Total pressure and centre on tank face AB

Reading of the figure: AB is the inclined side wall, vertical height 5 m and horizontal run 2 m, so its length is $L=\sqrt{5^{2}+2^{2}}=5.385$ m; width of tank $b=6$ m; A at the bottom, B at the top. The oil surface (level of B) is under a gauge pressure $p_0=19.62$ kPa. Oil $S=0.8$, $\gamma=0.8(9.81)=7.848$ kN/m³.

   B  p0 = 19.62 kPa
    \
     \   AB, L = 5.385 m
      \
   A   \ 5 m vertical

Equivalent oil head of the gauge pressure: $\dfrac{19.62}{7.848}=2.5$ m of oil above B.

Total pressure (centroid is 2.5 m below B):

$$A_{AB}=5.385\times6=32.31\ \text{m}^2$$ $$p_c=19.62+7.848(2.5)=39.24\ \text{kPa}$$ $$F=p_cA_{AB}=39.24\times32.31=1267.9\ \text{kN}$$

Centre of pressure. Distance of the centroid from the equivalent free surface along the slope, with $\sin\theta=5/5.385$:

$$\bar y=\frac{2.5+2.5}{\sin\theta}=5.385\ \text{m}$$ $$\bar y_{cp}-\bar y=\frac{L^{2}}{12\,\bar y}=\frac{29}{12(5.385)}=0.449\ \text{m}$$

The centre of pressure is $L/2+0.449=3.141$ m from B along AB (vertical depth below B $=3.141\sin\theta=2.917$ m).

Check with the pressure diagram. Pressures: at B, 19.62 kPa; at A, $19.62+7.848(5)=58.86$ kPa (trapezium):

$$F=\frac{19.62+58.86}{2}\times32.31=1267.9\ \text{kN}$$ $$x_{B}=\frac{L}{3}\cdot\frac{2(58.86)+19.62}{58.86+19.62}=\frac{5.385}{3}\times1.7500=3.141\ \text{m from B}$$

Answer: $F\approx1268$ kN, acting 3.14 m from B along the slope (2.92 m vertically below B).