# A car goes around a flat curve of radius 50 m at a speed of 14 ms^-1. What is the minimum coefficient of friction between the tires and the road for the car to make the turn?

Then teach the underlying concepts
Don't copy without citing sources
preview
?

#### Explanation

Explain in detail...

#### Explanation:

I want someone to double check my answer

16
Jul 3, 2016

Centripetal acceleration is given by $a = {v}^{2} / r$.

The minimum friction coefficient required is $0.4$.

(friction coefficients have no units)

#### Explanation:

To find a force we need to know something about a mass, and we haven't been told the mass of the car. Let's just call it '$m$' and leave it at that for the moment, because I have a feeling it will cancel out in the end.

The centripetal force is given by ${F}_{\text{cent}} = m a = \frac{m {v}^{2}}{r}$

We have values for the velocity and the radius, so:

${F}_{\text{cent}} = \frac{m \times {14}^{2}}{50} = 3.92 m$ $N$

The frictional force must be equal to or greater than this force in order for the car to successfully make it around the curve without sliding out.

The frictional force will be given by:

${F}_{\text{frict"=muF_"norm}}$

Where ${F}_{\text{norm}}$ is the normal force, equal to $m g$.

We can equate these two forces, the frictional force and the centripetal force:

${F}_{\text{cent"=F_"frict}}$

$3.92 m = \mu m g$

We can cancel out a factor of $m$ in both sides and rearrange to make $\mu$ the subject:

$\mu = \frac{3.92}{g}$

Substituting in the value $g = 9.8$ $m {s}^{-} 2$,

$\mu = \frac{3.92}{9.8} = 0.4$

• 26 minutes ago
• 26 minutes ago
• 27 minutes ago
• 31 minutes ago
• 53 seconds ago
• 4 minutes ago
• 4 minutes ago
• 13 minutes ago
• 14 minutes ago
• 21 minutes ago
• 26 minutes ago
• 26 minutes ago
• 27 minutes ago
• 31 minutes ago