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So according to my textbook applying the brakes hardly to essentially lock up the wheels causes skidding which is kinetic friction so the breaking distance is longer. However, if the brakes were "pumped" the card would not skid and it will be considered static friction so the braking distance is shorter. I UNDERSTAND that the coefficient of static friction has a greater magnitude than kinetic friction, so it makes sense as to why static has a greater deceleration and shorter brake distance. HOWEVER, I do not understand why skidding is considered kinetic friction and not static friction; and why "pumping" the brakes causes static friction. Can someone please explain to how they are so?

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2 Answers 2

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Kinetic friction is all about trying to stop one surface from skidding against another surface. When you have two things such as the wheel and the ground sliding against each other, this is kinetic friction. However, when the wheels are rotating, there is static friction between the ground and the wheel. This is because the wheel is rolling and not sliding against the ground. A point on the wheel only contacts the ground for a very very small instant per revolution, so this is static because there isn't sliding between the ground and the wheel.

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  • $\begingroup$ Oh, so would it be the same as in the case of like a block sliding down a ramp so the friction is considered kinetic? $\endgroup$
    – Prandals
    Jan 19, 2017 at 3:30
  • $\begingroup$ @Prandals yeah, so when you have a block down a ramp, the question you ask is, "is the object's point of contact moving relative to the surface?" The answer when sliding is yes. This is kinetic friction. If it is not moving relative to the ground, then it's static. $\endgroup$
    – rb612
    Jan 19, 2017 at 3:31
  • $\begingroup$ But I must note something confusing: when the rolling condition is met (that is, that $v=\omega*r$ there is no static friction acting. $\endgroup$
    – rb612
    Jan 19, 2017 at 3:37
  • $\begingroup$ one more thing; would pumping the brakes essentially be decelerating the angular velocity meaning that the wheel is still in pure roll so the static friction between the ground and the tire holds true? or am I misinterpreting this "pumping of the brakes"concept $\endgroup$
    – Prandals
    Jan 19, 2017 at 3:39
  • $\begingroup$ @Prandals this is what I understand. Static friction has a goal of meeting the rolling condition, so if angular velocity does not have the relationship anymore with its linear velocity, it isn't rolling and static friction tries to fix this by both 1) slowing down the object's translational motion and 2) applying a torque which causes an angular acceleration $\endgroup$
    – rb612
    Jan 19, 2017 at 3:43
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To add to rb612's answer: the important difference between static and kinetic friction is the difference in speed between the two surfaces. When a car isn't skidding the velocity of the tire is exactly zero at the contact point. The velocity at the contact point can be calculated by $\text{speed of car - rotation rate wheels}\times \text{radius of wheels}=V-\omega*r$.

Compare it to pushing against a static block. You can increase the force without moving the block until you reach maximum force $F_{max}$, at which point the block starts moving and experiences kinetic friction.

When you are braking you are exerting a torque on the wheel which will in turn exert a force on the contact surface. If this force is smaller than $F_{max}$ the velocity of the contact point will remain zero and the wheels will keep spinning. If the force exceeds $F_{max}$ the contact point will start sliding relative to the ground. The speed difference is now $V$ because the rotation rate has gone to zero. Since the speed difference is bigger than zero the wheels, and thus the car, will experience kinetic friction.

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