# A car of mass 1490 kg makes a 23.0 m radius turn at 7.85 m/s on flat ground. What is the (minimum) coefficient of static friction?

##### 1 Answer
Nov 3, 2016

Given

$m \to \text{mass of car} = 1490 k g$

$r \to \text{radius of circular turn} = 23 m$

$v \to \text{velocity of car on the turn"=7.85" m/s}$

Let

mu->"coefficient of static friction"=?

The centripetal force ${F}_{c}$ required for turning of the car of mass m in the circular path of radius r is given by

${F}_{c} = \frac{m {v}^{2}}{r}$

Here the force of static friction will provide the required centripetal force and resist the tendency of skidding and this force of static friction ${F}_{s}$ is given by

${F}_{s} = \mu \times \text{normal reaction}$

$\implies {F}_{s} = \mu \times m g$

By the condition of the problem

${F}_{s} \ge {F}_{c}$

$\implies \mu m g \ge \frac{m {v}^{2}}{r}$

$\implies \mu \ge {v}^{2} / \left(r \cdot g\right) = {7.85}^{2} / \left(23 \cdot 9.8\right)$

$\implies \mu \ge 0.27$