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UT Arlington PHYS 1443 - Lecture Notes

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Wednesday, June 28, 2006 PHYS 1443-001, Summer 2006Dr. Jaehoon Yu1PHYS 1443 – Section 001Lecture #16Wednesday, June 28, 2006Dr. Jaehoon Yu• Density and Specific Gravity• Fluid and Pressure• Absolute and Relative Pressure• Pascal’s Law• Buoyant Force and Archimedes’ Principle• Flow Rate and Continuity Equation• Bernoulli’s EquationWednesday, June 28, 2006 PHYS 1443-001, Summer 2006Dr. Jaehoon Yu2Announcements• Reading assignments– CH13 – 9 through 13 – 13• Final exam– Date and time: 8 – 10am, Friday, June 30– Location: SH103– Covers: Ch 9 – 13– No class tomorrowWednesday, June 28, 2006 PHYS 1443-001, Summer 2006Dr. Jaehoon Yu3Density and Specific GravityDensity, ρ (rho), of an object is defined as mass per unit volume ρ≡Unit? Dimension? 3/ mkg][3−MLSpecific Gravity of a substance is defined as the ratio of the density of the substance to that of water at 4.0 oC(ρH2O=1.00g/cm3).SG ≡Unit? Dimension? None None What do you think would happen of a substance in the water dependent on SG?1>SG1<SGSink in the waterFloat on the surface2substanceHOρρM VWednesday, June 28, 2006 PHYS 1443-001, Summer 2006Dr. Jaehoon Yu4Fluid and PressureWhat are the three states of matter?Solid, Liquid, and GasFluid cannot exert shearing or tensile stress. Thus, the only force the fluid exerts on an object immersed in it is the forces perpendicular to the surfaces of the object.AFP ≡How do you distinguish them?By the time it takes for a particular substance to change its shape in reaction to external forces.What is a fluid?A collection of molecules that are randomly arranged and looselybound by forces between them or by the external container.We will first learn about mechanics of fluid at rest, fluid statics. In what ways do you think fluid exerts stress on the object submerged in it?This force by the fluid on an object usually is expressed in the form of the force on a unit area at the given depth, the pressure, defined asNote that pressure is a scalar quantity because it’s the magnitude of the force on a surface area A.What is the unit and dimension of pressure?Expression of pressure for an infinitesimal area dA by the force dF isdFPdA=Unit:N/m2Dim.: [M][L-1][T-2]Special SI unit for pressure is Pascal2/11 mNPa ≡Wednesday, June 28, 2006 PHYS 1443-001, Summer 2006Dr. Jaehoon Yu5Example for PressureThe mattress of a water bed is 2.00m long by 2.00m wide and 30.0cm deep. a) Find the weight of the water in the mattress. The volume density of water at the normal condition (0oC and 1 atm) is 1000kg/m3. So the total mass of the water in the mattress is Since the surface area of the mattress is 4.00 m2, the pressure exerted on the floor ismPTherefore the weight of the water in the mattress is Wb) Find the pressure exerted by the water on the floor when the bed rests in its normal position, assuming the entire lower surface of the mattress makes contact with the floor.MWVρ=kg31020.1300.000.200.21000 ×=×××=mg=N431018.18.91020.1 ×=××=AF=Amg=341095.200.41018.1×=×=Wednesday, June 28, 2006 PHYS 1443-001, Summer 2006Dr. Jaehoon Yu6Variation of Pressure and DepthWater pressure increases as a function of depth, and the air pressure decreases as a function of altitude. Why?If the liquid in the cylinder is the same substance as the fluid, the mass of the liquid in the cylinder is MgAPPA−−0It seems that the pressure has a lot to do with the total mass of the fluid above the object that puts weight on the object.Let’s imagine a liquid contained in a cylinder with height h and the cross sectional area A immersed in a fluid of density ρ at rest, as shown in the figure, and the system is in its equilibrium.The pressure at the depth h below the surface of a fluid open to the atmosphere is greater than atmospheric pressure by ρgh.Therefore, we obtainAtmospheric pressure P0isPaatm510013.100.1 ×=P0APAMghMSince the system is in its equilibriumPVρ=Ahρ=AhgAPPAρ−−=00=ghPρ+=0Wednesday, June 28, 2006 PHYS 1443-001, Summer 2006Dr. Jaehoon Yu7Pascal’s Principle and HydraulicsA change in the pressure applied to a fluid is transmitted undiminished to every point of the fluid and to the walls of the container.The resultant pressure P at any given depth h increases as much as the change in P0. This is the principle behind hydraulic pressure. How?Therefore, the resultant force F2isWhat happens if P0is changed?PSince the pressure change caused by the the force F1applied on to the area A1is transmitted to the F2on an area A2.ghPPρ+=0This seems to violate some kind of conservation law, doesn’t it?d1d2F1A1A2F22FIn other words, the force gets multiplied by the ratio of the areas A2/A1and is transmitted to the force F2on the surface.No, the actual displaced volume of the fluid is the same. And the work done by the forces are still the same.2F11AF=22AF=121Fdd=112FAA=Wednesday, June 28, 2006 PHYS 1443-001, Summer 2006Dr. Jaehoon Yu8Example for Pascal’s PrincipleIn a car lift used in a service station, compressed air exerts a force on a small piston that has a circular cross section and a radius of 5.00cm. This pressure is transmitted by a liquid to a piston that has a radius of 15.0cm. What force must the compressed air exert to lift a car weighing 13,300N? What air pressure produces this force?PUsing the Pascal’s principle, one can deduce the relationship between the forces, the force exerted by the compressed air is1FTherefore the necessary pressure of the compressed air is122AFA=()()24320.051.33 10 1.48 100.15Nππ=××=×11AF=()Pa5231088.11048.1×=0.05×=πWednesday, June 28, 2006 PHYS 1443-001, Summer 2006Dr. Jaehoon Yu9Example for Pascal’s PrincipleEstimate the force exerted on your eardrum due to the water above when you are swimming at the bottom of the pool with a depth 5.0 m.We first need to find out the pressure difference that is being exerted on the eardrum. Then estimate the area of the eardrum to find out the force exerted on the eardrum.0PP −FSince the outward pressure in the middle of the eardrum is the same as normal air pressureEstimating the surface area of the eardrum at 1.0cm2=1.0x10-4m2, we obtainghWρ=Pa4109.40.58.91000 ×=××=()APP0−=N9.4100.1109.444≈×××≈−Wednesday, June 28, 2006 PHYS 1443-001, Summer 2006Dr. Jaehoon Yu10HdyyhExample for Pascal’s PrincipleWater is filled to a height H behind a dam of width w. Determine the resultant force exerted by the water on the dam.Since the water pressure varies as a


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UT Arlington PHYS 1443 - Lecture Notes

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