PHYSICS 220 Lecture 16 Fluids Lecture 16 Purdue

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PHYSICS 220 Lecture 16 Fluids Lecture 16 Purdue University, Physics 220 1

PHYSICS 220 Lecture 16 Fluids Lecture 16 Purdue University, Physics 220 1

States of Matter • Solid – Hold Volume – Hold Shape Fluids • Liquid

States of Matter • Solid – Hold Volume – Hold Shape Fluids • Liquid – Hold Volume – Adapt Shape Compressible/uncompressible • Gas – Adapt Volume – Adapt Shape Lecture 16 Purdue University, Physics 220 2

Pressure • Pressure = Fnormal / A – Scalar – Units: • Pascal (Pa)

Pressure • Pressure = Fnormal / A – Scalar – Units: • Pascal (Pa) = N/m 2 • Atmosphere: 1 atm = 101. 3 k. Pa • 1 atm ~ 10 N/cm 2 ~ 15 lb/in 2 • Force due to molecules of fluid colliding with container. Change in Impulse = Dp Lecture 16 Purdue University, Physics 220 3

Pascal’s Principle • A change in pressure at any point in a confined fluid

Pascal’s Principle • A change in pressure at any point in a confined fluid is transmitted everywhere in the fluid • Hydraulic Lift DP = F 1 / A 1 on right DP = F 2 / A 2 on left Since DP is same, set equal F 1 / A 1 = F 2 / A 2 F 2 = F 1 (A 2 / A 1) • Can get LARGE force! Volume is conserved A 1 d 1 = A 2 d 2 = d 1 A 1 / A 2 • Work is the SAME: W = Fd F 2 d 2 = F 1(A 2/A 1)*d 1 (A 1/A 2) = F 1 d 1 Lecture 16 Purdue University, Physics 220 4

Gravity and Pressure • Consider a small “piece” of the fluid Draw Forces on

Gravity and Pressure • Consider a small “piece” of the fluid Draw Forces on the fluid element y-direction: P 2 A – mg – P 1 A = 0 P 2 A – ( Ad)g – P 1 A = 0 P 2 – ( d)g – P 1 = 0 P 2 = P 1 + gd y x • Pressure under fluid P = Patmosphere + gd Basically “weight of air + weight of fluid” Lecture 16 Purdue University, Physics 220 5

Density • Mass density of an object of mass m and volume V •

Density • Mass density of an object of mass m and volume V • Density = Mass/Volume = m/V Units = kg/m 3 • Densities of some common things (kg/m 3) Water 1000 Ice 917 0. 917 Blood 1060 Lead 11, 300 Copper 8890 Mercury 13, 600 Aluminum 2700 Wood 550 Air 1. 29 Helium 0. 18 Uranium 19, 000 Lecture 16 1 g/cm 3 (floats on water) 1. 060 g/cm 3 (sinks in water) 11. 3 g/cm 3 8. 9 g/cm 3 13. 6 g/cm 3 2. 7 g/cm 3 0. 55 g/cm 3 (floats on water) 0. 00129 g/cm 3 0. 00018 g/cm 3 19. 0 g/cm 3 Purdue University, Physics 220 6

Pressure versus Depth • For a fluid in an open container: – The pressure

Pressure versus Depth • For a fluid in an open container: – The pressure is the same at a given depth, independent of shape of the container – The fluid level is the same in a connected container Lecture 16 Purdue University, Physics 220 7

i. Clicker • Two identical “light” containers are filled with water. The first is

i. Clicker • Two identical “light” containers are filled with water. The first is completely full of water, the second container is filled only ½ way. Compare the pressure each container exerts on the table. A) P 1 > P 2 B) P 1 = P 2 C) P 1 < P 2 P = F/A 1 = mg / A 2 Cup 1 has greater mass, but same area • Under water P = Patmosphere + g h Lecture 16 Purdue University, Physics 220 8

Atmospheric Pressure • • Basically weight of atmosphere! Air molecules are colliding with you

Atmospheric Pressure • • Basically weight of atmosphere! Air molecules are colliding with you right now! Pressure = 1 105 N/m 2 = 14. 7 lbs/in 2 Example 1: Disc r = 2. 4 in A = p r 2 = 18. 1 in 2 F=PA = (14. 7 lbs/in 2 )(18. 1 in 2) = 266 lbs • Example 2: Sphere r = 0. 1 m A = 4 p r 2 = 0. 125 m 2 F = 12, 000 N (over 2, 500 lbs)! Lecture 16 Purdue University, Physics 220 9

Atmospheric Pressure You buy a bag of potato chips in Lafayette and forget them

Atmospheric Pressure You buy a bag of potato chips in Lafayette and forget them (un-opened) under the seat of your car. You drive out to Denver Colorado to visit a friend for Thanksgiving, and when you get there you discover the lost bag of chips. The odd thing you notice right away is that the bag seems to have inflated like a balloon (i. e. it seems much more round and bouncy than when you bought it). How can you explain this ? Due to the change in height, the air is much thinner in Denver. Thus, there is less pressure on the outside of the bag of chips in Denver, so the bag seems to inflate because the air pressure on the inside is greater than on the outside. PD PL PU PU In Lafayette Lecture 16 Purdue University, Physics 220 In Denver 10

Pressure and Depth Barometer: a device to measure atmospheric pressure Pressure at points A

Pressure and Depth Barometer: a device to measure atmospheric pressure Pressure at points A and B is the same PA = Atmospheric Pressure PA = P B = 0 + g h = g h p 1=0 Measure h, determine patm Example: Mercury = 13, 600 kg/m 3 patm = 1. 05 x 105 Pa h p 2=patm A B h = 0. 757 m = 757 mm = 29. 80” (for 1 atm) Lecture 16 Purdue University, Physics 220 11

Question Suppose you have a barometer with mercury and a barometer with water. How

Question Suppose you have a barometer with mercury and a barometer with water. How does the height hwater compare with the height hmercury? A) B) C) D) hwater is much larger than hmercury hwater is a little smaller than hmercury hwater is much smaller than hmercury p 1=0 p 2=patm Lecture 16 Purdue University, Physics 220 h 12

Question Is it possible to stand on the roof of a five story (50

Question Is it possible to stand on the roof of a five story (50 foot) tall house and drink, using a straw, from a glass on the ground? A) No B) Yes p=0 pa h Evacuate the straw by sucking How high will water rise? no more than h = Pa/ g = 10. 3 m = 33. 8 feet no matter how hard you suck! Lecture 16 Purdue University, Physics 220 13

Lecture 16 Purdue University, Physics 220 14

Lecture 16 Purdue University, Physics 220 14

Manometer A device to measure gas pressure PB = PB’ = PC + gd

Manometer A device to measure gas pressure PB = PB’ = PC + gd P = PB - PC = gd The difference in mercury level d is a measure of the pressure difference. Lecture 16 Purdue University, Physics 220 15

i. Clicker A B Two dams of equal height prevent water from entering the

i. Clicker A B Two dams of equal height prevent water from entering the basin. Compare the net force due to the water on the two dams. A) FA > FB B) FA=FB C) FA< FB F = P A, and pressure is gh. Same pressure, same area same force even though more water in B! Lecture 16 Purdue University, Physics 220 16

Archimedes’ Principle • Determine force of fluid on immersed cube Draw FBD • FB

Archimedes’ Principle • Determine force of fluid on immersed cube Draw FBD • FB = F 2 – F 1 = P 2 A – P 1 A = (P 2 – P 1)A = gd. A = g. V • Buoyant force is weight of displaced fluid! A fluid exerts an upward buoyant force on a submerged object equal in magnitude to the weight of the volume of fluid displaced by the object Lecture 16 Purdue University, Physics 220 17

Sink or Float • The buoyant force is equal to the weight of the

Sink or Float • 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 • We can calculate how much of a floating object will be submerge • Object is in equilibrium FB = mg Lecture 16 y Purdue University, Physics 220 18

Archimedes’ Principle Does an object float or sink? F = ( fluid- solid) g.

Archimedes’ Principle Does an object float or sink? F = ( fluid- solid) g. V fluid > solid fluid = solid fluid < solid Lecture 16 F > 0 object rises F = 0 neutral buoyancy F < 0 object sinks Purdue University, Physics 220 19

Archimedes Example A cube of plastic 4. 0 cm on a side with density

Archimedes Example A cube of plastic 4. 0 cm on a side with density = 0. 8 g/cm 3 is floating in the water. When a 9 gram coin is placed on the block, how much does it sink below water surface? SF=ma Fb – Mg – mg = 0 Fb g Vdisp = (M+m) g h Vdisp = (M+m) / h A = (M+m) / Mg mg h = (M + m)/ ( A) = 4 x 4 x 4 x 0. 8 = (51. 2+9)/(1 x 4) = 3. 76 cm Lecture 16 M = plastic Vcube Purdue University, Physics 220 = 51. 2 g 20

Illustration y Lecture 16 Purdue University, Physics 220 21

Illustration y Lecture 16 Purdue University, Physics 220 21