Question about car braking and tire friction

In summary, static friction exists when there is a tendency of movement, as in the case of a body on a slope that does not move.
  • #1
jaumzaum
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Homework Statement
When designing a highway and its signaling system, it is necessary to consider variables that can interfere with the minimum distance required for a vehicle to stop. Consider a situation in which a car travels at a constant speed along a flat, horizontal road, with a fixed coefficient of static and dynamic friction and that, in a certain point, starts to brake, decelerating uniformly until it stops, without slipping. Disregard losses due to air resistance and friction between the mechanical components of the vehicle. Regarding the minimum braking distance, in the situations described, the following statements are made:
a) It increases proportionally to the car's mass.
b) It is inversely proportional to the coefficient of static friction.
c) It is not related to the acceleration of local gravity.
d) It is directly proportional to the square of the car's initial speed.
Relevant Equations
.
I'm a little confused about this question because I think the braking deceleration probably takes into account the braking system of the vehicle as well as the power of the motor. I don't understand how a car could actually brake without the "friction between the mechanical components of the vehicle". I don't understand how braking systems works, but I think it has something to do with trying to decelerate the wheels axle, and that can only be done by friction in my opinion.

Also, consider the following situation: a wheel is moving in a horizontal plane without slipping. In a stationary situation, where the v=wr, there v is the center of mass velocity, there is no friction being applied in the wheel, and it will move with a constant angular velocity and a constant center of mass velocity according to first Newton law.

If friction were applied to that wheel in a condition of no slipping, it would have to be static friction. But for friction to be applied, it would also need to exist another external force or torque (i.e. from the mechanical parts), as discussed above. With no external force or torque, there could be no friction.

So, judging the alternatives, I would say a) is a little confusing, because if we keep the "braking power" constant (the maximum force or torque that the braking system can apply in the wheels), and we increase the car mass, we would be reducing the deceleration, as well as increasing the distance.
b) makes sense, but only if we consider only friction going on.
c) also makes sense, also if we only consider friction going on.
d) the only that I would say is always correct, by Torrichelli.
 
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  • #2
jaumzaum said:
I'm a little confused about this question because I think the braking deceleration probably takes into account the braking system of the vehicle as well as the power of the motor. I don't understand how a car could actually brake without the "friction between the mechanical components of the vehicle". I don't understand how braking systems works, but I think it has something to do with trying to decelerate the wheels axle, and that can only be done by friction in my opinion.
I think you only need to consider the static friction between the wheels and the road. Note the " without slipping".
 
  • #3
willem2 said:
I think you only need to consider the static friction between the wheels and the road. Note the " without slipping".

But it's impossible to have only static friction without any external force. As the argument that I gave. Don't you agree?

If we didn't have any external force, we wouldn't have any friction.
 
  • #4
jaumzaum said:
But it's impossible to have only static friction without any external force. As the argument that I gave. Don't you agree?

If we didn't have any external force, we wouldn't have any friction.
Static friction is the external force; external to the car.
 
  • #5
jaumzaum said:
But for friction to be applied, it would also need to exist another external force or torque (i.e. from the mechanical parts), as discussed above
If I understand your concern is it that without friction in, for instance, the brake pads, the tires will be free-wheeling and there will be no friction with the road.

As I interpret the question, we are to disregard the mechanism by which the brakes are applied (drum brakes, disk brakes, engine braking, etc) and take the resulting rotation of the tires as a given: rolling without slipping with as much force as can be sustained without breaking loose and sliding.
 
  • #6
PeroK said:
Static friction is the external force; external to the car.

As I said, consider the following: We have a cylinder rolling without slipping in t=0 and no external force (except possibly friction) being applied. The velocity of its center of mass will be v. The angular velocity will be w. And v=wr as there is no slipping. The relative velocity of the point of contact and the ground is zero. So if we had any friction there, that friction must be static. We have static friction if and only if we have tendency of movement (i.e. if we assume friction force is zero, the body would start moving). For example, if a body is on a slope and it does not move, there is static friction because if we considered that there was not, the body would start to move. But, by the first Newton law, if we assume there is no friction in the case above (cylinder rolling), it will continue to roll with a constant CM velocity and a constant angular velocity. And the point of contact will have zero relative velocity, always. So there is no tendency of movement. And there couldn't be any friction if there was no external force besides the static friction.

Another argument is that, if a cylinder is rolling to the right without slipping, the velocity of the center of mass v is to the right, and the angular velocity is clockwise. Consider only a static friction force is able to stop it. For it to reduce the velocity v its direction should be be to the left. But any force to the left applied on the ground would produce a Torque in the clockwise direction, increasing the angular velocity of the cylinder, that will start to slip, because v decreased and w increased, which is an absurd. Cylinder stop almost entirely because of the action of air resistance. That's why when you roll a circular shape object it travels meters before stopping. If you move a flat body of the same material in the ground, it would travel centimetres, or even less. The first case doesn't have static friction going on, only air resistance.

And that's why when I said "external force" I meant "another force that is not friction". It's impossible for the wheel to brake without any other external force besides the friction, the braking system of the car must provide a friction force also for it to be able to brake.
 
  • #7
jaumzaum said:
And that's why when I said "external force" I meant "another force that is not friction". It's impossible for the wheel to brake without any other external force besides the friction, the braking system of the car must provide a friction force also for it to be able to brake.
Brakes are, of course, an internal force. For instance, pads on a disk. Both are part of the vehicle.

And cars, of course, are not cylinders. Unbalanced normal force on front and rear tires can provide a counter-torque for any torque arising from static friction.
 
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  • #8
jbriggs444 said:
If I understand your concern is it that without friction in, for instance, the brake pads, the tires will be free-wheeling and there will be no friction with the road.

Exactly!
jbriggs444 said:
As I interpret the question, we are to disregard the mechanism by which the brakes are applied (drum brakes, disk brakes, engine braking, etc) and take the resulting rotation of the tires as a given: rolling without slipping with as much force as can be sustained without breaking loose and sliding.

I also think this is what the creator of the question wanted, but it's very confusing to assume this if these considerations are not given in the question. Don't you agree?
 
  • #9
jaumzaum said:
As I said, consider the following: We have a cylinder rolling without slipping in t=0 and no external force (except possibly friction) being applied. The velocity of its center of mass will be v. The angular velocity will be w. And v=wr as there is no slipping. The relative velocity of the point of contact and the ground is zero. So if we had any friction there, that friction must be static. We have static friction if and only if we have tendency of movement (i.e. if we assume friction force is zero, the body would start moving). For example, if a body is on a slope and it does not move, there is static friction because if we considered that there was not, the body would start to move. But, by the first Newton law, if we assume there is no friction in the case above (cylinder rolling), it will continue to roll with a constant CM velocity and a constant angular velocity. And the point of contact will have zero relative velocity, always. So there is no tendency of movement. And there couldn't be any friction if there was no external force besides the static friction.

Another argument is that, if a cylinder is rolling to the right without slipping, the velocity of the center of mass v is to the right, and the angular velocity is clockwise. Consider only a static friction force is able to stop it. For it to reduce the velocity v its direction should be be to the left. But any force to the left applied on the ground would produce a Torque in the clockwise direction, increasing the angular velocity of the cylinder, that will start to slip, because v decreased and w increased, which is an absurd.Cylinder stop almost entirely because of the action of air resistance. That's why when you roll a circular shape object it travels meters before stopping. If you move a flat body of the same material in the ground, it would travel centimetres, or even less. The first case doesn't have static friction going on, only air resistance.

And that's why when I said "external force" I meant "another force that is not friction". It's impossible for the wheel to brake without any other external force besides the friction, the braking system of the car must provide a friction force also for it to be able to brake.

1) Braking is a little complicated, when you try to put it all together.

2) Your main conclusion that braking is impossible (without skidding) is false.

3) There is no other significant source of an external force apart from static friction. Static (or kinetic friction if you skid) is all you have.

4) All forces in the braking mechanism are internal forces and, by Newton's third law, cannot decelerate the car.
 
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  • #10
jbriggs444 said:
Brakes are, of course, an internal force. For instance, pads on a disk. Both are part of the vehicle.

And cars, of course, are not cylinders. Unbalanced normal force on front and rear tires can provide a counter-torque for any torque arising from static friction.

They are internal to the vehicle, but external to the wheel.
 
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  • #11
jaumzaum said:
I also think this is what the creator of the question wanted, but it's very confusing to assume this if these considerations are not given in the question. Don't you agree?
I was put off by the questioner wanting us to ignore internal forces:
Disregard losses due to [...] friction between the mechanical components of the vehicle
Any such frictional forces would be internal interactions and would be irrelevant in any case. Being instructed to ignore them is a waste of time. So I ignored that clause.

Without that clause in the question, the intent seems quite clear. We have a car braking to a stop. Braking systems are normally limited by tire traction, so we can ignore the brake mechanism entirely.
 
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  • #12
PeroK said:
2) Your main conclusion that braking is impossible (without skidding) is false.

I haven't said that braking is impossible without slipping (if it was, no car could brake without slipping). I just said that we need a force from the braking system plus another friction force. We cannot have only a friction force
 
  • #13
jaumzaum said:
I haven't said that braking is impossible without slipping (if it was, no car could brake without slipping). I just said that we need a force from the braking system plus another friction force. We cannot have only a friction force
I do not understand this claim. We have a force from the braking system: pad on disk. We have an external frictional force: road on tire. What third force are you contemplating?

Disk on axle? Axle on wheel? Wheel on tire? d'Alembert inertial force due to adoption of an accelerating frame? Something else?
 
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  • #14
jaumzaum said:
I haven't said that braking is impossible without slipping (if it was, no car could brake without slipping). I just said that we need a force from the braking system plus another friction force. We cannot have only a friction force
You need a braking system, of course. That system cannot directly contribute to decelerating the car.
 
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  • #15
jaumzaum said:
I haven't said that braking is impossible without slipping (if it was, no car could brake without slipping). I just said that we need a force from the braking system plus another friction force. We cannot have only a friction force
Is the same true of the initial acceleration of a vehicle? In that case, more explicitly, the force must be static fraction as air resistance cannot be used to accelerate.
 
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  • #16
jbriggs444 said:
I do not understand this claim. We have a force from the braking system: pad on disk. We have an external frictional force: road on tire. What third force are you contemplating?

None, only these two. As you said, we need a force of the pad on disk. But the question tells us explicitly to " Disregard losses due to friction between the mechanical components of the vehicle". The pads are, in my opinion, a mechanical component. That is what made me stuck. No car could break without the padding making a friction force on the disks. Also, if we consider the pad/disk force to be the limiting force, a greater mass will reduce the maximum possible acceleration without slipping (an the first alternative would be false).
 
  • #17
PeroK said:
You need a braking system, of course. That system cannot directly contribute to decelerating the car.
I agree. They do not contribute directly, but they are necessary.

PeroK said:
Is the same true of the initial acceleration of a vehicle? In that case, more explicitly, the force must be static friction as air resistance cannot be used to accelerate.
You need static friction in both cases, to accelerate and decelerate. But, in both cases, you will also need friction between the tires and another internal component of the car. Even though this second friction is a internal force to the car (it does not change the momentum), it's necessary for the movement.
 
  • #18
jaumzaum said:
None, only these two. As you said, we need a force of the pad on disk. But the question tells us explicitly to " Disregard losses due to friction between the mechanical components of the vehicle". The pads are, in my opinion, a mechanical component.
Fair enough. As I said, I took that bit as an unnecessary distraction and ignored it.

We are also told explicitly that the car brakes. An argument that the fine print does not permit braking flies in the face of the explicit statement that braking occurs. The explicit statement wins.

Also, if we consider the pad/disk force to be the limiting force, a greater mass will reduce the maximum possible acceleration without slipping (an the first alternative would be false).
Barring situations with long downhill grades, [decades old] drum brakes subject to fade or gas bubbles in the braking system, tire on pavement is essentially always the weakest link.
 
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  • #19
jaumzaum said:
You need static friction in both cases, to accelerate and decelerate. But, in both cases, you will also need friction between the tires and another internal component of the car. Even though this second friction is a internal force to the car (it does not change the momentum), it's necessary for the movement.
I think what you mean is that you need a force that takes away the KE, and static friction does no work. That could be kinetic friction in brake pads, or airbrakes, or regenerative braking...
 
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  • #20
jaumzaum said:
Homework Statement:: [...] Disregard losses due to air resistance and friction between the mechanical components of the vehicle. [...]

[...] I don't understand how a car could actually brake without the "friction between the mechanical components of the vehicle". [...]
By "friction between the mechanical components of the vehicle", they meant rolling resistance, which is usually ignored in simplification of that type of problem.

https://en.wikipedia.org/wiki/Rolling_resistance#Definitions said:
In the broad sense, specific "rolling resistance" (for vehicles) is the force per unit vehicle weight required to move the vehicle on level ground at a constant slow speed where aerodynamic drag (air resistance) is insignificant and also where there are no traction (motor) forces or brakes applied. In other words, the vehicle would be coasting if it were not for the force to maintain constant speed. This broad sense includes wheel bearing resistance, the energy dissipated by vibration and oscillation of both the roadbed and the vehicle, and sliding of the wheel on the roadbed surface (pavement or a rail).

They seem to have reduced "rolling resistance" to "bearing resistance", which is what I think they meant by "friction between the mechanical components of the vehicle". IMHO, not a good choice of words, if this is what they meant (and I can't think of anything else).
 
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Related to Question about car braking and tire friction

1. How does tire friction affect car braking?

Tire friction plays a crucial role in car braking. When the brakes are applied, the friction between the tires and the road surface helps to slow down the car. The more friction there is, the more effective the braking will be. However, too much friction can also cause the tires to lock up and skid, reducing the effectiveness of the brakes.

2. What factors can affect tire friction?

There are several factors that can affect tire friction, including the type of tire, the condition of the road surface, and the weight of the car. The type of tire and its tread pattern can greatly impact the amount of friction it can generate. A smooth road surface will provide more friction compared to a wet or icy surface. Additionally, a heavier car will require more friction to slow down compared to a lighter car.

3. How does tire pressure affect braking?

Tire pressure can significantly impact braking performance. If the tire pressure is too low, it can cause the tires to deform and reduce the contact area between the tires and the road surface. This will result in less friction and make it harder for the car to brake effectively. On the other hand, if the tire pressure is too high, it can cause the tires to lose traction and reduce the effectiveness of the brakes.

4. Can tire rotation affect braking?

Yes, tire rotation can affect braking. Regular tire rotation helps to ensure that all tires wear evenly, which can help maintain consistent levels of friction. If the tires are not rotated regularly, uneven wear can occur, leading to reduced friction and potentially affecting the car's braking performance.

5. How can I improve tire friction for better braking?

To improve tire friction for better braking, make sure to maintain proper tire pressure and regularly rotate the tires. It is also essential to choose the right type of tire for your car and driving conditions. Additionally, keeping the tires clean and free of debris can also help improve friction and braking performance.

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