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General Physics (PHY 2130)
Lecture 22
•  Solids and fluids
  density and pressure
  Pascal‘s principle and car lift
http://www.physics.wayne.edu/~apetrov/PHY2130/
Lightning Review
Last lecture:
1.  Rotational dynamics
  torque and angular momentum
  two equillibrium conditions
Review Problem: A figure skater stands on one spot on the ice (assumed frictionless) and
spins around with her arms extended. When she pulls in her arms, she reduces her
rotational inertia and her angular speed increases so that her angular momentum is
conserved. Compared to her initial rotational kinetic energy, her rotational kinetic energy
after she has pulled in her arms must be
1. the same.
2. larger because she’s rotating faster.
3. smaller because her rotational inertia is smaller.
Example:
Given:
Moments of inertia:
I1 and I2
Rotational kinetic energy is
KErot =
1 2 1
Iω = Lω
2
2
We know that (a) angular momentum L is conserved
and (b) angular velocity increases
Find:
K2 =?
Thus, rotational kinetic energy must increase!
 
Solids and Fluids
Question
What is a fluid?
1.
2.
3.
4.
A liquid
A gas
Anything that flows
Anything that can be made to change shape.
States of matter: Phase Transitions
ICE
WATER
Add
heat
STEAM
Add
heat
These are three states of matter (plasma is another one)
States of Matter
•  Solid
 
Has definite volume
 
Has definite shape
 
Molecules are held in specific
location by electrical forces and
vibrate about equilibrium positions
 
Can be modeled as springs
connecting molecules
•  Liquid
•  Gas
•  Plasma
States of Matter
•  Solid
 
Crystalline solid
 
 
Atoms have an ordered structure
Example is salt (red spheres are Na+
ions, blue spheres represent Cl- ions)
  Amorphous Solid
 
 
Atoms are arranged randomly
Examples include glass
•  Liquid
•  Gas
•  Plasma
States of Matter
•  Solid
•  Liquid
 
Has a definite volume
 
No definite shape
 
Exist at a higher temperature than solids
 
The molecules “wander” through the
liquid in a random fashion
 
The intermolecular forces are not
strong enough to keep the
molecules in a fixed position
•  Gas
•  Plasma
States of Matter
•  Solid
•  Liquid
•  Gas
•  Plasma
 
Has no definite volume
 
Has no definite shape
 
Molecules are in constant random motion
 
The molecules exert only weak forces on each other
 
Average distance between molecules is large
compared to the size of the molecules
States of Matter
•  Solid
•  Liquid
•  Gas
•  Plasma
 
Matter heated to a very high temperature
 
Many of the electrons are freed from the nucleus
 
Result is a collection of free, electrically charged ions
 
Plasmas exist inside stars or experimental reactors or
fluorescent light bulbs!
Is there a concept that helps to distinguish
between those states of matter?
Density
•  The density of a substance of uniform composition is
defined as its mass per unit volume:
m
ρ=
V
some examples:
4
Vsphere = π R 3
3
Vcylinder = π R 2 h
Vcube = a 3
•  The densities of most liquids and solids vary slightly with
changes in temperature and pressure
•  Densities of gases vary greatly with changes in
temperature and pressure (and generally 1000 smaller)
Units
SI
kg/m3
CGS
g/cm3 (1 g/cm3=1000 kg/m3 )
14
Fluids
•  Fluids: (liquids and gases) are materials that flow.
•  Fluids are easily deformable by external forces.
•  A liquid is incompressible. Its volume is fixed and is impossible to
change.
•  A liquid will flow to take the shape of the container that holds it. A gas
will completely fill its container.
15
Pressure
Pressure arises from the collisions between the particles of a fluid with
another object (container walls for example).
There is a momentum change
(impulse) that is away from the
container walls. There must be
a force exerted on the particle by
the wall.
By Newton’s 3rd Law, there is a
force on the wall due to the
particle.
Pressure
•  Pressure of fluid is the ratio of
the force exerted by a fluid on
a submerged object or on the
wall of the vessel to area
F
P≡
A
Units
SI
Pascal (Pa=N/m2)
Example: 100 N over 1 m2 is P=(100 N)/(1 m2)=100 N/m2=100 Pa.
Often: 1 atmosphere (atm) = 101.3 kPa = 1.013 x 105 Pa
17
Example: Someone steps on your toe, exerting a force of 500 N on an area of
1.0 cm2. What is the average pressure on that area in atmospheres?
First, let’s change centimeters into meters:
2
⎛ 1 m ⎞
−4
2
1.0 cm 2 ⎜
⎟ = 1.0 ×10 m
⎝ 100 cm ⎠
Then, using the definition of pressure, let’s determine pressure in
Pa and then change to atm:
F
500 N
Pav = =
A 1.0 × 10 − 4 m 2
1 atm
⎞
6
2 ⎛ 1 Pa ⎞⎛
= 5.0 × 10 N/m ⎜
⎟
2 ⎟⎜
5
⎝ 1 N/m ⎠⎝ 1.013 × 10 Pa ⎠
= 49 atm
Pressure and Depth
•  If a fluid is at rest in a container, all
portions of the fluid must be in
static equilibrium
•  All points at the same depth must
be at the same pressure
(otherwise, the fluid would not be in
equilibrium)
•  Three external forces act on the
region of a cross-sectional area A
External forces: atmospheric, weight, normal
∑F = 0
⇒ PA − Mg − P0 A = 0,
but : M = ρ V = ρ Ah, so : PA = P0 A + ρ Agh
P = P0 + ρ gh
ConcepTest 1
You are measuring the pressure at the depth of 10 cm in three
different containers. Rank the values of pressure from the
greatest to the smallest:
1.
2.
3.
4.
1-2-3
2-1-3
3-2-1
It’s the same in all three
10 cm
1
2
3
ConcepTest 1
You are measuring the pressure at the depth of 10 cm in three
different containers. Rank the values of pressure from the
greatest to the smallest:
1.
2.
3.
4.
1-2-3
2-1-3
3-2-1
It’s the same in all three
10 cm
1
2
3
Pressure and Depth equation
P = Po + ρgh
•  Po is normal
atmospheric pressure
•  1.013 x 105 Pa = 14.7 lb/
in2
•  The pressure does not
depend upon the shape
of the container
  Other
units of pressure:
76.0 cm of mercury
One atmosphere 1 atm =
1.013 x 105 Pa
14.7 lb/in2
Example:
Find pressure at 100 m below ocean surface.
Given:
masses: h=100 m
P = P0 + ρ H 2O gh, so
(
)(
)
P = 9.8 ×105 Pa + 103 kg m3 9.8 m s 2 (100 m)
≈ 10 6 Pa
Find:
P=?
(10 × atmospheric pressure)
 
Pascal’s Principle
•  A change in pressure applied to
an enclosed fluid is transmitted
undiminished to every point of
the fluid and to the walls of the
container.
•  The hydraulic press is an
important application of
Pascal’s Principle
F1 F2
P=
=
A1 A2
•  Also used in hydraulic brakes,
forklifts, car lifts, etc.
Since A2>A1, then F2>F1 !!!
24
ΔP at point 1 = ΔP at point 2
F1
F2
=
A1 A 2
⎛ A 2 ⎞
⎟⎟ F1
F2 = ⎜⎜
⎝ A1 ⎠
The work done pressing the
smaller piston (#1) equals
the work done by the larger
piston (#2).
F1d1 = F2d2
Using an hydraulic lift reduces the amount of force needed to lift a load, but
the work done is the same.
25
Example: Assume that a force of 500 N (about 110 lbs) is applied to the smaller
piston in the previous figure. For each case, compute the force on the larger
piston if the ratio of the piston areas (A2/A1) are 1, 10, and 100.
Using Pascal’s Principle:
A2 A1
1
10
100
F2
500 N
5000 N
50,000 N
26
Example: In the previous example, for the case A2/A1 = 10, it was found that
F2/F1 = 10. If the larger piston needs to rise by 1 m, how far must the smaller
piston be depressed?
Since the work done by both pistons is the same,
F2
d1 = d 2 = 10 m
F1
27
Example: Depressing the brake pedal in a car pushes on a piston with crosssectional area 3.0 cm2. The piston applies pressure to the brake fluid, which is
connected to two pistons, each with area 12.0 cm2. Each of these pistons presses a
brake pad against one side of a rotor attached to one of the rotating wheels. See
the figure for this problem. (a) When the force applied by the brake pedal to the
small piston is 7.5 N, what is the normal force applied to each side of the rotor?
The pressure in the fluid
Also,
Pr essure P =
P = Fb Ab .
Normal Force N
Area of the brake pad piston A
the normal force applied to each side of the rotor
A
12.0 cm 2
N = PA =
Fb =
(7.5 N) = 30 N
2
Ab
3.0 cm