The figure shows a vertical bar of width 50 millimeters fixed to the ceiling. An inclined plane labeled a a is shown and it is at an angle of 36 degrees with respect to the vertical and 54 degrees with respect to the horizontal. The load on the inclined surface has a magnitude of 26.286 kilo newton and is directed vertically downwards. This load is resolved into a vector of magnitude 21.266 kilo newton parallel to the inclined surface and a vector N perpendicular to the inclined surface and making an angle theta which equals 54 degrees with the vertical.

If the shear stress on section a-a is 25 MPa, then the shear force component V can be computed as:

τ=VAa-a       V=τ Aa-aV=(25 N/mm2)(850.65 mm2)  =21,266.25 N  =21.266 kN

The resultant axial force P is related to the shear component V by the sine function:

sin 54=VP       P=Vsin 54    P=21.266 kN0.80902=26.286 kN