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Unformatted text preview: MECH 411 Introduction to Fluid Mechanics October 30, 2004 American University of Beirut, Fall 2004-2005 Handout # PS2 Selected Problems from Problem Set 2 Problem 2.97 B A R θ x y y = R sin θ d F=- p r dS ˆ r = r = r = r = cos θ x + sin θ y ˆ ˆ ˆ dS = R d θ r ˆ θ B A R θ x y y = R sin θ d F=- p r dS ˆ d F=- p r dS ˆ r = r = r = r = cos θ x + sin θ y ˆ ˆ ˆ dS = R d θ r ˆ r ˆ θ The gauge pressure at y is p g ( y ) = p ( y ) − p a = ρg ( H − y ) = ρg ( H − R sin θ ) (1) The unit outward normal vector is ˆ n = ˆ r = cos θ ˆ x + sin θ ˆ y (2) The force due to pressure distribution is F = integraldisplay d F = integraldisplay S − p g ( y ) ˆ n d S = integraldisplay d F = − WR integraldisplay π π/ 4 ρg ( H − R sin θ ) (cos θ ˆ x + sin θ ˆ y ) dθ = − ρgWR parenleftBigg R 4 − H √ 2 parenrightBigg ˆ x + ρgWR parenleftBigg R parenleftbigg 3 π 8 + 1 4 parenrightbigg − H parenleftBigg 1 + 1 √ 2 parenrightBiggparenrightBigg ˆ y (3) Noting that the position vector on the gate is...
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This note was uploaded on 10/28/2009 for the course MECH 310 taught by Professor I.l during the Spring '08 term at American University of Beirut.
- Spring '08