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Subdivisions of matter

AP Physics B: Ch.11 - Fluid Mechanics. Subdivisions of matter. solids liquids gases rigid will flow will flow dense dense low density and incompressible and incompressible compressible. fluids. condensed matter.

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Subdivisions of matter

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  1. AP Physics B: Ch.11 - Fluid Mechanics. Subdivisions of matter solids liquids gases rigid will flow will flow dense dense low density and incompressible and incompressible compressible fluids condensed matter Q: what about thick liquids and soft solids?

  2. Fluid mechanics Ordinary mechanics Mass and force identified with objects Fluid mechanics Mass and force “distributed”

  3. Density and Pressure Densityr for element of fluid massDM volumeDV for uniform density massM volumeV units kg m-3

  4. Density and Pressure Pressurep force per unit area for uniform force units N m-2 or pascals(Pa) Atmospheric pressure at sea levelp0 on average 101.3 x103 Pa or 101.3 kPa Gauge pressurepg excess pressure above atmospheric p = pg +p0

  5. Density and Pressure atmospheric gauge Gauge pressurepg p = pg +p0 total pressure in excess of atmospheric typical pressures total gauge atmospheric 1.0x105 Pa 0 car tire 3.5x105 Pa 2.5x105 Pa deepest ocean 1.1x108 Pa 1.1x108 Pa best vacuum 10-12 Pa - 100 kPa

  6. Example The can shown has atmospheric pressure outside. The pump reduces the pressure inside to 1/4 atmospheric • What is the gauge pressure inside? • What is the net force on one side? to pump 30 cms 15 cms

  7. Fluids at rest (hydrostatics) surface is in equilibrium Hydrostatic equilibrium laws of mechanical equilibrium pressure just above surface is atmospheric, p0 hence, pressure just below surface must be same, i.e. p0

  8. Fluids at rest (hydrostatics) Hydrostatic equilibrium element of fluid surface area A height Dy laws of mechanical equilibrium S Fy =0 pA - (p+Dp)A - mg = 0 -DpA - rADyg = 0 (p+Dp)A Dp =- rgDy Dy Pressure at depth h at distance h below surface, pressure is larger byrgh mg = rADyg pA p = p0+rgh

  9. Question How far below surface of water must one dive for the pressure to increase by one atmosphere? What is total pressure and what is the gauge pressure, at this depth? ?

  10. Pascal’s principle The pressure at a point in a fluid in static equilibrium depends only on the depth of that point

  11. Pascal’s principle The pressure at a point in a fluid in static equilibrium depends only on the depth of that point Open tube manometer If h=6 cm and the fluid is mercury (r=13600 kg m-3) find the gauge pressure in the tank (ii) Find the absolute pressure if p0 =101.3 kPa

  12. Pascal’s principle The pressure at a point in a fluid in static equilibrium depends only on the depth of that point Barometer Find p0 if h=758 mm

  13. Pascal’s principle The pressure at a point in a fluid in static equilibrium depends only on the depth of that point A change in the pressure applied to an enclosed incompressible fluid is transmitted to every point in the fluid Hydraulic press alternative argument based on conservation of energy work out = work in volume moved is same on each side

  14. Archimedes’s principle When a body is fully or partially submerged in a fluid, a buoyant force from the surrounding fluid acts on the body. The force is upward and has a magnitude equal to the weight of fluid displaced. Fg imagine a hole in the water-a buoyancy force exists fill it with fluid of mass mf and equilibrium will exist Fb=mfg stone more dense than water so sinks Fb Fg wood less dense than water so floats now the water displaced is less -just enough buoyancy force to balance the weight of the wood Fb=Fg

  15. Fb Flotation volume immersed Vi total volume V For object of uniform density r Fb=Fg rfluid Vi g= rVg Vi/V = r/rfluid Fg Example 1 What fraction of an iceberg is submerged? (rice for sea ice =917 kg m-3 and rsea for sea water = 1024 kg m-3) Example 2 A “gold” statue weighs 147 N in vacuum and 139 N when immersed in salt water of density 1024 kg m-3 . What is the density of the “gold”?

  16. Fluid Dynamics The study of fluids in motion. Ideal Fluid 1. Steady flow Velocity of the fluid at any point fixed in space doesn’t change with time. This is called “ laminar flow”, and for such flow the fluid follows “streamlines”. 2. Incompressible We will assume the density is fixed. Accurate for liquids but not so likely for gases. 3. Inviscid “Viscosity” is the frictional resistance to flow. Honey has high viscosity, water has small viscosity. We will assume no viscous losses. Our approach will only be true for low viscosity fluids. turbulent laminar

  17. Equation of continuity Streamlines Conservation of mass in tube of flow means mass of fluid entering A1 in time Dt = mass of fluid leaving A2 in time Dt For incompressible fluid this means volume is also conserved. Volume entering and leaving in time Dtis DV DV = A1 v1 Dt=A2 v2 Dt Therefore A1 v1 = A2 v2 Equation of continuity (Streamline rule) tube of flow

  18. Bernoulli’s equation(Daniel Bernoulli, 1700-1782) For special case of fluid at rest (Hydrostatics!) For special case of height constant (y1=y2) The pressure of a fluid decreases with increasing speed

  19. Proof of Bernoulli’s equation Use work energy theorem work done by external force (pressure) =change in KE + change in PE W=DK+ DU Work done Change in KE Change in PE Note: same volume DV with mass Dm enters A1 as leaves A2in time Dt Work done at A1in timeDt (p1A1)v1 Dt =p1 DV

  20. Problem Titanic had a displacement of 43 000 tonnes. It sank in 2.5 hours after being holed 2 m below the waterline. Calculate the total area of the hole which sank Titanic.

  21. Examples of Bernoulli’s relation at work Venturi meter Aircraft lift

  22. Examples of Bernouilli’s relation at work “spin bowling”

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