The wing can be modeled as a simple cantilevered beam with a distributed load of 165 lb/ft and point force load of 4,500 lb.

The first step in solving this problem is to determine the reaction force and moment at the fixed end, A. This can be done more efficiently by replacing the distributed load with a point load of magnitude W_r at location x_r.

     
          = (165 lb/ft)(54 ft) = 8,910 lb

The location of the point load will be

     
          = (54 ft)/2 = 27 ft

     ΣF_y = 0
     A_y - 8,910 + 4,500 = 0
     A_y = 4,410 lb

     ΣM_A = 0
     M_r - 8,910 (27) + 4,500 (49) = 0
     M_r = 20,070 lb-ft

There will be two sets of shear and moment equations, one to the left of F and one to the right.

Section 1 FBD

After making the section cut, and developing a free-body diagram for the left side of the wing, sum the forces to get the shear equation.

     ΣF_y = 0
     4,410 - 165x - V₁ = 0
     V₁ = 4,410 - 165x

Summing the moments give the moment equation,

     ΣM_cut = 0
     M₁ + (165x) (x/2) + 20,070 - 4,410x = 0
     M₁ = -82.5x² + 4,410x - 20,070

Section 2 FBD

Once again, first make a cut in section 2 (left of point B). Next, construct a free-body diagram and then sum forces to get the shear equation.

     ΣF_y = 0
     4,410 - 165x + 4,500 - V₂ = 0
     V₂ = 8,910 - 165x

Summing the moments give the moment equation,

     ΣM_cut = 0
     M₂ + (165x) (x/2) + 20,070 - 4,410x
                - 4,500(x - 49) = 0
     M₂ = -82.5x² + 8,910x - 240,570

The shear equations for both sections can now be plotted to help identify the maximum and their locations. Both shear equations are linear lines, and shown at the left.

The maximum shear occurs at the far left where the wing is fixed to the aircraft body. The magnitude is 4,410 lb.

Like the shear, the two moment equations can be plotted to help identify the maximum internal moment. However, since the equations are quadratic curves, they are more difficult to plot than the linear lines for the shear.

There are a few techniques that can assist in plotting curves. If the second derivative is positive, then the curve is facing upward. For both equations, their second derivatives are negative, so the they face downward.

The maximum can be determined by equating the first derivative to zero. For the first section equation, that becomes,

     d(M₁)/dx = -165x + 4,410 = 0

     x = 26.73 ft

The maximum value at 26.73 ft is

     M_26.73 = -82.5(26.73)² + 4,410(26.73) - 20,070

     M_max = 38,860 lb-ft