A cubical tank of side $2 m$ is filled with $1.5 m$ of glycerine of specific gravity $1.6$. Find the force acting on the side of the tank when it is accelerated vertically upward at $5 m/s^2$

The pressure intensity at the bottom of the tank $P = \rho (g + a)h$

The average pressure intensity  $$\begin{gathered} P_{av}=\frac{0+P}{2} \\ =\frac{\rho (g+a)h}{2} \end{gathered}$$

The force on the side of the tank,
$F = P_{av} \times wetted area of wall$

$$\begin{gathered} =\frac{\rho (g+a)h}{2}\times \left(bh\right) \\ =\frac{\rho (g+a)b{h}^2}{2} \\ =\frac{1600(9.8 + 5) \times 2 \times 1.5^2}{2} \\ \simeq 53.316 kN. \end{gathered}$$