11/13/2015 9:49 AM 1 Chapter 9 Fluids and Buoyant Force In Physics, liquids and gases are collectively called fluids.

11/13/2015 9:49 AM 2 Fluids and Buoyant Force A fluid is a nonsolid state of matter in which the atoms or molecules are free to move past each other, as in a gas or a liquid.

11/13/2015 9:49 AM 3 Fluids and Buoyant Force Liquids differ from gases in that liquids have a definite volume, gases do not.

11/13/2015 9:49 AM 4 Fluids and Buoyant Force Recall, volume is the amount of space that an object occupies.

11/13/2015 9:49 AM 5 Fluids and Buoyant Force Of course, whether an object floats or sinks has little to do with size. It is the object’s mass density that matters.

11/13/2015 9:49 AM 6 Fluids and Buoyant Force Mass density (or just density) is the mass per unit volume of a substance.

11/13/2015 9:49 AM 7 Fluids and Buoyant Force The Mass density of fresh water at 4 o C and 1 atm of pressure is 1.00 g/cm 3 or 1.00X10 3 kg/m 3 Recall, 1 ml = 1 cm 3

11/13/2015 9:49 AM 8 Fluids and Buoyant Force That means that one liter (1000 ml) of fresh water at 4 o C and 1 atm of pressure has a mass of one kilogram (1000 g).

11/13/2015 9:49 AM 9 Fluids and Buoyant Force Scuba divers, submarines and many types of fish control their density to sink or rise in the water.

11/13/2015 9:49 AM 10 Fluids and Buoyant Force Formula for Mass Density

11/13/2015 9:49 AM 11

11/13/2015 9:49 AM 12 Fluids and Buoyant Force Example 1. If gold has a mass density of 19.3 g/cm 3 at STP (standard temperature and pressure), how large would a 3.75 g sample of pure gold be? Given: Find: Solution:  = 19.3 g/cm 3 m = 3.75 g V

11/13/2015 9:49 AM 13 Fluids and Buoyant Force You may have noticed that objects feel lighter in the water than they do in the air.

11/13/2015 9:49 AM 14 Fluids and Buoyant Force This is because the water exerts a buoyant force in the opposite direction as gravity.

11/13/2015 9:49 AM 15 Fluids and Buoyant Force Buoyant Force is a force that acts upward on an object submerged in a liquid or floating on a liquid’s surface.

11/13/2015 9:49 AM 16 Fluids and Buoyant Force Do you know the story of Archimedes, and how he discovered what has come to be known as “Archimedes Principle”?

11/13/2015 9:49 AM 17 Fluids and Buoyant Force Archimedes ( BC) is considered to be one of the greatest mathematicians of all times, and he is sometimes called the “father of mathematical physics.”

11/13/2015 9:49 AM 18 Fluids and Buoyant Force Hiero, the king of Syracuse, commissioned a craftsman to fashion a crown, and provided he with an exact amount of gold to make it.

11/13/2015 9:49 AM 19 Fluids and Buoyant Force The crown that Heiro received weighed the same amount as the gold he had provided, but he suspected the craftsman had used some silver instead of pure gold, and he asked Archimedes to prove it.

11/13/2015 9:49 AM 20 Fluids and Buoyant Force Archimedes was contemplating the problem while in the bath. He realized that by getting in the bath, he displaced an amount of water that was proportional to the part of his body that he submerged.

11/13/2015 9:49 AM 21 Fluids and Buoyant Force He is said to have become so excited that he ran naked through the streets, shouting “Eureka! Eureka!” (this translates to, “I have found it!”)

11/13/2015 9:49 AM 22 Fluids and Buoyant Force He formulated what is now known as Archimedes principle, and a way to test the composition of the crown.

11/13/2015 9:49 AM 23 Fluids and Buoyant Force Archimedes Principle – any object that is completely or partially submerged in a fluid experiences an upward buoyant force equal in magnitude to the weight of the fluid displaced by the object.

11/13/2015 9:49 AM 24 Fluids and Buoyant Force Formula for Buoyant Force

11/13/2015 9:49 AM 25 Fluids and Buoyant Force Example 2. A piece of an unknown mineral that is dropped into mercury displaces m 3 of Hg. Find the buoyant force on the mineral. (note,  of Hg = 13.6 x 10 3 kg/m 3 at STP.) Given: Find: Hint:  = 13.6 x 10 3 kg/m 3 V = m 3 FBFB so

11/13/2015 9:49 AM 26 Fluids and Buoyant Force Example 2. A piece of an unknown mineral that is dropped into mercury displaces m 3 of Hg. Find the buoyant force on the mineral. (note,  of Hg = 13.6 x 10 3 kg/m 3 at STP.) Given: Find:Solution:  = 13.6 x 10 3 kg/m 3 V = m 3 FBFB

11/13/2015 9:49 AM 27 Fluids and Buoyant Force If the buoyant force on an object is greater than the weight of the object, the object will bob to the surface.

11/13/2015 9:49 AM 28 Fluids and Buoyant Force For an object that is floating, the buoyant force is equal and opposite to the weight of the object.

11/13/2015 9:49 AM 29 Fluids and Buoyant Force In other words, if an object is less dense than the fluid that it rests in, it will float.

11/13/2015 9:49 AM 30 Fluids and Buoyant Force The density of the object determines how much of the object’s volume will be submerged below the surface.

11/13/2015 9:49 AM 31 Fluids and Buoyant Force Formula for Determining what Portion of a Floating Object will Submerge Sub=submerged V=Volume of entire object

11/13/2015 9:49 AM 32 Fluids and Buoyant Force Example 3. A 2.00 m 3 block of wood has a density of x 10 3 kg/m 3. If this object is placed in fresh water at 4 o C and 1 atm of pressure, what volume will be below the surface of the water? Given: Find: Solution:  object = x 10 3 kg/m 3 V = 2.00 m 3  fluid = x 10 3 kg/m 3 V sub

11/13/2015 9:49 AM 33 Fluids and Buoyant Force Notice, if an object is ½ as dense as water, then ½ of it will be under the water.

11/13/2015 9:49 AM 34 Fluids and Buoyant Force If an object is 90% as dense as water, than 90% of the object will be below the surface (AS IN AN ICEBERG).

11/13/2015 9:49 AM 35 Fluids and Buoyant Force If the density of the object is greater than the density of the fluid, it will continue to sink.

11/13/2015 9:49 AM 36 Fluids and Buoyant Force The apparent weight of a submerged object depends upon its density. This, so called “apparent weight” is really just the net force (F net ) in the y- axis.

11/13/2015 9:49 AM 37 Fluids and Buoyant Force Formulas for Determining the F Net of a Submerged Object Other

11/13/2015 9:49 AM 38 Fluids and Buoyant Force Example 4. Gold has a density of 19.3 x 10 3 kg/m 3. If a piece of gold with a volume of m 3 is completely submerged in water, how much will it seem to weigh? Given: Find:  object = 19.3 x 10 3 kg/m 3 V o = m 3  fluid = 1.00 x 10 3 kg/m 3 V f = m 3 F Net Hint: The Volume of the water is the same as the volume of the object.

11/13/2015 9:49 AM 39 Fluids and Buoyant Force Given: Find:  object = 19.3 x 10 3 kg/m 3 V o = m 3  fluid = 1.00 x 10 3 kg/m 3 V f = m 3 F Net Solution: Because V o = V f, we can rewrite our equation. DOWN (it will sink, not float!)

11/13/2015 9:49 AM 40 Another useful formula is Apparent weight = F g – F B Remember…F W and F g are the same…

11/13/2015 9:49 AM 41 Chapter 9 Section 9-1 Fluids and Buoyant Force Summary of Formulas

11/13/2015 9:49 AM 42 Chapter 9 Section 9-1 Fluids and Buoyant Force Summary of Formulas. Apparent weight = F g – F B

11/13/2015 9:49 AM 43 Fluid Pressure and Temperature An object that is submerged in water, or any other fluid, experiences pressure on all sides. You have likely noticed the effects of this pressure while swimming.

11/13/2015 9:49 AM 44 Fluid Pressure and Temperature Gases also exert pressure. Your body feels the weight of the atmosphere on it. Your ears also “pop” when you change altitude quickly.

11/13/2015 9:49 AM 45 Fluid Pressure and Temperature Pressure is the magnitude of the force on a surface per unit area. Movie Clip 

11/13/2015 9:49 AM 46 Fluid Pressure and Temperature Formula for Pressure

11/13/2015 9:49 AM 47 Fluid Pressure and Temperature Can you explain why a finger does not pop a balloon even when you exert more force on the balloon than you do with a needle?

11/13/2015 9:49 AM 48 Fluid Pressure and Temperature The SI unit of pressure is the pascal (Pa), which is equal to 1 N/m 2. Conversion Factor 1 Pa = 1 N/m 2

11/13/2015 9:49 AM 49 Fluid Pressure and Temperature Example 1. A cylinder-shaped column with a height of 2.0 m and a base radius of 0.50 m has a mass of 1250 kg. Find the pressure it exerts on the ground. Given: Find: Solution: m = 1250 kg h = 2.0 m r = 0.50 m P

11/13/2015 9:49 AM 50 Fluid Pressure and Temperature There are many other common units of pressure, including the atmosphere (atm) and millimeters of mercury (mm of Hg).

11/13/2015 9:49 AM 51 Fluid Pressure and Temperature Important Conversion Factors 1 atm = 760 mm of Hg = 760 torr = x 10 5 Pa

11/13/2015 9:49 AM 52 Fluid Pressure and Temperature Example 2. Convert 2.30 atm to Pa atmPa=2.33 x 10 5

11/13/2015 9:49 AM 53 Fluid Pressure and Temperature A barometer is an instrument that is used to measure atmospheric pressure.

11/13/2015 9:49 AM 54 Fluid Pressure and Temperature You have probably noticed that when you squeeze a balloon, it will bulge outwards in a different area. This can be explained by Pascal’s Principle.

11/13/2015 9:49 AM 55 Fluid Pressure and Temperature Pascal’s Principle – Pressure applied to a fluid in a closed container is transmitted equally to every point of the fluid and to the walls of the container.

11/13/2015 9:49 AM 56 Fluid Pressure and Temperature Formula for Pascal’s Principle

11/13/2015 9:49 AM 57 Fluid Pressure and Temperature Example 3. In a hydraulic car lift, the compressed air exerts a force on a piston with a base radius of 5.00 cm. The pressure is transmitted to a second piston with a base radius of 35.0 cm. How much force must be exerted on the first piston to lift a car with a mass of 1.40 x 10 3 kg on the second piston?

11/13/2015 9:49 AM 58 Fluid Pressure and Temperature Example 4. In a hydraulic car lift, the compressed air exerts a force on a piston with a base radius of 5.00 cm. The pressure is transmitted to a second piston with a base radius of 35.0 cm. How much force must be exerted on the first piston to lift a car with a mass of 1.40 x 10 3 kg on the second piston? Given: Find: Equation: r 1 = 5.00 cm r 2 = 35.0 cm m 2 = 1.40 x 10 3 kg F1F1

11/13/2015 9:49 AM 59 Fluid Pressure and Temperature Given: Find: Equation: r 1 = 5.00 cm r 2 = 35.0 cm m 2 = 1.40 x 10 3 kg F1F1 Solution:

11/13/2015 9:49 AM 60 Lesson 9-2 Fluid Pressure and Temperature Pressure varies with depth in a fluid. That is why your ears “pop” when you go up in an airplane or down in the water.

11/13/2015 9:49 AM 61 Lesson 9-2 Fluid Pressure and Temperature Atmospheric pressure (p atm ) is the pressure exerted by the atmosphere. It varies from day to day and from location to location.

11/13/2015 9:49 AM 62 Lesson 9-2 Fluid Pressure and Temperature Standard atmospheric pressure is equal to 1.0 atm or approximately x 10 5 Pa.

11/13/2015 9:49 AM 63 Lesson 9-2 Fluid Pressure and Temperature Gauge pressure (p gauge =  gh) is the pressure exerted by just the water, or other fluid. When calculating gauge pressure, leave the atmospheric pressure out of the calculation.

11/13/2015 9:49 AM 64 Lesson 9-2 Fluid Pressure and Temperature Gauge Pressure as a Function of Depth Where  = density of fluid, g = acceleration due to gravity and h = depth

11/13/2015 9:49 AM 65 Lesson 9-2 Fluid Pressure and Temperature Example 5. Calculate the gauge pressure at a point 10.0 m below the surface of the ocean. (note,  sea water = x 10 3 kg/m 3 ) Given: Find: Solution:  = x 10 3 kg/m 3 h = 10.0 m g = 9.81 m/s 2 p gauge

11/13/2015 9:49 AM 66 Lesson 9-2 Fluid Pressure and Temperature Absolute pressure (p) is the total pressure exerted by the water (or other fluid) and the atmosphere above the water.

11/13/2015 9:49 AM 67 Lesson 9-2 Fluid Pressure and Temperature Absolute Pressure as a Function of Depth Where p = absolute pressure, p 0 = pressure at point of comparison (usually atmospheric pressure),  = density of fluid, g = acceleration due to gravity and h = depth

11/13/2015 9:49 AM 68 Lesson 9-2 Fluid Pressure and Temperature Example 6. Calculate the absolute pressure experienced by a scuba diver 20.0 m below the sea (  = x 10 3 kg/m 3 ). Assume p atm = 1.0 x 10 5 Pa

11/13/2015 9:49 AM 69 Lesson 9-2 Fluid Pressure and Temperature Example 6. Calculate the absolute pressure experienced by a scuba diver 20.0 m below the sea (  = x 10 3 kg/m 3 ). Assume p atm = 1.0 x 10 5 Pa Given: Find:Solution: p 0 = 1.0 x 10 5 Pa h = 20.0 m  = x 10 3 kg/m 3 p

11/13/2015 9:49 AM 70 Lesson 9-2 Fluid Pressure and Temperature Summary of Formulas

11/13/2015 9:49 AM 71 Lesson 9-2 Fluid Pressure and Temperature Summary of Formulas 1 atm = 760 mm of Hg = 760 torr = x 10 5 Pa

11/13/2015 9:49 AM 72 Fluids in Motion The motion of a fluid can be described as either laminar (smooth) or turbulent (irregular).

11/13/2015 9:49 AM 73 Fluids in Motion When studying fluids, is helps to consider the behaviors of an ideal fluid.

11/13/2015 9:49 AM 74 Fluids in Motion An ideal fluid is an imaginary fluid that has no internal friction or viscosity, and is incompressible. The viscosity of a fluid is a measure of its internal resistance, or its resistance to flow.

11/13/2015 9:49 AM 75 Fluids in Motion Fluids with high viscosity, like molasses, flow slowly, due to internal friction.

11/13/2015 9:49 AM 76 Fluids in Motion The law of conservation of mass explains that the mass entering a section of pipe or tube must be the same as the mass that leaves.

11/13/2015 9:49 AM 77 Fluids in Motion The Continuity Equation area x speed in region 1 = area x speed in region 2 The Continuity Equation is an expression of the conservation of mass.

11/13/2015 9:49 AM 78 Fluids in Motion

11/13/2015 9:49 AM 79 Fluids in Motion Example 1. An ideal fluid is moving at 3.0 m/s through a section of pipe with a radius of 0.35 m. How fast will it flow through another section of the pipe with a radius of 0.20 m? HINT: A =  r 2

11/13/2015 9:49 AM 80 Fluids in Motion Example 1. An ideal fluid is moving at 3.0 m/s through a section of pipe with a radius of 0.35 m. How fast will it flow through another section of the pipe with a radius of 0.20 m? Given: 1 = 3.0 m/s r 1 = 0.35 m r 2 = 0.20 m Find: Formula: Solution: 2 A1A1 A2 A2

11/13/2015 9:49 AM 81 Fluids in Motion From example 1, we see that as the area of the pipe decreases, the velocity of the fluid becomes much greater.

11/13/2015 9:49 AM 82 Fluids in Motion Another property of fluids explains how the wings of an airplane produce lift.

11/13/2015 9:49 AM 83 Fluids in Motion To demonstrate Bernoulli’s principle for yourself, try blowing air over the top of a piece of paper.

11/13/2015 9:49 AM 84 Fluids in Motion Bernoulli’s Principle – The pressure in a fluid decreases as the fluid’s velocity increases.

11/13/2015 9:49 AM 85 Fluids in Motion The wings of an airplane are designed to allow air to flow quicker over the top than the bottom, resulting in lower pressure above the wings.

11/13/2015 9:49 AM 86 Fluids in Motion Bernoulli’s Equation pressure + kinetic energy per unit volume + gravitational potential energy per unit volume = constant along a given streamline.

11/13/2015 9:49 AM 87 Fluids in Motion Bernoulli’s Equation for Comparing a Fluid at Two Different Points pressure + kinetic energy per unit volume + gravitational potential energy per unit volume = pressure + kinetic energy per unit volume + gravitational potential energy per unit volume.

11/13/2015 9:49 AM 88 Important to know!!!!! IF there is no height difference, you can eliminate the If there is no area change, velocity stays the same IF there is no pressure change, P 1 = P 2

11/13/2015 9:49 AM 89 Fluids in Motion Example 2. Water is circulated through a system. If the water is pumped with a speed of 0.45 m/s under a pressure of 2.2 x 10 5 Pa from the first floor through a 6.0-cm diameter pipe, what will the pressure be on the next floor 4.0 m above? HINT: The key to this question is that the diameter of the pipe doesn’t change and, therefore, the velocity of the water remains constant.

11/13/2015 9:49 AM 90 Fluids in Motion If 1 = 2, then we can simplify our equation.

11/13/2015 9:49 AM 91 Fluids in Motion Example 2. Water is circulated through a system. If the water is pumped with a speed of 0.45 m/s under a pressure of 2.2 x 10 5 Pa from the first floor through a 6.0-cm diameter pipe, what will the pressure be on the next floor 4.0 m above? Given: P 1 = 2.2 x 10 5 Pa h1- h 2 = -4.0 m  water = 1000 kg/m 3 g = 9.81 m/s 2 Find:Formula: P2P2

11/13/2015 9:49 AM 92 Fluids in Motion Example 3. A water tank has a water level of 1.2 m. The spigot is located 0.40 m above the ground. If the spigot is opened fully, how fast will water come out of the spigot? HINT: The container is open to the atmosphere, so the pressure will be the same in both spots. We can assume that the velocity of the water at the top (v 2 ) is essentially zero.

11/13/2015 9:49 AM 93 Fluids in Motion The pressure of the fluid in both areas is the same, P 1 = P 2 so they cancel out. Also, the velocity at the top ( 2 ) is zero, so we cross out that expression.

11/13/2015 9:49 AM 94 Fluids in Motion Now, we can divide all expressions by  (the fluid doesn’t change, so density is the same throughout) and continue to isolate 1, the unknown.

11/13/2015 9:49 AM 95 Fluids in Motion Example 3. A water tank has a water level of 1.2 m. The spigot is located 0.40 m above the ground. If the spigot is opened fully, how fast will water come out of the spigot? Given: h 1 = 0.40 m h 2 = 1.2 m g = 9.81 m/s 2 Find:Formula: 2 Solution:

11/13/2015 9:49 AM 96 Summary of Formulas Fluids in Motion

11/13/2015 9:49 AM 97 Lesson 9-2 Fluid Pressure and Temperature The Kinetic Molecular Theory (KMT) of Gases also relates the temperature of a gas to the motion of its particles.

11/13/2015 9:49 AM 98 Lesson 9-2 Fluid Pressure and Temperature Temperature is a measure of the average kinetic energy of the particles of a substance. The SI unit of temperature is the Kelvin (K).

11/13/2015 9:49 AM 99 Lesson 9-2 Fluid Pressure and Temperature Converting between Celsius and Kelvin

11/13/2015 9:49 AM 100 Lesson 9-2 Fluid Pressure and Temperature Example 7. Convert 22.0 o C to Kelvin.