Problem 3.52 - A y p ⋅ ⌠ ⎮ ⎮ d ⋅ = 2 L cos θ...

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Problem 3.52 [Difficulty: 3] Given: Geometry of plane gate W h L = 3 m dF y L /2 w = 2 m Find: Minimum weight to keep it closed Solution: Basic equation F R A p d = dp dh ρ g = Σ M O 0 = or, use computing equations F R p c A = y' y c I xx Ay c + = Assumptions: static fluid; ρ = constant; p atm on other side; door is in equilibrium Instead of using either of these approaches, we note the following, using y as in the sketch Σ M O 0 = W L 2 cos θ () F y d = We also have dF p dA = with p ρ g h = ρ g y sin θ () = (Gage pressure, since p = p atm on other side) Hence W 2 L cos θ ()
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Unformatted text preview: A y p ⋅ ⌠ ⎮ ⎮ ⌡ d ⋅ = 2 L cos θ ( ) ⋅ y y ρ ⋅ g ⋅ y ⋅ sin θ ( ) ⋅ w ⋅ ⌠ ⎮ ⎮ ⌡ d ⋅ = W 2 L cos θ ( ) ⋅ A y p ⋅ ⌠ ⎮ ⎮ ⌡ d ⋅ = 2 ρ ⋅ g ⋅ w ⋅ tan θ ( ) ⋅ L L y y 2 ⌠ ⎮ ⌡ d ⋅ = 2 3 ρ ⋅ g ⋅ w ⋅ L 2 ⋅ tan θ ( ) ⋅ = Using given data W 2 3 1000 ⋅ kg m 3 ⋅ 9.81 × m s 2 ⋅ 2 × m ⋅ 3 m ⋅ ( ) 2 × tan 30 deg ⋅ ( ) × N s 2 ⋅ kg m ⋅ × = W 68 kN ⋅ =...
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