LECTURE 2
STATIC FLUIDS
Lecture Instructor: Kazumi Tolich
Lecture 2
2
! 
Reading chapter 15-3 to 15-5
!  Pressure
and depth
!  Pascal’s principle
!  Archimedes’ principle and buoyancy
Force and depth in static fluid
3
! 
! 
! 
Consider a cylinder with a cross-sectional area of A, filled with
a fluid with a density ρ.
The force pressing at the top is less than the force pressing at
the bottom because of the weight of the water.
At the surface, the pressure
must be at atmospheric
pressure.
Ftop = Pat A
Fbottom = Ftop + ρ ( Ah ) g
Pressure and depth
4
! 
Since the pressure is the force divided by area, if
the force increases as you go deeper, the pressure
also increases as you go deeper in the fluid.
P2 = P1 + ρ gh
Example: 1
5
! 
A diver swims to a depth
of d = 3.2 m in a
freshwater lake. What is
the increase in the force
pushing in on her
eardrum, compared to
what it was at the lake
surface? The area of the
eardrum is
A = 0.60 cm2.
Mercury barometer
6
! 
! 
A barometer can measure atmospheric
pressure using the variation of pressure with vacuum
depth of a fluid.
A glass tube is filled with a fluid with a
density ρ and placed upside-down in the
same fluid in a bowl.
Pat = ρ gh
! 
Mercury is often used because of its high
density.
1 atmosphere ≡ Pat ≡ 760 mmHg
Water level
7
Fluids flow until equilibrium is reached.
!  The pressure at the surface exposed to air is at Pat.
!  In equilibrium, the net force on each section of the
fluid is zero.
!  The pressure in the fluid
! 
Example: 2
8
! 
A U-shaped tube is filled mostly
with water, but a small amount of
oil has been added to both sides.
The density of water is
ρw = 1.00 × 103 kg/m3, and the
density of the oil is
ρo = 9.20 × 102 kg/m3. On the
left side of the tube, the depth of
the oil is dL = 3.00 cm, and on the
right side of the tube, the depth of
the oil is dR = 5.00 cm. Find the
difference in fluid level between
the two sides of the tube.
Clicker question: 1
9
Pascal’s principle
10
! 
! 
According to Pascal's principle,
an externally applied pressure changes the pressure
at every point in a confined liquid or gas by the same
amount.
An example of Pascal’s principle in action is hydraulic
lift.
F1 F2
ΔP =
=
A1 A2
A1d1 = A2 d2
Example: 3
11
! 
A hydraulic lift has two pistons
with different diameters initially
at the same height. The smaller
piston has a diameter of
D1 = 1.5 m, and the larger piston
has a diameter of D2 = 21 m.
a) 
b) 
If a weight with a mass
m = 2000 kg is placed on the
large piston, what weight on the
small piston will be required to
keep the pistons at the same
height?
Through what distance must the
large piston be pushed down to
raise the small piston a distance
of d1 = 3.5 m?
Demo: 1
12
! 
Hydraulic press
!  Demonstration
of Pascal’s principle
Buoyancy and Archimedes’ principle
13
The force exerted by a fluid on a body wholly or
partially submerged in it is called the buoyant force.
!  Archimedes’ principle:
A body wholly or partially submerged in a fluid is
buoyed up by a force equal to the weight of the
displaced fluid.
! 
Fb = ρfluid gV
V is the volume of the object in the fluid.
Demo: 2
14
! 
Archimedes’ principle
!  The
buoyant force is equal to the weight of the water
displaced.
Lifting a rock under water
15
Why is it easier to lift a rock under water?
!  The buoyant force is acting upward.
!  Since the density of water is much greater than that
of air, the buoyant force is much greater under the
water compared to in the air.
! 
Fb air = ρair gV
Fb water = ρwater gV
The crown and the nugget
16
Archimedes (287-212 BC) had been given the task of determining
whether a crown made for King Hieron II was pure gold.
In the above diagram, crown and nugget balance in air, but not in
water because the crown has a lower density.
Demo: 3
17
! 
Helium balloon in liquid nitrogen
!  Demonstration
of buoyancy and Archimedes’ principle
Floatation
18
! 
! 
! 
When an object floats, the buoyant force is equals its weight.
An object floats when it displaces an amount of fluid whose
weight is equal to the weight of the object.
An object that is denser than fluid can float if it displaces
enough fluid because of its shape.
Floating ice cubes
19
! 
! 
! 
! 
Why do ice cubes float in water?
When the ice cube is floating, the net force on it is zero.
The weight of the ice cube is equal in magnitude to the buoyant
force.
The density of ice is less than that of liquid water, so it displaces
a volume of water that is less than the volume of the ice.
W = mg = ρice gVentire ice
Fb = ρwater gVice in water
ρice gVentire ice = ρwater gVice in water
ρiceVentire ice = ρwaterVice in water
ρice < ρwater → Ventire ice > Vice in water
Clicker question: 2
20
Example: 4
21
a) 
b) 
What is the minimum area of
the top surface of a slab of
ice with thickness, h = 0.30 m,
floating on fresh water that
will hold up an automobile of
mass mc = 1100 kg sitting on
top? The density of water
and ice are
ρw = 0.998 × 103 kg/m3 and
ρi = 0.917 × 103 kg/m3,
respectively.
Does it matter where the car
is placed on the block of ice?
Submerged volume
22
! 
When a solid of volume Vs and density ρs floats in a
fluid of density ρf, the volume of the solid that is
submerged in the fluid, Vsub, is given by
ρs
Vsub = Vs
ρf
Example: 5
23
! 
You decide to enroll in a fitness
program. To determine your initial
fitness, your percentage of body
fat is measured. Assume the
average density of fat is
ρf = 0.90 × 103 kg/m3, and lean
tissue is ρl = 1.10 × 103 kg/m3.
Your apparent weight in water is
5.0% of your weight in air. What
fraction, f, of your body mass is
fat?