Chapter 5: Laws of Motion

Introduction

Key Points

• External force is needed to start or stop a body's motion.

• Examples of external forces:

  • Kicking a stationary football.

  • Wind moving tree branches or heavy objects.

  • Boat moving with river flow without rowing.

  • Magnet attracting iron nail from a distance (non-contact force).

  • Stone accelerating downward due to gravity.

• External forces can be contact or non-contact.

• No external force is needed to maintain uniform motion (Newton's First Law).

Solved Examples

• Ex 1: A football at rest is kicked with a force of 10 N. What causes it to move?
  - Force applied by the kick (external contact force) initiates motion.

• Ex 2: A stone falls from a building. What force causes acceleration?
  - Gravitational force (non-contact force) causes acceleration downward.

Practice Set

• Q1: Which of the following is a non-contact force?
  A) Friction
  B) Magnetic force
  C) Tension
  D) Applied force

• Q2: A boat moves downstream without rowing. What force causes this?
  A) Wind force
  B) River flow force
  C) Applied force
  D) Friction

• Q3: A stone is dropped from 20 m height. Which force acts on it?
  A) Normal force
  B) Gravitational force
  C) Frictional force
  D) Magnetic force

Answer Key

• Q1: B - Magnetic force acts at a distance without contact.

• Q2: B - River flow exerts force moving the boat downstream.

• Q3: B - Gravity pulls the stone downward causing acceleration.

Memory Aids

• External force = needed to start/stop motion.

• Contact force = direct touch; Non-contact = action at a distance.

• Uniform motion continues without external force.

Aristotle's Fallacy

Key Points

• Aristotle incorrectly stated that continuous force is needed to keep a body moving.

• In reality, force is only needed to overcome friction or other resistive forces.

• Galileo proposed that in absence of friction, uniform motion continues without force.

• Friction is an opposing force requiring external force to maintain motion in practice.

Solved Examples

• Ex 1: A block slides on a rough surface and stops. Why?
  - Friction opposes motion; no external force to counter it, so block stops.

• Ex 2: A puck slides on frictionless ice. What happens?
  - It continues moving at constant velocity without external force.

Practice Set

• Q1: According to Aristotle, what is required to keep a body moving?
  A) No force
  B) Continuous external force
  C) Only initial force
  D) Frictional force

• Q2: What force must be overcome to maintain motion on Earth?
  A) Gravity
  B) Friction
  C) Magnetic force
  D) Tension

• Q3: Galileo's concept of motion states:
  A) Motion stops without force
  B) Uniform motion continues without force
  C) Force is always needed
  D) Motion is impossible without friction

Answer Key

• Q1: B - Aristotle believed continuous force is needed to keep motion.

• Q2: B - Friction opposes motion and must be overcome.

• Q3: B - Galileo stated uniform motion continues without external force.

Memory Aids

• Aristotle: Force needed always (incorrect).

• Galileo: No force needed for uniform motion (correct).

• Friction = real-world force opposing motion.

Law of Inertia

Galileo's Observations

• Objects moving down an inclined plane accelerate.

• Objects moving up an inclined plane retard (decelerate).

• Motion on a horizontal plane shows no acceleration or retardation if frictionless.

• Conclusion: On a frictionless horizontal surface, an object moves with constant velocity.

Double Inclined Plane Experiment

• A ball released from rest on one incline rolls down and climbs the other.

• If friction is absent, final height ≈ initial height.

• Height does not depend on the shape of the incline.

Curved Track Motion

• Ball rolling upward slows down due to gravity.

• Ball rolling downward speeds up due to gravity.

• Direction and speed depend on slope and gravity.

Definition of Inertia

• If net external force is zero, a body at rest stays at rest.

• A body in motion continues with uniform velocity.

• Inertia = resistance to change in motion.

• Change in state requires external force.

Solved Examples

Ex 1: A ball rolls on a frictionless horizontal surface with velocity 5 m/s. What is its acceleration?

• Since no external force, acceleration = 0 m/s².

Ex 2: A ball released from height 2 m on one incline rolls up another incline. What is the final height reached (neglecting friction)?

• Final height = Initial height = 2 m.

Practice Set

Q1: An object on a frictionless horizontal surface moves with velocity 10 m/s. What force acts on it?

• A) Zero

• B) 10 N

• C) Depends on mass

• D) Cannot be determined

Q2: A ball rolls down a smooth incline and climbs another. If initial height is 3 m, final height is:

• A) Greater than 3 m

• B) Equal to 3 m

• C) Less than 3 m

• D) Zero

Q3: A ball rolls on a curved track. When moving upward, its speed:

• A) Increases

• B) Decreases

• C) Remains constant

• D) Becomes zero immediately

Answer Key

Q1: A) Zero — No external force means no acceleration.

Q2: B) Equal to 3 m — Energy conserved, height remains same without friction.

Q3: B) Decreases — Gravity opposes motion upward, reducing speed.

Memory Aids

• Inertia = resistance to change in motion.

• No net force → constant velocity or rest.

• Energy conservation on frictionless inclines → same height.

• Upward motion slows down; downward speeds up.

Newton's First Law

Law Statement

• A body at rest stays at rest unless acted upon by a net external force.

• A body in uniform motion continues in uniform motion unless acted upon by a net external force.

• Uniform motion: constant speed in a straight line.

Solved Examples

• Ex 1: A book lying on a table remains at rest. Why?
  Answer: No net external force acts; inertia keeps it at rest.

• Ex 2: A hockey puck slides on ice at constant speed. Why does it not stop?
  Answer: Negligible friction means no net force; it continues uniform motion.

Practice Set

• Q1: What happens to a body at rest if no external force acts?
  A) Moves randomly
  B) Remains at rest
  C) Accelerates
  D) Changes direction

• Q2: A car moving at constant velocity will:
  A) Speed up
  B) Slow down
  C) Continue at same velocity
  D) Stop immediately

• Q3: Which force changes the state of motion of a body?
  A) Balanced force
  B) Frictional force
  C) Net external force
  D) Gravitational force

Answer Key

• Q1: B; No net force means body remains at rest.

• Q2: C; Constant velocity implies no net force, so motion continues.

• Q3: C; Only net external force changes motion state.

Memory Aids

• "Law of Inertia": Objects resist change in motion.

• Uniform motion = constant velocity (speed + direction).

• No net force → no change in velocity.

Inertia

Definition and Properties

• Inertia: tendency of a body to resist change in its state of motion.

• Measured by mass; greater mass means greater inertia.

• Heavier objects require more force to change motion.

Solved Examples

• Ex 1: Why does a heavier object resist motion change more?
  Answer: Larger mass means greater inertia.

• Ex 2: Why does a lighter ball stop sooner?
  Answer: Less mass means less inertia, easier to change motion.

Practice Set

• Q1: Inertia depends on:
  A) Velocity
  B) Mass
  C) Force
  D) Acceleration

• Q2: Which has more inertia?
  A) 1 kg iron ball
  B) 2 kg iron ball
  C) 1 kg feather
  D) 0.5 kg stone

• Q3: To increase inertia, you must:
  A) Increase speed
  B) Increase mass
  C) Apply force
  D) Change direction

Answer Key

• Q1: B; Inertia is proportional to mass.

• Q2: B; 2 kg mass has greater inertia.

• Q3: B; Increasing mass increases inertia.

Memory Aids

• Inertia ∝ mass.

• More mass = more resistance to motion change.

• Think: "Heavy objects resist motion change more."

Types of Inertia

Categories

• Inertia of Rest: Resistance to change from rest to motion.
  Example: Mangoes fall when branch shakes.

• Inertia of Motion: Resistance to change from motion to rest.
  Example: Passenger falls forward when bus stops suddenly.

• Inertia of Direction: Resistance to change in direction of motion.
  Example: Mud thrown from rotating wheels stopped by mudguard.

Solved Examples

• Ex 1: Why does a passenger lurch forward when a bus stops?
  Answer: Inertia of motion resists sudden stop.

• Ex 2: Why do mangoes fall when branch shakes?
  Answer: Inertia of rest resists change from rest.

Practice Set

• Q1: Inertia of rest means:
  A) Resistance to start moving
  B) Resistance to stop moving
  C) Resistance to change direction
  D) Resistance to acceleration

• Q2: When a vehicle turns, mud flies off due to:
  A) Inertia of rest
  B) Inertia of motion
  C) Inertia of direction
  D) Gravity

• Q3: A passenger falls forward when bus stops because of:
  A) Inertia of rest
  B) Inertia of motion
  C) Inertia of direction
  D) Friction

Answer Key

• Q1: A; Inertia of rest resists starting motion.

• Q2: C; Inertia of direction resists change in direction.

• Q3: B; Inertia of motion resists stopping.

Memory Aids

• Rest → resist starting.

• Motion → resist stopping.

• Direction → resist turning.

Momentum

Definition and Formula

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

• Formula: p = mv

• Vector quantity: direction same as velocity.

• Measures quantity of motion.

• Introduced by Newton to quantify force effect.

Solved Examples

• Ex 1: Calculate momentum of 2 kg object moving at 3 m/s.
  Solution: p = 2 × 3 = 6 kg·m/s

• Ex 2: If velocity doubles, momentum:
  Solution: Doubles, since p ∝ v.

Practice Set

• Q1: Momentum is:
  A) Scalar
  B) Vector
  C) Unitless
  D) Force

• Q2: Momentum of 5 kg object moving at 4 m/s is:
  A) 9 kg·m/s
  B) 20 kg·m/s
  C) 1.25 kg·m/s
  D) 0.8 kg·m/s

• Q3: If mass is 3 kg and momentum is 15 kg·m/s, velocity is:
  A) 5 m/s
  B) 45 m/s
  C) 0.2 m/s
  D) 18 m/s

Answer Key

• Q1: B; Momentum has magnitude and direction.

• Q2: B; p = 5 × 4 = 20 kg·m/s.

• Q3: A; v = p/m = 15/3 = 5 m/s.

Memory Aids

• Momentum = mass × velocity.

• Vector quantity: direction matters.

• Double velocity → double momentum.

Newton's Second Law of Motion

Fundamental Concept

• Rate of change of linear momentum \( \vec{P} \) is proportional to applied force \( \vec{F} \).

• Mathematically: \( \vec{F} = \frac{d\vec{P}}{dt} \).

• For constant mass, \( \vec{P} = m\vec{v} \), so \( \vec{F} = m\frac{d\vec{v}}{dt} = m\vec{a} \).

• Acceleration \( a \) is directly proportional to net force \( F_{net} \) and inversely proportional to mass \( m \): \( a \propto \frac{F_{net}}{m} \).

• In SI units, constant of proportionality is 1: \( a = \frac{F_{net}}{m} \) or \( F_{net} = ma \).

• More general form when mass varies: \( F_{net,external} = \frac{d(mv)}{dt} = m\frac{dv}{dt} + v\frac{dm}{dt} \).

Solved Examples

• Ex 1: Calculate acceleration of a 5 kg object when a net force of 20 N is applied.

  • Use \( a = \frac{F}{m} = \frac{20}{5} = 4\, m/s^2 \).

• Ex 2: Find net force on a 10 kg object accelerating at \( 3\, m/s^2 \).

  • Use \( F = ma = 10 \times 3 = 30\, N \).

Practice Set

• Q1: What is the acceleration of a 2 kg object if a net force of 10 N acts on it?

  • A) 5 m/s²

  • B) 20 m/s²

  • C) 0.2 m/s²

  • D) 12 m/s²

• Q2: A force of 15 N causes an acceleration of 3 m/s². What is the mass of the object?

  • A) 5 kg

  • B) 45 kg

  • C) 18 kg

  • D) 12 kg

• Q3: An object of mass 4 kg accelerates at 2 m/s². If the mass suddenly increases to 6 kg without changing velocity, what is the new net force assuming velocity is constant?

  • A) 8 N

  • B) 0 N

  • C) 12 N

  • D) 4 N

Answer Key

• Q1: A) 5 m/s² — \( a = F/m = 10/2 = 5 \)

• Q2: A) 5 kg — \( m = F/a = 15/3 = 5 \)

• Q3: B) 0 N — Velocity constant means \( dv/dt = 0 \) and \( dm/dt = 0 \), so net external force is zero.

Memory Aids

• Force equals mass times acceleration: \( F = ma \).

• Acceleration increases with force, decreases with mass.

• Zero net force means zero acceleration (constant velocity or rest).

• For variable mass systems, use \( F = m\frac{dv}{dt} + v\frac{dm}{dt} \).

Newton's Third Law of Motion

Key Concepts

• For every action, there is an equal and opposite reaction.

• If body A exerts force \( \vec{F} \) on body B, then B exerts \( -\vec{F} \) on A.

• Action and reaction forces act along the same line but on different bodies.

• These forces never cancel each other as they act on different objects.

• Action-reaction pairs arise from contact or at a distance.

• Applicable whether bodies are at rest or in motion.

• Force always exists in pairs; isolated single force does not exist.

Solved Examples

Ex 1: A person pushes a wall with a force of 50 N. What is the force exerted by the wall on the person?

• Action force = 50 N (person on wall)

• Reaction force = -50 N (wall on person)

• Equal in magnitude, opposite in direction.

Practice Set

Q1: If a force of 10 N is applied by object A on object B, the force exerted by B on A is:

• A) 0 N

• B) 5 N

• C) 10 N opposite direction

• D) 10 N same direction

Q2: Two ice skaters push each other. Skater A exerts 30 N on B. What force does B exert on A?

• A) 0 N

• B) 15 N

• C) 30 N opposite direction

• D) 30 N same direction

Q3: A block of mass 2 kg rests on a table. The block exerts a force of 20 N on the table. What is the force exerted by the table on the block?

• A) 10 N upward

• B) 20 N upward

• C) 20 N downward

• D) 40 N upward

Answer Key

• Q1: C - Equal magnitude 10 N, opposite direction as per Newton's third law.

• Q2: C - Reaction force equals action force, opposite direction.

• Q3: B - Table exerts equal and opposite force of 20 N upward on block.

Memory Aids

• "Action = Reaction" (equal magnitude, opposite direction)

• Forces always come in pairs on different bodies.

• Action and reaction never cancel out.

Pseudo Force

Key Concepts

• Appears in non-inertial (accelerating) reference frames.

• Not a real force; imaginary force to apply Newton's laws in accelerating frames.

• Magnitude: \( F_{pseudo} = m a_f \), where \( a_f \) is acceleration of the frame.

• Direction opposite to acceleration of the reference frame.

• Does not arise from physical interaction.

• Newton's third law does not apply to pseudo forces.

• In non-inertial frame: \( \vec{F}_{ext} + \vec{F}_{pseudo} = m \vec{a} \)

Solved Examples

Ex 1: A block of mass 5 kg is observed from a frame accelerating at 2 m/s² to the right. Calculate the pseudo force on the block.

• Given: \( m = 5 \) kg, \( a_f = 2 \) m/s²

• \( F_{pseudo} = m a_f = 5 \times 2 = 10 \) N

• Direction opposite to frame acceleration (to the left).

Practice Set

Q1: A frame accelerates at 3 m/s² to the right. What is the pseudo force on a 4 kg object inside?

• A) 12 N right

• B) 12 N left

• C) 7 N right

• D) 7 N left

Q2: An observer in a non-inertial frame accelerating at 5 m/s² sees a 2 kg object. What is the magnitude and direction of pseudo force?

• A) 10 N opposite to acceleration

• B) 10 N same as acceleration

• C) 7 N opposite to acceleration

• D) 7 N same as acceleration

Q3: A block of mass 3 kg is on a smooth surface. A force \( \vec{F} = 15 \) N acts on it causing acceleration \( a = 5 \) m/s². An observer in a frame accelerating at 2 m/s² to the right observes the block. What is the net force on the block in observer's frame?

• A) 9 N right

• B) 21 N right

• C) 9 N left

• D) 21 N left

Answer Key

• Q1: B - \( F_{pseudo} = 4 \times 3 = 12 \) N opposite to acceleration (left).

• Q2: A - \( 2 \times 5 = 10 \) N opposite to acceleration.

• Q3: A - Net force = \( m(a - a_f) = 3(5 - 2) = 9 \) N right.

Memory Aids

• Pseudo force = mass × acceleration of frame.

• Direction opposite to frame acceleration.

• Only appears in accelerating frames.

Conservation of Momentum

Fundamental Principles

• Total linear momentum of an isolated system remains constant if no external force acts.

• Newton's second law: \( \vec{F} = \frac{d\vec{P}}{dt} \).

• For isolated system: \( \vec{F} = 0 \Rightarrow \frac{d\vec{P}}{dt} = 0 \Rightarrow \vec{P} = \text{constant} \).

• Mathematically: \( \vec{P}_1 + \vec{P}_2 + \cdots + \vec{P}_n = \text{constant} \).

Impulse and Momentum

• Impulse measures the total effect of a force applied over a short time.

• Impulse \( = \text{Force} \times \text{Time} \).

• For infinitesimal time \( dt \), impulse \( = \vec{F} dt \).

• For finite time interval \( t_1 \) to \( t_2 \), impulse \( = \int_{t_1}^{t_2} \vec{F} dt \).

• Impulse equals change in momentum: \( \vec{F} \Delta t = \Delta \vec{P} \) (Impulse-Momentum theorem).

Solved Examples

• Ex 1: A force of 10 N acts on a body for 0.5 s. Find the impulse.

  • Impulse = Force × time = 10 N × 0.5 s = 5 Ns.

• Ex 2: A force \( \vec{F} = 20 \hat{i} \) N acts from \( t=2s \) to \( t=4s \). Calculate impulse.

  • Impulse = \( \int_{2}^{4} 20 dt = 20 \times (4-2) = 40 \) Ns in \( \hat{i} \) direction.

Practice Set

• Q1: What is conserved in an isolated system with no external force?

  • A) Kinetic energy

  • B) Linear momentum

  • C) Potential energy

  • D) Angular velocity

• Q2: A force of 15 N acts on a particle for 0.2 s. What is the impulse?

  • A) 3 Ns

  • B) 30 Ns

  • C) 0.3 Ns

  • D) 75 Ns

• Q3: A force \( \vec{F} = (10 + 5t) \) N acts on a body from \( t=0 \) to \( t=2s \). Find the impulse.

  • A) 20 Ns

  • B) 30 Ns

  • C) 40 Ns

  • D) 50 Ns

Answer Key

• Q1: B) Linear momentum - It remains constant if no external force acts.

• Q2: A) 3 Ns - Impulse = Force × time = 15 × 0.2 = 3 Ns.

• Q3: C) 40 Ns - Impulse = \( \int_0^2 (10 + 5t) dt = [10t + \frac{5t^2}{2}]_0^2 = 20 + 10 = 30 \) Ns. (Correction: Actually 30 Ns, so option B is correct.)

Memory Aids

• Impulse = Force × Time (area under Force-Time graph).

• Impulse = Change in Momentum.

• Momentum conserved if external force = 0.

Equilibrium of a Particle

Key Concepts

• For two forces \( \vec{F_1} \) and \( \vec{F_2} \) acting on a particle, equilibrium requires \( \vec{F_1} = -\vec{F_2} \).

• For three concurrent forces \( \vec{F_1}, \vec{F_2}, \vec{F_3} \), equilibrium requires \( \vec{F_1} + \vec{F_2} + \vec{F_3} = 0 \).

• Concurrent forces act at the same point on a body.

• Resultant of concurrent forces in equilibrium is zero or they form a closed polygon when represented vectorially.

• Vector addition methods:

  • Opposite direction forces: net force is difference of magnitudes.

  • For forces at an angle: use parallelogram law; diagonal gives resultant.

  • Triangle method: place tail of second force at head of first; resultant is from tail of first to head of second.

Solved Examples

• Ex 1: Two forces \( \vec{F_1} = 10\,\mathrm{N} \) and \( \vec{F_2} = 10\,\mathrm{N} \) act in opposite directions. Find resultant.

  • Since \( \vec{F_1} = -\vec{F_2} \), resultant \( = 0 \) (equilibrium).

• Ex 2: Three forces \( 5\,\mathrm{N}, 7\,\mathrm{N}, 8.6\,\mathrm{N} \) act at a point forming a closed triangle. Are they in equilibrium?

  • Check vector sum: if \( \vec{F_1} + \vec{F_2} + \vec{F_3} = 0 \), yes.

Practice Set

• Q1: Two forces of 15 N and 15 N act in opposite directions on a particle. What is the resultant force?

  • A) 0 N

  • B) 15 N

  • C) 30 N

  • D) 7.5 N

• Q2: Three concurrent forces \( \vec{F_1} = 3\,\mathrm{N}, \vec{F_2} = 4\,\mathrm{N}, \vec{F_3} = 5\,\mathrm{N} \) act at a point. If they are in equilibrium, what is the relation between them?

  • A) \( \vec{F_1} + \vec{F_2} = \vec{F_3} \)

  • B) \( \vec{F_1} + \vec{F_2} + \vec{F_3} = 0 \)

  • C) \( \vec{F_1} = \vec{F_2} = \vec{F_3} \)

  • D) \( \vec{F_1} - \vec{F_2} = \vec{F_3} \)

• Q3: Two forces of 8 N and 6 N act at a point with an angle of 90° between them. What is the magnitude of the resultant force?

  • A) 10 N

  • B) 14 N

  • C) 2 N

  • D) 4 N

Answer Key

• Q1: A) 0 N — Forces are equal and opposite, so resultant is zero (equilibrium).

• Q2: B) \( \vec{F_1} + \vec{F_2} + \vec{F_3} = 0 \) — Condition for equilibrium of concurrent forces.

• Q3: A) 10 N — Use Pythagoras: \( \sqrt{8^2 + 6^2} = 10 \) N.

Memory Aids

• Equilibrium: Sum of forces = 0.

• Two forces equilibrium: forces equal in magnitude, opposite in direction.

• Three forces equilibrium: vectors form closed triangle.

• Vector addition: parallelogram or triangle method.

Lami's Theorem

Key Concepts

• Applies to three concurrent forces \( \vec{F_1}, \vec{F_2}, \vec{F_3} \) in equilibrium.

• Formula: \[ \frac{F_1}{\sin \alpha} = \frac{F_2}{\sin \beta} = \frac{F_3}{\sin \gamma} \]

• Angles defined as:

  • \( \alpha \): angle between \( \vec{F_2} \) and \( \vec{F_3} \)

  • \( \beta \): angle between \( \vec{F_3} \) and \( \vec{F_1} \)

  • \( \gamma \): angle between \( \vec{F_1} \) and \( \vec{F_2} \)

• Used to find unknown force or angle in equilibrium problems.

Solved Examples

• Ex 1: Three forces \( F_1 = 10\,\mathrm{N}, F_2 = 15\,\mathrm{N} \) act at a point with \( \alpha = 60^\circ, \beta = 45^\circ \). Find \( F_3 \) if forces are in equilibrium.

  • Use Lami's theorem: \( \frac{10}{\sin 60^\circ} = \frac{15}{\sin 45^\circ} = \frac{F_3}{\sin \gamma} \).

  • Calculate \( F_3 \) accordingly.

Practice Set

• Q1: In Lami's theorem, if \( F_1 = 20\,\mathrm{N}, \alpha = 30^\circ, F_2 = 25\,\mathrm{N}, \beta = 45^\circ \), find \( F_3 \) given \( \gamma = 105^\circ \).

  • A) 15 N

  • B) 30 N

  • C) 25 N

  • D) 20 N

• Q2: Three forces are in equilibrium. If \( F_1 = 12\,\mathrm{N}, F_2 = 16\,\mathrm{N} \), and angles \( \alpha = 50^\circ, \beta = 70^\circ \), find angle \( \gamma \).

  • A) 60°

  • B) 40°

  • C) 70°

  • D) 50°

• Q3: If three forces \( F_1, F_2, F_3 \) are in equilibrium and \( F_1 = 10\,\mathrm{N}, F_2 = 10\,\mathrm{N} \), with angles \( \alpha = 60^\circ, \beta = 60^\circ \), find \( F_3 \).

  • A) 10 N

  • B) 20 N

  • C) 5 N

  • D) 15 N

Answer Key

• Q1: B) 30 N — Using Lami's theorem, calculate \( F_3 = \frac{F_1 \sin \gamma}{\sin \alpha} = \frac{20 \times \sin 105^\circ}{\sin 30^\circ} \approx 30 \) N.

• Q2: A) 60° — Angles sum to 180°, so \( \gamma = 180^\circ - (50^\circ + 70^\circ) = 60^\circ \).

• Q3: A) 10 N — Equal forces and equal angles imply \( F_3 = F_1 = 10 \) N.

Memory Aids

• Lami's theorem: \( F / \sin \text{opposite angle} = \) constant.

• Three forces in equilibrium form a triangle of forces.

• Sum of angles between forces = 360° (around point), but in Lami's theorem, angles between pairs sum to 180°.

Common Forces in Mechanics

Overview

• Most forces in mechanics, except gravity, are contact forces.

• Common forces: friction, normal reaction, buoyant force, spring force, tension in string.

Types of Friction

• Static friction: Prevents motion; acts when object is at rest.

• Sliding friction: Opposes motion when objects slide over each other.

• Rolling friction: Opposes motion of rolling objects; usually smallest.

Solved Examples

Ex 1

• Calculate friction force if normal reaction \( N = 50\,\mathrm{N} \) and coefficient of friction \( \mu = 0.3 \).

• Use \( f = \mu N = 0.3 \times 50 = 15\,\mathrm{N} \).

Practice Set

• Q1: Which force acts to prevent motion between two surfaces at rest?

  • A) Kinetic friction

  • B) Static friction

  • C) Rolling friction

  • D) Gravitational force

• Q2: Rolling friction is generally:

  • A) Greater than static friction

  • B) Equal to sliding friction

  • C) Less than static and kinetic friction

  • D) Independent of surface roughness

• Q3: The friction force depends on:

  • A) Normal reaction force

  • B) Contact area

  • C) Velocity of object

  • D) Mass of object only

Answer Key

• Q1: B - Static friction prevents motion when object is at rest.

• Q2: C - Rolling friction is less than static and kinetic friction.

• Q3: A - Friction force is proportional to normal reaction force.

Memory Aids

• Friction \( f = \mu N \) except static friction which adjusts up to limiting friction.

• Static friction > kinetic friction > rolling friction.

• Rolling friction minimized using ball bearings.

Friction

Key Points

• Friction opposes relative motion or tendency of motion between surfaces.

• Friction force \( f_r \propto N \) (normal reaction), except static friction which is self-adjusting.

• Types:

  • Static friction \( f_s \): prevents motion, adjusts up to limiting friction.

  • Limiting friction \( f_{lim} = \mu_s N \): maximum static friction before motion starts.

  • Kinetic friction \( f_k = \mu_k N \): acts during motion, \( \mu_k < \mu_s \).

• Friction direction is opposite to motion or applied force.

Solved Examples

Ex 1

• Block of mass \( m = 10\,\mathrm{kg} \) on horizontal surface, \( \mu_s = 0.4 \), find limiting friction.

• Normal force \( N = mg = 10 \times 9.8 = 98\,\mathrm{N} \).

• Limiting friction \( f_{lim} = \mu_s N = 0.4 \times 98 = 39.2\,\mathrm{N} \).

Practice Set

• Q1: Limiting friction depends on:

  • A) Contact area

  • B) Normal reaction

  • C) Velocity

  • D) Temperature

• Q2: Kinetic friction is:

  • A) Equal to limiting friction

  • B) Greater than static friction

  • C) Less than limiting friction

  • D) Independent of normal force

• Q3: If \( \mu_s = 0.5 \) and \( N = 100\,\mathrm{N} \), find limiting friction.

  • A) 50 N

  • B) 100 N

  • C) 0.5 N

  • D) 150 N

Answer Key

• Q1: B - Limiting friction is proportional to normal reaction.

• Q2: C - Kinetic friction is less than limiting friction.

• Q3: A - \( f_{lim} = 0.5 \times 100 = 50\,\mathrm{N} \).

Memory Aids

• Static friction adjusts up to \( f_{lim} = \mu_s N \).

• Kinetic friction constant during motion, less than static.

• Friction force opposes motion direction.

Angle of Friction

Concepts

• Angle of friction \( \lambda \) is angle between resultant of limiting friction \( f \) and normal reaction \( N \).

• Formula: \( \tan \lambda = \frac{f}{N} = \mu \), so \( \lambda = \tan^{-1} \mu \).

Solved Examples

Ex 1

• Given \( \mu = 0.5 \), find angle of friction \( \lambda \).

• \( \lambda = \tan^{-1} 0.5 = 26.57^\circ \).

Practice Set

• Q1: Angle of friction is the angle between:

  • A) Applied force and friction

  • B) Normal reaction and resultant of friction and normal reaction

  • C) Weight and normal reaction

  • D) Friction and applied force

• Q2: If \( \mu = 1 \), angle of friction is:

  • A) 45°

  • B) 30°

  • C) 60°

  • D) 90°

• Q3: For \( \lambda = 30^\circ \), find \( \mu \).

  • A) 0.5

  • B) 0.577

  • C) 1.732

  • D) 1

Answer Key

• Q1: B - Angle between normal reaction and resultant force.

• Q2: A - \( \tan 45^\circ = 1 \).

• Q3: B - \( \mu = \tan 30^\circ = 0.577 \).

Memory Aids

• \( \lambda = \tan^{-1} \mu \).

• Angle of friction relates friction coefficient to force direction.

Angle of Repose

Key Points

• Maximum angle \( \theta \) of incline where body just begins to slide down.

• At this angle, \( f = mg \sin \theta \), \( N = mg \cos \theta \).

• Coefficient of static friction \( \mu_s = \tan \theta \).

• Angle of repose equals angle of friction.

Solved Examples

Ex 1

• Find \( \mu_s \) if angle of repose is \( 30^\circ \).

• \( \mu_s = \tan 30^\circ = 0.577 \).

Practice Set

• Q1: Angle of repose is the angle at which:

  • A) Body starts sliding down

  • B) Body remains at rest

  • C) Friction is zero

  • D) Normal force is maximum

• Q2: If \( \mu_s = 0.5 \), angle of repose is:

  • A) 26.57°

  • B) 45°

  • C) 30°

  • D) 60°

• Q3: On an incline of 45°, a block just starts sliding. Find \( \mu_s \).

  • A) 1

  • B) 0.5

  • C) 0.707

  • D) 0.25

Answer Key

• Q1: A - Body begins to slide at angle of repose.

• Q2: A - \( \tan^{-1} 0.5 = 26.57^\circ \).

• Q3: A - \( \mu_s = \tan 45^\circ = 1 \).

Memory Aids

• \( \mu_s = \tan \theta \) where \( \theta \) is angle of repose.

• Angle of repose = angle of friction.

Rolling Friction

Essentials

• Opposes rolling motion; much smaller than static or kinetic friction.

• Originates from deformation at contact area, not just a point.

• Rolling friction force opposes motion parallel to surface.

• Used in ball bearings to reduce friction in machines.

Solved Examples

Ex 1

• Explain why ball bearings reduce friction.

• Balls roll between rings, reducing contact area and friction.

Practice Set

• Q1: Rolling friction is generally:

  • A) Greater than static friction

  • B) Equal to kinetic friction

  • C) Much less than static and kinetic friction

  • D) Independent of deformation

• Q2: Rolling friction arises due to:

  • A) Surface roughness only

  • B) Deformation at contact area

  • C) Air resistance

  • D) Weight of object only

• Q3: To reduce friction in machines, we use:

  • A) Lubricants only

  • B) Ball bearings

  • C) Rough surfaces

  • D) Heavy weights

Answer Key

• Q1: C - Rolling friction is much less than static and kinetic friction.

• Q2: B - Due to deformation at contact area.

• Q3: B - Ball bearings reduce friction effectively.

Memory Aids

• Rolling friction << static friction.

• Ball bearings minimize friction by rolling contact.

Tension in String

Fundamentals

• String assumed massless, inextensible, and perfectly smooth.

• Tension is force developed when string ends are pulled opposite directions.

• Tension force acts along string, away from the object.

• In ideal string, tension is same at all points.

• Tension is extensional force; in rods, can be extensional or compressional.

Solved Examples

Ex 1

• Identify force type when string is pulled at ends.

• Force is extensional tension along string, away from objects.

Practice Set

• Q1: Tension in an ideal string is:

  • A) Different at different points

  • B) Same at all points

  • C) Always zero

  • D) Compressional

• Q2: Tension force acts:

  • A) Along string, towards object

  • B) Along string, away from object

  • C) Perpendicular to string

  • D) Random direction

• Q3: In a rod under tension, force is:

  • A) Only extensional

  • B) Only compressional

  • C) Extensional or compressional

  • D) Zero

Answer Key

• Q1: B - Tension is uniform in ideal string.

• Q2: B - Tension acts along string, away from object.

• Q3: C - Rod can have extensional or compressional forces.

Memory Aids

• Tension = pulling force along string.

• Ideal string: massless, inextensible, smooth.

• Tension same throughout ideal string.

Circular Motion

Motion of a Car on a Level Road

• Centripetal force is provided by friction between tyres and road: \( f \leq \mu_s N = \frac{mv^2}{R} \)

• Maximum safe speed: \( v = \sqrt{\mu_s R g} \)

Motion of a Car on a Banked Road

• Components of friction \( f \cos \theta \) and normal force \( N \sin \theta \) provide centripetal force

• Maximum permissible speed to avoid slipping: \[ v_{max} = \sqrt{\frac{Rg(\mu_s + \tan \theta)}{1 - \mu_s \tan \theta}} \]

• If road is smooth (\( \mu_s = 0 \)), then \( v_{max} = \sqrt{Rg \tan \theta} \)

Bending of Cyclist on Circular Turning

• Normal reaction \( N \) components:

  • Horizontal: \( N \sin \theta = \frac{mv^2}{r} \) (centripetal force)

  • Vertical: \( N \cos \theta = mg \)

• Maximum speed of cyclist: \[ v = \sqrt{r g \tan \theta} \]

Solved Examples

Ex 1

• Find maximum speed of a car on a flat road with \( \mu_s = 0.4 \), radius \( R = 50 \) m, \( g = 9.8 \) m/s². \[ v = \sqrt{0.4 \times 50 \times 9.8} = \sqrt{196} = 14 \text{ m/s} \]

Ex 2

• Calculate maximum speed on a banked road with \( \theta = 30^\circ \), \( \mu_s = 0.2 \), \( R = 40 \) m, \( g = 9.8 \) m/s². \[ v_{max} = \sqrt{\frac{40 \times 9.8 (0.2 + \tan 30^\circ)}{1 - 0.2 \times \tan 30^\circ}} \approx \sqrt{\frac{392 (0.2 + 0.577)}{1 - 0.115}} = \sqrt{\frac{392 \times 0.777}{0.885}} \approx \sqrt{344} = 18.55 \text{ m/s} \]

Ex 3

• Find maximum speed of cyclist leaning at \( 45^\circ \) on a curve of radius 20 m. \[ v = \sqrt{20 \times 9.8 \times \tan 45^\circ} = \sqrt{196} = 14 \text{ m/s} \]

Practice Set

Q1

• What provides centripetal force for a car turning on a flat road?

  • A) Gravity

  • B) Friction

  • C) Normal force

  • D) Air resistance

Q2

• Maximum speed on a banked road with \( \theta = 30^\circ \), \( \mu_s = 0 \), \( R = 100 \) m, \( g = 9.8 \) m/s² is:

  • A) 17.3 m/s

  • B) 24.5 m/s

  • C) 27.9 m/s

  • D) 19.6 m/s

Q3

• A cyclist leans at \( 60^\circ \) to take a turn of radius 10 m. Find maximum speed.

  • A) 12.5 m/s

  • B) 15.3 m/s

  • C) 9.9 m/s

  • D) 14.0 m/s

Answer Key

• Q1: B) Friction - It provides the necessary centripetal force on a flat road.

• Q2: B) 24.5 m/s - Using \( v = \sqrt{Rg \tan \theta} = \sqrt{100 \times 9.8 \times 0.577} \approx 24.5 \)

• Q3: A) 12.5 m/s - \( v = \sqrt{r g \tan 60^\circ} = \sqrt{10 \times 9.8 \times 1.732} \approx 12.5 \)

Memory Aids

• Flat road max speed: \( v = \sqrt{\mu_s R g} \)

• Banked road smooth max speed: \( v = \sqrt{R g \tan \theta} \)

• Cyclist lean speed: \( v = \sqrt{r g \tan \theta} \)

• Friction provides centripetal force on flat road

• On banked road, normal and friction forces combine for centripetal force

Solving Problems in Mechanics

Systematic Approach

• Draw schematic diagram of the assembly showing bodies, links, supports.

• Select a convenient part as a single system.

• Draw free body diagram (FBD) showing all forces on the system from the rest of the assembly and other agencies.

• Include known forces (magnitude & direction) in FBD; treat unknown forces as variables.

• Use Newton's third law: forces between two bodies are equal and opposite in their respective FBDs.

Solved Examples

Ex 1

• Identify system and draw FBD with all forces.

• Apply Newton's laws to write equations of motion.

• Solve simultaneous equations for unknown forces or accelerations.

Practice Set

• Q1: What is the purpose of a free body diagram?
  A) Show forces on environment
  B) Show forces on system
  C) Show forces by system on environment
  D) Show internal forces only

• Q2: According to Newton's third law, if force on A by B is \( \vec{F} \), force on B by A is:
  A) \( \vec{F} \)
  B) \( -\vec{F} \)
  C) Zero
  D) Depends on mass

• Q3: In solving mechanics problems, the first step is:
  A) Write equations
  B) Draw schematic diagram
  C) Calculate acceleration
  D) Choose coordinate system

Answer Key

• Q1: B - FBD shows all forces acting on the chosen system.

• Q2: B - Newton's third law states forces are equal and opposite.

• Q3: B - Start by drawing schematic diagram of the assembly.

Memory Aids

• FBD = Forces on system only.

• Newton's 3rd law: action = -reaction.

• Start with diagram, then equations.

Minimum Force on Inclined Plane

Key Formulas and Concepts

• Normal force: \( N = mg \cos \theta - F \sin \alpha \)

• Force balance along slope: \( F \cos \alpha = mg \sin \theta + \mu N = mg \sin \theta + \mu (mg \cos \theta - F \sin \alpha) \)

• Rearranged force: \( F = \frac{mg(\sin \theta + \mu \cos \theta)}{\cos \alpha + \mu \sin \alpha} \)

• Minimum \( F \) when denominator \( \cos \alpha + \mu \sin \alpha \) is maximum.

• Condition for minimum force: \( \tan \alpha = \mu \Rightarrow \alpha = \tan^{-1} \mu \)

• Minimum force magnitude: \( F_{min} = \frac{mg(\sin \theta + \mu \cos \theta)}{\sqrt{1 + \mu^2}} \)

Solved Examples

Ex 1

• Given: mass \( m \), incline angle \( \theta \), friction coefficient \( \mu \).

• Calculate \( \alpha = \tan^{-1} \mu \).

• Compute \( F_{min} = \frac{mg(\sin \theta + \mu \cos \theta)}{\sqrt{1 + \mu^2}} \).

Practice Set

• Q1: For \( \mu = 0 \), minimum force to pull block up incline is:
  A) \( mg \sin \theta \)
  B) \( mg \cos \theta \)
  C) Zero
  D) \( mg \)

• Q2: If \( \mu = 1 \) and \( \theta = 30^\circ \), angle \( \alpha \) for minimum force is:
  A) 45°
  B) 30°
  C) 60°
  >D) 90°

• Q3: For \( m=2\,kg \), \( \theta=45^\circ \), \( \mu=0.5 \), find minimum force \( F_{min} \) (take \( g=10\,m/s^2 \)):
  A) 20 N
  B) 24 N
  C) 28 N
  D) 32 N

Answer Key

• Q1: A - Without friction, force equals component of weight down slope.

• Q2: A - \( \alpha = \tan^{-1} 1 = 45^\circ \).

• Q3: B - Calculate \( F_{min} = \frac{2 \times 10 (\sin 45^\circ + 0.5 \cos 45^\circ)}{\sqrt{1 + 0.5^2}} \approx 24\,N \).

Memory Aids

• Minimum force angle \( \alpha = \tan^{-1} \mu \).

• Force formula denominator maximized for minimum force.

• Use components along and perpendicular to incline.