Friction and rolling resistance, and work done queries

  1. 1) when a wheel turns there is a forward acting friction but the 'frictional' force that opposes it is rolling resistance right? So when a wheel successfully turns and move does it mean the friction is greater than the rolling resistance?
    Then in a car the resisting force will be this rolling friction only? Because at the back wheels where there is no forfeit turning it, then which force acts on the back wheel to turn it, if so then won't the force be moving in the same direction as the wheel's motion and the rolling resistance? So how will the free body diagram of the back wheel look like?

    2) I learned that work done to bring an object up to eg 1m is the same for different steepness because the source will increase despite a smaller incline distance. But if I do that, when I minus the work done against gravity, I will get the same work done against friction (taking that there is no net work done). But if I take that work dine against friction and divide it by the displace, then the fictional force will vary from the different angle of inclines. So how can this be so?

    3) I'm not sure if I can use the formula net Work done=final KE-initial KE for all scenarios like in an incline situation. Can I do that?

    Thanks for the help everyone!
  2. jcsd
  3. tiny-tim

    tiny-tim 26,055
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    hi sgstudent! :smile:
    the engine exerts a torque on the back axle, which rotates the back wheels

    the static friction from the road makes the car move forward

    opposing this are the friction in the bearings, and the rolling resistance, which isn't friction but is basically the energy lost in the continual deforming of the tyre
    the free body diagram of the back wheel has a torque at the axle, and a forward friction force from the road, combining to give a forward acceleration

    (we don't normally put the rolling resistance or the friction in the bearings onto the free body diagram … the assumption is that they are already subtracted from the engine torque to give a net torque, which does go on the diagram)
    the work done against gravity is the same,

    thw work done against friction is different
    the work energy equation always works :wink:
  4. Hi tiny tam :) thanks for the help! But I thought the back wheel isn't powered by the engine so what does it mean it IDE turned by the torque of the wheel? So does it mean if I have a tire that has a higher friction will have a higher rolling resistance? Does rolling resistance improve stability of the car during wet weather like when I break the force can stop the car better? Thanks!
  5. tiny-tim

    tiny-tim 26,055
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    oh, if the engine drives the front wheels, then the friction from the road on the back wheels is opposite to the acceleration (and to the friction on the front wheels), see direction of friction :smile:
    rolling resistance has almost nothing to do with friction, see :wink:
  6. Oh okay! I read the rolling resistance already but I'm unsure what's the purpose of having high friction tires. I'm also not very sure about why the frictional force on the non-brake wheels is opposite to the frictional force. Because the wheel doesn't have any of its own force to turn. So I'm confused about why the fictional force will be in that direction. Thanks for the help!
    Last edited: Mar 30, 2012
  7. um.. so could anyone give me a clearer explanation on the non-driving wheel's frictional force? I don't really get why the force can produce the turning motion of the wheel. Also, if the wheel turns faster does it mean that the friction will also be faster? So, if i double the speed of a car will my friction be doubled too? Thanks for the help!
  8. The second question, static friction just stops something from moving up to a point. Kinetic friction, if I remember correctly, which is what we're talking about, is dependent on normal force and nothing else. So, if you push on the axle twice as hard with something, friction will be twice as strong. It's constant otherwise. To be very exact, a flux integral would be ideal for calculating normal force on "odd" surfaces, or it's what I'd use.
  9. tiny-tim

    tiny-tim 26,055
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    it doesn't!

    once the non-driving wheels have started turning (at a particular speed), they need no force to continue at that speed, do they? :smile:
  10. Correct. But, I think, if I'm getting the problem correct, there's torque being applied, whether we like it or not, because we have friction. This is a force/torque, so it will slow down. We need another force/torque to counter it.
  11. tiny-tim

    tiny-tim 26,055
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    nothing to do with helping rolling, except that it increases the speed at which skidding (sliding) starts :smile:
    from the pf library …​

    On the wheels of a non-skidding car:

    The wheels are rolling, and so the point of contact of each tyre with the road is instantaneously stationary, and static friction applies.

    There is also the "rolling friction", or rolling resistance, an additional small force, caused by deformation of the tyre, which in exam questions can usually be ignored.

    to find the direction of static friction, always ask "what would happen if there was no friction?" (for example, if one surface was ice) …

    If a brake is being applied to a particular pair of tyres (on the same axle), imagine that the road under those tyres suddenly becomes ice (but the road under the non-braked tyres remains normal): the braked tyres will go slower, but the car will stay the same speed (because there are no external braking forces on it): looking down from the window, you see the road going backward faster than the braked tyres are going backwards, so the road will try to drag those tyres backward with it

    In other words: the tendency is for the road to move backward relative to the braked tyres, so the friction on the road is forward, and the friction on the braked tyres is backward

    However, the non-braked tyres (usually the front, steering, tyres) will have a forward friction force from the road at the same time …

    To see this, imagine instead that the non-braked tyres (only) are on ice: then they will go at the same speed, but the car will go slower: looking down from the window, you see the road going backward slower than the non-braked tyres are going backward, so the road will try to drag those tyres forward.

    In other words: the tendency is for the road to move forward relative to the non-braked tyres, so the friction on the road is backward, and the friction on those tyres is forward.

    This of course is why 4-wheel braking slows a car faster than 2-wheel! :wink:

    Similarly, for a car being accelerated by its engine, the friction from the road on the driving tyres is forward, but on the non-driving tyres is backward.
  12. the part i don't quite understand is why the friction from non-driving tyres is backwards.. my teacher said that it was backwards in order to turn the wheel, but i dont understand it if the friction is going to advocate motion in the same direction then isn't it all 'wrong'? Thanks for the help tiny-tam really appreciate the help!
  13. tiny-tim

    tiny-tim 26,055
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    hi sgstudent! :smile:

    (just got up :zzz:)

    (btw, i'm tiny-tim, i was named after a famous character from dickens' "a christmas carol" :wink:)

    but it doesn't "advocate" forward motion of the car, it "advocates" forward rotation of the wheel, which means it has to be a backward force …

    there's nothing else to make it rotate forward!! :smile:

    (the forward rotation of the driving wheels comes from the engine)

    adapting a paragraph from above, if the car is accelerating forward

    To see this, imagine instead that the non-driving tyres (only) are on ice: then they will go at the same speed, but the car will go faster: looking down from the window, you see the road going backward faster than the non-driving tyres are going backward, so the road will try to drag those tyres backward.​
  14. Oh sorry tiny-Tim :) typed it out wrongly haha.
    So the non-driving wheels have an opposite friction to turn the wheels, but they don't have any forward friction? Also, in this link: I don't quite understand what direction the friction will be since its just rolling down. In this case would the wheels friction be like the barked ones or the non brakes ones? And why can't I include them into my work done against/by friction? I read somewhere that its because its static friction so it works only at each individual points so there is no distance like in the work done=force x distance formula. But then again, why would this be static friction and not kinetic friction/sliding friction?
    Thanks tiny-Tim!
  15. tiny-tim

    tiny-tim 26,055
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    your link says …
    The force of friction does not do work upon the cart because it acts upon the wheels of the cart and actually does not serve to displace either the cart nor the wheels.

    The friction force only serves to help the wheels turn as the cart rolls down the hill.

    Friction only does work upon a skidding wheel. ​

    … i've read something like this many times, but it's wrong! :frown:

    friction (from the road) does do work on a rolling wheel …

    in the obvious case of an accelerating car on a horizontal road, the only external horizontal force is the friction (from the road) …

    if it's not doing the work, what is??!! :smile:

    work done = force "dot" displacement of the point of application of the force, the point of application is the point of contact with the road, and that's moving!

    suppose the car has mass M, and the wheels have total mass m radius r and total moment of inertia I …

    if a horizontal force P pushes the car (on a horizontal road), and if the total friction force is F, and if the reaction force between the car and the wheels is R, then …

    P - F = (M+m)a

    Fr = Ia/r

    so P - (I/r2)a = (M+m)a,

    or P = (M+m+mr)a, where mr is the "rolling mass" I/r2

    in other words: either we must take F (the friction) into account, or we must add the "rolling mass" to the actual mass of the wheels and the car :wink:
  16. Oh but what direction will the friction be for this case? And if it is forward then will it be work done by friction? Thanks for the help tiny Tim!
  17. tiny-tim

    tiny-tim 26,055
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    i'm confused … which case? :confused:

    (and you know the drill here … you tell us what you think, and why, first :wink:)
  18. Like when a cart with wheels is pushed. This is what I came up with (assumptions): if the wheel is powered by literally turning it like in a generator then the friction is opposite to the turning direction. But if I just push the cart (not turning the wheel directly) or like in the non-driving wheel then the direction of friction will be opposite to the direction of total motion of the body. Is this right? Cos I'm taking O levels so I don't really understand all of these very well. We only covered more obvious stuff like kinetic friction in school. Thanks for the help :-)
  19. tiny-tim

    tiny-tim 26,055
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    it's simply torque = moment of inertia times angular acceleration …

    the torque has to be in the same direction as the angular acceleration

    if you push the cart, then looking from the right-hand side, the wheels will be accelerating clockwise,

    and the only external forces on the wheel are the weight (irrelevant), some friction from the axle (negligible), and the friction from the road

    so the friction from the road has to be clockwise, which means it's backwards :smile:
  20. oh okay! So the assumptions i took were right? Thanks for the help tiny-tim, thanks for the patience with me :smile:

    But then again, when the driving wheel turns looking at it from the right hand side, it turns clockwise, but my force is the opposite direction... thanks for the help tiny-tim! btw, i also have a dynamics question to ask hope you can help out there too! Thanks :wink:
    Last edited: Apr 1, 2012
  21. Hi tiny Tim I don't quite understand the net work done=final KE-initial KE. There's another formula which is net work done=change in energy so even when an object moves up an incline with constant speed, the two formulas collide. Thanks for the help!
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