Question

Calculate the contact force acting on the mass m placed over the inclined plane of friction coefficient 0.5 as shown:

Answer 1

As block is at rest there will be two component of weight:

Horizontal component, $=mgsin\theta$ and vertical component, $=mgcos\theta$

Here, normal reaction force will balance vertcal component, $mgcos\theta =R$ and frictional force will balance horizontal component, $=mgsin\theta$

So resultant of normal reaction and frictional force will be, $F_c=\sqrt{f^2+R^2}⟹F_c=\sqrt{{\left(mgsin\theta \right)}^2+{\left(mgcos\theta \right)}^2}=mg$