A body of mass 10kg is placed on foxed rough horizontal surface. Coefficient of static friction is 0.6 and coefficient of kinetic friction is 0.6 and coefficient of kinetic friction is 0.4. If horizontal force 20Nis applied on the body find out the frictional force (`g=10ms^(-2))`

Explanation

Limiting friction
`f_(S)=mu_(S)N=0.6xx10xx10=60N`
Kinetic friction `=0.4xx10xx10=40N`
The force applied on the body is F=20N. Since the body remains at rest, frictional force is static in nature and its value is `f_(s)=20N`.If we extend the discussion and make F=40N. `f_(s)` will be 40 N and the body remains at rest. If F=59N then `f_(s)` will be 59N and the body remains at rest. Static frictional force can take any volue between zero and 60N. Please note that in the formula `f_(s)=mu_(s)N,f_(s)` is maximum static frictional force or limiting friction. Its important to understand that the term static friction and limiting friction are not exactly the same. One should note that limiting friction is static in nature, but every static friction is not limiting friction. If `Fgt60N` the body slides on the surface, Hence the friction force involved will be kinetic frictional force and its magnitude will be `f_(k)=mu_(k)N`.`f_(k)=0.4xx10xx10=40N`
In Fig. 1.4 (a) (b), the block is subjected to three forces mg, `P and R^(.) (R^(.)` is the resultant of normal reaction N and frictional force f) and hence they must be concurrent.
`R^(.)=sqrt(f^(2)+N^(2)), tanphi_(s)=f_(max)/N=(mu_(s)N)/N, tanphi_(s)=mu_(s).........(1.2)`
where `phi_(s)` is known as the angle of friction which represents the angle between the resultant `R^(.)` and the normal reaction N. It is to be remembered that as long as `thetaltphi_(s)`, the block is in equilibrium and when `thetaltphi_(s)`, the motion is about to begin.