一公路上有一圓弧形彎道 係為54km/hr的車速設計

1. 54 km/h = 15 m/s
Let R be the normal reaction acting on the car
R.cos(theta) = mg where m is the mass of the car
R.sin(theta) = m(15)^2/150

Dividing, tan(theta) = (15^2)/150g = 0.15 (take g = 10 m/s^2)
Angle (theta) = 8.53 degrees

2. Frictional force Ff = u.R where u is the coefficient of friction
But R = mg
Hence, Ff = u.mg

Therefore, u.mg = m(15^2)/150
u = 15^2/150g = 0.15