Laws of Motion: theory | Physics Circle

CONCEPT

Inertia, Momentum & F = ma

In one line
A body changes its motion only when a net external force acts on it, and the force equals the rate of change of momentum.

INERTIA

4.1.1 Inertia

Inertia is the property by which a body cannot change its state of rest or uniform motion by itself.
It is not a physical quantity: no unit, no dimension.
It depends only on mass, not on velocity.
A point mass is a body whose size is negligible compared with the distance it covers.

Inertia of	Example
rest	Passenger thrown back when a bus starts; dust leaves a beaten carpet; coin falls in a glass when the card is flicked
motion	Passenger falls forward when a bus stops; athlete runs before a long jump
direction	Stone flies off along the tangent when the string breaks; mud leaves a wheel along the tangent

Lower body moves with the bus; upper body stays back

Bullet on a window: very short contact, so only a small part of the glass moves ⇒ a clean hole.
Ball on a window: longer contact ⇒ the whole pane cracks.

Mass M hanging between two strings

Sudden jerk on B: inertia of M stops the force reaching A ⇒ B breaks.
Steady pull on B: tension in A is more by Mg ⇒ A breaks.

NEWTON’S FIRST LAW

4.1.2 Newton’s first law

A body stays at rest or in uniform straight-line motion unless an external force acts on it.
Net force zero ⇒ acceleration zero.
No force is needed to keep a body moving uniformly in a straight line.
The first law defines force and inertia.

LINEAR MOMENTUM

4.1.3 Linear momentum

p = mv

Vector, along the velocity.
Unit: kg m s⁻¹. Dimension: [MLT⁻¹].
Same momentum: the lighter body moves faster.

v₁v₂ = m₂m₁

p ∝ v, p ∝ m, and v ∝ 1m for fixed p

NEWTON’S SECOND LAW

4.1.4 Newton’s second law

F = dpdt = m dvdt = ma

Force and acceleration are in the same direction.
F = ma holds when mass stays constant.

mass changing: F = m dvdt + v dmdt

UNITS OF FORCE

4.1.5 Units of force

[F] = [MLT⁻²]

Unit	Value
1 N	1 kg m s⁻² = 10⁵ dyne
1 kgf	9.8 N
1 gf	980 dyne

USEFUL RESULTS

4.1.6 Useful results from F = ma

same force on two masses: a = a₁a₂a₁ + a₂

stopped in distance s: a = u²2s

v² = kx ⇒ 2a = k, a constant

Find a first, then multiply by m.
Two forces at an angle: find the resultant first, then divide by m.
Momentum after falling a height h: p = m√(2gh), so p ∝ √h.

Practise PYQs on this topicInertia, First & Second Laws · 44 questions with solutions

CONCEPT

Force Vectors, Spring & Variable Forces

In one line
Force is a vector: apply F = ma separately along each axis. When the force varies, integrate.

KINDS OF FORCE

4.2.1 Kinds of force

Contact forces: friction, normal reaction, tension, viscous force.
Field forces: gravitation, electric, magnetic.
Conservative: gravitational, electric, elastic. Non-conservative: friction, viscous.

Force	Strength order
nuclear	strongest
electromagnetic	next
weak	next
gravitational	weakest

Normal reaction is perpendicular to the surface

WHAT A FORCE DOES

4.2.2 What a force does

Situation	Effect
force too small	body stays at rest
force along motion	speed rises
force against motion	speed falls
force ⟂ velocity (uniform circle)	only direction changes
force at another angle	both change

COMPONENTS

4.2.3 Components: one equation per axis

F resolved along x and y

F = F_xî + F_yĵ + F_zk̂

F_x = ma_x · F_y = ma_y · F_z = ma_z

Motion along each axis is independent of the others.
A force along z changes only v_z.
Given position r(t): differentiate twice to get a, then F.
Given force F: divide by m, then use v = u + at and r = ut + ½at² for each axis.

a = v − ut ⇒ F = m v − ut

SPRING FORCE

4.2.4 Spring force: Hooke’s law

Restoring force is opposite to the extension

F = −kx

k = spring constant, unit N m⁻¹.
Negative sign: force opposes the extension.
Same spring: extension is proportional to the tension.

cut spring: k l = constant ⇒ k₁k₂ = l₂l₁

VARIABLE FORCE

4.2.5 Variable force

Force depends on	Example	Method
time	charge in alternating field	a = F ÷ m, then integrate
position	spring, gravitation	v dv = a dx
velocity	viscous drag, magnetic force	separate v and t

JEE · NEET

Corner

Force depends on time

Use a = F ÷ m, then integrate: v = ∫a dt and x = ∫v dt.
Area under the F–t graph gives the change in momentum.

Force depends on position

a = v dvdx

Separate the variables: v dv = a(x) dx.
v = c√x means a = c²2, a constant.

Force depends on velocity

F = −kv ⇒ v = mgk(1 − e^−ktm)

Falling through a viscous medium: speed rises and settles at the terminal value mgk.
Drag F = −kv²: use 1v − 1v₀ = ktm.
Ball thrown up with constant air drag: ascent is quicker than descent.

Practise PYQs on this topicInertia, First & Second Laws · 44 questions with solutions

CONCEPT

Equilibrium, Third Law & Frames

In one line
Forces in equilibrium add up to zero. Forces always come in action–reaction pairs on different bodies.

CONCURRENT FORCES

4.3.1 Concurrent forces

ΣF = 0

ΣF_x = 0 · ΣF_y = 0 · ΣF_z = 0

Equilibrium: no change in the state of rest or uniform motion.
Three forces in equilibrium form a closed triangle, taken in order.
If a particle at rest has F₂ and F₃ balancing F₁, removing F₁ gives a = F₁ ÷ m.

Sides in order: F₁ + F₂ + F₃ = 0

LAMI’S THEOREM

4.3.2 Lami’s theorem

α, β, γ are the angles between the forces

F₁sin α = F₂sin β = F₃sin γ

Each force goes with the sine of the angle between the other two.

STRINGS IN EQUILIBRIUM

4.3.3 Strings in equilibrium

Body held by two strings

T₁ cos θ₁ = T₂ cos θ₂

T₁ sin θ₁ + T₂ sin θ₂ = mg

Horizontal pull F at a point of a rope, rope at θ with the vertical:

F = mg tan θ · T = mgcos θ = Fsin θ

Turning effect: the door needs F × (distance from the hinge) to stay the same.

NEWTON’S THIRD LAW

4.3.4 Newton’s third law

F_AB = −F_BA

Forces occur in pairs; a single isolated force is not possible.
Action and reaction act on different bodies, so they never cancel.
True whether the bodies are at rest or moving.

Book pushes the table; table pushes the book

Walking: the ground pushes you forward

Swimming, gun recoil, a rebounding ball.
Walking is hard on sand or ice: the reaction is small.

FRAMES OF REFERENCE

4.3.5 Frames of reference

Inertial	Non-inertial
at rest or uniform velocity	accelerated
Newton’s laws hold	Newton’s laws need a pseudo force
bus at constant velocity, lift at constant speed	car on a curve, lift accelerating, plane taking off, merry-go-round, train slowing down

Earth can be taken as inertial for a falling apple, but not for the motion of planets.

PSEUDO FORCE

4.3.6 Pseudo force

Bob in a truck accelerating to the right

F_pseudo = −ma_frame

tan θ = ag

The bob tilts opposite to the acceleration.
It acts only for an observer in the accelerated frame.

Practise PYQs on this topicEquilibrium, Third Law & Frames · 25 questions with solutions

CONCEPT

Impulse

In one line
A large force acting for a short time is measured by its total effect: impulse, which equals the change in momentum.

IMPULSE

4.4.1 Impulse of a force

I = ∫F dt = F_av Δt

Vector, along the force.
Unit: N s or kg m s⁻¹. Dimension: [MLT⁻¹].
Impulsive force rises from zero to a peak and falls back to zero.

IMPULSE–MOMENTUM THEOREM

4.4.2 Impulse–momentum theorem

I = Δp = p₂ − p₁

F_av = ΔpΔt

Hit back along the same line: use Δv = v + u, because the direction reverses.
Ball rebounding from a wall with the same speed:

Δp = 2mv

n balls at once: F = 2nmvt

Dropped from height h₁, rebounds to h₂:

a_av = √(2gh₁) + √(2gh₂)Δt

Hits a wall at angle θ with the wall and returns at the same angle: only the normal part changes.

Δp = 2mv sin θ (θ with the wall)

Turning with the same speed: force is along Δv = v₂ − v₁, not along v₂.

F–t GRAPH

4.4.3 F–t graph: area gives impulse

I = ½ F t for a triangular graph

Area under the curve = F_av Δt

Area under the F–t graph = impulse = change in momentum.
Starting from rest: v = area ÷ m.
Given F(t): integrate over the time it acts.

TIME OF CONTACT

4.4.4 Time of contact

Longer contact means a smaller force

F_av ∝ 1Δt (for fixed Δp)

A fielder draws the hands back: longer contact, less force.
Jumping on sand is safe; china wares are packed in straw.

Practise PYQs on this topicImpulse & Momentum Conservation · 29 questions with solutions

CONCEPT

Momentum Conservation, Machine Gun & Rocket

In one line
With no external force, total momentum of a system stays constant. Force is what changes momentum per second.

CONSERVATION LAW

4.5.1 Conservation of momentum

m₁v₁ + m₂v₂ + … = constant

F_ext = 0 ⇒ total p is constant.
Momentum of each part may change; the sum does not.
It is equivalent to Newton’s third law.
The law holds in every inertial frame.

RECOIL OF A GUN

4.5.2 Recoil of a gun

Bullet forward, gun backward

v_G = −m_Bm_G v_B

Negative sign: recoil is opposite to the bullet.
Heavier gun ⇒ smaller recoil.
Held tight to the shoulder: gun and shooter act as one larger mass.

EXPLOSION

4.5.3 Explosion

Vector sum of momenta is zero

equal parts at 90°: m₃V = m v √2

Find each mass from the ratio first.
Heaviest part flies opposite to the resultant of the other two.

MACHINE GUN

4.5.4 Machine gun

F = n m v

n_max = F_maxmv

Put m in kg.
Each bullet carries momentum mv away from the gun in one shot.

CONVEYOR AND JETS

4.5.5 Conveyor belt and jets

F = v dmdt

Sand falling on a belt moving at constant v: F = Mv where M is in kg per second.
Water stopped by a palm: F = ρAv².

ROCKET

4.5.6 Rocket in short

Gases go down, rocket goes up

thrust F = u dmdt

initial thrust = m(g + a)

u = speed of exhaust gases.
Rocket needs no air: it pushes on its own gases.

JEE · NEET

Corner

Rocket, in full

m₀ = initial mass, m = mass at time t, m_r = mass of the empty rocket, u = exhaust speed.

a = um dmdt − g

v = u ln(m₀m) − gt

burnt-out speed: v_b = u ln(m₀m_r)

Thrust F = u dmdt (gravity neglected).
Lift-off with acceleration a: u dmdt = m(g + a).
Just lifting off: u dmdt = mg.

Mass sweeping onto a body

F = v dmdt

Sand on a conveyor belt, dust on a satellite, water hitting a palm.
Water jet: F = ρAv² if it stops; F = 2ρAv² if it bounces back.

Practise PYQs on this topicImpulse & Momentum Conservation · 29 questions with solutions

CONCEPT

Free-Body Diagram & Apparent Weight

In one line
A weighing machine reads the normal reaction R, not mg. R changes with the acceleration of the lift.

FREE-BODY DIAGRAM

4.6.1 Free-body diagram

All forces on one block, nothing else

Draw one body by itself.
Show every force acting on it, none it exerts on others.
Choose an axis along the acceleration.
Write ΣF = ma along it and ΣF = 0 across it.

APPARENT WEIGHT

4.6.2 Apparent weight in a lift

Actual weight mg stays the same; reading R changes

Lift	Equation	Reading R
at rest	R − mg = 0	mg
constant velocity	R − mg = 0	mg
accelerating up, a	R − mg = ma	m(g + a) > mg
accelerating up, g	R − mg = mg	2mg
accelerating down, a	mg − R = ma	m(g − a) < mg
free fall, a = g	mg − R = mg	zero (weightless)
down, a > g	mg − R = ma	negative: body sticks to the ceiling

What counts is the direction of acceleration, not of velocity.
Going up and slowing down is the same as accelerating down.
Spring balance in a lift reads the apparent weight.
A perfect physical balance is not affected: both pans change alike.

TWO READINGS

4.6.3 Two readings: finding real weight

mg = R_up + R_down2

R_upR_down = g + ag − a

w₁w₂ = gg − a (rest : going down)

CABLE TENSION

4.6.4 Cable tension

T = (M + m)(g + a) going up

T = (M + m)(g − a) going down

Use the total mass of the lift and people.
Acceleration from tension: Ma = T − Mg; positive means upward.

Body in a lift dropped: lift observer sees g − a; ground observer sees g.
Floor yielding under a load: R = m(g − a).

CLIMBING A ROPE

4.6.5 Climbing a rope

Motion	Tension
climbing up with a	m(g + a)
climbing down with a	m(g − a)
uniform speed	mg

Rope breaks if T exceeds its limit.
Rope limit T_max < mg: climbing down is the only choice, so that m(g − a) ≤ T_max.

BALLOON

4.6.6 Balloon dropping mass

descending to rising at a: x = 2mag + a

Upthrust U stays the same when ballast is dropped.
Write one equation before and one after, then eliminate U.

Practise PYQs on this topicApparent Weight & Lift · 25 questions with solutions

CONCEPT

Contact Forces, Strings & Massive Rope

In one line
Treat the whole system first to get a. Then take one part to get the force between the parts.

BLOCK ON A SMOOTH FLOOR

4.7.1 Block on a smooth floor

Pull: R falls. Push: R rises. Acceleration is the same

Case	R	a
horizontal pull	mg	F ÷ m
pull at θ above	mg − F sin θ	F cos θm
push at θ below	mg + F sin θ	F cos θm

SMOOTH INCLINE

4.7.2 Block on a smooth incline

Only mg sin θ acts along the plane

a = g sin θ · R = mg cos θ

t = √(2lg sin θ) · v = √(2gh)

Same height: t ∝ 1sin θ.
Same length: t ∝ 1√(sin θ).
A lift at uniform velocity makes no change.

BLOCKS IN CONTACT

4.7.3 Blocks in contact

Contact force moves the block in front

a = Fm₁ + m₂

f = m₂Fm₁ + m₂

Three blocks, pushed at m₁: f₁₂ = (m₂ + m₃)Fm₁ + m₂ + m₃ and f₂₃ = m₃Fm₁ + m₂ + m₃.
Contact force = (mass in front) × a.

BLOCKS JOINED BY STRINGS

4.7.4 Blocks joined by strings

String pulls everything behind it

a = Fm₁ + m₂ + m₃

T₁ = m₁Fm₁ + m₂ + m₃ · T₂ = (m₁ + m₂)Fm₁ + m₂ + m₃

Tension = (mass behind) × a.
Vertical pull: T = (mass below)(g + a).
Constant speed: net force on every block is zero.

MASSIVE ROPE

4.7.5 Massive rope

a = FM + m

force on block: T = MFM + m

at mid-point: T = (2M + m)F2(M + m)

Tension changes along the rope, largest at the end pulled.
Rod joining two blocks is treated the same way: count half its mass.

SPRING BALANCE

4.7.6 Spring balance

Reading = tension in the spring

Two light balances in series both read the full weight.
Balance at rest or in uniform motion reads mg.
Pulley efficiency: output work ÷ input work.

JEE · NEET

Corner

Tension in a massive rope, along its length

Tension falls towards the free end

T = F(L − xL)

pulled at both ends: T = F₁(1 − xL) + F₂ xL

x is measured from the end where F is applied.
Rope hanging by its own weight: T at distance x from the lower end = ML x g.

Spring cut suddenly

A spring cannot change its length at once, so its force stays the same just after a string is cut.
A string force drops to zero at once.
Spring cut into lengths: k l stays constant.

Practise PYQs on this topicConnected Bodies & Pulleys · 51 questions with solutions

CONCEPT

Pulley Systems

In one line
A light string over a smooth pulley has the same tension on both sides. Connected bodies share one acceleration.

TWO HANGING MASSES

4.8.1 Two hanging masses: Atwood machine

Heavier mass goes down

a = m₂ − m₁m₂ + m₁ g

T = 2m₁m₂m₁ + m₂ g

force on the support: T₂ = 2T = 4m₁m₂m₁ + m₂ g

Equal masses: a = 0, T = mg.
Extra mass m on one of two blocks M: driving force mg, total mass 2M + m.

a = gn ⇒ m₂m₁ = n + 1n − 1

THREE HANGING MASSES

4.8.2 Three hanging masses

a = (m₂ + m₃) − m₁m₁ + m₂ + m₃ g

T₁ = 2m₁(m₂ + m₃)m₁ + m₂ + m₃ g · T₂ = 2m₁m₃m₁ + m₂ + m₃ g

4.8.3 Block on a table, other hanging

Smooth table

a = m₂gm₁ + m₂ · T = m₁m₂gm₁ + m₂

4.8.4 Block on an incline, other hanging

Smooth incline

a = m₂ − m₁ sin θm₁ + m₂ g

T = m₁m₂(1 + sin θ)m₁ + m₂ g

If the system is at rest: T = m₂g = m₁g sin θ.
Then the plane pushes m₁ with R = m₁g cos θ.

DOUBLE INCLINE

4.8.5 Double incline

String over the top

a = m₂ sin β − m₁ sin αm₁ + m₂ g

T = m₁m₂(sin α + sin β)m₁ + m₂ g

BLOCK BETWEEN TWO PULLEYS

4.8.6 Block between two pulleys

a = m₁ − m₂m₁ + m₂ + M g

HANGING CHAIN

4.8.7 Chain over a pulley

a = (L − l) − lL g

Chain of length L with l on one side.
Treat the two parts as two masses.

THREAD WITH ACCELERATION

4.8.8 Thread lifted or lowered

T_upT_down = g + ag − a

HOW TO SOLVE

4.8.9 How to solve pulley problems

Draw a free-body diagram for each block.
Mark a in the direction of likely motion.
Write one equation per block; add to eliminate T.

JEE · NEET

Corner

Movable pulley

A movable pulley halves the acceleration

a₁ = 2a₂ ⇒ a₁ = 2m₂g4m₁ + m₂

T = 2m₁m₂g4m₁ + m₂

Length of string stays fixed: this gives the relation between accelerations.
Block on the fixed string side moves twice as far.

Pulley with mass M and radius R

a = m₁ − m₂m₁ + m₂ + M2 g

Tensions on the two sides are now different.

Accelerating wedge (pseudo force)

Block at rest on a wedge pushed sideways

a = g tan θ

R = mgcos θ

Wedge free to slide: acceleration of the block relative to the wedge is

a_rel = (M + m) g sin θM + m sin²θ

Practise PYQs on this topicConnected Bodies & Pulleys · 51 questions with solutions

CONCEPT

Friction Basics

In one line
Friction opposes relative motion or its tendency. Static friction adjusts itself up to a limit; kinetic friction is steady.

TYPES OF FRICTION

4.9.1 Types of friction

Block at rest: friction F = applied force P

Static: no motion yet. F = P, a self-adjusting force.
No pull on a block at rest: friction is zero.
Limiting: maximum static friction, just before motion.
Kinetic: during motion; stays the same at any speed.

F_l = μ_sR · F_k = μ_kR

μ_k < μ_s

μ has no unit and no dimension.
μ depends on the materials and nature of the surfaces.
μ does not depend on the area of contact.
Rolling friction < kinetic friction < limiting friction.

GRAPH

4.9.2 Graph of friction against applied force

OA static, A limiting, BC kinetic

OA: static friction rises with the applied force.
A: limiting friction.
BC: kinetic friction, flat.
So a–t graph for F = kt: zero first, then a jump up and a straight line.

FRICTION CAUSES MOTION

4.9.3 Friction can cause motion

Walking: foot pushes the ground back; friction pushes you forward.
Cycling: friction on the rear wheel is forward, on the front wheel backward.

Block moves with the truck only through friction

a_max = μg

If ma < μmg, friction = ma, not μmg.
Advantages: walking, brakes, writing, belts.
Disadvantages: heat, wear, loss of energy.
Reduce it with polishing, lubricants, ball bearings, streamlining.

ANGLE OF FRICTION

4.9.4 Angle of friction

S is the force from the surface

tan θ = μ

S = √(F² + R²)

mg ≤ S ≤ mg√(1 + μ²)

BLOCK ON A WALL

4.9.5 Block on a wall

R = F · μF ≥ W

Friction holds up the weight.
Extra upward push: friction can reverse.
Maximum upward force: mg + μF.

PULL AND PUSH

4.9.6 Pull and push at an angle

Case	R	To just move
pull	mg − P sin α	P = μmgcos α + μ sin α
push	mg + P sin α	P = μmgcos α − μ sin α

Pull is easier: R is smaller, so friction is smaller.
Moving block: ma = P cos α − μR.

STOPPING

4.9.7 Stopping by friction

Retardation a = μg

S = u²2μg = p²2μm²g

t = uμg

Distance does not depend on mass (given u).
Bag on a moving belt: slip distance relative to the belt is the same formula with v.

HANGING MASS

4.9.8 Hanging mass, rough table

Friction on m₁ opposes the pull of m₂

just about to move: m₂ = μm₁

moving: a = m₂g − μm₁gm₁ + m₂

T = m₁m₂g(1 + μ)m₁ + m₂

JEE · NEET

Corner

Rolling friction

F_rolling = μ_r Rr

μ_r has the dimension of length.
Rolling friction is much less than sliding friction.

Minimum force at the best angle

tan α = μ · P_min = μmg√(1 + μ²)

Pulling is easier than pushing, because a pull reduces R.

Boy pulling his own plank

Rope pulled by a person standing on the body: draw the free-body diagram of the person and the body separately.

Practise PYQs on this topicFriction Basics · 38 questions with solutions

CONCEPT

Friction on an Incline

In one line
On a rough slope, friction acts opposite to the motion or its tendency. At the angle of repose, tan α = μ.

ANGLE OF REPOSE

4.10.1 Angle of repose

At the limit: F = mg sin α, R = mg cos α

μ = tan α

Angle of repose = angle of friction.
Constant-velocity slide: μ_k = tan θ.
Then the contact force from the plane = mg, straight up.
Block at rest below the limit: f = mg sin θ, not μR.
g changes (rocket, lift) but μ = tan θ stays: g cancels.

θ ≤ α: block stays, net force = 0

SLIDING DOWN

4.10.2 Sliding down a rough incline

a = g(sin θ − μ cos θ)

Friction acts up the plane.
Smooth plane: a = g sin θ.

SLIDING UP

4.10.3 Sliding up a rough incline

Gravity and friction both oppose

a = g(sin θ + μ cos θ)

S = u²2g(sin θ + μ cos θ)

Up is slower to last, down takes longer: t_down > t_up.

FORCE ALONG THE PLANE

4.10.4 Force along the plane

Aim	Force
just move up	mg sin θ + μmg cos θ
just stop sliding down	mg sin θ − μmg cos θ
accelerate up at a	mg sin θ + μmg cos θ + ma

F_up − F_down = 2μmg cos θ

at 45°: F_upF_down = 1 + μ1 − μ

Range of force keeping it at rest: F_down to F_up.
If 2 N down and 10 N up are the limits: f = 6 N and mg sin θ = 4 N.

ROUGH AND SMOOTH

4.10.5 Rough vs smooth: time of descent

Same length, n times the time

μ = tan θ (1 − 1n²)

at 45°: μ = 1 − 1n²

n = 2 ⇒ μ = 0.75.
50% more time means n = 1.5.

HALF ROUGH, HALF SMOOTH

4.10.6 Half rough, half smooth

μ = 2 tan θ

Energy from the full slope = friction work on the rough half.

4.10.7 Hanging mass, rough incline

to start moving up: m₂ = m₁(sin θ + μ cos θ)

m₂m₁ = sin θ + μ cos θ

WORK AGAINST FRICTION

4.10.8 Work against friction

up a slope: W = mgs(sin θ + μ cos θ)

on a level surface: W = μmgs

JEE · NEET

Corner

Curved surface

Block stays at rest as long as dydx ≤ μ.
Find x from dydx = μ, then put it in y(x).

Force at an angle on the incline

up the plane: P = W sin(θ + λ)cos(α − θ)

to stop sliding: P = W sin(λ − θ)cos(α + θ)

θ = angle of friction, λ = slope, α = angle of P with the plane.

Practise PYQs on this topicFriction on an Incline · 33 questions with solutions

CONCEPT

Two-Body Friction

In one line
Friction between two surfaces gives the upper block its acceleration. It can do this only up to a_max = μg.

BLOCK ON BLOCK

4.11.1 Block on block

Friction drives the upper block

a_max = μ_sg

F_max = μ_s(m + M)g

Up to F_max both move together.
With a rough floor, add the floor friction:

F_max = (μ_s + μ′)(m + M)g

HANGING CHAIN

4.11.2 Hanging chain

Weight of hanging part = friction on the rest

μ = l′l − l′ · l′ = μlμ + 1

INSECT IN A BOWL

4.11.3 Insect in a bowl

tan θ = μ at the highest point

h = r[1 − 1√(1 + μ²)]

BLOCK ON A CART

4.11.4 Block on a cart

Normal reaction R = ma

a_min = gμ

F_min = (M + m)gμ

Blocks pressed on a wall: treat them as one to find the friction from the wall.

ROTOR

4.11.5 Rotor

Friction balances the weight

ω_min = √(gμr)

JEE · NEET

Corner

Force on the upper block

F ≤ μ_smg: both move together.
F > μ_smg: A slips.

a_A = F − μ_kmgm · a_B = μ_kmgM

A falls off B in t = √(2La_rel)

Force on the lower block

together if F ≤ μ_s(m + M)g

beyond that: a_A = μ_kg, a_B = F − μ_kmgM

Friction between the blocks is what moves the upper block.
Floor also rough: replace F by F − μ′(m + M)g.

Two blocks released by a spring

S₁S₂ = (m₂m₁)²

Momenta are equal and opposite, so S falls as the mass squared.

Practise PYQs on this topicTwo-Body Friction · 13 questions with solutions

EXAM-DAY CARD

Just see this before the exam

F = dpdt = ma

second law

I = FΔt = Δp

impulse

F = nmv

machine gun

R = m(g ± a)

lift

a = m₂ − m₁m₂ + m₁ g

pulley

F_l = μR, μ = tan α

friction

FORMULA SHEET

Whole chapter on one page

Laws & equilibrium

p = mv, F = dpdt = ma

F = −kx, k l = constant

F₁sin α = F₂sin β = F₃sin γ

F = mg tan θ, T = mgcos θ

Impulse & momentum

I = ∫F dt = Δp, rebound: 2mv

v_G = −m_Bm_G v_B

n = Fmv, F = v dmdt

Thrust = u dmdt = m(g + a)

Lift & rope

Up: R = m(g + a)

Down: R = m(g − a)

mg = R_up + R_down2

T = (M + m)(g + a)

Connected bodies

a = FΣm, f = (mass in front)a

T = (mass behind)a

a = m₂ − m₁m₂ + m₁ g

T = 2m₁m₂m₁ + m₂ g

Friction

F_l = μ_sR, F_k = μ_kR

tan θ = μ, a_max = μg

S = u²2μg, t = uμg

m₂ = μm₁

Incline & two bodies

a = g(sin θ ∓ μ cos θ)

μ = tan θ(1 − 1n²)

F_max = μ(m + M)g

l′ = μlμ + 1, a_min = gμ

TOP MISTAKES

Avoid these = easy marks

MISTAKE ✗

Ball turned back: Δv = v − u

Lift reading always equals mg

Block at rest: friction = μR

Action and reaction cancel

CORRECT ✓

Direction reverses: Δv = v + u

R = m(g ± a), up or down

Static: friction = applied force

They act on different bodies