Key relationship
N cosθ − f sinθ = mg; N sinθ + f cosθ = mv²/r; no friction: v² = rg tanθ; level road: v² ≤ μrg.
Physics · Class 11
Level and banked roads
A vehicle needs inward net force mv²/r. On a level road static friction supplies it. On a banked road the tilted normal contributes. At the design speed no friction is needed; below or above it friction acts in opposite directions. Grip is possible only when the required friction is within μN.
Open the interactive experiment →Banking gives the normal force an inward component.
வளைவுப் பாதையில் உள்நோக்கிய நிகர விசை தேவை. சாய்வான சாலையின் செங்குத்து எதிர்வினைக்கு உள்நோக்கிய கூறு உண்டு. தேவையான உராய்வு μN வரம்புக்குள் இருக்க வேண்டும்.
Key notes, relationships and model limits
Remember
A vehicle needs inward net force mv²/r. On a level road static friction supplies it. On a banked road the tilted normal contributes. At the design speed no friction is needed; below or above it friction acts in opposite directions. Grip is possible only when the required friction is within μN.
Common trap
Centripetal force is the inward net force, not an additional force to add to the diagram.
Model limits
Steady circular turn; road bank rises away from the centre; f positive down the bank. Point vehicle, no aerodynamic forces or suspension. Outside the friction bound the diagram reports failure, not a simulated skid.
What you will learn
Resolve banked-road forces and check static friction feasibility on level and banked roads.
Try the experiment
Compare level road with the frictionless design speed. Raise speed until the grip condition fails.
Think of it this way
A tilted bowl guides a bead inward through the surface’s normal force.
Worked example
For r = 50 m and θ = 30°, v = √(50 × 9.81 × tan30°) = 16.83 m/s without friction.
Why this matters
A feasibility check is essential: force equations can demand more friction than a road can provide.
Before you start
Vector components · velocity · SI units
Check your understanding
At the frictionless design speed on a banked road, which force supplies the inward component?
- An extra centrifugal force in the inertial frame
- The normal reaction
- Gravity alone
Read the explanation
The normal reaction is tilted and has an inward horizontal component; gravity is vertical.
Course connection: CBSE XI Chapter 4; NEET/Main Unit 3; JEE Advanced Mechanics. Apply each exam’s stated assumptions.