Physics 151 Lecture 29 Todays Agenda l Todays

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Physics 151: Lecture 29 Today’s Agenda l Today’s topics çFluids under static conditions, Ch.

Physics 151: Lecture 29 Today’s Agenda l Today’s topics çFluids under static conditions, Ch. 14. 1 through 14. 4 çPressure çPascal’s Principle (hydraulic lifts etc. ) çArchimedes’ Principle (floatation) Physics 151: Lecture 29, Pg 1

See text: 14. 1 Fluids l At ordinary temperature, matter exists in one of

See text: 14. 1 Fluids l At ordinary temperature, matter exists in one of three states çSolid - has a shape and forms a surface çLiquid - has no shape but forms a surface çGas - has no shape and forms no surface l What do we mean by “fluids”? çFluids are “substances that flow”…. “substances that take the shape of the container” çAtoms and molecules are free to move. çNo long range correlation between positions. Physics 151: Lecture 29, Pg 2

See text: 14. 1 Fluids l What parameters do we use to describe fluids?

See text: 14. 1 Fluids l What parameters do we use to describe fluids? çDensity units : kg/m 3 = 10 -3 g/cm 3 r(water) = 1. 000 x 103 kg/m 3 = 1. 000 g/cm 3 r(ice) = 0. 917 x 103 kg/m 3 = 0. 917 g/cm 3 r(air) = 1. 29 kg/m 3 = 1. 29 x 10 -3 g/cm 3 r(Hg) = 13. 6 x 103 kg/m 3 = 13. 6 g/cm 3 Physics 151: Lecture 29, Pg 3

Fluids l What parameters do we use to describe fluids? çPressure units : 1

Fluids l What parameters do we use to describe fluids? çPressure units : 1 N/m 2 = 1 Pa (Pascal) 1 bar = 105 Pa 1 mbar = 102 Pa 1 torr = 133. 3 Pa l 1 atm = 1. 013 x 105 Pa = 1013 mbar = 760 Torr = 14. 7 lb/m 2 (=PSI) Any force exerted by a fluid is perpendicular to a surface of contact, and is proportional to the area of that surface. çForce (a vector) in a fluid can be expressed in terms of pressure (a scalar) as: n A Physics 151: Lecture 29, Pg 4

See text: 14. 2 Pressure vs. Depth Incompressible Fluids (liquids) l l l When

See text: 14. 2 Pressure vs. Depth Incompressible Fluids (liquids) l l l 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! p 0 y 1 p 1 F 1 y 2 A p 2 mg F 2 Consider an imaginary fluid volume (a cube, face area A) çThe sum of all the forces on this volume must be ZERO as it is in equilibrium: F 2 - F 1 - mg = 0 Physics 151: Lecture 29, Pg 5

See text: 14. 2 Pressure vs. Depth l For a fluid in an open

See text: 14. 2 Pressure vs. Depth l For a fluid in an open container pressure same at a given depth independent of the container l Fluid level is the same everywhere in a connected container, assuming no surface forces l Why is this so? Why does the pressure below the surface depend only on depth if it is in equilibrium? F Imagine a tube that would connect two regions at the same depth. F If the pressures were different, fluid would flow in the tube! F However, if fluid did flow, then the system was NOT in equilibrium since no equilibrium system will spontaneously leave equilibrium. Physics 151: Lecture 29, Pg 6

Lecture 29, ACT 1 Pressure l What happens with two fluids? ? Consider a

Lecture 29, ACT 1 Pressure l What happens with two fluids? ? Consider a U tube containing liquids of density r 1 and r 2 as shown: çCompare the densities of the liquids: A) r 1 < r 2 B) r 1 = r 2 d. I r 2 r 1 C) r 1 > r 2 Physics 151: Lecture 29, Pg 7

Example l l A U-tube of uniform crosssectional area, open to the atmosphere, is

Example l l A U-tube of uniform crosssectional area, open to the atmosphere, is partially filled with mercury. Water is then poured into both arms. If the equilibrium configuration of the tube is as shown in Figure on the right, with h 2 = 1. 00 cm. Determine the value of h 1. Physics 151: Lecture 29, Pg 8

Example l Figure on the right shows Superman attempting to drink water through a

Example l Figure on the right shows Superman attempting to drink water through a very long straw. With his great strength he achieves maximum possible suction. The walls of the tubular straw do not collapse. l (a) Find the maximum height through which he can lift the water. Physics 151: Lecture 29, Pg 9

See text: 14. 2 Pascal’s Principle l l So far we have discovered (using

See text: 14. 2 Pascal’s Principle l l So far we have discovered (using Newton’s Laws): çPressure depends on depth: Dp = rg. Dy 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. l Pascal’s Principle explains the working of hydraulic lifts çi. e. the application of a small force at one place can result in the creation of a large force in another. çDoes this “hydraulic lever” violate conservation of energy? » Certainly hope not. . Let’s calculate. Physics 151: Lecture 29, Pg 10

See text: 14. 2 Pascal’s Principle l Consider the system shown: çA downward force

See text: 14. 2 Pascal’s Principle l Consider the system shown: çA downward force F 1 is applied to the piston of area A 1. çThis force is transmitted through the liquid to create an upward force F 2. çPascal’s Principle says that increased pressure from F 1 (F 1/A 1) is transmitted throughout the liquid. l F 2 > F 1 : Have we violated conservation of energy? ? Physics 151: Lecture 29, Pg 11

See text: 14. 2 Pascal’s Principle l Consider F 1 moving through a distance

See text: 14. 2 Pascal’s Principle l Consider F 1 moving through a distance d 1. çHow large is the volume of the liquid displaced? çThis volume determines the displacement of the large piston. l Therefore the work done by F 1 equals the work done by F 2 We have NOT obtained “something for nothing”. Physics 151: Lecture 29, Pg 12

Lecture 29, ACT 2 a Hydraulics l Consider the systems shown to the right.

Lecture 29, ACT 2 a Hydraulics l Consider the systems shown to the right. çIn each case, a block of mass M is placed on the piston of the large cylinder, resulting in a difference d. I in the liquid levels. çIf A 2 = 2´A 1, compare d. A and d. B. A) d. A=(1/2)d. B B) d. A = d. B d. A A 1 M A 10 d. B A 2 M A 10 C) d. A = 2 d. B Physics 151: Lecture 29, Pg 13

Lecture 29, ACT 2 b Hydraulics l Consider the systems shown to the right.

Lecture 29, ACT 2 b Hydraulics l Consider the systems shown to the right. çIn each case, a block of mass M is placed on the piston of the large cylinder, resulting in a difference d. I in the liquid levels. çIf A 10 = 2´A 20, compare d. A and d. C. d. A A 10 d. C A 1 A) d. A = (1/2)d. C B) d. A = d. C M M A 20 C) d. A = 2 d. C Physics 151: Lecture 29, Pg 14

See text: 14. 4 Archimedes’ Principle l Suppose we weigh an object in air

See text: 14. 4 Archimedes’ Principle l Suppose we weigh an object in air (1) and in water (2). çHow do these weights compare? W 1 < W 2 W 1 = W 2 W 1 > W 2 çWhy? » Since the pressure at the bottom of the object is greater than that at the top of the object, the water exerts a net upward force, the buoyant force, on the object. W 1 W 2 ? Physics 151: Lecture 29, Pg 15

See text: 14. 4 Archimedes’ Principle l W 1 W 2 ? The buoyant

See text: 14. 4 Archimedes’ Principle l W 1 W 2 ? The buoyant force is equal to the difference in the pressures times the area. Archimedes: The buoyant force is equal to the weight of the liquid displaced. l The buoyant force determines whether an object will sink or float. How does this work? Physics 151: Lecture 29, Pg 16

See text: 14. 4 Sink or Float? l l l The buoyant force is

See text: 14. 4 Sink or Float? l l l The buoyant force is equal to the weight of the liquid that is displaced. If the buoyant force is larger than the weight of the object, it will float; otherwise it will sink. y FB mg We can calculate how much of a floating object will be submerged in the liquid: çObject is in equilibrium Animation Physics 151: Lecture 29, Pg 17

See text: 14. 4 The Tip of the Iceberg l What fraction of an

See text: 14. 4 The Tip of the Iceberg l What fraction of an iceberg is submerged? y FB mg Physics 151: Lecture 29, Pg 18

Lecture 29, ACT 3 Buoyancy l A) A lead weight is fastened to a

Lecture 29, ACT 3 Buoyancy l A) A lead weight is fastened to a large styrofoam block and the combination floats on water with the water level with the top of the styrofoam block as shown. çIf you turn the styrofoam+Pb upside down, what happens? It sinks B) C) Pb styrofoam D) styrofoam Pb Physics 151: Lecture 29, Pg 19

See text: 14. 4 ACT 3 -A More Fun With Buoyancy l Two cups

See text: 14. 4 ACT 3 -A More Fun With Buoyancy l Two cups are filled to the same level with water. One of the two cups has plastic balls floating in it. çWhich cup weighs more? (A) Cup I (B) Cup II (C) the same Cup II (D) can’t tell Physics 151: Lecture 29, Pg 20

See text: 14. 4 ACT 3 -B Even More Fun With Buoyancy l A

See text: 14. 4 ACT 3 -B Even More Fun With Buoyancy l A plastic ball floats in a cup of water with half of its volume submerged. Next some oil (roil < rball < rwater) is slowly added to the container until it just covers the ball. l oi water l çRelative to the water level, the ball will: (A) move up (B) move down (C) stay in same place Physics 151: Lecture 29, Pg 21

Recap of today’s lecture l Chapter 14. 1 -4 çPressure çPascal’s Principle çArchimedes Principle

Recap of today’s lecture l Chapter 14. 1 -4 çPressure çPascal’s Principle çArchimedes Principle Physics 151: Lecture 29, Pg 22