Page 1Physics 207 – Lecture 21Physics 207: Lecture 21, Pg 1Chapter 15, Fluids & ElasticityThis is an actual photo of an iceberg, taken by a rig manager for Global Marine Drilling in St. Johns, Newfoundland. The water wascalm and the sun was almost directly overhead so that the diverPhysics 207: Lecture 21, Pg 2Lecture 21Goals:Goals:••Chapter 15Chapter 15 Understand pressure in liquids and gases Use Archimedes’ principle to understand buoyancy Understand the equation of continuity Use an ideal-fluid model to study fluid flow. Investigate the elastic deformation of solids and liquids••AssignmentAssignment HW10, Due Wednesday, Apr. 14th Thursday: Read all of Chapter 16Page 2Physics 207 – Lecture 21Physics 207: Lecture 21, Pg 3Fluids At ordinary temperature, matter exists in one of three states Solid - has a shape and forms a surface Liquid - has no shape but forms a surface Gas - has no shape and forms no surface What do we mean by “fluids”? Fluids are “substances that flow”…. “substances that take the shape of the container” Atoms and molecules are free to move. No long range correlation between positions.Physics 207: Lecture 21, Pg 4Fluids An intrinsic parameter of a fluid Density Vm=ρunits :kg/m3= 10-3g/cm3 ρ(water) = 1.000 x 103kg/m3 = 1.000 g/cm3ρ(ice) = 0.917 x 103kg/m3 = 0.917 g/cm3ρ(air) = 1.29 kg/m3 = 1.29 x 10-3g/cm3ρ(Hg) = 13.6 x103kg/m3 = 13.6 g/cm3ρ(W or Au) = 19.3 x103kg/m3 = 19.3 g/cm3Page 3Physics 207 – Lecture 21Physics 207: Lecture 21, Pg 5FluidsnFˆAp=rA Any force exerted by a fluid is perpendicular to a surface of contact, and is proportional to the area of that surface. Force (a vector) in a fluid can be expressed in terms of pressure (a scalar) as:A Fp = Another parameter: PressurenˆPhysics 207: Lecture 21, Pg 6What is the SI unit of pressure?A. PascalB. AtmosphereC. BernoulliD. YoungE. p.s.i.Units : 1 N/m2= 1 Pa (Pascal)1 bar = 105Pa1 mbar = 102Pa1 torr = 133.3 Pa1 atm = 1.013 x105Pa= 1013 mbar= 760 Torr= 14.7 lb/ in2(=PSI)Page 4Physics 207 – Lecture 21Physics 207: Lecture 21, Pg 7 When the pressure is small, relative to the bulk modulus of the fluid, we can treat the density as constant independent of pressure:incompressible fluid For an incompressible fluid, the density is the same everywhere, but the pressure is NOT! p(y) = p0- y g ρ Gauge pressure (subtract p0) pGauge= p(y) - p0Pressure vs. DepthIncompressible Fluids (liquids)y1y2Ap1p2F1F2mg0pF2= F1+ m g = F1+ ρVgF2/A = F1/A + ρVg/Ap2= p1- ρg yPhysics 207: Lecture 21, Pg 8Pressure vs. Depth For a uniform fluid in an open container pressure same at a given depth independent of the containerp(y)y Fluid level is the same everywhere in a connected container, assuming no surface forcesPage 5Physics 207 – Lecture 21Physics 207: Lecture 21, Pg 9Pressure Measurements: Barometer Invented by Torricelli A long closed tube is filled with mercury and inverted in a dish of mercury The closed end is nearly a vacuum Measures atmospheric pressure as 1 atm = 0.760 m (of Hg)Physics 207: Lecture 21, Pg 10Exercise Pressure What happens with two fluids?? Consider a U tube containing liquids of density ρ1and ρ2 as shown:Compare the densities of the liquids:(A) ρ1< ρ2(B) ρ1= ρ2(C) ρ1> ρ2ρρρρ1ρρρρ2dIPage 6Physics 207 – Lecture 21Physics 207: Lecture 21, Pg 11Exercise Pressure What happens with two fluids?? Consider a U tube containing liquids of density ρ1and ρ2 as shown: At the red arrow the pressure must be the same on either side. ρ1x = ρ2(d1+ y) Compare the densities of the liquids:(A) ρ1< ρ2(B) ρ1= ρ2(C) ρρρρ1> ρρρρ2ρρρρ1ρρρρ2dIyPhysics 207: Lecture 21, Pg 12Archimedes’ Principle: A Eureka Moment Suppose we weigh an object in air (1) and in water (2).How do these weights compare? W2?W1W1< W2W1= W2W1> W2 Buoyant force is equal to the weight of the fluid displacedPage 7Physics 207 – Lecture 21Physics 207: Lecture 21, Pg 13The Golden Crown In the first century BC the Roman architect Vitruvius related a story of how Archimedes uncovered a fraud in the manufacture of a golden crown commissioned by Hiero II, the king of Syracuse. The crown (corona in Vitruvius’s Latin) would have been in the form of a wreath, such as one of the three pictured from grave sites in Macedonia and the Dardanelles. Hierowould have placed such a wreath on the statue of a god or goddess. Suspecting that the goldsmith might have replaced some of the gold given to him by an equal weight of silver, Hiero asked Archimedes to determine whether the wreath was pure gold. And because the wreath was a holy object dedicated to the gods, he could not disturb the wreath in any way. (In modern terms, he was to perform nondestructive testing). Archimedes’solution to the problem, as described by Vitruvius, is neatly summarized in the following excerpt from an advertisement: The solution which occurred when he stepped into his bath and caused it to overflow was to put a weight of gold equal to the crown, and known to be pure, into a bowl which was filled with water to the brim. Then the gold would be removed and the king’s crown put in, in its place. An alloy of lighter silver would increase the bulk of the crown and cause the bowl to overflow. From http://www.math.nyu.edu/~crorres/Archimedes/Crown/CrownIntro.htmlPhysics 207: Lecture 21, Pg 14Archimedes’ Principle Suppose we weigh an object in air (1) and in water (2). How do these weights compare? W2?W1W1< W2W1= W2W1> W2 Why?Since the pressure at the bottom of the object is greater than that at the top of the object, the water exerts a net upward force, the buoyant force, on the object.Page 8Physics 207 – Lecture 21Physics 207: Lecture 21, Pg 15Sink or Float? The buoyant force is equal to the weight of the liquid that is displaced. If the buoyant force is larger than the weight of the object, it will float; otherwise it will sink.F mgBy We can calculate how much of a floating object will be submerged in the liquid: Object is in equilibriummgFB=objectobjectliquidliquidVgVg⋅⋅=⋅⋅ρρliquidobjectobjectliquidVVρρ=Physics 207: Lecture 21, Pg 16Bar Trickpiece of rockon top of iceWhat happens to the water level when the ice melts?A. It risesB. It stays the sameC. It
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