DENSITY OF AN IRREGULAR SHAPED OBJECT

MASS 9/28/2017 AP PHYSICS 2 DENSITY UNIT 1 FLUID STATICS AND DYNAMICS CHAPTER 10 FLUIDS AT REST 0.12 0.1 0.08 0.06 0.04 0.02 0 MASS vs. VOLUME y = 1000x 0 0.00002 0.00004 0.00006 0.00008 0.0001 0.00012 VOLUME m = ρ V Density ( ) = slope. slope of a diagonal straight line is constant. m = ρ V ρ = m V kg m 3 DENSITY Density: Slope of mass vs. volume of graph. The ratio of the mass m of a substance divided by the volume V of that substance. Density Mass describe solid objects that have real boundaries. For gases, a much more useful physical quantity than mass of the individual particles is the mass of one unit of volume density. stays the same If volume increases, then density: If mass increases, then density: stays the same Tip DENSITY OF AN IRREGULAR SHAPED OBJECT 1

AMAZING 9 LAYER DENSITY TOWER DENSITY DENSITY OF WATER (4 C) 1000 kg/m 3 SPECIFIC GRAVITY DENSITY 1 The specific gravity of a substance is defined as the ratio of the density of that substance to the density of water at 4 C (1,000 kg/m 3 ). Specific gravity is just a ratio (no units). Substance A has a density of 6 g/cm 3 and substance B has a density of 9 g/cm 3. In order to obtain equal masses of these two substances, what must be the ratio of the volume of A to the volume of B? Platinum ρ = Acetone ρ = Hydrogen ρ = What is the specific gravity of: 21,450 kg m 3 791 kg m 3 21.45 : : 0.791 0.090 kg 0.00009 : m 3 ρ A = m A V A ρ B = m B V B m B = m B ρ A V A = ρ B V B V A V B = ρ A ρ B V A V B = 3 2 DENSITY 2 DENSITY 3 A hollow sphere of negligible mass and radius R is completely filled with a liquid so that its density is ρ. You now enlarge the sphere so its radius is 2R and completely fill it with the same liquid. a. What is the density of the enlarged sphere? b. what must be the ratio of the mass of the first sphere to the mass of the second sphere? ρ 1 = ρ 2 m 1 m 2 = 1 8 A styrofoam sphere of radius R has a density ρ. You now carefully compress the sphere so its radius is R/2. What is the density of the compressed sphere? ρ 2 = 8ρ 1 2

DENSITY 4 PRESSURE Estimate the density of a person. Pressure: 1.Simplify and diagram. 2.Represent mathematically. m = 80 kg A = 0.3 m x 0.1 cm h = 1.8 m ρ = 1481 kg m 3 P = F A N m 2 Pa The ratio of the force F that a fluid exerts perpendicular to a surface area. Pressure is caused by fluid particles colliding elastically with objects in contact with the fluid. The SI unit of pressure is the Pascal (Pa), where 1 Pa = 1 N/m 2. P = F A PRESSURE N m 2 Pa Pressure math model: Operational definition (calculations) Does not explain what causes pressure Macroscopic approach. PRESSURE 1 When a heavy metal block is supported by a cylindrical vertical post of radius R, it exerts a force F on the post. If the radius of the post is increased to 2R: A) what force does the block now exert on the post? B) what is the new average pressure? F 1 = F 2 P 2 = P 1 4 PRESSURE 2 When a box rests on a round sheet of wood on the ground, it exerts an average pressure P on the wood. If the wood is replaced by a sheet that has half the diameter of the original piece, what is the new average pressure? P 2 = 4P 1 ATMOSPHERIC PRESSURE Earth exerts a force on air molecules. The weight of these air molecules all around us (area) is called the atmospheric pressure. Measurements show that the pressure of the atmospheric air at sea level is on average: P atm = 101,000 N m 2 P atm =101,000 Pa Atmosphere: 1.0 atm = P atm =101,000 Pa 3

PRESSURE 3 Estimate the total force that air exerts on the front side of your body, assuming that the pressure of the atmosphere is constant. F AonO P = P A A = 1.8 m x 0.3 m P = 101,000 Pa F AonO P = 54,540 N Pressure exerted by a fluid Pressure exerted by a fluid Take a water bottle and poke four holes at the same height along its perimeter. Parabolic-shaped streams of water shoot out of the holes. The water inside must push out perpendicular to the wall of the bottle. Because the four streams are identically shaped, the pressure at all points at the same depth in the fluid is the same. Pascal's first law An increase in the pressure of a static, enclosed fluid at one place in the fluid, causes a uniform increase in pressure throughout the fluid. Pascal's first law at a microscopic level Particles inside a container move randomly in all directions. When we push harder on one of the surfaces of the container, the fluid becomes compressed. The molecules near that surface collide more frequently with their neighbors, which in turn collide more frequently with their neighbors. The extra pressure exerted at one surface quickly spreads, such that soon there is increased pressure throughout the fluid. Glaucoma A person with glaucoma has closed drainage canals. The buildup of fluid causes increased pressure throughout the eye, including at the retina and optic nerve, which can lead to blindness. 4

PASCAL S FIRST LAW APPLIED TO THE HYDRAULIC LIFT Pressure changes uniformly throughout the liquid, so the pressure under piston 2 is the same as the pressure under piston 1 if they are at the same elevation. HYDRAULIC LIFT: Examples Force piston 1 on liquid Force liquid on piston 2 P 1 = P 2 F 1onL A 1 = F Lon2 A 2 PROBLEM A hydraulic lift has a small piston with surface area 0.0020 m 2 and a larger piston with surface area 0.20 m 2. Piston 2 and the car placed on piston 2 have a combined mass of 1800 kg. P 1 = P 2 P = F A Conservation of Energy W 1 = W 2 W = F d What is the minimal force that piston 1 needs to exert on the fluid to slowly lift the car? (write an expression for force first) How far down will you need to push the piston in order to lift the car 20 cm? (write an expression for distance first) 5

SOLUTION W 1 = W 2 P1 = P 2 F 1 = F F 2 1 d 1 = F 2 d 2 A 1 A 2 F 1 m g = A 1 A 2 F 1 = m g A 1 A 2 F 1 = 176. 4 N d 1 = F 2 d 2 F 1 d 1 = m g d 2 m g A 1 A 2 d 1 = A 2 d 2 A 1 d 1 = 20m..... QUICKLY! OUTCOME Bottle is open Mr. Largo has a soda bottle filled with water. The bottle has been punctured with 3 tacks. The bottle is open. Predict (draw) what will happen when Mr. Largo removes all tacks at the same time. OUTCOME : Experiments...... Yay! Experiment #1 Two tacks on each side of the plastic bottle, one above the other. Pascal's first law fails to explain this pressure variation at different depths below the surface. 6

Experiments...... Yay! Experiments...... Yay! Experiment #2 Fill the bottle with water to the same distance above the bottom tack as it was filled above the top tack in experiment 1. Experiment #3 Repeat experiment 1 with a thinner or wider bottle with the water level initially same distance above the top tack in experiment 1. We also see that the pressure at a given depth is the same in all directions. The pressure of the liquid at the hole depends only on the height of the liquid above the hole, and not on the mass of the liquid above...... QUICKLY! Testing our model of pressure in a liquid Mr. Largo has a soda bottle filled with water. The bottle has been punctured with 2 tacks The bottle is closed. Predict what will happen when Mr. Largo removes the bottom tack. P atm + P depth Predict what will happen when Mr. Largo then removes the top tack. Testing our model of pressure in a liquid Why does pressure vary at different levels? Stream stops 7

Why does pressure vary at different levels? Why does pressure vary at different levels? F Bbelow on B5 Draw a force diagram (system is book 5) Mind the length of your arrows. F EonB5 F Babove on B5 Quantifying pressure change with depth Quantifying pressure change with depth F Fbelow on C Draw a force diagram (system is cylinder of fluid C) Mind the length of your arrows. F EonB5 F Fabove on C FLUID AT REST Quantifying pressure change with depth Quantifying pressure change with depth F Fbelow on C FLUID AT REST Object is in equilibrium What PHYSICS concept do we use when we have a force diagram? F Fbelow on C Fy = 0 Newton s second law! F EonB5 F Fabove on C Fx = 0 a = 0 m/s 2 a = F m F EonB5 F Fabove on C v = 0 m/s Apply Newton s second law to this force diagram. v = constant 8

a = F m NEWTON S SECOND LAW a = 0 m/s 2 (fluid is at rest) F FBonC F FAonC F EonC = 0 ΣF = 0 F EonC = g m PASCAL S SECOND LAW: VARIATION OF PRESSURE WITH DEPTH The pressure P in a static fluid at position y can be determined in terms of the pressure P 0 at position y 0 : F FBonC F FAonC g m = 0 m = V F FBonC F FAonC g FLUID V = 0 V = A h F FBonC F FAonC g FLUID A h = 0 P 1 P 2 g FLUID h = 0 P = F/A = fluid = density of the fluid g = 9.8 N/kg (positive) Variation of pressure with depth P 1 = P 2 + g FLUID h : Identify P, P 0 P 0 PASCAL S PARADOX When water is poured into the apparatus below, the water level was the same in all parts despite the shapes and the mass of water. Why do you think this happens? P = P 0 + P depth P = P atm + fluid g (h 0 h) P Measuring pressure Pascal s 2 nd law: All points inside the fluid experience the same pressure at the same depth. 9

SIMPLE FLUID BAROMETER Measuring atmospheric pressure Its been known since Galileo's time that a pump consisting of a piston in a long cylinder that pulls up water. Pascal s 2 nd law: All points inside the fluid experience the same pressure at the same depth. Fluid height Suction pumps where used to pump waters out of wells and to remove water from flooded mines. Water cannot not be pumped more than 10.3 m high. Finding P atm P 2 = P 3 + fluid g h Use Pascal s second law and find the pressure at point 2? (Assumption: You do not know the magnitude of the atmospheric pressure). Note: in the picture, point 2 is at the bottom and point 3 at the top. P 2 = 0 + 1000 9.8 (10.3) P 2 = 100,940 Pa P 2 = P 1 = P atmosphere Use Pascal s second law and find the maximum height that mercury can be pumped. Write an expression for the height before finding its magnitude. Assume: mercury = 13600 kg/m 3 P 1 = P 2 + ρ fluid g h h = P 1 P 2 ρ fluid g h = 100,000 0 13600 9.8 h = 0.750 m 10

PHYSICS UNDERSTANDING Pressure is often measured and reported in mm Hg and the magnitude of atmospheric pressure is 760 mm Hg. The atmospheric pressure (101,000 N/m 2 ) can push mercury of density (13,560 kg/m 3 ) 760 mm up a column, and it can push water of density (1,000 kg/m 3 ) 10,300 mm up a column. THE BAROMETER This is a mercury barometer, developed by Evangelista Torricelli to measure atmospheric pressure of mercury is such that the pressure in the tube at the surface level is 1 atm. Therefore. The height of the column, pressure is often quoted in millimeters (or inches) of mercury. SOLUTION BAROMETER - b P 0 =? P = P 0 + FLUID g h P 0 = P - FLUID g h P 0 = 100,000 13600 9.8 0.736 P 0 = 0 Pa P 0 = 1905 Pa P 0 = 71345 Pa Find the pressure at the top of each column of mercury. Assume: mercury = 13600 kg/m 3 and P atm = 100,000 Pa P = P ATM Same height Same pressure P 0 = 100,000 98094 P 0 = 1905 Pa SUCTION CUPS The pressure at the top of the water in a city s gravity fed reservoir is atmospheric pressure. Determine the pressure at the faucet of a home 30 m below the reservoir. P = 394,000 Pa When a suction cup is pushed onto a glossy, non textured, non porous surface, the air underneath is pushed out and a vacuum is created. As there is now no air pressure under the suction cup, only atmospheric pressure is being exerted on the outside of the cup, so pushing it down onto the surface. Friction keeps the cup from sliding 11

SUCTION CUP A square suction cup is held against the ceiling, and the air is pressed out from inside it. SUCTION CUP F atm on SC A 5 kg kitten hangs by a rope from the suction cup (maximum mass the suction cup can withstand) 1. Draw a Force diagram (suction cup, cat, rope is the system) 2. Write an expression for the side L of the suction cup. 3. Calculate the side L of the suction cup. 4. If the side L is tripled, how many kitties can hang from the suction cup? Suction cup + Cat + string F E on SC F atm on SC F E on SC = 0 SUCTION CUP MANOMETER F atm on SC F E on SC = 0 P atm A g m = 0 P atm L 2 g m = 0 L = 0.022 m L = g m P atm 9 kitties P balloon See the classroom set up and find the pressure at the top of the column of water (point P). P 0 What is the pressure of the gas at point P if the water raises 40 cm? Sorry I don t have the answer, I change the balloon every time! Same height Same pressure P = P atm P = 96080 Pa 12

OBSERVATIONAL EXPERIMENTS 1. Divide your whiteboard in 4 sections. 2. Mr. Largo is about to show you four experiments. 3. You must draw a FD for each experiment. 4. Include the magnitude of each force in each FD. 5. Let s go... Everybody to the front of the room! 1. Mr. Largo hangs a mass at the bottom of a spring scale. 2. Mr. Largo hangs a mass at the bottom of a spring scale. Then he dips half of the mass in water. 3. Mr. Largo hangs a mass at the bottom of a spring scale. Then he dips the mass in water (water barely covers the top of the mass). 4. Mr. Largo hangs a mass at the bottom of a spring scale. Then he dips the mass completely in water. 9.2 N 9.0 N 9.0 N 13

9.0 N THE MAGNITUDE OF THE FORCE THE FLUID EXERTS ON A SUBMERGED OBJECT The force exerted by the fluid pushing up on the bottom is greater than the force exerted by the fluid pushing down on the top of the block. a = F m a = 0 m/s 2 (fluid is at rest) F SonB + F WonB F EonB = 0 F WonB = F EonB F SonB To calculate the magnitude of the upward buoyant force exerted by the fluid on the block, we start from Pascal s second law to determine the upward pressure of the fluid on the bottom surface of the block, compared to the downward pressure of the fluid on the top surface of the block. P A = P 0 A + fluid g h A P = F/A F = P A F = F 0 + fluid g V BUOYANT FORCE (F WonO ) V = volume of fluid displaced = density of the fluid F WonO = FLUID V disp g ARCHIMEDES' PRINCIPLE: THE BUOYANT FORCE A stationary fluid exerts an upward buoyant force (F WonO ) on an object that is totally or partially submerged in the fluid. The magnitude of this force is the product of the fluid density FLUID, the volume of the fluid V DISP that is displaced by the object, and the gravitational constant g 14

NOTES: density pressure If the object is completely submerged in the fluid, the volume V in the equation is just the whole volume of the object. If the object floats, the volume V in the equation is the volume submerged in the fluid. Pascal s 2 nd Law Buoyant Force You hold on your hand a 50 kg rock of density 2200 kg/m 3 that is completely submerged in water of density 1000 kg/m 3. A) Sketch B) Force Diagram C) Find the magnitudes of: F EonR 490 N F WonR 222.7 N F HonR 267.3 N ρ water ρ duckie = Duckie is the system 1. Draw a force diagram 2. Use Newton s second Law and find an expression for the ratio of densities: ρ water ρ duckie V duckie V water displaced BUOYANT FORCE (F wono ) If OBJECT < FLUID, then the object floats partially submerged. If OBJECT = FLUID, then the object remains wherever it is placed totally submerged at any depth in the fluid. You have four objects at rest, each of the same volume. Object A is partially submerged, and objects B,C, and D are totally submerged in the same container of liquid. A) Determine if the object floats, sinks or remains in place (inside the water). If OBJECT > FLUID, then the object sinks at increasing speed until it reaches the bottom of the container. B) Determine if water is >, <, or = to the cube 15

Cube A ρ F ρ C = V c V FD ρ F > ρ c Cube floats above water Find the minimum mass required for a uniform sphere of radius 10 cm to sink (have nothing above the surface) in fresh water. Cube B Cube C ρ F ρ C = V c V FD ρ F ρ C = V c V FD ρ F = ρ c ρ F = ρ c Cube floats submerged Cube floats submerged F WonS = F EonS ρ w g V = g m ρ w V = m m = ρ w 4 3 πr3 Cube D ρ F ρ C = V c V FD ρ F < ρ c Cube sinks m = ρ w 4 3 πr3 m = 4.19 kg You stand on a bathroom scale. Suppose your mass is 70.0 kg and your density is 970 kg/m 3. Determine the reduction of the scale reading (F SonP ) due to air. Assume density of air 1.29 kg/m 3 A) Force diagram. B) Find the magnitudes of: F EonYOU F EonY = 686 N F AIRonYOU (Buoyant Force) F FonY = 0.912 N F SonYOU F FonY = 685.088 N Pretend for a moment that you are Archimedes, who needs to determine whether a crown made for the king is pure gold or some less valuable metal. You find that the force that a string attached to a spring scale exerts on the crown is 25.0 N when the crown hangs in air and 22.6 N when the crown hangs completely submerged in water. FD + N 2 nd What is the buoyant force exerted on the crown? F FonC = 2.4 N What is the mass of the crown? m = 2.55 kg What is the volume of the crown? V = 0.000245 m 3 What is the density of the crown? = 10,416 kg/m 3 Compare the crown to the density of gold ( GOLD = 19,300 kg/m 3 ). Is the crown made of gold? Crown is not made of gold, what a rip off!!! When analyzing a sample of ore, a geologist finds that it weighs 2.00 N in air and 1.13 N when immersed in water. Assume the ore sinks in water. A) Sketch B) FD + N 2 nd D) Volume of water displaced E) Density of ore V = 0.000089 m 3 = 2298.85 kg/m 3 16

A typical human being has a density of 900 kg/m 3. as this typical human floats in sea water (1250 kg/m 3 ) A) FD + N 2 nd B) what percent of his/her body is above water? a) 10% ρ human ρ sea water c) 72% A 3.6 kg goose floats on a lake with 40% of its body below the 1000 kg/m 3 water level.. A) FD + N 2 nd B) Find the density the goose. g = 400 kg/m 3 ρ goose ρ water b) 28% d) 90% A 3.0m x 1.0m rectangular plastic container that is 1.0 m high, has a mass of 1500 kg. The container floats in fresh water (density 1000 kg/m 3 ), partially submerged. A) FD + N 2 nd B) Write an expression for the depth of the container submerged in water. C) Find the depth of the container submerged in water. An empty 400 kg life raft of cross-sectional area 2.0 m x 3.0 m has its top edge 0.36 m above the waterline. How many 75-kg passengers can the raft hold before water starts to flow over its edges? The raft is in seawater of density 1025 kg/m 3. h submerged = m ρ F A h = 0.5 m n = 24 passengers At a pool party, you pull an inflated beach ball under water by tying a light string around it. The ball has a mass m and an overall specific gravity of 0.25. While you hold the string, and therefore the ball, under water motionless, what is the tension in the string? a) F T = gm b) F T = 2gm c) F T = 3gm d) F T = 4gm F WonB = F EonB + F T F T = F WonB F EonB F T = ρ W V WD g m B g F T = g ρ W V WD m B - SOLUTION F T = g ρ W V B m B F T = g F T = g m B ρ W m B ρ B m B ρ W ρ B 1 F T = g m B 1000 250 1 F T = g m B 4 1 V WD = V B F T = 3 g m B 17

Up, up and away! - SOLUTION The average balloon has a volume of 0.01 m 3. The density of helium is 0.164 kg/m 3. The density of air is 1.3 kg/m 3. How many balloons would you need to attach to an average house (50,000 kg) in order for it to float? Hint: System = balloons F AonB = F EonB + F HonB ρ air V B g = ngm B + gm H ρ air nv B = nm B + m H ρ air nv B = nv B ρ Helium + m H ρ air nv B nv B ρ Helium = m H nv B ρ air ρ Helium = m H m H n = V B ρ air ρ Helium n = 4 M 18