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

`mu_(s)=0.5`
`mu_(k)=0.45`
Find the acceleration, force of friction and contact force on the body by the plane when the driving force is `(g=10m//s^(2))`
(i) 40 N
(ii) 60 N

Explanation

FBD of the body :

Normal reaction N = mg = 100 N
Limiting friction on the body , ` f _(L) = mu_(s) N = 0*5 xx 100 N = 50 N `
(i) F = 40 N is less than the limiting fiction 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))= 107.7 N `

(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*45 xx 100 N = 45 N `
FBD :

` :. "Acceleration of the body is " `
` a = (F- f_(k))/m`
` = (60 - 45)/10`
` = 1*5 "m//s^(2)`
Contact force, ` C= sqrt(f^(2) + N^(2))= sqrt(f_(k)^(2)+ N^(2) ) = sqrt((45)^(2)+ (100)^(2))cong 109.7 N`