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I Where does static friction come from in ABS?

  1. Aug 26, 2018 #1
    Ok, so assume we have a wheel on the road travelling to the left. The wheel spins CCW and also has translation to the left.

    If we let the wheel roll, assuming no slip, then the translation velocity to the left will counteract the rotational velocity to the right at the bottom of the wheel, so the section of the wheel touching the ground is not displaced in that moment.

    So this means that the little part of the wheel is sitting on top of that piece of ground for a little moment... and then static friction? I'm confused here because isn't static friction a reaction to an opposing force? If that piece of wheel is just sitting there, like I'm sitting in a chair, what "pushing force" is that static friction trying to oppose?
     
  2. jcsd
  3. Aug 26, 2018 #2
    I think there are a few layers to your question, so let me answer in pieces.

    First, static friction is a force of constraint. It is equal and opposite to whatever force is applied until the force is too great and the surface slips.

    Second, yes, up until the force grows large enough that the contact patch slips you are dealing with static friction.

    Third, yes as you describe it there could be moments when there is no force trying to make the contact patch slip, so the opposing force of static friction is also 0. However that is not the usual condition. There is always something trying to change the motion. If the car is coasting, air resistance is trying to slow the car, friction in the axles and bearings, too. There are forces in turning a corner or stepping on the accelerator and static friction keeps the tire “stuck” to the road. In turn that frictional reaction force is what accelerates, decelerated, and turns the car.

    Since your post is titled with ABS I presume you are most interested in braking. The brakes try to stop the wheel from turning, but as long as the braking torque doesn’t exceed the maximum torque available from static friction the wheel keeps turning. Your post seems to suggest that as long as the tires don’t slip there is no friction force, but that is wrong. Static friction provides a rearward force that prevents the wheels from stopping under braking, and that rearward force is what stops the car. The force can be as large as the static friction allows but above that limit the tire starts slipping. (and the ABS detects the slipping and eases off the brakes)
     
  4. Aug 26, 2018 #3

    tnich

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    The title of your post mentions ABS. Torque applied by the brakes results in a static frictional reaction force on the tires. The ABS reduces the braking force when necessary to prevent the tires from slipping.
     
    Last edited: Aug 26, 2018
  5. Aug 26, 2018 #4
    Can you go step by step how the static friction "exists" if the contact patch is not accelerating to the left, and how the static friction prevents the wheels from stopping? If anything, it seems like the static friction helps to oppose leftwards translational acceleration, but doesn't do anything to impede a car otherwise.
     
  6. Aug 27, 2018 #5
    At time 0 the car and the wheel are traveling to the left at v. The wheel is rotating counter-clockwise at w = v/(2 pi r). Through friction on the brake pads the brakes resist the motion of the wheel applying a clockwise torque T. To keep the wheel rolling without slipping the friction with the ground must apply an equal and opposite counter-clockwise torque -T = F r implying that through static friction the ground is pushing the contact patch to the right with a force F = -T/r. That force is what stops the car.
     
  7. Aug 27, 2018 #6
    It's becoming clearer, but now my question is what is enforcing the no-slip condition, and how does this stop the car? If the brake pads create CW torque, and the ground creates CCW torque of same magnitude, which is required to maintain no-slip, then essentially no torque is acting on the wheels so the wheels don't slow down. The ground torque works to turn the wheel, not stop the car, unless the wheels can no longer move independently of the car. But then the question still remains... what says the ground has to maintain no-slip?
     
  8. Aug 27, 2018 #7

    A.T.

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    No, in general it opposes (prevents) slippage. Only in the most trivial cases you can identify a single force that is balanced by the static friction. But in general it's the sum of all forces that determines how much static friction is required to prevent slippage.

    It's not the same magnitude, because the wheel has some rotational inertia too.
     
    Last edited: Aug 27, 2018
  9. Aug 27, 2018 #8
    Yes, that is the instantaneous condition I’ve described, no torque on the tire, so it doesn’t change speed, but that is only approximately correct. To achieve that condition I have a net force on the car. The car slows down, and the next instant everything is a little slower. Naturally the instantaneous condition shouldn’t be exactly no torque, but rather there will be a little residual torque so that the wheel slows down at the same rate the car decelerates. It’s only a little more difficult to solve correctly, but I didn’t want to confuse the issue.
     
  10. Aug 27, 2018 #9
    Static friction is a force of constraint. It is equal and opposite to whatever tangential force is applied. This is no different than rods or ropes or walls in statics problems. You lean on a wall. How hard does the wall push on you? Well, it CAN’T move, so it pushes back on you exactly however hard you push on it. The only difference with static friction is that there is a limit. It won’t allow movement and will match any tangential force until that force exceeds a limit and the contact slips.

    Put your hand on a table with a decent normal force. Now apply some tangential force to try and make it slip. You can put a little force or a lot of force but in all cases your hand doesn’t slip. It isn’t moving so the sum of the forces must be zero. Just as the table is providing whatever equal and opposite normal force is necessary to keep your hand from going through the table, static friction must be providing whatever equal and opposite force is necessary to keep your hand from slipping. However, push hard enough and eventually your hand will slip.
     
  11. Aug 27, 2018 #10

    A.T.

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    The normal force on a wall has a limit too.
     
  12. Aug 27, 2018 #11
    True! And I have leaned on walls where the limit was, unfortunately, less than the applied force.
     
  13. Aug 27, 2018 #12

    russ_watters

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    You aren't trying to stop the wheels, you are trying to stop the car. So to within an arbitrarily low limit, yes the net torque on the wheels can be zero.
    It's probably helpful to think of the rolling case as being not fundamentally different from the sliding case.
    It doesn't have to, it just stops the car faster that way.
     
  14. Aug 27, 2018 #13

    Mister T

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    If you try to slide to the left, there will be a static friction force pushing you towards the right. You push to the left on the chair, the chair pushes to the right on you. That's an action-reaction pair of forces. Action-reaction pairs of forces don't oppose each other because they act on different objects.

    Likewise, the tire pushes on the road and the road pushes on the tire.
     
  15. Aug 27, 2018 #14
    But the tire isn't pushing on the road because the that contact patch has no velocity from the road's perspective, right?

    So let me try to compare the contact patch that is not moving relative to the ground, and a block sitting on the road. I push the block with a force lower than the maximum static friction, and the block doesn't move because it feels a force of the same magnitude pushing back on it. But ONLY until I push it does the block feel the counteracting static friction.

    Is this not the same with the contact patch? The contact patch should feel no static friction force against it until a translational force is acting on the contact patch, right? So I don't see how static friction slows the car down when it should only exhibit itself when translational force is introduced.
     
  16. Aug 27, 2018 #15

    russ_watters

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    Did you really mean to say that? When leaning against a wall, your velocity is zero, but you are applying forces, right?
    Yes... but I don't see how this tells you anything useful about a car. Again: you are trying to stop the car, not the wheel. Whether the contact patch is stopped (rolling wheel) or moving (locked up wheel) doesn't tell you much about stopping the car (only different in magnitude).

    You have stated repeatedly that the torques on the wheel are (roughly) balanced, so it doesn't slow down. That's correct! So can we talk about the car now? That's what you really want to know about!

    Try this: a car with locked tires is like a sliding block, right? Why does it stop?
    The static friction force is translational, so this statement seems circular and empty. I'm not sure what you are trying to say.
     
  17. Aug 27, 2018 #16
    Sorry, I meant to say the contact patch doesn't seem to be applying any forces against the road. It has no velocity relative to the road so it's not sliding, so there's no kinetic friction against it... And there's no static friction against it since there's no force leftwards.

    A car with locked tires slides, and the kinetic friction pushes in the opposite direction to slow it down until it stops.

    I'm trying to understand HOW the static friction comes into play. ABS keeps the tires from slipping, so it doesn't go into the kinetic friction regime. But to manage it, it wants to keep the contact patch with no velocity wrt road. So now it's in the static friction regime, which would show if you were to apply a translation force on the contact patch to the left.
     
  18. Aug 27, 2018 #17
    No relative motion does NOT mean no force! If you push on a wall the wall pushes back. Neither you nor the wall moves indicating the NET force on each of you is zero, but you are each exerting a force on each other. Somewhere there is a balancing force.

    In the case of the tire the road exerts a torque on the tire, but the brake pad applied an opposing torque. The road applies a linear force on the tire. The momentum of the car applie an opposing linear force through the axle. The net effect is that the frictional force from the road is translated to the car and opposes the cars momentum so the car slows down by F=ma

    Let me give another example. You are on ice skates facing a wall. You use a broom stick to push off of the wall. You push on the end of the broom stick with force F. The broomstick pushes on the wall with force F. The wall pushes back on the broom stick with force F. The broom stick pushes on you with force F. You accelerate away from the wall. Here there is no NET force on the broom stick because it gets pushed from both ends but there are definitely forces on the broom stick (and reaction forces from the broom stick). It does not move, but forces are being transmitted through the broom stick. You effectively push off the wall via the stationary broom stick.

    The tire is like the broom stick. The car pushes off the road via the quasi stationary wheel. The wheel and tire have opposed and (nearly) cancelling forces on them, but those forces are real and they do stop the car.
     
  19. Aug 27, 2018 #18

    russ_watters

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    How does that follow from this:
    You seem to be confusing summing to zero with being zero. 1-1=0, but neither 1 nor 1 are zero.
    Great! So what makes you think anything has changed when not locked? Hint: how many torques are there and by what type of friction do they arise? What has really changed?
    Yes, with non locked up braking there is a static friction force applied to the wheel by the road, which stops the car. Or if preferred....
     
  20. Aug 27, 2018 #19
    Sorry, is the bolded part a typo?
     
  21. Aug 27, 2018 #20

    russ_watters

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    It's not. Again: The forces on the wheel are equal and opposite. But if you want to know how the car stops, not the wheel, you need to start describing the forces acting on the car!
     
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