Banked Curve Safety in Wet Weather
A highway engineer in a hilly district is checking a curve of radius 48 m banked at 10° for a posted speed of 60 km/h. The road surface is dry in winter but often wet in monsoon. Tests give a coefficient of static friction of 0.60 on dry asphalt and 0.25 on wet asphalt. The engineer must decide whether the posted speed is safe in both conditions and whether a warning sign is needed in rain. A student models a 1200 kg car as a point mass and neglects air resistance and rolling friction. The car travels at constant speed around the curve. The engineer is unsure whether the bank angle alone is enough at 60 km/h and wants a calculation before recommending a lower wet-weather speed. The lane width is 3.5 m and there is no superelevation runoff.
1. Calculate the maximum speed, in km/h, at which the car can take the curve without sliding on (a) dry asphalt and (b) wet asphalt. Show the force balance that gives your expression.
Read the worked answer
Take the car as a particle. On the bank, the normal reaction N is perpendicular to the road and friction f acts along the road. For the maximum speed the car is about to slide up the bank, so f points down the slope. Resolving vertically: N cosθ = mg + f sinθ. Resolving horizontally toward the centre: N sinθ + f cosθ = mv²/r. At the limit f = μN. Eliminating N and f gives v_max² = rg(tanθ + μ)/(1 − μ tanθ).
Here r = 48 m, g = 9.8 m/s², θ = 10°, tanθ = 0.1763, rg = 470.4 m²/s².
Dry, μ = 0.60: numerator = 0.1763 + 0.60 = 0.7763; denominator = 1 − 0.60(0.1763) = 0.8942; ratio = 0.8682; v_max = sqrt(470.4 × 0.8682) = 20.2 m/s = 72.8 km/h.
Wet, μ = 0.25: numerator = 0.4263; denominator = 0.9559; ratio = 0.4460; v_max = sqrt(470.4 × 0.4460) = 14.5 m/s = 52.1 km/h.
The posted 60 km/h is below the dry limit but above the wet limit.
2. At the posted 60 km/h on the wet road, determine the minimum coefficient of static friction required. Compare with the measured 0.25 and state what advice the engineer should give.
Read the worked answer
At v = 60 km/h = 16.67 m/s, v²/(rg) = (16.67²)/(470.4) = 0.5908.
The required friction coefficient follows from rearranging the same force balance: μ_req = [v²/(rg) − tanθ]/[1 + (v²/(rg))tanθ] = (0.5908 − 0.1763)/(1 + 0.5908 × 0.1763) = 0.4145/1.1042 = 0.375.
The wet road supplies only about 0.25, so at 60 km/h the car would need 50% more friction than is available. The engineer should not rely on the 60 km/h sign in rain; a wet-weather advisory of about 50 km/h is consistent with the calculated 52 km/h limit, leaving a small margin.
3. Explain why the 10° bank alone is not sufficient at 60 km/h, and calculate the zero-friction design speed of the curve.
Read the worked answer
The zero-friction design speed is the speed at which no friction is needed: N sinθ = mv²/r and N cosθ = mg, so v² = rg tanθ. With rg = 470.4 and tanθ = 0.1763, v = sqrt(82.94) = 9.11 m/s = 32.8 km/h.
The 10° bank is therefore matched to about 33 km/h, not 60 km/h. At 60 km/h the car needs a large inward force that the horizontal component of N alone cannot supply, so friction must point down the slope. On a wet road the available friction is insufficient, which is why the bank angle alone cannot make the posted speed safe.