I've been doing physics problems regarding cars for a while. I understand that there is a static friction (which appears when the wheel is rolling) and kinetic friction (which appears when the wheel is sliding). However, the way I'm visualizing it, static friction between the tire and road should not stop a car. In fact, when I asked this question to my teacher a long time ago, he said that it is actually the friction between the axle and the wheel that stops the car, and the road friction actually helps the car to move. But I know that when the car slips, the friction decreases and thus stopping time increases. How could this possible be linked to the axle? What is going on?
Newtonian Mechanics – What Force Stops a Car?
free-body-diagramfrictionnewtonian-mechanicsrotational-dynamics
Related Solutions
I'm not sure I fully understand what you're asking, but rolling resistance and static friction are very different.
Rolling resistance tends to be a catch all term for the energy dissipated in the many moving parts as a vehicle moves. Most of this is probably viscous drag due to oil in the bearings, gearbox etc.
Static friction is the force required to make two surfaces slide over each other, but as long as the surfaces remain static and don't skid there is no energy dissipated.
Take your example of towing a car. Suppose you tow it at a constant speed for 1 metre and suppode you have to pull with a force of 100N to do this, then the work you've done is just force times distance or 100J. Since the car was moving at a constant speed no energy was used to accelerate it, so the 100J went into heating up the oil in the bearings and gearbox etc. It's this energy dissipation that is responsible for the "rolling resistance" of 100N.
The static friction in this example is between the tyres and the road. However as long as the tyres don't skid no energy is dissipated so the static friction doesn't affect the force you feel as you try to tow the car. If you reduce the friction between the tyres and the road, e.g. tow the car on wet ice, then at some point the tyres will start skidding instead of gripping the road. When this happens the wheels don't rotate and effectively the car behaves as a single solid object. Now the energy is dissipated in the contact patch between the tyres and the road and the force you need to tow the car at a steady speed depends on how much energy is dissipated. The slipperier the surface the less energy is dissipated so the less force is needed to tow the car.
The slip ratio depends on the speed for the car you would calculate based circumferential speed of the wheel in the frame of the car (angular velocity of wheel times radius), and the actual linear speed of the car.
The reason that you get slip at even the smallest forces results not from the fact that the tire is slipping against the ground, but that the tire is elastic. Let's see how this could happen.
To measure the slip, lets put little green splotches of die on the circumference of the tire spaced 1cm apart. From this we can tell how much the wheel has rotated. Now imagine a car that is accelerating. What happens to the tires? Well the road is providing a force on the tire. What does that do to the bottom of the tire? Well just imagine a stationary tire that can't rotate and you apply a force tangent to the tire. This will cause the tire to deform and the part of the tire you are apply the force to will get scrunched up in the direction of the force. Now if you force the tire to rotate against the force, the scrunched up part will go to where the "ground" (the thing applying the force) is.
This means that our little green splotches, instead of being 1 cm apart, they will be .8 cm apart. Suppose there are a total of 11 splotches. Then by the time tire turns enough for each splotch gets to its original position, the wheel has rotated a full revolution. On the other hand, the car has only moved 8cm, because each of the splotches is .8 cm apart and there are 11 of them (so 10 intervals). Now when we compare this 8cm that the car has actually moved while the wheel rotated once to the full circumference of the wheel, which is 10 cm, we conclude that the wheel has slipped.
Since there will always be some scrunching given a non-zero tangent force, you will always get slip for a nonzero tangent force.
Of course this scrunching goes to zero in the limit of an infinitely rigid wheel, which is the sort of wheel used in physics homework problems.
Now for high enough slip ratios, the wheel will actually slide across the pavement, but until you get to the this point, static friction is still in play, so the car is accelerating from static friction.
Best Answer
Imagine a car just going along at constant speed. Do your free-body diagram. The net force on the car has to be zero. So the wheels have to be exerting on the ground (net at least) only a vertical force.
Now imagine the car decreasing speed. There has to be a force opposite to the velocity. Free-body time again. The force on the ground has to include a force component that opposes the motion of the car. The wheels are the part touching the ground, so they must be supplying that force.
Remember your Newton's laws. To stop the car must be acted on by an external force. If you call the wheels "part of" the car, then the stopping force has to be applied by the ground. Meaning the wheels have to push back exactly as hard.
If you call the wheels "not part of" the car, then you can describe it as the braking mechanism applying a force to the wheels. Then it's "the wheel's problem" what it does with that force. But in that case, the car is stopped by friction between the brake mechanism and the wheel. It may be that your teacher is trying to get you to think that way.