A generic diagram of a slope in a jointed rock mass that is threatened by a planar block slide is shown in the sketch. A minimum safety factor of 1.05 is required. Although not shown bench height is 60 ft (18.3 m). Determine if FS =1.05 is possible, given that Mohr–Coulomb failure criteria apply, the joints constitute 79% of the potential shear failure surface, slide block weight is 1.351(10⁷) lbf/ft (198.6 MN per meter) of thickness and: 

1 slope height                          H = 540 ft (165 m);

2 failure surface angle             α = 32◦;

3 slope angle                           β = 45◦;

4 friction angle (rock)          φr = 33◦;

5 cohesion (rock)                     cr = 2870 psi (19.8 MPa);

6 friction angle (joint)             φj = 28◦;

7 cohesion (joint)                    cj = 10.0 psi (0.069 MPa);

8 specific weight                     γ = 158pcf (25.0kN/m³);

9 tension crack depth              hc = 50.0 ft (15.2 m);

10 water table depth               hw = 60.0 ft (18.3 m);

11 seismic coefficient              ao = 0.15;

12 surcharge                             σ = 0.0 psf (0.0 kPa).

If FS < 1.05 will drainage allow the safety factor objective to be achieved?