Physics 231 Lecture 24
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1 Physics 31 Lecture 4 Main points of today s lecture: uoyancy y y ernouli s equation 1 P g V W F displaced f displaced gy v P gy v P f f f Viscous flow Poisseuille s law d Av F L P P R t V 8 1 4
2 ulk modulus example A steel block with a volume of 8 cm 3 is dropped in the ocean and comes to rest at the bottom, which is 3 km below the surface. 1) What is the pressure at the bottom of the ocean? ) y what amount V is the volume decreased. (The ulk modulus for steel is 1.6x10 11 N/m. The density of water is 1000kg/m 3. 1 atm. =1.01x10 5 N/m ) a) x10 5 N/m 1) Pbot Ptop gh b) 4x10 5 N/m 5 bot 3 c) x10 6 N/m P 1.0x10 Pa 1000kg / m 9.8m / s 3000m d) 3x10 7 N/m P 7 bot 3.0x10 N/m V ) P V V V P V P P bot top 3 8cm 1.6x10 N / m x10 cm 3x10 N / m 1x10 N / m 7 5
3 Reading Quiz. Gauge pressure is true pressure. Explain. A. larger than. smaller than C. the same as D. None of the above Gauge pressure Comes from the definition P PP gauge atmosphere Slide 13-8
4 Reading Quiz 3. The buoyant force on an object submerged in a liquid depends on A. the object s mass.. the object s volume. C. the density of the liquid. D. both and C. Slide 13-10
5 uoyancy Consider the small volume in the vessel at the right. If the volume is filled with the same fluid in the vessel, it will be in equilibrium. Thus: V F W 0 V in _ box displaced where W watervin _ boxg, therefore F V g water displaced Now, replace the volume of fluid with the same volume of some other material. There will still be the same buoyant force pushing up on the block of material. Example: A kidney weighs 5.7 N in air. If the kidney is completely submerged in water, its apparent weight is 1.6 N. Determine the specific gravity of the kidney. (specific gravity = kidney / water ) watervkidneyg Apparent weight T W F W(1 F / W) W1 kidneyvkidneyg Apparent weight W 1 water / kidney 1 water / kidney Apparent weight/w water / kidney 1Apparent weight/w 1-1.6/ Specific gravity / 1/ kidney water T F
6
7 Example A block of wood floats on the surface of a lake with 60% of its volume below the surface of the water. What is the density of the wood? F V g F water displaced wood wood W V g W watervdisplacedg woodvwood also; V displaced 0.6 V wood 0.6V\ g \ V\ g \ water wood wood wood g wood kg /m water 3
8 Quiz A metal lblock k( ( 786 steel =7.86x10 3 kg/m 3 )fl floats on the top of a pool of mercury ( mercury =1.35x10 4 kg/m 3 ). The fraction of the volume of the block which lies below the surface is a) ) b) 0.5 c) 0.58 d) 0.4 e) 0.75 Vdisplaced fbelowvsteel W F ; W steel Vsteelg; F mercury fbelow Vsteelg steelvsteelg mercuryfbelowvsteelg steel / mercury fbelow x10 kg / m /1.35x10 kg / m.58 f below Hint: mercury plays the role of water. The uoyancy force comes from displacing the mercury. Equate the weight of the steel to the buoyancey force from the mercury.
9 Does the water level go up or down when the anchor is lowered to the bottom of the lake? When resting on the bottom V displaced =V anchor a) The water level goes up. b) The water level goes down. c) The water level goes neither up nor down. When floating in the boat w V displaced g= anchor V anchor g V displaced = anchor / w V anchor
10 Conceptual question A lead weight is fastened on top of a large solid piece of Styrofoam that floats in a container of water. ecause of the weight of the lead, the water line is flush with the top surface of the Styrofoam. If the piece of Styrofoam is turned upside down so that the weight is now suspended underneath it, a) the arrangement sinks. b) the water line is below the top surface of the Styrofoam. c) the water line is still flush with the top surface of the Styrofoam. Same displaced water in both figures!
11 Conceptual quiz A lead weight is fastened to a large solid piece of Styrofoam that t floats in a container of water. ecause of the weight of the lead, the water line is flush with the top surface of the Styrofoam. If the piece of Styrofoam is turned upside down, so that the weight is now suspended underneath it, the water level in the container a) rises. b) drops. c) remains the same. If an object floats: F = w gv displaced =weight of object. The weight doesn t change. V displaced doesn t change. The level doesn t change.
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1/7 2009/11/14 上 午 11:10 Manage this Assignment: Chapter 16 Due: 12:00am on Saturday, July 3, 2010 Note: You will receive no credit for late submissions. To learn more, read your instructor's Grading Policy