is it frictional force greater than centrifugal force to avoid skidding of vehicle?

Answer 1

Actually, it only has to be equal.

Answer 2

when the car doesnt skid...it DOES mean that the frictional force is equal tothe centrfiugal force.
friction provides the centripetal force for the car to turn that balances the centrifugal force. in places where the cars go in high speed as in a highway or more on a racing track when it is known that the friction alone will be unable to provide enough centripetal force at that speed, banking is done. this means the road is bent so that the downward force because of the mass of the car increases...this helps in not skidding at high speeds. ( as friction is independent on speed and surface area of contact and solely relies on the weight or more correctly the downward force applied)

Answer 3

When you solve questions that involve rotation, you don't assume a new force called centrifugal force. However, you just draw a Free Body Diagram with all the forces (weight downwords, normal force normal to the surface, friction opposite to the direction of motion.). Then, you analyze each of those forces into components either pointing to the center of rotation or normal to this direction. The sum of forces pointing to the center of rotation should equal mv^2/r. These forces are called centrifugal forces.
Assume a road banked by an angle theta. We draw a FBD which will be a triangle with one 90deg angle. The forces action will be:
- W=mg downwords
- N perpendicular to the surface of the road (this is the reaction of the surface of the road to the car weight)
- F Friction, parallel to the surface of the road going inwords towards the center of rotation.
We then analyze N to N sin (theta) towards the center of rotation and N cos(theta) opposite to the Weight.
F will be analyzed into F cos(theta) towards the center and F sin(theta) parallel to the weight.
Then we applying Newton's second low in both directions noting that the resultant force pointing to the center is mv^2/r.
We get:
N sin(theta) + F cos (theta) = m v^2/r
of course, F= U.N where U is the frictin coefficient. It follows that:
N sin(theta) + N.U cos (theta) = m v^2/r

In order to stay on the right track, the radius resulting from the above equation should match the radius of the turn.
In other words: the left side of the above equations are constant for the driver, he can't change them. If he increases his spead, r will increase accordingly so that m v^2/r will stay constant.
Of course, N can be calculated by writing the equation of the other direction which will be:
N cos(theta) = mg + F sin (theta) or,
N cos(theta) = mg + U.N. sin (theta)

Answer 4

It is bcoz of frictional force every one is walking on the earth.
Frictional force and Centrifugal force should be equal so that every one can stand without skidding.

Answer 5

No, frictional force is only grater than and equal to centrifugal force.