Analysis
When the front and back walls (the trapezoidal faces) are removed, the water is free to spill. However, immediately after removal, the water retains its shape momentarily. The unbalanced force acting on the tank is due to the pressure exerted by the water on the remaining slanted side walls.
Because the tank tapers (trapezoidal cross-section), the pressure forces on the side walls have an unbalanced component along the longitudinal axis. We can calculate this by integrating the pressure force over the “difference” in area between the wide end ($l_2$) and the narrow end ($l_1$).
Derivation
The net force $F$ is equivalent to the hydrostatic force on the remaining $(l_2 – l_1)$ portion
The total force is given by:
$$F = \frac{1}{6} \rho g h^2 (l_2 – l_1)$$Using Newton’s Second Law, $F = ma$, where $m$ is the mass of the tank walls/base.
$$a = \frac{F}{m}$$