Question

A motor van weighing 4400 kg rounds a level curve of radius 200 m on an unbanked road at 60 km/hr. What should be the minimum value of coefficient of friction to prevent skidding? At what angle the road should be banked for this velocity?

Answer 1

Hint : The frictional force provides the centripetal force for the circular motion of the motor van. The coefficient of friction can be given as the tangent of the angle of banking.

Formula used: In this solution we will be using the following formula;
${\mu _{\min }} = \dfrac{{{a_c}}}{g}$ where ${\mu _{\min }}$ is the minimum coefficient of friction to prevent skidding, ${a_c}$ is centripetal acceleration during the bend, and $g$ is acceleration due to gravity.
${a_c} = \dfrac{{{v^2}}}{r}$ where ${a_c}$ is the centripetal acceleration of a body bending round a circle, $v$ is the speed of the body circulating, and $r$ is the radius of the circle created by the path of the body.
$\mu = \tan \theta$ where $\mu$ is still the coefficient of friction, and $\theta$ is the banking angle.

Complete step by step answer:
A motor van is said to be rounding a bend at a particular speed, the minimum coefficient of friction required to make that possible. The formula for minimum coefficient of friction is given as
${\mu _{\min }} = \dfrac{{{a_c}}}{g}$ but the centripetal acceleration is given as
${a_c} = \dfrac{{{v^2}}}{r}$ where $v$ is the speed of the body circulating, and $r$ is the radius of the circle created by the path of the body.
Hence, from known values, the centripetal acceleration is
${a_c} = \dfrac{{{{16.7}^2}}}{{200}}$ (since 60 km/hr is 16.7 m/s)
Hence, ${a_c} = 1.39445m/{s^2}$
Then inserting this value into ${\mu _{\min }} = \dfrac{{{a_c}}}{g}$
Then
${\mu _{\min }} = \dfrac{{1.39445}}{{9.81}} = 0.1421$
Then the angle which will provide the coefficient of friction can be given by the formula
$\mu = \tan \theta$ where $\mu$ is still the coefficient of friction, and $\theta$ is the banking angle.
Hence,
$\theta = {\tan ^{ - 1}}\mu$
By inserting the value of $\mu$ we have
$\theta = {\tan ^{ - 1}}0.1421$
$\Rightarrow \theta = 8.08^\circ$ or about $8^\circ 4'$ since 60 arc minute make one degree.

Note:
For clarity, the formula ${\mu _{\min }} = \dfrac{{{a_c}}}{g}$ can be proven as follows:
The centripetal force required to round the bend must be provided by the frictional force between the tires and the road, hence,
$m{a_c} = \mu mg$ ,
Hence, by cancelling the mass and making $\mu$ subject of the formula, we have
$\mu = \dfrac{{{a_c}}}{g}$ .