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Lecture 19Lecture 19Fluids: density, pressure,
Pascal’s principle and Buoyancy.Water tower
Pascal’s vases
Hydraulic press
Barometer
What is a fluid?
Fluids are “substances that flow”…. “substances that take the shape of the container”
Atoms and molecules must be free to move .. No long range correlation between positions (e.g., not a crystal).
Gas or liquid… or granular materials (like sand)
In a fluid:
Density, pressure
Density:
Units:
Pascal Pa = 1 N/m2
psi = lb/in2
atmosphere 1 atm = 1.013 × 105 Pabar 1 bar = 105 Pa
Ex: Pure water: 1000 kg/m3
Vm
Pressure: Fp
A
Particles are always moving
i.e., exerting (perpendicular) forces on surfaces
F
i.e., hitting surfaces
Surface of area A
Atmospheric pressure
DEMO: Piston and weight
The atmosphere of Earth is a fluid, so every object in air is subject to some pressure.
At the surface of the Earth, the pressure is
patm ~ 1.013 x 105 Pa = 1 atm
Area of a hand ~ 200 cm2 = 0.02 m2
atm ~ 2000 N on your hand due to air!F p A
Vacuum gun
Sealed tube, air pumped out
Ping-pong ball
What happens if we punch a little hole on one side?
DEMO: Vacuum gun
Length of tube L ~ 3 mMass of ball m ~ 3 gRadius of tube R ~ 2 cm
W = ΔKE
FL ∼ (patmR2)L =12
mv 2− 0
v ∼ √2patmR2L
m∼300 km/s
Pressure vs. depth
Fbottom
Ftop
mg
Imaginary box of fluid with density ρ with bases of area A
and height h
bottom topp p gh
m Ah
hFor floating object, net force must be zero!
bottom topF F mg
DEMO: Plastic tube
with cover
Example: How deep under water is p = 2 atm ?
h =pbottom−ptop
ρfluidg=
1.01×105 Pa(103 kg/m3) (9.81 m/s2)
= 10.3 m
(i.e. 1 atm is produced by a 10.3 m high column of water)
Called guage pressure
Pbottom/top =Fbottom/top
A
Fluid in an open container
Pressure is the same at a given depth, independent of the shape of the container. p(y)
y
Fluid level is the same everywhere in a connected container (assuming no surface forces)
•A
•B
If liquid height was higher above A than above B
If liquid height was higher above A than above B
pA > pB pA > pB Net force
Net force
Net flow
Net flow
This is not equilibrium!This is not equilibrium!
DEMO: Pascal’s vases
ACT: U tube
Two liquids Y and G separated by a thin, light piston (so they cannot mix) are placed in a U-shaped container. What can you say about their densities?
A. ρG < ρY
B. ρG = ρY
C. ρG > ρY
YG
•A
•B
Water towers
Water towers are a common sight in the Midwest… because it’s so flat!
h
house atm waterp p hg
So physics sucks, but how much?
Your physics professor sucks on a long tube that rises out of a bucket of water. He can get the liquid to rise 5.5 m (vertically). What is the pressure in His mouth at this moment?
A. 1 atm
A. 0.67 atm
B. 0.57 atm
C. 0.46 atm
D. 0 atm
h
x A
x B
So physics sucks, but how much?
Your physics professor sucks on a long tube that rises out of a bucket of water. He can get the liquid to rise 5.5 m (vertically). What is the pressure in His mouth at this moment?
A. 1 atm
A. 0.67 atm
B. 0.57 atm
C. 0.46 atm
D. 0 atm
DEMO: Sucking
through a hose
water atmmouth
atm watermouth
5 3 3 2 10 Pa 10 kg/m 9.8 m/s 5.5 m 46100 Pa 0.46 atm
p gh pp p gh
h
x A
x B
Pascal’s principle
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 is most often applied to incompressible fluids (liquids):
Increasing p at any depth (including the surface) gives the same increase in p at any other depth
Hydraulic chamber
F2 can be very large…
No energy is lost:
F1
A1
=F2
A2
F2 =A2
A2
F1
W = F1d1=(F2
A1
A2)(d2
A2
A1)=F2d2
Incompressible fluid: A1d1 = A2 d2
ACT: Hydraulic chambers
In each case, a block of mass M is placed on the piston of the large cylinder, resulting in a difference di between the liquid levels. If A2 = 2A1, then:
A1 A10
A2 A10
M
MdB
dA
A. dA < dB
B. dA = dB
C. dA > dB
Measuring pressure with fluids
Barometer Measures absolute pressure Top of tube evacuated (p = 0) Bottom of tube submerged into pool of mercury
open to sample (p) Pressure dependence on depth:
vacuum
p=0
h atmosphere
p=p0
Sample at p
Sample at p
hh
Vacuum p = 0
Vacuum p = 0
pp patmpatm
∆h∆h
patm
h
p
Manometer Measures gauge pressure: pressure relative to
atmospheric pressure. Pressure dependence on depth: Δ h =
p−patm
ρHg g
A unit for pressure760 mm Hg = 760 torr = 1 atm
h =p
ρHgg
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?
vacuum
stopper
A. Water spurts out of the side tube.
B. Air flows in through the side tube.
C. Nothing, the system was in equilibrium and remains in equilibrium.
Buoyancy and the Archimedes’ principle
ybottom
ytop
hA
A box of base A and height h is submerged in a liquid of density ρ.
bottom topAp Ap
atm bottom atm topA p gy A p gy
A hg
Archimedes’ principle: The liquid exerts a net force upward called buoyant force whose magnitude is equal to the weight of the displaced liquid.
direction upVg
topbottomF F F
Ftop
Fbottom
Net force by liquid:
In-class example: Hollow sphere
A hollow sphere of iron (ρFe = 7800 kg/m3) has a mass of 5 kg. What is the maximum diameter necessary for this sphere to be completely submerged in water? (ρwater = 1000 kg/m3)
A. It will always be submerged.
B. 0.11 m
C. 0.21 m
D. 0.42 m
E. It will always only float.
In-class example: Hollow sphere
A hollow sphere of iron (ρFe = 7800 kg/m3) has a mass of 5 kg. What is the maximum diameter necessary for this sphere to be completely submerged in water? (ρwater = 1000 kg/m3)
A. It will always be submerged.
B. 0.11 m
C. 0.21 m
D. 0.42 m
E. It will always only float.
FB
mg
The sphere sinks if FB< mg
maxMaximum diameter 2 0.21 mR
ρwater43
πR3 g < mg ⇒ R <3√
3m4πρwater
= 0.106 m
What if the sphere is solid?!
Density rule
A hollow sphere of iron (ρFe = 7800 kg/m3) has a mass of 5 kg. What is the maximum diameter necessary for this sphere to be fully submerged in water? (ρwater = 1000 kg/m3) Answer: R = 0.106 m.
And what is the average density of this sphere?
3
sphere water33
5 kg 1000 kg/m4 4 0.106 m3 3
m
R
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 (will not be completely submerged)
DEMO: Frozen helium
balloon
This is why you cannot sink in the Dead Sea
(ρDead Sea water = 1240 kg/m3 , ρhuman body = 1062 kg/m3 ) !
Buoyancy (in the Dead Sea)
ACT: Styrofoam and lead
A piece of lead is glued to a slab of Styrofoam. When placed in water, they float as shown.
What happens if you turn the system upside down?
A
Pb
styrofoam
Pb
styrofoam
Pb
styrofoam
B
C. It sinks.
1 2
ACT: Floating wood
Two cups have the same level of 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.
ACT: Aluminum and lead
Two blocks of aluminum and lead with identical sizes are suspended from the ceiling with strings of different lengths and placed inside a bucket of water as shown. In which case is the buoyant force greater?
A. Al
B. Pb
C. It’s the same for both
Al
Pb
ceiling
ACT: Wooden brick
When a uniform wooden brick (1 m x 1 m x 2 m) is placed horizontally on water, it is partially submerged and the height of the brick above the water surface is 0.5 m. If the brick was placed vertically, the height of brick above the water would be:
A. 0.5 m
B. 1.0 m
C. 1.5 m.
0.5 m