Friction and Newton's laws in car braking systems

[lemonxx]

TL;DR Summary

when a car brakes slowly, where does the static friction point relative to the motion?

When tires lock, the tires exert a forward force on the ground and the ground exerts a reaction force (kinetic friction) on the tires in the opposite direction. But if the car brakes slowly, the tires are still rotating and so they exert a backward force on the ground, and the ground exerts a forward force (static friction) on the tires? if that's the case, how come the static friction decelerates the car more efficiently than if it is slipping, considering one acts forward with the direction of motion, while the other opposes the motion?

 

[russ_watters]

But if the car brakes slowly, the tires are still rotating and so they exert a backward force on the ground, and the ground exerts a forward force (static friction) on the tires?

That isn't the case. The engine and brakes are torquing the wheels in opposite directions.

 

[jack action]

But if the car brakes slowly, the tires are still rotating and so they exert a backward force on the ground, and the ground exerts a forward force (static friction) on the tires?

No.

how come the static friction decelerates the car more efficiently than if it is slipping,

The typical relationship between the slip ratio and the friction coefficient is like this:

You get a maximum friction coefficient with a slip ratio ratio of about 20%.

This is because of the concept of static and kinetic friction:

One can see the similarities between the two graphs.

When the tire is slipping, part of the tire patch is static (no motion with respect to the ground) and part is kinetic (sliding on the ground). The higher the slip ratio, the larger the kinetic portion of the tire patch. The following is for lateral forces but the same concept applies to longitudinal forces:

At higher slip angles portions of the tire patch are sliding, and you get less increase in lateral force with an increase of slip angle. This is called the transition region. As the curve tops out, more of the contact patch is sliding and the tire produces less lateral force. After the peak of the curve, lateral force can fall off 30% within a few degrees of extra slip angle. At these high slip angles most of the contact patch is sliding, producing a lot of heat and wear.
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By the "adhesive" area, one should read the "static" area, and by the "slip" area one should read the "kinetic" area.

Ah! Here, I found a representation of those areas in rolling:

The applied torque to the tire - magnitude and direction - will vary the size of those areas.

 

[Lnewqban]

But if the car brakes slowly, the tires are still rotating and so they exert a backward force on the ground, and the ground exerts a forward force (static friction) on the tires?

Welcome, @lemonxx !

When brakes are applied with enough strength, and the velocity of the car is steadily decreasing, it is because there is a net force from the pavement acting on the contact patches of the wheels.
According to Newton's second law, in what direction that force vector should be pointing?

When the tires exert a backward force on the ground, they must simultaneously exert a forward force on the chassis of the car; therefore, the car accelerates forward.

 

[lemonxx]

That isn't the case. The engine and brakes are torquing the wheels in opposite directions.

what does that mean exactly? my main issue is that i don’t understand why static friction changes direction when braking as opposed to driving normally

 

[lemonxx]

Welcome, @lemonxx !

When brakes are applied with enough strength, and the velocity of the car is steadily decreasing, it is because there is a net force from the pavement acting on the contact patches of the wheels.
According to Newton's second law, in what direction that force vector should be pointing?

When the tires exert a backward force on the ground, they must simultaneously exert a forward force on the chassis of the car; therefore, the car accelerates forward.

it should be backwards to decelerate the car, but i don’t get why or how considering the wheels are still rotating

 

[A.T.]

i don’t understand why static friction changes direction when braking as opposed to driving normally

The intuitive way to find the direction of static friction, is to look at the relative motion (slipping) that would happen at the contact if the contact was frictionless. Static friction is always opposite to that potential relative motion (slipping), while its magnitude follows from Newtons 2nd Law: Whatever is needed to prevent the relative motion (slipping) within the limits given by the normal force and static friction coefficient.

 

[russ_watters]

what does that mean exactly? my main issue is that i don’t understand why static friction changes direction when braking as opposed to driving normally

The static friction with the ground comes from the torque applied by the engine or brakes, not the direction of motion. When the engine is torquing forwards the static friction is pushing the ground backwards. When the brakes are applying a torque backwards the static friction is pushing the ground forwards (corrected). The force applied to the car is in the direction of acceleration and the force applied to the ground is in the opposite direction.

If you don't believe that, consider cases where the car is standing still: brakes or engine holding a car on a hill (up or down). Torque applied by the engine at the moment acceleration starts(forwards or backwards). You can't say the friction force applied to the ground is in the opposite direction from the motion when the car is standing still.

Also consider the case where the car is moving but coasting; no torque applied to the wheels, no no static friction with the ground(other than a tiny amount from internal friction and rolling resistance).

 

[A.T.]

The static friction with the ground comes from the torque applied by the engine or brakes, not the direction of motion. When the engine is torquing forwards the static friction is pushing the ground backwards. When the brakes are applying a torque backwards the static friction is pushing the ground backwards forwards.

I think you meant "pushing the ground forwards" at the end during braking.

 

[jack action]

Here is an image of a tire accelerating. The vehicle goes towards the left with the wheel rotating counterclockwise. Notice how the rubber is "pushed" in front and "pulled" in the back:

It is more difficult to find a wheel braking where the tire deformation is so obvious but in the following image, you can see the deceleration induced by a pothole. The vehicle is still going towards the left with its wheel rotating counterclockwise. But now the rubber is "pulled" in front and it is "pushed" in the back:

Here are the videos where those images were taken from:

​

So, the same direction of rotation but opposite twisting torques are applied.

 

[lemonxx]

Here is an image of a tire accelerating. The vehicle goes towards the left with the wheel rotating counterclockwise. Notice how the rubber is "pushed" in front and "pulled" in the back:

View attachment 353277​

It is more difficult to find a wheel braking where the tire deformation is so obvious but in the following image, you can see the deceleration induced by a pothole. The vehicle is still going towards the left with its wheel rotating counterclockwise. But now the rubber is "pulled" in front and it is "pushed" in the back:

View attachment 353279​

Here are the videos where those images were taken from:

​

So, the same direction of rotation but opposite twisting torques are applied.

i finally got it. thank you so much!