Chapter 9 Fluids and Buoyant Force

Size: px

Start display at page:

Download "Chapter 9 Fluids and Buoyant Force"

Norman Anderson
5 years ago
Views:

1 Chapter 9 Fluids and Buoyant Force In Physics, liquids and gases are collectively called fluids. 3/0/018 8:56 AM 1

2 Fluids and Buoyant Force Formula for Mass Density density mass volume m V water 1000 kg 3 m 3/0/018 8:56 AM

3 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. 3/0/018 8:56 AM 3

4 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. 3/0/018 8:56 AM 4

5 Fluids and Buoyant Force Formula for Buoyant Force F B F w (displaced fluid) m g f magnitude of buoyant force weight of fluid displaced 3/0/018 8:56 AM 5

6 Fluids and Buoyant Force Example 1. 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: f = 13.6 x 10 3 kg/m 3 V f = m 3 Find: Hint: F B m so m V V so F m g ρ V g 3/0/018 8:56 AM 6 B f f f

7 Fluids and Buoyant Force Example 1. 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: f = 13.6 x 10 3 kg/m 3 V f = m 3 B f F B Find: F m g Solution: ρ V g f (13.6 x 10 3 f kg/m 3 )(0.057 m )(9.81m/s 3/0/018 8:56 AM 7 3 ) 7600 N

8 Fluids and Buoyant Force If the buoyant force on a submerged object is greater than the weight of the object, the object will bob to the surface. 3/0/018 8:56 AM 8

9 Fluids and Buoyant Force For an object that is FLOATING ALREADY, the buoyant force is equal and opposite to the weight of the object. 3/0/018 8:56 AM 9

10 Fluids and Buoyant Force Formula for Determining what Portion of a Floating Object will Submerge Sub=submerged V V sub o object fluid V o =Volume of entire object 3/0/018 8:56 AM 10

11 Fluids and Buoyant Force Example. A.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: object = x 10 3 kg/m 3 V o =.00 m 3 fluid = x 10 3 kg/m 3 Find: Solution: V (.00 m )(0.500 x 10 kg/m V sub Vo object (.00 m )(0.500 x 10 kg/m ) 3 Vsub 3 3 fluid 1.00 x 10 3 kg/m 1.00 m 3 fluid 1.00 x 10 kg/m 3/0/018 8:56 AM 11 )

12 Fluids and Buoyant Force The apparent weight of a submerged object depends upon its density. The apparent weight is really just the net force (F net ) in the y-axis (down). 3/0/018 8:56 AM 1

13 Formulas for Determining the F Net and apparent weight of a Submerged Object F Net F F g B - F F B B F apparent weight, so g -apparent weight F W F Net V g f f o V g o F Net m g f m g o Other useful Formula (submerged!) F F g B o f F F weight in air 3/0/018 8:56 AM 13 w g

14 Fluids and Buoyant Force Example 3. 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? Hint: The Volume of the water displaced is the same as the volume of the object. Given: 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 Find: F Net 3/0/018 8:56 AM 14

15 Fluids and Buoyant Force Given: 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 Find: F Net Solution: F Net ( - )Vg f o (1.00 x N - Because V o = V f, we can rewrite our equation. 3 kg/m 3 F Net V g V g ( - )Vg f x x 10 3/0/018 8:56 AM 15 3 f N kg/m 3 o o )(0.450 m DOWN (it will sink, not float!) 3 f o )(9.81m/s )

16 Problems 9A Problem 1: Use the difference between weight in air and weight in fluid to find F B, then use other useful formula to find the densities of the other liquid. Repeat whole process for second liquid! Problem 4: Volume of a sphere 4/3πr 3 3/0/018 8:56 AM 16

17 Happy Wednesday! Please take out 9A for me to check! 3/0/018 8:56 AM 17

18 Fluid Pressure and Temperature Pressure is the magnitude of the force on a surface per unit area. 3/0/018 8:56 AM 18

19 Fluid Pressure and Temperature Formula for Pressure p F A pressure force area 3/0/018 8:56 AM 19

20 Fluid Pressure and Temperature The SI unit of pressure is the pascal (Pa), which is equal to 1 N/m. p F A Conversion Factor 1 Pa = 1 N/m 3/0/018 8:56 AM 0

21 Fluid Pressure and Temperature Example 1. A cylinder-shaped column with a height of.0 m and a base radius of 0.50 m has a mass of 150 kg. Find the pressure it exerts on the ground. Given: m = 150 kg h =.0 m r = 0.50 m Find: P Solution: F A mg r (150 kg)(9.81 m/s (0.50 m) p x 10 Pa 3/0/018 8:56 AM 1 )

22 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. 3/0/018 8:56 AM

23 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. 3/0/018 8:56 AM 3

24 Fluid Pressure Formula for Pascal s Principle p increase F A 1 1 F A 3/0/018 8:56 AM 4

25 Fluid Pressure Example. 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? 3/0/018 8:56 AM 5

26 Fluid Pressure Example. 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: F 1 Equation: r 1 = 5.00 cm r = 35.0 cm m = 1.40 x 10 3 kg A1 r1 F1 x F x mg A r 3/0/018 8:56 AM 6

27 Fluid Pressure Given: Find: F 1 Equation: Solution: r 1 = 5.00 cm r = 35.0 cm m = 1.40 x 10 3 kg A1 r1 F1 x F x m1g A r F 1 A A 1 x F (0.05m) r1 r x m g 1 3 x (1.40 x 10 kg)(9.81 m/s ) 80 N (0.35m) 3/0/018 8:56 AM 7

28 Problems 9B pg 37 3/0/018 8:56 AM 8

29 Fluid Pressure Pressure varies with depth in a fluid. That is why your ears pop when you go up in an airplane or drive up mountains or you feel pressure on your head deep down in the water. 3/0/018 8:56 AM 9

30 Fluid Pressure Atmospheric pressure (p atm ) is the pressure exerted by the atmosphere. It varies from day to day and from location to location. 3/0/018 8:56 AM 30

31 Fluid Pressure Standard atmospheric pressure is equal to 1.0 atm or approximately x 10 5 Pa. 3/0/018 8:56 AM 31

32 Fluid Pressure 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. 3/0/018 8:56 AM 3

33 Fluid Pressure Gauge Pressure as a Function of Depth p gh gauge Where = density of fluid, g = acceleration due to gravity and h = depth 3/0/018 8:56 AM 33

34 Fluid Pressure p Example 1. Calculate the gauge pressure at a point 10.0 m below the surface of the ocean. (note, sea water = 1.05 x 10 3 kg/m 3 ) Given: Find: Solution: gauge gh = 1.05 x 10 3 kg/m 3 h = 10.0 m g = 9.81 m/s p gauge (1.05 x 10 3 kg/m 3 )(9.81m/s )(10.0 m) 1.0 x 10 5 Pa 3/0/018 8:56 AM 34

35 Fluid Pressure Absolute pressure (p) is the total pressure exerted by the water (and/or other fluid(s)) and the atmosphere above the water. 3/0/018 8:56 AM 35

36 Fluid Pressure Absolute Pressure as a Function of Depth p p gh 0 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 3/0/018 8:56 AM 36

37 Fluid Pressure Example. Calculate the absolute pressure experienced by a scuba diver 0.0 m below the sea ( = 1.05 x 10 3 kg/m 3 ). Assume p atm = 1.0 x 10 5 Pa 3/0/018 8:56 AM 37

38 Fluid Pressure p Example. Calculate the absolute pressure experienced by a scuba diver 0.0 m below the sea ( = 1.05 x 10 3 kg/m 3 ). Assume p atm = 1.0 x 10 5 Pa Given: Find: p 0 p gh (1.0 x x 10 p 0 = 1.0 x 10 5 Pa h = 0.0 m = 1.05 x 10 3 kg/m Pa) Pa Solution: (1.05 x 10 kg/m )(9.81m/s )(0.0 m) 3/0/018 8:56 AM

39 Problems 9C pg. 330 You could layer many fluids and calculate the pressure if you know their depths and their densities. Ex: water, alcohol, oil, atmosphere Atmosphere oil alcohol water 3/0/018 8:56 AM 39

40 Happy Thursday! Please take out 9C for me to check! 3/0/018 8:56 AM 40

41 Fluids in Motion When studying fluids, it helps to consider the behaviors of an ideal fluid. 3/0/018 8:56 AM 41

42 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. 3/0/018 8:56 AM 4

43 Fluids in Motion Bernoulli s Principle The pressure in a fluid decreases as the fluid s velocity increases. 3/0/018 8:56 AM 43

44 Fluids in Motion To demonstrate Bernoulli s principle, try blowing air over the top of a piece of paper. 3/0/018 8:56 AM 44

45 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. 3/0/018 8:56 AM 45

46 3/0/018 8:56 AM 46

47 3/0/018 8:56 AM 47

48 3/0/018 8:56 AM 48

49 3/0/018 8:56 AM 49

50 3/0/018 8:56 AM 50

51 More Class Demos Floating Balls T-peed Trash Bag Vacuum in a bell jar 3/0/018 8:56 AM 51

52 Fluids in Motion Another property of fluids explains how the wings of an airplane produce lift. 3/0/018 8:56 AM 5

53 Paper Airplanes Look at the sheet of airplane designs. Read their specifications and then: Rate them 1-6 for air time (1 long, 6 short) Then pick one to construct and we ll test them in the hallway. 3/0/018 8:56 AM 53

54 Bernoulli Demo OrTAJjg beach ball sg3vhkm - tp 3/0/018 8:56 AM 54

55 Happy Friday! Please complete your airplane construction Steve Spangler Makes Ellen Float Kevin Delaney makes a cloud Kevin Delaney and Pressure 3/0/018 8:56 AM 55

56 3/0/018 8:56 AM 56

57 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. Fluids in Motion 3/0/018 8:56 AM 57

58 Fluids in Motion The Continuity Equation A ν A ν 1 1 area x speed in region 1 = area x speed in region The Continuity Equation is an expression of the conservation of mass. A ν flowrate ( in / sec) 3/0/018 8:56 AM 58 _ 3 m

59 Fluids in Motion A ν A ν 1 1 3/0/018 8:56 AM 59

60 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 0.0 m? HINT: A = r 3/0/018 8:56 AM 60

61 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 0.0 m? Given: n 1 = 3.0 m/s r 1 = 0.35 m r = 0.0 m Find: n Formula: ν r1 ν1 (0.35 m) (3.0 m/s) 9. m/s r (0.0 m) 3/0/018 8:56 AM 61 Solution: r 1 ν1 r ν A 1 A

62 From example 1, we see that as the area of the pipe decreases, the velocity of the fluid becomes much greater. Fluids in Motion 3/0/018 8:56 AM 6

63 Fluids in Motion Bernoulli s Equation for Comparing a Fluid at Two Different Points 1 P n gh P n gh pressure + KE per unit volume + gravitational PE per unit volume = pressure + KE per unit volume + gravitational PE per unit volume. 1 3/0/018 8:56 AM 63

64 Important to know!!!!! IF there is no height difference, you can eliminate the gh andgh 1 If there is no area change, velocity stays the same throughout Since the fluid is the same throughout, the density is the same throughout. IF there is no pressure change (i.e. both ends open to atmospheric pressure), P 1 = P 3/0/018 8:56 AM 64

Fluids in Motion Example. Water is circulated through a system. If the water is pumped with a speed of 0.45 m/s under a pressure of. x 10 5 Pa from the first floor through a 6.

65 Fluids in Motion Example. Water is circulated through a system. If the water is pumped with a speed of 0.45 m/s under a pressure of. 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. 3/0/018 8:56 AM 65

66 Fluids in Motion If n 1 = n, then we can simplify our equation. 1 P n gh P n gh P gh P gh P P g(h - h 1 1 ) 3/0/018 8:56 AM 66

67 Fluids in Motion Example. Water is circulated through a system. If the water is pumped with a speed of 0.45 m/s under a pressure of. 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 =. x 10 5 Pa h 1 - h = -4.0 m water = 1000 kg/m 3 g = 9.81 m/s Find: P P 1 P g(h. x h Formula: ) Pa (1000 kg/m 3 P P1 g(h 1 - h ) )(9.81m/s )(-4.0 m) x 10 3/0/018 8:56 AM 67 Pa

Fluids in Motion Example 3. A water tank has a water level of 1. 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?

68 Fluids in Motion Example 3. A water tank has a water level of 1. 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. Also, We can assume that the velocity of the water at the top (v ) is essentially zero. 3/0/018 8:56 AM 68

69 Example 3. A water tank has a water level of 1. 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? The pressure of the fluid in both areas is the same, P 1 = P so they cancel out. Also, the velocity at the top (n ) is zero, so we cross out that expression. 1 P n gh P n gh n 1 gh 1 gh 3/0/018 8:56 AM 69

70 Example 3. A water tank has a water level of 1. m. The spigot is located 0.40 m above the ground. If the Fluids spigot is opened in fully, Motion how fast will water come out of the spigot? Now, we can divide all expressions by (the fluid doesn t change, so density is the same throughout) and continue to isolate n 1, the unknown. 1 n n n n n 1 gh gh 1 gh gh 1 gh - gh g(h - h 1 3/0/018 8:56 AM 70 1 ) g(h - h1)

71 Fluids in Motion Example 3. A water tank has a water level of 1. 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 = 1. m g = 9.81 m/s Find: n Formula: n 1 g(h - h1 ) Solution: n 1 (9.81m/s )(1. m m) 4.0 m/s 3/0/018 8:56 AM 71

72 Problems 9 D pg /0/018 8:56 AM 7

73 Problems 9 D and section review pg. 337 For # 3 in the section review, gauge pressure can be found by solving for P 1 P as a single quantity 3/0/018 8:56 AM 73

Fluid Mechanics. Liquids and gases have the ability to flow They are called fluids There are a variety of LAWS that fluids obey

Fluid Mechanics. Liquids and gases have the ability to flow They are called fluids There are a variety of LAWS that fluids obey Fluid Mechanics Fluid Mechanics Liquids and gases have the ability to flow They are called fluids There are a variety of LAWS that fluids obey Density Regardless of form (solid, liquid, gas) we can define