Problem *3.116[Difficulty: 2]Given:Rectangular container with constant accelerationFind:Slope of free surfaceSolution:Basic equationIn componentsxp∂∂−ρgx⋅+ρax⋅=yp∂∂−ρgy⋅+ρay⋅=zp∂∂−ρgz⋅+ρaz⋅=We haveayaz=0=gxg sinθ()⋅=gyg−cosθ⋅=gz0=Hencexp∂∂−ρg⋅sinθ⋅+ρax⋅=(1)yp∂∂−ρg⋅cosθ⋅−0=(2)zp∂∂−0=(3)From Eq. 3 we can simplify from ppxy, z, =toxy, =Hence a change in pressure is given bydpxp∂∂dx⋅yp∂∂dy⋅+=At the free surface p = const., sodp0=xp∂∂dx⋅yp∂∂dy⋅+=ordydxxp∂
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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.