Resolve Force Into Normal and Shear Components  
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The figure shows a steel bar of height 100 millimeters. A section inclined at an angle of 55 degrees with respect to the horizontal is shown. The load on the left end of the bar is 40 kilo newton directed leftwards. The load on the inclined surface is P and it is directed rightwards. Load P is resolved into a vector V parallel to the inclined surface and a vector N perpendicular to the inclined surface. Vector N is at an angle of theta which equals 35 degrees with respect to the horizontal.

The 40 kN horizontal force will be resolved into force components acting perpendicular (N) and parallel (V) to section a-a.

Therefore, the angle between horizontal force P and the normal force component N is 35∘.

The two force components can now be calculated:

N=(40 kN) cos 35∘=32.766 kNV=(40 kN) sin 35∘=22.943 kN