1. Solid & Fluid Dynamics
Physics

2. Solids & Fluids Contents
Overview of the four physical states of Matter
Solids, liquids, and gases
Solid Mechanics
Deformation of Solids
Fluid Mechanics
Density & Pressure
Buoyant Forces: Archimedes’ Principle
Fluids in Motion
Bernoulli Equation
Application of Fluid Dynamics

4. Intermolecular Forces hold molecules together
Instantaneous Dipoles that are created by constantly moving electrons.

5. Comparisons of the Three States of Matter

6. Three States of Matter Shape
Gases have no shape because of little attractive forces and independent movement. Liquids take the shape of their container but do not expand readily because of attractive forces. Solid molecules have definite shape and are held in fixed position.

7. States of Matter Density
Section 4 Changes of State
States of Matter Density
Density is mass per unit volume and indicates the closeness of particles in a sample of matter.
Gas Liquid Solid
Low High High

8. Three States of Matter Particle Energy
Differences in attractive forces slow down particle movement.
Gases- high kinetic energy because of low attraction between particles.
Liquids- moderate kinetic energy and attraction
Solids- low kinetic energy and high attractive forces.

9. Factors that effect a GAS
The quantity of a gas, n, in moles
The temperature of a gas, T, in Kelvin (Celsius degrees + 273)
The pressure of a gas, P, in pascals
The volume of a gas, V, in cubic meters

10. Gas Law #1 – Boyles’ Law (complete TREE MAP)
“The pressure of a gas is inverse related to the volume”
Moles and Temperature are constant

11. Gas Law #2 – Charles’ Law
“The volume of a gas is directly related to the temperature”
Pressure and Moles are constant

12. Gas Law #3 – Gay-Lussac’s Law
“The pressure of a gas is directly related to the temperature”
Moles and Volume are constant

13. Gas Law #4 – Avogadro’s Law
“The volume of a gas is directly related to the # of moles of a gas”
Pressure and Temperature are constant

14. Gas Law #5 – The Combined Gas Law
You basically take Boyle’s Charles’ and Gay-Lussac’s Law and combine them together.
Moles are constant

15. Solids & Fluids Contents
Solid Mechanics
Deformation of Solids
Stress, strain, and
Young’s Modulus: Elasticity of Length
Shear modulus: Elasticity of Shape
Bulk Modulus: Volume Elasticity

16. Atomic Arrangement of a solid
Crystalline Solid
Very structured atomic arrangement.
Ex: sodium chloride (salt)
Amorphous Solid
Randomly arranged atoms
Ex: glass

17. Solids: vibrating atoms
Temperature is related to the average kinetic energy of the particles in a substance.
Vibration is slight, essentially fixed
Atomic attraction is electrical
Solids are elastic
KE = mv2 2

18. Deformation of a Solid Solids are elastic
Application of a external force can…
Deform a solid
Break a solid
Removal of external force
Solid returns to original shape
Unless you surpass the elastic limit.
Video

19. Stress & Strain
Demo
Stress & strain are the terms used to discuss the elastic properties of a solid.
Stress (σ)- is the force per unit area causing deformation
Strain (s)- is a measure of the amount of the deformation
Hooke’s Law:
K = springness
Δx = displacement
Elastic modulus Y: proportionality constant.
It is analogous to the spring constant k.
=Stress & strain are the terms used to discuss the elastic properties of a solid.
Y is the stiffness of a material. It is determined experimentally
A material having a large is very stiff, therefore difficult to deform.

20. Elastic Modulus Elastic modulus “Y” is like the spring constant.
stiffness of a material.
Large elastic modulus = very stiff
Small elastic modulus = not stiff
Three types of stress related to this expression
Tensile “Y” - pulling apart, force is perpendicular to cross-section
Shear “S”- pushing apart, force is parallel to cross-section
Bulk “-B”- squeezing force

21. Elastic Modulus Values

22. Overview

23. Stress & Strain
Elastic Modulus is determine in the lab and is unique to the material.
Elastic Modulus Y is determine in the lab and is unique to the material.

24. Tensile Stress: Young’s Modulus
Elasticity of length
A- is cross-sectional area
F- force perpendicular to cross-section
Y- young’s modulus
Lo- original length’
ΔL – change in length
Units:
Add information on tensile stress

25. Tensile Stress: Young’s Modulus
Elasticity of length
Strain
Consider the metal bar. When an external force is applied perpendicular cross-sectional area the atomic bonds of the metal, by their nature, resist the distortion (“stretching”).
The bar is said to experience tensile stress.(pulling)
Strain is the ratio of change in length and original length.
Add information on tensile stress

26. Tensile Strength Breaking Point

27. Pause for a Cause
A vertical steel beam in a building supports a load of 6.0 X 104 N.
A) If the length of the beam is 4.0 m and its cross-sectional area is 8.0 X 10-3 m2, find the distance the beam is compressed along its length.
B) Find the maximum load that the beam can support.
F = 6.0 X 104 N
A = 8.0 X 10-3 m2
Lo = 4.0 m
Y = 20 X 1010 Pa
ΔL =?
F = 6.0 X 104 N
A = 8.0 X 10-3 m2
Lo = 4.0 m
Y = 20 X 1010 Pa
ΔL =?

28. Pause for a Cause Look up Tensile Strength limit from chart for steel
A vertical steel beam in a building supports a load of 6.0 X 104 N.
A) If the length of the beam is 4.0 m and its cross-sectional area is
8.0 X 10-3 m2, find the distance the beam is compressed along its length.
B) Find the maximum load that the beam can support.
Look up Tensile Strength limit from chart for steel
F = 6.0 X 104 N
A = 8.0 X 10-3 m2
Lo = 4.0 m
Y = 20 X 1010 Pa
ΔL =?
F = 6.0 X 104 N
A = 8.0 X 10-3 m2
Lo = 4.0 m
Y = 20 X 1010 Pa
ΔL =?

29. Pause for a Cause
Determine the elongation of the metal rod if it is under a tension of 5.8 X 103 N.
F = 5.8 X 103 N
A = π r2
LoCu = 2.6 m
LoAl = 1.3 m
Ycu = 11 X 1010 Pa
YAl = 7.0 X 1010 Pa
ΔL =?
F = 6.0 X 104 N
A = 8.0 X 10-3 m2
Lo = 4.0 m
Y = 20 X 1010 Pa
ΔL =?

30. Shear Modulus: rigidness
Elasticity of shape
A- is cross-sectional area
F- force parallel to cross-section
S- Shear modulus
h- height of object
Δx – distance displaced
Units:

31. Shear Modulus: rigidness
Elasticity of shape
When a force is applied parallel to one its faces while the other is held fixed.
Ex: A rectangular block under shear stress would become a parallelogram.
Force must be parallel to the cross-sectional area.
Ex: book

32. Shear Modulus: Pause for a Cause
A 125 kg linebacker of makes a flying tackle at vi = 4.00m/s on a stationary quarterback of mass 85 kg. The linebackers helmet makes solid contact with the quarterbacks femur.
a) What is the speed vf of the two athletes immediately after contact? Assume this is a Perfectly inelastic collision from the point of impact.
b) If the collision last for s, estimate the average force exerted on the quarterbacks femur.
c) If the cross-sectional area of the quarterback’s femur is 5.0 x 10-4 m2, calculate the shear stress exerted on the bone in the collision.

33. Shear Modulus: Practice
A 125 kg linebacker of makes a flying tackle at vi = 4.00m/s on a stationary quarterback of mass 85 kg. The linebackers helmet makes solid contact with the quarterbacks femur.
a) What is the speed vf of the two athletes immediately after contact? Assume this is a Perfectly inelastic collision from the point of impact.

34. Shear Modulus: Practice
A 125 kg linebacker of makes a flying tackle at vi = 4.00m/s on a stationary quarterback of mass 85 kg. The linebackers helmet makes solid contact with the quarterbacks femur.
b) If the collision last for s, estimate the average force exerted on the quarterbacks femur.

35. Shear Modulus: Practice
A 125 kg linebacker of makes a flying tackle at vi = 4.00m/s on a stationary quarterback of mass 85 kg. The linebackers helmet makes solid contact with the quarterbacks femur.
c) If the cross-sectional area of the quarterback’s femur is 5.0 x 10-4 m2, calculate the shear stress exerted on the bone in the collision.
The average shear stress of an athletes femur is 7x107 Pa, so his did not leg break?

36. Bulk Modulus: compressibility
Elasticity of volume
ΔP- volume stress
F/A
B- Bulk modulus, always -
V- original volume
ΔV- change in volume
Units:

37. Bulk Modulus: compressibility
Elasticity of volume
This is a deformation due to uniform squeezing.
All the external forces are perpendicular to every surface and are evenly distributed.
Ex: Deep sea diving
An object under this stress will experience a deformation of volume.

38. Bulk Modulus: Pause for a Cause
A solid lead sphere of volume 0.50 m3, dropped in the ocean, sinks to a depth of 2.0 x 103 m (1 mile), where the pressure increases by 2.0 x 107 Pa. Lead has a bulk modulus of
4.2 x 1010 Pa. What change is the change in volume of the lead sphere?

39. Fluid Mechanics - Hydrostatics

40. Fluids Contents Fluid Mechanics Density & Pressure Pressure with Depth
Pressure Measurements
Buoyant Forces: Archimedes’ Principle
Fluids in Motion

41. Density: Quick Quiz
Suppose you have one cubic meter of gold, two cubic meters of silver, and six cubic meters of aluminum.
Rank each of them by mass, from smallest to largest.
Specific gravity- is the ratio of an objects density to the density of water at 4˚ C (1.0 X 103 kg/m3)

42. Density
The 3 primary states have a distinct density, which is defined as mass per unit of volume.
Density is represented by the Greek letter, “RHO”, r
Specific gravity- is the ratio of an objects density to the density of water at 4˚ C (1.0 X 103 kg/m3)
Specific gravity- is the ratio of an objects density to the density of water at 4˚ C (1.0 X 103 kg/m3)

43. Common Densities

44. Pause for a Cause A water bed is 2.0 m on each side an 30.0 cm deep.
(a) Find its weight if the density of water is 1000 kg/m3.
(b) Find the pressure the that the water bed exerts on the floor. Assume that the entire lower surface of the bed makes contact with the floor.
1.2 m3
1200 kg
11760 N
2940 N/m2

45. Why fluids are useful in physics?
Typically, liquids are considered to be incompressible.
That is once you place a liquid in a sealed container you can DO WORK on the FLUID as if it were an object.
The PRESSURE you apply is transmitted throughout the liquid and over the entire length of the fluid itself.
The only stress a fluid can exert on a submerged object is compression.

46. What is a Fluid?
By definition, a fluid is any material that is unable to withstand a static shear stress. Unlike an elastic solid which responds to a shear stress with a recoverable deformation, a fluid responds with an irrecoverable flow.
The only stress a fluid can exert is compression on a submerged object.
What kind of stress is that?
The stress experienced on a submerged object is always perpendicular to all surfaces.

47. Hydrostatic Pressure Video1 Video2
Suppose a Fluid (such as a liquid) is at REST, we call this HYDROSTATIC PRESSURE
Two important points
• A fluid will exert a pressure in all directions
• A fluid will exert a pressure perpendicular to any surface it compacts
The only stress a fluid can exert on a submerged object is compression.
Notice that the arrows on TOP of the objects are smaller than at the BOTTOM. This is because pressure is greatly affected by the DEPTH of the object. Since the bottom of each object is deeper than the top the pressure is greater at the bottom.

48. Pressure
One of most important applications of a fluid is it's pressure- defined as a Force per unit Area
Atmospheric pressure: Is defined as the amount of pressure exert on an object due to the weight of the air from the object to outer space.
The boiling point of liquids is dependant on the atmospheric pressure.
English
PSI: pound/inch2

49. Pressure
One of most important applications of a fluid is it's pressure- defined as a Force per unit Area
Blood Pressure: is the measure of how much pressure one heart beat exerts on the walls of your vascular system
How is it measured?
English
PSI: pound/inch2

50. Pressure: Example
If you tried to support your total weight (F=mg), on a bed of one nail. Your weight would be divided by the tiny area of the tip of the nail.

51. Pressure vs. Depth
Key Points
All portions of the fluid must be in static equilibrium
All points at the same depth must be at the same pressure
Since P = F/A if the pressure was greater on the left of the container so the force would be greater.
If the Force was greater left of the block, the block would accelerate to the right.

52. Pressure vs. Depth Fwater -Fabove + mg = 0 The weight of the object mg
Suppose we had an object submerged in water. If we were to draw an FBD for this object we would have three forces
The weight of the object mg
The force of the water above
The force of the water pressing up
If the object does not move then the sum of all forces is zero.
What would that equation look like? mg
Fwater
Fwater -Fabove + mg = 0

53. Pressure vs. Depth Fwater -Fatm + mg = 0
Suppose we had an object submerged in water with the top exposed to the atmosphere. If we were to draw an FBD for this object we would have three forces
The weight of the object mg
The force of the atmosphere pushing down
The force of the water pressing up
If the object does not move then the sum of all forces is zero.
What would that equation look like? mg
Fwater
Fwater -Fatm + mg = 0

54. Pressure vs. Depth
But recall, pressure is force per unit area. So if we solve for force we can insert our new equation in.
Note: Solving pressure for force gives us
Note: Now consider mass m. solving density for mass gives
Note: Now consider volume V. Solving volume for height gives.
Note: Dived out all of the areas A
Note: The initial pressure in this case is atmospheric pressure, which is a CONSTANT. Po = 1x105 N/m2

55. A closer look at Pressure vs. Depth
Depth below surface
Initial Pressure – May or MAY NOT be atmospheric pressure
ABSOLUTE PRESSURE
Gauge Pressure = CHANGE in pressure or the DIFFERENCE in the initial and absolute pressure

56. Pause for Cause
a) Calculate the absolute pressure at an ocean depth of 1000 m. Assume that the density of water is 1000 kg/m3 and that Po= 1.01 x 105 Pa (N/m2).
b) Calculate the total force exerted on the outside of a 30.0 cm diameter circular submarine window at this depth.
9.9x106 N/m2
2.80 x 106 N

57. Pause for Cause
In a huge oil tanker, salt water has flooded an oil tank to a depth of 5.00 m. On top of the water is a layer of oil 8.00m deep. The oil has a density of g/cm3. Find the pressure at the bottom of the tank if the density of salt water is 1025 kg/m3.
(1 m)3 = (100 cm)3 = 106 cm3

58. A closed system The idea behind this is called PASCAL’S PRINCIPLE
If you take a liquid and place it in a system that is CLOSED like plumbing for example or a car’s brake line, the PRESSURE is the same everywhere.
Since this is true, if you apply a force at one part of the system the pressure is the same at the other end of the system. The force, on the other hand MAY or MAY NOT equal the initial force applied. It depends on the AREA.
You can take advantage of the fact that the pressure is the same in a closed system as it has MANY applications.
The idea behind this is called PASCAL’S
PRINCIPLE
-thank you for hydraulics Pascal
Pascal’s Principle: A change in the pressure applied to an enclosed fluid is transmitted undiminished to every point of the fluid and to the walls of the container.
Pascal’s Principle: A change in the pressure applied to an enclosed fluid is transmitted undiminished to every point of the fluid and to the walls of the container.

59. Pascal’s Principle

60. Another Example - Brakes
In the case of a car's brake pads, you have a small initial force applied by you on the brake pedal. This transfers via a brake line, which had a small cylindrical area. The brake fluid then enters a chamber with more AREA allowing a LARGE FORCE to be applied on the brake shoes, which in turn slow the car down.

61. Pause for a Cause: Pascal’s Principle
What force must the small piston of radius 5.00 cm exert on the large piston of radius 15.0 cm to lift a 13,300 N car?
r1 = 5.00 cm
r2 = 15.0 cm
F2 = 13,300 N

62. Buoyancy Demo
The principle affecting objects submerged in fluids was discovers by Greek mathematician and natural philosopher Archimedes.
Principle: Any object immersed completely or partially in a fluid it is buoyed UPWARD by a force with a magnitude equal to the weight of the fluid displaced by the object.
When the object is placed in fluid it DISPLACES a certain amount of fluid. If the object is completely submerged, the VOLUME of the OBJECT is EQUAL to the VOLUME of FLUID it displaces.

63. Archimedes's Principle
" An object is buoyed up by a force equal to the weight of the fluid displaced."
In the figure, we see that the difference between the weight in AIR and the weight in WATER is 3 lbs. This is the buoyant force that acts upward to cancel out part of the force. If you were to weight the water displaced it also would weigh 3 lbs.

64. Archimedes's Principle: “Eureka!”
The bathtub epiphany! -According to legend, Archimedes was asked by King Hieron to determine whether the king’s crown was made of only pure gold, or merely a gold alloy (mix).
It had to be done without damaging the crown.
-The solution came while taking a bath. Archimedes realized he felt lighter when submerged in water. As the story goes he was so excited he ran naked through the streets yelling “Eureka!” “Eureka!” -Greek for, “I have found it.”

65. Archimedes's Principle

66. Pause for a Cause
A bargain hunter purchases a "gold" crown at a flea market. After she gets home, she hangs it from a scale and finds its weight in air to be 7.84 N. She then weighs the crown while it is immersed in water (density of water is 1000 kg/m3) and now the scale reads 6.86 N. Is the crown made of pure gold if the density of gold is 19.3 x 103 kg/m3?
Sub (1) into (2) in terms of mg

67. Pause for a Cause: Archimedes
The weight in air = 7.84 N
The weight in water = 6.86 N
рwater = 1000 kg/m3
The density of pure Gold is known to be 19.3 X 103 kg/m3. Determine the density of the Gold crown and determine if the crown is made of pure Gold.
The weight in air = 7.84 N
The weight in water = 6.86 N
The density of pure Gold is known to be 19.3 X 103 kg/m3

68. Fluid Flow
Up till now, we have pretty much focused on fluids at rest. Now let's look at fluids in motion
It is important that you understand that an IDEAL FLUID:
Is non viscous (meaning there is NO internal friction)
Is incompressible (meaning its Density is constant)
Its motion is steady and NON – TURBULENT
A fluid's motion can be said to be STREAMLINE, or LAMINAR.
The path itself is called the streamline.
By Laminar, we mean that every particle moves exactly along the smooth path as every particle that follows it.
If the fluid DOES NOT have Laminar Flow it has TURBULENT FLOW in which the paths are irregular and called EDDY CURRENTS (angular momentum).

69. Mass Flow Rate Consider a pipe with a fluid moving within it.
The volume of the blue region is the AREA times the length.
Length is velocity times time
Density is mass per volume
Putting it all together you have MASS FLOW RATE. A L v A v L

70. What happens if the Area changes?
The first thing you MUST understand is that MASS is NOT CREATED OR DESTROYED!
IT IS CONSERVED. v2 A2
L1=v1t
L2=v2t
The MASS that flows into a region = The MASS that flows out of a region. v1 A1
Using the Mass Flow rate equation and the idea that a certain mass of water is constant as it moves to a new pipe section:
example: The pipe delivery your house hold water supply from the meter to the house a 1 inch pipe. However the pipe delivery the water from your hot water heater is usually a ½ pipe.
If the fluid is incompressible then the density wont change
We have the Fluid Flow Continuity equation

71. Equation of Continuity
What do we mean by “good water pressure.”
The mass of the fluid is conserved
How would you expect the velocity of flow to change according to the cross-sectional area of the pipe?
Pause for Cause: What is the initial velocity of the fluid at A1 if?

72. Pause for a Cause
The speed of blood in the aorta is 50 cm/s and this vessel has a radius of 1.0 cm. If the capillaries have a total cross sectional area of 3000 cm2, what is the speed of the blood in them?
0.052 cm/s

73. Volume Flow Rate Consider a pipe with a fluid moving within it.
The product of area* velocity (m3/s) is called flow rate
Important
Since fluids are not compressible the volume of fluid the enters a tube in a given time interval is equal to the amount of fluid that leaves the tube over the same time interval.

74. Pause for a Cause
A water hose 2.50 cm in diameter is used by a gardener to fill a 30.0 liter bucket (One liter = 1000 cm3). The gardener notices that is takes 1.00 min to fill the bucket. A nozzle with an opening of cross-sectional area cm3 is then attached. The nozzle is held so that water is projected horizontally from a point 1.00 m above the ground. Over what horizontal distance can the water be projected?

75. Bernoulli's Principle
The Swiss Physicist Daniel Bernoulli, was interested in how the velocity changes as the fluid moves through a pipe of different area. He especially wanted to incorporate pressure into his idea as well. Conceptually, his principle is stated as: " If the velocity of a fluid increases, the pressure decreases and vice versa."
The velocity can be increased by pushing the air over or through a CONSTRICTION
Consequence of energy conservation as applied to an ideal fluid
A change in pressure results in a NET FORCE towards the low pressure region.

76. Bernoulli's Principle
Funnel
Ping pong
Ball
Constriction

77. Bernoulli's Principle
The constriction in the Subclavian artery causes the blood in the region to speed up and thus produces low pressure. The blood moving UP the LVA is then pushed DOWN instead of up causing a lack of blood flow to the brain. This condition is called TIA (transient ischemic attack) or “Subclavian Steal Syndrome.
One end of a gopher hole is higher than the other causing a constriction and low pressure region. Thus the air is constantly sucked out of the higher hole by the wind. The air enters the lower hole providing a sort of air re-circulating system effect to prevent suffocation.

78. Bernoulli's Equation Let’s look at this principle mathematically.
X = L
F1 on 2
-F2 on 1
Work is done by the blue section of water applying a force on the red section.
Formula for work is?
According to Newton’s 3rd law, the red section of water applies an equal and opposite force back on the first.
Consequence of energy conservation as applied to an ideal fluid

79. Bernoulli's Equation v2 A2 y2
L1=v1t
L2=v2t y1 v1 A1
ground
Work is also done by GRAVITY as the water travels a vertical displacement UPWARD. As the water moves UP the force due to gravity is DOWN. So the work is NEGATIVE (Potential Energy).

80. Bernoulli's Equation Change in velocity is? KINETIC ENERGY!
Part of the work goes into changing the velocity. Ex: the volume that passes through A1 in a time interval Δt equals the volume that passes through A2 in the same time interval.
Change in velocity is?
KINETIC ENERGY!

81. Bernoulli's Equation
Put it all together now.
Mass = ρV
Divided out volume
Rearrange like terms
Swiftly moving fluids exert less pressure than do slowly moving fluids
Swiftly moving fluids exert less pressure than do slowly moving fluids

82. Bernoulli's Equation Moving everything related to one side results in:
What this basically shows is that Conservation of Energy holds true within a fluid and that if you add the PRESSURE, the KINETIC ENERGY (in terms of density) and POTENTIAL ENERGY (in terms of density) you get the SAME VALUE anywhere along a streamline.

83. Pause for a Cause
Water circulates throughout the house in a hot-water heating system. If the water is pumped at a speed of 0.50 m/s through a 4.0 cm diameter pipe in the basement under a pressure of 3.0 atm, what will be the flow speed and pressure in a 2.6 cm-diameter pipe on the second floor 5.0 m above?
1 atm = 1x105 Pa
1.183 m/s
2.5x105 Pa(N/m2) or 2.5 atm

84. Pause for a Cause
Consider what we know
A large pipe with cross-sectional area of 1.00 m2 descends 5.00 m and narrows to 0.500m2, where it terminates in a valve at point (1). If the pressure at (2) is atmospheric pressure, and the valve is opened wide and water allowed to flow freely, find the speed of the water leaving the pipe.
1 atm = 1x105 Pa
Sub for V2
Consider what we know
Both pressures are ATM
Solve for V1

85. Pause for a Cause
Consider what we know
A large pipe with cross-sectional area of 1.00 m2 descends 5.00 m and narrows to 0.500m2, where it terminates in a valve at point (1). If the pressure at (2) is atmospheric pressure, and the valve is opened wide and water allowed to flow freely, find the speed of the water leaving the pipe.
1 atm = 1x105 Pa
Subtract Po out
Multiply the 2 out divide the ρ out
Get V1 together
Consider what we know
Factor V1 out
= 11.4m/s
Solve for V1

86. Note:
The denominator goes away if the ratio of A1 and A2 is very large
Ex:
A1 = 0.5 m2
A2 = 1000 m2
Consider what we know
General Relativity

87. Pause for a Cause
1 atm = 1x105 Pa
A nearsighted sheriff fires at a cattle rustler with his trusty six-shooter. Fortunately for the rustler, the bullet misses him and penetrates the town water tank, causing a leak. A2 is 1000 m2 and A1 is 0.5 m2. a) If the top of the tank is open to the atmosphere, determine the speed at which water leaves the hole when the water level is m above the hole. b) Where does the stream hit the ground if the hole is 3.00 m above the ground?
Assume the ratio between A1/A2 = 1
Solve for v1
Assume the ratio between A1/A2 = 1
= 3.13 m/s

88. Pause for a Cause
1 atm = 1x105 Pa
A nearsighted sheriff fires at a cattle rustler with his trusty six-shooter. Fortunately for the rustler, the bullet misses him and penetrates the town water tank, causing a leak. a) If the top of the tank is open to the atmosphere, determine the speed at which water leaves the hole when the water level is m above the hole. b) Where does the stream hit the ground if the hole is 3.00 m above the ground?
= 3.13 m/s
b) Projectile motion
Assume the ratio between A1/A2 = 1
= 2.45 m

89. Cohesion & Adhesion continued
The force of attraction between unlike charges in the atoms or molecules of substances are responsible for cohesion and adhesion.
Cohesion is the clinging together of molecules/atoms within a substance. Ever wonder why rain falls in drops rather than individual water molecules? It’s because water molecules cling together to form drops.
Adhesion is the clinging together of molecules/atoms of two different substances. Adhesive tape gets its name from the adhesion between the tape and other objects. Water molecules cling to many other materials besides clinging to themselves.
continued

90. Cohesion & Adhesion (cont.)
The meniscus in a graduated cylinder of water is due to the adhesion between water molecules the sides of the tube. The adhesion is greater than the cohesion between the water molecules.
The reverse is true about a column of mercury: Mercury atoms are attracted to each other more strongly than they are attracted to the sides of the tube. This causes a sort of “reverse meniscus.”
H2O Hg

91. Why molecules “cling” + O _ H
To understand why molecules cling to each other or to other molecules, lets take a closer look at water. Each blue line represents a single covalent bond (one shared pair of electrons). Two other pairs of electrons also surround the central oxygen atom. The four electron pairs want to spread out as much as possible, which _
gives H2O its bent shape. It is this shape that account for water’s unusual property of expanding upon freezing.
The shared electrons are not shared equally. Oxygen is more electronegative than hydrogen, meaning this is an unequal tug-o-war, where the big, strong oxygen keeps the shared electrons closer to itself than to hydrogen. The unequal sharing, along with the electron pairs not involved in sharing, make water a polar molecule. Water is neutral, but it has a positive side and a negative side. This accounts for water’s cohesive and adhesive nature as well as its ability to dissolve so many other substances.

92. Why molecules “cling” (cont.)
The dashed lines represent weak, temporary bonds between molecules. Water molecules can cling to other polar molecules besides them-selves, which is why water is a good solvent. Water won’t dissolve nonpolar molecules, like grease, though. (Detergent molecules have polar ends to attract water and nonpolar ends to mix with the grease.) Nonpolar molecules can attract each other to some extent, otherwise they couldn’t exist in a liquid or solid state. This attraction is due to random asymmetries in the electron clouds around the nuclei of atoms.

93. Capillary Action
How do trees pump water hundreds of feet from the ground to their highest leaves? Why do paper towels soak up spills? Why does liquid wax rise to the tip of a candle wick to be burned? Why must liquids on the space shuttle be kept covered to prevent them from crawling right out of their containers?! These are all examples of capillary action--the movement of a liquid up through a thin tube. It is due to adhesion and cohesion.
Capillary action is a result of adhesion and cohesion. A liquid that adheres to the material that makes up a tube will be drawn inside. Cohesive forces between the molecules of the liquid will “connect” the molecules that aren’t in direct contact with the inside of the tube. In this way liquids can crawl up a tube. In a pseudo-weightless environment like in the space shuttle, the “weightless” fluid could crawl right out of its container.
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94. Capillary Action (cont.)
The setups below looks just like barometers, except the tubes are open to the air. Since the pressure is the same at the base and inside the tube, there is no pressure difference to support the column of fluid. The column exists because of capillarity. (Barometers must compen-sate for this effect.) The effect is greater in thin tubes because there is more surface area of tube per unit of weight of fluid: The force supporting fluid is proportional to the surface area of the tube, 2  r h, where h is the fluid height. The weight of the fluid in the tube is proportional to its volume,  r 2 h. If the radius of the tube is doubled,
the surface area doubles (and so does the force supporting the fluid), but the volume quadruples (as does the weight). Note: if the fluid were mercury, rather than rise it be depressed by the tube.

95. Surface Tension
Ever wonder why water beads up on a car, or how some insects can walk on water, or how bubbles hold themselves together? The answer is surface tension: Because of cohesion between its molecules, a substance
tends to contract to the smallest area possible. Water on a waxed surface, for example, forms round beads because in this shape, more weak bounds can be formed between molecules than if they were arranged in one flat layer. The drops of water are flattened, however, due to their weight. Cohesive forces are greater in mercury than in water, so it forms a more spherical shape. Cohesive forces are weaker in alcohol than in water, so it forms a more flattened shape.
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mercury
water
alcohol

96. Surface Tension (cont.)
Below the surface a molecule in fluid is pulled in all directions by its neighbors with approximately equal strength, so the net force on it is about zero. This is not the case at the surface. Here the net force on a molecule is downward. Thus, the layer of molecules at the surface are slightly compressed. This surface tension is strong enough in water to support objects denser than the itself, like water bugs and even razorblades (so long as the blade is laid flat on the water so that more water molecules can help support its weight).
Surface tension can be defined as the force per unit length holding a surface together. Imagine you’re in a water balloon fight. You have one last balloon, but it’s got a slash in it, so you tape it up and fill it
with water. The surface tension is the force per unit length the tape must exert on the balloon to hold it together. A bubble is similar to the water balloon. Rather than tape, the bubble is held together by the cohesive forces in the bubble.