Question

what should be the coefficient of friction between the tyres and the road, when a car is travelling at 60km/h makes a level turn of radius 40m?

Answer 1

Given,
Speed of car=60 km/h=60000/3600 m/s=16.7 m/s
Radius of turn=40 m
$$\begin{gathered} \mathrm{The}\mathrm{maximum}\mathrm{centripetal}\mathrm{force}\mathrm{that}\mathrm{friction}\mathrm{can}\mathrm{provide}\mathrm{here},\mathrm{is} \\ \mathrm{F}={\mathrm{\mu }}_{\mathrm{s}}\mathrm{N}={\mathrm{\mu }}_{\mathrm{s}}\mathrm{W}={\mathrm{\mu }}_{\mathrm{s}}\mathrm{mg} \\ \mathrm{Thus}, \\ \mathrm{Centripetal}\mathrm{force}=\mathrm{Frictional}\mathrm{force} \\ \frac{{\mathrm{mv}}^2}{\mathrm{r}}={\mathrm{\mu }}_{\mathrm{s}}\mathrm{mg} \\ \mathrm{or}{\mathrm{\mu }}_{\mathrm{s}}=\frac{{\mathrm{v}}^2}{\mathrm{rg}} \\ =\frac{16.7^2}{40\times 9.8} \\ =\frac{278.89}{392}=0.71 \end{gathered}$$