Question

A car weighing 15000 N rounds a curve of 60 m radius banked at angle 30°. Find the friction force acting on tires when the car is travelling at 100 kph. The coefficient of friction force between the tires and the road is 0.90.

Answer 1

Given: weight W = 15000 N ⇒ m = W/g = 15000/9.8 = 1530.61 kg.
v = 100 km/h = 27.7778 m/s, r = 60 m, θ = 30° (sin30 = 0.5, cos30 = 0.8660, tan30 = 0.57735).

From force balance on a banked curve:
N sinθ + f cosθ = m v^2 / r
N cosθ − f sinθ = mg

Eliminating N gives
f = m cosθ (v^2/r − g tanθ).

Compute:
v^2/r = (27.7778^2)/60 = 12.8601 m/s^2
g tanθ = 9.8·0.57735 = 5.6598 m/s^2
Difference = 7.2003 m/s^2

So
f = 1530.61·0.8660·7.2003 ≈ 9.54 × 10^3 N.

Direction: since v > the “no-friction” (ideal) speed, the friction acts down the plane (i.e., toward the center of the curve).

Check max static friction: N = (mg + f sinθ)/cosθ ≈ 2.2834×10^4 N ⇒ μN = 0.9·N ≈ 2.055×10^4 N > f, so no slipping.

Answer: f ≈ 9.54 kN, down the slope (toward the center).