Problem 3.63

# Problem 3.63 - Problem 3.63 Given Geometry of rectangular...

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Problem 3.63 [Difficulty: 3] Given: Geometry of rectangular gate F 1 D L y’ F 2 Find: Depth for gate to open Solution: Basic equation dp dh ρ g = Σ M z 0 = Computing equations F R p c A = y' y c I xx Ay c + = I xx bD 3 12 = Assumptions: Static fluid; ρ = constant; p atm on other side; no friction in hinge For incompressible fluid p ρ g h = where p is gage pressure and h is measured downwards The force on the vertical gate (gate 1) is the same as that on a rectangle of size h = D and width w Hence F 1 p c A = ρ g y c A = ρ g D 2 D w = ρ g w D 2 2 = The location of this force is y' y c I xx c + = D 2 wD 3 12 1 × 2 D × + = 2 3 D = The force on the horizontal gate (gate 2) is due to constant pressure, and is at the centroid
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## This note was uploaded on 10/19/2011 for the course EGN 3353C taught by Professor Lear during the Fall '07 term at University of Florida.

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