Hi. There seems to be a bit of confusion about the solution without a proper explanation being given.
Firstly, note that the net force acting on the side panel due to the presence of the fluid within the tank is the product of the pressure of the fluid and the area of the side panel.
You should know that the fluid pressure is a function of its depth:
Pfluid=ρfluidgh.
The net force acting on the side panel can then be given as a surface integral:
F=∫l=01.25∫y=00.6P(l,h) dl dh F=∫l=01.25∫y=00.6ρfluidgh dl dh F=[[2ρfluidgh2l]l=01.25]h=00.6 ⟹F=2207.25 N.
Since the pressure varies linearly from the top of the tank to the bottom of the tank, so does the force. The force distribution forms a triangular profile and since the centroid of a triangle lies at a distance of a third of its base length from its right angle, this is where the centre of pressure also lies and where the resultant force acts.
There is a force of unknown magnitude acting on the side panel in the w-axis direction due to the clasp. You need to take moments about the bottom of the tank for the side panel about the l-axis. It is advisable to take moments about this location to eliminate the unknown reactions from the hinges. Note that the hinges do not generate a reaction moment as per the definition of a hinge. You know: the force and distance from the hinges of the force due to fluid pressure, and the distance from the hinges of the force from the clasp. Hence, you can solve for the only unknown of the force from the clasp.