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B Why does the car skid when brakes are applied too hard?

  1. Oct 12, 2017 #1
    When the car is accelerating lets assume the force on it is ma. In order to stop this force we have to apply the brakes which apply a force in the opposite direction. But this force should not exceed ma. It is because the additional force will act on the tyres and create an additional torque. This will make the car skid. Am I correct?
     
  2. jcsd
  3. Oct 12, 2017 #2

    davenn

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    the car skids because the wheels/tyres lock up and then loose traction with the road surface

    this is why anti-skid breaking was invented to overcome that problem

    a comical, but sorta realistic, example of how it works

    upload_2017-10-12_20-6-35.png



    you can google -- Anti-lock Breaking System -- for more info


    Dave
     
  4. Oct 12, 2017 #3

    A.T.

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    The cannot stop immediately, so you need slippage somewhere: at the brakes or at the tire. Which of them will slip depends on the ratios of friction and radii.
     
  5. Oct 12, 2017 #4

    sophiecentaur

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    You have chosen a particularly confusing scenario here, I think. You have your foot down hard on the accelerator and the wheels would spin but you apply the brakes to stop it happening. What is happening is that you are applying a force (a torque, actually) with the (internal) brakes in the opposite direction which results in a lower torque being applied the wheels - so less force between tyre and road. Forces and torques add up vectorially so one torque can subtract from another.
    The limiting friction for the tyres is lower than the limiting friction for the brakes in most modern cars. This was not the case with older cars with drum brakes all round. You could only lock the wheels when the road surface was particularly bad and, however hard you stamped on the brakes in an old Morris Minor (for example) you inexorably drifted towards a collision with no wheel locking if you had been travelling too fast. Needless to say, wheel spin was only possible on ice or loose gravel.
     
  6. Oct 12, 2017 #5

    CWatters

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    As sophiecentaur said, that's an unusual set up. Normally you aren't accelerating and braking at the same time. Aside: Someone is bound to point out that going around a corner is also an acceleration but it's bad practice to brake on bends.

    No. Braking and accelerating (in a straight line) create forces acting in opposite directions. So skidding is NOT a result of these two forces exceeding some limit or being negative or anything like that.

    Car wheels skid/slip when the net force between the car and the road is greater than the maximum value of static friction (=μN) between tyre and road. So if we assume the car is travelling in a straight line then...

    If the car is accelerating at "a" then the wheels will spin if ma > μN.
    If the car is decelerating at "d" then the wheels will skid if md > μN.

    If the car is turning as well as accelerating/braking then the net acceleration acting on the car is the vector sum of it's tangential acceleration (eg a or d) and the centripetal acceleration (mv2/r). In other words you can use the available friction force to brake and/or steer but only up to the limit. So driving around a corner the maximum braking force available is reduced.
     
  7. Oct 13, 2017 #6


    but my point is that if the brakes are applied to reduce the net force on the car then how will it suddenly exceed the force applied by static friction if earlier the case want so ??
     
  8. Oct 13, 2017 #7

    A.T.

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    When brakes are applied at constant speed speed, the magnitude of the net force increases.
     
  9. Oct 13, 2017 #8
    how does that happen
     
  10. Oct 13, 2017 #9

    sophiecentaur

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    What does Newton's First Law tell you?
     
  11. Oct 13, 2017 #10

    CWatters

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    The brakes are applied to increase the force on the car. The harder you press the more braking force there is up to the limit due to static friction.
     
  12. Oct 13, 2017 #11

    CWatters

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    The brakes try and make the wheel skid, increasing the friction force between wheel and road.
     
  13. Oct 13, 2017 #12

    A.T.

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    How would the car brake, without a net force on it?
     
  14. Oct 13, 2017 #13
    so yea i got why the net force increases in magnitude due to friction.... so the wheels skid... it might sound stupid but i wanna know then why doesn't the car skid when we accelerate too much? why does it skid only in the presence of excess force due to friction??? I m really sorry if i asked a stupid question
     
  15. Oct 13, 2017 #14

    A.T.

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    It does.
     
  16. Oct 13, 2017 #15

    I like Serena

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    Indeed it does. It's just that most cars can't accelerate fast enough to reach that point (on a dry road), but they can brake hard enough.
    On snow or ice it's easy though. Accelerating will make the tires skid pretty quick.
    (If you want to try it out, please do so on a large empty parking lot. ;))
     
    Last edited: Oct 13, 2017
  17. Oct 13, 2017 #16

    FactChecker

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    What happened when the car accelerated, and the associated Fa=MAa, does not have any effect on the question of whether it will skid when brakes are applied. They are separate problems. In addition, the way to stop the acceleration force is to take your foot off the gas, not to put on the brakes. Suppose you take your foot off the gas. Then Fa becomes 0. Now apply the brakes. The brakes give a different force, Fb=MAb. If Fb exceeds the tire capabilities, the tires will skid.
     
  18. Oct 13, 2017 #17

    CWatters

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    It's inertia that provides the force, it's friction that is employed to counter it.

    As others have said. Cars do skid when accelerating...

    If they try and accelerate too fast in a straight line the wheels spin/slip. Many road cars can do it in the wet. My 10 year old people carrier can do it on a dry road.

    Remember that velocity is a vector so changing direction (turning or going around a corner) is also an acceleration. This means friction limits how fast you can go around a corner. If you go too fast the centripetal force required (mv2/r) can exceed the maximum friction force (μN) and the wheels start to skid sideways causing the radius of the turn to be greater than the radius of the bend in the road (eg you skid off the road).
     
  19. Oct 13, 2017 #18

    russ_watters

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    Sorry, but this is kind of a mess:
    No. In order to stop the acceleration, you just lift your foot off the gas pedal, which stops the engine from producing a driving torque. This scenario has nothing to do with stopping the car; it only results in the car coasting.
    Since the scenario you presented has nothing to do with stopping a car, this doesn't either: f=ma, sure, but it isn't the engine torque you are opposing when stopping the car, it is the inertia of the car. So you can apply whatever force you want to achieve whatever deceleration you want (or vice versa) up to the point where the friction on the tires can no longer apply that force.
     
  20. Oct 13, 2017 #19

    russ_watters

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    You seem to be missing Newton's first law: when the car is in motion at constant speed, the net force is zero. So that isn't why the brakes are applied. The brakes are applied to create a net force on the car.
    Because they are totally unrelated.
     
  21. Oct 13, 2017 #20

    I like Serena

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    I'm sorry, but this doesn't look quite correct to me.
    Inertia doesn't provide a force. Instead it resists a force, which is how we measure mass.
    Brakes provide friction from the car to the wheels and onto the road, which in turn generates a force of friction (equal and opposite) from the road to the wheels and onto the car. Inertia resists that friction, which leads to skidding if the force is greater than the maximum static friction.
     
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