# Why is the direction of static friction towards centre in case of circular turning? Here , a car turning on a circular track. I have marked all the kinds of friction , velocity happening in the diagram.

For the front wheels ,

1. VT (Tangential velocity) ,Vr(Radial velocity) , VR(Tangential + radial = Resultant velocity)

2. fs1 & fs2 is the static friction happening on the front wheels i.e in direction opposite to the direction of velocity since we know friction happens in direction opposite to motion of body.

3. Direction of acceleration 4. For back wheels ,

resultant velocity is in front direction & static friction backwards.

Q1 What I think :

Ft=Fs2=0. Therefore , static friction acts in radial direction because static friction at back of wheel = 0 ?

Q2 I think I am wrong because if I find to find total inward force. Then l it will be = centripetal + static friction.

Credits of the picture goes toMr.Kamran. I have just marked the directions.

• So what is the problem? Dec 25, 2021 at 13:28
• @RameezUlHaq I’m not sure I’m correct. No one has ever given me this explanation. I have thought of this on my own. So , want to cross check. Neither the textbooks explain like this. Dec 25, 2021 at 14:04
• @RameezUlHaq What do you think sir ? Dec 25, 2021 at 14:13

When a car moves in a circular trajectory a centrifugal force is ""compelling" the car to maintain its straight line. In order for the car to move in the circular trajectory a centripetal force is required.

Sine the motion is radially outward, (and the friction is opposite to the direction of the motion) the friction that develops is inwards.

UPDATE this is just a short comment about tire dynamics and slip ratio. The wheel does not act like a non deformable body. The front tire in a Ackermann steering, deforms in the section that is in contact with the road. The end results is the slip angle Figure: Slip angle illustration (race dynamics)

Using the concept of slip angle is possible to determine longituninal and transverse velocities for the front wheel.

• K. Is what I have assumed correct ? Also , I do agree with you in terms of explaining. Is it possible if u could explain using equation & derive the desired result. It will be really helpful since that is what I have been trying to do so far. Dec 25, 2021 at 14:09
• @S.M.T it's already said exactly with plain words. If you want to see it as equations check this image (nothing what's not already said) i.stack.imgur.com/LeOJJ.jpg Equation pair 1 gives the coordinates if the rotation happens around x=y=0. Eq2 contains them differentiated once. That's the velocity. Eq3 gives the acceleration component; it's Eq2 differentiated. Acceleration vector has the opposite direction of the placement vector. Any force which deflects the vehicle from straight line has the direction of the acceleration. if the mass stays constant (by Isaac Newton) Dec 25, 2021 at 15:34
• @user287001 Do you think what I have written in my Q correct ? Dec 25, 2021 at 16:32
• @S.M.T four wheel vehicle actually cannot be properly seen as a point of mass. If it runs along a circle with constant velocity without spinning more nor less than what's needed to keep one side straight towards the centerpoint we can decide that the resultant force conveyed by the wheels is towards the center and the total toque of the same forces relative to any point must be zero. How the force distributes between the wheels depends on the design and material properties of the vehicle and how one drives it. There's nothing elementary in the details. Dec 25, 2021 at 18:31
• @S.M.T. Adding to user287001 comment, I need to point out that what is happening to a tire during turning is quite complicated. If you are really interested in this subject you need to read about tyre dynamics and slip ratio. Dec 25, 2021 at 20:08