Now for vehicles moveing at 40 kmh^{1 }(11.1 ms^{1})
The centripetal force F_{c} = mv^{2}/r becomes less and as the car tends to slip down the slope a frictional force F acting up the place opposes it.
F can be resolved into can be resolved into F _{v} and F _{h}
F_{v } = F Sin θ and F_{h } = F Cos θ
Recall F = μR so F_{v } = μR Sin θ and F_{h } = μR Cos θ
So if the car is not moving vertically then by Newton's first law:
R_{v } + F_{v } = W
R_{v } + μR Sin θ = mg
R Cos θ + μR Sin θ = mg
mg = R Cos θ + μR Sin θ Eq5
Horizontally F_{c} = mv^{2}/r = R_{h }  F_{h}
mv^{2}/r = R_{h }  μR Cos θ
mv^{2}/r = R Sin θ  μR Cos θ Eq6
Dividing Eq6 by Eq5 we get:
mv^{2}/rmg = (R Sin θ  μR Cos θ) / (R Cos θ + μR Sin θ)
v^{2}/rg = R (Sin θ  μ Cos θ) / R (Cos θ + μ Sin θ)
v^{2}/rg = (Sin θ  μ Cos θ) / (Cos θ + μ Sin θ)
(11.1 ms^{1})^{2} / (200m) (9.81 ms^{2}) = (Sin θ  μ Cos θ) / ( Cos θ + μ Sin θ)
6.3 X 10^{2} = (Sin θ  μ Cos θ) / ( Cos θ + μ Sin θ)
(6.3 X 10^{2}) ( Cos θ + μ Sin θ) = (Sin θ  μ Cos θ)
{(6.3 X 10^{2}) ( Cos θ)] + [(6.3 X 10^{2}) ( μ Sin θ)] = (Sin θ)  (μ Cos θ)
{(6.3 X 10^{2}) ( Cos θ)] + [(6.3 X 10^{2}) ( μ Sin θ)] + (μ Cos θ) = (Sin θ)
((6.3 X 10^{2}) (μ Sin θ) + (μ Cos θ) = (Sin θ)  (6.3 X 10^{2}) ( Cos θ)
μ [(6.3 X 10^{2}) Sinθ + Cos θ)] = (Sin θ)  (6.3 X 10^{2}) ( Cos θ)
μ = [(Sin θ)  (6.3 X 10^{2}) ( Cos θ)] / [(6.3 X 10^{2}) Sinθ + Cos θ)]
μ = [(Sin 8.1°)  (6.3 X 10^{2}) ( Cos 8.1°)] / [(6.3 X 10^{2}) Sin 8.1° + Cos 8.1°)]
μ = 7.86 X 10^{2}
