Problem :

A block of 10N rests on a plane inclined 45o. In addition a horizontal force of 10N is applied to the block. What is the normal force applied by the inclined plane?

We solve the problem by drawing a free body diagram, and resolving all force vectors into components parallel and perpendicular to the plane:

Solution 1
The component of the gravitational force perpendicular to the plane is given by:

FGâä¥ = FGsin 45o = 10 sin 45o = 7.07N    

Similarly, the component of the applied force perpendicular to the plane is:

Fâä¥ = F sin 45o = 10 sin 45o = 7.07N    

Thus the normal force on the block is simply the sum of the two perpendicular forces, or 14.14N.