Problem 3.123 - Problem *3.123 [Difficulty: 3] Given:...

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Problem *3.123 [Difficulty: 3] Given: Cubical box with constant acceleration Find: Slope of free surface; pressure along bottom of box 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 x a x = g x 0 = a y 0 = g y g = a z 0 = g z 0 = Hence x p SG ρ a x = (1) y p SG ρ g = (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 + = (4) At the free surface p = const., so dp 0 = x p dx y p dy + = or dy dx x p y p = a x g = 0.25 g g = Hence at the free surface dy dx 0.25 = The equation of the free surface is then y x 4 C + = and through volume conservation the fluid rise in the rear balances the fluid fall in the front, so at the midpoint the free surface has not moved from the rest position For size L8 0 c m = at the midpoint x L 2 = y L 2 = (box is half filled) L 2 1 4 L 2 C + = C 5 8 L = y 5 8 L x 4 = Combining Eqs 1, 2, and 4
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