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Q.What is the need for banking of Road? Find expression for the maximum velocity required for a Car on a Banked road by taking into account the force of friction for safe turn.

(OR)
State Newton's second law of motion. Show that Newton's second law is the real law of motion.
Himachal HpboseHPBOSE Himachal Pradesh Class 11 Board Exam 2026Subjective· 5mImportance★★★★★
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Banking supplies part of the centripetal force through the normal reaction (reducing reliance on friction); including friction at the point of maximum speed, vmax = √[ rg(μ + tanθ) / (1 − μ tanθ) ].

Need for banking of roads: When a vehicle goes around a curved, unbanked (flat) road, the necessary centripetal force (mv²/r, directed towards the centre of the curve) is provided ENTIRELY by friction between the tyres and the road. At high speeds, or when friction is low (wet or icy roads), the available friction may not be enough, and the vehicle skids outward.

To reduce reliance on friction alone, the outer edge of the road is raised above the inner edge — this is called banking. On a banked road, the normal reaction N (perpendicular to the road surface) has a horizontal component directed towards the centre of the curve, which supplies part (or all) of the required centripetal force. This:

  • allows the vehicle to negotiate the curve safely even at higher speeds,
  • reduces wear and tear on tyres, and
  • reduces the chance of skidding, especially when friction is low.

Derivation of maximum safe speed (with friction): Consider a car of mass m going around a banked curve of radius r, banking angle θ, with coefficient of (limiting/static) friction μ between tyres and road. At the maximum possible speed, the car is on the verge of skidding outward (up the incline), so friction f = μN acts down along the incline, opposing the outward slide.

Resolving forces:

Vertical direction (no vertical acceleration):

N cosθ = mg + f sinθ = mg + μN sinθ

N (cosθ − μ sinθ) = mg ...(1)

Horizontal (centripetal) direction (net force = mv²/r):

N sinθ + f cosθ = mv²/r

N sinθ + μN cosθ = mv²/r

N (sinθ + μ cosθ) = mv²/r ...(2)

Dividing equation (2) by equation (1):

(sinθ + μ cosθ) / (cosθ − μ sinθ) = v² / (rg)

Dividing numerator and denominator on the left by cosθ:

(tanθ + μ) / (1 − μ tanθ) = v² / (rg)

vmax² = rg × (μ + tanθ) / (1 − μ tanθ)

vmax = √[ rg(μ + tanθ) / (1 − μ tanθ) ]

…

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