Banked Curve on a Wet Road
A road engineer is checking a curve of radius 75 m banked at 10° to the horizontal. The design must keep a 1200 kg car on the road when the surface is wet, with coefficient of static friction μ_s = 0.35. On dry asphalt μ_s = 0.80. The car's speedometer reads 65 km/h as it enters the curve. A sensor records that the normal force on the outer tyres is larger than on the inner tyres. Midway around, the driver applies the brakes gently, reducing the speed while continuing around the same curve. The car has four wheels, and you may treat it as a point mass at its centre of mass for the friction limit, but not for the normal-force distribution. Atmospheric and rolling resistances are negligible.
1. Calculate the maximum speed at which the car can negotiate the curve on the wet road without slipping, and compare it with the speedometer reading.
Read the worked answer
For the maximum speed without slipping, friction acts down the slope. Resolving vertically and radially:
N cosθ - f sinθ = mg N sinθ + f cosθ = m v²/r.
At the limit f = μN. Dividing the radial equation by the vertical equation gives
v² = rg (tanθ + μ)/(1 - μ tanθ).
With r = 75 m, θ = 10°, μ = 0.35: tan10° = 0.1763, numerator = 0.5263, denominator = 1 - 0.35(0.1763) = 0.9383, rg = 75(9.81) = 735.8 m²/s².
v² = 735.8(0.5263/0.9383) = 412.8, v = 20.3 m/s ≈ 73 km/h.
The speedometer reading 65 km/h = 18.1 m/s is below this, so the car can stay on the wet curve. Since μ > tanθ, there is no nonzero minimum speed; the car could even be stationary without slipping.
2. Explain why the normal force on the outer tyres is greater, and what this implies about which tyres are more likely to lose grip.
Read the worked answer
The car's centre of mass must accelerate horizontally toward the centre of the curve. The contact normal forces must both provide the centripetal force and prevent the car from rolling outward. Taking moments about the centre of mass, the outer tyres must have a larger normal force than the inner tyres to produce the inward roll moment that balances the outward pseudo-torque in the car frame. Because the friction limit at each tyre is μN, the inner tyres have a smaller grip limit than the outer tyres. In a hard corner the inner tyres are therefore more likely to reach their limit first, even though the outer tyres carry more of the load.
3. The driver brakes gently mid-corner. Use a friction-circle argument to explain why this increases the risk of sliding, even though the car is slowing down.
Read the worked answer
Each tyre's friction force is limited by a friction circle of radius μN. Before braking, most of the available friction is used laterally to provide centripetal force. When the driver brakes, each tyre also needs longitudinal friction to decelerate the car. The vector sum of lateral and longitudinal friction cannot exceed μN, so the lateral force available becomes sqrt((μN)² - F_brake²), which is smaller. In addition, braking transfers load from the rear to the front, reducing the rear normal forces and hence the rear lateral grip. The car is therefore more likely to slide or rotate at the rear even though its speed is falling.