# When a car's non-driving wheels are turned, what is the frictional force vector that actually causes the vehicle to turn in that direction? [duplicate]

I am almost a little embarrassed to be asking this question since my education and experience is in mechanical engineering. However, I've drawn a few diagrams, but am still a little puzzled by this question... In my diagrams, the vectors I've drawn don't seem to actually produce a resultant force that would actually push the vehicle one way or the other when a turn is desired....

From everyday experience, the short answer is that the actual forces that cause a vehicle to turn are due to frictional forces. Anyone that has driven a car on reduced traction roads due to slush and ice can attest that those conditions result in a car usually not wanting to turn in the direction the steering wheels are turned. Furthermore it isn't hard to imagine a car already in motion on a totally frictionless surface - turning the front wheels one way or the other would not make the vehicle turn in that direction at all...

For this case, please only consider a car with rear wheel drive and front wheel steering. It seems rather obvious that in vehicle configurations where the steering wheels are also driving wheels (that is, front wheel drive, AWD, or 4WD), then the vehicle turns in the direction the steering wheels are turned because the actual forces the road surface exerts on the wheels are turned relative to the centerline and CG of the vehicle.

In a car with front steering and rear driving wheels, every part of the car forward of the rear driving wheels is being pushed forward by the rear driving wheels. The car wants to go in a straight line.

Imagine that the car is in motion going forward and that the front wheels are turned only very slightly to left or right (a few degrees left or right of directly straight ahead), what force vectors and resultant actually produce the 'kick' that causes the entire front end of the vehicle to turn in the desired direction? It is obvious that a rolling tire wants to continue rolling and that turning the wheel relative to its initial straight-ahead path would produce a scuffing situation - a high friction situation compared to rolling.

In a related situation, we can examine when a car is not in motion, but stopped. When a vehicle is at a complete stop and the front wheels are turned in the direction of an intended turn and obviously not rolling in the forward direction, what are the force vectors that result in the front of the vehicle turning in the direction of intended motion?

In my diagrams, I've isolated and examined the forces on the steering tires' contact patch with the road, but still cannot resolve a frictional force vector that would explain why the front end of the car turns... I've even considered the gyroscopic effect of rotating wheels and thought that it might be due to a force that is exerted on a rotating wheel that has its rotating axis deflected when the wheels are steered - again I wasn't able to find anything and the gyroscopic restoring force would not explain how a stationary turned wheel causes a vehicle to turn anyway...

• Not clear what you are asking. Can you upload one of your force diagrams and ask about that? Commented Aug 8, 2018 at 12:24
• @sammygerbil Thank you for the suggestion, I added in my diagram here to help you better understand my questions... Commented Aug 8, 2018 at 13:40

what force vectors and resultant actually produce the 'kick' that causes the entire front end of the vehicle to turn in the desired direction?

This force is a component of the force of the axle pushing on the wheel, aligned with the plane of the wheel. The friction force acting normal to the plane of the wheel just cancels the second component of the force from the axle, normal to the plane of the wheel.

When a force is acting on an unconstrained body, the body accelerates in the direction of the force.

When the movement of the body is blocked from some direction by an obstacle, a component of the force acting in that direction is canceled by the normal reaction from the obstacle. This reduces the original force to a component tangential to the surface of the obstacle. As a result, the body will accelerate in that (tangential) direction - not along the direction of the original force.

This could be illustrated with a block on an inclined plane. Without the plane on the way, the block, under the force of gravity, would fall straight down. With the plane, as an obstacle, the component of the weight normal to the plane is canceled by the normal reaction of the plane and the remaining component of the weight, parallel to the plane, makes the block slide down along the plane.

If we replace the block by the wheel, the weight by the push from the axle and the normal reaction of the plane by the friction force exerted on the wheel by the road, we can apply the same force diagram and explain why the wheel moves in the direction it is pointing to and why the car turns.

• thank you for providing this very intuitive explanation. It really helped me understand what I was struggling with. Commented Aug 9, 2018 at 12:38
• @PhilosophStein Happy to help.
– V.F.
Commented Aug 9, 2018 at 12:40

One way to think about this is to imagine your car front wheels driving in a rut on a dirt road that causes the car to veer ( similar to a train track). Now imagine that a road surface is made of many smaller ruts (friction) that cause the car to do the same thing when the wheel is misaligned (steered) with the direction of travel. The road pushes on the rubber, the rubber pushes on the rim, the rim pushes on the hub, etc etc.

Imagine the car is moving in Y direction. The wheels are rotating with angular speed such that $V=wR$ so that they do not slip over the road.

Now the road becomes frictionless. The wheals are still rotating with the same speed, so there is still no slipping. In the frame of reference of the vehicle the velocity of the road pavement and velocity of the lower part of the tires is the same.

Now we turn the wheels. The velocity of the pavement and velocity of the lower parts of the wheels are different now. Velocity of the pavement is directed along the Y. The velocity of the the rubber on tire (the part that touches road) has some X component.

Relative velocity of rubber over pavement has some X component. If now there appears some small friction between rubber and pavement, the friction force would have X component. And this is the force that turns the vehicle.