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: 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.