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From most graph, when the tire doesn't slip, then the friction is zero, for example see the below image

Why there is no friction when there is no slip? As the car stand still on slope, there is no slip but still there is friction holding the car against gravity

wheel slip

user2174870
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  • If you search tyre slip, every graph is similar to this, i.e very low friction at low slip ratio – user2174870 Mar 20 '14 at 14:16
  • @JohnRennie The work done by the frictional force from the road must even be non-zero, otherwise the car would be losing kinetic energy to friction in the air. – Brian Moths Mar 21 '14 at 02:11
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    possible duplicate of Does a tire need to slip to generate force? – Rick Jul 13 '15 at 16:39
  • The confusion comes from how slip% is defined. It's a ratio and when a car is parked the numerator and denominator are both zero and slip% is undefined. – Rick Jul 13 '15 at 17:06
  • Can you please provide a link to your source? – WhatRoughBeast Jul 13 '15 at 18:57
  • @WhatRoughBeast If that was to me, I didn't have a particular soruce at the time (just been studying tire and vehicle dynamics for about 8 years) but here's a nice one that has similar graphs and good definitions: http://www-cdr.stanford.edu/dynamic/WheelSlip/SlmillerGerdesACC.pdf – Rick Jul 14 '15 at 12:00
  • @Rick - Since you cannot provide a source, I presume you generated the curve yourself, and you got the vertical axis wrong. Please pay close attention to the link you sent me. The vertical axis is force, not friction, and the wet/dry curves represent behavior for two different values of friction (roads get slippery when wet). So your question (based on the figure) should be "why is there no slip when there is no force?", and that one is dead easy. Try Googling "force slip curve" and compare the results with "friction slip curve". – WhatRoughBeast Jul 14 '15 at 14:28
  • @WhatRoughBeast Sorry, I think your source question was meant for the OP, but I responded unsure who you were asking. I assumed the "$Friction [\mu]$" axis was the total friction force divided by the normal load, which makes it proportional to the friction force so the distinction is mostly irrelevant. I did just google both, but they mostly turned up the same results and the graphs that were unique to each search still all looked pretty much the same. What differences was the OP or I supposed to notice? – Rick Jul 14 '15 at 20:27
  • @WhatRoughBeast Also, the OP specifically mentions the car parked on a hill scenario where the force is not zero but the absolute slip is zero. That question is still confusing regardless of whether it's a "friction slip curve" or a "force slip curve" the question is only resolved when you realize it's a "friction/force vs slip ratio curve" – Rick Jul 14 '15 at 20:31
  • @Rick - Sorry, I did mean to respond to the OP. My apologies. When you realize that each curve represents behavior for a different friction, his question becomes obvious: in the parked condition, there is no wheel motion and therefore no slip (until the grade gets steep enough that the car falls off the hill). – WhatRoughBeast Jul 14 '15 at 23:55
  • @WhatRoughBeast what do you mean my "a different friction"? – Rick Jul 15 '15 at 12:21
  • @Rick - Sorry, a condition of different coefficient of friction. The upper curve is "Dry", and indicates a dry interface between road and tire. The lower curve is "Wet" and indicates a wet condition. Effective tire friction is lower under wet conditions than when dry. This is why the dry curve requires more force (the vertical axis) for a given slip angle, than the wet curve. Again, please find the source of your curves. The vertical axis is force, not friction. Differences in friction are implicit in the dry/wet labels of the two curves. – WhatRoughBeast Jul 15 '15 at 13:41
  • Let us continue this discussion in chat. – Rick Jul 15 '15 at 14:02
  • @rick - this question is the original - the duplicate is the one you linked! it was asked after this question was asked.... – tom Jul 22 '15 at 11:27
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    @tom I'm aware this question was asked first, but I thought the other question was both asked more clearly, and answered more clearly. As such I thought it would be much more useful to mark this question as the duplicate to direct people to the better question/answer. This is the generally accepted practice I believe: http://meta.stackexchange.com/a/10844/298895 – Rick Jul 22 '15 at 14:28
  • @rick thanks for explaining.... didn't realize the ettiquette about that... – tom Jul 22 '15 at 22:28

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For an (idealized) perfectly round wheel on a perfectly smooth road, there is only a single point of contact between the wheel and the road at any given time. If you were to plot the motion of a single point on the wheel's surface as it goes around and then touches the ground, you would see that it follows a curve called a cycloid. The picture in that wikipedia article explains it better than I possibly could.(*) As you can see from the image, the point on the wheel's surface is actually changing directions as it touches the road, so at that point in time its instantaneous velocity is zero. Because it is stationary relative to the road, there is no kinetic friction.

However, there can still be static friction, such as if you're driving the car around a curve. In that case, it's static friction on the wheel that prevents you from slipping and keeps you following the curved path. (Or static friction plus a contribution from gravity if the curve is banked.)

There is also static friction between the wheels and the road that causes the car to accelerate in the first place. (I'm assuming it starts from rest.) If friction between wheels and ground were zero, the wheels would spin in place but the car would never go anywhere.

(*) The picture makes it very clear, but if you prefer a verbal explanation: The wheel as a whole is moving forward (relative to the road), but when the point on the wheel's surface is at the bottom of its rotation, it's moving backward relative to the center of the wheel. The result is that the point on the surface of the wheel is stationary (relative to the road) when it's at the bottom of its rotation.

Tim Goodman
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  • In fact, the all of a realworld tire's contact area with the road has zero velocity :-) – Carl Witthoft Mar 20 '14 at 12:45
  • I still can't get it.Is that a no slip tire can not move a car against external force, e.g wind? – user2174870 Mar 20 '14 at 14:04
  • I'm sorry, I don't understand the comment. "No slip" for a tire means "rotating in the normal way", such that a point on the tire's surface follows the cycloid curve. You can still have static friction in this case, and this is needed to prevent slipping when accelerating (e.g. going around a turn, speeding up, slowing down, etc.) I suppose that would resist wind as well. – Tim Goodman Mar 20 '14 at 14:28
  • Thanks, I can get it, but still not clear. Pls consider this example, a geared pinion rolls above the rack without slip, but it still can provide "kinetic friction" force (ala acceleration). A wheel is a bit different but could surface roughness provide acceleration without slip like rack and pinion? – user2174870 Mar 20 '14 at 15:07
  • Hmm, I don't think a rack and pinion requires kinetic friction either. Kinetic friction arises when one surface is sliding against another. Looking at the animation in this article, for example, I don't see anything I'd describe as sliding: http://en.wikipedia.org/wiki/Rack_and_pinion – Tim Goodman Mar 20 '14 at 15:16
  • @CarlWitthoft Unless the tire is exerting no traction force the tire-patch is always divided into regions of no-slip and slip, so not all of the contact area is zero velocity. – Rick Jul 13 '15 at 16:57
  • @TimGoodman in a rack an pinion system the contact point is not always directly on the line between the centers of rotation and thus the surfaces must slide against one another. The movement is extremely small, however this kinetic friction is where the power loss comes from in gears and why gears need lubrication. – Rick Jul 13 '15 at 17:03
  • It seems like this answer is claiming that slip and slip% are always zero for tires. If not, how does slip effect frictional force? – Rick Jul 13 '15 at 17:28
  • @Rick, I understood the question to be asking about the case where there is no slip. "Why there is no friction when there is no slip?" In that case (for an idealized, wheel, i.e., a circle rolling on a plane) there can be no kinetic friction due to the contact between the wheel and the road. There can still be static friction, as I note. There can also be kinetic friction at the axle, but not at the point of contact between the wheel and road (for idealized tires.) If there is slip, then there would be kinetic friction. – Tim Goodman Jul 13 '15 at 17:57
  • @TimGoodman So you interpreted the vertical axis of the question's graph as Kinetic friction component that doesn't include the static friction component? – Rick Jul 13 '15 at 18:02
  • @Rick Yes, otherwise I don't see how the plot makes sense. It's certainly not the case that the static coefficient of friction is zero when there's zero slippage. As the questioner notes, you can have static friction even when standing completely still on a slope. – Tim Goodman Jul 13 '15 at 18:13
  • @TimGoodman the plot is of total friction force on the y-axis but the x-axis is slip% not just slip, so as a car approaches a stand still slip% approaches 0/0. 0/0 can take on any value depending on the context. I think it's more appropriate for the axis to be flipped as it's really a certain force that will cause a certain slip% but then the graph ceases to be a "function" so most people prefer it the way it is. If it were oriented that direction we'd see that a car parked on a hill would have a non-zero slip% which is entirely accurate if non-intuitive. – Rick Jul 13 '15 at 18:23
  • @TimGoodman for a source http://www-cdr.stanford.edu/dynamic/WheelSlip/SlmillerGerdesACC.pdf – Rick Jul 14 '15 at 11:59
  • @Rick The frictional force described in that PDF (given by eq. 2) is kinetic friction, which goes to zero when the rotation of the wheel keeps up with it's motion (i.e., when the numerator in eq. 1, and thus the slip, is zero). But the OP asked about the case of a car "standing still on a slope". In that case there is static friction. The PDF and the plot aren't wrong by any means, they're just describing a moving vehicle, not one at rest. – Tim Goodman Jul 16 '15 at 19:24
  • @Rick Actually even in the case of a moving vehicle on level ground, if there is no slip then there is static friction between the wheel and the road (because the point of contact has zero instantaneous velocity in the horizontal direction). It's this friction that converts the rotation of the wheels into forward motion. If there were no static friction, then you couldn't have forward motion without slip. But the OP asked about a car on a slope, where there's static friction even at rest to resist gravity. – Tim Goodman Jul 16 '15 at 19:38
  • While I'm pedantically correcting myself, I should say that you could have forward motion even in the zero-friction case if you were to get out and push. :) But if your engine is just spinning the wheels, then it's only because of friction between the wheels and the ground that this gets turned into forward motion. And if there's no slip, then it's static friction. – Tim Goodman Jul 16 '15 at 20:04
  • @TimGoodman the only way to have zero slip%* is to have zero friction. That's the whole premise of this question. If you'd like to know more about why this is the case it has been answered here quite well. If you'd like to look at a simplistic way to model this behavior I explain some math behind it here – Rick Jul 17 '15 at 14:00
  • Usually zero slip means zero slip% but when the vehicle isn't moving then slip% can take on any value. this is why the parked car example can't be used to refute the claim (which is correct) that zero slip% requires zero frictional force. The reason pure static friction can't occur while the tire is in motion is explained in those two links.
    • – Rick Jul 17 '15 at 14:05
  • @Rick, thanks for the links, but I'm not sure where you disagree with me on. I agree with the answer you linked to, which says that you get slip at small forces because the tire is elastic, and also that unless the wheel is sliding across the surface, the car is accelerating from static friction. – Tim Goodman Jul 20 '15 at 18:27
  • I tried to be clear that I was simplifying by ignoring the elasticity of the wheel - which, I agree, is too big a simplification to explain the OP's graph, but not too big to explain the role of static friction, which is what the OP seemed to be missing. Do you disagree that a moving wheel can experience static friction (because the tangential velocity of the wheel at the point of contact with the road is zero)? – Tim Goodman Jul 20 '15 at 18:29
  • @Rick Actually, on a re-read, maybe I get what you're saying. An idealized, perfectly elastic wheel would move due to static friction while experiencing zero slip, whereas a real-world elastic wheel would also move due to static friction (at low enough speeds that it wasn't sliding), but would have a non-zero slip at any speed. So for physical wheels the graph makes sense reading it as the full frictional force, not just a kinetic friction. My objection that S (eq. 1 in your PDF) could be zero and you still are moving won't apply if you consider the wheel to be elastic. I buy it. – Tim Goodman Jul 20 '15 at 18:38
  • @Rick So while what I said was true for a perfectly rigid wheel (and I do think the OP was perhaps confused about static friction since they mention a car standing still), the graph describes a wheel world elastic tire. Thank you for your patient clarification on this point. – Tim Goodman Jul 20 '15 at 18:40
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    @TimGoodman Glad to have helped clarify. I found your comments delightfully thought provoking. – Rick Jul 20 '15 at 19:49
  • I can't resist pointing out the amusing typo in my last comment. I meant "a real-world elastic tire", not "a wheel world elastic tire". – Tim Goodman Jul 20 '15 at 19:52