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Friction round a corner

  1. Nov 16, 2008 #1
    Im trying to model a car going round a corner and the speed at which it will overturn (assuming no slip)
    Is it correct that when turning there are lateral friction forces that oppose the inertial 'centrifugal' force? i.e. towards the centre of curvature
    When the front (driving) wheel are turned at an angle does this force still act perpendicular to the direction of the wheel?
    Also, how much effect would there be from friction resisting the direction of travel?
    and i know that the friction on a 'non driven' wheel opposes the direction of travel but, is it opposite for a 'driving' wheel?

    Any help would be appreciated
  2. jcsd
  3. Nov 17, 2008 #2


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    Welcome to PF!

    Hi paton51! Welcome to PF! :smile:
    Yes. :smile:
    If the car is not changing speed (as distinct from direction), then the centripetal friction force is always perpendicular to the velocity of the car.
    Almost none … that's air resistance, and the internal resistance of the bearings etc inside the car.
    No … friction from the ground (on a 'non driven' wheel) only stops the wheel from slipping … it doesn't impede the forward movement.
    Friction from the ground again stops the wheel from slipping, but also provides forward acceleration. :smile:
  4. Nov 17, 2008 #3
    Paton51, in the real world there is a significant amount of drag induced when the car is cornering. The tire operates at a slip angle, which means that the wheel must be turned more than the path of the tire would seem to indicate. The resultant drag is enough to partially substitute for braking, and needs to be countered with the application of throttle during cornering in order to maintain a set speed.
  5. Nov 17, 2008 #4
    Ok thaks guys, i'll see how i get on now with this info :)
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