This road is designed so that when the vehicle velocity is 60 kmh^{1} (16.7 ms^{1}) the centripetal force is generated ONLY by the normal reaction i.e. there is no friction between the road and the vehicle.
Since the car is not moving vertically then by Newton's first law:
F _{v } = W
F_{v } = mg
Now F_{v } = F Cos θ
So F Cos θ = mg Eq1
Centripetal force (F_{c}) is the unbalanced F_{h } = F Sin θ
F_{h } = F Sin θ
F Sin θ = mv^{2}/r Eq2
Dividing E2 by Eq1 we get:
F Sin θ / F Cos θ = (mv^{2}/r) / mg
Tan θ = (v^{2}/rg)
Tan 40° = (133 ms^{1})^{2}/[(r) (9.81 ms^{2})
r = (133 ms^{1})^{2} / [(tan 40°) (9.81 ms^{2})
r = 2.14 X 10^{3} m
