find the hydrostatic force

Question

A tank full of water is designed with ends in the shape of an isosceles triangle with height 13 meters and width, at the top, of 14 meters. Find the hydrostatic force on one end of the tank in newtons.

Answer 1

Force on a submerged surface = pressure at centroid * area ...

F= 9807*13/3(.5*13*14)= 3.87E6 Newtons

Answer 2

Let

H = 13 m be the height of the side,

B = 14 m be the length of its base at the top,

s = 10³ kg/m³ be the density of water,

g = 9.81 m/s² be gravitational acceleration.

At depth h from the top, consider a very thin horizontal strip of thickness dh.

That strip is the base of another isosceles triangle with height H-h.

Those two triangles are similar, and the smaller tringle has features reduced

by the factor (H-h)/H.

Therefore the length of the strip is B(H-h)/H.

And the area of the strip is thus B(H-h)dh/H

The pressure at depth h is sgh.

The force F(h) acting on that strip is the pressure times the area:

F(h) = sghB(H-h)dh/H

The force acting on the whole side is then the sum over all such strips,

that is, the definite integral from h=0 to h=H over F(h).

Let's first do the indefinite integral

∫sghB(H-h)dh/H =

sgB/H ∫h(H-h)dh =

sgB/H (∫hHdh - ∫h²dh) =

sgB/H (Hh²/2 - h³/3)

Now the definite integral between 0 and H is

sgB/H (H³/2 - H³/3) = sgBH²/6

Substituting actual numbers, the total force on the side is

10³×9.81×14×13²/6 = 3.9×10⁶ N