Problem 3.63[Difficulty: 3]Given:Geometry of rectangular gateF1DLy’ F2Find:Depth for gate to openSolution:Basic equationdpdhρg⋅=ΣMz0=Computing equationsFRpcA⋅=y'ycIxxAyc⋅+=IxxbD3⋅12=Assumptions:Static fluid; ρ= constant; patmon other side; no friction in hingeFor incompressible fluidpρg⋅h⋅=where p is gage pressure and h is measured downwardsThe force on the vertical gate (gate 1) is the same as that on a rectangle of size h = D and width wHenceF1pcA⋅=ρg⋅yc⋅A⋅=ρg⋅D2⋅D⋅w⋅=ρg⋅w⋅D2⋅2=The location of this force isy'ycIxxc⋅+=D2wD3⋅121⋅×2D×+=23D⋅=The force on the horizontal gate (gate 2) is due to constant pressure, and is at the centroid
This is the end of the preview. Sign up
access the rest of the document.
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.