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Lecture 16: Fluids I

Lecture 16: Fluids I. Pressure Pascal’s Principle Density Archimedes’ Principle. States of Matter. Solid Hold Volume Hold Shape Liquid Hold Volume Adapt Shape Gas Adapt Volume Adapt Shape. Fluids. Pressure. Force due to molecules of fluid colliding with container. Impulse = D p

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Lecture 16: Fluids I

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  1. Lecture 16:Fluids I • Pressure • Pascal’s Principle • Density • Archimedes’ Principle

  2. States of Matter • Solid • Hold Volume • Hold Shape • Liquid • Hold Volume • Adapt Shape • Gas • Adapt Volume • Adapt Shape Fluids

  3. Pressure • Force due to molecules of fluid colliding with container. • Impulse = Dp • Average Pressure = F / A

  4. Atmospheric Pressure • Basically: weight of the atmosphere! • Air molecules are colliding with you right now! • Pressure = 1x105 N/m2 = 14.7 lbs/in2! • This great pressure often goes unnoticed because the same pressure is on the other side of the object (for example, the same pressure on the outside and inside of an “empty” bottle).

  5. Pascal’s Principle • A change in pressure at any point in a confined fluid is transmitted everywhere in the fluid. • Hydraulic Lift • DP = F1 / A1 on right • DP = F2 / A2 on left • Since DP is same, set equal • P1 = P2 • F1 / A1= F2 / A2 • F2 = F1 (A2 / A1) • Can get LARGE forces!

  6. Density • Density = Mass/Volume •  = M / V • Units: kg/m3 • Densities of some common things (kg/m3) • Water 1000 • ice 917 (floats on water) • blood 1060 (sinks in water) • lead 11,300 • Copper 8890 • Mercury 13,600 • Aluminum 2700 • Wood 550 • air 1.29 • Helium 0.18

  7. Gravity and Pressure • Consider a small “piece” of the fluid. • Draw FBD • F = ma • P2 A – mg – P1 A = 0 • P2 A – (rAd)g – P1 A = 0 • P2 – (rd)g – P1 = 0 • P2 = P1 + rgd • Pressure under fluid P = Patmosphere + rgd • Basically: “weight of air + weight of fluid” y x

  8. Barometers: Pressure and Depth p1 = 0 h p2 = patm • Pressure at points A and B is the same. • PA = Atmospheric Pressure • PA = PB • PA = 0 + r g h • PA = r g h A B

  9. Archimedes’ Principle • Determine force of fluid on immersed cube • Draw FBD • FB = F2 – F1 • = P2 A – P1 A • = (P2 – P1)A • = r g d A • FB = r g V • Buoyant Force is weight of the displaced fluid!

  10. Summary • Pressure is force exerted by molecules “bouncing” off container: • P = F/A • Gravity/weight effects pressure: • P = P0 + rgd • Buoyant force is “weight” of displaced fluid: • FB = r g V

  11. Buoyant Force Example • A block of wood (w = 550 kg/m3) with a volume of 0.2 m3 is floating on water (H2O = 1000 kg/m3). How big of a block of lead (l = 11,300 kg/m3) would need to be tied to the wood to completely submerge it underwater? • Draw a FBD: • F = ma • FBw + FBl – Fgw – Fgl = 0 FBw wood FBl Fgw lead Fgl

  12. Buoyant Force Example • A block of wood (w = 550 kg/m3) with a volume of 0.2 m3 is floating on water (H2O = 1000 kg/m3). How big of a block of lead (l = 11,300 kg/m3) would need to be tied to the wood to completely submerge it underwater? • Recall: • FB = fluid Vobject g • Fg = mg = object Vobject g • Simplify: • H2OVwg + H2OVlg – wVwg – lVlg = 0 wood lead

  13. Buoyant Force Example • A block of wood (w = 550 kg/m3) with a volume of 0.2 m3 is floating on water (H2O = 1000 kg/m3). How big of a block of lead (l = 11,300 kg/m3) would need to be tied to the wood to completely submerge it underwater? • Solve for Vl: wood lead = 0.0087 m3

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