Question

A turn of radius $20m$ is banked for the vehicles going at a speed of $36km/h$. If the coefficient of static friction between the road and the tyre is $0.4$. What are the possible speed of a vehicle so that is neither slips down nor skids up then?

• A. 14.74
• B. 66
• C. 88
• D. 99

Answer 1

The correct option is A 14.74

$N=\frac{m{v}^2}{rsin\theta }+mgcos\theta$.

$mgsin\theta -\frac{m{v}^2}{rcos\theta }=\mu N$

(Difference symbol because if centrifugal greater then friction stops skid outward where as if component of weight greater then friction stops slip inwards)

Solve above 2 equation we get:

$v^2/gr=(tan\theta -\mu )/(1+\mu tan\theta )or(tan\theta +\mu )/(1-\mu tan\theta )$

Put value of $g,r,\mu ,tan\theta =\frac{{36}^2}{rg}$

$v=4.082m/sor15m/s$

multiply by $18/5$ to get kph

$v=14.7kphor54kph$