A cubical tank has sides of 1.5 m. It contains water for the lower 0.6 m depth. The upper remaining part is filled with oil of specific gravity 0.9. Calculate for one vertical side of the tank total pressure.

$$ P_D = \rho_1 g h_1 $$ $$ P_D = 900 \times 9.81 \times 0.9 $$ $$ P_D = 7946.1 \, \text{N/m}^2 $$ $$ P_B = P_D + \rho_2 g h_2 $$ $$ P_B = 7946.1 + (1000 \times 9.81 \times 0.6) $$ $$ P_B = 7946.1 + 5886 $$ $$ P_B = 13832.1 \, \text{N/m}^2 $$

$$ F_1 \text{ (from top triangle ADE)} = (\frac{1}{2} \times P_D \times h_1) \times b $$ $$ F_1 = (\frac{1}{2} \times 7946.1 \times 0.9) \times 1.5 $$ $$ F_1 = 5363.6 \, \text{N} $$ $$ F_2 \text{ (from rectangle BDEF)} = (P_D \times h_2) \times b $$ $$ F_2 = (7946.1 \times 0.6) \times 1.5 $$ $$ F_2 = 7151.5 \, \text{N} $$ $$ F_3 \text{ (from bottom triangle EFC)} = (\frac{1}{2} \times (P_B - P_D) \times h_2) \times b $$ $$ F_3 = (\frac{1}{2} \times 5886 \times 0.6) \times 1.5 $$ $$ F_3 = 2648.7 \, \text{N} $$

$$ F_{total} = F_1 + F_2 + F_3 $$ $$ F_{total} = 5363.6 + 7151.5 + 2648.7 $$ $$ F_{total} = 15163.8 \, \text{N} $$

$$ \text{Moment}_1 = F_1 \times (\frac{2}{3}h_1) $$ $$ \text{Moment}_1 = 5363.6 \times (\frac{2}{3} \times 0.9) $$ $$ \text{Moment}_1 = 5363.6 \times 0.6 = 3218.16 \, \text{N-m} $$ $$ \text{Moment}_2 = F_2 \times (h_1 + \frac{h_2}{2}) $$ $$ \text{Moment}_2 = 7151.5 \times (0.9 + \frac{0.6}{2}) $$ $$ \text{Moment}_2 = 7151.5 \times 1.2 = 8581.8 \, \text{N-m} $$ $$ \text{Moment}_3 = F_3 \times (h_1 + \frac{2}{3}h_2) $$ $$ \text{Moment}_3 = 2648.7 \times (0.9 + \frac{2}{3} \times 0.6) $$ $$ \text{Moment}_3 = 2648.7 \times 1.3 = 3443.31 \, \text{N-m} $$

$$ M_{total} = \text{Moment}_1 + \text{Moment}_2 + \text{Moment}_3 $$ $$ M_{total} = 3218.16 + 8581.8 + 3443.31 $$ $$ M_{total} = 15243.27 \, \text{N-m} $$

$$ h^* = \frac{M_{total}}{F_{total}} $$ $$ h^* = \frac{15243.27}{15163.8} $$ $$ h^* \approx 1.005 \, \text{m} $$