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

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

Purdue University, Physics 220 Pressure Pressure = Fnormal / A Scalar Units: Pascal (Pa) = N/m2 Atmosphere: 1 atm = 101.3 kPa 1 atm ~ 10 N/cm2 ~ 15 lb/in2 Force due to molecules of fluid colliding with container. Change in Impulse = Dp Lecture 16 Purdue University, Physics 220

Purdue University, Physics 220 Pascal’s Principle A change in pressure at any point in a confined fluid is transmitted everywhere in the fluid Hydraulic Lift DP = F1 / A1 on right DP = F2 / A2 on left Since DP is same, set equal F1 / A1 = F2 / A2 F2 = F1 (A2 / A1) Can get LARGE force! Volume is conserved A1d1 = A2d2 d2 = d1 A1 / A2 Work is the SAME: W = Fd F2 d2 = F1(A2/A1)*d1 (A1/A2) = F1 d1 Lecture 16 Purdue University, Physics 220

Purdue University, Physics 220 Gravity and Pressure Consider a small “piece” of the fluid Draw Forces on the fluid element y-direction: P2 A – mg – P1 A = 0 P2 A – (rAd)g – P1 A = 0 P2 – (rd)g – P1 = 0 P2 = P1 + rgd Pressure under fluid P = Patmosphere + rgd Basically “weight of air + weight of fluid” y x Lecture 16 Purdue University, Physics 220

Purdue University, Physics 220 Density Mass density of an object of mass m and volume V Density = Mass/Volume  = m/V Units = kg/m3 Densities of some common things (kg/m3) Water 1000 1 g/cm3 Ice 917 0.917 g/cm3 (floats on water) Blood 1060 1.060 g/cm3 (sinks in water) Lead 11,300 11.3 g/cm3 Copper 8890 8.9 g/cm3 Mercury 13,600 13.6 g/cm3 Aluminum 2700 2.7 g/cm3 Wood 550 0.55 g/cm3 (floats on water) Air 1.29 0.00129 g/cm3 Helium 0.18 0.00018 g/cm3 Uranium 19,000 19.0 g/cm3 Lecture 16 Purdue University, Physics 220 1

Purdue University, Physics 220 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

Purdue University, Physics 220 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) P1 > P2 B) P1 = P2 C) P1 < P2 Under water P = Patmosphere + r g h P = F/A = mg / A Cup 1 has greater mass, but same area Example w/ container of water? Double water double weight, calculate using density, see how A disappears. 1 2 Lecture 16 Purdue University, Physics 220

Purdue University, Physics 220 Atmospheric Pressure Basically weight of atmosphere! Air molecules are colliding with you right now! Pressure = 1105 N/m2 = 14.7 lbs/in2 Example 1: Disc r = 2.4 in A = p r2 = 18.1 in2 F = P A = (14.7 lbs/in2 )(18.1 in2) = 266 lbs Example 2: Sphere r = 0.1 m A = 4 p r2 = 0.125 m2 F = 12,000 N (over 2,500 lbs)! Lecture 16 Purdue University, Physics 220

Purdue University, Physics 220 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. PL PU In Lafayette PU PD In Denver Lecture 16 Purdue University, Physics 220

Purdue University, Physics 220 Pressure and Depth Barometer: a device to measure atmospheric pressure Pressure at points A and B is the same PA = Atmospheric Pressure PA = PB = 0 + r g h = r g h h p2=patm p1=0 Measure h, determine patm Example: Mercury  = 13,600 kg/m3 patm = 1.05 x 105 Pa  h = 0.757 m = 757 mm = 29.80” (for 1 atm) A B Lecture 16 Purdue University, Physics 220

Purdue University, Physics 220 Question Suppose you have a barometer with mercury and a barometer with water. How does the height hwater compare with the height hmercury? A) hwater is much larger than hmercury B) hwater is a little larger than hmercury C) hwater is a little smaller than hmercury D) hwater is much smaller than hmercury h p2=patm p1=0 Lecture 16 Purdue University, Physics 220

Purdue University, Physics 220 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 = 10.3m = 33.8 feet no matter how hard you suck! Lecture 16 Purdue University, Physics 220

Purdue University, Physics 220 Lecture 16 Purdue University, Physics 220

Purdue University, Physics 220 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. Start from beginning. Lecture 16 Purdue University, Physics 220

Purdue University, Physics 220 iClicker 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 rgh. Same pressure, same area same force even though more water in B! Lecture 16 Purdue University, Physics 220

Archimedes’ Principle Determine force of fluid on immersed cube Draw FBD FB = F2 – F1 = P2 A – P1 A = (P2 – P1)A = r g d A = r 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

Purdue University, Physics 220 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 y Lecture 16 Purdue University, Physics 220

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 Lecture 16 Purdue University, Physics 220

Purdue University, Physics 220 Archimedes Example A cube of plastic 4.0 cm on a side with density = 0.8 g/cm3 is floating in the water. When a 9 gram coin is placed on the block, how much does it sink below water surface? S F = m a Fb – Mg – mg = 0 r g Vdisp = (M+m) g Vdisp = (M+m) / r h A = (M+m) / r h = (M + m)/ (r A) = (51.2+9)/(1 x 4 x 4) = 3.76 cm mg Fb Mg h Transparency M = rplastic Vcube = 4x4x4x0.8 = 51.2 g Lecture 16 Purdue University, Physics 220

Purdue University, Physics 220 Illustration y Lecture 16 Purdue University, Physics 220