**1. The problem statement, all variables and given/known data**

An automatic valve consists of a 225 X 225 mm square plate that is pivoted about a horizontal axis through A located at a distance h = 90 mm above the lower edge. Determine the depth (d) of the water for which the valve will open." Given: h = 90 mm, y = 225 mm

*Answer given in the back of the book = 300 mm*

Diagram: http://ift.tt/1hq3Rv9

**2. Relevant equations**

F = Pressure(P) * Area(A)

F = ∫∫_{A} P * dA

Pressure P = ρgh, where

g = 9.81 m/s^2

h = depth in meters

ρ = density = 10^3 kg/m^3

**3. The attempt at a solution**

My theory: If the force on the valve above the hinge equals the force on the valve below the hinge, the valve will begin to open.

Force on Top: F_{TOP} = [itex]\int ^{225}_{0}\int ^{225}_{90} \rho g(d-y) dydx[/itex]

Simplified: F_{TOP} = 255[itex]\int ^{225}_{90} \rho g(d-y) dy[/itex]

Force on bottom: F_{BOT} = [itex]\int ^{225}_{0}\int ^{90}_{0} \rho g(d-y) dydx[/itex]

Simplified: F_{BOT} = 255[itex]\int ^{90}_{0} \rho g(d-y) dy[/itex]

[itex]\Rightarrow[/itex] F_{TOP} = F_{BOT}

Equate: 255[itex]\int ^{225}_{90} \rho g(d-y) dy[/itex] = 255[itex]\int ^{90}_{0} \rho g(d-y) dy[/itex]

Simplify: [itex]\int ^{225}_{90} (d-y) dy[/itex] = [itex]\int ^{90}_{0} (d-y) dy[/itex]

Integrate: (dy – [itex]\frac{y^2}{2}[/itex]|[itex]\stackrel{225}{90}[/itex]) = (dy – [itex]\frac{y^2}{2}[/itex]|[itex]\stackrel{90}{0}[/itex])

[itex]\Rightarrow[/itex] 135d – 21262.5 = 90d – 4050

[itex]\Rightarrow[/itex] 45d = 17212.5

[itex]\Rightarrow[/itex] d = 382.5 mm

Answer given in book = 300 mm

**4. Notes**

I know there are many ways to solve this problem. I would like to see a solution involving calculus because I believe that is where all other solution methods are derived from. Thanks in advance.

Captain1024

http://ift.tt/1gXVewI