A block of mass 5kg is placed on a rough horizontal plane whose coefficient of friction is `0.3`. Find the least horizontal force needed to move the block along the plane and the resultant of normal reaction and the force applied.

Answer

`51.5N`, `16^(@)`

Explanation

`R=sqrt(F^(2)+N^(2))=sqrt(mu_(s)^(2)m^(2)g^(2)+m^(2)g^(2))=sqrt(((0.3^(2)xx5^(2)+5^(2))9.8^(2))=51.5N`. `tantheta=F//N=0.287theta_(1)` `theta=16^(@)`