Dr. J. Charles Hicks jchicks@cox.net
Summer Session I 2009

Physics 2C is a catch-all class
Fluids – Pressure, Archimedes’ Principle, Continuity, and
Bernoulli’s Principle
Thermodynamics – Temperature dependence of the bulk properties of matter and resulting energy transfers. First and Second Laws of Thermodynamics – Heat engines, the Carnot Cycle, and Entropy.
Wave Motion - Properties and mathematical description of waves.
Optics – Both geometrical and physical optics (small and large wavelengths)

http://physics.ucsd.edu/students/courses/weblinks/
The website will contain:
• Syllabus
• Announcements
• Problem assignments
• Solutions to the problem assignments as well as solutions to the quizzes and final.
• Lecture slides in a pdf format.
I will not send out class emails, so it is your responsibility to check the website frequently. It should be helpful. The syllabus is intended as a guideline and the progress of the course determines what material is covered on any given quiz. Quiz questions will be related to material covered in lectures the preceding week.

Problems will be assigned twice per week as per the syllabus. Additionally there will be two problems assigned each week to be turned in and graded. These problems are due by the end of class on Friday (week 1 the due date is
Monday, July 6). Each of the problems to be graded will be worth 10 points so that the total maximum score from these problems will be 100 points. If it is to your advantage, this total will replace your worst quiz score.
Closed-book quizzes will be given on Monday every week.
A 3x5 cheat card will be allowed. The necessary equations will be written on the blackboard or in the quiz. Any requests for formulas will usually be honored and written on the board for everyone.
Solutions and scores will be posted on the web. Your overall quiz grade will be computed from your highest
4/5 scores which includes the 4 quizzes + the total score on the problems that are turned in. This total will count 2/3 toward your final grade. Your score on the final will contribute the remaining 1/3.
THERE WILL BE NO MAKE-UP QUIZZES FOR ANY REASON WHAT-SO-EVER!!

Fluid is matter that flows under influence of external forces (?)
Under shear stress the net force is zero! Solids experience some finite deformation while
“Fluids flow under shear”!

P=F/A
Pressure is a scalar but force is a vector?
The force exerted against any surface is always normal to that surface. It is the normal to the surface that determines the direction of the force!

Condition for hydrostatic equilibrium – constant pressure throughout the fluid volume (in the absence of a gravitational field).
Variation of pressure creates net force in the direction of decreasing pressure.

No net force!
A solid would be in equilibrium.
What about a fluid?

Density measures the mass per unit volume.
SI units are kg / m 3
,
ρ
= m / V [ kg / m 3 ]
Pressure measures the normal force per unit area exerted by a fluid.
SI units are N / m 2 or Pascal Pa .
[
2 =
]
As an example atmospheric pressure is P atm
= 1.013 x 10 5 Pa
There are compressible and incompressible fluids.

Mass – 63 metric tons ( 63,000 kg )
Caterpillar area – 6 m 2
Ground pressure:
Proboscis – ~ 0.1
mm in diameter.
Pressure of 10 5 Pa corresponds to
Equivalent mass is m = F/g ~ 10 -4 kg = 0.1 g

Equivalent mass is m = F/g ~ 10 -4 kg = 0.1 g
A mosquito weighs about 2 mg = 0.002
g .
An ant can lift 50 × its weight.
If the same is true for a mosquito, it can produce a force equivalent to
50 × .002 = .1
g and exert the same pressure as an Abrahams tank

Of course they do, since they apply everywhere unless the things get relativistic or quantum.
What is the problem with the fluid mechanics then?
Why do not we study it together with the regular mechanics?
The trouble is we have to deal with a continuum… where the notions of a “body” is difficult to define and it is not always straightforward where the forces come from.
What do we do?
For appropriate description of continuum mechanics we have got to introduce new notions and parameters.

nd
Assume that the pressure only changes in the x direction
For a fluid element with sides
Newton’s 2 nd law becomes
“ So we see that it is the negative pressure gradient that results in the acceleration of a fluid element ”

Force from above –
Force from below – (
)
Gravitational force (pulling down) dF g
= ρ gA dh balance
PA
+ ρ gA dh
=
( P
+ dP ) A
ρ gA dh
= dP
AdP =>
Note that h measures increasing depth.
dh
= ρ g

dP
= ρ g dh
This is a differential equation, albeit a very simple one. How do we calculate the actual pressure?
We integrate the equation
P
=
P
0
+ ρ gh
And end up getting an integration constant, P
0
.
What is this constant about?
Mostly pain in the neck!..
In order to find pressure, P , in a given point, we need to know the pressure P
0 in some reference point, h
0
.
Then we compare the depth between the two points and calculate P as:

In order to find pressure, P , in a given point, we need to know the pressure P
0 in some reference point.
What might this reference point be?
Surface of water with P
0
= P atm is the usual candidate.
Ocean (or cup).
P
=
P atm
+ ρ w gh
Here h is the depth.
P
=
Atmosphere
P atm
− ρ air gh
Here h is the altitude.
Pressure drops by about 120 Pa every 10 m
Pressure grows by about 1 atm ( 10 5 Pa ) every 10 m

h
When I got my tooth extracted I was warned not to drink through a straw. Why so?
The pressure of liquid at the top of the straw is
P top
=
P atm
− ρ gh
For you to drink the pressure inside your mouth should drop below P top
.
This implies a large pressure difference between the blood vessels in your gums and your mouth, which may open the wound.

(a) At what depth is the pressure twice atmospheric pressure?
(b) What is the pressure at the bottom of the 11.3
km deep
Mariana’s Trench?
Demonstration with column and plate.

Find the force of water on a pool wall that is 15 m wide and 3 m deep.
Substituting the values P
0
= 1.0 x 10 5 Pa and
= 1000 kg/m 3 leads to
F= 5.2 x 10 6 N .
What is the direction of this force?
How will the force on a dam depend on the size of the lake?

Static fluid pressure does not depend on the shape, total mass or surface area of the liquid.
volume = V = hA weight = mg
The most remarkable thing about this expression is what it does not include.
The expression for hydrostatic pressure is easy to see for the straight, unobstructed column, but not obvious for the cases of different geometry which are shown.
Demonstration of equilibrium tubes.

Static fluid pressure does not depend on the shape, total mass or surface area of the liquid.
The total force acting on the bottom of the vessel is
The most remarkable thing about the expression for pressure is what it does not include.
The expression for hydrostatic pressure is easy to see for the straight, unobstructed column, but not obvious for the cases of different geometry which are shown.
The pressure force exerted on the bottom may be strikingly different from the weight of the liquid!
F
=
P
⋅
A
P
= ρ w gh

The pressure force exerted on the bottom may be strikingly different from the weight of the liquid!
What about Newton? What supports the weight of the liquid?
The weight of the liquid is: d
1 h
2 d
2 h
1
However PA at the bottom + PA at the lip is
We see that F = W as it had to!! The lower wall of the cylinder supports the weight of the fluid on the lip.
What about the other containers on the previous slide?

Static fluid pressure does not depend on the shape, total mass or surface area of the liquid.
volume = V = hA weight = mg
The pressure on the sides of the containers results in a force that has a vertical component. It is this vertical component plus the force from the pressure at the bottom that supports the total weight of the liquid!

A couple of additional things that we should know about fluids that are different than solid objects
The first is Pascal’s law:
“A pressure increase anywhere in the fluid results in the same pressure increase throughout the fluid.”
As we shall see soon this is the basis for hydraulic systems

A multiplication of force can be achieved by the application of fluid pressure according to Pascal's principle , which for the two pistons implies
P
1
=P
2
This allows the lifting of a heavy load with a small force, as in an auto hydraulic lift … but of course there can be no multiplication of work , so in an ideal case with no frictional loss:
W in
= W out

As an example, a small piston that exerts a force of and supports a car of mass 3200 kg .
The diameter of the piston that supports the car is 120 cm .
What is the diameter of the smaller piston in order to maintain equilibrium?
From Pascal’s principle the pressure throughout is the same, hence

Pressure of the air, P atmosphere
, drives mercury into the hollow tube.
There is no air and no pressure inside the tube.
Therefore P atmosphere is balanced by hydrostatic pressure of the mercury.
P atmosphere
ρ
=
ρ gh
Hg
= 13600 kg/m 3
P atm
= 1.013 x 10 5 Pa h = P atm
/
g = .760
m= 760 mm
What if we use wine rather than mercury?

Now there is a fluid (gas) under pressure in the reservoir.
The DIFFERENCE between the pressure inside the reservoir,
P res
, and P atmosphere is now balanced by hydrostatic pressure of the mercury.
P res
- P atmosphere
=
gh

P
=
P atm
− ρ gh
1 h
1 P
=
P atm h
2
P
=
P atm
+ ρ gh
2
Pressure in the static case.
A flow causes redistribution of pressures.
P
=
P atm