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static friction
Really? A coefficient of friction of 7? Unheard of. Probably 0.7 Use F = mv^2 / r where F will be the centripetal force as well as the frictional force, (mu)mg. Solve for v.
No steering wheel won't provide the centripetal only the friction between the tyre and road provides the necessary centripetal. Steering would simply turn the wheels of the car to the desired direction.
No, as far as movement goes, friction is not required. Too much friction would slow a car to a halt, remove it and it'd glide along. Granted you need friction to steer, but not to move. As far as engines go, friction is what wears them down over time. The key to making more efficient cars is less friction.
I believe that the answer to this lies in the forces acting on the bike, namely the the frictional force, the turn itself is made by the wheels, which create friction sideways. The centripetal force acting in the turn is the friction, imagine a motorcycle driver that doesn't lean into the curve; because he doesn't lean, the friction is not strong enough to keep the driver from going in a straight line. However, when he leans, the static friction from the road increases, therefore the centripetal force increases. Note: There is an aspect of this question that also relates to Torque.
static friction
Friction is the force that makes it possible to stop a car - a friction between wheels and road surface is a stopping force there. Also, friction is necessary for maneuvering - if the wheels slipped, you wouldn't be able to make a turn.
When the wheels of a car turn, friction causes the wheels to push back on the road, which in turn causes the road to push the car forward. Without friction the wheels would just spin in place. You can sometimes see this occur on icy roads, where there is very little friction.
Automobile brakes and clutches require friction. We also need friction between the tyres and the road when we accelerate or when we brake or when we turn a corner.
Really? A coefficient of friction of 7? Unheard of. Probably 0.7 Use F = mv^2 / r where F will be the centripetal force as well as the frictional force, (mu)mg. Solve for v.
7