Problem 3.116 - Problem*3.116[Difficulty 2 Given...

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Problem *3.116 [Difficulty: 2] Given: Rectangular container with constant acceleration Find: Slope of free surface Solution: Basic equation In components x p ρ g x + ρ a x = y p ρ g y + ρ a y = z p ρ g z + ρ a z = We have a y a z = 0 = g x g sin θ () = g y g cos θ = g z 0 = Hence x p ρ g sin θ + ρ a x = (1) y p ρ g cos θ 0 = (2) z p 0 = (3) From Eq. 3 we can simplify from pp x y , z , = to x y , = Hence a change in pressure is given by dp x p dx y p dy + = At the free surface p = const., so dp 0 = x p dx y p dy + = or dy dx x p
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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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