Physics 151: Lecture 29
Today’s Agenda

Today’s topics
Fluids under static conditions, Ch. 14.1 through 14.4
Pressure
Pascal’s Principle (hydraulic lifts etc.)
Archimedes’ Principle (floatation)
Physics 151: Lecture 29, Pg 1
See text: 14.1
Fluids

At ordinary temperature, matter exists in one of three states
Solid - has a shape and forms a surface
Liquid - has no shape but forms a surface
Gas - has no shape and forms no surface

What do we mean by “fluids”?
Fluids are “substances that flow”…. “substances that
take the shape of the container”
Atoms and molecules are free to move.
No long range correlation between positions.
Physics 151: Lecture 29, Pg 2
See text: 14.1
Fluids

What parameters do we use to describe fluids?
Density

m
V
units :
kg/m3 = 10-3 g/cm3
(water) = 1.000 x103 kg/m3
= 1.000 g/cm3
(ice)
= 0.917 x103 kg/m3
= 0.917 g/cm3
(air)
= 1.29 kg/m3
= 1.29 x10-3 g/cm3
(Hg)
= 13.6 x103 kg/m3
= 13.6 g/cm3
Physics 151: Lecture 29, Pg 3
Fluids

What parameters do we use to describe fluids?
Pressure
F
p
units :
1 N/m2 = 1 Pa (Pascal)
1 bar = 105 Pa
1 mbar = 102 Pa
1 torr = 133.3 Pa

A
1atm = 1.013 x105 Pa
= 1013 mbar
= 760 Torr
= 14.7 lb/m2 (=PSI)
Any force exerted by a fluid is perpendicular to a surface of
contact, and is proportional to the area of that surface.
Force (a vector) in a fluid can be expressed in terms of
pressure (a scalar) as:
n
F  pAnˆ
A
Physics 151: Lecture 29, Pg 4
See text: 14.2
Pressure vs. Depth
Incompressible Fluids (liquids)



When the pressure is much less
than the bulk modulus of the fluid,
we treat the density as constant
independent of pressure:
incompressible fluid
For an incompressible fluid, the
density is the same everywhere,
but the pressure is NOT!
p
0
y1
p1
F1
y2
A
p2
mg F2
Consider an imaginary fluid volume (a cube, face area A)
The sum of all the forces on this volume must be ZERO as
it is in equilibrium: F2 - F1 - mg = 0
F2  F1  p2 A  p1 A
mg  ( y 2  y1 )Ag
p2  p1  g ( y 2  y1 )
Physics 151: Lecture 29, Pg 5
See text: 14.2
Pressure vs. Depth

For a 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

Why is this so? Why does the
pressure below the surface
depend only on depth if it is in
equilibrium?
 Imagine a tube that would connect two regions at the same depth.
 If the pressures were different, fluid would flow in the tube!
 However, if fluid did flow, then the system was NOT in equilibrium
since no equilibrium system will spontaneously leave equilibrium.
Physics 151: Lecture 29, Pg 6
Lecture 29, ACT 1
Pressure

What happens with two fluids??
Consider a U tube containing liquids of
density 1 and 2 as shown:
Compare the densities of the liquids:
A) 1 < 2
B) 1 = 2
dI
2
1
C) 1 > 2
Physics 151: Lecture 29, Pg 7
Example


A U-tube of uniform crosssectional area, open to the
atmosphere, is partially filled
with mercury. Water is then
poured into both arms. If the
equilibrium configuration of
the tube is as shown in Figure
on the right, with h2 = 1.00 cm.
Determine the value of h1.
Physics 151: Lecture 29, Pg 8
Example

Figure on the right shows
Superman attempting to drink
water through a very long
straw. With his great strength
he achieves maximum possible
suction. The walls of the
tubular straw do not collapse.

(a) Find the maximum height
through which he can lift the
water.
Physics 151: Lecture 29, Pg 9
See text: 14.2
Pascal’s Principle


So far we have discovered (using Newton’s Laws):
Pressure depends on depth: p = 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.

Pascal’s Principle explains the working of hydraulic lifts
i.e. the application of a small force at one place can result
in the creation of a large force in another.
Does this “hydraulic lever” violate conservation of energy?
» Certainly hope not.. Let’s calculate.
Physics 151: Lecture 29, Pg 10
See text: 14.2
Pascal’s Principle

Consider the system shown:
A downward force F1 is applied
to the piston of area A1.
This force is transmitted through
the liquid to create an upward
force F2.
Pascal’s Principle says that
increased pressure from F1
(F1/A1) is transmitted throughout
the liquid.
F1
F
 2
A1
A2

F1
F2
d2
d1
A1
A2
A2
F2  F1
A1
F2 > F1 : Have we violated conservation of energy??
Physics 151: Lecture 29, Pg 11
See text: 14.2
Pascal’s Principle

Consider F1 moving through a
distance d1.
How large is the volume of the
liquid displaced?
V1  d1 A1
F1
d2
d1
This volume determines the
displacement of the large piston.
V2  V1
d2
A1
A2
A1
 d1
A2
W2  F2 d 2  F1

F2
A2
A
d1 1  W1
A1
A2
Therefore the work done by F1 equals the work done by F2
We have NOT obtained “something for nothing”.
Physics 151: Lecture 29, Pg 12
Lecture 29, ACT 2a
Hydraulics

Consider the systems shown to the
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.
If A2 = 2A1, compare dA and dB.
A) dA=(1/2)dB
B) dA = dB
dA
A1
M
A10
dB
A2
M
A10
C) dA = 2dB
Physics 151: Lecture 29, Pg 13
Lecture 29, ACT 2b
Hydraulics

Consider the systems shown to the
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.
If A10 = 2A20, compare dA and dC.
dA
A1
A10
dC
A1
A) dA = (1/2)dC
B) dA = dC
M
M
A20
C) dA = 2dC
Physics 151: Lecture 29, Pg 14
See text: 14.4
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 > W 2
W1
W2?
Physics 151: Lecture 29, Pg 15
See text: 14.4
Archimedes’ Principle

W2?
W1
The buoyant force is equal to the difference
in the pressures times the area.
FB  ( p2  p1 )  A  g(y2 - y1)A
FB   liquidgVliquid  Mliquid  g  Wliquid
Archimedes:
The buoyant force is equal to the
weight of the liquid displaced.
F
y1
1
y2
p
1

The buoyant force determines
whether an object will sink or float.
How does this work?
A
p
2
F
2
Physics 151: Lecture 29, Pg 16
See text: 14.4
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:
Object is in equilibrium
FB  mg
 liquid  g  Vliquid  object  g  Vobject
Vliquid
Vobject
object

 liquid
Animation
Physics 151: Lecture 29, Pg 17
See text: 14.4
The Tip of the Iceberg

What fraction of an iceberg is submerged?
Vliquid
Vobject
object

 liquid
y
FB mg
Vwater
ice
917 kg/m3


 90%
3
Vice
water 1024 kg/m
Physics 151: Lecture 29, Pg 18
Lecture 29, ACT 3
Buoyancy

A 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.
If you turn the styrofoam+Pb upside
down, what happens?
A) It sinks
B)
styrofoam
Pb
C)
styrofoam
Pb
Pb
styrofoam
D)
styrofoam
Pb
Physics 151: Lecture 29, Pg 19
See text: 14.4
ACT 3-A
More Fun With Buoyancy

Two cups are filled to the same level
with water. One of the two cups has
plastic balls floating in it.
Which cup weighs more?
(A) Cup I
(B) Cup II
(C) the same
Cup I
Cup II
(D) can’t tell
Physics 151: Lecture 29, Pg 20
See text: 14.4
ACT 3-B
Even More Fun With Buoyancy

A plastic ball floats in a cup of water with
half of its volume submerged. Next some
oil (oil < ball < water) is slowly added to
the container until it just covers the ball.
water

Relative to the water level, the ball
will:
(A) move up
(B) move down
(C) stay in same place
Physics 151: Lecture 29, Pg 21
Recap of today’s lecture

Chapter 14.1-4
Pressure
Pascal’s Principle
Archimedes Principle
Physics 151: Lecture 29, Pg 22