Physics of ABS (anti lock brakes) ? help..?

Find the difference in stopping distance for an average car going at reasonable speed on snow with an anti-lock brake system over the same car going without an ABS ???





you can assume numbers to explain the question, and formulas.|||Weird question...





The main advantage of anti-lock brakes is actually that it allows you to continue steering through the skid, but it still also has more braking power.





The braking power difference stems from the difference between static and kinetic friction coefficients. When two objects are touching, and not moving, their friction is greater, but as soon as you overcome the friction and begin to slide, there is actually less friction then when the object is holding still.





Ever try to push something really heavy, and it took a lot of force to get it moving, but not all that much to keep it moving? Same thing is happening.





When a tire is rolling, there is actually no sliding, or kinetic friction. A portion of the tire and ground make contact, and then that portion of the tire is lifted up and replaced with a new portion (basically, the wheel is rolling, not sliding) so a rolling tire has much more friction.





When you brake and skid, your tires lock and you slide. This uses the kinetic friction coefficient which is lower then the static one. Anti-lock brakes detect a slide and make the wheels start rolling again, so you use the static coefficient.





For numbers, I believe rubber on asphalt has a static coefficient of 0.8, and a kinetic of 0.6 (could be wrong on those, doing them from memory). So anti-lock brakes could stop in 3/4 the distance that non anti-lock brakes could.





Let's make up coefficients of 0.5 and 0.3 for static and kinetic friction between rubber and snow (or look them up, doesn't matter).





Let's assume the ability for the car to brake is denoted by the variable x (a function of weight and speed). And with anti-lock brakes it stops in 100 feet. Then you'd have x/(0.5) = 100





This means x = 50





So without anti-lock brakes, you'd have 50/(0.3) = 166.7 feet.





This is really, REALLY hand-wavy and assuming many things, but again, it's a strange question, not sure how detailed it wants you to be.





If you have a feeling it wants more math then this then, well, hopefully my explanation at least helped with the theory behind it.|||THis is about the difference between static friction and sliding friction.


Assume the car with ABS keeps the tires turning and the car without locks them up and skids.


Since KE = W = FD, D will go as one over F


THe Fs in the same ratio as the coefficients of friction, as g and the car's mass are constant