Question

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?

Answer 1

Given

`m->"mass of car"=1490kg`

`r->"radius of circular turn"=23m`

`v->"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=(mv^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=muxx"normal reaction"`

`=>F_s=muxxmg`

By the condition of the problem

`F_s>=F_c`

`=>mumg>=(mv^2)/r`

`=>mu>=v^2/(r*g)=7.85^2/(23*9.8)`

`=>mu>=0.27`