Coefficient 45821

What is the smallest radius a turn must have for a car to enter safely without exceeding a speed of 50 km/h? The coefficient of shear friction between the tires and the surface is 0.4.

Correct answer:

r =  7.8655 m

Step-by-step explanation:

$$\begin{gathered} v=50\text{km/h}\to \text{m/s}=50:3.6\text{m/s}=13.88889\text{m/s} \\ g=9.81{\text{m/s}}^2 \\ k=0.4 \\ {F}_g=mg \\ {F}_o=m{v}^2/r \\ {F}_{o}k=F_g \\ m{v}^2/rk=mg \\ {v}^2/rk=g \\ r=\frac{v^2\cdot k}{g}=\frac{13.8889^2\cdot 0.4}{9.81}=7.8655\text{m} \end{gathered}$$

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