Question

Is Friction Independent of actual Area of Contact?

Answer 1

The Normal reaction of any body is independent of its area of contact.Normal reaction of a body depends on the mass of the object.Friction is a force that opposes the motion between any surfaces that are in contact. Friction works in the direction opposite to that of motion.

Complete answer:
Friction helps in walking, sliding, applying brakes to stop a moving vehicle, writing on a notebook or blackboard, drilling a nail into a wall etc.Formula of friction is $f=\mu N$ where $f$ is force of friction, $\mu$ is coefficient of friction and $N$ is Normal reaction.

The coefficient of friction is a measure of the amount of friction existing between two surfaces. The Normal reaction depends on the mass of the body and it is independent of area of contact. As Friction happens due to the interatomic forces at the minute regions of atomic contact. The contact occurs only between peaks which have irregular distribution in contacting surfaces, therefore the contact area is much smaller than the projected area.

It can be explained with the help of an example. Suppose there are 3 objects of the same nature and same mass of $10$ kg but different in shape. Object $1$ is a cube, object $2$ is a cone and object $3$ is a square plate(cuboid). These are connected individually with a spring balance. Now one by one each object is pulled and each object starts sliding when the spring balance shows a reading of $20$ N. Thus, friction is the same in all cases as it is independent of area of contact.The area of contact on atomic scale is a very small fraction of total surface area.

Therefore yes, friction is independent of area of contact.

Note: Friction depends only on the nature of surfaces in contact and weight of the body. If weight is different then value of friction is also different.Friction helps us in walking.The friction between the shoes and the floor stops us from slipping, similarly, the friction between the car and the road prevents the car from sliding.