Laws of Motion

4.1 INTRODUCTION

**Force** is an external agent required to change the state of motion of a body. Common experience shows that:

A body at rest requires a force to start moving (e.g., kicking a football)

A moving body requires a force to stop (e.g., friction stopping a rolling ball)

Forces can act through contact (hand pushing an object) or at a distance (gravity, magnetism)

The fundamental question is: **What governs the motion of bodies?** This chapter answers this through Newton's Laws of Motion.

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4.2 ARISTOTLE'S FALLACY

**Aristotle's Law of Motion (Incorrect):** An external force is required to keep a body in motion.

**Historical Context:**

Greek philosopher Aristotle (384–322 B.C.) believed that continuous force is needed to maintain motion

Example: Arrow in flight stays airborne because air pushes it forward

This view seemed intuitive from everyday experience (toy car needs continuous pulling)

**The Flaw in Aristotle's Reasoning:**

When a toy car is dragged, it appears to need continuous force. However:

The car experiences **friction force** opposing its motion (by floor)

The applied force balances friction: **F_applied = F_friction**

Net external force = 0, so the car moves with constant velocity

**Corollary:** If friction were absent, no external force would be needed for uniform motion

**Why Aristotle Failed:**

He observed bodies on Earth with friction present everywhere

He coded practical experience (with friction) as a universal law

He did not imagine an ideal frictionless world

**Conclusion:** Aristotle confused the need to overcome friction with the need for force itself

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4.3 THE LAW OF INERTIA

**Galileo's Revolutionary Insight (17th Century):**

Galileo conducted experiments on inclined planes to understand motion without friction:

**Experiment 1: Double Inclined Plane**

A ball released from rest on one incline rolls down and climbs the opposite incline

With friction: Final height < Initial height

**Ideal case (frictionless):** Final height = Initial height

**Experiment 2: Changing Slope**

When the slope of the second incline decreases, the ball still reaches the same height but travels a longer distance

**Limiting case:** When slope = 0 (horizontal plane), the ball would travel **infinite distance**

In other words: **The ball's motion would never cease on a frictionless horizontal plane**

**Galileo's Conclusion:**

The **state of rest** and the **state of uniform linear motion** are equivalent because:

Both imply **zero net external force**

Both imply **zero acceleration**

**No force is needed to maintain uniform motion** — only to overcome friction

**Definition of Inertia:**

**Inertia** = Property of a body to resist change in its state of motion

A body at rest stays at rest

A body in motion stays in motion (same velocity)

Unless an external force acts on it

**The Law of Inertia (Galileo's Statement):**

If the net external force on a body is zero, the body continues in its state of rest or uniform linear motion indefinitely.

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4.4 NEWTON'S FIRST LAW OF MOTION

**Newton's First Law (Statement):**

Every body continues to be in its state of rest or of uniform motion in a straight line unless compelled by some external force to act otherwise.

**Mathematical Form:**

If **F_net = 0**, then **a = 0**

Acceleration is zero only when net external force is zero

**Two Practical Scenarios:**

**Scenario 1: Known that F_net = 0**

Example: A spaceship far from all objects with rockets off

**Conclusion:** Acceleration = 0; if moving, velocity remains constant

**Scenario 2: Known that object is unaccelerated**

Example: A book resting on a table

**Observation:** Book is at rest (a = 0)

**Conclusion by First Law:** F_net must be 0

Therefore: **Weight (W) = Normal Force (R)**

**Critical Distinction:**

*Incorrect reasoning:* "Since W = R, forces cancel, so the book is at rest"

*Correct reasoning:* "Since the book is at rest, by Newton's First Law, net force must be zero. Therefore, R must equal W."

**Common Misconception:** The First Law is often misunderstood as requiring force to maintain motion. The truth is: **No net force is needed for uniform motion; force is needed to change motion.**

**Real-Life Example: Motion in a Bus**

When a bus suddenly accelerates from rest:

Passengers are "thrown backward" relative to the bus

Why? Our body has inertia — it tends to remain at rest (stationary state)

While our feet move with the bus (friction acts), our upper body resists acceleration

The deformable nature of the human body creates this effect

Solution: Muscular forces restore the body to move with the bus

Similarly, when a bus suddenly stops:

Our feet stop due to friction with the floor

Upper body continues forward due to inertia

We're thrown forward until muscular forces stop us

**Example 4.1 (Astronaut in Space):**

*Problem:* An astronaut is separated from a spaceship accelerating at 100 m/s² in interstellar space. What is the astronaut's acceleration immediately after separation?

*Solution:*

No nearby stars exert gravitational force

Spaceship's gravitational pull is negligible

**Net external force = 0**

**By Newton's First Law: a = 0**

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4.5 NEWTON'S SECOND LAW OF MOTION

Momentum

**Definition:**

Momentum **p** = **m v** (mass × velocity)

**SI Unit:** kg⋅m⋅s⁻¹ or N⋅s

**Nature:** Vector quantity (direction of velocity)

Why Momentum Matters: Key Observations

**Observation 1: Mass Dependence**

A loaded truck requires much greater force than a small car to reach the same speed in the same time

Heavy objects need greater opposing force to stop

**Conclusion:** Mass significantly affects force requirements

**Observation 2: Velocity Dependence**

A high-speed bullet pierces tissue; low-speed bullet causes minimal damage

For given mass, greater speed requires greater opposing force to stop

**Conclusion:** Velocity matters in force effects

**Observation 3: Time Dependence (Cricket Catch Example)**

**Experienced cricketer:** Draws hands backward, takes longer time to stop ball (smaller force needed, less pain)

**Novice:** Hands fixed, tries to stop ball instantly (large force needed, hands hurt)

**Same change in momentum, different time intervals → different forces required**

**Conclusion:** Force depends on **rate of change** of momentum, not just the change itself

**Observation 4: Direction Change**

Rotating a stone in a horizontal circle at constant speed

Magnitude of momentum stays constant, but **direction changes continuously**

Hand must apply force through string to change momentum direction

Greater speed or smaller radius = greater force needed (greater rate of change of momentum)

**Conclusion:** Force is needed to change direction of momentum, even with constant magnitude

Newton's Second Law (Statement)

**The rate of change of momentum of a body is directly proportional to the applied force and takes place in the direction in which the force acts.**

Derivation of F = ma

Consider a body of mass **m** with:

Initial velocity: **u**

Final velocity: **v**

Time interval: Δt

Applied force: **F**

**Step 1:** Initial momentum = **p₁ = m u**

**Step 2:** Final momentum = **p₂ = m v**

**Step 3:** Change in momentum = Δp = m(v - u) = m⋅Δv

**Step 4:** Rate of change of momentum = Δp/Δt = m(Δv/Δt)

**Step 5:** By definition, Δv/Δt = **a** (acceleration)

**Step 6:** Therefore, rate of change of momentum = **m a**

**By Newton's Second Law:**

$$F \propto \frac{\Delta p}{\Delta t}$$

$$F = k \cdot \frac{\Delta p}{\Delta t} = k \cdot m a$$

The proportionality constant **k = 1** in SI units, so:

$$\boxed{F = m a}$$

**Vector Form:** **F** = m **a**

Alternative Statement of Second Law

**From F = ma, we derive:**

$$a = \frac{F}{m}$$

Or in terms of momentum:

$$F = \frac{dp}{dt}$$

**Interpretation:**

Acceleration produced in a body is directly proportional to applied force

Acceleration is inversely proportional to mass

Direction of acceleration = Direction of applied force

Units and Dimensions

**Newton (N):** SI unit of force

1 N = 1 kg⋅m⋅s⁻²

**Definition:** Force that produces acceleration of 1 m/s² in a mass of 1 kg

**Dimensional Formula:** [F] = M L T⁻²

Important Insights

**Newton's Second Law separates inertial mass from force:**

**Inertial mass (m):** Resistance of body to acceleration (property of body)

**Force (F):** External agent causing acceleration

Not all forces are visible; gravity acts at a distance

**The mass is always positive:**

Same force on different masses produces different accelerations inversely proportional to mass

Lighter objects accelerate more easily

**Net force matters:**

Only the **net (total) external force** determines acceleration

Individual forces can balance (net = 0), then a = 0

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4.6 NEWTON'S THIRD LAW OF MOTION

**Newton's Third Law (Statement):**

To every action, there is an equal and opposite reaction. The forces act on different bodies.

**Mathematical Form:**

If body A exerts force **F_AB** on body B, then body B exerts force **F_BA** on body A, such that:

$$\boxed{F_{AB} = -F_{BA}}$$

**Or:** **F_action = -F_reaction**

**Critical Points:**

1. **Action and reaction are equal in magnitude but opposite in direction**

2. **Action and reaction act on different bodies**

They cannot cancel each other

We cannot add them together in force balance equations

3. **Forces act simultaneously** — there is no time delay between action and reaction

4. **Same type of force** — if action is gravitational, reaction is gravitational; if normal force, reaction is normal force

Examples of Action-Reaction Pairs

**Example 1: Book on Table**

Table pushes book upward with force **R** (action)

Book pushes table downward with force **R** (reaction)

Both forces are normal forces, equal magnitude, opposite direction

They act on **different bodies** (book and table), so they don't cancel

Book's weight ≠ reaction force; weight and normal force balance book's weight

**Example 2: Walking**

Person pushes ground backward with force (action)

Ground pushes person forward with force (reaction)

Equal magnitude, opposite direction

Person accelerates forward due to reaction force from ground

**Example 3: Rocket Launch**

Rocket pushes exhaust gases downward with force (action)

Exhaust gases push rocket upward with force (reaction)

Rocket accelerates upward despite gravity

**Example 4: Swimming**

Swimmer pushes water backward (action)

Water pushes swimmer forward (reaction)

Key Distinction: Force Pairs vs. Balanced Forces

| **Action-Reaction Pair** | **Balanced Forces** |

|---|---|

| Equal, opposite forces | Equal, opposite forces |

| Act on **different bodies** | Act on **same body** |

| Cannot cancel each other | Cancel each other (net = 0) |

| Always present together | Present only in equilibrium |

| Example: Book weight (down) and table normal (up) | Not an action-reaction pair |

**Correct Analysis of Book on Table:**

Weight (W) and Normal force (R) are **not** an action-reaction pair

Both act on the book

They balance: W = R (equilibrium)

**Actual action-reaction pairs:**

Earth pulls book down (action); book pulls Earth up (reaction)

Table pushes book up (action); book pushes table down (reaction)

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4.7 CONSERVATION OF MOMENTUM

Statement of Conservation Law

**If the net external force on a system of bodies is zero, the total momentum of the system remains constant.**

$$\boxed{p_{initial} = p_{final} \quad \text{or} \quad \Delta p = 0}$$

Derivation

Consider a system of **n** bodies with **no net external force** (F_external = 0):

**From Newton's Second Law:**

$$F_{net} = \frac{dp_{total}}{dt}$$

If F_net = 0:

$$\frac{dp_{total}}{dt} = 0$$

Therefore:

$$p_{total} = \text{constant}$$

**For two bodies colliding (internal forces only):**

Body 1 exerts force **F₁₂** on body 2

Body 2 exerts force **F₂₁** on body 1

By Newton's Third Law: **F₁₂ = -F₂₁**

**In time interval Δt:**

Change in momentum of body 1: Δp₁ = F₂₁⋅Δt

Change in momentum of body 2: Δp₂ = F₁₂⋅Δt

$$\Delta p_1 + \Delta p_2 = (F_{21} + F_{12}) \Delta t = 0$$

$$p_1 + p_2 = \text{constant}$$

$$m_1 v_1 + m_2 v_2 = m_1 v_1' + m_2 v_2'$$

**Where:**

v₁, v₂ = initial velocities

v₁', v₂' = final velocities

Conditions for Conservation

1. **Net external force = 0**

2. System must be **isolated** (no external forces acting)

3. Internal forces (between bodies in system) can exist; they don't affect total momentum

Real-Life Examples

**Example 1: Gun Recoil**

A person fires a gun:

**System:** Gun + Bullet (no external horizontal force)

**Initial momentum:** 0 (both at rest)

**Final momentum:** m_bullet⋅v_bullet + m_gun⋅v_gun = 0

$$m_{bullet} v_{bullet} = -m_{gun} v_{gun}$$

**Gun recoils backward because momentum must be conserved.**

**Example 2: Rocket in Space**

**System:** Rocket + Fuel (no external force in space)

Rocket ejects exhaust gases backward

By conservation of momentum, rocket moves forward

Total momentum remains zero (both initially at rest)

**Example 3: Collision on Frictionless Surface**

Two bodies collide elastically:

**Before collision:** p = m₁u₁ + m₂u₂

**After collision:** p = m₁v₁ + m₂v₂

**Momentum conserved:** m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

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4.8 EQUILIBRIUM OF A PARTICLE

**Definition:** A particle is in **equilibrium** when:

Net external force = 0

Acceleration = 0

Velocity is constant (may be zero or non-zero)

Types of Equilibrium

**Static Equilibrium:**

Body is at rest

Velocity = 0

Net force = 0

**Dynamic Equilibrium:**

Body moves with constant velocity

Acceleration = 0

Net force = 0

Condition for Equilibrium

$$\sum \vec{F} = 0$$

**In component form (2D):**

$$\sum F_x = 0 \quad \text{and} \quad \sum F_y = 0$$

**In component form (3D):**

$$\sum F_x = 0, \quad \sum F_y = 0, \quad \sum F_z = 0$$

Method of Free Body Diagram (FBD)

**Procedure:**

1. Isolate the body (or particle)

2. Draw all external forces acting on it

3. Choose a coordinate system

4. Resolve forces into components

5. Apply equilibrium condition: ΣF = 0 in each direction

Example: Book on Inclined Plane

**Setup:** Book of mass **m** on smooth incline at angle **θ**

**Forces:**

1. Weight: W = mg (vertically downward)

2. Normal force: N (perpendicular to incline)

**Component along incline (parallel):**

$$mg \sin\theta = 0 \quad \text{(if at rest or moving with constant velocity)}$$

But if no other force acts parallel to incline, the book will accelerate.

**To maintain equilibrium (constant velocity up the incline):**

Applied force F must be such that: **F = mg sinθ** (up the incline)

**Component perpendicular to incline:**

$$N = mg\cos\theta$$

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4.9 COMMON FORCES IN MECHANICS

4.9.1 Gravitational Force

**Definition:** Attractive force between any two masses.

$$F_g = \frac{GMm}{r^2}$$

**Where:**

G = 6.67 × 10⁻¹¹ N⋅m²⋅kg⁻²

M, m = masses

r = distance between centers of mass

**Near Earth's surface (r ≈ R_E):**

$$F_g = mg$$

**Where:**

g = 9.8 m/s² (acceleration due to gravity)

**Weight W = mg** (SI unit: Newton)

**Characteristics:**

Always attractive

Acts along line joining centers of mass

Proportional to product of masses

Inversely proportional to square of distance

4.9.2 Normal Force (N)

**Definition:** Contact force perpendicular to surface, preventing one body from penetrating another.

**Characteristics:**

Acts perpendicular to surface

Self-adjusting force (adjusts to prevent penetration)

Cannot push (only push or become zero)

Reaction to applied normal force

**Example: Block on Horizontal Surface**

If no external force acts on block: **N = mg**

If downward force F acts on block: **N = mg + F**

If upward force F acts: **N = mg - F** (if F < mg)

4.9.3 Tension (T)

**Definition:** Contact force in a string, rope, or cable, along its length.

**Characteristics:**

Acts along the string/rope

Always pulling (never pushing)

Assumed same throughout (if string is massless and pulley frictionless)

Same on both ends of string

**Example: Hanging Mass**

If mass m hangs from string: **T = mg** (at rest)

If mass accelerates upward with acceleration a: **T = m(g + a)**

If mass accelerates downward with acceleration a: **T = m(g - a)**

4.9.4 Friction

**Definition:** Contact force opposing relative motion (or tendency to move) between surfaces.

**Static Friction (f_s):**

Acts when surfaces are in contact but not sliding

Prevents relative motion

Magnitude varies: **0 ≤ f_s ≤ μ_s N**

**Maximum static friction = μ_s N**

**Where μ_s** = coefficient of static friction

**Kinetic Friction (f_k):**

Acts when surfaces are sliding relative to each other

Always opposes direction of relative motion

**f_k = μ_k N** (constant magnitude)

**Where μ_k** = coefficient of kinetic friction

**Important Relations:**

**μ_k < μ_s** (kinetic friction less than maximum static friction)

Both are dimensionless

Both depend on nature of surfaces, not on normal force magnitude alone

**Characteristics:**

Always opposes motion (or tendency to move)

Acts parallel to surfaces in contact

Cannot exist without normal force

**Real-Life Example: Car on Road**

When car accelerates forward:

Wheels push road backward (action)

Road pushes wheels forward (reaction) — this is friction

Friction accelerates the entire car forward

Without friction, wheels would spin, and car wouldn't move

**Direction Rule:** Friction opposes the relative motion (or tendency). If car tends to slip backward relative to ground, friction acts forward on car.

4.9.5 Spring Force

**Definition:** Restoring force exerted by a spring when compressed or extended.

**Hooke's Law:**

$$F_s = -kx$$

**Where:**

k = spring constant (N/m)

x = displacement from natural length (m)

Negative sign indicates restoring nature (opposite to displacement)

**SI Unit:** Newton (N)

**Spring constant units:** N⋅m⁻¹

**Characteristics:**

Proportional to displacement

Opposes displacement (restoring)

Different springs have different k values

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4.10 CIRCULAR MOTION

Circular Motion Definition

**Circular motion** is motion of a body along a circular path with center at fixed point.

Uniform Circular Motion

**Definition:** Motion in circular path with constant speed.

**Key Parameters:**

**Speed (v):**

Magnitude of velocity along tangent

Constant in uniform circular motion

**v = 2πr/T** where T = period

**Velocity (instantaneous):**

Direction: along tangent to circle (continuously changing)

Magnitude: constant

**Angular velocity (ω):**

$$\omega = \frac{v}{r} = \frac{2\pi}{T} = 2\pi f$$

**SI Unit:** rad⋅s⁻¹

**Where f** = frequency (revolutions per unit time)

Centripetal Acceleration

**Definition:** Acceleration directed toward center of circle, necessary to change velocity direction.

**Derivation:**

Velocity magnitude constant, but direction changes

Rate of change of velocity = acceleration

This acceleration is centripetal

$$a_c = \frac{v^2}{r} = \omega^2 r = \omega v$$

**SI Unit:** m⋅s⁻²

**Direction:** Always toward center

**Characteristics:**

Not zero (despite constant speed)

Perpendicular to velocity

Causes circular motion

Centripetal Force

**Definition:** Net force toward center of circle required to produce centripetal acceleration.

**From Newton's Second Law:**

$$F_c = m a_c = \frac{mv^2}{r} = m\omega^2 r$$

**SI Unit:** Newton (N)

**Direction:** Toward center of circular path

**Nature:**

Centripetal force is not a new force type

It's the net force component toward center

Could be provided by tension, friction, normal force, gravity, or combination

Examples of Centripetal Force

**Example 1: Stone Rotating on String**

Centripetal force provided by **tension in string**

If string breaks: stone flies off tangentially (no centripetal force)

**Example 2: Car on Circular Road**

Centripetal force provided by **friction between tires and road**

Friction acts toward center of curve

Maximum speed determined by: **v_max = √(μgr)**

**Example 3: Moon Orbiting Earth**

Centripetal force provided by **gravitational force**

Gravity provides the centripetal force: **F_g = F_c**

**Example 4: Vertical Circular Motion (Loop-the-Loop)**

At **top of loop:**

Weight and normal force both point toward center (downward)

**mg + N = mv²/r**

**N = m(v²/r - g)**

Minimum speed for contact: v_min = √(gr)

At **bottom of loop:**

Normal force points toward center (upward), weight downward

**N - mg = mv²/r**

**N = m(v²/r + g)** (normal force larger)

Important Insight: Direction of Force vs. Acceleration

In circular motion:

**Force is centripetal (toward center)**

**Velocity is tangential (along tangent)**

Force and velocity are perpendicular

**This is why circular motion is possible with perpendicular force and velocity**

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4.11 SOLVING PROBLEMS IN MECHANICS

General Strategy

**Step 1: Understand the Problem**

Identify all bodies involved

List given data and find quantities

Identify physical situation (equilibrium, acceleration, circular, etc.)

**Step 2: Draw Free Body Diagram (FBD)**

Isolate each body

Show all forces acting on the body

Label forces with magnitudes and directions

Use appropriate symbols (T, N, f, W, F, etc.)

**Step 3: Choose Coordinate System**

Usually horizontal (x) and vertical (y) for linear motion

Choose convenient direction (often direction of motion)

For inclined plane, use parallel and perpendicular to plane

**Step 4: Resolve Forces into Components**

Break each force into x and y components

Use trigonometry: F_x = F cosθ, F_y = F sinθ

**Step 5: Apply Newton's Laws**

**For equilibrium:** ΣF_x = 0 and ΣF_y = 0

**For acceleration:** ΣF_x = ma_x and ΣF_y = ma_y

**Step 6: Solve Equations**

Solve system of equations for unknowns

Use kinematic equations if needed: v² = u² + 2as, etc.

**Step 7: Check Answer**

Verify units

Check if answer is reasonable

Check limiting cases

Common Problem Types

**Type 1: Equilibrium Problems**

Book on table, hanging mass, body on incline

Apply: ΣF = 0

**Type 2: Linear Motion with Constant Force**

Block on horizontal surface, motion on incline

Apply: F_net = ma

**Type 3: Connected Bodies**

Pulley systems, Atwood machine

Treat each body separately, use constraint relations

**Type 4: Circular Motion**

Apply: F_c = mv²/r toward center

**Type 5: Collision and Momentum**

Apply: Conservation of momentum if no external force

Worked Example: Block on Incline with Friction

**Problem:** A block of mass 5 kg is placed on a smooth incline at angle 30°. Find the acceleration of the block down the incline.

**Solution:**

**Step 1:** m = 5 kg, θ = 30°, g = 10 m/s² (assuming)

**Step 2: Free Body Diagram**

Weight W = mg (vertically downward)

Normal force N (perpendicular to incline)

Incline is smooth (frictionless)

**Step 3: Coordinate System**

x-axis: along incline (positive downward)

y-axis: perpendicular to incline (positive away from incline)

**Step 4: Resolve Forces**

Weight component along incline: W_∥ = mg sinθ = 5 × 10 × sin30° = 50 × 0.5 = 25 N

Weight component perpendicular: W_⊥ = mg cosθ = 5 × 10 × cos30° = 50 × 0.866 = 43.3 N

Normal force: N (perpendicular to incline)

**Step 5: Apply Newton's Second Law**

Along incline (x-direction):

$$mg\sin\theta = ma$$

Perpendicular to incline (y-direction):

$$N = mg\cos\theta$$

**Step 6: Solve for acceleration**

$$a = g\sin\theta = 10 \times \sin30° = 10 \times 0.5 = 5 \text{ m/s}^2$$

**Answer:** Acceleration = 5 m/s² down the incline

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DIMENSIONAL ANALYSIS (Quick Reference)

**Momentum:** [p] = M L T⁻¹

**Force:** [F] = M L T⁻²

**Weight:** [W] = M L T⁻²

**Normal Force:** [N] = M L T⁻²

**Friction:** [f] = M L T⁻²

**Centripetal Force:** [F_c] = M L T⁻²

**Spring constant:** [k] = M T⁻²

All forces have same dimensions (as expected from F = ma).

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SI UNITS (Quick Reference)

| Quantity | Symbol | SI Unit | Equivalent |

|---|---|---|---|

| Force | F | Newton (N) | kg⋅m⋅s⁻² |

| Momentum | p | kg⋅m⋅s⁻¹ | N⋅s |

| Weight | W | Newton (N) | kg⋅m⋅s⁻² |

| Acceleration | a | m⋅s⁻² | — |

| Mass | m | kilogram (kg) | — |

| Angular velocity | ω | rad⋅s⁻¹ | s⁻¹ |

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SUMMARY OF LAWS

**Newton's First Law:** F_net = 0 ⟹ a = 0 (rest or uniform motion)

**Newton's Second Law:** F = ma (force produces acceleration proportional to mass)

**Newton's Third Law:** F_action = -F_reaction (equal, opposite, different bodies)

**Conservation of Momentum:** If F_external = 0, then p_total = constant

**Circular Motion:** F_c = mv²/r (centripetal force toward center)