A body of mass 10 kg is kept on a horizontal floor of coefficent of static friction `mu_(s)=0.5` and coefficient of kinetic friction `mu_(k)=0.45` as shown in figure.

Find the acceleration, force of friction and contact force on the body by the plane when the driving force is `(g=10ms^(-2))`
(i) 40 N (ii)60N

Explanation

FBD of the body

Normal reaction `N=mg=100N`
Limiting friction on the body, `f_(L)=mu_(s)N=0.5xx100N=50N`
(i) F=40 N is less than the limiting friction so the body is static. So ,a=0
Force of friction acting on the body is static friction, f= Driving force= 40 N
Contact force is the resultant of force of friction and normal reaction. So
`C=sqrt(f^(2)+N^(2))=sqrt((40)^(2)+(100)^(2))=107.7N`

(ii) F=60 N is greater than the limiting friction on the body, so body will start moving. Force of friction acting on the body
= kinetic friction `=mu_(k)N=0.45xx100N=45N`
FBD of the body

`therefore `Acceleration of the body is
`a=(F-f_(k))/(m)=(60-45)/(10)=1.5 ms^(-2)`
Contact force, `C=sqrt(f^(2)+N^(2))=sqrt(f_(k)^(2)+N^(2))`
`=sqrt((45)^(2)+(100)^(2))~~109.7 N`