Transcript Slide 1

This Week
Liquids and Gases
Pressure
How do they lift your car for service?
Atmospheric pressure
We’re submerged!
How can you drink a Coke?
Archimedes ! Eureka!!!
Balloons of all sizes
Bubbles of all sizes
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Liquids and Gases
The unit of volume is
the meter cubed , m3,
which is a very large
volume. Very often we use
cm3 = cc. Other everyday
units are gallons, quarts,
pints
As we know liquids and gases act very differently than solids.
Liquids and gases have mass but their constituent atoms are not bound
so that 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.
A volume of gas can have it’s volume changed fairly easily. Both have the
property of being able to flow, for example water and gas lines in a house.
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Pressure
A volume of liquid or gas has mass and F = ma is still a good law
except a force at a point on the surface of water only moves the
water near that point
So generally we work with liquids and gases in containers
and exert forces over a surface.
We define pressure as
P = F/A
that is the force divided by the area over
which the force acts.
Any change in pressure is transmitted uniformly
throughout the liquid.
Units are N/m2 and 1 N/m2 = 1Pascal
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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 50lbs can lift 5000lbs.
Of course if we push F1 down a distance h1 then the F2
Side will only rise h1 A1/A2 because h1A1 = h2A2, that is the
displaced volumes must be the same.
Work done = F1h1 = F2h2
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Atmospheric pressure
If I 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 as does the pressure.
F
g
In the example shown the force on the face
of the bottom brick would be 6mg and the
pressure = 6mg/A.
At the earths surface we are supporting
a column of air which exerts a force and
because it is a gas it exerts an equal pressure
In all directions.
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Density
If one takes two objects of exactly the same volume made of
different materials they have different weights. So we define
a useful quantity
Density ρ = mass/unit volume, kg/m3 or grams/cc so the
mass of an object is ρV and the weight 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
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Density and pressure
g
If I take a container of liquid the pressure at
any depth is the weight of the water above
that depth divided by the area.
It is very useful to define
Mass/unit volume ρ = kg/m3.
Very often we use grams/cubic centimeter
1 gram/cm3 = 1000kg/m3
Water is 1 gram/cc
W = mg = ρAhg
Pressure P = W/A = ρgh and the pressure is
the same at all depths increasing with depth.
This is the pressure increase from the surface
which is also subject to atmospheric pressure
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Effects of Atmospheric Pressure
In our everyday lives we do not feel
we are supporting a very large weight
because it is the environment in which we
developed.
No Air
If we immerse a tube in a liquid
So that all the air bubbles are removed and then
we raise the tube to a vertical position we find g
that atmospheric pressure supports 76cm
of mercury or 32 feet of water.
Knowing the weight of the Mercury and the
area of the tube we find that
1 Atmosphere = 1.013 105 Pa
P is the same
This is equal to ρgh for the Mercury
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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
 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 and the bends.
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Suction cup
Question: What is the minimum area of a
suction cup that can be used to lift 100kg.
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.
F
PA
mg
1.013 x 105 x A = 100 x 9.8
A = 9.67/1000m2
Which would be a circle of radius just over 2
inches
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Archimedes Principle
If an object is lowered into a liquid
the volume it occupies was being supported by an
upward force that exactly balanced the weight of
the same volume of liquid so the object will feel the
same upward buoyant force.
T
Fb
g
buoyant force = weight of liquid displaced
this is true for objects that are immersed and for
objects that float. So 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.
mg
T + B = W = mg
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Floating and buoyant force
If a floating object is submerged to a depth h
then the pressure on the bottom is ρgh and the
upward force is ρghA but hA is the volume of
liquid displaced, V, and ρgV is the weight of
liquid displaced so the
Fb = weight of liquid displaced = weight of object
ρliquid x Vdisplaced x g = ρobject x Vobject x g
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
g
Fb
T
g
Fb = ρg(db – dt)A
= ρgV = weight of liquid displaced.
T + Fb = weight of the object = mg
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Pressure and Volume of a Gas
In the apparatus shown the pressure at
point A is the same as at point B so the
pressure exerted by the gas is equal to
ρgh.
If we add more mercury we can
measure both pressure and volume
and if the temperature does not change
we find
PV = constant
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g
h
A
A
B
B
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Summary of Chapter 9
P = F/A Pascals
F1/A1 = F2/A2
Work done = F1h1 = F2h2
1 Atmosphere = 1.013 105 Pa
and will support
76cm of mercury
32 feet of water
No Air
g
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Liquids
P = W/A = ρgh
Water is 1 gram/cc
1 gram/cm3 = 1000kg/m3
T
buoyant force = the weight of liquid displaced
Fb
T + FB = W = mg
g
For a floating object T = 0
mg
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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
32 feet of water so the force on 1 square foot
is ~ 2000 lbs!
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PA
F
F
If two cups are
pushed together,
does this make
it twice as
difficult to
separate ? 16
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.
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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.
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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
A
B
The increase in pressure ρgh is
measured by the difference in
height of the liquid in the U tube.
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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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
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2B-08 Buoyant Force
Compare the Buoyant Force between two
cylinders of equal volume and different mass.
T
T
Which object
experiences the
greater buoyant
force, the
heavier one or
lighter one ?
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.
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2B-09 Archimedes I
T
FB
What happens to
the reading on
the upper scale
when the block is
lowered into the
beaker of water ?
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
T3
T2
FB
B
A
Mg
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.
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2A-08 Buoyancy of Air
Investigating the Buoyant force resulting from Air
Does the air
exert a buoyant
force ?
ρairgVa
ρairgVb
mag
mbg
Setting the sum of torques on equal-arm balance
about pivot equal to zero, we have in the presence
of air:
mag – ρairgVa = ρairgVb – mag
Vb > Va implies mb > ma which is demonstrated in
vacuum
IF SENSITIVE WEIGHING OF AN OBJECT IS REQUIRED,
UNEQUAL BUOYANT FORCES COULD AFFECT THE RESULTS.
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2B-02 Pascal's Vases
Examining the Pressure at the bottom of
differently shaped vessels filled with fluid
If the different
vessels were filled
to the same height,
how does the
pressure differ at
each vessel base ?
P = ρgh
THE PRESSURE IS DEPENDENT ONLY ON THE HEIGHT h
AND NOT THE VOLUME OR SHAPE OF THE VESSEL.
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2B-03 Water Seeks Own Level
Liquid pressure
depends on the
“height” of the
liquid column.
But how is this
height measured ?
Investigating the Dependence of Pressure on Height
The slanted cylinder
and twisted cylinder
hold a longer “total
length” of water. But in
each case the vertical
height is the same.
P = ρgh
LIQUID PRESSURE DEPENDS ONLY ON VERTICAL HEIGHT
(MEASURED STRAIGHT DOWN THAT IS PARALLEL TO g).
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2D-01 Hydraulic Press
Is it possible to
crush a 2x4 only
with the force
exerted by one
hand ?
Using the uniformity of pressure within a
liquid as a mechanical advantage
Within the Hydraulic fluid the pressure is uniform:
F1/A1 = F2/A2  F1/F2 = A1/A2
Pump piston diameter = 0.5 in
Lift piston diameter = 1.25 in
F1= 6.25*F2
This is not enough mechanical advantage to
crush the wood. How is it done?
Remember, the lever arm also gives a mechanical
advantage. We have: F1≈ (16)*6.25*F2 = 100*F2
THE INPUT FORCE IS GREATLY ENHANCED BY THE HYDRAULIC
FLUID AND BY THE LEVERAGE GAINED USING THE LONG
HANDLE.
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2D-03 Pascal's Principle (Wine Bag)
Creating a Pressure to exert a Force
Is it possible to
lift a heavy
object using
only your
breath ?
P = F/A  F = PA
A pressure applied over an area
can generate a large force.
BY CREATING A PRESSURE WITHIN A VESSEL, A PRESSURE IS
EXERTED ON ALL THE WALLS OF THE VESSEL. THIS MEANS A
FORCE = PA WHICH IS LARGE ENOUGH TO LIFT A PERSON
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2A-05 Pressure Force Milk Jug
Effect of atmospheric pressure on a partially evacuated milk bottle
What will happen
to the container
if it is sealed
after putting in a
small amount of
very hot water?
The hot liquid heats the air
which expands and some
leaves the container. The
hot liquid also causes
vapor to form in the jug
which then cools and
condenses.
Liquid takes up a much smaller volume than steam. As the
steam cools and condenses a partial vacuum forms and the
pressure drops. Atmospheric pressure then crushes it.
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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. small
heels the pressure will be very large
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
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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 biggest 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
M/m = (PA)car/(Pa)bike so Ma/mA = Pcar /Pbike the ratio of the areas is
smaller than the ratio of the weights so
Pbike is higher.
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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
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
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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 32 feet high
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 you have a smaller column of
air to support so the height would decrease
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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
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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.
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
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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
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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.
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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
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Ch 9 E 6
Hydraulic system: A2 = 50 A1
F2 = 6000 N. What is F1?
F2 = 6000N.
A1
A2
Pressure is the same just underneath each piston
F1/A1 = F2/A2
F2/F1 = A2/A1 = 50A1/A1 = 50
F1 = F2/50 = 6000/50 = 120N
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Ch 9 E 8
T = constant, P1 = 10 kPa, V1 = 0.6 m3.
P2 = 90 kPa, V2 = ?
P1 = 10 kPa
V1 = 0.6 m3
P2 = 90 kPa
V2 = ?
P1V1 = P2V2
(10 kPa)(0.6 m3)/(90 kPa) = V2 = 1/15 = 0.067 m3
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Ch 9 E 12
Boat displaces 2.5 m3 of water.
Density of water H2O = 1000 kg/m3.
a) What is the mass of water displaced?
b) What is the buoyant force?
a) Mass of fluid displaced
(mFD) = volume x density of fluid.
MFD = VFDH2O = (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
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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
a) v1A1 = v2A2
(0.5 m/s)(A1) = v2(¼A1)
v2 = 2 m/s
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Ch 9 E 16
Wing has a cross sectional area A = 10 m2. Wing
experiences Lift = 60000 N.
What is the difference in air pressure b/w top and
bottom of wing?
Pt
A = 10m2
Pb
60000 N
a) P = F/A , Pb – PA = F/A = 60000N/10m2 = 6000Pa
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Ch 9 CP 2
Water density = H2O = 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 = V = (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
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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
1m
3m
b) Fnet = Fb – W = 0
Fb = 11760 N
W
c) Fb = H2O Vg (see Ch 9 E 12)
Fb/H2Og = 11760N/(1000 kg/m3)(9.8 m/s2) = V = 1.2 m3
d) V = LWh = (3m)(1.5m)h = 1.2 m3
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Fb
h = 0.27 m
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Balloons
FB
mg
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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.
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.
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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
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