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?

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?

1 Answer

Impact of this question

4039 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-03T00:35:08

Given

$m \to mass of car = 1490 k g$ $m->"mass of car"=1490kg$

$r \to radius of circular turn = 23 m$ $r->"radius of circular turn"=23m$

$v \to velocity of car on the turn = 7.85 m/s$ $v->"velocity of car on the turn"=7.85" m/s"$

Let

$μ \to coefficient of static friction = ?$ $mu->"coefficient of static friction"=?$

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

$F_{c} = \frac{m v^{2}}{r}$ $F_c=(mv^2)/r$

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

$F_{s} = μ \times normal reaction$ $F_s=muxx"normal reaction"$

$\Rightarrow F_{s} = μ \times m g$ $=>F_s=muxxmg$

By the condition of the problem

$F_{s} \geq F_{c}$ $F_s>=F_c$

$\Rightarrow μ m g \geq \frac{m v^{2}}{r}$ $=>mumg>=(mv^2)/r$

$\Rightarrow μ \geq \frac{v^{2}}{r \cdot g} = \frac{{7.85}^{2}}{23 \cdot 9.8}$ $=>mu>=v^2/(r*g)=7.85^2/(23*9.8)$

$\Rightarrow μ \geq 0.27$ $=>mu>=0.27$

Science

Math

Humanities

... and beyond

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

Related questions

Impact of this question