find the force of one end of the tank in newtons

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 in NEWTONS on one end of the tank.

Answer 1

It has been a long time since I took calculus, so go over this carefully to be sure it makes sense.

Place a coordinate system centered on the vertex of the triangle. The differential area of the right half of the triangle is xdy. The pressure at the depth where the differential area is can be written as (13 - y) * 997 kg/m³ x 9.8 ms^-2. The differential force on this differential area is (13 - y) * 997 kg/m³ * 9.8 ms^-2 * xdy. The equation for the right side of the triangle is y = mx or y = 13/7x or x = 7y/13. Substitute this into the force equation for x to get

dF = [(13 - y) * 997 * 9.8 * 7y/13] dy

Integrate dF over y from y = 0 to 13. This should give you the total force for the right half. The overall total force is twice the force calculated.

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 some depth h from the top, consider a very thin horizontal strip of size dh.

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

Those two triangles are similar, and the smaller triangle 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 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