The body of mass 400 g slides on a rough horizontal surface. If the frictional force is 3.0 N, find (a). the angle made by the contact force on the body with the vertical and (b). the magnitude of the contact force. Take `g=10 m/s^2`

Explanation

Let the contact force on the block by the surface be `F_(c )` which makes an angle `lambda ` with the vertical (shown figure )

The component of `F_( c ) ` perpendicular to the contact surface is the normal force N and the component of F parallel to the surface is the frictionf. As the surface is horizontal, N is vertically upward. For vertical equilibrium.
`N =Mg = ( 0.400 kg ) ( 10 m//s^(2)) = 4.0 N `
The magnituded of the contact force is
`F = sqrt( N^(2) +f^(2)) = sqrt((4.0 N )^(2) + ( 3.0 N )^(2)) =5.0 N `