Fluids and Elasticity
Chapter 15
Density (r)
•
•
•
•
•
r = mass/volume
Rho ( r) – Greek letter for density
Units - kg/m3
Specific Gravity = Density of substance
Density of water (4oC)
Unitless ratio
Ex: Lead has a sp. Gravity of 11.3 (11.3 times
denser than water
Ex: 1
Estimate the mass of air in this classroom
Pressure
•
•
•
•
Force per unit area
P = F/A
Unit - N/m2 (Pascal)
The larger the area, the less the pressure
– Shoeshoes
– Elephant feet
– Bed of nails
Fluid Pressure
• A fluid exerts the same pressure in all directions
at a given depth
• P = Po + rgh
• The atmosphere is a fluid
• Po often 1 atm (101.3 kPa)
Pressure: Example 1
A water storage tank is 30 m above the water faucet
in a house. Calculate the pressure at the faucet:
We will neglect the atmospheric pressure since it is
the same at the tank and at the surface
DP = rgh = (1000 kg/m3)(9.8 m/s2)(30 m)
DP = 29,000 kgm2/m3s2 = 29,000 kg m/s2m2
DP = 29,000 N/m2
Pressure: Example 2
The Kraken can live at a depth of 200 m. Calculate
the pressure the creature can withstand (neglect
atmospheric pressure)
Pressure: Example 2
DP = rgh = (1000 kg/m3)(g)(200 m)
DP = 1.96 X 106 N/m2
Atmospheric Pressure
• 1 atm = 1.013 X 105 N/m2 = 101.3 kPa
• 1 bar = 1 X 105 N/m2 (used by meteorologists)
• Gauge pressure
P = Patm + PG
Absolute pressure atmospheric pressure Gauge pressure
• We usually measure gauge pressure
• Ex: A tire gauge reads 220 kPa, what is the
absolute pressure?
P = Patm + PG
P = 101.3 kPa + 220 kPa = 321 ka
Straw Example
You can pick up soda in a
straw using your finger. Why
doesn’t the soda fall out?
Another Straw Example
What pushes soda up a straw when you drink
through it?
Torrecilli’s Work
P = Patm + PG
P = Patm + rgh
What is the highest column of water
that the atmosphere can support?
P = Patm + rgh
0 = 1.013 X 105 N/m2 + (1000kg/m3)(9.8m/s2)(h)
h = 10.3 m
• No vacuum pump can pump more than ~30 feet
Try the same calculation with mercury
P = Patm + rgh
0 = 1.013 X 105 N/m2 + (13,600kg/m3)(9.8m/s2)(h)
h = 0.760 m (760 mm)
1 atm = 760 mm Hg (760 torr)
Can an astronaut attach suction cups to the boots of
his spacesuit to help him climb around the space
shuttle while in space?
Pascal’s Principle
• Pressure applied to a confined fluid increases the
pressure the same throughout
Pin = Pout
Fin
Ain
=
Fout
Aout
for pistons:
DF = rg(A1 + A2)d2
Pascal’s Principle: Ex 1
A hydraulic lift can produce 200 lb of force. How
heavy a car can be lifted if the area of the lift is
20 times larger that the input of the Force?
Fin
Ain
=
Fout
Aout
Fout = Fin Aout
Ain
=
(200 lb) (20) = 4000 lbs
1
Pascal’s Principle: Ex 1
A hydraulic lift has the car rest ona 25 cm pipe.
The lift the car, compressed air pushes on a 6.0
cm pipe.
a. Calculate the force needed to lift a 1300 kg car.
b. Calculate how much the air pressure force must
be increased to lift the car 2.0 m.
Buoyancy
• Buoyancy
– The “lift” provided by water
– Objects weight less in water than out
• Caused by pressure differential between top and
bottom of an object.
Fbouyant = rgV
Derivation of the Buoyancy Formula
Fb = F2 – F1
P = F/A
F = PA
F = rghA
Fb = rgh2A – rgh1A
Fb = rgA(h2 - h1)
Fb = rgV
Archimedes Principle
“The bouyant force on an
object equals the weight
of fluid displaced by the
object”
w’ = weight of an object in
water (or any liquid)
w’ = mg - Fb
Buoyancy: Example 1
A 7000-kg ancient statue lies at the bottom of the
sea. Its volume is 3.0 m3. How much force is
needed to lift it?
Fb = rgV
Fb = (1000 kg/m3)(9.8 m/s2)(3.0m3)
Fb = 2.94 X 104 kg-m/s2
Fb = 2.94 X 104 N
Fb
mg
w’ = mg - Fb
w’ = (7000 kg)(9.8m/s2) - 2.94 X 104 N
w’ = 3.92 X 104 N
Say, isn’t w’ just the sum of the forces?
Yep.
Fb
SF = w’
mg
Buoyancy: Example 2
Archimedes tested a crown for the king. Out of
water, it masses 14.7 kg. In water, it massed 13.4
kg. Was the crown gold?
w’ = mcrg – Fb
w’ = mcrg – rgVcr
(13.4 kg)(g) = (14.7 kg)(g) – (1000 kg/m3)(g)(Vcr)
131 N = 144 N – (9800 kg/ms2)(Vcr)
Vcr = 0.00133 m3
Now we can calculate the density of the crown:
r = m/V = 14.7 kg/0.00133 m3
r = 11,053 kg/m3
Gold’s density is about 19,000kg/m3. This is much
closer to lead.
Example 3
A cube of wood that is 10 cm on a side is held
underwater by tying a string to the cube and the
bottom on the container. The wood has a density
of 700 kg/m3.
a. Draw a free body diagram showing all the forces
on the block.
b. Calculate the force of bouyancy
c. Calculate the tension in the string.
Floating
• Objects that are less dense than water will float
• Part of the object will be above the water line
• A case of static equilibrium
SF = 0
Fb
mg
Floating: Example 1
A 1200 kg log is floating in water. What volume of
the log is under water?
SF = 0
SF = 0 = mg – Fb
mg = Fb
mg = rgVlog
Vlog = mg
rg
Fb
mg
Vlog = m
r
(Hey, the g’s cancel!)
Vlog = 1200 kg
1000 kg/m3
=
1.2 m3
Floating: Example 1
A wooden raft has a density of 600 kg/m3, an area
of 5.7 m2, and a volume of 0.60 m3. How much
of the raft is below water in a freshwater lake?
Let’s first calculate the mass of the raft:
r = m/V
m = rV = (600 kg/m3)(0.60 m3) = 360 kg
Now we can worry about the raft.
SF = 0
SF = 0 = mg – Fb
mg = Fb
mg = rgVsubmerged
mg = rghsubmergedA
mg = rghsubmergedA
m = rhsubmergedA
(Hey, the g’s cancelled!)
hsubmerged = m/rA
hsubmerged =
360 kg
= 0.063 m
(1000 kg/m3)(5.7 m2)
Floating: Example 3
Suppose a continent is floating on the mantle rock.
Estimate the height of the continent above the
mantle (assume the continent is 35 km thick).
SF = 0 = mg – Fb
0 = mcg – rmangVc(submerged)
mcg = rmangVc(submerged)
mc = rmanVc(submerged)
We don’t know the mass of the continent
rc = mc/Vc(total)
mc = rcVc(total)
mc = rmanVc(submerged)
mc = rmanVc(submerged)
mc = rcVc(total)
rmanVc(submerged) = rcVc(total)
Vc(submerged) = rc
Vc(total)
rman
=
(2800 kg/m3)
(3300 kg/m3)
= 0.85
This means that 85% of the continent is submerged,
and only 15% is above:
(0.15)(35 km) = 5.25 km
Floating: Ex 4
A block is placed in water and 5.8 cm is
submerged. The same block is placed in an
unknown liquid and 4.6 cm is submerged.
Calculate the density of the unknown liquid.
Assume the same face of the block pointed
downward in both cases (A).
Fluid Flow
• Laminar Flow – Smooth, streamline flow
(laminar means “in layers”)
• Turbulent Flow – erratic flow with “eddies”
• Viscosity – Internal friction of a liquid
– High viscosity = slow flow
– Viscosity is NOT the same as density
Equation of Continuity
A1v1 = A2v2
A = Area of a pipe
v = velocity of the liquid
Equation of Continuity
v1A1 = v2A2
• Fluid will flow faster through a smaller opening
• Placing your finger over a hose opening.
The term “vA” is the “volume rate of flow”
A = m2
v = m/s
vA = m3/s
Q = vA
Eqn. Of Continuity: Example 1
A garden hose has a radius of 1.00 cm and the
water flows at a speed of 0.80 m/s. What will be
the velocity if you place your finger over the hose
and narrow the radius to 0.10 cm?
A1 = pr2 = (3.14)(0.01 m)2 = 3.14 X 10-4 m2
A2 = pr2 = (3.14)(0.001 m)2 = 3.14 X 10-6 m2
A1v1 = A2v2
v2 = A1v1
A2
v2 = (3.14 X 10-4 m2)(0.80 m/s) = 80 m/s
(3.14 X 10-6 m2)
Eqn. Of Continuity: Example 2
A water hose 1.00 cm in radius fills a 20.0-liter
bucket in one minute. What is the speed of water
in the hose?
A1 = pr2 = (3.14)(1 cm)2 = 3.14 cm2
Remember that Av is volume rate of flow.
A2v2=20.0 L 1 min 1000 cm3 = 333 cm3/s
1 min 60 s
1L
A1v1 = A2v2
v1 = A2v2/A1
v1 = 333 cm3/s =
3.14 cm2
160 cm/s or 1.60 m/s
Eqn. Of Continuity: Example 3
A sink has an area of about 0.25 m2. The drain has
a diameter of 5 cm. If the sink drains at 0.03 m/s,
how fast is water flowing down the drain?
Ad = pr2 = (p)(0.025 m)2 = 1.96 X 10-3 m3
Advd = Asvs
vd = Asvs/Ad=[(0.25 m2)(0.03 m/s)]/(1.96 X 10-3 m3)
vd = 3.82 m/s
Eqn. Of Continuity: Example 4
The radius of the aorta is about 1.0 cm and blood
passes through it at a speed of 30 cm/s. A typical
capillary has a radius of about 4 X 10-4 cm and
blood flows through it at a rate of 5 X 10-4 m/s.
Estimate how many capillaries there are in the
human body.
Aava = NAcvc
(N is the number of capillaries)
Aa = pr2 = (3.14)(0.01 m)2 = 3.14 X 10-4 m2
Ac = pr2 = (3.14)(4 X 10-6 cm)2 = 5.0 X 10-11 cm2
N = Aava/ Acvc
N = (3.14 X 10-4 m2)(0.30 m/s) = ~ 4 billion
(5.0 X 10-11 cm2)(5 X 10-4 m/s)
Eqn. Of Continuity: Example 5
How large must a heating duct be to replenish the
air in a room 300 m3 every 15 minutes? Assume
air moves through the vent at 3.0 m/s.
Advd = Arvr
Arvr = volume rate of flow:
Arvr = 300 m3 1 min = 0.333 m3/s
15 min 60 s
Ad = Arvr/vd
Ad = 0.333 m3/s = 0.11 m2
3.0 m/s
Bernoulli’s Principle
The velocity and
pressure of a fluid are
inversely related.
Why does a shower curtain sometimes “attack” a
person taking a shower?
What will happen to closed windows during a
tornado? Will they blow in or out?
Applications of Bernoulli’s Principle
1. Airplane wing
Applications of Bernoulli’s Principle
2. Prairie Dog Burrows
1. Air moves faster (lower pressure) at the top
2. Draws air through the burrow
3. The exact same thing happens with our chimneys
Applications of Bernoulli’s Principle
3. Spray Paint
Flow of air (low pressure)
Applications of Bernoulli’s Principle
4. Dime in a cup
Bernoulli’s Equation
Pt + ½rvt2 + rgyt = Pb + ½rvb2 + rgyb
(note: you often have to use the Eqn. Of Continuity
in these situations:)
A1v1 = A2v2
• Often useful when you have both a change in
height and area:
Pipe from a water
reservoir to a house
Pipe from a house into
a sewer pipe
If there is no change in altitude, the equation
simplifies:
Pt + ½rvt2 + rgyt = Pb + ½rvb2 + rgyb
Pt + ½rvt2 = Pb + ½rvb2
Bernoulli’s Equation: Example 1
A water heater in the basement of a house pumps
water through a 4.0 cm pipe at 0.50 m/s and 3.0
atm. What will be the flow speed and pressure
through a 2.6 cm spigot on the second floor, 5.0
m above?
3.0 atm
1.013 X 105 N/m2 = 3.0 X 105 N/m2
1 atm
Flow speed:
Atvt = Abvb
vt = Abvb/At
vt = pr2bvb
pr2t
vt = r2bvb =
r2t
(Remember A = pr2)
(Hey, the p’s cancel!)
(0.02 m)2(0.50m/s) = 1.2 m/s
(0.013 m)2
Now the pressure:
Pt + ½rvt2 + rgyt = Pb + ½rvb2 + rgyb
Pt + ½(1000)(1.2)2 + (1000)(9.8)(5) = 3.0X105 +
½(1000)(0.50)2 + (1000)(9.8)(0)
Pt = 2.5 X 105 N/m2
Bernoulli’s Equation: Example 2
A drunken redneck shoots a hole in the bottom of
an aboveground swimming pool. The hole is 1.5
m from the top of the tank. Calculate the speed
of the water as it comes out of the hole.
yt = 1.5 m
yb = 0 m
The top of the pool is a much larger area than the
hole. We will assume that the vt = 0.
Pt + ½rvt2 + rgyt = Pb + ½rvb2 + rgyb
Pt + rgyt = Pb + ½rvb2 + rgyb
Also, both the top and the hole are open to the
atmosphere, so Pt = Pb
Pt + rgyt = Pb + ½rvb2 + rgyb
rgyt = ½rvb2 + rgyb
Set the bottom of the pool as yb = 0.
rgyt = ½rvb2 + rgyb
rgyt = ½rvb2
vb2 = 2rgyt/r
vb2 = 2gyt
vb2 = (2)(9.8m/s2)(1.5 m)
vb = 5.42 m/s
Example 3
A hydroelectric dam is 200 m above the power
plant. The inlet hose at the top has a diameter of
100 cm, and the outlet hose to the turbine has a
diameter of 50 cm.
a. Calculate the speed of the water into the turbine
(both are open to the atmosphere)
Elasticity: Hooke’s Law
• Hooke’s Law – usually
used with a spring
• Can consider anything to be
like a spring
• F = kDL (F=kxspring)
• k = proportionality (spring)
constant
• Can’t stretch things forever
Elastic region – material will still bounce back
Plastic region – material will not return to original
length (but has not broken)
F = kDL
This is only linear
in the proportional
region
HOPE I DON’T
REACH MY
ELASTIC LIMIT!
Elastic Region: Young’s
Modulus(E)
Stress = Force
Area
=
F
A
Strain = Change in length =
Original length
DL
Lo
Y = stress
strain
Y = F/A
DL/Lo
or
F = Y DL
A
Lo
Young’s Modulus: Example 1
A 1.60 m long steel piano wire has a diameter of
0.20 cm. How great is the tension in the wire if it
stretches 0.30 cm when tightened?
A = pr2 = p(0.0010 m)2 = 3.1 X 10-6 m2
F = Y DL
A
Lo
F = AY DL
Lo
F = AY DL
Lo
F = (3.1 X 10-6m2)(200 X 1011N/m2)(0.0030 m)
1.60 m
F = 1200 N
Young’s Modulus: Example 2
A steel support rod of radius 9.5 mm and length 81
cm is stretched by a force of 6.2 X 104 N (about 7
tons).
a. Calculate the stress.
b. Calculate the elongation.
Area = pr2 = p(0.0095 m2
Stress = Force = 6.2 X 10-4m = 2.2 X 108 N/m2
Area
2.84 X 10-4 m2
F = YDL
A
Lo
DL = FL =
YA
(6.2 X 104 N)(0.81m)
(200 X 109 N/m2)(2.84 X 10-4 m2)
DL = 8.84 X 10-4 m = 0.89 mm
Young’s Modulus: Example 2
A 2.0 m long, 1.0 mm diameter wire is suspended.
Hanging a 4.5 kg mass stretches the wire length
by 1.00 mm.
a. Calculate Young’s modulus
b. Identify the material from the table.
The Three Types of Stress
Stretching
Squeezing
Horizontal
Other Modulus’
Shear Modulus – Used for shear stress
Bulk Modulus – Used for even compression on all
sides (an object when submerged)
Fracture
• Breaking Point
• Uses
– Tensile Strength – Stretching
– Compressive Strength – under a load
– Shear Stress – Shearing
• Safety Factor – reciprocal that is multiplied by
the tensile strength
• Ex: A safety factor of 3 means you will only use
1/3 of the maximum stress
Fracture: Example 1
A concrete column 5 m tall will have to support 1.2
X 105 N (compression). What area must it have
to have a safety factor of 6?
Max stress = (1/6)(compressive strength)
Max stress = (1/6)(20 X 106 N/m2)= 3.3X106 N/m2
Stress = F
A
Stress = F
A
Area = F =
Stress
(1.2 X 105 N) = 3.64 X 10-2 m
3.3 X 106N/m2
How much will the column compress under the
load?
F = EDL
A
Lo
DL = FL =
EA
(1.2 X 105 N)(5 m)
(20 X 109 N/m2)(3.64 X 10-2 m2)
DL = 8.3 X 10-4 m = 0.83 mm
Fracture: Example 2
Spider-man’s webbing has
a tensile strength of 600
X 106 N/m2 and he
wishes to use a safety
factor of 3. What is the
diameter of the webbing
if the maximum force at
the bottom of a swing is
1500 N?
Maximum Stress
(1/3)(600 X 106 N/m2) = 200 X 106 N/m2
Stress = Force
Area
Area = Force
Stress
=
1500 N
= 7.5 X 10-6m2
200X106N/m2
Area = pr2
r = (A/p)1/2 = 1.55 X 10-3 m or 1.55 mm
Diameter = 3.10 mm
Concrete
• Concrete is much stronger under compression
than tension
– Tensile Strength – 2 X 106N/m2
– Compressive Strength – 20 X 106N/m2
• Prestressed concrete – rods or mesh are stretched
when the concrete is poured. Released after
concrete dries.
• Now under compression