JEE Main Physics Laws of Motion 2027: Friction, Banking & Pulleys Made Simple
Free-body diagrams, Atwood machines, friction and banked roads, taught with worked examples so Laws of Motion questions stop feeling random.
Edurack
September 28, 2026

Ninety percent of Laws of Motion mistakes happen before any calculation starts: a missing force, a wrong direction, a forgotten friction arrow. The fix is one habit. Draw the free-body diagram first. This chapter feeds Work-Energy, Rotation and Electrostatics, so a strong base here lifts your whole Physics score.
If you can draw the forces correctly, the equation almost writes itself.
Chapter at a Glance
| Snapshot | Detail |
|---|---|
| NTA unit | Unit 3 of 20 — Laws of Motion |
| Priority (trend-based) | High |
| Typical question style | Multi-body MCQs with friction, pulleys and circular-motion numericals |
| Best first step | Draw a free-body diagram for every problem, no exceptions |
Priority reflects past-paper trends, not an official NTA weightage.
What the NTA Syllabus Covers
- Force and inertia, Newton's first, second and third laws, momentum and impulse
- Conservation of linear momentum and its applications, equilibrium of concurrent forces
- Static and kinetic friction, laws of friction, rolling friction
- Dynamics of uniform circular motion, centripetal force, vehicle on a level road and on a banked road
Master These Topics
1. Newton's Second Law and the Atwood Machine
The real form of the second law is F = dp/dt, which reduces to F = ma for constant mass. Impulse equals change in momentum: J = FΔt = Δp.
Worked example (Atwood machine): Masses m₁ = 3 kg and m₂ = 2 kg hang over a light frictionless pulley (g = 10).
- Acceleration:
a = (m₁ − m₂)g / (m₁ + m₂) = 1 × 10 / 5 = 2 m/s² - Tension:
T = 2m₁m₂g / (m₁ + m₂) = 2 × 3 × 2 × 10 / 5 = 24 N
Check: for the heavier block, T = m₁(g − a) = 3 × 8 = 24 N. Always verify with the second block.
Lift problems: apparent weight is N = m(g + a) when the lift accelerates upward and m(g − a) when it accelerates downward. A 60 kg person in a lift accelerating up at 2 m/s² feels 60 × 12 = 720 N.
2. Friction: Static, Kinetic and the Angle of Repose
Static friction adjusts itself up to a limit: f_s ≤ μ_s N. Once sliding starts, kinetic friction is f_k = μ_k N, usually smaller than the static limit.
- On a rough incline, a block just starts sliding when
tanθ = μ_s. That angle is the angle of repose. - Friction is not always opposite to motion. It opposes relative motion (or the tendency for it) between surfaces.
Trap: Never write f = μN when the block is not on the verge of slipping. Static friction equals whatever value keeps the block at rest, up to the maximum.
3. Circular Motion: Level Roads and Banked Roads
Friction supplies the centripetal force on a level road, so the maximum safe speed is v = √(μrg). With μ = 0.4 and r = 40 m: v = √(0.4 × 10 × 40) = √160 ≈ 12.6 m/s.
On a banked road of angle θ with no friction needed, v = √(rg tanθ). With friction, the limiting speed is v² = rg (tanθ + μ) / (1 − μ tanθ). Banking lets vehicles turn at higher speeds without depending on tyre grip alone.
4. Conservation of Linear Momentum
With no external force on a system, total momentum stays constant. This explains gun recoil: a 2 kg gun firing a 20 g bullet at 300 m/s recoils at 0.02 × 300 / 2 = 3 m/s.
Common Traps to Avoid
- Skipping the free-body diagram and losing a force.
- Applying
f = μNwhen friction is static but below its limit. - Forgetting the pseudo force when working in an accelerating frame such as a lift.
- Mixing up the angle of banking with the angle of repose.
60-Second Revision Sheet
F = dp/dt, impulseJ = Δp- Atwood:
a = (m₁ − m₂)g/(m₁ + m₂),T = 2m₁m₂g/(m₁ + m₂) - Angle of repose:
tanθ = μ_s - Level road:
v_max = √(μrg); banked (no friction):v = √(rg tanθ)
Your Study Plan
- Day 1: free-body diagrams for 15 mixed problems, no calculation.
- Day 2: pulleys and connected bodies, including wedge and incline setups.
- Day 3: friction on inclines and block-on-block problems.
- Day 4: circular motion, banking and pseudo-force problems in a timed set.
Practice Laws of Motion Questions Free → (opens in a new tab)
Continue Your Physics Journey
- Previous chapter: Kinematics
- Next chapter: Work, Energy & Power
- All 20 JEE Main Physics chapters
- Complete JEE Main Syllabus 2027 guide
Frequently Asked Questions
Why does Laws of Motion feel harder than Kinematics?
Because you must choose which forces to include and set up equations yourself. A free-body diagram removes most of that confusion.
Is rolling friction important for JEE Main?
It is listed in the syllabus but rarely tested numerically. Know the concept: rolling friction is much smaller than sliding friction.