centripetal-forcefriction
Why is friction not reliable for circular turn on horizontal road for centripetal force when high speed and sharp curve are involved?
So banking of road has to be present to introduce a new force $N \cdot \cos\theta$ where $N$ is the normal contact force to the banked road.
Also why does an overload truck turn over on a banked road at high speed and sharp curve?
(1) is correct, but (2) and (3) are not. Consider a car turning counterclockwise around a track that is banked inwards. There are three forces acting on the vehicle: the weight of the car $\vec{W}$, the normal force $\vec{N}$, and the force of friction $\vec{f}$. Breaking these up into Cartesian coordinates gives:
$$\vec{W} = -mg\hat{y} \\ \vec{N} = N \left(\cos\phi \hat{y} - \sin\phi \hat{x}\right) \\ \vec{f} = f \left(-\sin\phi \hat{y} - \cos\phi \hat{x}\right)$$
Getting the direction correct for the friction is a bit subtle. We know from the $\phi \rightarrow 0$ limit that the friction has to point roughly in the direction of the turn, and also be perpendicular to the normal force, so for these coordinates, it must point down and to the left.
Since the car is moving in a circle in the horizontal plane, the forces in the $\hat{y}$ direction must cancel. Setting these equal to each other and considering the limiting case $f = \mu_s N$ gives an equation that can be solved for $N$. Then you can sum up the horizontal forces an insert this value to get the centripitel force.
What you need for circular motion is Centripetal Force. Definition:
Centripetal force is a force that makes a body follow a curved path: its direction is always orthogonal to the velocity of the body, toward the fixed point of the instantaneous center of curvature of the path. Centripetal force is generally the cause of circular motion.
If the road is flat, the centripetal force is provided by Friction between the tyres of the car and the road. This image show how:
If the there is no friction and the road is flat, the car would not be able to turn, it would keep sliding in the same direction.
So, if there is no friction, there has to be some force that can provide that necessary centripetal force. If the road is banked, the horizontal component of the Normal vector of the car that is going towards the centre of turn can act as the centripetal force. Hence, the car can turn on a banked road even without friction. Here is how:
Best Answer
There are some good reasons why you should not take a sharp turn at high speeds.
1) On a flat road, the force of static friction is what provides the centripetal force to accelerate you through a curve. Unfortunately, there is a maximum value for static friction that depends greatly on the mass of the vehicle. The heavier it is, the more static friction you can use to get you through a turn. The centripetal force required to get you through the turn is proportional to the inverse of the radius of the turn and the velocity of the vehicle squared. This means that sharp turns at high speeds require very large centripetal forces to complete. If you are in a light vehicle, the force required to make a high-speed turn may be more than static friction can provide. The result, in this case, is that the wheels will slip and slide outward from the turn and you'll quickly find out how your vehicle handles off-road (if nothing worse). Even a heavy vehicle shouldn't be used for this because the force required scales proportional to the mass of the vehicle as does the maximum static friction, which means if the light vehicle would skid, adding weight to it changes nothing.
2) The static friction available to you also depends on the contact area of the wheels (and the material, etc). Having more wheels means more static friction. Freight trucks usually have many large wheels to provide extra contact area and a higher maximum static friction. However, the mass distribution can play an important role. The mass distribution of the vehicle determines where the center of mass is. If you take a sharp turn at high speeds with enough friction to get through the turn but your center of mass is not very close to the ground, then you could have problems. The friction forces that make you turn act on the bottom of the wheels. If your center of mass is above this, that makes these forces produce a net torque on the vehicle. The higher the center of mass, the more torque these forces apply to the vehicle. Since the vehicle is free to rotate (or pivot) around the point where the outside wheels meet the road, once the torque from friction overcomes the torque due to gravity on the inside half of the vehicle, it's going to rotate. The result is the vehicle rolls onto its side (and possibly keeps rolling). And remember, the centripetal force is proportional to the square of the velocity and the inverse of the radius of the turn. So the faster it is and sharper the turn, the easier it is to roll the vehicle.