## Friction Force And Circular Motion

It's not that hard to explain. When the rear tires (friction forces) push on the ground, they also push on the front tires (rigid body constraint through the chassis, axles, etc.). The front tires are turned at an angle, however; so unless you're doing something unsafe, the friction forces on the front tires constrain them and the motion of the entire car. This constraint is only perpendicular to the front tires, however, and they, as well as the rest of the car, are free to move along the angle they are turned to. So the car turns. If you're doing something risky given the driving conditions, however, the car might skid, tilt, or even flip over.

 Quote by Bashyboy Well, how does a tire rotate, wouldn't it have to overcome some resistive force?
The tire rotate because of the axle, the crank, and ultimately the combustion of gasoline inside the cylinder.

 Quote by Bashyboy Okay, that sounds good. So does the friction have an equal and opposite force? Or do not all forces have to have equal and opposite forces?
In Newtonian mechanics/model, all forces are paired. The friction pushes on both the tire and the Earth, but in opposite directions.
 In circular motion we need force to accelerate the car to the center. On banking curve, the normal force supply the force. On level curve, the only force is from static force(since radius remains constant). It has limited supply of force = μsMg If more force is needed due to increase in speed or reduction of radius and static friction has reached its maximum value then the force to keep the car in circular motion not met thus 'overshoot' occurs.

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