Liquids and Gases
The unit of volume is the meter
cubed , m3, which is a very large
volume. Very often we use
cm3 = cc, or Litres = 103cc
Everyday units (U.S.) are
gallons, quarts, pints
1 qt = 2 lbs H2O, pt.=lb.
1 L = 2.2 lbs H2O = 1 kg H2O
As we know liquids and gases (fluids) act very differently than solids.
Liquids and gases have mass but their constituent atoms are not rigidly
bound. Therefore each part of the liquid or gas can move.
The atoms of a liquid are more tightly bound so a liquid can be kept in an
open container whereas gas usually requires a closed container.
Liquids, like solids are not very compressible, that is, it is difficult to
change the volume. We call liquids and solids “condensed matter”.
The volume of a gas can be changed fairly easily by changing the
pressure. Both liquids and gases have the property of being able to flow,
Hence the name “fluid”.
For example water and gas lines in a house.
Week 8
Physics 214
1
ANNOUNCEMENT
BONUS and MAKEUP POLICY
ONE-HALF OF YOUR FINAL EXAM SCORE WILL
REPLACE YOUR LOWEST HOUR-EXAM SCORE, IF
THAT IS TO YOUR ADVANTAGE (i.e. if that amount
is MORE than your lowest hour-exam score)
This will also be the “makeup” score for an
EXCUSED missed hour-exam.
Example: T1 65 points
T2 75
Final 160
160/2=80, 80 replaces worst score so T1 becomes 80
Week 8
Physics 214
2
Pressure
A volume of liquid or gas has mass and F = ma is still a good law.
But if we exert a force at a point on, say, the surface of water in a
container, only the water near the point moves, and that
movement is quite complicated.
So generally we work with liquids and gases in containers
and exert forces over a surface.
We define pressure as
P = F/A
“force per unit area”
that is, the force divided by the area over which the force
acts.
Any change in pressure is transmitted uniformly
throughout a liquid.
Units are N/m2
Week 8
1 N/m2 = 1Pascal
Physics 214
3
Hydraulic Jack
If the liquid levels on
each side are the same height
then the pressure just below
each piston is the same and
F1/A1 = F2/A2 so F2 = F1A2/A1
If we make A2/A1 = 100 then an
F1 force of 50 lbs can lift 5000 lbs.
Of course if we push F1 down a distance h1 then the F2
Side will only rise h2=h1A1/A2 because h1A1 = Vol. = h2A2, that is,
the displaced volumes must be the same. So h2=h1/100
Work done = F1h1 = F2h2 Like a lever, the small force works over
the long distance and vice versa.
Week 8
Physics 214
4
Ch 9 E 6
Hydraulic system: A2 = 50 A1
F2 = 7000 N. What is F1?
(neglect height difference)
F2 = 7000N.
A1
A. 50N
B. 7000N
C. 350,000N
A2
D. 3,500N
E. 140N
F1/A1= Pressure = F2/A2 =F2/50A1 then multiply both sides by A1
So F1 = 7000/50 = 140 N
.
Week 8
Physics 214
5
Atmospheric pressure
If you stack a pile of bricks each brick has
to support all the bricks above it. So as
We go down the stack the force increases,
and so does the pressure.
In the example shown the force on the face
of the bottom brick would be 6mg and the
pressure = 6mg/A, where A is the area of contact
m
g
F
At the earth’s surface we are under
a column of air which exerts a weight force.
Because it is a fluid the air exerts an equal pressure
In all directions.
This pressure comes from the weight of air above
a given horizontal area.
Week 8
Physics 214
6
Density
If one takes two objects of exactly the same
volume made of different materials they usually
have different weights. So we define a useful
quantity
ρ (rho)
Density ρ = mass/unit volume, [kg/m3] or grams/cc
so the mass of an object is M= ρV
and the weight is W= ρVg
If an object is put into a container of liquid
it will float if ρobject is less than ρliquid
It will sink if ρobject is greater than ρliquid
Example: wood in water vs. rock in water
Week 8
Physics 214
7
Density and pressure
In a container of liquid the pressure at any
depth is the weight of the column of water
above that depth divided by the horizontal
area of the column. (Relative to the
surface, where the pressure is typically
one atmosphere.)
g
Mass/unit volume ρ in units kg/m3.
Other times we use grams/cubic
centimeter
1 gram/cm3 = 1000kg/m3
ρwater is 1 gram/cc or
1 Ton/m3 , where
we mean Metric Ton (= 103 kg)
Column weighs W = mg = ρVg = ρAhg
Pressure
P = W/A = ρgh.
P increases with depth.
At a given depth, pressure is uniform
(and omnidirectional.)
Start with atmospheric pressure at the top
surface of the liquid, then add with depth.
Week 8
Physics 214
8
Effects of Atmospheric Pressure
In our everyday lives we do not feel that
we are supporting a very large weight
No Air,
because it is the environment in which we
Vacuum,
developed. Also the pressure is almost the
same inside our lungs, and inside our ears, as it is P=0
outside us.
If we fill a tube with a liquid so that all the air bubbles
are removed and then we put the tube in a vertical
position we find that atmospheric pressure supports
76cm of mercury or 10.13m ≈ 33 feet of water
Knowing the weight of the Mercury (or water) and the
area of the tube we find that
1 Atmosphere = 1.013 105 Pa = 101.3 kPa
This is equal to ρgh for the Mercury, so if we know h,
and ρHg=13.534 gm/cc (and g), we can calculate Patm
Week 8
Physics 214
g
P is the same
9
Everyday examples
™ Suction cups – remove the air and the atmosphere
holds it in place.
™ Drinking through a straw – create a partial vacuum
in your mouth and the atmospheric pressure
pushes the fluid up the straw
™ Impossible to suck water higher than 33 ft, can’t
have less than a vacuum at the top (except in tree
sap channels, where water is IN TENSION –
effectively a negative pressure)
™ Air pressure is lower at higher altitudes – water
boils at a lower temperature
™ Pressure is higher the deeper you go in the ocean –
leads to more nitrogen being absorbed in blood,
and the bends. Also raises boiling point of water
(hot vents at midocean ridges, weird critters there)
Week 8
Physics 214
10
Suction cup
Question: What is the minimum area of a
suction cup that can be used to lift 100kg = 220
lbs.
The suction cup will stay attached to the block
providing PA is greater than F/A. In practice
there would only be a partial vacuum under the
cup so the cup would break away from the
block at a lower force and pressure.
1.013x105 N/m2 x A = W = 100kgx9.8m/s2 = 980N
A = 980/101,300(N/m2)/N = .00967 m2 or 96.7cm2
This would be a circle of diameter just over 4
inches
Week 8
Physics 214
F
PA
mg
11
Archimedes Principle
If an object is lowered into a liquid it occupies a
volume. That volume was previously some of the
liquid, and it was sitting still in equilbrium.
That volume of liquid was being supported by an
upward force that exactly balanced its weight.
So the object will feel an upward (buoyant) force
equal to the weight of liquid displaced.
buoyant force = weight of liquid displaced
T
Fb
g
mg
This is true for objects that are immersed, and for
objects that float. A boat made of steel can float
because it can displace a volume of liquid greater
than it’s own weight. A large volume of the boat is air
so the average density is less than that of water.
For an immersed object supported by string Tension
T + FB = W = mg
Week 8
Physics 214
12
Floating and buoyant force
If a floating object has a flat bottom at a depth h then the
EXTRA pressure on the bottom is ρgh [relative to the air
pressure at the surface] and the NET upward force is
ρghA but hA is the volume of liquid displaced, V, and
ρgV is the weight of liquid displaced so the
upward force = weight of liquid displaced
If an object has a density larger than the liquid it will sink.
Suppose the top is at a depth of dt and the bottom at db.
Then there is downward pressure on the top ρgdt and an
upward pressure on the bottom ρgdb so the net buoyant
force, upward of course, is
g
Fb
T
Fb = ρg(db – dt)A
= ρgV = weight of liquid displaced.
T + Fb = weight of the object = mg
Result is more general than for flat tops and
bottoms.
Week 8
Physics 214
g
Fb
13
Floating and buoyant force
Water ice is LESS dense than liquid water
(by about 10%). Water is a very ususual
substance.
Upward force = weight of liquid displaced
So ice floats in water with about 10% of its
volume above the water level.
If the red “ice cube” melts, THE WATER
LEVEL WILL NOT CHANGE.
If floating sea ice melts (like the Arctic) the
sea level will not change.
If ice ON LAND melts, (Greenland, West
Antarctic) sea level WILL rise. ~20 feet
each!!
There’s enough ice on land right now to
raise sea level more than 250 feet.
It wouldn’t even have to melt, if it slid off
into the sea and floated!! (Greenland??
***** Break for web pages *****
Week 8
Physics 214
g
Fb
14
Ch 9 CP 4
Wooden boat: 3m x 1.5m x 1m that carries five people.
Total mass of boat and people equals 1200 kg.
a) What is total weight?
b) What is buoyant force required to float?
c) What volume of water must be displaced to float?
d) How much of the boat underwater?
a) W = Mg = 1200 kg (9.8 m/s2)
W = 11760 N
1 .5
b) Fnet = Fb – W = 0
Fb = 11760 N
1m
m
3m
W
c) Fb = ρH2O Vg (see Ch 9 E 12)
Fb/ρH2Og = 11760N/(1000 kg/m3)(9.8 m/s2) = V = 1.2 m3
Easier way: boat displaces 1.2 T, water has m/V = 1T/m3
d) V = LWh = (3m)(1.5m)h = 1.2 m3 h = 0.27 m
Week 8
Physics 214
Fb
15
Ch 9 E 12
Boat displaces 2.5 m3 of water.
Density of water ρH2O = 1000 kg/m3.
a) What is the mass of water displaced?
b) What is the buoyant force?
A. 24500N
D. 9800N
B. 2500N
E. 24.5N
C. 1000N
a) Mass of fluid displaced
(mFD) = volume x density of fluid.
MFD = VFDρH2O = (2.5 m3)(1000 kg/m3) = 2500 kg
Fb = WFD
b) Buoyant force equals weight of fluid displaced.
Fb = WFD = mFD g = (2500 kg)(9.8 m/s2) = 24500 N
Week 8
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Ch 9 CP 2
Water density = ρH2O = 1000 kg/m3. Depth of swimming pool = 3m.
a) What is the volume of a column of water 3m deep and cross
sectional area 0.5 m2?
b) What is its mass?
c) What is its weight?
d) What is the excess pressure exerted on the pool bottom?
e) Compare to atmospheric pressure.
a) V = Ad = (0.5 m2)(3m) = 1.5 m3
b) M = Vρ = (1.5 m3)(1000 kg/m3) = 1500 kg
c) W = Mg = (1500 kg)(9.8 m/s2) = 14700 N
0.5m2
3m
d) P = F/A = 14700N/0.5m2 = 29400 Pa
e) Atmospheric Pressure is about 100 kPa
P is about 30 kPa
P/Atm = (29400 Pa)/(1.013 x 105 Pa) = 0.29
Week 8
Physics 214
17
Pressure and Volume of a Gas
In the apparatus shown the pressure at
point A is the same as at point B.
First example: volume V1 is at
atmospheric pressure.
Second example, more mercury is
added. The extra height of mercury
adds ρgh to the pressure at (new) point
A. The new volume, V2, is less. It feels
a total pressure of 1 Atmosphere + ρgh
g
h
A
A
B
B
If the temperature does not change we
find experimentally
PV = constant
Double the pressure, the volume
shrinks to half as much, for example.
Week 8
Physics 214
18
Fluids in motion
Liquid flow in a pipe: speed x cross
sectional area is constant due to
“conservation of volume”
(incompressibility)
A1v1 = A2v2
Volume flow, per unit time, doesn’t
change, so the narrow channel has
faster flow than the wide part.
A1
v1
v2
Streamlined gas flow also tends to
conserve volume, so the same eqn.
A2
Week 8
19
Physics 214
Fluids in motion
Fluid flow – Bernoulli’s Principle
PV (Pressure x Volume) has units of
energy, and represents Potential
Energy stored in a given volume of
fluid.
Bernoulli realized that if the flow is
streamlined, and we can neglect
drag, the MechEnergy of the fluid,
which is KE + PE, is conserved.
Faster liquid has greater KE,
therefore it MUST have lower PE and
hence, lower Pressure. DEMOS
(F/A)V : [N/m2]x m3 = [Nm] = [Joules]
Week 8
Physics 214
A1
v1
v2
A2
20
Fluids in motion
Bernoulli’s principle has many applications.
It can generate “lift” on a wing if the curved wing forces the air to flow
faster over the top surface than over the bottom surface.
It can cause water to be drawn up into an open tube, and then atomized
by the air flow (paint sprayer).
It can measure speed, via a “Pitot tube” exposed to the fluid-stream
around an aircraft (or boat) – the reduced pressure is taken via a tube to
be compared with still air pressure.
Air flow speed INSIDE a tube can similarly be measured.
Tap water flow past an opening can pull a partial vacuum.
Demo of air flow crosswise on a rotating cylinder (similar to baseball
phenomenon of the “curveball” – which happens when the ball SPINS.
Week 8
Physics 214
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Ch 9 E 16
Wing has a horizontal area A = 10 m2. Wing
experiences Lift = 60000 N.
What is the difference in air pressure btw. top and
bottom of wing?
Pt
g (downward)
A = 10m2
Pb
60000 N
A. 1.01x105Pa
B. 6x105Pa
D. 6x103 Pa
C. 6x104 Pa
E. 6x102 Pa
a) P = F/A , Pb – PA = F/A = 60000N/10m2 = 6000Pa
Week 8
Physics 214
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22
Q11 If you filled an airtight balloon at the top of a mountain,
would the balloon expand or contract as you descend the
mountain?
It would contract because the atmospheric pressure would
increase and the pressure inside the balloon would increase to
balance this change
Q12 When you go over a mountain pass in an automobile,
your ears often “pop” both on the way up and on the way
down. How can you explain this effect?
As you go up or down the atmospheric pressure changes and the
popping is the inner ear adjusting to the pressure change – air flows
out or in via the eustacean tubes (if they aren’t stopped up.)
Week 8
Physics 214
23
Questions Chapter 9
Q1 Is it possible for a 100-lb woman to exert a greater pressure on
the ground than a 250-lb man? Explain.
Yes. The pressure will be mg/A so if A is small e.g. narrow
heels, the pressure will be very large
A knife cuts well for similar reasons, the pressure under the sharp
blade is very high since the edge is exceedingly thin and has small
contact area.
Q3 The same force is applied to two cylinders that contain air.
One has a piston with a large area, and the other has a piston with
a small area. In which cylinder will the pressure be greater?
The pressure is F/A so the one with the smallest A
Week 8
Physics 214
24
Q4 A penny and a quarter are embedded in the concrete bottom of
a swimming pool filled with water. Which of these coins
experiences the greater downward force due to water pressure
acting on it?
Each coin has to support the weight of water in a vertical column
so the quarter has the bigger force F =PA
Q5 Why are bicycle tires often inflated to a higher pressure
than automobile tires, even though the automobile tires must
support a much larger weight?
Once again the upward force has to support the weight so
F = mg but F = PA where A is the area of the tire on the road. So if
the bike tire has a relatively very small area compared to the tire
area on the road, compared to the weight ratio, then
Pbike would need to be higher. Say the contact area ratio is 1% vs
weight ratio (bike+rider)/car is 2% then the bike tire needs twice
the pressure.
Week 8
Physics 214
25
Q6 The fluid in a hydraulic system pushes against two pistons, one
with a large area and the other with a small area.
A. Which piston experiences the greater force due to fluid pressure
acting on it?
B. When the smaller piston moves, does the larger piston move
through the same distance, a greater distance, or a smaller distance
than the smaller piston?
A. The pressure is the same and F = PA so the larger piston has
the larger force.
B. The work done is the same so the small piston moves the most.
Equivalently, to “conserve liquid volume” the smaller piston
moves the larger distance.
Q8 When a mercury barometer is used to measure atmospheric
pressure, does the closed end of the tube above the mercury
column usually contain air?
No it needs to be a vacuum
Week 8
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26
Q9 Could we use water instead of mercury to make a barometer?
What advantages and disadvantages would be associated with the
use of water?
The height of the liquid depends on the density. So one can use
water but the column would be 33 feet high, non toxic, but more
easily evaporates than Hg (but Hg DOES have a vapor pressure!
Q10 If you climbed a mountain carrying a mercury barometer,
would the level of the mercury column in the glass tube of the
barometer increase or decrease (compared to the mercury
reservoir) as you climb the mountain?
The pressure decreases because there is less mass of air above
you. So the height of the Mercury column would decrease
Week 8
Physics 214
27
Q15 Is it possible for a solid metal ball to float in mercury?
The upward force is the weight of liquid displaced and the
downward force is the weight of the ball. If the density of the
liquid is greater than that of the ball it will float. Steel or copper
or lead will float, gold will sink (and amalgamate with the
mercury, but that’s another story.)
Q16 A rectangular metal block is suspended by a string in a
breaker of water so that the block is completely surrounded by
water. Is the water pressure at the bottom of the block equal to,
greater than, or less than the water pressure at the top of the
block?
The pressure is ρgh so the pressure is higher at the bottom. The
difference in pressure provides the upward force on the block
Week 8
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Q19 A large bird lands on a rowboat that is floating in a swimming
pool. Will the water level in the pool increase, decrease, or remain
the same when the bird lands on the boat?
The buoyant force is the weight of liquid displaced so to
support a larger weight more liquid is displaced and the level
rises
Q20 A rowboat is floating in a swimming pool when the anchor
is dropped over the side. When the anchor is dropped, will the
water level in the swimming pool increase, decrease, or remain
the same?
When the anchor is in the boat it’s whole weight is supported and
the amount of water displaced balances that weight. When it is
thrown overboard it sinks and only displaces it’s volume so the
water level falls
Week 8
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Q22 If an object has the same density as water, will the object float
to the top, sink to the bottom, or take neither course?
Providing the object and water are incompressible the object
will stay at whatever depth it is placed. It will not sink or rise.
Week 8
Physics 214
30
Ch 9 E 4
Pressure of gas in piston = 300 N/m2.
Area of Piston = 0.2m2.
What is force exerted by piston on gas?
A = 0.2m2
p = 300 N/m2
P = F/A, F = PA = 300 N/m2 (0.2 m2) = 60 N
Week 8
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Ch 9 E 8
Temp. = constant, Amount of gas is constant
P1 = 10 kPa, V1 = 0.6 m3.
P2 = 90 kPa, V2 = ?
P1 = 10 kPa
V1 = 0.6 m3
P2 = 90 kPa
V2 = ?
diagram is
not to scale
A. 0.067 m3
B. 0.111 m3
C. 9.0 m3
D. 5.4 m3
E. 6.7 m3
P1V1 = P2V2 ; V2 =(P1/P2)V1 =(10/90)(0.6 m3) = 0.067 m3
.
Week 8
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Ch 9 E 14
Stream moves at v1 = 0.5 m/s in cross sectional area A1.
Stream reaches point where A2 = ¼ A1.
What is v2?
V2
V1
A2
A1
v1A1 = v2A2
but 4A2 = A1
(0.5 m/s)(4A2) = v2(A2)
v2 = 2 m/s
Week 8
Physics 214
33
Balloons
FB
mg
Week 8
Any object in the atmosphere is subject
to a buoyant force and Archimedes law
applies so if the buoyant force is greater
than the weight of an object it will rise.
So since the material of a balloon has a
density greater than air then the balloon
must be filled with a gas having a
density less than air. In practice balloons
either use Helium or hot air. Hydrogen
works too, but is flammable and
explosive.
As the balloon rises the buoyant force
decreases and the balloon will float at
constant altitude when the buoyant force
is equal to the weight.
Physics 214
34
Surface tension
The molecules and atoms in a liquid are
continually in motion so that a molecule at
the surface can escape and this is
evaporation. However a molecule at the
surface feels an attractive force pulling it
back into the liquid and this is surface
tension. This is the reason that one can
form bubbles and water drops
The capiilary action is caused by “wetting”
forces between the edge of the liquid
surface and the walls of the tube.
Week 8
Physics 214
35
Surface tension
The pressure inside a bubble is bigger for
smaller bubbles – this extra pressure is
proportional to the sharpness of curvature
of the surrounding surface.
Which means proportional to 1/Radius of
curvature.
So big bubbles are very soft and floppy,
while tiny bubbles are very round and
springy.
Week 8
Physics 214
36
Summary of Chapter 9
P = F/A Pascals or N/m2
P = 1 Atm + ρgh
F1/A1 = F2/A2
Work done = F1h1 = F2h2
1 Atmosphere = 1.013 105 Pa
= 101.3 kPa (kiloPascals)
and will support
76cm of mercury
33 feet of water
Week 8
No Air
g
Physics 214
37
Summary: Liquids
P = F/A = ρgh + pressure at top of liquid
Water density is 1 gram/cc
1 gram/cm3 = 1000kg/m3 = 1 Ton/m3
T
buoyant force = the weight of liquid displaced
Fb
T + FB = W = mg
T = external support force
For a floating object T = 0
Week 8
g
mg
Physics 214
38
Summary Fluids in motion
Liquid flow in a pipe: speed x cross
sectional area is constant due to
“conservation of volume”
(incompressibility)
A1v1 = A2v2
A1
v1
v2
A2
Week 8
Physics 214
39
Summary: Bernoulli’s Principle
WHEN A FLUID SPEEDS UP, ITS
PRESSURE DROPS.
A1
P
v1
Faster-flowing liquid has greater KE,
therefore it MUST have lower PE and
hence, lower Pressure.
PV = (F/A)V
[N/m2]x m3 = [Nm] = [Joules]
v2
p
This is true not only for flow in
tubes, but for wings, sails, boat
keels, etc. (these “free” objects can
experience “lift”)
A2
Week 8
40
Physics 214
2A-01 Suction Cups
How does a
suction cup work ?
How does a suction cup ‘hold on’ to objects?
PA
Holding the suction cup by
itself I only have to support the
weight since the force due to
atmospheric pressure acts on
the top and bottom of the cup.
If I place it on a surface and
exclude all the air the cup is
held to the surface by a force
due to atmospheric pressure
of 1.013x105 Pascals per
square meter F = PAA
Remember atmospheric pressure can support
33 feet of water so the force on 1 square foot
is ~ 2000 lbs!
Week 8
Physics 214
PA
F
F
If two cups are
pushed together,
does this make it
twice as difficult
to separate ?
41
2B-08 Buoyant Force
Compare the Buoyant Force between two immersed
cylinders of equal volume and different mass.
T
Which object
experiences the
greater buoyant
force, the
heavier one or
lighter one ?
T
FB
Mg
Scale reads
tension in cord:
T = Mg – FB
Mg = FB + T
Mg
We find Fb to be exactly equal
for both masses
BUOYANT FORCE DOES NOT DEPEND ON THE MATERIAL OF
THE OBJECT DISPLACING THE FLUID. THE BUOYANT FORCE
DEPENDS ONLY ON THE VOLUME OF FLUID DISPLACED.
Week 8
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42
Otto von Guericke 1602 - 1686 was the “inventor” of
the nothing we now call a vacuum.
Von Guericke created a
vacuum by attaching
two hemispheres and
then evacuating the air
from the resultant
sphere. Von Guericke
demonstrated the force
of the vacuum before
the German emperor
Ferdinand III by having
two teams of horses
attempt to disengage
the hemispheres.
Week 8
Physics 214
43
2A-03 Vacuum Demos
Effects of Vacuum on objects made largely of air or air pockets.
Why do the
balloons burst in
the vacuum ?
Why do the
marshmallows
get bigger in
vacuum ?
Do the balloons
burst in vacuum
differently then they
normally burst ?
What will happen when
the marshmallows are
returned to normal
pressure ?
Can you guess what happens when
Shaving Cream is placed in vacuum ?
AN AIR-POCKET/BALLOON WILL EXPAND WHEN THE PRESSURE IS REDUCED
AND IT WILL DEFLATE WHEN THE PRESSURE IS INCREASED. SO BALLOONS
WILL EXPAND AS THEY RISE IN THE ATMOSHPERE AND THE EXPANSION OF
A PARTIALLY EVACUATED CAN IS USED IN BAROMETERS.
Week 8
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44
2B-05 Pressure Forces in Liquids
An open ended cylinder kept shut by liquid pressure
What happens as
the submerged
cylinder filled with
air is filled with
water ?
There are two forces acting
on the plate. It’s weight
down and PA up. When PA
exceeds the weight the
cylinder stays intact
In this situation the plate
has to now support the
weight of the water and
when the weight of the
water plus plate exceeds PA
the cylinder opens
Air
PA
Water
PA
THE LIQUID PRESSURE DEPENDS ONLY ON DEPTH P = ρgh.
THE UPWARD FORCE DEPENDS ON THE AREA
F = PA
Week 8
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45
2B-04 Liquid Pressure
Investigating Pressure in different directions within a liquid in equilibrium.
What will happen
to the reading on
the manometer as
the sensor is
rotated ?
PRESSURE IS NOT A VECTOR. IT ACTS
EQUALLY IN ALL DIRECTIONS
g
h
A
B
The increase in pressure ρgh is
measured by the difference in
height of the liquid in the U tube.
A
B
IF A LIQUID IS IN EQUILIBRIUM, THE
FORCES ACTING AT A POINT CANNOT
HAVE A PREFERENTIAL DIRECTION
OR THE LIQUID WOULD MOVE.
AT ANY GIVEN POINT IN A STATIONARY LIQUID, THE PRESSURE IS
THE SAME IN ALL DIRECTIONS.
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Physics 214
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2B-09 Archimedes I
T
What happens to
the reading on the
upper scale when
the block is
lowered into the
beaker of water ?
FB
Mg
Should anything
happen to the
reading on the
lower scale ?
The fluid exerts a
buoyant force on the
block, which reduces
the tension on the
cord. The reading on
the scale is lowered.
T = Mg – FB
Since the fluid exerts a force on the
block, the block exerts an EQUAL
and OPPOSITE force on the fluid.
EVEN THOUGH THE BLOCK DOES NOT ‘TOUCH’ THE LOWER SCALE, THE FORCE ON THE
FLUID DUE TO THE BLOCK IS TRANSMITTED TO THE SCALE. THE REDUCTION IN READING ON
THE UPPER SCALE IS EXACTLY EQUAL TO THE INCREASE IN READING ON THE LOWER
SCALE. IF THE CONTAINER WAS FULL SO THAT WHEN THE BLOCK WAS INSERTED THE
VOLUME THE BLOCK DISPLACED SPILLS OUT OF THE CONTAINER THEN THE BOTTOM SCALE
WOULD NOT CHANGE.
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2B-10 Archimedes II
Use a scale to establish the relationship between the Buoyant
Force on an object and the Weight of Fluid Displaced by the object
T1
FB
B
A
Mg
T3
T2
C
Mg
FB
WDF
Mg
A. The block is not immersed
T1 = Mg
B. The block is immersed but the liquid runs out
T2 = Mg – FB
C. The displaced liquid is poured into the can
T3 = Mg – FB + WDF
T1 is found to equal T3 which means that the bouyant force FB is
equal to WDF the Weight of the displaced Fluid
THE BUOYANT FORCE IS EQUAL TO THE WEIGHT OF THE FLUID DISPLACED.
Week 8
Physics 214
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