Travelling at a speed of 36 km/hr , a car is taking a turn on a...

In the rotating frame of the car, the block experiences a centrifugal force $F_c = \dfrac{mv^2}{R}$ directed radially outwards.

Let $\theta$ be the angle between the normal to the plate and the radius of the circular road. The centrifugal force can be resolved into two components relative to the vertical plate. The component perpendicular to the plate provides the normal force $N = F_c \cos\theta$, and the component parallel to the plate is the driving force $F_{\parallel} = F_c \sin\theta$.

The maximum static frictional force between the block and the plate is f_\max = \mu N = \mu F_c \cos\theta.

The block will just start sliding when the parallel component of the centrifugal force equals the maximum static frictional force:

$F_c \sin\theta = \mu F_c \cos\theta$

$\tan\theta = \mu$

Given $\mu = 0.58$, we get:

$\tan\theta = 0.58$

Since $\tan 30^\circ = \dfrac{1}{\sqrt{3}} \approx 0.577$, the angle at which the block starts sliding is $\theta = 30^\circ$.

Answer: $30^\circ$