Centripetal Force
Acceleration of a body moving in a circle of radius R with uniform speed Ï…is Ï…2/R directed towards the centre. According to second law, the force f providing this acceleration is
Motion of a Car on a Level Road
Three forces act on a car:

The weight of the car

Normal reaction, N

Frictional force, f
N  mg = 0
N = mg
which is independent of the mass of the car.
The above equation shows that for a given value of Î¼_{s} and R, there is a maximum speed of circular motion of the car possible.
Motion of a Car on a Banked Road
We can reduce the contribution of friction to the circular motion of the car if the road is banked. Since there is no acceleration along the vertical direction, the net force along this direction must be zero. Hence,
N cos q = mg + f sin Î¸
The centripetal force is provided by the horizontal components of N and f.
But
Thus to obtain Ï…_{max}, we put,
N cos Î¸=mg + Î¼_{s} N sin Î¸
Substituting the value of N in the above equation,
The maximum possible speed of a car on a banked road is greater than that on a flat road.
Î¼_{s }= 0, then
At this speed, frictional force is not needed at all to provide the necessary centripetal force. Driving at this speed on a banked road will cause little wear and tear of the tyres. The same equation also tells us that for Ï…< Ï…0,the frictional force will be up the slope and that a car can be parked only if .
N cos q = mg + f sin Î¸
But
Thus to obtain Ï…_{max}, we put,
N cos Î¸=mg + Î¼_{s} N sin Î¸
Î¼_{s }= 0, then
Solving Problems in Mechanics
In mechanics, many a time, the problems do not involve merely single body under the action of a single force or a number of forces. In certain problems, one may come across a system consisting of a number of bodies exerting forces on each other through various kinds of supports, connecting strings, etc. In addition, other forces such as the gravitational force, frictional force, etc may also be acting between bodies.
Under such circumstances, each body is to be considered separately. Equation of motion for each body is to be obtained, taking into account all the forces acting on it and then equating the net force acting on the body to its mass times the acceleration produced
A diagram for each body of the system depicting all the forces on the body by the remaining part of the system is called the free body diagram.
 Draw a diagram showing schematically the various parts of the assembly of bodies, the links, supports etc.
 Choose a convenient part of the assembly as one system.
 Draw a separate diagram which shows this system and all the forces on the system by the remaining part of the assembly. Include also the forces on the system by other agencies. Do not include the forces on the environment by the system. A diagram of this type is known as Free Body Diagram.
 In a free body diagram, include information about forces that are either given or you are sure of. The rest should be treated as unknowns to be determined using laws of motion.
 if necessary, follow the same procedure for another choice of the system. in doing so, employ the Newton's third law of motion. That is, if in the freebody diagram of A, the force on A due to B is shown as F, then in the free body diagram of B, the force on B due to A should be shown as F.