1
$\begingroup$

I'm reading physics last year at a Swedish Gymnasium (equivalent to High School/College) and have a problem understanding a concept in the following text.

Translated text:

A car with the mass $m$ drives with constant velocity in a horizontal curve with radius $r$. The Swedish Road Administration has unfortunately forgotten to cant the road.

$\ldots$ (some calculations).

$F_\mu=m\frac{v^2}{r}$

This formula shows that it only is the friction that keeps the car on the road. The car loses grip of the road and starts to slide when the velocity gets so big that $F_\mu=\mu N=\mu mg$ where $\mu$ is the coefficient of friction for sliding friction.

I do not understand why the formula itself shows that it is only the friction that keeps the car on the road. It may be the case that I miss some concepts in formulae intepretation.

EDIT: This might be an answer to my question, but it does not explain how the formula itself shows that it only is the friction that keeps e.g. a car on the road.

$\endgroup$
3
  • 1
    $\begingroup$ The right hand side of the equation is the centripetal force, which is the force responsible for keeping the car on a circular path. If the car wants to take a turn in the road, it needs some kind of force that provides this centripetal force for it: this job is done by friction. That's why the road needs a coefficient of friction of $\mu = \frac{v^2}{g r}$, where $v$ is the velocity in tangential direction. $\endgroup$
    – DomDoe
    Mar 5, 2018 at 9:44
  • $\begingroup$ Think about it, what other forces (other than friction) do you think acts towards the center? $\endgroup$ Mar 5, 2018 at 16:08
  • 1
    $\begingroup$ The formula itself doesn't show that only friction keeps the car on the road. Presumably, "... (some calculations)" claimed to be deriving all the forces acting on the car, and found that $F_\mu$ was the only centripetal one. So you could argue that the the formula along with the claim that it represents all forces acting on the car is what shows that only friction is keeping the car on the road. When the road is canted, there is an additional component of the normal force itself along the direction of acceleration with the curving road. $\endgroup$
    – Mike
    Mar 5, 2018 at 16:29

1 Answer 1

1
$\begingroup$

A car with the mass $m$ drives with constant velocity in a horizontal curve with radius $r$.

It is not the velocity which is constant it is the speed which is constant as the direction of motion of the car changes as the car travels in a circular path.


To change the velocity of the car which in this case means changing the direction of travel of the car a force must be applied on the car - the car undergoes a centripetal acceleration.

If the road is not banked then the force acting on the car must be a frictional force on the car due to the road on which the car is travelling.
If there was no friction the direction of travel of the car would not change - think of hitting a patch of ice on the road.

The static frictional force $F_{\rm static}$ is related to the normal reaction $N$ as follows $F_{\rm static} \le \mu_{\rm static}N = \mu_{\rm static}mg$ where $m$ is the mass of the car noting that the static frictional force can vary up to a maximum value of $\mu_{\rm static}mg$.

In the situation that you have described that static frictional force is the only horizontal force capable of producing a centripetal acceleration $\dfrac{v^2}{r}$ where $v$ is the speed of the car and $r$ the radius of the bend.

Using Newton's second law, $F=ma$, gives $\mu_{\rm static}mg = \dfrac{mv_{\rm maximum}^2}{r}$.

At a lower speed than $v_{\rm maximum}$ the frictional force is less than $\mu_{\rm static}mg$ and the car goes aroung the bend.
If the speed of the car exceeds $v_{\rm maximum}$ the frictional force is insufficient to keep the the car going round the bend of radius $r$ and the car starts going along a path which has a larger radius of curvature ie the car cannot follow the bend and moves outwards.

It is sometimes the case that when this happens the car suddenly starts to skid there being relative movement between the tyres and the road.
When this happens kinetic friction is the important parameter and as it is smaller than static friction the frictional force drops dramatically possibly leading to a situation where the car is out of the control of the driver - the car is travelling almost in a straight line.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.