0
$\begingroup$

There are a few things in car Physics that keep confusing me and I hope you can help me improve my understanding. The first thing I would like to discuss is braking.
I understand that when a car is accelerating, the wheels are pushing backwards on the asphalt which results in a towing force speeding the car up (N3 pair). However, when the brakes are applied and the braking force does not exceed a certain value, the wheels are still rotating - and so still pushing back on the asphalt - the same story. Yet all of the diagrams I found on the Internet depict the force vector with its head against the movement of the car, which makes perfect sense intuitively, but I would really like to know what this change in direction is caused by.

My hunch is that it must have something to do with the braking pads being engaged, they generate a lot of friction on the rotor, changing the car's kinetic energy into heat. But, when I attempt to draw a free-body diagram, there is nowhere the inner force could be drawn.

Thanks for investing your time in advance!

$\endgroup$
2
$\begingroup$

When a car is braking, the wheels are pushing on the asphalt in opposite direction then when accelerating. So it is not the same story.

Acceleration: the force is generated by the engine which is applying torque on the shaft, which is turning the wheels. The friction at the contact point between the wheels and the road gives rise to the forward force.

Braking: during braking, the engine is usually not actively applying torque to the shaft, so the forward force is gone. On the other hand, the brake pads are applying force on the rotor, slowing it down. This has the opposite effect on the wheel rotation: they are not forced to turn faster, but slower. This (+ friction) gives rise to decelerating force at the contact point with the road.

$\endgroup$
1
$\begingroup$

Your hunch is somewhat correct, for details see below link.

And about the free body diagram, I don't think the internal forces need to be drawn. The fact is that, the stopping of rotation of wheels doesn't stop the car itself.

(Also the cars kinetic energy is not related to the rotation of the wheels, rather it depends on the moving car itself.)

When the car moves at constant velocity the wheels don't experience any friction from the ground because the wheels rotate at a speed which is equal to the speed of the car.

You might have seen this expression: $$v=ωr$$

This is a condition for pure rotation, where there is no friction acting on the wheels. However when $$ωr<v$$ the wheels experience friction in the direction opposite to that of the motion, which is the reason for the car slowing down.

How Car Brake Works

$\endgroup$
  • $\begingroup$ You said they don't experience any friction, what about rolling friction? Does it not blur the image? $\endgroup$ – ILoveChess Jul 19 '17 at 10:03
  • $\begingroup$ Rolling friction is experienced only when the car is accelerating. Here I have taken the car's velocity to be constant. $\endgroup$ – Anjan Jul 19 '17 at 10:11
  • $\begingroup$ And even when break is applied the wheel stops rotating within a short time. After which there is no point in taking rolling friction. After the wheel stop rotating, you might as well replace the wheels with cubical blocks. $\endgroup$ – Anjan Jul 19 '17 at 10:17
1
$\begingroup$

The no slip condition is $v=r \omega$, where $v$ is the linear speed of the centre of the tyre/wheel/car, $r$ is the radius of the wheel and $\omega$ the angular speed of the wheel, will result in the frictional force between the tyre and the road at the point of contact being zero on a horizontal road.

When the engine tries to make the wheel rotate faster there is the potential of relative movement between the tyre and the road at the point of contact so a static frictional force acts on the tyre which increases the linear speed (and angular speed) of the tyre/wheel/car.
This is the static frictional force acting in the forward direction.

When the brakes try and slow the rotation of the wheel down again a static frictional force acts on the tyre which reduces the linear speed (and angular speed) of the tyre/wheel/car.
That static frictional force acts in the opposite direction to the motion of the car.

So the static frictional force between the tyre and the car always tries to make sure there is no relative movement between the tyre and the road at the point of contact.

If there is slipping between the tyre and the road (either during acceleration or braking) then kinetic friction takes over to try and reduce the relative motion between the tyre and the road at the point of contact.

So in simple terms your have an number of forces acting on the car including the weight of the car, the normal reaction of the road on the car, the frictional forces between the tyres of the car and the road abd the frictional forces between the air and the car.

To get the full picture you also need to consider the torques which are acting on the car which will make the front of the car lift a little when accelerating and drop a little when it is braking.

Too great an acceleration will produce torques which can cause problems.

$\endgroup$
0
$\begingroup$

Imagine the road floating. Braking would push the road forward, and accelerating would push the road backwards.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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