Monday, Apr. 19, 2004 PHYS 1441-004, Spring 2004Dr. Jaehoon Yu1PHYS 1441 – Section 004Lecture #21Monday, Apr. 19, 2004Dr. Jaehoon Yu• Buoyant Force and Archimedes’ Principle • Flow Rate and Equation of Continuity• Bernoulli’s Equation• Simple Harmonic MotionQuiz next Wednesday, Apr. 28!Monday, Apr. 19, 2004 PHYS 1441-004, Spring 2004Dr. Jaehoon Yu2Announcements• Quiz stats– Average : 48.4/90 Î 53.8/100– Top score: 80/90– Previous quiz scores: 38.2, 41, 57.9• Next quiz on Wednesday, Apr. 28– Covers: ch. 10.4 – ch.11, including all that are covered in the lecture• Mid-term grade one-on-one discussion– 11 have not done this yet– One more opportunity todayMonday, Apr. 19, 2004 PHYS 1441-004, Spring 2004Dr. Jaehoon Yu3Buoyant Forces and Archimedes’ PrincipleWhy is it so hard to put a beach ball under water while a piece of small steel sinks in the water?The water exerts force on an object immersed in the water. This force is called Buoyant force.How does the Buoyant force work?Let‘s consider a cube whose height is h and is filled with fluid and at its equilibrium. Then the weight Mg is balanced by the buoyant force B.This is called, Archimedes’ principle. What does this mean?The magnitude of the buoyant force always equals the weight of the fluid in the volume displaced by the submerged object.BBMghAnd the pressure at the bottom of the cube is larger than the top by ρgh.P∆Therefore,Where Mg is the weight of the fluid.gF=Mg=AB /=ghρ=BPA∆=ghAρ=Vgρ=BgF=Vgρ=Mg=Monday, Apr. 19, 2004 PHYS 1441-004, Spring 2004Dr. Jaehoon Yu4More Archimedes’ PrincipleLet’s consider buoyant forces in two special cases. Let’s consider an object of mass M, with density ρ0, is immersed in the fluid with density ρf.Case 1: Totally submerged objectThe total force applies to different directions depending on thedifference of the density between the object and the fluid.1. If the density of the object is smaller than the density of the fluid, the buoyant force will push the object up to the surface.2. If the density of the object is larger that the fluid’s, the object will sink to the bottom of the fluid.What does this tell you?The magnitude of the buoyant force isBMghBThe weight of the object isgFTherefore total force of the system isFVgfρ=Mg=Vg0ρ=gFB−=()Vgf 0ρρ−=Monday, Apr. 19, 2004 PHYS 1441-004, Spring 2004Dr. Jaehoon Yu5More Archimedes’ PrincipleLet’s consider an object of mass M, with density ρ0, is in static equilibrium floating on the surface of the fluid with density ρf, and the volume submerged in the fluid is Vf.Case 2: Floating objectSince the object is floating its density is always smaller than that of the fluid. The ratio of the densities between the fluid and the object determines the submerged volume under the surface.What does this tell you?The magnitude of the buoyant force isBMghBThe weight of the object isgFTherefore total force of the system isFSince the system is in static equilibriumgVffρgVffρ=Mg=gV00ρ=gFB −=gVgVff 00ρρ−=gV00ρ=fρρ00VVf=0=Monday, Apr. 19, 2004 PHYS 1441-004, Spring 2004Dr. Jaehoon Yu6Example for Archimedes’ PrincipleArchimedes was asked to determine the purity of the gold used in the crown. The legend says that he solved this problem by weighing the crown in air and in water. Suppose the scale read 7.84N in air and 6.86N in water. What should he have to tell the king about the purity of the gold in the crown? In the air the tension exerted by the scale on the object is the weight of the crown airTIn the water the tension exerted by the scale on the object is waterTTherefore the buoyant force B isBSince the buoyant force B isBThe volume of the displaced water by the crown iscVTherefore the density of the crown iscρSince the density of pure gold is 19.3x103kg/m3, this crown is either not made of pure gold or hollow. mg=N84.7=Bmg−=N86.6=waterairTT −=N98.0=gVwwρ=gVcwρ=N98.0=wV=gNwρ98.0=34100.18.9100098.0m−×=×=ccVm=gVgmcc=gVc84.7=334/103.88.9100.184.7mkg×=××=−Monday, Apr. 19, 2004 PHYS 1441-004, Spring 2004Dr. Jaehoon Yu7Example for Buoyant ForceWhat fraction of an iceberg is submerged in the sea water?Let’s assume that the total volume of the iceberg is Vi. Then the weight of the iceberg Fgiis Since the whole system is at its static equilibrium, we obtaingiFLet’s then assume that the volume of the iceberg submerged in the sea water is Vw. The buoyant force B caused by the displaced water becomes BgViiρTherefore the fraction of the volume of the iceberg submerged under the surface of the sea water isiwVVAbout 90% of the entire iceberg is submerged in the water!!!gViiρ=gVwwρ=gVwwρ=wiρρ=890.0/1030/91733==mkgmkgMonday, Apr. 19, 2004 PHYS 1441-004, Spring 2004Dr. Jaehoon Yu8Flow Rate and the Equation of ContinuityStudy of fluid in motion: Fluid DynamicsIf the fluid is water: •Streamline or Laminar flow: Each particle of the fluid follows a smooth path, a streamline•Turbulent flow: Erratic, small, whirlpool-like circles called eddy current or eddies which absorbs a lot of energyTwo main types of flowWater dynamics?? Hydro-dynamics Flow rate: the mass of fluid that passes a given point per unit time/mt∆∆since the total flow must be conserved1mt∆=∆11Vtρ∆=∆11 1Altρ∆=∆111Avρ111Avρ=1mt∆=∆Equation of Continuity2mt∆∆222AvρMonday, Apr. 19, 2004 PHYS 1441-004, Spring 2004Dr. Jaehoon Yu9Example for Equation of ContinuityHow large must a heating duct be if air moving at 3.0m/s along it can replenish the air every 15 minutes, in a room of 300m3volume? Assume the air’s density remains constant.Using equation of continuity111Avρ=Since the air density is constant11Av =Now let’s imagine the room as the large section of the duct1A=221/Altv=21Vvt=⋅23000.113.0 900m=×222Avρ22Av221Avv=Monday, Apr. 19, 2004 PHYS 1441-004, Spring 2004Dr. Jaehoon Yu10Bernoulli’s EquationBernoulli’s Principle: Where the velocity of fluid is high, the pressure is low, and where the velocity is low, the pressure is high. Amount of work done by the force, F1, that exerts pressure, P1, at point 11W=Work done by the gravitational force to move the fluid mass, m, from y1to y2is11Fl∆=11 1PAl∆Amount of work done on the other section of the fluid is2W=3W=()21mg y y−−22 2PAl−∆Monday, Apr. 19, 2004 PHYS 1441-004, Spring 2004Dr. Jaehoon Yu11Bernoulli’s Equation cont’dThe net work done on the fluid isW =11 1PAl=∆From the work-energy principle2212mvSince
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