Physics 121: Lecture 20 Today's Agenda Announcements Homework 8: due Friday Nov. 11@6: 00 PM Chap8:#7,22,28,33,35,44,45,50,54,61,and65 Today's topics Fluid and solids Pressure measurement Archimedes principle Fluids in motion Physics 121: Lecture 20, Pg
Physics 121: Lecture 20, Pg 1 Physics 121: Lecture 20 Today’s Agenda Announcements Homework 8: due Friday Nov. 11 @ 6:00 PM. Chap. 8: # 7, 22, 28, 33, 35, 44, 45, 50, 54, 61, and 65. Today’s topics Fluid and solids Pressure measurement Archimedes’ principle Fluids in motion
Fluids: Chapter 9 At ordinary temperature, matter exists in one of three states Solid -has a shape and forms a surface Liquid -has no shape but forms a surface Gas-has no shape and forms no surface What do we mean by fluids"? F| uids are“ substances that flow”..."“ substances that take the shape of the container Atoms and molecules are free to move No long range correlation between positions Physics 121: Lecture 20, Pg 2
Physics 121: Lecture 20, Pg 2 Fluids : Chapter 9 At ordinary temperature, matter exists in one of three states Solid - has a shape and forms a surface Liquid - has no shape but forms a surface Gas - has no shape and forms no surface What do we mean by “fluids”? Fluids are “substances that flow”…. “substances that take the shape of the container” Atoms and molecules are free to move. No long range correlation between positions
Fluids What parameters do we use to describe fluids? Density units kg/m3=10-3 g/cm3 Pressure units 1 N/m2= 1 Pa(Pascal 1 bar =105 Pa 1 mbar= 102 Pa 1 torr 133.3Pa Any force exerted by a fluid is perpendicular to a surface of contact, and is proportional to the area of that surface Force(a vector)in a fluid can be expressed in terms of pressure (a scalar)as F=pAn A Physics 121: Lecture 20, Pg 3
Physics 121: Lecture 20, Pg 3 Fluids What parameters do we use to describe fluids? Density Pressure units : kg/m3 = 10-3 g/cm3 units : 1 N/m2 = 1 Pa (Pascal) 1 bar = 105 Pa 1 mbar = 102 Pa 1 torr = 133.3 Pa F = pAn ˆ A n Any force exerted by a fluid is perpendicular to a surface of contact, and is proportional to the area of that surface. Force (a vector) in a fluid can be expressed in terms of pressure (a scalar) as:
Fluids Bulk modulus △ B △∨/V) LIQUID: incompressible (density almost constant GAS: compressible(density depends a lot on pressure) Pb Gas(STP) HO Steel 10410 10 10 10 101l Bulk modulus(Pa=N/m2) Physics 121: Lecture 20, Pg
Physics 121: Lecture 20, Pg 4 Fluids Bulk Modulus ( V / V ) p B − = LIQUID: incompressible (density almost constant) GAS: compressible (density depends a lot on pressure) Bulk modulus (Pa=N/m2 ) Gas (STP) H2O Steel Pb
Pressure vs Depth Incompressible fluids (liquids) When the pressure is much less than the bulk modulus of the fluid we treat the density as constant y1 independent of pressure incompressible fluid A For an incompressible fluid the density is the same everywhere mg F2 but the pressure is not Consider an imaginary fluid volume(a cube, face area A) The sum of all the forces on this volume must be zero as it is in equilibrium: F2-F1-mg=0 mg=Py、01A F=p2A →|p2=p1+pg(y2-y y,)Ag Physics 121: Lecture 20, Pg 5
Physics 121: Lecture 20, Pg 5 When the pressure is much less than the bulk modulus of the fluid, we treat the density as constant independent of pressure: incompressible fluid For an incompressible fluid, the density is the same everywhere, but the pressure is NOT! Pressure vs. Depth Incompressible Fluids (liquids) Consider an imaginary fluid volume (a cube, face area A) The sum of all the forces on this volume must be ZERO as it is in equilibrium: F2 - F1 - mg = 0 y 1 y2 A p 1 p 2 F1 F 2 mg 0 p F2 − F1 = p2 A − p1 A mg = ( y2 − y1 )Ag p p g( y y ) 2 = 1 + 2 − 1
Pressure VS Depth For a fluid in an open container pressure same at a given depth independent of the container p(y Fluid level is the same everywhere in a connected container, assuming no surface fo rces Why is this so? Why does the pressure below the surface depend only on depth if it is in equilibrium Imagine a tube that would connect two regions at the same depth If the pressures were different, fluid would flow in the tube However, if fluid did flow, then the system was Not in equilibrium since no equilibrium system will spontaneously leave equilibrium Physics 121: Lecture 20, Pg 6
Physics 121: Lecture 20, Pg 6 If the pressures were different, fluid would flow in the tube! However, if fluid did flow, then the system was NOT in equilibrium since no equilibrium system will spontaneously leave equilibrium. Pressure vs. Depth For a fluid in an open container pressure same at a given depth independent of the container p(y) y Fluid level is the same everywhere in a connected container, assuming no surface forces Why is this so? Why does the pressure below the surface depend only on depth if it is in equilibrium? Imagine a tube that would connect two regions at the same depth
Lecture 20. ACT 1 Pressure What happens with two fluids? Consider a U tube containing liquids of density p, and p, as shown Compare the densities of the liquids A)P1< p2 B)P1=p2 C)Pl Physics 121: Lecture 20, Pg 7
Physics 121: Lecture 20, Pg 7 Lecture 20, ACT 1 Pressure What happens with two fluids?? Consider a U tube containing liquids of density 1 and 2 as shown: Compare the densities of the liquids: A) 1 2 1 2 dI
Pascals Principle So far we have discovered (using Newtons Laws) Pressure depends on depth:△p=pg△y Pascal's Principle addresses how a change in pressure is transmitted through a fluid Any change in the pressure applied to an enclosed fluid is transmitted to every portion of the fluid and to the walls of the containing vessel Pascal's Principle explains the working of hydraulic lifts e. the application of a small force at one place can result in the creation of a large force in another Does this"hydraulic lever"violate conservation of energy? ) Certainly hope not. Let's calculate Physics 121: Lecture 20, Pg 8
Physics 121: Lecture 20, Pg 8 Pascal’s Principle So far we have discovered (using Newton’s Laws): Pressure depends on depth: p = gy Pascal’s Principle addresses how a change in pressure is transmitted through a fluid. Any change in the pressure applied to an enclosed fluid is transmitted to every portion of the fluid and to the walls of the containing vessel. Pascal’s Principle explains the working of hydraulic lifts i.e. the application of a small force at one place can result in the creation of a large force in another. Does this “hydraulic lever” violate conservation of energy? »Certainly hope not.. Let’s calculate
Pascal's Principle Consider the system shown a downward force F1 is applied to the piston of area A1 This force is transmitted through d he liquid to create an upward force f Pascals principle says that increased pressure from F Anout A A2 F,Al is transmitted through the liquid F A A F2>F1: Have we violated conservation of energy?? Physics 121: Lecture 20, Pg 9
Physics 121: Lecture 20, Pg 9 Pascal’s Principle Consider the system shown: A downward force F1 is applied to the piston of area A1 . This force is transmitted through the liquid to create an upward force F2 . Pascal’s Principle says that increased pressure from F1 (F1 /A1 ) is transmitted throughout the liquid. F F 1 2 d 2 d 1 A1 A2 2 2 1 1 A F A F = 1 2 2 1 A A F = F F2 > F1 : Have we violated conservation of energy??
Pascals Principle F Consider F1 moving through a distance d1 How large is the volume of the ↑d liquid displaced? △V=dA This volume determines the displacement of the large piston A △V2=△Vm A A 2=F2d2= Therefore the work done by F1 equals the work done by We have NoT obtained"something for nothing Physics 121: Lecture 20, Pg 10
Physics 121: Lecture 20, Pg 10 Pascal’s Principle Consider F1 moving through a distance d1 . How large is the volume of the liquid displaced? F F 1 2 d 2 d 1 A1 A2 V1 = d1A1 V2 = V1 2 1 2 1 A A d = d Therefore the work done by F1 equals the work done by F2 We have NOT obtained “something for nothing”. 1 2 1 1 1 2 2 2 2 1 W A A d A A W = F d = F = This volume determines the displacement of the large piston
