1. Lecture 16Purdue University, Physics 2201 Lecture 16 Fluids PHYSICS 220

2. Lecture 16Purdue University, Physics 2202 States of Matter Solid –Hold Volume –Hold Shape Liquid –Hold Volume –Adapt Shape Gas –Adapt Volume –Adapt Shape Fluids Compressible/uncompressible

3. Lecture 16Purdue University, Physics 2203 Pressure Pressure = F normal / A –Scalar –Units: Pascal (Pa) = N/m 2 Atmosphere: 1 atm = 101.3 kPa 1 atm ~ 10 N/cm 2 ~ 15 lb/in 2 Force due to molecules of fluid colliding with container. Change in Impulse =  p

4. Lecture 16Purdue University, Physics 2204 y Illustration

5. Lecture 16Purdue University, Physics 2205 Atmospheric Pressure Basically weight of atmosphere! Air molecules are colliding with you right now! Pressure = 1  10 5 N/m 2 = 14.7 lbs/in 2 Example 1: Disc r = 2.4 in A =  r 2 = 18.1 in 2 F = P A = (14.7 lbs/in 2 )(18.1 in 2 ) = 266 lbs Example 2: Sphere r = 0.1 m A = 4  r 2 = 0.125 m 2 F = 12,000 N (over 2,500 lbs)!

6. Lecture 16Purdue University, Physics 2206 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. PLPL PUPU In Lafayette PUPU PDPD In Denver

7. Lecture 16Purdue University, Physics 2207 Pascal’s Principle A change in pressure at any point in a confined fluid is transmitted everywhere in the fluid Hydraulic Lift  P = F 1 / A 1 on right  P = F 2 / A 2 on left Since  P 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 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

8. Lecture 16Purdue University, Physics 2208 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 1 g/cm 3 Ice 917 0.917 g/cm 3 (floats on water) Blood 1060 1.060 g/cm 3 (sinks in water) Lead 11,300 11.3 g/cm 3 Copper 8890 8.9 g/cm 3 Mercury 13,600 13.6 g/cm 3 Aluminum 2700 2.7 g/cm 3 Wood 550 0.55 g/cm 3 (floats on water) Air 1.29 0.00129 g/cm 3 Helium 0.18 0.00018 g/cm 3 Uranium 19,000 19.0 g/cm 3 Density

9. Lecture 16Purdue University, Physics 2209 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 Pressure under fluid P = P atmosphere +  gd Basically “weight of air + weight of fluid” y x

10. Lecture 16Purdue University, Physics 22010 iClicker 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 Under water P = P atmosphere +  g h 1 2 P = F/A = mg / A Cup 1 has greater mass, but same area

11. Lecture 16Purdue University, Physics 22011 Barometer: a device to measure atmospheric pressure Pressure at points A and B is the same P A = Atmospheric Pressure P A = P B = 0 +  g h =  g h h p 2 =p atm p 1 =0 AB Pressure and Depth Measure h, determine p atm Example: Mercury  = 13,600 kg/m 3 p atm = 1.05 x 10 5 Pa  h = 0.757 m = 757 mm = 29.80” (for 1 atm )

12. Lecture 16Purdue University, Physics 22012 Suppose you have a barometer with mercury and a barometer with water. How does the height h water compare with the height h mercury ? A) h water is much larger than h mercury B) h water is a little larger than h mercury C) h water is a little smaller than h mercury D) h water is much smaller than h mercury h p 2 =p atm p 1 =0 Question

13. Lecture 16Purdue University, Physics 22013 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 h p a p=0 Evacuate the straw by sucking How high will water rise? no more than h = P a /  g = 10.3m = 33.8 feet no matter how hard you suck!

14. Lecture 16Purdue University, Physics 22014

15. Lecture 16Purdue University, Physics 22015 Manometer A device to measure gas pressure P B = P B’ = P C +  gd  P = P B - P C =  gd The difference in mercury level d is a measure of the pressure difference. Gauge Pressure: P gauge = P abs - P atm

16. Lecture 16Purdue University, Physics 22016 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

17. Lecture 16Purdue University, Physics 22017 iClicker Two dams of equal height prevent water from entering the basin. Compare the net force due to the water on the two dams. A) F A > F B B) F A =F B C) F A < F B A B F = P A, and pressure is  gh. Same pressure, same area same force even though more water in B!

18. Lecture 16Purdue University, Physics 22018 Archimedes’ Principle Determine force of fluid on immersed cube –Draw FBD F B = F 2 – F 1 = P 2 A – P 1 A = (P 2 – P 1 )A =  g d 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

19. Lecture 16Purdue University, Physics 22019 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 F B = mg y

20. Lecture 16Purdue University, Physics 22020 Archimedes’ Principle Does an object float or sink? F = (  fluid -  solid ) gV  fluid >  solid F > 0 object rises  fluid =  solid F = 0 neutral buoyancy  fluid <  solid F < 0 object sinks

21. Lecture 16Purdue University, Physics 22021 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? mg FbFb Mg  F = m a F b – Mg – mg = 0  g V disp = (M+m) g V disp = (M+m) /  h A = (M+m) /  h = (M + m)/ (  A) = (51.2+9)/(1 x 4 x 4) = 3.76 cm M =  plastic V cube = 4x4x4x0.8 = 51.2 g h