Physics 207 – Lecture 21
Lecture 21
Goals:
• Chapter 15
Understand pressure in liquids and gases
Use Archimedes’ principle to understand buoyancy
Understand the equation of continuity
Use an ideal-fluid model to study fluid flow.
Investigate the elastic deformation of solids and liquids
• Assignment
HW9, Due Wednesday, Apr. 8th
Thursday: Read all of Chapter 16
Physics 207: Lecture 21, Pg 1
Pressure vs. Depth
Incompressible Fluids (liquids)
p
In many fluids the bulk modulus is
such that we can treat the density as
a constant independent of pressure:
0
y1
F1
y2
p
1
An incompressible fluid
A
p
2
For an incompressible fluid, the
density is the same everywhere, but
the pressure is NOT!
p(y) = p0 - y g ρ = p0 + d g ρ
Gauge pressure (subtract p0,
pressure 1 atm, i.e. car tires)
mg F2
F2 = F1+ m g
= F1+ ρVg
F2 /A = F1/A + ρVg/A
p2 = p1 - ρg y
Physics 207: Lecture 21, Pg 2
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Physics 207 – Lecture 21
Pressure vs. Depth
For a uniform fluid in an open container pressure same at
a given depth independent of the container
y
p(y)
Fluid level is the same everywhere in a connected
container, assuming no surface forces
Physics 207: Lecture 21, Pg 3
Pressure Measurements: Barometer
Invented by Torricelli
A long closed tube is filled with mercury
and inverted in a dish of mercury
The closed end is nearly a vacuum
Measures atmospheric pressure as
1 atm = 0.760 m (of Hg)
Physics 207: Lecture 21, Pg 4
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Physics 207 – Lecture 21
Exercise
Pressure
What happens with two fluids??
Consider a U tube containing liquids of
density ρ1 and ρ2 as shown:
ρ2
At the red arrow the pressure must be the
same on either side. ρ1 x = ρ2 (d1+ y)
Compare the densities of the liquids:
(A) ρ1 < ρ2
(B) ρ1 = ρ2
dI
ρ1 y
(C) ρ1 > ρ2
Physics 207: Lecture 21, Pg 5
Archimedes’ Principle: A Eureka Moment
Suppose we weigh an object in air (1) and
in water (2).
W1
W 2?
How do these weights compare?
W1 < W2
W1 = W2
W1 > W2
Buoyant force is equal to the weight of the
fluid displaced
Physics 207: Lecture 21, Pg 6
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Physics 207 – Lecture 21
The Golden Crown
In the first century BC the Roman architect Vitruvius related a story of how
Archimedes uncovered a fraud in the manufacture of a golden crown
commissioned by Hiero II, the king of Syracuse. The crown (corona in
Vitruvius’s Latin) would have been in the form of a wreath, such as one of the
three pictured from grave sites in Macedonia and the Dardanelles. Hiero
would have placed such a wreath on the statue of a god or goddess.
Suspecting that the goldsmith might have replaced some of the gold given to
him by an equal weight of silver, Hiero asked Archimedes to determine
whether the wreath was pure gold. And because the wreath was a holy
object dedicated to the gods, he could not disturb the wreath in any way. (In
modern terms, he was to perform nondestructive testing). Archimedes’
solution to the problem, as described by Vitruvius, is neatly summarized in
the following excerpt from an advertisement:
The solution which occurred when he stepped into his bath and caused it to
overflow was to put a weight of gold equal to the crown, and known to be
pure, into a bowl which was filled with water to the brim. Then the gold would
be removed and the king’s crown put in, in its place. An alloy of lighter silver
would increase the bulk of the crown and cause the bowl to overflow.
From http://www.math.nyu.edu/~crorres/Archimedes/Crown/CrownIntro.html
Physics 207: Lecture 21, Pg 7
Archimedes’ Principle
Suppose we weigh an object in air (1) and in water (2).
How do these weights compare?
W1 < W2
W1 = W2
Why?
Since the pressure at the bottom
of the object is greater than that at
the top of the object, the water
exerts a net upward force, the
buoyant force, on the object.
W1 > W2
W1
W 2?
Physics 207: Lecture 21, Pg 8
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Physics 207 – Lecture 21
Sink or Float?
The buoyant force is equal to the weight of
the liquid that is displaced.
If the buoyant force is larger than the
weight of the object, it will float; otherwise
it will sink.
y
FB mg
We can calculate how much of a floating object will be
submerged in the liquid:
F B = mg
Object is in equilibrium
ρ liquid ⋅ g ⋅ V liquid = ρ object ⋅ g ⋅ V object
V liquid
V object
=
ρ object
ρ liquid
Physics 207: Lecture 21, Pg 9
Bar Trick
What happens to the water level when the ice melts?
Expt. 1
A. It rises
Expt. 2
B. It stays the same
piece of rock
on top of ice
C. It drops
Physics 207: Lecture 21, Pg 10
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Physics 207 – Lecture 21
Exercise
V1 = V2 = V3 = V4 = V5
m1 < m2 < m3 < m4 < m5
What is the final position of each block?
Physics 207: Lecture 21, Pg 11
Exercise
V1 = V2 = V3 = V4 = V5
m1 < m2 < m3 < m4 < m5
What is the final position of each block?
But this
Not this
Physics 207: Lecture 21, Pg 12
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Physics 207 – Lecture 21
Home Exercise Buoyancy
A small lead weight is fastened to a large
styrofoam block and the combination floats on
water with the water level with the top of the
styrofoam block as shown.
Pb
styrofoam
If you turn the styrofoam + Pb upside-down,
What happens?
(A)
It sinks
(B)
styrofoam
(C)
styrofoam
(D)
Pb
Pb
styrofoam
Pb
Active Figure
Physics 207: Lecture 21, Pg 13
Exercise
Buoyancy
A small lead weight is fastened to a large
styrofoam block and the combination floats on
water with the water level with the top of the
styrofoam block as shown.
Pb
styrofoam
If you turn the styrofoam + Pb upside-down,
What happens (assuming density of Pb > water)?
(A)
It sinks
(B)
styrofoam
(C)
styrofoam
Pb
Pb
(D)
styrofoam
Pb
Active Figure
Physics 207: Lecture 21, Pg 14
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Physics 207 – Lecture 21
Home Exercise
More Buoyancy
Two identical cups are filled to the same level
with water. One of the two cups has plastic
balls floating in it.
Cup I
Cup II
Which cup weighs more?
(A)
Cup I
(C) the same
(B) Cup II
(D) can’t tell
Physics 207: Lecture 21, Pg 15
Exercise
More Buoyancy
Two identical cups are filled to the same level
with water. One of the two cups has plastic
balls floating in it.
Cup I
Cup II
Which cup weighs more?
(A)
Cup I
(B) Cup II
(C) the same
(D) can’t tell
Physics 207: Lecture 21, Pg 16
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Physics 207 – Lecture 21
Home Exercise
Even More Buoyancy
oi
A plastic ball floats in a cup of water with
half of its volume submerged. Next some oil
(ρ
ρoil < ρball < ρwater) is slowly added to the
container until it just covers the ball.
water
Relative to the water level, the ball will:
Hint 1: What is the buoyant force of the part in
the oil as compared to the air?
(A) move up
(B) move down
(C) stay in same place
Physics 207: Lecture 21, Pg 17
Pascal’s Principle
So far we have discovered (using Newton’s Laws):
Pressure depends on depth: ∆p = ρ g ∆y
Pascal’s Principle addresses how a change in pressure is
transmitted through a fluid.
Any change in the pressure applied to an enclosed
fluid is transmitted to every portion of the fluid and to
the walls of the containing vessel.
Physics 207: Lecture 21, Pg 18
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l
Physics 207 – Lecture 21
Pascal’s Principle in action:
Hydraulics, a force amplifier
F1
Consider the system shown:
A downward force F1 is applied
to the piston of area A1.
F2
d2
d1
This force is transmitted through
the liquid to create an upward
force F2.
A1
A2
Pascal’s Principle says that
increased pressure from F1
(F1/A1) is transmitted
throughout the liquid.
F2 > F1 with conservation of energy
Physics 207: Lecture 21, Pg 19
Exercise
Consider the systems shown on right.
In each case, a block of mass M is
placed on the piston of the large
cylinder, resulting in a difference di
in the liquid levels.
A
If A2 = 2 A1, how do dA and dB compare?
dA
A10
1
V10 = V1 = V2
dA A1 = dB A2
dA A1 = dB 2A1
dB
A2
M
M
A10
dA = dB 2
Physics 207: Lecture 21, Pg 20
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Physics 207 – Lecture 21
Home Example Hydraulics: A force amplifier
Aka…the lever
Consider the system shown on right.
Blocks of mass M are placed on
the piston of both cylinders.
The small & large cylinders
displace distances d1 and d2
Compare the forces on disks A1 &
A2 ignoring the mass and weight of
the fluid in this process
W2 = - M g d2 = - F2 d2
M
d1
A1
M
A2
V1 = V2 = A1 d 1 = A2 d 2
W1 = M g d1 = F1 d1
d 1 / d 2 = A2 / A1
W1 + W 2 = 0 = F1 d1 - F2 d2
F2 d2 = F1 d1
F2 = F1 d1/ d2= F1 A2 / A1 implying P1 = P2 = F1 /A1 = F2
/A2
Physics 207: Lecture 21, Pg 21
Fluids in Motion
To describe fluid motion, we need
something that describes flow:
Velocity v
There are different kinds of fluid flow of varying complexity
non-steady
/ steady
compressible / incompressible
rotational
/ irrotational
viscous
/ ideal
Physics 207: Lecture 21, Pg 22
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Physics 207 – Lecture 21
Types of Fluid Flow
Laminar flow
Each particle of the fluid
follows a smooth path
The paths of the different
particles never cross each
other
The path taken by the
particles is called a
streamline
Turbulent flow
An irregular flow
characterized by small
whirlpool like regions
Turbulent flow occurs when
the particles go above some
critical speed
Physics 207: Lecture 21, Pg 23
Types of Fluid Flow
Laminar flow
Each particle of the fluid
follows a smooth path
The paths of the different
particles never cross each
other
The path taken by the
particles is called a
streamline
Turbulent flow
An irregular flow
characterized by small
whirlpool like regions
Turbulent flow occurs when
the particles go above some
critical speed
Physics 207: Lecture 21, Pg 24
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Physics 207 – Lecture 21
Ideal Fluids
Streamlines do not meet or cross
A2
A1
Velocity vector is tangent to streamline
v1
Volume of fluid follows a tube of flow
bounded by streamlines
v2
Streamline density is proportional to
velocity
Flow obeys continuity equation
Volume flow rate (m3/s)
Q = A·v
is constant along flow tube.
A1v1 = A2v2
Follows from mass conservation if flow is incompressible.
Mass flow rate is just ρ Q (kg/s)
Physics 207: Lecture 21, Pg 25
Exercise
Continuity
A housing contractor saves some
money by reducing the size of a
pipe from 1” diameter to 1/2”
diameter at some point in your
house.
v1
v1/2
Assuming the water moving in the pipe is an ideal fluid,
relative to its speed in the 1” diameter pipe, how fast is the
water going in the 1/2” pipe?
(A) 2 v1
(B) 4 v1
(C) 1/2 v1
(D) 1/4 v1
Physics 207: Lecture 21, Pg 26
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Physics 207 – Lecture 21
Exercise
Continuity
v1
(A) 2 v1
(B) 4 v1
(C) 1/2 v1
v1/2
(D) 1/4 v1
For equal volumes in equal times then ½ the diameter implies ¼
the area so the water has to flow four times as fast.
But if the water is moving 4 times as fast then it has
16 times as much kinetic energy.
Something must be doing work on the water
(the pressure drops at the neck) and we recast the work as
P ∆V = (F/A) (A ∆x) = F ∆x
Physics 207: Lecture 21, Pg 27
Conservation of Energy for
Ideal Fluid
Recall the standard work-energy relation W net = ∆K = Kf – Ki
Apply this principle to a finite fluid volume of ∆V and mass
∆m = ρ ∆V (here W net is work done on the fluid)
Net work by pressure difference over ∆x with ∆x1 = v1 ∆t)
Wnet = F1 ∆x1
–
F2 ∆x2
= (F1/A1) (A1∆x1) – (F2/A2) (A2 ∆x2)
=
P1 ∆V1 – P2 ∆V2
v1
v1/2
and ∆V1 = ∆V2 = ∆V (incompressible)
Wnet = (P1– P2 ) ∆V
Cautionary note: This work is not internal to the fluid (it is incompressible). It
actually reflects a piston or some other moving barrier off to the left
displacing the fluid and a second barrier on the right maintaining the lower
pressure. For the left barrier on the left the force is in the direction of
displacement and on the right the force will be opposite the displacement.
In the chapter 17 we will be using an ideal gas as a working fluid and, if the
volume of the piston is fixed, then no work is done
Physics 207: Lecture 21, Pg 28
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Physics 207 – Lecture 21
Conservation of Energy for
Ideal Fluid
W
= (P1– P2 ) ∆V and
P1
W
= ½ ∆m
v22
– ½ ∆m
P2
v12
= ½ (ρ ∆V) v22 – ½ (ρ ∆V) v12
(P1– P2 ) = ½ ρ v22 – ½ ρ v12
P1+ ½ ρ v12 = P2+ ½ ρ v22 = const.
and with height variations:
Bernoulli Equation P1+ ½ ρ v12 + ρ g y1 = constant
Physics 207: Lecture 21, Pg 29
Lecture 21
• Question to ponder:
or float?
Does heavy water (D2O) ice sink
• Assignment
HW9, due Wednesday, Apr. 8th
Thursday: Read all of Chapter 16
Physics 207: Lecture 21, Pg 30
Page 15