Raymond A. Serway Chris Vuille Chapter Nine Solids and Fluids States of Ma<er •  Solid, liquid, gas •  Plasma •  This chapter introduces basic properDes of solids and liquids –  Includes some properDes of gases IntroducDon Solids •  Have definite volume •  Have definite shape •  Molecules are held in specific locaDons –  By electrical forces •  Vibrate about equilibrium posiDons •  Can be modeled as springs connecDng molecules SecDon 9.1 More About Solids •  External forces can be applied to the solid and compress the material –  In the model, the springs would be compressed •  When the force is removed, the solid returns to its original shape and size –  This property is called elasDcity SecDon 9.1 Crystalline Solid •  Atoms have an ordered structure •  This example is salt –  Gray spheres represent Na+ ions –  Green spheres represent Cl-­‐ ions SecDon 9.1 Amorphous Solid •  Atoms are arranged almost randomly •  Examples include glass SecDon 9.1 Liquid •  Has a definite volume •  No definite shape •  The molecules “wander” through the liquid in a random fashion –  The intermolecular forces are not strong enough to keep the molecules in a fixed posiDon SecDon 9.1 Gas Has no definite volume Has no definite shape Molecules are in constant random moDon The molecules exert only weak forces on each other •  Average distance between molecules is large compared to the size of the molecules •
•
•
•
SecDon 9.1 Plasma •  Gas heated to a very high temperature •  Many of the electrons are freed from the nucleus •  Result is a collecDon of free, electrically charged ions SecDon 9.1 Types of Ma<er •  Normal ma<er –  About 5% of total ma<er •  Dark ma<er –  Affects the moDon of stars in galaxies –  May be as much as 25% of total ma<er •  Dark energy –  Accounts for acceleraDon of the expansion of the universe –  May be as much as 70% of all ma<er SecDon 9.1 Density •  The density of a substance of uniform composiDon is defined as its mass per unit volume: •  SI unit: kg/m3 (SI) –  Oden see g /cm3 (cgs) •  1 g /cm3 = 1000 kg/m3 SecDon 9.2 Specific Gravity •  The specific gravity of a substance is the raDo of its density to the density of water at 4° C –  The density of water at 4° C is 1000 kg/m3 •  Specific gravity is a dimensionless quanDty SecDon 9.2 Pressure •  The force exerted by a fluid on a submerged object at any point is perpendicular to the surface of the object •  The average pressure P is the force divided by the area SecDon 9.2 Pressure and Weight of Water: (a) Calculate the weight of a cylindrical column of water with height h = 40.0 m and radius r = 1.00 m. (b) Calculate the force exerted by air on a disk of radius 1.00 m at the water’s surface. (c) What pressure at a depth of 40.0 m supports the water column? Pressure and Weight of Water: (a) Calculate the weight of a cylindrical column of water with height h = 40.0 m and radius r = 1.00 m. (b) Calculate the force exerted by air on a disk of radius 1.00 m at the water’s surface. (c) What pressure at a depth of 40.0 m supports the water column? (a) W = mg = ρVg = (1000 kg/m3 )(πr 2 h )(9.8 m/s2 ) = 1.23 × 10 6 N
(b) Pdown =
(c) Pup =
Fdown
⇒ Fdown = Pdown A = (1.01 × 10 5 N/m2 )(πr 2 ) = 3.173 × 10 5 N
A
Fup
A
Fup = Fdown + W ⇒ Pup = Pdown +
ρVg
W
= Pdown +
= Pdown + ρhg
A
A
Pup = 1.01 × 10 5 N/m2 + (1000 kg/m3 )(40m)(9.8 m/s2 ) = 4.93 × 10 5 N
€
DeformaDon of Solids •  All objects are deformable •  It is possible to change the shape or size (or both) of an object through the applicaDon of external forces •  When the forces are removed, the object tends to its original shape –  An object undergoing this type of deformaDon exhibits elas/c behavior SecDon 9.3 ElasDc ProperDes •  Stress is the force per unit area causing the deformaDon •  Strain is a measure of the amount of deformaDon •  The elas*c modulus is the constant of proporDonality between stress and strain –  For sufficiently small stresses, the stress is directly proporDonal to the strain –  The constant of proporDonality depends on the material being deformed and the nature of the deformaDon SecDon 9.3 ElasDc Modulus •  stress = elasDc modulus x strain •  The elasDc modulus can be thought of as the sDffness of the material –  A material with a large elasDc modulus is very sDff and difficult to deform SecDon 9.3 Young’s Modulus: ElasDcity in Length •  The bar is stressed –  Its length is greater than Lo –  The external force is balanced by internal forces •  Tensile stress is the raDo of the external force to the cross-­‐secDonal area –  Tensile is because the bar is under tension SecDon 9.3 Young’s Modulus, cont. •  SI unit of stress is Pascal, Pa –  1 Pa = 1 N/m2 •  The tensile strain is the raDo of the change in length to the original length –  Strain is dimensionless •  The elasDc modulus is called Young’s modulus SecDon 9.3 ElasDc Behavior Graph •  It is possible to exceed the elas/c limit of the material –  No longer directly proporDonal –  Ordinarily does not return to its original length •  If stress conDnues, it surpasses its ul/mate strength –  The ulDmate strength is the greatest stress the object can withstand without breaking SecDon 9.3 Built to Last: A verDcal steel beam in a building supports a load of 6.0 × 10 4 N. (a) If the length of the beam is 4.0 m and its cross-­‐secDonal area is 8.0 × 10-­‐3 m2, find the distance the beam is compressed along its length. (b) What maximum load in newtons could the steel beam support before failing? The equation relating tensile stress
(a)
ΔL
F
to tensile strain
is :
L0
A
6.0 × 10 4 N)( 4.0 m)
(
ΔL
F L0
F
−4
=Y
⇒ ΔL =
=
=
1.5
×
10
m
L0
A Y (8.0 × 10 −3 m2 )(2.00 × 1011 Pa)
A
(b) The maximum pressure the vertical steel beam can support under compression
is referred to as the compression strength and from Table 9.3 is Pcomp = 5 × 10 8 Pa
Pcomp =
€
Fmax
⇒ Fmax = Pcomp A = 4.0 × 10 6 N
A
Shear Modulus: ElasDcity of Shape •  Forces may be parallel to one of the object’s faces •  The stress is called a shear stress –  Defined as the raDo of the magnitude of the parallel force to the area of the face being sheared •  The shear strain is the raDo of the horizontal displacement and the height of the object •  The shear modulus is S SecDon 9.3 Shear Modulus, EquaDons •  •  S is the shear modulus •  A material having a large shear modulus is difficult to bend SecDon 9.3 Football Injuries: A defensive lineman of mass M = 125 kg makes a flying tackle at vi = 4.00 m/s on a staDonary quarterback of mass m =85.0 kg, and the lineman’s helmet makes solid contact with the quarterback’s femur. (a) What is the speed vf of the two athletes immediately ader contact? Assume a linear inelasDc collision. (b) If the collision lasts for 0.100 s, esDmate the average force exerted on the quarterback’s femur. (c) If the cross-­‐secDonal area of the quarterback’s femur is equal to 5.00 × 10-­‐4 m2, calculate the shear stress exerted on the bone in the collision. (a) Conservation of momentum
M
v = 2.38 m/s
Mv i = (m + M)v f ⇒ v f =
m+M i
(b) The force on the quarterback's femur is equal to the
change in the momentum of the linesman
F=
(
)
Δp p f − pi M v f − v i
=
=
= 2.03 × 10 3 N
Δt
Δt
Δt
(c)
2.03 × 10 3 N
6
The shear stress is =
-4
2 = 4.06 × 10 Pa
5.00 × 10 m
The equation that relates shear stress
F
Δx
=S
A
h
€
Δx
F
to shear strain
is :
h
A
Bulk Modulus: Volume ElasDcity •  Bulk modulus characterizes the response of an object to uniform squeezing –  Suppose the forces are perpendicular to, and act on, all the surfaces •  Example: when an object is immersed in a fluid •  The object undergoes a change in volume without a change in shape SecDon 9.3 Bulk Modulus, cont. •  Volume stress, ΔP, is the raDo of the change in the magnitude of the applied force to the surface area –  This is also a change in pressure •  The volume strain is equal to the raDo of the change in volume to the original volume SecDon 9.3 Bulk Modulus, final •  A material with a large bulk modulus is difficult to compress •  The negaDve sign is included since an increase in pressure will produce a decrease in volume –  B is always posiDve •  The compressibility is the reciprocal of the bulk modulus SecDon 9.3 Lead Ballast Overboard: Ships and sailing vessels oden carry lead ballast in various forms, such as bricks, to keep the ship properly oriented and upright in the water. Suppose a ship takes on cargo and the crew jersons a total of 0.500 m3 of lead ballast into water 2.00 km deep. Calculate (a) the change in the pressure at that depth and (b) the change in volume of the lead upon reaching the bo<om. Take the density of sea water to be 1.025 × 103 kg/m3, and take the bulk modulus of lead to be 4.2 × 1010 Pa. (a) The change in pressure can be found from :
Pbot - Ptop = ρgh = (1.025 × 10 3 kg)(2. × 10 3 m)(9.8m/s2 ) = 2.01 × 10 7 Pa
(b) The equation that relates the bulk stress ΔP to the bulk strain
ΔV
is :
V0
7
3
ΔPV0 (2.01 × 10 Pa)(9.5m )
ΔV
⇒ ΔV =
=
= −2.4 × 10 −4 m3
ΔP = -B
10
-B
−4.2 × 10
V0
€
VariaDon of Pressure with Depth •  If a fluid is at rest in a container, all porDons of the fluid must be in staDc equilibrium •  All points at the same depth must be at the same pressure –  Otherwise, the fluid would not be in equilibrium –  The fluid would flow from the higher pressure region to the lower pressure region SecDon 9.4 Pressure and Depth •  Examine the darker region, assumed to be a fluid –  It has a cross-­‐secDonal area A –  Extends to a depth h below the surface •  Three external forces act on the region SecDon 9.4 Pressure and Depth equaDon •  •  Po is normal atmospheric pressure –  1.013 x 105 Pa = 14.7 lb/in.2 •  The pressure does not depend upon the shape of the container SecDon 9.4 Oil and Water: In a huge oil tanker, salt water has flooded an oil tank to a depth of h2 = 5.00 m. On top of the water is a layer of oil h1 = 8.00 m deep, as in the cross-­‐secDonal view of the tank in the Figure. The oil has a density of 0.700 g /cm3. Find the pressure at the bo<om of the tank. (Take 1,025 kg/m3 as the density of salt water.) Oil and Water: In a huge oil tanker, salt water has flooded an oil tank to a depth of h2 = 5.00 m. On top of the water is a layer of oil h1 = 8.00 m deep, as in the cross-­‐secDonal view of the tank in the Figure. The oil has a density of 0.700 g /cm3. Find the pressure at the bo<om of the tank. (Take 1,025 kg/m3 as the density of salt water.) Because there are fluids of different density present we calculate the pressure
at the interfaces first.
P0 = 1.01 × 10 5 Pa
5
P1 = P0 + ρ oil gh1 = 1.01 × 10 Pa + 0.7
10 −3 kg
(10 m)
−2
(9.8 m/s )(8.00m) ⇒
2
3
P1 = 1.56 × 10 5 Pa
P2 = P1 + ρ water gh2 = 1.56 × 10 5 Pa + 1,025
P2 = 2.06 × 10 5 Pa
€
kg
2
3 (9.8 m/s )(5.00m) ⇒
m
Pascal’s Principle •  A change in pressure applied to an enclosed fluid is transmi<ed undiminished to every point of the fluid and to the walls of the container. –  First recognized by Blaise Pascal, a French scienDst (1623 – 1662) SecDon 9.4 Pascal’s Principle, cont •  The hydraulic press is an important applicaDon of Pascal’s Principle •  Also used in hydraulic brakes, forklids, car lids, etc. SecDon 9.4 Pressure Measurements: Manometer •  One end of the U-­‐
shaped tube is open to the atmosphere •  The other end is connected to the pressure to be measured •  If P in the system is greater than atmospheric pressure, h is posiDve –  If less, then h is negaDve SecDon 9.5 Absolute vs. Gauge Pressure •  The pressure P is called the absolute pressure –  Remember, P = Po + ρgh •  P – Po = ρgh is the gauge pressure SecDon 9.5 Car Lift: In a car lift used in a service station, compressed air exerts a force on
a small piston of circular cross section having a radius of r1 = 5.00 cm. This
pressure is transmitted by an incompressible liquid to a second piston
of radius r2 = 15.0 cm. (a) What force must the compressed air exert on the
small piston in order to lift a car weighing 13,300 N? Neglect the weights of the
pistons. (b) What air pressure will produce a force of that magnitude? (c) Show
that the work done by the input and output pistons is the same.
Car Lift: In a car lift used in a service station, compressed air exerts a force on
a small piston of circular cross section having a radius of r1 = 5.00 cm. This
pressure is transmitted by an incompressible liquid to a second piston
of radius r2 = 15.0 cm. (a) What force must the compressed air exert on the
small piston in order to lift a car weighing 13,300 N? Neglect the weights of the
pistons. (b) What air pressure will produce a force of that magnitude? (c) Show
that the work done by the input and output pistons is the same.
We apply Pascal's principle to solve this problem :
The pressure applied by the small piston is transmitted by the
incompressible liquid equaly to the second piston.
2
⎛ 5.00cm ⎞ 2
F1
F2
F2 ⎛ r1 ⎞
(a) P1 = P2 ⇒
=
⇒ F1 = A1
= ⎜ ⎟ F2 = ⎜
⎟ 13,300N
⎝ 15.0cm ⎠
A1
A2
A2 ⎝ r2 ⎠
(b) P1 =
F1
F
= 12
A1 πr1
(c) W1 = F1d1 =
F1A1d1 F1V1
=
A1
A1
F2 A2d 2 F2V2
=
A2
A2
Since the fluid is incompressible V1 = V2 so W1 = W2
W2 = F2d 2 =
€
Archimedes' Principle •  Any object completely or parDally submerged in a fluid is buoyed up by a force whose magnitude is equal to the weight of the fluid displaced by the object SecDon 9.6 Buoyant Force •  The upward force is called the buoyant force •  The physical cause of the buoyant force is the pressure difference between the top and the bo<om of the object SecDon 9.6 Buoyant Force, cont. •  The magnitude of the buoyant force always equals the weight of the displaced fluid SecDon 9.6 Archimedes’ Principle: Totally Submerged Object •  The upward buoyant force is B=ρfluidVobjg •  The downward gravitaDonal force is w=mg=ρobjVobjg •  The net force is B-­‐w=(ρfluid-­‐ρobj)Vobjg SecDon 9.6 Totally Submerged Object •  The net force is B-­‐w=
(ρfluid-­‐ρobj)Vobjg •  If the object is less dense than the fluid the object experiences a net upward force SecDon 9.6 Totally Submerged Object, 2 •  The net force is B-­‐w=
(ρfluid-­‐ρobj)Vobjg •  If the object is more dense than the fluid the object experiences a net downward force SecDon 9.6 Archimedes’ Principle: FloaDng Object •  The object is in staDc equilibrium •  The upward buoyant force is balanced by the downward force of gravity •  Volume of the fluid displaced corresponds to the volume of the object beneath the fluid level SecDon 9.6 Archimedes’ Principle: FloaDng Object, cont •  The forces balance •  –  Neglects the buoyant force of the air SecDon 9.6 Pressure Measurements: Barometer •  Invented by Torricelli (1608 – 1647) •  A long closed tube is filled with mercury and inverted in a dish of mercury •  Measures atmospheric pressure as ρgh SecDon 9.5 Pressure Values in Various Units •  One atmosphere of pressure is defined as the pressure equivalent to a column of mercury exactly 0.76 m tall at 0o C where g = 9.806 65 m/s2 •  One atmosphere (1 atm) = –  76.0 cm of mercury –  1.013 x 105 Pa –  14.7 lb/in2 SecDon 9.5 Gold Crown: A bargain hunter purchases a “gold” crown at a flea market. Ader she gets home, she hangs it from a scale and finds its weight to be 7.84 N (Figure a). She then weighs the crown while it is immersed in water, as in Figure b, and now the scale reads 6.86 N. Is the crown made of pure gold? Gold Crown: A bargain hunter purchases a “gold” crown at a flea market. Ader she gets home, she hangs it from a scale and finds its weight to be 7.84 N (Figure a). She then weighs the crown while it is immersed in water, as in Figure b, and now the scale reads 6.86 N. Is the crown made of pure gold? The weight of the crown when it's hanging in the air is :
Wair = mcrown g (1)
When the crown is in the water :
Wwater + Fbuoy - mcrown g = 0 ⇒ Wwater + mwater g − mcrown g = 0 ⇒
Wwater + ρ waterVcrown g − mcrown g = 0 ⇒
Wwater − Wair + ρ waterVcrown g = 0 ⇒
Wwater − Wair + ρ water
mcrown g
=0⇒
ρ crown
Wwater − Wair + ρ water
Wair
=0
ρ crown
ρ crown
1000 kg/m3 )( 7.84 N)
(
ρ water Wair
=
=
= 8000 kg/m3
Wair − Wwater
0.98 N
but ρ gold = 19.3 × 10 3 kg/m3 (don't by gold at a flea market)
€
Floating Down the River: A raft is constructed of wood having a density of
6.00 × 102 kg/m3. Its surface area is 5.70 m2, and its volume is 0.60 m3. When
the raft is placed in fresh water as in the Figure, to what depth h is the bottom
of the raft submerged?
Floating Down the River: A raft is constructed of wood having a density of
6.00 × 102 kg/m3. Its surface area is 5.70 m2, and its volume is 0.60 m3. When
the raft is placed in fresh water as in the Figure, to what depth h is the bottom
of the raft submerged?
The buoyancy force is equal to the weight of the raft.
The buoyancy force is also equal to the weight of the displaced water :
FB = mraft g ⇒ mdw g = mraft g
ρ waterVdw = ρ raftVraft ⇒ ρ water Araft h = ρ raftVraft
6.00 × 10 kg/m )0.60m
(
h=
(1000 kg/m )(5.70m )
2
3
3
€
2
3
= 0.063m
ρ raftVraft
⇒h=
⇒
ρ water Araft
Floating in two fluids: A 1.00 × 103-kg cube of aluminum is placed in
a tank. Water is then added to the tank until half the cube is immersed. (a)
What is the normal force on the cube? (See Figure a.) (b) Mercury is now
slowly poured into the tank until the normal force on the cube goes to zero.
(See Figuere b) How deep is the layer of mercury?
Floating in two fluids: A 1.00 × 103-kg cube of aluminum is placed in
a tank. Water is then added to the tank until half the cube is immersed. (a)
What is the normal force on the cube? (See Figure a.) (b) Mercury is now
slowly poured into the tank until the normal force on the cube goes to zero.
(See Figuere b) How deep is the layer of mercury?
(a) Apply Newton's 2nd Law ∑ F = 0
n + Fbuoy = mcube g (1)
Fbuoy = mdw g = ρ H 2 0Vdw g = ρ H 2 0
ρ AL =
mcube
m
⇒ Vcube = cube
Vcube
ρ AL
Vcube
g (2)
2
(3)
(1)^(2)^(3) ⇒ n = mcube g − Fbuoy = mcube g − ρ H 2 0
⎛ ρ H 0 ⎞
mcube
g = mcube g⎜1 − 2 ⎟
n = mcube g − ρ H 2 0
2 ρ AL
⎝ 2 ρ AL ⎠
Vcube
g⇒
2
Floating in two fluids: A 1.00 × 103-kg cube of aluminum is placed in
a tank. Water is then added to the tank until half the cube is immersed. (a)
What is the normal force on the cube? (See Figure a.) (b) Mercury is now
slowly poured into the tank until the normal force on the cube goes to zero.
(See Figuere b) How deep is the layer of mercury?
(b) Apply Newton's 2nd Law ∑ F = 0
There is buoyancy from Hg and water
Fbuoy, Hg + Fbuoy, H 2 0 = mcube g ⇒
mdHg g + mdw g = mcube g ⇒ ρ HgVdHg g + ρ H 2 0Vdw g = mcube g ⇒
ρ Hg Ah Hg g + ρ H 2 0 Ahwater g = mcube g = ρ AL Ahg (4)
h
hwater = ,(water height is half the height of the cube) (5)
2
⎛ m ⎞
⎛ m ⎞1/ 3 ⎛ 1 × 10 3 kg ⎞1/ 3
3
cube
cube
= 0.718m (6)
and from (3) Vcube = h = ⎜
⎟ ⇒ h = ⎜
⎟ = ⎜
3
3 ⎟
⎝ 2.7 × 10 kg/m ⎠
⎝ ρ AL ⎠
⎝ ρ AL ⎠
h
(4)^(5)^(6) ⇒ ρ Hg h Hg + ρ H 2 0 = ρ AL h
2
h
ρ AL h − ρ H 2 0
0.718m (2.7 × 10 3 kg/m3 ) − (1 × 10 3 kg/m3 )(0.5)
2 =
h Hg =
= 0.116m
ρ Hg
13.6 × 10 3 kg/m3
[
€
]
Fluids in MoDon: Streamline Flow •  Streamline flow –  Every parDcle that passes a parDcular point moves exactly along the smooth path followed by parDcles that passed the point earlier –  Also called laminar flow •  Streamline is the path –  Different streamlines cannot cross each other –  The streamline at any point coincides with the direcDon of fluid velocity at that point SecDon 9.7 Streamline Flow, Example •  Streamline flow shown around an auto in a wind tunnel SecDon 9.7 Fluids in MoDon: Turbulent Flow •  The flow becomes irregular –  Exceeds a certain velocity –  Any condiDon that causes abrupt changes in velocity •  Eddy currents are a characterisDc of turbulent flow SecDon 9.7 Turbulent Flow, Example •  The smoke first moves in laminar flow at the bo<om •  Turbulent flow occurs at the top SecDon 9.7 Fluid Flow: Viscosity •  Viscosity is the degree of internal fricDon in the fluid •  The internal fricDon is associated with the resistance between two adjacent layers of the fluid moving relaDve to each other SecDon 9.7 CharacterisDcs of an Ideal Fluid •  The fluid is nonviscous –  There is no internal fricDon between adjacent layers •  The fluid is incompressible –  Its density is constant •  The fluid moDon is steady –  The velocity, density, and pressure at each point in the fluid do not change with Dme •  The fluid moves without turbulence –  No eddy currents are present –  The elements have zero angular velocity about its center SecDon 9.7 EquaDon of ConDnuity •  For a steady flow the conservaDon of mass leads to: ρ1A1v1 = ρ2A2v2 For an incompressible fluid: A1v1 = A2v2 •  The product of the cross-­‐
secDonal area of a pipe and the fluid speed is a constant •  The product Av is called the flow rate, Av=ΔV/Δt SecDon 9.7 Daniel Bernoulli •  1700 – 1782 •  Swiss physicist and mathemaDcian •  Wrote Hydrodynamica •  Also did work that was the beginning of the kineDc theory of gases SecDon 9.7 Bernoulli’s EquaDon •  Relates pressure to fluid speed and elevaDon •  Bernoulli’s equaDon is a consequence of ConservaDon of Energy applied to an ideal fluid •  Assumes the fluid is incompressible and nonviscous, and flows in a nonturbulent, steady-­‐state manner SecDon 9.7 Bernoulli’s EquaDon, cont. •  States that the sum of the pressure, kineDc energy per unit volume, and the potenDal energy per unit volume has the same value at all points along a streamline SecDon 9.7 ApplicaDons of Bernoulli’s Principle: Measuring Speed •  Shows fluid flowing through a horizontal constricted pipe •  Speed changes as diameter changes •  Can be used to measure the speed of the fluid flow •  Swidly moving fluids exert less pressure than do slowly moving fluids SecDon 9.7 ApplicaDons of Bernoulli’s Principle: Venturi Tube •  The height is higher in the constricted area of the tube •  This indicates that the pressure is lower SecDon 9.7 ApplicaDon – Golf Ball •  The dimples in the golf ball help move air along its surface •  The ball pushes the air down •  Newton’s Third Law tells us the air must push up on the ball •  The spinning ball travels farther than if it were not spinning SecDon 9.8 ApplicaDon – Atomizer •  A stream of air passing over an open tube reduces the pressure above the tube •  The liquid rises into the airstream •  The liquid is then dispersed into a fine spray of droplets SecDon 9.8 ApplicaDon – Vascular Flu<er •  The artery is constricted as a result of accumulated plaque on its inner walls •  To maintain a constant flow rate, the blood must travel faster than normal •  If the speed is high enough, the blood pressure is low and the artery may collapse SecDon 9.8 ApplicaDon – Airplane Wing •  The air speed above the wing is greater than the speed below •  The air pressure above the wing is less than the air pressure below •  There is a net upward force –  Called li? •  Other factors are also involved Niagara Falls: Each second, 5,525 m3 of water flows over the 670-­‐m-­‐wide cliff of the Horseshoe Falls porDon of Niagara Falls. The water is approximately 2 m deep as it reaches the cliff. EsDmate its speed at that instant. Niagara Falls: Each second, 5,525 m3 of water flows over the 670-­‐m-­‐wide cliff of the Horseshoe Falls porDon of Niagara Falls. The water is approximately 2 m deep as it reaches the cliff. EsDmate its speed at that instant. m3
V
= 5525
The volume flow rate is :
s
Δt
3
V AΔl
1
V
m
=
= Av ⇒ v =
= 5525
= 4.2m/s
Δt
Δt
ΔtA
s (670m)(2m)
€
Watering a Garden: A water hose 2.50 cm in diameter is used by a gardener to fill a 30.0-­‐liter bucket. (One liter = 1,000 cm3.) The gardener noDces that it takes 1.00 min to fill the bucket. A nozzle with an opening of cross-­‐secDonal area 0.500 cm2 is then a<ached to the hose. 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? The volume flow rate for an incompressible fluid is :
ΔV
= Av
Δt
3
ΔV 3. × 10 4 cm3
−4 m
=
= 5 × 10
60s
s
Δt
ΔV
m3
5 × 10 −4
s
v = Δt =
2 = 1.02m/s
−2
A
π (1.25 × 10 m)
When a nozzle of cross - sectional area A1 is attached the new velocity will satisfy the equation :
2
−2
Av π (1.25 × 10 m)
Av = A1v1 ⇒ v1 =
=
(1.02m/s) = 10.01m/s
2
A1
0.5(10 −2 m)
Now we apply the equations of motion in 2D :
1
1
2y 0
y = y 0 + v oy t - gt 2 ⇒ 0 = y 0 − gt 2 ⇒ t =
2
2
g
x = vt = v
€
2(1.00m)
2y 0
= (10.01m/s)
= 4.52m
g
9.8m/s2
Shoot-­‐Out at the Old Water Tank: A nearsighted sheriff fires at a ca<le rustler with his trusty six-­‐shooter. Fortunately for the rustler, the bullet misses him and penetrates the town water tank, causing a leak (Fig. 9.32 on page 304). (a) If the top of the tank is open to the atmosphere, determine the speed at which the water leaves the hole when the water level is 0.500 m above the hole. (b) Where does the stream hit the ground if the hole is 3.00 m above the ground? Shoot-­‐Out at the Old Water Tank: A nearsighted sheriff fires at a ca<le rustler with his trusty six-­‐shooter. Fortunately for the rustler, the bullet misses him and penetrates the town water tank, causing a leak (Fig. 9.32 on page 304). (a) If the top of the tank is open to the atmosphere, determine the speed at which the water leaves the hole when the water level is 0.500 m above the hole. (b) Where does the stream hit the ground if the hole is 3.00 m above the ground? Bernoulli's equation is :
1
P + ρv 2 + ρgy = constant
2
We apply this equations at points 2 and 1:
1
1
P0 + ρv 22 + ρgy 2 = P0 + ρv12 + ρgy1 ⇒
2
2
A
A1v1 = A2v 2 ⇒ v 2 = 1 v1 ≈ 0
A2
1 2
v = gy 2 − gy1 ⇒ v1 = 2g( y 2 − y1 ) = 3.13m/s
2 1
1
2y 0
0 = y 0 − gt 2 ⇒ t =
= 0.78s
2
g
x = vt = ( 3.13m/s)(0.78s) = 2.44m
Fluid Flow in a Pipe: A large pipe with a cross-­‐secDonal area of 1.00 m2 descends 5.00 m and narrows to 0.500 m2, where it terminates in a valve at point 1 (see Figure). If the pressure at point 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. Fluid Flow in a Pipe: A large pipe with a cross-­‐secDonal area of 1.00 m2 descends 5.00 m and narrows to 0.500 m2, where it terminates in a valve at point 1 (see Figure). If the pressure at point 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. Bernoulli's equation is :
1
P + ρv 2 + ρgy = constant
2
We apply it at points 2 and 1:
1
1
P2 + ρv 22 + ρgy 2 = P1 + ρv12 + ρgy1
2
2
A
A2v 2 = A1v1 ⇒ v 2 = 1 v1
A2
1
1
P2 + ρv 22 − P1 − ρv12 = ρgy1 − ρgy 2 = −ρgh ⇒
2
2
2
⎛
⎞
1 ⎜ 2 ⎛ A1 ⎞ 2 ⎟
P2 − P1 − ρ⎜ v1 − ⎜ ⎟ v1 ⎟ = −ρgh ⇒
2 ⎝
⎝ A2 ⎠ ⎠
2
1 2 ⎡ ⎛ A1 ⎞ ⎤
2gh
⇒ v1 = 11.4m/s
P2 − P1 − ρv1 ⎢1 − ⎜ ⎟ ⎥ = −ρgh ⇒ v12 =
2 ⎤
⎡
2
A
⎛
⎞
⎝
⎠
⎢⎣
⎥⎦
2
A
⎢1 − ⎜ 1 ⎟ ⎥
⎢⎣ ⎝ A2 ⎠ ⎥⎦
since P2 = P1 = P0
LiD on an Airfoil: An airplane has wings, each with area 4.00 m2, designed so that air flows over the top of the wing at 245 m/s and underneath the wing at 222 m/s. Find the mass of the airplane such that the lid on the plane will support its weight, assuming the force from the pressure difference across the wings is directed straight upward. LiD on an Airfoil: An airplane has wings, each with area 4.00 m2, designed so that air flows over the top of the wing at 245 m/s and underneath the wing at 222 m/s. Find the mass of the airplane such that the lid on the plane will support its weight, assuming the force from the pressure difference across the wings is directed straight upward. Bernoulli's equation is :
1 2
P + ρv + ρgy = constant
2
Apply above and below the wing :
1
1
Pb + ρv b2 = Pa + ρv a2
2
2
1
Pb − Pa = ρ(v a2 − v b2 )
2
F
1
P = ⇒ F = PA = ρA(v a2 − v b2 ) = 27712N
A
2
Ftotal = 2 × 27712N = m plane g ⇒ m plane = 5656 kg
€
Sailing Upwind: How can a sailboat accomplish the seemingly impossible task of sailing into the wind? Surface Tension •  Net force on molecule A is zero –  Pulled equally in all direcDons •  Net force on B is not zero –  No molecules above to act on it –  Pulled toward the interior of the fluid SecDon 9.9 Surface Tension, cont •  The net effect of this pull on all the surface molecules is to make the surface of the liquid contract •  Makes the surface area of the liquid as small as possible –  Example: Water droplets take on a spherical shape since a sphere has the smallest surface area for a given volume SecDon 9.9 Surface Tension on a Needle •  Surface tension allows the needle to float, even though the density of the steel in the needle is much higher than the density of the water •  The needle actually rests in a small depression in the liquid surface •  The verDcal components of the force balance the weight SecDon 9.9 Surface Tension, EquaDon •  The surface tension in a film of liquid is defined as the raDo of the magnitude of the surface tension force to the length along which the force acts: •  SI unit: N/m •  In terms of energy, any equilibrium configuraDon of an object is one in which the energy is a minimum SecDon 9.9 Measuring Surface Tension •  The force is measured just as the ring breaks free from the film •  –  The 2 L is due to the force being exerted on the inside and outside of the ring SecDon 9.9 A Closer Look at the Surface of Liquids •  Cohesive forces are forces between like molecules •  Adhesive forces are forces between unlike molecules •  The shape of the surface depends upon the relaDve size of the cohesive and adhesive forces SecDon 9.9 Liquids in Contact with a Solid Surface – Case 1 •  If the adhesive forces are greater than the cohesive forces then the liquid clings to the walls of the container •  The liquid “wets” the surface SecDon 9.9 Liquids in Contact with a Solid Surface – Case 2 •  If the cohesive forces are greater than the adhesive forces then the liquid curves downward •  The liquid does not “wet” the surface SecDon 9.9 Contact Angle •  The angle, φ, between the solid surface and a line drawn tangent to the liquid at the surface is called the contact angle •  In a, φ > 90° and cohesive forces are greater than adhesive forces •  In b, φ < 90° and adhesive forces are greater than cohesive forces SecDon 9.9 Capillary AcDon •  Capillary acDon is the result of surface tension and adhesive forces •  The liquid rises in the tube when adhesive forces are greater than cohesive forces •  At the point of contact between the liquid and the solid, the upward forces are as shown in the diagram SecDon 9.9 Capillary AcDon, cont. •  Here, the cohesive forces are greater than the adhesive forces •  The level of the fluid in the tube will be below the surface of the surrounding fluid SecDon 9.9 Capillary AcDon, final. •  The height to which the fluid is drawn into the tube is given by: –  h will also be the distance to the depressed surface SecDon 9.9 Walking on Water: Many insects can literally walk on water, using surface tension for their support. To show this is feasible, assume the insect’s “foot” is spherical. When the insect steps onto the water with all six legs, a depression is formed in the water around each foot, as shown in Figure 9.43a. The surface tension of the water produces upward forces on the water that tend to restore the water surface to its normally flat shape. If the insect’s mass is 2.0 × 1025 kg and the radius of each foot is 1.5 × 1024 m, find the angle θ.
Walking on Water: Many insects can literally walk on water, using surface tension for their support. To show this is feasible, assume the insect’s “foot” is spherical. When the insect steps onto the water with all six legs, a depression is formed in the water around each foot, as shown in Figure 9.43a. The surface tension of the water produces upward forces on the water that tend to restore the water surface to its normally flat shape. If the insect’s mass is 2.0 × 1025 kg and the radius of each foot is 1.5 × 1024 m, find the angle θ.
For the insect to balance the total vertical force from
tension must equal the weight of the insect.
The surface tension is :
F
γ H2 0 =
⇒ F = 2πrγ H 2 0
2πr
The total verticle force from all 6 legs is :
Ftotal = 12πrγ H 2 0 cosθ = minsg ⇒ cosθ =
minsg
⇒
12πrγ H 2 0
(2.0 × 10 kg)(9.8m/s ) = 0.47 ⇒ θ = cos
cosθ =
12π (1.5 × 10 m)(0.073N/m)
−5
−4
2
−1
(0.47) = 62 
As the weight of the insect increases, the angle θ increases
as well, up to the point where θ = 0, at which point the insect sinks.
€
Rising Water: Find the height to which water would rise in a capillary tube with
a radius equal to 5.0 × 10-5 m. Assume the contact angle between the water
and the material of the tube is small enough to be considered zero.
Rising Water: Find the height to which water would rise in a capillary tube with
a radius equal to 5.0 × 10-5 m. Assume the contact angle between the water
and the material of the tube is small enough to be considered zero.
The forces between like molecules are called cohesive and the forces
between different molecules adhesive. In the case of water rising in
a glass tude the adhesive forces between the glass molecules and the
water molecules are stronger than the forces between the water molecules.
F
2πr
The verticle component of the force is :
Fv = γ 2πrcosφ = mg = ρ H 2 0Vg ⇒
The surface tension is γ =
γ 2πrcosφ = ρ H 2 0πr 2 hg ⇒
h=
€
2γ
2γcosφ
≅
= 0.30m
ρ H 2 0 rg ρ H 2 0 rg
Viscous Fluid Flow •  Viscosity refers to fricDon between the layers •  Layers in a viscous fluid have different velociDes •  The velocity is greatest at the center •  Cohesive forces between the fluid and the walls slow down the fluid on the outside SecDon 9.9 Coefficient of Viscosity •  Assume a fluid between two solid surfaces •  A force is required to move the upper surface •  η is the coefficient of viscosity •  SI units: N . s/m2 •  cgs units are Poise –  1 Poise = 0.1 N.s/m2 SecDon 9.9 Poiseuille’s Law •  Gives the rate of flow of a fluid in a tube with pressure differences SecDon 9.9 Reynold’s Number •  At sufficiently high velocity, a fluid flow can change from streamline to turbulent flow –  The onset of turbulence can be found by a factor called the Reynold’s Number, RN –  If RN = 2000 or below, flow is streamline –  If 2000 <RN<3000, the flow is unstable –  If RN = 3000 or above, the flow is turbulent SecDon 9.9 Transport Phenomena •  Movement of a fluid may be due to differences in concentraDon –  As opposed to movement due to a pressure difference –  ConcentraDon is the number of molecules per unit volume •  The fluid will flow from an area of high concentraDon to an area of low concentraDon •  The processes are called diffusion and osmosis SecDon 9.10 Diffusion and Fick’s Law •  Molecules move from a region of high concentraDon to a region of lower concentraDon •  Basic equaDon for diffusion is given by Fick’s Law •  D is the diffusion coefficient SecDon 9.10 Diffusion •  ConcentraDon on the led is higher than on the right of the imaginary barrier •  Many of the molecules on the led can pass to the right, but few can pass from right to led •  There is a net movement from the higher concentraDon to the lower concentraDon SecDon 9.10 Osmosis •  Osmosis is the movement of water from a region where its concentraDon is high, across a selecDvely permeable membrane, into a region where its concentraDon is lower –  A selec/vely permeable membrane is one that allows passage of some molecules, but not others SecDon 9.10 Blood Transfusion: A paDent receives a blood transfusion through a needle of radius 0.20 mm and length 2.0 cm. The density of blood is 1 050 kg/m3. The bo<le supplying the blood is 0.500 m above the paDent’s arm. What is the rate of flow through the needle? We use Poiseuille's Law to determine the flow rate :
4
ΔV πR ( P2 − P1 )
=
(1)
8ηL
Δt
The pressure at the top of the bootle and the end of the needle is the atmospheric pressure.
If the pressure at the front of the needle is P2 and the pressure at the end
of the needle (that goes into the arm) is P1 we have :
P2 − P1 = P2 − Patm = ρblood gh (2)
ΔV πR 4 ρblood gh
(1)^(2) ⇒
=
= 6 × 10 −8 m3 /s
Δt
8ηL
In the absence of viscosity :
1
1
P2 + ρblood v 22 + ρ blood gh = P1 + ρblood v12 ⇒
2
2
1
ρblood gh = ρ blood v12 ⇒ v1 = 2gh = 2.(9.8m/s2 )(0.5m) = 3.131 m/s
2
2
ΔV
= Av1 = πR 2v1 = π (0.2 × 10 −3 m) ( 3.131m/s) = 3.9 × 10 −7 m3 /s
Δt
€
Turbulent Flow of Blood: Determine the speed at which blood flowing through an artery of diameter 0.20 cm will become turbulent. The Reynolds number RN is given by :
ρvd
, where ρ is the density of the fluid, d is the diameter of the tube and η is the viscosity of the fluid.
RN =
η
The flow departs from streamline flow and becomes turbulent for RN > 3000.
v turb =
€
(RN)η
ρd
(3.00 × 10 )(2.7 × 10 N s/m ) = 3.9m/s
=
(1.05 × 10 kg/m )(0.2 × 10 m)
3
3
2
−3
3
−2
MoDon Through a Viscous Medium •  When an object falls through a fluid, a viscous drag acts on it –  The force of resistance depends on the shape and velocity of the falling object •  The resisDve force on a small, spherical object of radius r falling through a viscous fluid is given by Stoke’s Law: SecDon 9.10 Motion Through a Viscous Fluid. Calculate the terminal velocity of the sphere.
Motion Through a Viscous Fluid. Calculate the terminal velocity of the sphere.
The magnitude of the resistive force Fr on a sphere of radius r moving through a fluid
of viscosity η with a velocity v is : Fr = 6πrηv
Initially when the sphere is released in the water Fr = 0. As the velocity begins to increase
so does the resistive force until the sum of all the forces acting on the sphere becomes zero.
At this point the sphere moves with a constant terminal velocity of vt :
W = Fr + Fbuoy ⇒ ρsphVg = 6πrηv t + ρ f Vsph g ⇒ v t =
vt =
€
(
2r 2 g ρsph − ρ f
9η
)
(
Vg ρsph − ρ f
6πrη
)
4 3
πr g ρsph − ρ f
3
=
⇒
6πrη
(
)
Terminal Velocity, General •  Stokes’ Law will not work if the object is not spherical •  Assume the resisDve force has a magnitude given by Fr = k v –  k is a coefficient to be determined experimentally •  The terminal velocity will become SecDon 9.10 SedimentaDon Rate •  The speed at which materials fall through a fluid is called the sedimentaIon rate –  It is important in clinical analysis •  The rate can be increased by increasing the effecDve value of g –  This can be done in a centrifuge SecDon 9.10 Centrifuge •  High angular speeds give the parDcles a large radial acceleraDon –  Much greater than g –  In the equaDon, g is replaced with ω2r SecDon 9.10 Centrifuge, cont •  The parDcles’ terminal velocity will become •  The parDcles with greatest mass will have the greatest terminal velocity •  The most massive parDcles will se<le out on the bo<om of the test tube first SecDon 9.10