find the force of newtons

Question

A dam on a river is built so that the wall facing the water is shaped like the region above the curve  and below the line . (Here, distances are measured in meters.) The water level is  meters below the top of the dam. Find the force (in Newtons) exerted on the dam by water pressure. (Water has a density of , and the acceleration of gravity is  .)

Answer 1

Let

H = 280m - 22m = 258m be the height of the water,

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

g = 9.8 m/s² be gravitational acceleration.

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

The length of the strip can be obtained from

y = 0.8x⁴

x = (y/0.8)^1/4

If we set y = H-h, then x = ((H-h)/0.8)^1/4.

Therefore the length of the strip, denoted L(h):

L(h) = 2 ((H-h)/0.8)^1/4 = 2 (5(H-h)/4)^1/4

For future use, let's precompute the indefinite integrals

∫L(h)dh = -2 (4/5)² (5(H-h)/4)^5/4 = -2 (4/5)² (5(H-h)/4) (5(H-h)/4)^1/4 = -(4/5)(H-h)L(h) (1)

∫(∫L(h)dh)dh = 2(4/5)³(4/9)(5(H-h)/4)^9/4 = (4/5)³(4/9)(5(H-h)/4)²L(h) = (4/5)(4/9)(H-h)²L(h) (2)

The area of the strip is L(h)dh.

The pressure at depth h is sgh.

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

F(h) = sghL(h)dh

The force acting on the whole dam 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

∫F(h)dh = ∫sghL(h)dh = sg ∫hL(h)dh

Let's do the integral by parts following the formula

∫uv' = uv - ∫u'v, where

u = h, u' = 1

v' = L(h), v = ∫L(h)dh = -(4/5)(H-h)L(h) (by (1))

Thus

∫F(h)dh =

sg ∫hL(h)dh = sg (-(4/5)h(H-h)L(h) - ∫(∫L(h)dh)dh) =

sg ((-(4/5)h(H-h)L(h) -(4/5)(4/9)(H-h)²L(h)) = (by (2))

sg(-4/5)(H-h) (hL(h) + (H-h)L(h)) (3)

When computing the definite ∫F(h)dh from h = 0 to h=H,

then for h=H the formula (3) is 0.

Therefore the whole definite integral is

sg(4/5)H²L(0) = (8/5)sgH²(H/0.8)^1/4

Substituting actual numbers, the force on the dam is

(8/5)×10³×9.8×258²×(258/0.8)^1/4 = 4423×10⁶N