Respuesta :
Answer:
F = 1.65 10⁸ N
Explanation:
In this pressure problem we have to use the definition of pressure
P = dF / dA
dF = P dA
we already have the expression for force and the pressure in a liquid is
P = Po + rho g (H-y)
Where Po is the atmospheric pressure acting on both sides of the dam, whereby its contribution is canceled and (H-y) is the distance from the surface
Let's look for an expression for the area differential
A = xy
dA = dx dy
y = 0.3 x²
x = √(y/0.3)
Let's build our equation with these expressions and integrate between the initial limit where the height is measured from the bottom of the dam y = 0, x = 0 to the upper limit, let's call it H = 200m, x = RA y / 0.3 and F = 0
∫ dF = ∫ (ρ g (H-y) dx dy
-F = ρ g [∫∫ H dy dx - ∫∫ ydy dx]
F = ρ g [∫ H x dy - ∫ x2√2 / 2 dy
Let's evaluate between the limits of integration
F = ρ g [∫ (H (√y /√0.3) dy - ∫ y/0.3 1/2 dy
F = ρ g (H /√0.3 ∫ √y dy - 1 /0.6 ∫ y dy)
Let's do the second integral
F = ρ g (H/√0.3 [tex]y^{3/2}[/tex]) 2/3 - 1/0.6 y2 / 2)
F = ρ g (2H/3√0.3 [tex]y^{3/2}[/tex]) - 1/1.2 y2 )
We evaluate at the limits
Y = 0
Y = 38 m
F = 1000 9.8 (2 200/3√3 √38³ - 1/1.2 38²)
F = 9800 (18032 - 1203)
F = 1.65 10⁸ N
