Laws of Motion— Notes, Formulas & Revision

Every body continues in its state of rest or uniform motion in a straight line unless acted on by a net external force.

This is the law of inertia — bodies resist changes to their motion.

Inertia is measured by mass: larger mass → more resistance to change.

There is no distinction between 'at rest' and 'moving at constant velocity' — both are zero-net-force states.

F_net = ma — net force equals mass times acceleration.

More generally: F = dp/dt where p = mv is momentum.

Direction of acceleration = direction of net force.

Doubling the force doubles the acceleration; doubling the mass halves it.

To every action there is an equal and opposite reaction.

Action and reaction act on DIFFERENT bodies — they never cancel each other out.

The pair always has: same magnitude, opposite direction, same line of action, same type (both gravitational, both normal, etc.)

Rocket propulsion, walking, and swimming all rely on the third law.

Newton's First Law (Inertia): A body at rest stays at rest, and a body in motion continues in uniform motion, unless acted on by a net external force.

Newton's Second Law: F = ma — the net force on a body equals its mass times acceleration. This is the workhorse of JEE mechanics.

Newton's Third Law: For every action, there is an equal and opposite reaction. The forces act on DIFFERENT bodies.

The concept of inertial frames is fundamental — Newton's laws hold only in non-accelerating reference frames.

Free Body Diagrams (FBDs) are the most important tool. Always isolate the body and draw ALL forces acting on it.

In connected body problems (Atwood machine, pulleys), use constraint equations: acceleration of connected bodies is related.

Pseudo forces appear in non-inertial frames. In a frame accelerating at 'a', a pseudo force = −ma acts on every body.

The normal force is NOT always equal to mg. It depends on the situation (inclines, lifts, circular motion).

A Free-Body Diagram shows all external forces on a single object, represented by arrows from the object.

Standard forces: weight (mg, down), normal (N, ⊥ to surface), friction (f, parallel to surface opposing motion), tension (T, along string), applied force.

Choose coordinate axes to simplify — align with incline for ramp problems.

Then apply ΣF = ma along each axis independently.

Static friction f_s adjusts to match applied force, up to a maximum: f_s ≤ μ_s N.

Once motion begins, kinetic friction takes over: f_k = μ_k N (constant).

Generally μ_k < μ_s — which is why it's harder to start moving than to keep moving.

Friction acts parallel to the surface, opposing (actual or impending) relative motion.

Friction is a contact force that opposes the relative motion (or tendency of motion) between two surfaces.

Static friction (fₛ) acts when there is no relative motion. It adjusts up to a maximum value: fₛ(max) = μₛN.

Kinetic friction (fₖ) acts when surfaces slide. It is constant: fₖ = μₖN, and always μₖ ≤ μₛ.

On an inclined plane at angle θ: component of gravity along incline = mg sin θ, normal force N = mg cos θ.

A block stays stationary on an incline if tan θ ≤ μₛ. The angle of repose θᵣ = tan⁻¹(μₛ).

Friction does negative work on sliding objects, converting kinetic energy to heat.

In JEE problems, always draw a Free Body Diagram (FBD) and identify the direction of friction carefully.

Rolling friction is much smaller than sliding friction — this is why wheels are efficient.

On a frictionless incline at angle θ: a = g sinθ (down the slope), N = mg cosθ.

With kinetic friction μ: a = g(sinθ − μ cosθ).

Block stays at rest if tanθ ≤ μ_s. Angle of repose: θ_r = tan⁻¹(μ_s).

Going up: a = −g(sinθ + μ cosθ) (friction adds to gravity along slope).

Two masses m₁ and m₂ over a massless, frictionless pulley.

a = (m₂ − m₁)g / (m₁ + m₂), with heavier mass falling.

Tension T = 2m₁m₂g / (m₁ + m₂).

For equal masses, a = 0 and T = mg.

Two blocks m₁ and m₂ pulled by force F via an inextensible string share the same acceleration.

a = F / (m₁ + m₂).

Tension in string = m₁·a (the force needed to accelerate the rear block).

Works for any number of blocks connected in series — just divide by total mass.

Apparent weight = N (normal force on a person).

Elevator accelerating up: N = m(g+a) → heavier feeling.

Accelerating down: N = m(g−a) → lighter feeling.

Free fall (a = g): N = 0 → weightlessness.

Constant velocity or at rest: N = mg (true weight).

On a road banked at angle θ, the horizontal component of N provides centripetal force.

Ideal speed (no friction needed): v_ideal = √(rg tanθ).

With friction μ: v_max = √(rg(tanθ+μ)/(1−μ tanθ)); v_min = √(rg(tanθ−μ)/(1+μ tanθ)).

Banking reduces wear on tyres by not relying on friction alone.

A force F at angle θ has rectangular components Fx = F cosθ, Fy = F sinθ.

Components depend on the chosen axes — pick axes that simplify the problem (along motion / along incline).

Magnitude is recovered as F = √(Fx² + Fy²); direction as tanθ = Fy/Fx.

Angle of repose θ_r is the maximum incline angle at which a block stays at rest.

Above θ_r the down-incline component mg sinθ exceeds maximum static friction μ_s mg cosθ.

Equating gives θ_r = tan⁻¹(μ_s) — independent of mass.

Frictionless wedge of mass M on smooth floor; block of mass m on the incline.

Both accelerate: block down the incline AND wedge horizontally (recoil).

Solve using Newton's laws in ground frame, or pseudo-force in wedge frame.

Mass on string in horizontal circle: tension T provides the centripetal force.

T = mv²/r = mω²r — increases with v² (or ω²) and decreases with r.

Cut the string and the mass flies off tangentially (Newton's 1st law).