1. Chapter 13 Fluids (Liquids and Gases)
What is a fluid
Density, pressure, of gases and liquids
Static fluids p =p0 +rgh
Mercury barometer,open tube manometer
Pascal’s principle
Archimedes principle
Surface tension and adhesive forces
Fluids in motion
Equation of continuity
Bernoulli’s Equation
about soap bubbbles

2. Fluids: liquids and gases Fluids conform to the container that holds them and we normally talk about their density and pressure
Liquids are Incompressible
Gases are very compressible
Uniform Density
ρ=m/v kg/m3
Pressure (Static)
P=F/A N/m2 or Pa
Scalar- independent of orientation
Atmospheric pressure
1atm 1.01 x 105 N/m2
760 mm of Hg or Torr
14.7 lb/in2

3. What are all the forces acting on this book? Put a book on the table.

4. What is the force acting downward at level 1? at level 2 ?
Static Pressure
What is the force acting downward at level 1?
at level 2 ?
Level 1 F = Weight of Atmosphere
Level 2 F = Weight of Atmosphere + Weight of water
The pressure is just the weight of the water plus the weight of the air per square meter

5. What is the absolute and gauge pressure at a 2 meter depth in the water? First find the atmospheric pressure P0.
Or N/m2
(Gauge Pressure)

6. Atmospheric Pressure Bands represent the volume of the
Less
Dense
Bands represent the volume of the
atmosphere that contains 1000 kg
of air as you go up in altitude.
The density of air gets thinner and
thinner as you go up in altitude.
Density = 1.25 kg/m3 at sea level
Pressure of atmosphere at sea level
= 100,000 N/m2 (Pascal or Pa) or
= 14.7 lbs/in2
= bs/ft2
= 1 ton/ft2

7. What is a Mercury Barometer?
This is the upward acting force due to the pressure at the bottom of the tube.
This will support an amount of mercury equal to the weight
So or
Solving for h,
We get or
How high would a column of water go?

8. Pascal Vases
Shows that the pressure in a liquid depends on the height of the liquid, not it's volume

9. Pascal’s principle: A change in pressure applied to an enclosed incompressible fluid is transmitted undiminished to every portion of the fluid and to the walls of its container
Change in pressure =
Vf=VF
Work done d D
SHOW Pascal vases DEMO
The work done by both systems is the same.
The advantage is that a given force over a long distance can be transformed
to a larger force over a shorter distance.

10. Archimedes principle
The stone displaces a volume of water equal to its volume.
The upwardly directed buoyant force is equal to the weight
of the displaced fluid or water. This is the surprising thing.
Floating: If the buoyant force is equal to the weight mg
of the object, Then it is said to float.
If it sinks then you can define its apparent weight.
apparent weight = actual weight - buoyant force
Buoyancy of air
Weight of air
SHOW Archimedes principle I
Archimedes Demo
Example 26 chpt 14 ed 7

11. Archimedes principle
To show that the buoyant force exerted on an object is equal to the weight of the displaced water (or fluid).

12. Weight of Thumb Stick thumb in the water and weigh it.
Remove thumb and stick it in the water and weigh it again .
Can you account for the readings.

13. Archimedes Principle Yes The difference is the buoyant force,
which is the weight of the displaced fluid.
In this case it is the weight of water whose volume is the same as your thumb.
Archimedes Principle

14. Example
A .1 m x .3 m x 0.5 m open box is floating in a pool of water. The box sinks 0.2 m into the water before floating. How big is the buoyant force?
Buoyant force = Density of water x Volume of displaced water x g
Density = 1000 kg/m3
Volume = 0.1 m x .3m x .2m = m3
g = 9.81 m/s2
= 1000 x x 9.81 = 58.9N
Weight
Buoyant Force
0.2
What is the boxes weight?

15. If you push the box all the way in the water, what is the buoyant force now?
Buoyant force = Density of water x Volume of displaced water x g
Density = 1000 kg/m3
Volume = 0.1 m x 0.3 m x 0.5m = m3
g = 9.81 m/s2
= 1000 x x 9.81 =147.2 N
Buoyant Force
0.5
Weight
What happens if you let it go? Net F= = 88.3 N

16. Chapter 14 Ed 7 Problem 26E
Problem 36 Ejected ball from water
Problem 20 “Sucking” up lemonade

17. Air effects
Buoyancy of air
Weight of Air

18. Why does Helium filled Balloon move upward in air?
A rubber balloon filled with helium
Weight of Helium + rubber = W
(Weighs less than the air it displaces )
Weight of displaced air = Bouyant force = B
An upward net force in air = F
F= B - W
Balloons average density < density air
Air B He F
Demonstration: Helium Balloon in Air
Demonstration: Helium Balloon in Helium
Demonstration: Helium Soap Bubbles
Demonstration: Methane Soap Bubbles W
Net F = B - W = ma

19. Which way does a helium filled balloon move in your car when you accelerate?

20. Irrotational flow(no vortices)
Laminar Flow
Steady flow
Incompressible flow
Non-viscous flow
Irrotational flow(no vortices)
An example of viscous flow

21. Fluid flowing in a pipe high pressure low velocity
Streamline or velocity line
Low Pressure
High Velocity
low pressure
high velocity
Legend
More dense lines means
high velocity
More red means lower pressure
Lowest velocity
Highest Pressure

22. Equation of continuity:
= Volume flow rate
Mass flow rate= density x
Volume flow rate.
Show water stream
narrowing in diameter
as it falls.

23. Bernoulli’s Equation Conservation of energy in a volume of fluid
SHOW Air flow dynamics

24. Problem 45 ed 6
A water pipe having a 2.5 cm diameter carries water into the
basement of a house at a speed of 0.9 m/s and at a pressure of
170 kPa. If the pipe tapers to 1.2 cm and rises to the second floor
7.6 m above the input point, what are (a) the speed and
(b) the water pressure at the second floor.

25. Problem 45 ed 6
A water pipe having a 2.5 cm diameter carries water into the
basement of a house at a speed of 0.9 m/s and at a pressure of
170 kPa. If the pipe tapers to 1.2 cm and rises to the second floor
7.6 m above the input point, what are (a) the speed and
(b) the water pressure at the second floor.

26. Special case of Bernoulli Equation y =0
Venturi tube demo explanation using air

27. Venturi Tube and Qualitative Verification of Bernoulli's Equation
To demonstrate operation of a venturi tube and to verify variation of gas velocity with area of tube.
Show different pressures in flow tube by flowing air through system and noting pressure relationships by the level of colored liquid in the tubes.

28. Air Flow Around Different Objects
Curve ball may be demonstrated by using the thrower and Styrofoam ball. The ball curves in the direction of the leading edge spin.
Ping-pong ball may be supported on an air jet held vertical and also may be held in a funnel with an air jet.
The cardboard cylinder is wrapped with a long length of rubber band and pulled back. It is released in such a manner that it spins out from the bottom. If it is released with enough energy, it will describe a loop in the air.
The plexiglass plates can be held together with quite an impressive force by blowing air through the nozzle and bringing the plates together.

29. Water Tower 10 m 2 m p(total) =p0 (air)+ρgh(Water)
What is the pressure at the top?
What is the pressure here?
2 m

30. Water Tower p = 100,000 +rgh rgh = 1000 x 9.81 x h = 9810 x h 2 m 10 m
Top p = 100, 000 Pa (h = 0)
2 m deep
p = 100, x = 119,620 Pa
12 m deep
p = 100, x 12
= 217,720 Pa
Does not depend on shape of pipe

31. Water tank problem
What is the speed of the water through the narrow hole of area a? The water tank as large area A.
Apply Bernoulli’s equation to the top of the water level where y=h
and at the hole y= 0 where the water is ejected.
Cancel out p0
From the continuity eq.
Since A/a >>1, neglect (a/A)2
compared to 1

32. Cartesian Diver
To illustrate the transmission of pressure in a liquid.

33. What is the percentage alcoholic content in a drink that would sink the ice cubes?

34. ConcepTest 11.1 Balancing Rod
1) 1/4 kg
2) 1/2 kg
3) 1 kg
4) 2 kg
5) 4 kg
A 1 kg ball is hung at the end of a rod 1 m long. If the system balances at a point on the rod 0.25 m from the end holding the mass, what is the mass of the rod ?
1kg 1m
[CORRECT 5 ANSWER]

35. ConcepTest 11.2 Mobile ? 1) 5 kg 2) 6 kg 3) 7 kg 4) 8 kg 5) 9 kg 1 kg
A (static) mobile hangs as shown below. The rods are massless and have lengths as indicated. The mass of the ball at the bottom right is 1 kg. What is the total mass of the mobile ?
1 kg
1 m
3 m
2 m ?
[CORRECT 5 ANSWER]

36. ConcepTest 11.3a Tipping Over I
1) all
2) 1 only
3) 2 only
4) 3 only
5) 2 and 3
A box is placed on a ramp in the configurations shown below. Friction prevents it from sliding. The center of mass of the box is indicated by a blue dot in each case. In which case(s) does the box tip over ? 1 2 3

37. ConcepTest 11.3b Tipping Over II
Consider the two configurations of books shown below. Which of the following is true?
1) case 1 will tip
2) case 2 will tip
3) both will tip
4) neither will tip
1/2
1/4 1 2

38. ConcepTest 11.1 Balancing Rod
1) 1/4 kg
2) 1/2 kg
3) 1 kg
4) 2 kg
5) 4 kg
A 1 kg ball is hung at the end of a rod 1 m long. If the system balances at a point on the rod 0.25 m from the end holding the mass, what is the mass of the rod ?
The total torque about the pivot must be zero !! The CM of the rod is at its center, 0.25 m to the right of the pivot. Since this must balance the ball, which is the same distance to the left of the pivot, the masses must be the same !!
1 kg X
CM of rod
same distance
mROD = 1 kg

39. ConcepTest Mobile
1) 5 kg
2) 6 kg
3) 7 kg
4) 8 kg
5) 9 kg
A (static) mobile hangs as shown below. The rods are massless and have lengths as indicated. The mass of the ball at the bottom right is 1 kg. What is the total mass of the mobile ?
Use torques in two steps: (1) find the big mass on the bottom left (lower rod only). (2) use the entire lower rod assembly (with two masses) to find the mass on top right. Finally, add up all the masses.
1 kg
1 m
3 m
2 m ?

40. ConcepTest 11.3a Tipping Over I
1) all
2) 1 only
3) 2 only
4) 3 only
5) 2 and 3
A box is placed on a ramp in the configurations shown below. Friction prevents it from sliding. The center of mass of the box is indicated by a blue dot in each case. In which case(s) does the box tip over ?
The torque due to gravity acts like all the mass of an object is concentrated at the CM. Consider the bottom right corner of the box to be a pivot point If the box can rotate such that the CM is lowered, it will !! 1 2 3

41. ConcepTest 11.3b Tipping Over II
Consider the two configurations of books shown below. Which of the following is true?
1) case 1 will tip
2) case 2 will tip
3) both will tip
4) neither will tip
The CM of the system is midway between the CM of each book. Therefore, the CM of case #1 is not over the table, so it will tip.
1/2
1/4 1 2