Question

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

• A. $20$ km/hr
• B. $30$ km/hr
• C. $40$ km/hr
• D. $50$ km/hr

Answer 1

Answer: (A), (B), (C)

$40$ km/hr

Speed of vehicle (v) = 36 km/hr = 10 m/sec
Let $\theta$ be the angle at which the road is banked.

Let us find out all the forces that are acting on this moving vehicle

(i) Mg in downward direction

(ii) N (normal reaction) acting perpendicular to the surface of road

(iii) pseudo force acting in radially outward direction

(iv) frictional force acting parallel to the road, either in inward or outward direction

Case 1:
To avoid slipping of the vehicle
In this case friction will act in outward direction


Let us now resolve all the forces in the direction along the surface and perpendicular to the surface.

Now we get ,
$N=Masin\theta +Mgcos\theta$.......(i)

$Mgsin\theta =\mu N+Macos\theta$...(ii)

To calculate $\theta$ we know

$tan\theta =\frac{a}{g}$

$tan\theta =\frac{v^2}{rg}$

$tan\theta =\frac{(10{)}^2}{20g}$ = 0.5........(iii)

Put (iii) in (ii) and (i) we get
$V_{min}=\frac{10}{\sqrt{6}}$ m/s = 14.7 km/h

Case 2:

To avoid vehicle from skidding up

In this case friction will act inward direction.



Again resolve the forces along the surface and perpendicular to the surface.

$\mu N+mgsin\theta =m{a}_{3}cos\theta$.......(v)

$N=m{a}_{3}sin\theta +mgcos\theta$.....(vi)

$a_3=\frac{v_{max}^2}{r}$...(vii)
on solving (v), (vi) and (vii) this we get

$v_{max}$= 44.1 km/h

$\therefore$ Options A, B and C are correct