Information about Chapter15 a

Published on March 11, 2014

Author: johnryanrizal

Source: slideshare.net

HOT AIRHOT AIR BALLOONS useBALLOONS use heated air, which isheated air, which is less dense than theless dense than the surrounding air, tosurrounding air, to create an upwardcreate an upward buoyant force.buoyant force. According to Archi-According to Archi- medes’ Principle,medes’ Principle, the buoyant force isthe buoyant force is equal to the weightequal to the weight of the air displacedof the air displaced by the balloon.by the balloon. Fluids at RestFluids at Rest Paul E. Tippens

Objectives: After completing thisObjectives: After completing this module, you should be able to:module, you should be able to: • Define and apply the concepts ofDefine and apply the concepts of densitydensity andand fluid pressurefluid pressure to solve physical problems.to solve physical problems. • Define and apply concepts ofDefine and apply concepts of absoluteabsolute,, gaugegauge,, andand atmosphericatmospheric pressures.pressures. • StateState Pascal’s lawPascal’s law and apply for input andand apply for input and output pressures.output pressures. • State and applyState and apply Archimedes’ PrincipleArchimedes’ Principle to solveto solve physical problems.physical problems.

Mass DensityMass Density 2 kg, 4000 cm3 Wood 177 cm3 45.2 kg ; mass m Density volume V ρ= = Lead: 11,300 kg/mLead: 11,300 kg/m33 Wood: 500 kg/mWood: 500 kg/m33 4000 cm3 Lead Same volume 2 kg Lead Same mass

Example 1:Example 1: The density of steel isThe density of steel is 7800 kg/m7800 kg/m33 .. What is the volume of aWhat is the volume of a 4-kg4-kg block of steel?block of steel? 4 kg 3 4 kg ; 7800 kg/m m m V V ρ ρ = = = V = 5.13 x 10-4 m3V = 5.13 x 10-4 m3 What is the mass if the volume is 0.046 m3 ? 3 3 (7800 kg/m )(0.046 m );m Vρ= = m = 359 kgm = 359 kg

Relative DensityRelative Density The relative density ρr of a material is the ratio of its density to the density of water (1000 kg/m3 ). Steel (7800 kg/m3 ) ρr = 7.80 Brass (8700 kg/m3 ) ρr = 8.70 Wood (500 kg/m3 ) ρr = 0.500 Steel (7800 kg/m3 ) ρr = 7.80 Brass (8700 kg/m3 ) ρr = 8.70 Wood (500 kg/m3 ) ρr = 0.500 Examples:Examples: 3 1000 kg/m x r ρ ρ =

PressurePressure Pressure is the ratio of a force F to the area A over which it is applied: Pressure ; Force F P Area A = = A = 2 cm2 1.5 kg 2 -4 2 (1.5 kg)(9.8 m/s ) 2 x 10 m F P A = = P = 73,500 N/m2P = 73,500 N/m2

The Unit of Pressure (Pascal):The Unit of Pressure (Pascal): A pressure of one pascal (1 Pa) is defined as a force of one newton (1 N) applied to an area of one square meter (1 m2 ). 2 1 Pa = 1 N/mPascal: In the previous example the pressure was 73,500 N/m2 . This should be expressed as: P = 73,500 PaP = 73,500 Pa

Fluid PressureFluid Pressure A liquid or gas cannot sustain a shearing stress - it is only restrained by a boundary. Thus, it will exert a force against and perpendicular to that boundary. • The force F exerted by a fluid on the walls of its container always acts perpendicular to the walls. Water flow shows ⊥ F

Fluid PressureFluid Pressure Fluid exerts forces in many directions. Try to submerse a rubber ball in water to see that an upward force acts on the float. • Fluids exert pressure in all directions. F

Pressure vs. Depth in FluidPressure vs. Depth in Fluid Pressure = force/area ; ; mg P m V V Ah A ρ= = = Vg Ahg P A A ρ ρ = = h mgArea • Pressure at any point in a fluid is directly proportional to the density of the fluid and to the depth in the fluid. P = ρgh Fluid Pressure:

Independence of Shape and Area.Independence of Shape and Area. Water seeks its own level, indicating that fluid pressure is independent of area and shape of its container. • At any depth h below the surface of the water in any column, the pressure P is the same. The shape and area are not factors. • At any depth h below the surface of the water in any column, the pressure P is the same. The shape and area are not factors.

Properties of Fluid PressureProperties of Fluid Pressure • The forces exerted by a fluid on the walls ofThe forces exerted by a fluid on the walls of its container are always perpendicular.its container are always perpendicular. • The fluid pressure is directly proportional toThe fluid pressure is directly proportional to the depth of the fluid and to its density.the depth of the fluid and to its density. • At any particular depth, the fluid pressure isAt any particular depth, the fluid pressure is the same in all directions.the same in all directions. • Fluid pressure is independent of the shape orFluid pressure is independent of the shape or area of its container.area of its container. • The forces exerted by a fluid on the walls ofThe forces exerted by a fluid on the walls of its container are always perpendicular.its container are always perpendicular. • The fluid pressure is directly proportional toThe fluid pressure is directly proportional to the depth of the fluid and to its density.the depth of the fluid and to its density. • At any particular depth, the fluid pressure isAt any particular depth, the fluid pressure is the same in all directions.the same in all directions. • Fluid pressure is independent of the shape orFluid pressure is independent of the shape or area of its container.area of its container.

Example 2.Example 2. A diver is locatedA diver is located 20 m20 m belowbelow the surface of a lake (the surface of a lake (ρρ = 1000 kg/m= 1000 kg/m33 ).). What is the pressure due to the water?What is the pressure due to the water? h ρ = 1000 kg/m3 ∆P = ρgh The difference in pressure from the top of the lake to the diver is: h = 20 m; g = 9.8 m/s2 3 2 (1000 kg/m )(9.8 m/s )(20 m)P∆ = ∆P = 196 kPa∆P = 196 kPa

Atmospheric PressureAtmospheric Pressure atm atm h Mercury P = 0 One way to measure atmospheric pressure is to fill a test tube with mercury, then invert it into a bowl of mercury. Density of Hg = 13,600 kg/m3 Patm = ρgh h = 0.760 m Patm = (13,600 kg/m3 )(9.8 m/s2 )(0.760 m) Patm = 101,300 PaPatm = 101,300 Pa

Absolute PressureAbsolute Pressure Absolute Pressure:Absolute Pressure: The sum of the pressure due to a fluid and the pressure due to atmosphere. Gauge Pressure:Gauge Pressure: The difference between the absolute pressure and the pressure due to the atmosphere: Absolute Pressure = Gauge Pressure + 1 atmAbsolute Pressure = Gauge Pressure + 1 atm h ∆P = 196 kPa 1 atm = 101.3 kPa ∆P = 196 kPa 1 atm = 101.3 kPa Pabs = 196 kPa + 101.3 kPa Pabs = 297 kPaPabs = 297 kPa

Pascal’s LawPascal’s Law Pascal’s Law: An external pressure applied to an enclosed fluid is transmitted uniformly throughout the volume of the liquid. FoutFin AoutAin Pressure in = Pressure outPressure in = Pressure out in out in out F F A A =

Example 3.Example 3. The smaller and larger pistons ofThe smaller and larger pistons of a hydraulic press have diameters ofa hydraulic press have diameters of 4 cm4 cm andand 12 cm12 cm. What input force is required to. What input force is required to lift alift a 4000 N4000 N weight with the output piston?weight with the output piston? Fout Fin AouttAin ;in out out in in in out out F F F A F A A A = = 2 2 (4000 N)( )(2 cm) (6 cm) inF π π = 2 ; 2 D R Area Rπ= = F = 444 NF = 444 N Rin= 2 cm; R = 6 cm

Archimedes’ PrincipleArchimedes’ Principle • An object that is completely or partially submerged in a fluid experiences an upward buoyant force equal to the weight of the fluid displaced. 2 lb 2 lb The buoyant force is due to the displaced fluid. The block material doesn’t matter.

Calculating Buoyant ForceCalculating Buoyant Force FB = ρf gVf Buoyant Force: h1 mg Area h2 FB The buoyant force FB is due to the difference of pressure ∆P between the top and bottom surfaces of the submerged block. 2 1 2 1; ( )B B F P P P F A P P A ∆ = = − = − 2 1 2 1( ) ( )B f fF A P P A gh ghρ ρ= − = − 2 1 2 1( ) ( ); ( )B f fF g A h h V A h hρ= − = − Vf is volume of fluid displaced.

Example 4:Example 4: A 2-kg brass block is attached toA 2-kg brass block is attached to a string and submerged underwater. Find thea string and submerged underwater. Find the buoyant force and the tension in the rope.buoyant force and the tension in the rope. All forces are balanced: mg FB = ρgV T Force diagram FB + T = mg FB = ρwgVw 3 2 kg ; 8700 kg/m b b b b b b m m V V ρ ρ = = = Vb = Vw = 2.30 x 10-4 m3 Fb = (1000 kg/m3 )(9.8 m/s2 )(2.3 x 10-4 m3) FB = 2.25 NFB = 2.25 N

Example 4 (Cont.):Example 4 (Cont.): A 2-kg brass block isA 2-kg brass block is attached to a string and submerged underwater.attached to a string and submerged underwater. Now find the the tension in the rope.Now find the the tension in the rope. mg FB = ρgV T Force diagram FB + T = mg T= mg - FB FB = 2.25 NFB = 2.25 N T = (2 kg)(9.8 m/s2 ) - 2.25 N T = 19.6 N - 2.25 N T = 17.3 NT = 17.3 N This force is sometimes referred to as the apparent weight.

Floating objects:Floating objects: When an object floats, partially submerged, the buoyant force exactly balances the weight of the object. FB mg FB = ρf gVf mx g = ρxVx g ρf gVf = ρxVx g ρf Vf = ρxVx ρf Vf = ρxVxFloating Objects: If Vf is volume of displaced water Vwd, the relative density of an object x is given by: Relative Density: x wd r w x V V ρ ρ ρ = =

Example 5:Example 5: A student floats in a salt lakeA student floats in a salt lake with one-third of his body above the surface.with one-third of his body above the surface. If the density of his body is 970 kg/mIf the density of his body is 970 kg/m33 , what, what is the density of the lake water?is the density of the lake water? 1/3 2/3 Assume the student’s volume is 3 m3 . Vs = 3 m3 ; Vwd = 2 m3 ; ρs = 970 kg/m3 ρw Vwd = ρsVs ρw Vwd = ρsVs 3 w3 32 m ; 3 m 2 s wd s w s V V ρ ρ ρ ρ = = = 3 w 3 3(970 kg/m ) 2 2 sρ ρ = = ρw = 1460 kg/m3ρw = 1460 kg/m3

Problem Solving StrategyProblem Solving Strategy 1. Draw a figure. Identify givens and what is to be1. Draw a figure. Identify givens and what is to be found. Use consistent units for P, V, A, andfound. Use consistent units for P, V, A, and ρρ.. 2.2. Use absolute pressure PUse absolute pressure Pabsabs unless problemunless problem involves a difference of pressureinvolves a difference of pressure ∆∆P.P. 3.3. The difference in pressureThe difference in pressure ∆∆P is determined byP is determined by the density and depth of the fluid:the density and depth of the fluid: 2 1 m F ; = ; P = V A P P ghρ ρ− =

Problem Strategy (Cont.)Problem Strategy (Cont.) 4.4. Archimedes’ Principle: A submerged or floatingArchimedes’ Principle: A submerged or floating object experiences anobject experiences an buoyant forcebuoyant force equal to theequal to the weight of the displaced fluid:weight of the displaced fluid: 4.4. Archimedes’ Principle: A submerged or floatingArchimedes’ Principle: A submerged or floating object experiences anobject experiences an buoyant forcebuoyant force equal to theequal to the weight of the displaced fluid:weight of the displaced fluid: B f f fF m g gVρ= = 5. Remember: m, r and V refer to the displaced fluid. The buoyant force has nothing to do with the mass or density of the object in the fluid. (If the object is completely submerged, then its volume is equal to that of the fluid displaced.)

Problem Strategy (Cont.)Problem Strategy (Cont.) 6.6. For a floating object, FFor a floating object, FBB isis equal to the weight of thatequal to the weight of that object; i.e., the weight of theobject; i.e., the weight of the object is equal to the weight ofobject is equal to the weight of the displaced fluid:the displaced fluid: 6.6. For a floating object, FFor a floating object, FBB isis equal to the weight of thatequal to the weight of that object; i.e., the weight of theobject; i.e., the weight of the object is equal to the weight ofobject is equal to the weight of the displaced fluid:the displaced fluid: xorx f x f fm g m g V Vρ ρ= = FB mg

SummarySummary ; mass m Density volume V ρ= = 3 1000 kg/m x r ρ ρ = Pressure ; Force F P Area A = = 2 1 Pa = 1 N/mPascal: P = ρgh Fluid Pressure:

Summary (Cont.)Summary (Cont.) FB = ρf gVf Buoyant Force:Buoyant Force:Archimedes’Archimedes’ Principle:Principle: in out in out F F A A =Pascal’sPascal’s Law:Law:

CONCLUSION: Chapter 15ACONCLUSION: Chapter 15A Fluids at RestFluids at Rest

Chapter 15, Title 11, United States Code is a section of the United States bankruptcy code that deals with jurisdiction. Under Chapter 15 a representative ...

Read more

Ancillary and Other Cross-Border Cases Chapter 15 is a new chapter added to the Bankruptcy Code by the Bankruptcy Abuse Prevention and Consumer Protection ...

Read more

DEFINITION of 'Chapter 15' A chapter under the U.S. Bankruptcy Code, added to foster a cooperative environment in international insolvencies. Chapter 15's ...

Read more

If you’re having trouble, want to report a bug, provide a suggestion, or just want to say hello — please fill out the form below.

Read more

Oxford University Press USA publishes scholarly works in all academic disciplines, bibles, music, children's books, business books, dictionaries, reference ...

Read more

Lets Play Skylanders Trap Team: Chapter 15 - Skyhighlands w/ Tae Kwon Crow (Dad & Mom Face Cam) - Duration: 1:00:50. by TheSkylanderBoy AndGirl ...

Read more

CHAPTER 15. PROTECTED HANDICAPPED STUDENTS. Sec. 15.1. Purpose. 15.2. Definitions. 15.3. General. 15.4. Annual notice. 15.5. School district initiated ...

Read more

995 Chapter 15 Quality Assurance Chapter Overview Section 15A The Analytical Perspective—Revisited Section 15B Quality Control Section 15C Quality Assessment

Read more

Chapter 15: Caching. A fundamental trade-off in dynamic Web sites is, well, they’re dynamic. Each time a user requests a page, the Web server makes all ...

Read more

## Add a comment