Question

A car is moving on a road with a radius of curvature $70\text{m}$ banked at an angle of ${45}^{\circ }$. The coefficient of static friction between the road and the wheels of the car is ${\mu }_s=0.15$. Take $g=10{\text{m/s}}^2$. If the speed of the car is $20\text{m/s}$, then will the car skid ? If so, in what direction?

• A. The car will not skid.
• B. The car will skid up the incline.
• C. The car will skid down the incline.
• D. Skidding of car depends on the mass of the car.

Answer 1

The correct option is C The car will skid down the incline.

For case of $v_{max}$, car will have a tendency to skid up. Therefore, friction $f_{max}$ will act down the incline.


Writing equations of circular dynamics, we get:
$N\sin \theta +f_{max}\cos \theta =\frac{m{v_{max}}^2}{r}...\left(i\right)$
where $f_{max}={\mu }_{s}N....\left(ii\right)$

and $N\cos \theta -f_{max}\sin \theta =mg.....\left(iii\right)$

Using eqn $\left(i\right),\left(ii\right),\left(iii\right)$:
$v_{max}=\sqrt{\left(\frac{{\mu }_s+\tan \theta }{1-{\mu }_s\tan \theta }\right)rg}$
$\Rightarrow v_{max}=\sqrt{\left(\frac{0.15+\tan {45}^{\circ }}{1-0.15\times \tan {45}^{\circ }}\right)70\times 10}$
$\Rightarrow v_{max}=30.77\text{m/s}$

Hence the car will not slide up the incline ($\because v<v_{max}$)

For the case of $v_{min}$, car will have a tendency to slip down. Hence, friction will act up the plane:


Using equations of circular dynamics:
$N\sin \theta -f_{max}\cos \theta =\frac{m{v_{min}}^2}{r}....\left(i\right)$
where $f_{max}={\mu }_{s}N...\left(ii\right)$
$N\cos \theta +f_{max}\sin \theta =mg.......\left(iii\right)$

From Eq. $\left(i\right)$, $\left(ii\right)$, $\left(iii\right)$ :-
$v_{min}=\sqrt{\frac{rg(\tan \theta -{\mu }_s)}{1+{\mu }_s\tan \theta }}\text{m/s}$
$=\sqrt{\left(\frac{\tan {45}^{\circ }-0.15}{1+0.15\times \tan {45}^{\circ }}\right)70\times 10}$
$=22.74\text{m/s}$

Since the speed of the car $20\text{m/s}<$ the minimum speed ($v_{min}$), the car will skid down the incline.