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physicsyd.au/teach_res/jp/fluids/wfluids.htm

http://www.physics.usyd.edu.au/teach_res/jp/fluids/wfluids.htm. http://www.physics.usyd.edu.au/teach_res/jp/fluids/. web notes: lect3.ppt buoyancy.pdf surface.pdf – not examinable. BUOYANCY - FLOATING AND SINKING. Why do ice cubes float on water?

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physicsyd.au/teach_res/jp/fluids/wfluids.htm

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  1. http://www.physics.usyd.edu.au/teach_res/jp/fluids/wfluids.htmhttp://www.physics.usyd.edu.au/teach_res/jp/fluids/wfluids.htm http://www.physics.usyd.edu.au/teach_res/jp/fluids/ web notes: lect3.ppt buoyancy.pdf surface.pdf – not examinable

  2. BUOYANCY - FLOATING AND SINKING Why do ice cubes float on water? Why does a hot air balloon rise?

  3. spring tension FT buoyant force FB FG weight FG = FT + FB FB FT fixed

  4. Archimedes' Principle When an object is completely or partially immersed in a fluid, the fluid exerts an upward force on the object equal to the weight of the fluid displaced by the object. buoyant force FB

  5. Floating: partially submerged Weight of object < weight of fluid that can be displaced by object Volume of displaced water < volume of object Weight of liquid displaced by partially submerged object = weight of object Water displaced buoyant force FB FG weight

  6. buoyant force FB FG weight Floating: fully submerged Weight of object = weight of fluid displaced by object Water Volume of displaced water = displaced volume of object Static equilibrium Some fish can remain at a fixed depth without moving by storing gas in their bladder. Submarines take on or discharge water into their ballast tanks to rise or dive

  7. Why can ships float?

  8. A 200 tonne ship enters a lock of a canal. The fit between the sides of the lock is so tight that the weight of the water left in the lock after it closes is much less than the ship's weight. Can the ship float?

  9. Why does an object float? FLOATING: weight of object = buoyant force + FB FG

  10. Object partially submerged: object floats on surface top A bottom ro rF h mass of fluid displaced mF volume of fluid displaced VF mass of object mo volume of object Vo Vo > VF

  11. buoyant force FB FG weight + Object floating on surface Origin of buoyant force FB = (pbottom – ptop) A FB = (patm + rF g h – patm) A FB = rFg h A =rF VF g FB = mF g FG = FB mo g = mF g mo = mF ro Vo = rF VF Vo > VF ro < rF volume of fluid displaced VF = h A

  12. Object fully submerged: object floats under surface top rF A bottom h ro w mass of fluid displaced mF volume of fluid displaced VF mass of object mo volume of object Vo Vo = VF

  13. buoyant force FB FG weight + Object floating under surface Origin of buoyant force FB = (pbottom – ptop) A FB = (patm + rF g h – patm – rFg (h - w)) A FB = rF g w A = rF Vo g = rF VF g FB = mF g FG = FB mo g = mF g mo = mF ro Vo = rF VF Vo = VF ro = rF If o > F FB – FG < 0  object will accelerate downward and sink to the bottom.

  14. cups weigh the same the level is the same

  15. When the oil is poured over the object it displaces some oil. This means it feels a buoyant force from the oil in addition to the buoyant force from the water. Therefore it rises higher

  16. A Cartesian diver can be made by completely filling an empty soft drink bottle with water. A medicine dropper is then placed in the bottle so that a small air pocket is trapped at the top of the dropper. When the bottle is squeezed, the medicine dropper sinks. When the bottle is release, the diver rises. Why does the dropper sink when the bottle is squeezed? Your answer should include a list of the important physical quantities involved, the names and descriptions of the physical principles involved.

  17. Solution Pascals Principle– when pressure is applied to an enclosed fluid, the pressure is transmitted undiminished to every point in the fluid and to every point on the walls of the container.  the pressure exerted on the air pocket is increased Boyle’s Lawp V = constant T = constant p increased  V decreased Density  = m / V V decreased increased  average density of the dropper has increased Archimedes Principle – When an object is floating in a fluid, the fluid exerts an upward force on the object called the buoyancy force. This buoyant force is equal to the weight of the fluid displaced by the body * average density of object  density of surrounding fluid  object floats * average density of object > density of surrounding fluid  object sinks Hence by squeezing the bottle, the average density of the dropper is increased and becomes larger than the surrounding fluid and the dropper sinks.

  18. Flift + FB = FG Flift + FB m a = 0 m = 470 kg FG = 4.61x103 N Vclam = 0.350 m3 g = 9.8 m.s-2 water = 1.0x103 kg.m-3 FG

  19. Vdisplaced = Vclam F = 0 Flift + FB = FG Flift = ? N FB = watergVdisplaced = watergVclam FG = m g Flift = FG – FB = m g - watergVclam Flift = (470)(9.8) – (103)(9.8)(0.350) N Flift = 1.2103 N

  20. A ring weighs 6.32710-3 N in air and 6.03310-3 N when submerged in water. What is the volume of the ring? What is the density of the ring? What is the ring made of? FT + FB water air FT FG FT = FG FT + FB = FG

  21. A ring weighs 6.32710-3 N in air and 6.03310-3 N when submerged in water. What is the volume of the ring? What is the density of the ring? What is the ring made of? FT + FB air FT water FT + FB = FG FG FT = FG FB know FT and FG know FT Archimedes’ Principle FB = weight of water displaced FB = rFVringg weight of ring FG and mass of ring mring rring= mring / Vring Vring = 3.0x10-8 m3rring = 2.2x104 kg.m-3 maybe gold

  22. Practice Problem – see pdf file buuoyancy.pdf A wooden raft has a density of 0.500103 kg.m-3 and dimensions of 3.05 m  6.10 m  0.305 m. How deep does it sink into the water when unloaded? What is the maximum number of 70 kg people can the raft carry before it sinks?

  23. Practice Problem – see pdf file buuoyancy.pdf A hydrometer is a simple instrument used to measure the density of a liquid by measuring how deep it sinks in a liquid. A particular hydrometer was made of a uniform cylindrical rod that could float vertically in a liquid. The length of the rod was 0.250 m and its cross sectional area was 2.0010-4 m2. The mass of the rod was 4.5010-2 kg. To calibrate the hydrometer it is placed into water that had a density of 1.000103 kg.m-3. How far from the bottom end of the rod should a mark of 1.000 be placed to indicate the relative density of the water? The hydrometer sinks to a depth of 0.229 m when placed into an alcohol solution. What is the density of the alcohol solution?

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