Buoyancy and Archimedes` principle

Buoyancy and Archimedes’
principle
Fluid statics
• What is a fluid?
Density Pressure
• Fluid pressure and depth
Pascal’s principle
• Buoyancy
Archimedes’ principle
Fluid dynamics
• Reynolds number
• Equation of continuity
• Bernoulli’s principle
• Viscosity and turbulent flow
• Poiseuille’s equation
Lecture 3
Dr Julia Bryant
web notes: Fluidslect3.pdf
buoyancy.pdf
surface.pdf – not examinable
1
http://www.physics.usyd.edu.au/teach_res/jp/fluids/wfluids.htm
Buoyancy
•  When a solid object is wholly
or partly immersed in a fluid,
the fluid molecules are
continually striking the
submerged surface of the
object. "
•  The forces due to these
impacts can be combined into
a single force the buoyant
force which counteracts the
weight."
T = W - FB
T
FB
W
2
If Fb > Fg body floats.
If Fb < Fg body sinks.
A body floats in any
liquid with density
ρfluid > ρbody
Fb
Fg
Fb > Fg
DEMO
Fb < Fg
3
How high will it float?
- What fraction of an iceberg is under water?
Water expands on freezing
by 10%.
Density of ice is 0.9g/cm3
Fraction of iceberg
submerged is
ρice / ρ water= 0.9/1.0
Therefore 90% of the
iceberg is submerged.
4
How can we
measure the
force due to
buoyancy?
DEMO
5
Archimedes’ Principle
•  When an object is immersed in a
fluid, there is an upward buoyant
force equal to the weight of the
volume of fluid displaced by the
object.
•  This applies to either full or partial
immersion (i.e. a “sinking” or
“floating” object)
T = W - FB
T
FB
m
W
6
Hung
Tension
FB
T = W - FB
Weight
Sunk
Contact
FB
FC = W - FB
Water
displaced
Weight
7
Floating:
buoyant
force
equals
Water
displaced
weight
Floating: fully submerged
Some fish can remain at a fixed
depth without moving by storing
gas in their bladder.
Water
displaced
Submarines take on or discharge
water into their ballast tanks.
8
You are floating
around a pool drinking
a bottle of Kahlua.
FB
mg
If you accidently drop
the bottle into the
water (with the lid on),
how much of the
Kahlua would you
have had to have
drunk for the bottle
to float back up to
you?
9
How much of the Kahlua would you
have had to have drunk for the bottle
to float back up to you?
FB > mg
What is the volume of the bottle and it’s
contents?
FB
Full bottle has 350mls of liquid
= 350cm3 = 3.5x10-4m3 with a specific gravity
1.15
Small amount of air in the top
=πR2h= π x (1.5cm)2 x 2cm = 14 cm3 = 14 ml
mg
How much glass is there?
375g (empty bottle) with density of
2500kg.m3 ==> V=1.5x10-4m3 =150cm3
Total volume = 350 + 14 + 150cm3
= 514cm3 =5.14x10-4m3
10
Liquid before any is drunk
VKah= 350cm3 = 3.5x10-4m3
ρKah = 1150 kg.m-3
Air in the top Vair=14 ml =1.4x10-5 m3
Glass
Vglass= 150cm3 with ρglass = 2500Kg.m3
Total volume =5.14x10-4m3
FB
Buoyant force FB = ρwater V g
= 1000 x 5.14x10-4 x 9.8 = 5.037 N
mg = (ρair Vair + ρKah VKah + ρglass Vglass ) x g
mg
but some has been drunk
Vinside bottle = Vair + Vkah_before = 3.5x10-4 + 1.4x10-5 m3 = 3.64x10-4 m3
mg = [1.21 x (3.64x10-4 - VKah ) + 1150 x VKah + 2500 x 1.5x10-4] x 9.8 N
= 0.00432 + 11258.14 VKah + 3.675 N
The bottle will float if FB > mg
5.037 > 0.00432 + 11258.14 VKah + 3.675
==> VKah < 1.21x10-4 m3
or VKah < 121ml
(equivalent mass to 139mls water)
11
What is wrong with the picture of the ship
on the left?
12
Why can ships float?
A steel ship can encompass a great deal of empty space
and so have a large volume and a relatively small density."
Weight of ship = weight of water displaced"
13
A 200 tonne ship enters a lock of a canal.
The fit between the sides of the lock is so
tight that the weight of the water left in the
lock after it closes is much less than the
ship's weight. Can the ship float?
14
The buoyant force is equal to the weight of the water displaced, not
the water actually present. The missing water that would have filled
the volume of the ship below the waterline is the displaced fluid.
Volume of water displaced. This
volume is not necessarily the
volume present.
Weight of ship = weight of water displaced
15
Why is buoyancy due to the weight of fluid displaced?
An object floats because of the pressure difference between the
top and bottom of the object.
top"
A!
ρF"
h!
ρo"
bottom"
F = (pbottom – ptop) A
F = (patm + ρF g h – patm) A
p = p0 + ρ g h
F = ρF g h A = ρF VF g
16
F = mF g = Weight of displaced fluid =>Buoyant force
Hydrometer
Hydrometer can be used to measure fluid density.
A hydrometer floats due to buoyancy.
Higher fluid density => higher buoyant force
FB = ρ V g
ρwater = 1000 kg.m-3
ρkerosene = 817 kg.m-3
DEMO
17
Hydrometer - example
If a hydrometer was a rod that had a length of 0.250 m, cross sectional
area was 2.00×10-4 m2, and mass of 4.50×10-2 kg,
(a) How far from the bottom end of the rod should a mark of 1.000 be
placed to indicate the relative density of the water (density 1.000×103
kg.m-3)?
(b) If the hydrometer sinks to a depth of 0.229m when placed into an
alcohol solution. What is the density of the alcohol solution? A
Let h be the height to which it is submerged.
FB = ρAhg upwards Fg= mg downwards
ρAhg = mg
In water:
h= m =
4.50×10-2
= 0.225m
(ρA)
(1.000×103)(2.00×10-4)
In alcohol solution:
ρ  = m =
4.50×10-2
= 983 kg.m -3
(Ah) (2.00×10-4)(0.229)
0.250m
h
18
?
A
Two cups are filled to the same level.
One cup has ice cubes floating on it.
Which weighs more?"
B
Two cups are filled to the same level.
One of the cups has ice cubes floating
in it. When the ice melts, in which cup
is the level higher?
Cup 1
Cup 2
19
?
A
Two cups are filled to the same level.
One cup has ice cubes floating on it.
Which weighs more?"
Cups weigh the same.
• Weight of the ice cubes is equal to the buoyant force. "
• The buoyant force is equal to the weight of the water displaced by the
ice cubes. "
• This means that the weight that the ice cubes add to the cup is exactly
what an amount of water that is equal to that submerged volume of ice
cubes would add. "
Cup 1
Cup 2
mg FB
20
?
A
Two cups are filled to the same level.
One cup has ice cubes floating on it.
Which weighs more?"
B
Two cups are filled to the same level.
One of the cups has ice cubes floating
in it. When the ice melts, in which cup
is the level higher?
Cup 1
Cup 2
21
?
B
Two cups are filled to the same level.
One of the cups has ice cubes floating
in it. When the ice melts, in which cup
is the level higher?
The level is the same.
• The weight of the ice cubes is equal to the weight of the water that
would fill the submerged volume of the cubes. "
• When the cubes melt into the water the volume of melted water is
exactly equal to the volume of water that the cubes were displacing."
Cup 1
Cup 2
22
A giant clam has a
mass of 470 kg and a
volume of 0.350 m3
lies at the bottom of a
freshwater lake. How
much force is needed
to lift it at constant
velocity?
23
m = 470 kg
Vclam = 0.350 m3
g = 9.8 m.s-2
FT + FB = FG"
FT = ? N
m"
FT + FB"
ΣF = 0
a = 0"
ρwater = 1.0x103 kg.m-3
FG"
FT = FG - FB
FT = m g - ρwater g Vdisplaced
FT = m g - ρwater g Vclam
FT = (470)(9.8) – (103)(9.8)(0.350) N
FT = 1.2×103 N
= force to lift 70 full 750ml Kahlua bottles in air!
24
A ring weighs 6.327×10-3 N in air and
6.033×10-3 N when submerged in water.
What is the volume of the ring?
What is the density of the ring?
What is the ring made of?
air
water
25
Weighs 6.327×10-3 N in air and 6.033×10-3 N in water.
Volume? density? What is the ring made of?
air
FTair = FG
water
Archimedes’ Principle
FB = weight of water displaced
FB = ρF Vring g
FTwater + FB = FG
Decrease in weight due to buoyant force "
FB = FGair – FGwater = 6.327×10-3 - 6.033×10-3 = 0.294x10-3 N "
Find the volume of the ring:"
Since the ring is fully submerged"
VR = FB/ρF g =(0.294x10-3) / {(103 )(9.8)} m3 "
= 3.00x10-8 m3"
Find the density of the ring:"
ρR = mR / VR = Fgair /(g VR)=(6.327x10-3) / {(9.8)(3.0x10-8 )} kg.m-3
= 21.5x103 kg.m-3"
What is the ring made of?"
VR = 3.0x10-8 m3 ρR = 2.2x104 kg.m-3 maybe gold
"
26
REYNOLDS NUMBER
A British scientist Osborne Reynolds (1842 –
1912) established that the nature of the flow
depends upon a dimensionless quantity, which
is now called the Reynolds number Re.
Re = ρ v L / η
ρ
density of fluid
v
average flow velocity over the cross
section of the pipe
L
characteristic dimension
η
viscosity
pascal 1 Pa = 1 N.m
Re = ρ v L / η
newton 1 N = 1 kg.m.s
[Re] ≡ [kg.m-3] [m.s-1][m]
1 Pa.s = kg.m.s . m .s
[Pa.s]
≡ kg x m x m x s2.m2 = [1]
m3 s
kg.m.s
Re is a dimensionless number
-2
-2
-2
-2
As a rule of thumb, for a flowing fluid
Re < ~ 2000 laminar flow
~ 2000 < Re < ~ 3000 unstable laminar
to turbulent flow
Re > ~ 2000 turbulent flow
Consider an IDEAL FLUID
Fluid motion is very complicated. However, by
making some assumptions, we can develop a
useful model of fluid behaviour. An ideal fluid is
Incompressible – the density is constant
Irrotational – the flow is smooth, no turbulence
Nonviscous – fluid has no internal friction (η=0)
Steady flow – the velocity of the fluid at each
point is constant in time.
Consider the average motion of
the fluid at a particular point in
space and time."
An individual fluid element will
follow a path called a flow line."
Streamlines"
- in steady flow,
a bundle of
streamlines
makes a flow
tube.
Steady flow is when the pattern of
flow lines does not change with
time.
v
Velocity of particle is
tangent to streamline"
Streamlines cannot cross