Buoyancy and Archimedes principle. Buoyancy. Archimedes Principle. ρ fluid > ρ body T = W - F B F B. A body floats in any liquid with density

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1 Buoyancy and Archimedes principle Fluid statics What is a fluid Density! Pressure! Fluid pressure and depth Pascal s principle Buoyancy Archimedes principle Lecture 3 Dr Julia Bryant buoyancy.pdf Fluid dynamics! Reynolds number Equation of continuity! surface.pdf not examinable Bernoulli s principle Viscosity and turbulent flow Poiseuille s equation These lecture notes: Another resource: 1 Buoyancy When a solid object is wholly or partly immersed in a fluid, the fluid molecules are continually striking the submerged surface of the object. The forces due to these impacts can be combined into a single force the buoyant force which counteracts the weight. T = W - T W 2 If F b > F g body floats. If F b < F g body sinks. A body floats in any liquid with density ρ fluid > ρ body How can we measure the force due to buoyancy F b F b < F g F g F b > F g 3 4 Archimedes Principle When an object is immersed in a fluid, there is an upward buoyant force equal to the weight of the volume of fluid displaced by the object. This applies to either full or partial immersion (i.e. a sinking or floating object) T = W - T m W Hung Sunk Tension W a t e r d i s p l a c e d Contact T = W - Quiz 5 6 Weight Weight F C = W -

2 Floating: Water displaced Floating: fully submerged Water displaced buoyant force equals weight 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. mg You are floating around a pool drinking a bottle of Kahlua. If you accidently drop the bottle into the water (with the lid on), how much of the Kahlua would you have had to have drunk for the bottle to float back up to you 7 8 How much of the Kahlua would you have had to have drunk for the bottle to float back up to you > mg What is the volume of the bottle and it s contents Full bottle has 350mls of liquid = 350cm 3 = 3.5x10-4 m 3 with a specific gravity 1.15 Small amount of air in the top =πr 2 h= π x (1.5cm) 2 x 2cm = 14 cm 3 = 14 ml Liquid before any is drunk V Kah = 350cm 3 = 3.5x10-4 m 3 ρ Kah = 1150 kg.m -3 Air in the top V air =14 ml =1.4x10-5 m 3 Glass V glass = 150cm 3 with ρ glass = 2500Kg.m 3 Total volume =5.14x10-4 m 3 Buoyant force = ρ water V g = 1000 x 5.14x10-4 x 9.8 = N mg mg = (ρ air V air + ρ Kah V Kah + ρ glass V glass ) x g but some has been drunk V inside bottle = V air + V kah_before = 3.5x x10-5 m 3 = 3.64x10-4 m 3 mg = [1.21 x (3.64x V Kah ) x V Kah x 1.5x10-4 ] x 9.8 N = V Kah N How much glass is there The bottle will float if > mg 375g (empty bottle) with density of 2500kg.m mg 3 ==> V=1.5x10-4 m 3 =150cm > V Kah ==> V Kah < 1.21x10-4 m 3 or V Kah < 121ml Total volume = cm 3 9 (equivalent mass to 139mls water) 10 = 514cm 3 =5.14x10-4 m 3 What is wrong with the picture of the ship on the left Why can ships float 11 A steel ship can encompass a great deal of empty space and so have a large volume and a relatively small density. Weight of ship = weight of water displaced 12

3 The buoyant force is equal to the weight of the water displaced, not the water actually present. The missing water that would have filled the volume of the ship below the waterline is the displaced fluid. 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 Volume of water displaced. This volume is not necessarily the volume present. Weight of ship = weight of water displaced Why is buoyancy due to the weight of fluid displaced An object floats because of the pressure difference between the top and bottom of the object. top A! Hydrometer Hydrometer can be used to measure fluid density. A hydrometer floats due to buoyancy. Higher fluid density => higher buoyant force ρ F h! F = (p bottom p top ) A ρ o bottom = ρ V g ρ water = 1000 kg.m -3 ρ kerosene = 817 kg.m -3 F = (p atm + ρ F g h p atm ) A F = ρ F g h A = ρ F V F g p = p 0 + ρ g h F = m F g = Weight of displaced fluid =>Buoyant force Hydrometer - example If a hydrometer was a rod that had a length of m, cross sectional area was m 2, and mass of kg, (a) How far from the bottom end of the rod should a mark of be placed to indicate the relative density of the water (density kg.m -3 ) One cup has ice cubes floating on it. Which weighs more Quiz (b) If the hydrometer sinks to a depth of 0.229m when placed into an alcohol solution. What is the density of the alcohol solution A Let h be the height to which it is submerged. = ρahg upwards F g = mg downwards ρahg = mg 0.250m In water: h = m = = 0.225m (ρa) ( )( ) h In alcohol solution: ρ = m = = 983 kg.m -3 (Ah) ( )(0.229) 17 18

4 One cup has ice cubes floating on it. Which weighs more Cups weigh the same. Weight of the ice cubes is equal to the buoyant force. The buoyant force is equal to the weight of the water displaced by the ice cubes. This means that the weight that the ice cubes add to the cup is exactly what an amount of water that is equal to that submerged volume of ice cubes would add. One of the cups has ice cubes floating in it. When the ice melts, in which cup is the level higher mg One of the cups has ice cubes floating in it. When the ice melts, in which cup is the level higher The level is the same. The weight of the ice cubes is equal to the weight of the water that would fill the submerged volume of the cubes. When the cubes melt into the water the volume of melted water is exactly equal to the volume of water that the cubes were displacing. A giant clam has a mass of 470 kg and a volume of m 3 lies at the bottom of a freshwater lake. How much force is needed to lift it at constant velocity m = 470 kg V clam = m 3 F T = N g = 9.8 m.s -2 ρ water = 1.0x10 3 kg.m -3 F T + = F G m F T + ΣF = 0 a = 0 A ring weighs N in air and 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 F G F T = F G - F T = m g - ρ water g V displaced F T = m g - ρ water g V clam F T = (470)(9.8) (10 3 )(9.8)(0.350) N F T = N = force to lift 70 full 750ml Kahlua bottles in air! 23 air water 24

5 Weighs N in air and N in water. Volume density What is the ring made of air F Tair = F G water F Twater + = F G Archimedes Principle = weight of water displaced = ρ F V ring g Decrease in weight due to buoyant force = F Gair F Gwater = = 0.294x10-3 N Find the volume of the ring: Since the ring is fully submerged V R = /ρ F g =(0.294x10-3 ) / {(10 3 )(9.8)} m 3 = 3.00x10-8 m 3 Find the density of the ring: ρ R = m R / V R = F gair /(g V R )=(6.327x10-3 ) / {(9.8)(3.0x10-8 )} kg.m -3 = 21.5x10 3 kg.m -3 What is the ring made of V R = 3.0x10-8 m 3 ρ R = 2.2x10 4 kg.m -3 maybe gold 25 REYNOLDS NUMBER A British scientist Osborne Reynolds ( ) established that the nature of the flow depends upon a dimensionless quantity, which is now called the Reynolds number R e = inertial forces ~ ρ v 2/ L R e = ρ v L / η viscous forces ~ ηv/l 2 At low Reynolds number, viscous effects dominate. ρ density of fluid v average flow velocity over the cross section of the pipe L characteristic dimension η viscosity R e = ρ v L / η [R e ] [kg.m -3 ] [m.s -1 ][m] [Pa.s] kg x m x m x s 2.m 2 = [1] m 3 s kg.m.s R e is a dimensionless number As a rule of thumb, for a flowing fluid pascal 1 Pa = 1 N.m -2 newton 1 N = 1 kg.m.s -2 1 Pa.s = kg.m.s -2. m -2.s R e < ~ 2000 laminar flow ~ 2000 < R e < ~ 3000 unstable laminar to turbulent flow R e > ~ 2000 turbulent flow Consider an IDEAL FLUID Fluid motion is very complicated. However, by making some assumptions, we can develop a useful model of fluid behaviour. An ideal fluid is Incompressible the density is constant Irrotational the flow is smooth, no turbulence Nonviscous fluid has no internal friction (η=0) Steady flow the velocity of the fluid at each point is constant in time. Consider the average motion of the fluid at a particular point in space and time. An individual fluid element will follow a path called a flow line. Steady flow is when the pattern of flow lines does not change with time. Streamlines - in steady flow, a bundle of streamlines makes a flow tube. v Velocity of particle is tangent to streamline Streamlines cannot cross