PHYSICS 220 States of Matter Lecture 16 Fluids Fluids Solid Hold Volume Hold Sape Liquid Hold Volume Adapt Sape Gas Adapt Volume Adapt Sape Lecture 16 Purdue University, Pysics 220 1 Lecture 16 Purdue University, Pysics 220 2 Pressure Pressure = F normal / A Scalar Units: Pascal (Pa) = N/m 2 Atmospere: 1 atm = 101.3 kpa = 1.013 x 10 5 Pa 1 atm ~ 10 N/cm 2 ~ 15 lb/in 2 Force due to molecules of fluid colliding wit container. Cange in Impulse =!p Illustration!p! $ mv average vertical force = f = y # $ y = y!t!t " % ' & ' y Lecture 16 Purdue University, Pysics 220 3 Lecture 16 Purdue University, Pysics 220 4
Atmosperic Pressure Basically weigt of atmospere! Air molecules are colliding wit you rigt now! Pressure = 1"10 5 N/m 2 = 14.7 lbs/in 2 Example 1: Plate 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: Spere r = 0.1 m A = 4 # r 2 = 0.125 m 2 F = 12,000 N (over 2,500 lbs)! Lecture 16 Purdue University, Pysics 220 5 Atmosperic Pressure You buy a bag of potato cips in Lafayette and forget tem (un-opened) under te seat of your car. You drive out to Denver Colorado to visit a friend for Tanksgiving, and wen you get tere you discover te lost bag of cips. Te odd ting you notice rigt away is tat te bag seems to ave inflated like a balloon (i.e. it seems muc more round and bouncy tan wen you bougt it). How can you explain tis? Due to te cange in eigt, te air is muc tinner in Denver. Tus, tere is less pressure on te outside of te bag of cips in Denver, so te bag seems to inflate because te air pressure on te inside is greater tan on te outside. P U P L In Lafayette Lecture 16 Purdue University, Pysics 220 6 P U P D In Denver Demo 2A - 05 Hot water steam in jug wen closed Steam condensed wit cooling reducing te internal pressure and jug crumpled Pascal s Principle A cange in pressure at any point in a confined fluid is transmitted everywere in te fluid Hydraulic Lift!P = F 1 / A 1 on rigt!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 te 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, Pysics 220 8
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 tings (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) add to crib seet 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! = m V Lecture 16 Purdue University, Pysics 220 9 Gravity and Pressure Consider a small piece of te fluid Draw Forces on te 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 atmospere + $gd Basically weigt of air + weigt of fluid Lecture 16 Purdue University, Pysics 220 10 y x Problem Two identical ligt containers are filled wit water. Te first is completely full of water, te second container is filled only! way. Compare te pressure eac container exerts on te table. A) P 1 > P 2 B) P 1 = P 2 C) P 1 < P 2 Pressure and Dept Barometer: a device to measure atmosperic pressure Pressure at points A and B is te same P A = Atmosperic Pressure P A = P B = 0 + $ g = $ g p 1 =0 P = F/A = mg / A 1 2 Cup 1 as greater mass, but same area Under water P = P atmospere + $ g Measure, determine p atm Example: Mercury $ = 13,600 kg/m 3 p atm = 1.01 x 10 5 Pa % = 0.757 m = 757 mm = 29.80 (for 1 atm) p 2 =p atm A B Lecture 16 Purdue University, Pysics 220 11 Lecture 16 Purdue University, Pysics 220 12
iclicker Suppose you ave a barometer wit mercury and a barometer wit water. How does te eigt water compare wit te eigt mercury? Question Is it possible to stand on te roof of a five story (50 foot) tall ouse and drink, using a straw, from a glass on te ground? A) water is muc larger tan mercury B) water is a little larger tan mercury C) water is a little smaller tan mercury D) water is muc smaller tan mercury P a =!g so, = P a!g! mercury = 13.6! water water = 13.6 mercury p 1 =0 p 2 =p atm Lecture 16 Purdue University, Pysics 220 13 A) No B) Yes P a =!g = P a!g Evacuate te straw by sucking How ig will water rise? no more tan = P a /$g = 10.3m = 33.8 feet no matter ow ard you suck! Lecture 16 Purdue University, Pysics 220 14 p=0 p a Manometer A device to measure gas pressure P B = P B = P C +!gd "P = P B - P C =!gd Te difference in mercury level d is a measure of te pressure difference. Gauge Pressure: P gauge = P abs - P atm Lecture 16 Purdue University, Pysics 220 15 Lecture 16 Purdue University, Pysics 220 16
Pressure versus Dept iclicker For a fluid in an open container: Te pressure is te same at a given dept, independent of sape of te container Te fluid level is te same in a connected container A B Two dams of equal eigt prevent water from entering te basin. Compare te net force due to te water on te two dams. A) F A > F B B) F A =F B C) F A < F B Lecture 16 Purdue University, Pysics 220 17 F = P A, and pressure is $g. Same pressure, same area same force even toug more water in B! Lecture 16 Purdue University, Pysics 220 18 Arcimedes 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 weigt of displaced fluid! A fluid exerts an upward buoyant force on a submerged object equal in magnitude to te weigt of te volume of fluid displaced by te object Lecture 16 Purdue University, Pysics 220 19 Lecture 16 Purdue University, Pysics 220 20
Sink or Float Te buoyant force is equal to te weigt of te liquid tat is displaced If te buoyant force is larger tan te weigt of te object, it will float; oterwise it will sink We can calculate ow muc of a floating object will be submerge Object is in equilibrium F B = mg y Arcimedes 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 V displaced V object =! object! liquid Lecture 16 Purdue University, Pysics 220 21 Lecture 16 Purdue University, Pysics 220 22 Arcimedes Example A cube of plastic 4.0 cm on a side wit density = 0.8 g/cm 3 is floating in te water. Wen a 9 gram coin is placed on te block, ow muc does it sink below water surface? & F = m a F b Mg mg = 0 F b $ g V disp = (M+m) g V disp = (M+m) / $ A = (M+m) / $ = (M + m)/ ($ A) = (51.2+9)/(1 x 4 x 4) = 3.76 cm Mg mg M = $ plastic V cube = 4x4x4x0.8 = 51.2 g Lecture 16 Purdue University, Pysics 220 23