Force of Friction
Whenever two rough surfaces are in contact, sliding between the surfaces is opposed by the force of friction which the surfaces exert on each other. The force of friction acts parallel to the surfaces in contact and on both the surfaces.
Static friction : If the tendency to slide against each other is too small to cause actual sliding motion, the force of friction is called as `force of static friction'. The magnitude of this force balances the net applied force. Hence if there is no sliding between the surfaces,
Force of static friction = Net applied force parallel to the surfaces.
Critical point (maximum static friction) : If the sliding between the surfaces is about to begin, the static friction is at its maximum value which is equal to µsN, where N = normal reaction between the surfaces and µs = coefficient of static friction. In this situation, we say that the surfaces are at their point of sliding and are exerting a force µsN on each other so as to oppose sliding.
Kinetic friction : If actual sliding is taking place between the surfaces, the force of friction is called as force of kinetic friction or the force of sliding friction (fk).
fk = µkN where µk = coefficient of kinetic friction
Note : Force of friction on a body always acts against the sliding tendency.
Illustration :
A block lying on a horizontal surface is pulled by a force of 0.1 N but the block does not move i.e., it remains at rest. To analyse the frictional force on the block we proceed as follows :
As the block remains at rest, the force of static friction is balancing the 0.1 N force (the applied force). So the frictional force = 0.1 N.
Illustration :
A block of weight 100 N lying on a horizontal surface just begins to move when a horizontal force of 25 N acts on it. Determine the coefficeint of static friction.
As the 25 N force brings the block to the point of sliding, the frictional force = µsN.
From force diagram : N = 100 N
µsN = 25 
µs = 0.25.
Illustration :
A block lying on a inclined plane has a weight of 50 N. It just begins to slide down when inclination of plane with the horizontal is 30º. find µs.
The block reaches the point of sliding when the plane makes an angle of 30º with the horizontal.
Hence in this situation, frictional force = µsN
Balancing the forces :
N = W cos 30º
µsN = W sin 30º
µs W cos 30º = W sin 30º
µs = 1/
Note : Minimum angle for which a block starts sliding down an inclined plane is known as angle of repose.
Illustration :
A block of weight 100 N lying on a horizontal surface is pushed by a force F acting at an angle 30º with horizontal. For what value of F will the block begin to move if µs = 0.25?
Consider the force diagram of the block at the moment when it is just to start moving.
Balancing the forces :
N = mg + F sin 30º
F cos 30º = µs N
F cos 30º =
µs (mg + F sin 30º)
F 
