An object of mass $m=\sqrt{8}kg$  rests on an inclined plane that makes angle $\theta ={45}^0$  with the horizontal floor. What minimum force (in N), parallel to the base of the incline must be applied to the object in order to move it. The coefficient of static friction between the object and the plane is ${\mu }_s=1.25$

There are four forces acting on the object; a normal force N perpendicularly away from the incline, gravity mg vertically downward, the applied force F in the direction of motion and a static frictional force f before the object begins to move. Since the object is about to slide along the plane, the frictional force must be at its maximum value,  $f={\mu }_{s}N$. The sum of the force components perpendicular to the incline is zero, so, 
$N=mg\cos \theta \Rightarrow f={\mu }_{s}mg\cos \theta$        ……………..(1)
On the other hand, the force components parallel to the surface of the incline are sketched in the following free-bogy diagram.

Note that the frictional force must make some angle $\varphi$  as drawn because it initially balances the other two forces on this diagram,
 $F=f\cos \varphi$                         ……………..(2)