A house-top water tank is made of flat plates and is full to the brim. Its height is twice that of any side. The ratio of total thrust force on the bottom of the tank to that on any side will be:

Concept:

Whenever a static mass of fluid comes into contact with a surface the fluid exerts force upon that surface. The magnitude of this force is known as the hydrostatic force or total pressure force.

The magnitude of hydrostatic force is given by F = ρ g h̅̅  A N

Where ρ = Density of the fluid, h̅ = Depth of center of gravity of the surface from free liquid surface, A = Area of the surface

Calcualtion:

Given tank has the height twice as the width of the tank. 

Pressure force acting on the vertical surface:

The center of gravity of the vertical surface from the free surface will be

\(\bar h = \frac{{2x}}{2} = x\;m\)

Area of the vertical surface = \(2x \times x = 2{x^2}\)

Now the Pressure force \(\rho \times g \times x \times 2{x^2} = 2\rho g{x^3}\)

Now, the force on the bottom surface:

The center of gravity of the bottom surface from the free liquid surface = hieht of the tank = 2x

Area of the bottom surface = \(x \times x = {x^2}\)

Now the pressure force = \(\rho \times g \times 2x \times {x^2} = 2\rho g{x^3}\)

Therefore the ratio of the forces will be one as the forces on the bottom side and the vertical sides are equal.