Solution :
Given data is :
Density of the milk in the tank, [tex]$\rho = 1020 \ kg/m^3$[/tex]
Length of the tank, x = 9 m
Height of the tank, z = 3 m
Acceleration of the tank, [tex]$a_x = 2.5 \ m/s^2$[/tex]
Therefore, the pressure difference between the two points is given by :
[tex]$P_2-P_1 = -\rho a_x x - \rho(g+a)z$[/tex]
Since the tank is completely filled with milk, the vertical acceleration is [tex]$a_z = 0$[/tex]
[tex]$P_2-P_1 = -\rho a_x x- \rho g z$[/tex]
Therefore substituting, we get
[tex]$P_2-P_1=-(1020 \times 2.5 \times 7) - (1020 \times 9.81 \times 3)$[/tex]
[tex]$=-17850 - 30018.6$[/tex]
[tex]$=-47868.6 \ Pa$[/tex]
[tex]$=-47.868 \ kPa$[/tex]
Therefore the maximum pressure difference in the tank is Δp = 47.87 kPa and is located at the bottom of the tank.
