Pressure Profile

For the force equilibrium in vertical direction, the normal reaction by ground (N) should balance weight of the object (Mg).
The normal reaction force is distributed on the complete base. Let us assume a strip of width x at a distance x from the left face as shown. The force acting on the strip will be pressure at this strip times the area of the strip and the total normal reaction can be calculated by integrating this force from 0 to L.
$N = \int_0^L {(\alpha x + \beta )Wdx}$ $\begin{gathered} Mg = W(\alpha \frac{{{L^2}}}{2} + \beta L) \\ \alpha L + 2\beta = \frac{{2Mg}}{{WL}} = 100 & ...(1) \\ \end{gathered}$

For the rotatory equilibrium, the torque of all forces about the side AB should also equal 0.

$\begin{gathered} {\tau _N} = {\tau _F} + {\tau _{Mg}} \\ \int_0^L {(\alpha x + \beta )Wxdx = F\frac{H}{2} + Mg\frac{L}{2}} \\ W(\alpha \frac{{{L^3}}}{3} + \beta \frac{{{L^2}}}{2}) = F\frac{H}{2} + Mg\frac{L}{2} \\ 2\alpha L + 3\beta = \frac{3}{{W{L^2}}}(FH + MgL) = 157.5 & ...(2) \\ \end{gathered}$

Solving equations (1) and (2)

$\begin{gathered} \beta = 42.5 \\ \alpha = 7.5 \\ \frac{\beta }{\alpha } = 5.67 \\ \end{gathered}$