# Will do air later p p 0 g y pressure and buoyancy eg

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Unformatted text preview: t includes the force contribution from the weight of all the ﬂuid above it. Well known to divers: pressure increases as you dive deeper underwater. The pressure goes up by 1atm for every 10.3m of water depth. y p( y ) Assuming constant dy extra weight of water in the range dy p(y + dy ) = p(y ) + ρgdy Monday, September 30, 2013 density (incompressible) good approx for water, not air. Will do air later. p = p 0 + ρg ∆ y Pressure and Buoyancy E.g. try to submerge a water polo ball underwater: Water p = p 0 + ρg ∆ y Ball pressure a little bigger here than on top net force: Side pressures cancel, bigger pressure below than above, so net buoyant force up. Fb = − Around 250 BC Monday, September 30, 2013 pda = 0 Buoyant force from bigger force on bottom Buoyancy, cont. King to Archimedes: how pure is the gold in my crown? Archimedes mulled, and came to the idea in his bath. Buoyant force: upward, equal to weight of the ﬂuid displaced by the body. E.g. mship g = ρH2 O g Vdisplaced Vdisplaced = mship /ρH2 O Determines how much volume is hidden below the water surface in the picture. Monday, September 30, 2013 Icebergs W = gVtotal ρice What fraction is above the surface? ρice = 917 kg/m3 ρsea Fb = gVsub ρsea Monday, September 30, 2013 Fb = W water (Vtotal − Vsub )/Vtotal = 1 − ( water = 1024 kg/m3 ρice ρsea water ) ≈ 10% Fluid Flow If a ﬂuid is (approximately) incompressible (like water), dm = ρA1 v1 dt = ρA2 v2 dt A1 v 1 = A2 v 2 compressible case: Monday, September 30, 2013 ρ1 A1 v1 = ρ2 A2 v2 Bernoulli Equation Expresses conservation of energy Work done on ﬂuid element: dW = p1 A1 dx1 − p2 A2 dx2 dW = (p1 − p2 )dV (suppose incompressible) 1 2 2 Change of its kinetic energy: dK = ρdV (v2 − v1 ) 2 Change of its potential energy: dU = ρdV g (y2 − y1 ) dW = dK + dU Monday, September 30, 2013 12 p + ρgy + ρv = const. 2 Bernoulli, physically 12 p + ρgy + ρv = const. 2 Monday, September 30, 2013 Large tank with small hole example hole 12 p + ρgy + ρv = const. 2 (Approx.) Since small hole, and big tank. Top: p = patm , y = h, v = 0. On side away from the hole Bottom: p = patm + ρgh, y = 0, v = 0. Hole: ￿ p = patm , y = 0, v = 2gh Equilibrium with outside at the hole. Monday, September 30, 2013 Same velocity as an object dropped from a height h....
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