Note: My question is duplicate of the following

  1. Direction of friction when a car turns

  2. Why does friction cause a car to turn?

I've gone through many related questions especially the first. As I understand the static friction is always opposite to the force applied on the object as shown:
image http://www.school-for-champions.com/science/images/friction-slide_kinetic.gif
But in the case when front wheels of a vehichal are turned the force of static friction is not opposite to the applied force. For example consider a car accelerating forward. The net force on the car is in forward direction which is provided from the rear tyres, if eventually break is pressed static friction(assuming tyres aren't skidding) comes into picture. This friction should be and is opposite in direction to the direction of force applied by the rear tyres. When the front tyres are turned the direction of static friction is changed(radially inward) means the direction of static friction is not opposite to the direction of applied force as shown:
image 2

Question: Is the force of static friction is always opposite to the applied force ? If not then what determines its direction?

  • $\begingroup$ Definitely not opposite to direction of force, it depends on the tendency of direction of motion of point of contact, if you push a tyre from top then the bottom point tends to move backwards and then the friction is also forward and in the same direction as that of applied force. $\endgroup$ Dec 28, 2013 at 14:41
  • $\begingroup$ What about net force ? In my example only 2 forces are applied, one weight of body, which conveniently does not account for direction in this case and the other force applied at the top of a wheel, the tire tends to turn and the bottom point tends to move in opposite direction of applied force, so friction is applied in the direction of force to resist motion of bottom point. $\endgroup$ Dec 28, 2013 at 14:47
  • $\begingroup$ Read your question again, you have asked is friction always opposite to applied force ? I just explained it is not, why all this skidding about ? $\endgroup$ Dec 28, 2013 at 14:54
  • 2
    $\begingroup$ Isn't this broadly the same question as physics.stackexchange.com/q/87976 $\endgroup$ Dec 28, 2013 at 15:04
  • $\begingroup$ @rijulgupta can you cite any reference from where you got this example. If friction is in same direction then work done by friction will be positive and this would be like creating energy from friction which is not possible thermodynamically. >On my comment about skidding: If there's net force at the bottom of the tyre then the bottom will accelerate hence rolling will be converted into skidding(assuming tyre was at rest initially). $\endgroup$
    – user31782
    Dec 29, 2013 at 2:37

2 Answers 2


Static friction always opposes relative motion at the point of contact.

There are two cases possible:

1)It orients itself in direction and magnitude in such a way that the relative acceleration of the contact point is zero.

2)If this is not possible(such as in friction is too small to prevent motion),it tries to minimize the relative acceleration.


enter image description here

Red: Direction of motion of the top of the wheel relative to road

Orange: Direction of motion of the bottom of the wheel relative to road

Green: Direction of force of friction acting on the wheel

If the wheels are still rolling: friction at the ground/wheel interface simply opposes the rolling of the wheel, slowing it down. In the reference frame of the chassis, this friction has both x & y components for any nonzero angle of steer, and also acts on the chassis because the wheel is connected to it by axle (thereby allowing the car to simultaneously advance and turn).

If the wheels are locked due to braking, then it doesn't matter what their orientation is, as a per the extreme example in the second diagram, and the reason why frozen roads are so dangerous.

  • 1
    $\begingroup$ The "friction has both x & y components"... is the y component providing the centripetal force? Also, as there is an x component, will magnitude of velocity increase(this goes against what I learned in uniform circular motion)? $\endgroup$
    – Eliza
    Dec 28, 2013 at 17:55

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