acceleration ac Acceleration towards the centre = v²/r m/s²
v = ωr ω = 2π / T = 2πf ac = v² / r = ω²r Fc = mv² / r = mω²r Number of revolutions, n = t / T

Examples of Uniform Circular Motion

Key Concept: In uniform circular motion, the speed is constant but the velocity is not constant (because direction changes). Therefore, uniform circular motion is always an accelerated motion, and the acceleration is the centripetal acceleration directed towards the centre.

10) Difference Between Linear Motion and Circular Motion

Linear Motion Circular Motion
Body moves along a straight line Body moves along a circular path
Direction does not change (if moving straight) Direction changes continuously
Velocity can be constant if speed and direction both are constant Velocity changes even if speed is constant
No centripetal force required Centripetal force is required
Acceleration is along the line of motion Acceleration is towards the centre (radial)
Parameters: displacement, velocity, acceleration Parameters: angular displacement, angular velocity, angular acceleration

11) Top Topics for ICSE Exams

The following topics are frequently asked in ICSE Class 10 Physics board exams:

12) Solved Numericals

Numerical 1 — Moment of Force

A force of 25 N is applied on a spanner to tighten a nut. If the perpendicular distance from the line of action of the force to the nut is 15 cm, find the moment of force.

Given: F = 25 N, d = 15 cm = 0.15 m τ = F × d = 25 × 0.15 τ = 3.75 N m (clockwise or anticlockwise depending on direction)

Numerical 2 — Principle of Moments

A uniform metre scale of mass 80 g is pivoted at the 30 cm mark. A mass of 60 g is suspended at the 10 cm mark. Where should a mass of 100 g be suspended to balance the scale?

Weight of scale = 80 g acts at 50 cm mark (centre) Pivot at 30 cm mark Clockwise moment of scale weight = 80 × (50 - 30) = 80 × 20 = 1600 g cm Anticlockwise moment of 60 g = 60 × (30 - 10) = 60 × 20 = 1200 g cm Net clockwise = 1600 - 1200 = 400 g cm (needs to be balanced) Let 100 g be suspended at distance x on right of pivot 100 × x = 400 → x = 4 cm ∴ 100 g at 30 + 4 = 34 cm mark

Numerical 3 — Centripetal Force

A stone of mass 0.5 kg is whirled in a horizontal circle of radius 1.5 m with a speed of 6 m/s. Calculate the centripetal force acting on the stone.

Given: m = 0.5 kg, r = 1.5 m, v = 6 m/s Fc = mv² / r Fc = (0.5 × 6²) / 1.5 Fc = (0.5 × 36) / 1.5 Fc = 18 / 1.5 Fc = 12 N

Numerical 4 — Uniform Circular Motion

A body completes 20 revolutions in 10 seconds on a circular path of radius 2 m. Find its angular speed, linear speed, and centripetal acceleration.

Given: n = 20 revolutions, t = 10 s, r = 2 m Frequency, f = n/t = 20/10 = 2 Hz Time period, T = 1/f = 0.5 s Angular speed, ω = 2πf = 2 × 3.14 × 2 = 12.56 rad/s Linear speed, v = ωr = 12.56 × 2 = 25.12 m/s Centripetal acceleration, ac = v²/r = (25.12)²/2 = 630.91/2 = 315.46 m/s²

Numerical 5 — Equilibrium and COG

A uniform rod of length 2 m and weight 40 N is pivoted at one end. It is kept horizontal by a vertical force at the other end. Calculate the magnitude of the force.

Weight of rod = 40 N acts at centre = 1 m from pivot Moment of weight = 40 × 1 = 40 N m (clockwise) Let the vertical force at the other end be F Distance from pivot = 2 m Anticlockwise moment = F × 2 For equilibrium: F × 2 = 40 ∴ F = 20 N

13) Important Formulas — Quick Recap

Force: F = ma Moment of force (Torque): τ = F × d Principle of moments: Σ Anticlockwise moment = Σ Clockwise moment Centripetal force: Fc = mv²/r Centrifugal force: F = mv²/r (outward, pseudo force) Linear speed: v = ωr Angular speed: ω = 2π/T = 2πf Centripetal acceleration: ac = v²/r = ω²r Frequency: f = 1/T Number of revolutions: n = t/T

14) Quick Revision Points

For ICSE board exams, practice numerical problems on the principle of moments and centripetal force. These are the most common 3-4 mark questions. Also be thorough with the definitions and differences (translational vs rotational, centripetal vs centrifugal, stable vs unstable vs neutral).
Written by
Asad Abdulllah
ICSE 93.4% · 2026
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