Lecture 29 (Walker: ) Fluids II April 13, 2009

Transcription

1 Physics 111 Lecture 29 (Walker: ) Fluids II April 13, 2009 Lecture 29 1/32 Pressure in Fluids Pressure is the same in every direction in a fluid at a given depth; if it were not, the fluid would flow. Lecture 29 2/32

2 Pressure versus Depth in Fluid There is an increase in fluid pressure as we descend further below the top surface. This is due to the increasing weight of fluid us. For an open container of cross-sectional area A and height h filed with fluid of density ρ F bottom = P at A+ ρahg P bottom = F bottom /A Lecture 29 3/32 Pressure vs. Depth The same relation works between any two points separated by vertical distance h. This relation is valid for any liquid whose density does not change with depth. Lecture 29 4/32

3 Example: Find pressure at 100 m below ocean surface. P = P at +ρ W gh P = 101 kpa+(10 3 kg/m 3 )(9.8 N/kg)(100m) = 1081 kpa (about 10 times P at ) Lecture 29 5/32 Pressure and Depth Pressure-depth relation is true in any container where the fluid can flow freely the pressure at the same height will be the same everywhere. Fluid Seeks It s Own Level Lecture 29 6/32

4 Question a) b) c) Three containers a), b), c) each have the same bottom surface area, A, and are each filled with water to the same height h. Pressure as a function of depth = P atm + ρgh. Which container has the greatest pressure of water pushing on the bottom surface? Lecture 29 7/32 Pascal s Principle If an external pressure is applied to a confined fluid, the pressure at every point within the fluid increases by that amount. This principle is used, for example, in hydraulic lifts: Lecture 29 8/32

5 Pascal s Principle: Work-Energy Force F 1 needed on piston of area A 1 = 0.04m 2 to lift 5000N car on piston of area A 2 = 4 m 2? Need pressure P 2 = 5000N/4m 2 = P 1 = F 1 /(0.04m 2 ) Thus F 1 = (0.04m 2 /4m 2 )(5000N)=50N Lecture 29 9/32 What distance d 1 does F 1 need to push down to raise car a distance d 2 =0.1m? Volume of fluid displaced on left, V=d 1 A 1, equals volume increase on right, V=d 2 A 2. So, d 1 =d 2 A 2 /A 2 = 10m. Work done by F 1 is W 1 = F 1 d 1 is same as work done by F 2 : W 2 =F 2 d 2 = d 2 (A 2 F 1 /A 1 )=(d 2 A 2 )(F 1 /A 1 ) =( d 1 A 1 )(F 1 /A 1 )= F 1 d 1 = W 1 Energy Conservation - Small force F 1 pushing through a large displacement d 1 is converted to large force F Lecture 29 2 pushing 10/32 through small displacement d 2.

6 Pascal s Principle in Hydraulic Brakes Lecture 29 11/ Archimedes Principle and Buoyancy A fluid exerts a net upward force on any object it surrounds, called the buoyant force. This force is due to the increased pressure at the bottom of the object compared to the top. Lecture 29 12/32

7 Buoyancy and Archimedes Principle This is an object submerged in a fluid. There is a net force on the object because the pressures at the top and bottom of it are different. The buoyant force is found to be the upward force on the same volume of water: where V is object volume Lecture 29 13/32 Archimedes Principle An object completely immersed in a fluid experiences an upward buoyant force equal in magnitude to the weight of fluid displaced by the object. Buoyant force is the same for a totally submerged object of any size, shape, or density The buoyant force is exerted by the fluid Whether an object sinks or floats depends on relationship between buoyant force and weight Lecture 29 14/32

8 Applications of Archimedes Principle An object floats when it displaces an amount of fluid equal to its weight. Some fraction of object volume will be submerged. Lecture 29 15/32 Iceberg: ρ(ice)<ρ(water) An Iceberg is floating in the ocean. As the iceberg melts, does the ocean level 1) rise, 2) sink, or 3) stay the same. h As ice, the iceberg displaces a volume of water equal in mass to the iceberg s mass. Once it is melted, the iceberg still displaces a volume of water equal in mass to the original iceberg (melting doesn t change the iceberg mass). Lecture 29 16/32

9 Archimedes Principle: Totally Submerged Object The upward buoyant force is F b =ρ fluid g V obj The downward gravitational force is W = mg = ρ obj g V obj The net force is F b -W=(ρ fluid -ρ obj ) gv obj F b Depending on the direction of the net force, the object will either float up or sink! F b Lecture 29 17/32 The net force is F b -W=( W=(ρ fluid The object is less dense than the fluid ρ fluid <ρ obj The object experiences a net upward force fluid -ρ obj obj ) g V obj The object is more dense than the fluid ρ fluid >ρ obj The net force is downward, so the object accelerates downward F b F b Lecture 29 18/32

10 Archimedes Principle: Floating Object For a floating object, the fraction that is submerged depends on the densities of the object and of the fluid. Lecture 29 19/32 Submerged Volume of Floating Object Object of volume V obj and density ρ obj floating in fluid of density ρ f. Weight of object W obj = ρ obj V obj g must equal F b = ρ f V sub g Thus: V sub = V obj (ρ obj /ρ f ) For example, object with density 20% that of water will be only 20% submerged. Lecture 29 20/32

11 Applications of Archimedes Principle An object made of material that is denser than water can float only if it has indentations or pockets of air that make its average density less than that of water. Lecture 29 21/32 Buoyancy and Archimedes Principle If an object of density less than that of water is placed under water, there will be an upward net force on it, and it will rise until it is partially out of the water. Lecture 29 22/32

12 Buoyancy and Archimedes Principle This principle also works in the air; this is why hot-air and helium balloons rise. Lecture 29 23/32 End of Lecture 29 For Wednesday, April 15, read Walker Homework Assignment 15b is due at 11:00 PM on Wednesday, April 15. Lecture 29 24/32