Lecture 03 archimedes. fluid dynamics.

1. Lecture 3
Archimedes principle.
Fluid dynamics.

2. ACT: Side tube
A sort of barometer is set up with a tube that has a side tube
with a tight fitting stopper. What happens when the stopper is
removed?
A. Water spurts out of the side
tube.
vacuum
stopper
B. Air flows in through the side
tube.
C. Nothing, the system was in
equilibrium and remains in
equilibrium.
DEMO:
Side tube

3. Buoyancy and the Archimedes’ principle
A box of base A and height h is
submerged in a liquid of density ρ.
ytop
Net force by liquid:
ΣF = Fbottom − Ftop
= Apbottom − Aptop
A
ybottom
(
= A ( patm + ρ gybottom ) − A patm + ρ gy top
= A ρhg = ρVg
Ftop
)
h
Fbottom
direction up
Archimedes’s principle: The liquid exerts a net force
upward called buoyant force whose magnitude is equal to
the weight of the displaced liquid.

4. In-class example: Hollow sphere
A hollow sphere of iron (ρFe = 7800 kg/m3) has a mass of 5 kg. What is the
minimum diameter necessary for this sphere not to sink in water?
(ρwater = 1000 kg/m3)
A. It will always sink.
FB
B. 0.11 m
C. 0.21 m
mg
D. 0.42 m
E. It will always float.
The sphere sinks if mg > FB
mg > ρwater
R <
3
4
πR 3g
3
3m
= 0.106 m
4πρwater
Minimum diameter = 2R ≥ 0.21 m

5. Density rule
DEMO:
Frozen helium
balloon
A hollow sphere of iron (ρFe = 7800 kg/m3) has a mass of 5 kg. What is the
minimum diameter necessary for this sphere not to sink in water
( ρwater
= 1000 kg/m3) ? Answer: R = 0.106 m.
And what is the average density of this sphere?
5 kg
m
ρsphere =
=
= 1000 kg/m3 = ρwater
3
4
4
πR 3
π ( 0.106 m )
3
3
An object of density ρobject placed in a fluid of density ρfluid
• sinks if ρobject > ρfluid
• is in equilibrium anywhere in the fluid if ρobject = ρfluid
• floats if ρobject ρfluid
This is why you float on the sea (1025 kg/m3) but not on a pool (1000 kg/m3) …

6. ACT: Styrofoam and lead
A piece of lead is glued to a slab of
Styrofoam. When placed in water, they float
as shown.
Pb
styrofoam
What happens if you turn the system upside
down?
styrofoam
styrofoam
Pb
Pb
A
C. It sinks.
B
The displaced volume in both cases must be the same (volume of water
whose weight is equal to the weight of the lead+Styrofoam system)

7. ACT: Floating wood
Two cups are filled to the same level with water. One of the two
cups has a wooden block floating in it. Which cup weighs more?
A. Cup 1
B. Cup 2
C. They weigh the same.
1
2
Cup 2 has less water than cup 1.
The weight of the wood is equal to the weight of the missing
liquid (= “displaced liquid”) in 2.
DEMO:
Bucket of water
with wooden block

8. Attraction between molecules
Wood floats on water because it is less dense than water. But a paper clip (metal,
denser than water!) also floats in water… (?) .
Molecules in liquid attract each other (cohesive forces that
keep liquid as such!)
Very small attraction by
air molecules.
On the surface: Net
force on a molecule is
inward.
In the bulk: Net force
on a molecule is zero.
…And this force is
compensated by the
incompressibility of
the liquid.

9. Surface tension
Overall, the liquid doesn’t “like” surface molecules because
they try to compress it.
Liquid adopts the shape that minimizes the surface area.
Any attempt to increase this area is opposed by a restoring force.
The surface of a liquid behaves like an elastic membrane.
The weight of the paper clip is small
enough to be balanced by the elastic
forces due to surface tension.

10. Drops and bubbles
Water drops are spherical
(shape with minimum area for a
given volume)
Adding soap to water decreases surface tension. This is useful to:
• Force water through the small spaces between cloth fibers
• Make bubbles! (Large area and small bulk)

11. How wet is water?
Molecules in a liquid are also attracted to the medium it is in contact
with, like the walls of the container (adhesive forces).
Water in a glass
Fadhesive > Fcohesive
Water in wax- or
teflon-coated glass
Fadhesive < Fcohesive
Or: surface tension in air/liquid interface is larger/smaller than surface tension in
wall/liquid interface

12. Fluid flow
Laminar flow:
no mixing
between layers
Turbulent flow:
a mess…

13. Dry water, wet water
Within the case of laminar flow:
Slower near the walls
Faster in the center
Same speed
everywhere
Real (wet) fluid:
friction with walls and
between layers (viscosity)
Ideal (dry) fluid:
no friction (no viscosity)

14. Flow rate
Consider a laminar, steady flow of an ideal, incompressible fluid
at speed v though a tube of cross-sectional area A
v dt
A
dV = Avdt
Volume flow rate
dV
= Av
dt
Mass flow rate
dm
= ρAv
dt

15. Continuity equation
The mass flow rate must be the same at any point along the tube
(otherwise, fluid would be accumulating or disappearing somewhere)
ρ1Av1 = ρ2Av2
1
2
v2
A2
v1 A1
ρ1
If fluid is incompressible
(constant density):
ρ2
Av1 = Av2
1
2

16. Example: Garden hose
When you use your garden faucet to fill your 3 gallon watering can, it takes
15 seconds. You then attach your 1.5 cm thick garden hose fitted with a
nozzle with 10 holes at the end. You turn on the water, and 4 seconds later
water spurts through the nozzle. When you hold the nozzle horizontally at
waist level (1 m from the ground), you can water plants that are 5 m away.
a) How long is the hose?
b) How big are the openings in the nozzle?
Volume flow rate
dV
= Ahosevhose
dt
dV 3 gallons 3.785 liters
1 m3
=
= 7.6 × 10 −4 m3 /s
dt
15 s
1 gallon 1000 liter
vhose
dV
−4
3
dt = 7.6 × 10 m /s = 1.1 m/s
=
Ahose π 1.5 × 10 −2 m 2
(
Length of hose = vhoset = ( 1.1 m/s ) ( 4 s ) = 4.3 m
)

17. When you use your garden faucet to fill your 3 gallon watering can, it takes 15
seconds. You then attach your 1.5 cm thick garden hose fitted with a nozzle with 10
holes at the end. You turn on the water, and 4 seconds later water spurts through the
nozzle. When you hold the nozzle horizontally at waist level (1 m from the ground), you
can water plants that are 5 m away.
a) How long is the hose?
b) How big are the openings in the nozzle?
Ahosevhose = Anozzlevnozzle
2
∅2 vhose = 10∅nozzlevnozzle
hose
We use kinematics to determine vnozzle:
x = 0 + vnozzlet
t =
g 2
t
2
2
 x 
g
0 =h +x − 
÷
2  vnozzle ÷


x
h
vnozzle
x
0 = h + vnozzlet −
∅nozzle
→
∅hose vhose
( 1.5 cm )
=
=
10 vnozzle
10
vnozzle =
gx
=
2( h + x )
2
( 9.8 m/s ) ( 5 m)
2
2 ( 6 m)
1.1 m/s
= 0.073 cm = 0.73 mm
4.5 m/s
2
= 4.5 m/s