A block of 10N rests on a plane inclined 45^{o}. 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:

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

F_{Gâä¥} = F_{G}sin 45^{o} = 10 sin 45^{o} = 7.07N

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

F_{âä¥} = F sin 45^{o} = 10 sin 45^{o} = 7.07N

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