Where does static friction come from in ABS?

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SUMMARY

The discussion centers on the role of static friction in the context of an Anti-lock Braking System (ABS) during vehicle braking. It is established that static friction acts as a force of constraint, opposing any applied force until it reaches its limit, at which point slipping occurs. The interaction between the braking torque applied by the brake pads and the counteracting torque from static friction is crucial for maintaining wheel rotation without slipping. This dynamic is essential for effective braking, as static friction provides the necessary force to decelerate the vehicle while preventing wheel lockup.

PREREQUISITES
  • Understanding of static friction and its role in mechanics
  • Basic knowledge of vehicle dynamics and braking systems
  • Familiarity with torque and rotational motion concepts
  • Knowledge of the function and operation of ABS in vehicles
NEXT STEPS
  • Research the mechanics of static friction in detail
  • Study the principles of vehicle dynamics during braking
  • Learn about the operation and components of Anti-lock Braking Systems (ABS)
  • Explore the effects of tire-road interaction on vehicle control
USEFUL FOR

Automotive engineers, physics students, and professionals involved in vehicle safety and braking system design will benefit from this discussion.

  • #61
yosimba2000 said:
O
How could the car ever stop in a no-loss situation?
With no brakes and no other losses, the car will never stop, per Newton's First Law.
Ok, so a perfectly slippery road and the car was in motion beforehand. Makes sense so far then.
The road should not be thought of as perfectly slippery, otherwise we can't run your scenarios on it. The losses we're referring to include things like wind resistance, rolling resistance and drivetrain friction.
 
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  • #62
russ_watters said:
The road should not be thought of as perfectly slippery, otherwise we can't run your scenarios on it. The losses we're referring to include things like wind resistance, rolling resistance and drivetrain friction.

Ok
 
  • #63
yosimba2000 said:
It all makes sense except for how the car translates without tangential force.

In this thread there have been several very similar exchanges that lead me to suspect you are confused about something very fundamental: Newton’s first law of motion. A body will continue in its present state of uniform motion until acted on by an external force. Once the car is moving at uniform speed you don’t need a force to keep it moving. It will keep moving all by itself. Once the wheel is spinning you don’t need a torque to keep it spinning. It will keep spinning all by itself.

This is counter intuitive to many people (including Aristotle) because we know in the real world nothing just keeps moving. However that is because we can’t easily eliminate all the outside forces. However we can do a pretty good job of minimizing them and in that case things will keep moving for a long time. My car will take 10 seconds to slow from 65 to 60 mph when coasting and that is mostly wind resistance. The bearing friction and rolling resistance are impressively small. The voyager spacecraft is at a point where even the gravity of the sun is weak, and it is going to keep on flying away forever without needing any additional force.

Force is not required for motion. Force is required to change motion.
 
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  • #64
yosimba2000 said:
Ok, so a perfectly slippery road and the car was in motion beforehand. Makes sense so far then.
No! The road does not need to be slippery! That’s what the wheels are for, the axle bearings need to be slippery.
 
  • #65
yosimba2000 said:
How could the car ever stop in a no-loss situation?
With regenerative braking you store the energy internally, instead of dissipating and loosing it.
 
  • #66
At the risk of getting things even more confused, try this.
The following is a bit simplified but hopefully easy enough to follow.
  • Car weighs 4000 pounds, so the load on each tire is 1000 pounds. Static Coefficient Of Friction (COF) between tire and road is 0.6.
  • Car coasting left, tires turning CCW matching car speed. Result is no sliding between tires and road.
  • Brakes are partially applied, meaning the brake pads/shoes are sliding on the brake rotor/drum, just for discussion let's say the brakes are applied sufficient to supply 500 pounds of retarding force at the outer diameter of the tire.
  • The car is still moving left taking the tires along with it. Due to the retarding force of the brakes, there is now 500 a pound sliding force on each tire due to the inertia of the moving car. Since the tire patch is not sliding on the road, there can be up to 1000 x 0.6 = 600 pounds of sliding force before the tire starts to slide. Result is the car slows.
The energy taken from the moving car to slow it is converted to heat at the brake pads/shoes.

Now let's investigate what happens if the brakes are applied more forcefully.
  • Same vehicle as above but we take into account the Sliding Coefficient Of Friction (COF) between tire and road as 0.4. Static COF remains 0.6
  • This time the brakes are applied sufficient to supply 700 pounds of retarding force at the outer diameter of the tire.
  • The car is still moving left taking the tires along with it. Due to the retarding force of the brakes, there is 700 a pound sliding force on each tire due to the inertia of the moving car. The 700 pound sliding force experienced by the tire patch now exceeds the 600 pounds of static friction between tire and road, so the tire stops rotating and starts sliding.
  • Note that due to the Sliding COF, being only 0.4, the tires will continue sliding until the retarding force of the brakes is decreased to less than 400 pounds at the outer diameter of the tire.

The above also explains why it is better to avoid locking the brakes in a panic stop. Once the tires start sliding, they supply less retarding force than if they kept rolling. That's something that is hard to remember in a panic situation!

Hope this helps.

Cheers,
Tom

p.s. ABS works by sensing the rotational speed of each tire during braking. If a tire slows appreciatively in relation to the others, the ABS reduces the hydraulic pressure to the brake on that tire until it stops sliding.
 
  • #67
Is it not about time one of you posted a diagram showing the relevant forces?
 
  • #68
AZFIREBALL said:
Is it not about time one of you posted a diagram showing the relevant forces?
Yes! The OP should.
 

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