9.4 A. Density The density of a substance of uniform composition is defined as its mass per unit volume (or mass:volume ratio)

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1 Page 1 of 24 AP Physics Mechanics Chapter 9-10 Fluid Mechanics - Text Ch. 9 - Select topics Reading pp textbook HW -- #15,17,18,19,20,22,29,32,34,35,36,39,40 Ch 10 - Select topics Reading pp textbook HW -- #1,4,5,6,7,13,15,16,20 Ch 9 Fluid Statics 9.4 Density and Pressure 9.4 A. Density The density of a substance of uniform composition is defined as its mass per unit volume (or mass:volume ratio) m V Note: assume uniform density for all fluids in this chapter Units of Density: mks: kg/m 3, cgs: g/cm 3, chemistry: g/ml **For conversions remember the density of water** Density of water = 1 g/cm 3 = 1 g/ml = 1 kg/l = 1000 kg/m 3 Note: it s no coincidence that the density of water is 1 g/cm 3. Here is original SI definition of the gram: 1 gram = the absolute weight of a volume of pure water equal to the cube of the hundredth part of a metre (1 cm 3 ), and at the temperature of melting ice"

2 ρ water = 1000 kg/m 3 (at 4⁰C, 1 atm) ρ ice = 917 kg/m 3 ρ air = 1.29 kg/m 3 ρ helium =.179 kg/m 3 Page 2 of B. Specific gravity (S) Specific gravity (S) = the ratio of an object s density to the density of water. S object water IF ρ object > ρ water then S > 1 and the object sinks in water. IF ρ object < ρ water then S < 1 and the object floats in water. IF ρ object < ρ air the object floats in air IF ρ object > ρ air the object sinks in air #1) An object has density of 3 x 10 3 kg /m 3. Find its specific gravity. [3]

3 Page 3 of 24 Ex 2) Here s one to ponder: A pint s a pound the world around. Is it true? Well that depends Some approximate English Metric Conversions: 1 L= 2.11 US pints 1 L = 1.76 Imperial pint 1 US pint = 16 fl. oz 1 Imperial pint = 20 oz 1L 1kg 2.2lb 1L 1kg 2.2lb 1USP int x x x 1. 04lb 1P int x x x 1. 25lb 2.11P int 1L 1kg 1.76Pint 1L 1kg A better saying might be: a liter s a kilogram in every land. Done. #3) What is the weight of a gallon of milk? [8.3 lb] 4Quart.94L 1kg 2.2lb 1USGalx x x 8. 3lb 1USGal 1quart 1L 1kg 9.4 C. Pressure Pressure is defined as the ratio of the force to area P F A Units: the N/m 2, a.k.a. the Pa (Pascal), thus 1 N/m 2 = 1 Pa #4) Find the pressure exerted by 1 leg of your chair on the floor. [5 x 10 7 Pa or 70 psi, assuming r=1 cm and m=60 kg]

4 Page 4 of 24 How can this device be used to measure fluid pressure? Note: The force of the fluid at any point on a submerged object is perpendicular to the surface of the object. Note: we do not say the pressure is perpendicular, as pressure is a SCALAR.

5 Page 5 of 24 #5) Hmmm. Water beds. Does anyone still have one? a) Find the weight (in Newtons) of a king-sized water bed that is 2 m x 2 m x 0.3 m : [12000 N] b) Find the pressure the floor exerts on the bed. [3000 Pa] c) If bed is tilted on its side, find the pressure the floor exerts on the bed. [20000 Pa]

6 Page 6 of A) Variation of Pressure with Depth Consider this: A) If a fluid is at rest in a container, all portions of the fluid must be in equilibrium. Therefore: B) All points at the same depth must be at Box of water the same pressure. C) As depth increases, pressure increases. ΣF y = F depth - F air - mg = 0 N or: F depth = F air + mg and since F=PA: Box of water PA depth = PA air + mg and since m = ρv=ρah: PA depth = PA air + (ρah)g canceling like terms: P absolute = P air + ρgh #6) An ehrlenmeyer flask and a beaker, both at sea level, are filled to the same level with water. Point P is at the same depth in both containers. Compare the absolute pressure at point P in each container. Pressure is only affected by height of the liquid above a point, and is not affected by the shape of the container.

7 Page 7 of 24 #7) a) Calculate the absolute pressure at a pool depth of 4 m. Remember: ρ water = 1000 kg/m 3 p atmospheric = 15psi or kpa) (1.413 x 10 5 Pa or kpa or 21 psi) b) What would a Pressure Gauge read at this depth? (4.0x10 4 Pa or 6 psi) What is the relationship between gauge pressure and absolute pressure? Think: what does an air gauge read on a bike tire with NO AIR in it? 0 psi or 0 Pa A Pressure gauge always reads gauge pressure, which is: Gauge Pressure = Absolute pressure Air Pressure And since: P absolute = P air + ρgh ρgh = P absolute - P air Thus: Gauge Pressure = ρgh Note: at 4m, P abs = 21 psi. What if the depth was the deepest SCUBA Dive on record of m P=3.28 x 10 6 Pa or 486 psi. Why don t you get crushed under this pressure? Because your body is made of fluids and exert an equal force outward. The air spaces however, shrink in size because air is compressible, making it eventually impossible to breath at very low depths, thus limiting how far one can SCUBA. The better question is what would happen if you REMOVED the air pressure around the body? (Total Recall Mars Scene) Now go home and do Homework - #15, 17, 18, 19, 20, 22 (ρ ethanol = 806 kg/m 3 )

8 Page 8 of Pressure Measurements with pressure gauges A. The open-tube manometer (Mahna, Mahna) Remember: Gauge Pressure = ρgh P abs > P air P abs = P air P abs < P air P abs = P air + ρgh P abs = P air - ρgh Note: AP B problems deal with + gauge pressure, only B. The Barometer (Draw one be sure to point out P Abs = 0, thus P air = P gauge ) Air pressure of 1.013x10 3 Pa can support a column of mercury of what height? (FBD) ΣF = 0 N = F atm - F Hg or F atm = F Hg F atm = m Hg g and since F=PA P atm A = ρ Hg Vg and since V = Ah P atm A = ρ Hg Ahg P atm = ρ Hg gh since ρ Hg = x 10 3 kg/m 3 h = (1.013x10 5 Pa)/( x 10 3 kg/m 3 )( m/s 2 ) =.7598 mhg

9 9.5 B) Pascal s Principle (Blaise Pascal ) Page 9 of 24 Pressure applied to an enclosed fluid is transmitted undiminished to every point of the fluid and to the walls of the containing vessels (hydraulic press). Derive equation: P 1 = P 2 F 1 = F 2 or F 2 = A 2 F 1 Greater area = greater force A 1 A 2 A 1 Note: V displaced1 = V displaced2 A 1 h 1 = A 2 h 2 or h 2 = A 1 h 1 Greater area = less height A 2 F 1 h 1 = F 2 h 2 (See Video Pascal s Principle- Hydraulics)

10 Page 10 of 24 #8) What is the minimum force exerted on the smaller piston to hold a 1500-kg car if the diameters of the pistons have a 4:1 ratio? (938 N) Now go home and do HW -- #29

11 Page 11 of Buoyant Forces - Archimedes Principle Why is it easy to lift someone in the pool? When using the ladder to climb out of the pool, why do you feel so much heavier (and it s not because your Speedo is wet) Archimedes principle- any body that completely or partially submerged in a fluid is buoyed up by a force equal to the weight of the fluid displaced by the body. Question: Why is there an upward force on an object immersed in a liquid? Since gauge pressure INCREASES with depth, the deeper surface of the object experiences a greater force. For a box of water: ΣF y = 0 N = F 2 F 1 mg or F 2 F 1 = mg F B is defined as F 2 F 1 F B = mg Box of water since F B = PA and P = ρg h F B = (ρg h)a = ρvg = mg In Summary: F B = ρ f Vg = m f g

12 Page 12 of 24 What if the box of water is replaced with a steel object of same volume? F B is still the weight of water displaced, But the steel object will NOT be in equilibrium. ΣF y = F B m steel g or ΣF y = ρ f Vg ρ o Vg Box of steel Case I: Sinking object (ρ o > ρ f ) V f displaced = V o #9) A 1 cm x 1 cm x 1 cm cube of silver is placed in the middle of a large cup of water. (ρ silver =10.5 x 103 kg/m 3 ) a)find the buoyant force on the silver block. F B = ρ f Vg = m f g (0.01 N) b) Does it float or sink comparing forces acting on block? (Hint: FBD) w=m o g = ρ o Vg (0.105 N) c) Describe the motion of the block initially. It accelerates. Find the block's acceleration. (9.05 m/s 2 ) d) If the block is now replaced by a gold block of the same dimension find the buoyant force on the block. F B = ρ f Vg = m f g (0.01 N) e) What happens when it reaches the bottom of the cup? (buoyant force?)

13 Case II: Floating object (ρ o < ρ f ) Page 13 of 24 When an object is floating in a fluid, part of it is submerged. In this case: V f displaced < V o ΣF y = 0 N = F B - m o g or F B = m o g #10) A wooden raft with density ρ raft = 600 kg/m 3 is floating on the water. The dimensions of the raft are 2.5m x 2.5m x 0.1m a) Find buoyant force. Since floating: F B = m o g = ρ o V o g (3750 N) b) Find depth that the raft is below the water level. F B = ρ f V f g = ρ f (Ah)g (0.06m) See any ratio relationship? Since: F B = m o g = ρ o V o g when floating and F B = ρ f V f g always, then: ρ o V o g= ρ f V f g or ρ o V o = ρ f V f (6/10) (1) = (1) (6/10) The density of the raft is 6/10x the density of water, thus it displaces 6/10x its volume in water.

14 Page 14 of 24 #11) a)a person pushes a hollow metal sphere (radius=5 cm) filled with air below the surface of water. Determine force required to hold the sphere below the surface. (for simplicity, assume skin of sphere has negligible mass) (5.24 N) Holding it means ΣF = 0 N, F ext= F B = ρ f V f g b) As the sphere is pushed deeper, ideally, the buoyant force (inc,dec,rs) Although P abs increases with depth, F B = ρ f V f g = (ρ f g h)a. The h is the same! c) In reality, if the incompressible hollow metal sphere is replaced with a balloon, and the balloon is pushed deeper and deeper in the water, the buoyant force will. (Dec, Inc, or RS?) Think: when submerged: F B = ρ f V f g As depth increases, Pressure increases, thus DECREASING the volume of the balloon. Since this displaces less fluid, F B decreases. #12) Estimate a person s buoyant force in air. (0.903 N) ρ f =1.29kg/m 3 Assume m = 70 kg and ρ o = 1000 kg/m 3 Thus: V o = m/ρ o 70/1000 = m 3 F B = ρ f V f g

15 Page 15 of 24 #13) And now the legend behind Archimedes Archimedes( B.C.), The staff scientist for King Hiero II of Syracuse, uncovered a fraud in the manufacture of a golden crown commissioned by the King, like the one to the right. Suspecting that the goldsmith might have replaced some of the gold given to him by an equal weight of silver, Hiero asked Archimedes to determine whether the wreath was pure gold. Because the wreath was a holy object dedicated to the gods, Archimedes could not disturb the wreath in any way. The story goes that Archimedes happened to go to the bath, and on getting into a tub observed that the more his body sank into it the more water ran out over the tub. As this pointed out the way to explain the case in question, he jumped out of the tub and rushed home naked, crying with a loud voice that he had found what he was seeking; he shouted repeatedly in Greek, "Eureka, eureka." meaning "I have found (it), I have found (it)." How did Archimedes supposedly solve this problem? Submerge crown of mass m in water, collect displaced water. Submerge equal mass of pure gold in water, collect displaced water. If Volume displaced is the same, then V C = V Au, and m C =m Au, then ρ C =ρ Au Note: V C > V Au and m C =m Au, so ρ C <ρ Au and Goldsmith was killed. Galileo in 1586 felt that this method would not have been precise enough to show such a small difference in density. He offered this technique as more precise: Given: Spring Scale Reading in Air = 7.84 N Scale Reading in the water = 6.86 N Determine the density of the crown. (m =.784 kg, FB= 0.98 N, V=9.8 x 10-5 m 3, ρ=7960 kg/m 3 NOT GOLD!) Now go home and do HW -- #32, 34, 35, 36, 39, 40 LAB: Density - irregularly shaped object

16 Page 16 of 24 AP Physics Mechanics Chapter 10 Fluid Dynamics - Ch 10 - Select topics Reading pp textbook HW -- #1,4,5,6,7,13,15,16,20 Moving Fluids are really cool. Does a curve ball curve? Does a rising fastball actually rise? Why are knuckle balls so hard to hit? Don t like baseball? How does an airplane fly? During a hurricane, should you open windows a crack to equalize pressure? To understand fluid dynamics, we need to assume something: That we are dealing with an IDEAL FLUID A. Ideal Fluid assumptions 1) Fluid is non viscous no internal friction between adjacent fluid levels 2) Fluid is incompressible -- density is constant 3) Fluid motion is steady velocity, density and pressure at each point in fluid does not change in time 4) Fluid flow is streamlined or laminar. This means it moves without turbulence each element of the fluid has zero angular velocity about its center (no eddy currents) Note: liquids are nearly incompressible, and gases are compressible, but when not confined can be treated as incompressible (such as wind)

17 Page 17 of B. Equation of Continuity for an ideal fluid moving through a pipe of changing diameter: the FLOW RATE (Volume/Time) must be constant (since fluid incompressible). Therefore the volume flowing through cross-section A 1 must be the same as the volume of fluid flowing through cross-section A 2 in the same amount of time. What does this mean about the velocity at A 1 and A 2? Velocity 1(through narrower pipe ) > Velocity 2(through wider pipe). Derive the equation of continuity: Vol 1 Vol 2 t t Becomes: The equation of continuity is: A1 x1 A2 x 2 t t Av A v And since: v x t

18 Page 18 of 24 #1) A pipe under the sink with a 2-cm diameter is reduced to 1-cm diameter to bring water to the dishwasher. If water enters the larger section at 100 cm/s, what is the speed as it exits the smaller hose? Demo: Fun with the garden hose (400 cm/s) #2) A garden hose with a 2-cm diameter is used to fill a 20-liter bucket. If it takes 60 s to fill the bucket, what is the speed at which water leaves the hose? (1 ml = 1 cm 3 ) (use A 1 v 1 = Vol 2 /t) (V=20,000 cm 3, 318 cm/s) Now go home and do HW - # 1, 4, 5

19 Page 19 of C. Bernoulli s Equation - The sum of the pressure, the kinetic energy per unit volume, and the potential energy per unit volume, has the same value at all points along a streamline. Bernoulli s Equation derived: W 1 = +F x 1 = (P 1 A 1 ) x 1 = P 1 V W 2 = -F x 2 = -(P 2 A 2 ) x 2 = -P 2 V Note: F 1 is + due to water flow from the left. F 2 is due to work done by gravity being negative. Wnet = KE + PE P 1 V -P 2 V = (½mv 2 2 -½mv 1 2 )+ (mgh 2 - mgh 1 ) Divide all terms by V: P 1 V -P 2 V = (½mv ½mv 1 2 )+ (mgh 2 - mgh 1 ) Since ρ = m/v: V V V V V V P 1 -P 2 = (½ρv ½ρv 2 1 )+ (ρgh 2 - ρgh 1 ) P 1 + ½ρv ρgh 1 = P 2 +½ρv ρgh 2 Moving terms for 1 & 2 to same side: Or... P + ½ρv 2 + ρgh = Constant

20 Page 20 of 24 Bernoulli s Equation states that: As a fluid moves through a pipe of varying cross sections and elevations, the Pressure and velocity will change along the pipe, but the SUM of the pressure, KE per unit volume, and the PE per unit volume will remain constant. P 1 + ½ρv ρgh 1 = P 2 + ½ρv ρgh 2 Things to consider: 1) When using Bernoulli's equation, it is important to choose 2 points on your diagram and label them points A and B. 2) If the fluid speed increases (assume same height), then the pressure must decrease. Therefore, swiftly moving fluids exert LESS pressure than slowly moving fluids. Practical applications of Bernoulli s Equation: 1. The Venturi Tube- This leads to # 2-3 (Do demo 1-3 first) DEMO 1: Blow under and over a piece of paper. Which gives it lift? DEMO 2: Air over 8.5" x 11" paper DEMO 3: Air between two balloons, or leaf blower and two bowling balls

21 Page 21 of 24 2) Water aspirators in the AP Chem room that produce a partial vacuum using the kinetic energy from the faucet water pressure 3) Household plumbing sink trap and roof ventilation system 4) The Magnus Force Which is the fastball? The Curve ball? How is a Slider thrown so that it curves horizontally? How is a knuckleball thrown? DEMO 4: Big Air Bag DEMO 5: Styrofoam Ball in air stream DEMO 6: Defying Gravity Ping Pong Ball upsidedown in funnel THEN DISCUSS AIRPLANE WING

22 Page 22 of 24 #3) A large open tank filled with liquid of density ρ leaves the bottom of the vessel through a small hole. (Assume A2 >> A1, then fluid level drops very slowly and v 2 =0m/s) a) Find the speed that the liquid emerges from the hole. (Toricelli's Law) b) Find horizontal distance the water travels DEMO: (Toricelli s Law)

23 Page 23 of 24 More Problems with Torricelli s Law: #4) A college student opens a beverage can and notices that, for some reason, there is a small pinhole on the side of the can and the beverage shoots out as shown. Find h. and determine if the can was full when he opened it. (6.96 cm, yes) #5) If wind blows at 30 m/s over the roof of a house, what is the pressure difference between air inside and outside? (Assume the roof is VERY thin) ( P=581Pa or.09 psi) So, During a hurricane, should you open windows a crack to equalize pressure? Now go home and do HW -- #6, 7, 13, 15, 16, 20

24 Page 24 of 24 Remember: Continuity equation and Bernoulli s equation are for when a fluid is moving. In the Bernoulli s equation, what happens when both v 1 = v 2 = 0? The equation boils down to: ΔP = ρg h LAB ACTIVITY: Toricelli s law. Where does the water stream hit the ground? Place cup on ground! (assume that the speed when it leaves the hole is about 90% of the theoretical speed due to viscosity/surface tension/friction) Ch 9-10 and Mechanics. Done.

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