2. Physics 151: Lecture 29, Pg 1 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)

3. Physics 151: Lecture 29, Pg 2 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. See text: 14.1

4. Physics 151: Lecture 29, Pg 3 Fluids l What parameters do we use to describe fluids? çDensity See text: 14.1 units : kg/m 3 = 10 -3 g/cm 3  (water) = 1.000 x10 3 kg/m 3 = 1.000 g/cm 3  (ice) = 0.917 x10 3 kg/m 3 = 0.917 g/cm 3  (air) = 1.29 kg/m 3 = 1.29 x10 -3 g/cm 3  (Hg) = 13.6 x10 3 kg/m 3 = 13.6 g/cm 3

5. Physics 151: Lecture 29, Pg 4 Fluids A n l 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: units : 1 N/m 2 = 1 Pa (Pascal) 1 bar = 10 5 Pa 1 mbar = 10 2 Pa 1 torr = 133.3 Pa 1atm = 1.013 x10 5 Pa = 1013 mbar = 760 Torr = 14.7 lb/m 2 (=PSI) l What parameters do we use to describe fluids? çPressure

6. Physics 151: Lecture 29, Pg 5 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 l For an incompressible fluid, the density is the same everywhere, but the pressure is NOT! Pressure vs. Depth Incompressible Fluids (liquids) l 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 y 1 y 2 A F 1 F 2 mg p 1 p 2 0 p See text: 14.2

7. Physics 151: Lecture 29, Pg 6 FIf the pressures were different, fluid would flow in the tube! FHowever, if fluid did flow, then the system was NOT in equilibrium since no equilibrium system will spontaneously leave equilibrium. 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? FImagine a tube that would connect two regions at the same depth. See text: 14.2

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

9. Physics 151: Lecture 29, Pg 8 Example l A U-tube of uniform cross- sectional 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. l Determine the value of h 1.

10. Physics 151: Lecture 29, Pg 9 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.

11. Physics 151: Lecture 29, Pg 10 Pascal’s Principle l So far we have discovered (using Newton’s Laws):  Pressure depends on depth:  p =  g  y l 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. See text: 14.2

12. Physics 151: Lecture 29, Pg 11 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?? See text: 14.2

13. Physics 151: Lecture 29, Pg 12 Pascal’s Principle l Consider F 1 moving through a distance d 1. çHow large is the volume of the liquid displaced? l Therefore the work done by F 1 equals the work done by F 2 We have NOT obtained “something for nothing”. See text: 14.2 çThis volume determines the displacement of the large piston.

14. Physics 151: Lecture 29, Pg 13 A1A1 A 10 M dAdA A2A2 M dBdB A) d A =(1/2)d B B) d A = d B C) d A = 2d B Lecture 29, ACT 2a 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.  If A 2 = 2  A 1, compare d A and d B.

15. Physics 151: Lecture 29, Pg 14 A1A1 A 20 M dCdC A1A1 A 10 M dAdA A) d A = (1/2)d C B) d A = d C C) d A = 2d C Lecture 29, ACT 2b 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.  If A 10 = 2  A 20, compare d A and d C.

16. Physics 151: Lecture 29, Pg 15 Archimedes’ Principle l Suppose we weigh an object in air (1) and in water (2). çHow do these weights compare? W2?W2? W1W1 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. See text: 14.4

17. Physics 151: Lecture 29, Pg 16 l The buoyant force is equal to the difference in the pressures times the area. W2?W2? W1W1 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? Archimedes’ Principle See text: 14.4

18. Physics 151: Lecture 29, Pg 17 Sink or Float? l The buoyant force is equal to the weight of the liquid that is displaced. l If the buoyant force is larger than the weight of the object, it will float; otherwise it will sink. Fmg B y l We can calculate how much of a floating object will be submerged in the liquid: çObject is in equilibrium See text: 14.4 Animation

19. Physics 151: Lecture 29, Pg 18 The Tip of the Iceberg l What fraction of an iceberg is submerged? Fmg B y See text: 14.4

20. Physics 151: Lecture 29, Pg 19 Lecture 29, ACT 3 Buoyancy l 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? styrofoam Pb A) It sinks C)B)D) styrofoam Pb

21. Physics 151: Lecture 29, Pg 20 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? Cup I Cup II (A) Cup I (B) Cup II (C) the same (D) can’t tell See text: 14.4

22. Physics 151: Lecture 29, Pg 21 ACT 3-B Even More Fun With Buoyancy A plastic ball floats in a cup of water with half of its volume submerged. Next some oil (  oil <  ball <  water ) is slowly added to the container until it just covers the ball. l çRelative to the water level, the ball will: water oil (A) move up (B) move down (C) stay in same place See text: 14.4

23. Physics 151: Lecture 29, Pg 22 Recap of today’s lecture l Chapter 14.1-4 çPressure çPascal’s Principle çArchimedes Principle