Question

What is the centripetal force acting on a car of weight $15000\text{N}$ moving on a frictionless road banked at an angle of ${30}^{\circ }$?

• A. $1500\text{N}$
• B. $15000\sqrt{3}\text{N}$
• C. $75000\text{N}$
• D. $5000\sqrt{3}\text{N}$

Answer 1

The correct option is D $5000\sqrt{3}\text{N}$

From the FBD of a car moving on a banked circular track:


We know,

Centripetal force is provided by the horizontal component of the normal reaction.

So we can write
i.e $N\sin \theta =\frac{m{v}^2}{r}$ Where, r is the radius of circular path
and $N\cos \theta =mg$

According to the question

The weight of car is given to be $15000\text{N}$ and $\theta ={30}^{\circ }$
$\therefore N\cos \theta =15000$

Or,
$\Rightarrow N=\frac{15000}{\frac{\sqrt{3}}{2}}$

Therefore, centripetal force given by:
$F_c=N\sin \theta =\frac{15000}{\frac{\sqrt{3}}{2}}\times \frac{1}{2}$
$=5000\sqrt{3}\text{N}$

Hence the centripetal force is $5000\sqrt{3}N$