1. Hydrostatics and Fluid Dynamics
Density: mass per unit volume m
Density
ρ = V
ex. Find the mass of a block of iron 10.0 cm x 5.0 cm x 7.5 cm.
The density of iron is 7.8 g/cm3.
V = l w h
= ( 10.0 cm )( 5.0 cm )( 7.5 cm )
= 375 cm3 m
m = ρ V
= ( 7.8 g/cm3 )( 375 cm3 )
ρ = V
m = g

2. For most substances, density
decreases with increased
temperature
; not water
Ice: ρ = g/cm3
As ice melts, hydrogen bonds
break, and volume decreases
Max density at 4oC: ρ = 1.00 g/cm3

3. specific gravity: the ratio of a substance's density to that of
water at 4oC
ex. Density of aluminum = 2.7 g/cm3
specific gravity
of aluminum
2.7 g/cm3 =
= 2.7
1.0 g/cm3
Aluminum is 2.7 times as dense as water

4. So specific gravity of block is 0.67
ex.:
A block of wood, 12.0 cm x 10.0 cm x 6.0 cm, is placed in water. At equilibrium, 2.0 cm of the largest face is above water. What is the specific gravity of the wood?
2 cm
What fraction of the
block is under water?
6 cm
12 cm 2 3
Convert to percent:
67%
So specific gravity of block is 0.67
Density = g/cm3
Convert to kg/m3

5. Density = g/cm3
Convert to kg/m3
1000 g = 1 kg
100 cm = 1 m
0.67 g kg
100 cm
100 cm
100 cm
cm3
1000 g m m m
= 670 kg/m3

6. Pressure:
force per unit area F
P = A
ex. Your foot has an approximate area of 300 cm2. If you have a mass
of 65.0 kg, what is the pressure on the soles of your feet? F
Wt.
m g
( 65 kg )( 9.8 m/s2 )
P = = = = A A A
300 cm2
P = N/cm2

7. Pressure in Liquids As you swim down into water,
the pressure on your body increases
The pressure is a result of the
weight of the water above you
* The pressure at equal depths within a uniform liquid is equal
So the pressure 2 meters below the surface of the Aquatic Center
pool is the same as the pressure 2 meters below the surface of
Lake Michigan
(but not the same as the pressure 2 meters below the surface of the
Pacific Ocean, because salt water is a different liquid)

8. F m g P = = A A m ρ = m = ρ V V V = A h ρ V g ρ (A h) g P = = A A
ρ =
m = ρ V V
V = A h
ρ V g
ρ (A h) g
P = = A A
ρ = density
h = depth
Pressure
in liquids
P = ρ g h

9. ex. What is the pressure at a depth of 150 m in the ocean?
The density of seawater is g/cm3
, or 1025 kg/m3
P = ρ g h
= ( 1025 kg/m3 )( 9.8 m/s2 )( 150 m )
P = x 106 N/m2
Another pressure unit:
lb/in2, or psi
1 lb/in2 = 6895 N/m2
1.51 x 106 N/m2
1 lb/in2
= lb/in2
6895 N/m2
Check this out:

10. Pressure in Gases Kinetic Theory of Gases:
- Particles in a gas are in constant motion
- Particles collide with each other and the
walls of the container
- Particles exert a force when they collide,
and cause pressure
* Pressure in gases results from collisions
At room temperature, gas molecules
undergo 109 collisions per second

11. Atmospheric Pressure We live at the bottom of an ocean of air
9 km up and you have 70% below you
(height of Mt. Everest)
38 km up and you have 99% below you
(height of “space jump”)

12. Torricelli’s Barometer
Fill tube with mercury; cap
and invert into dish of
mercury; remove cap
Weight of air pushes mercury
up the tube
At equilibrium, 76.0 cm of mercury would be above level in dish
(Pressure will vary slightly as different weather systems
move through )
Definition: 1 atmosphere = inches Hg
= 760 mm Hg
= N/m2 = Pa
= lb/in2

13. Gauge Pressure
Gauges measure difference between tire's inside pressure
(pushing outward) and atmospheric pressure (pushing inward)
Absolute pressure is sum of two:
Absolute
Pressure
PA = atmospheric pressure
PG = gauge pressure
P = PA + PG

14. Gauge Pressure
ex. Inflate a car tire to 32.0 psi. What is absolute pressure?
P = PA + PG
= psi psi
P = psi

15. ex. A scuba diver dives to a depth of 15.0 m in the ocean.
Determine the absolute pressure at that depth.
P = ρ g h
ρseawater = kg/m3
P = ( 1025 kg/m3 )( 9.8 m/s2 )( 15.0 m )
P = N/m2
Absolute Pressure = Water pressure + Atmospheric Pressure
= N/m N/m2
Absolute Pressure = x 105 N/m2

16. Pascal's Principle and Hydraulics
Pressure applied to a confined fluid increases
the pressure throughout by the same amount
Apply force to
small piston
Force exerts
pressure on
fluid
Pressure exerts
force on large
piston
Pressure is transmitted
throughout fluid

17. the small piston of the hydraulic system below.
ex. A hydraulic lift is used to lift a car. A force of 800 N is exerted on
the small piston of the hydraulic system below.
(a) Find the pressure that small piston exerts on the fluid. F2
F1 = 800 N
A1 = 100 cm2
A2 = 1500 cm2 F1
800 N
P1 = =
P1 = N/cm2 A1
100 cm2

18. the small piston of the hydraulic system below.
ex. A hydraulic lift is used to lift a car. A force of 800 N is exerted on
the small piston of the hydraulic system below.
(b) Find the force exerted on the large piston. F2
F1 = 800 N
A1 = 100 cm2
A2 = 1500 cm2
P1 = 8.00 N/cm2
P2 = 8.00 N/cm2
By Pascal’s Principle, P2 = P1 = N/cm2 F2
Then P2 =
F2 = P2 A2
= ( 8.00 N/cm2 )( 1500 cm2 ) A2
F2 = N

19. the small piston of the hydraulic system below.
ex. A hydraulic lift is used to lift a car. A force of 800 N is exerted on
the small piston of the hydraulic system below.
(c) Find the mass of the car that can be lifted by this system.
F2 = N
F1 = 800 N
m = ?
A1 = 100 cm2
A2 = 1500 cm2
P1 = 8.00 N/cm2
P2 = 8.00 N/cm2
F2 = Wt. of car
= m g F2 N
m = =
= m = kg g
9.8 m/s2

20. the small piston of the hydraulic system below.
ex. A hydraulic lift is used to lift a car. A force of 800 N is exerted on
the small piston of the hydraulic system below.
(d) If car is to be raised 2.0 m, how far must small piston move?
A1 = 100 cm2 V2
h2 = 2.0 m
A2 = 1500 cm2 V1
h1 = ?
V1 = V2
V = A h
A1 h1 = A2 h2
A2 h2
( 1500 cm2 )( 2.0 m )
h1 = =
= h1 = 30 m A1
100 cm2

22. Other Phenomena of Liquids
Result from the interplay between cohesive and adhesive forces
cohesive forces cause liquid particles to stick to each other
adhesive forces cause liquid particles to stick to other things
Water droplets assume a spherical
shape due to cohesive forces
Spheres have the largest
volume to surface area ratio
of any shape (cubes,
tetrahedrons, etc.)

23. Other spheres in nature:

24. Cohesive forces also cause surface tension
Water forms a "skin"
at the surface
Surfactants take away the
surface tension; key ingredient
in laundry detergents

25. Water also exhibits adhesive properties
Responsible for things getting “wet”
As object is pulled out of water, cohesive
forces compete with adhesive forces;
things get wet because adhesive forces
are greater than cohesive forces

26. When water is put into a narrow tube, adhesive forces
cause it to climb up the sides of the tube; known as capillary action
A meniscus forms in a graduated
cylinder due to adhesion; volume is
read at bottom of meniscus

27. Other examples:

28. Could this work as a generator of energy?

29. Water is
adhesive
Mercury is
cohesive

30. Buoyant Force and Archimedes Principle
Fill a tank with water
1 kg
Consider a 1-kg “piece”
of the water
9.8 N
buoyant force
Wt.
Water weighs 9.8 N
Water just below the piece exerts an upward
"buoyant“ force of 9.8 N to produce equilibrium

31. Buoyant Force and Archimedes Principle
Now replace the piece with a
stone of equal volume, but
not of equal mass
1.5 kg
Let mass of stone = 1.5 kg
9.8 N
buoyant force
The weight of the rock
outside of the water = 14.7 N
14.7 N
The water still exerts a buoyant force of 9.8 N, as if the
"piece” of water was still there
If the rock was released, it would accelerate to the bottom

32. Buoyant Force and Archimedes Principle
Now replace the piece with a
stone of equal volume, but
not of equal mass
4.9 N
1.5 kg
Let mass of stone = 1.5 kg
9.8 N
buoyant force
The weight of the rock
outside of the water = 14.7 N
14.7 N
If suspended from a spring scale, what would it read,
in newtons?
14.7 N N = 4.9 N

33. Archimedes Principle: The buoyant force exerted on a body immersed
in a fluid is equal to the weight of the fluid
displaced by the body.
ex. A block of metal weighs 15.4 N in air. When submerged in
water, it weighs 9.7 N.
(a) What is the buoyant force exerted by the water?
FB = N N
= FB = 5.7 N FB
9.7 N

34. ex. A block of metal weighs 15.4 N in air. When submerged in
water, it weighs 9.7 N.
(b) What is the volume of water displaced by the metal?
Wt. of water = FB = 5.7 N
Wt.
5.7 N
Wt. = m g
m = = g
9.8 m/s2
= kg
= 582 g m
ρ =
ρ = g/cm3 V
V ρ = m FB m
582 g
9.7 N
V = =
= V = 582 cm3 ρ
1.00 g/cm3

35. ex. A block of metal weighs 15.4 N in air. When submerged in
water, it weighs 9.7 N.
(c) What is the density of the metal?
Wt. in air = N
Vm = Vw = 582 cm3 m
m = ?
ρ = V
Wt.
15.4 N
m = =
= kg g
9.8 m/s2
15.4 N
= g m
1570 g
ρ = =
= ρ = 2.7 g/cm3 V
582 cm3

36. Archimedes Principle:
The buoyant force exerted on a body immersed
in a fluid is equal to the weight of the fluid
displaced by the body.
Things That Float:
Lower boat
into water

37. Archimedes Principle:
The buoyant force exerted on a body immersed
in a fluid is equal to the weight of the fluid
displaced by the body.
Things That Float:
Lower boat
into water

38. Archimedes Principle:
The buoyant force exerted on a body immersed
in a fluid is equal to the weight of the fluid
displaced by the body.
Things That Float:
Lower boat
into water
Volume of
water displaced
Volume of boat
below water line =

39. Archimedes Principle:
The buoyant force exerted on a body immersed
in a fluid is equal to the weight of the fluid
displaced by the body.
Things That Float:
Buoyant Force FB
Lower boat
into water
Weight of
water displaced =
Buoyant Force

40. If the weight of the displaced water equals the weight of the
object, the object will float.
Things That Float:
Buoyant Force FB
Lower boat
into water
Weight of
water displaced =
Buoyant Force

43. Fluids in Motion
Consider a flowing river
; enters a narrow canyon
What happens to the rate of flow?
Water must speed up

44. Fluids in Motion
Volumes must be equal A1 V1 A2 V2 x2 x1
V1 = V2
V = A x
A1 x1 = A2 x2
A1 v1 t = A2 v2 t
x = v t
Continuity
Equation
A1 v1 = A2 v2

45. Now put pipes into the flow, with the valves closed

46. When valves are opened, lateral pressure pushes
water up pipes, but not an equal amount
Bernoulli’s Principle: A fast-moving fluid exerts less lateral
pressure than a slow-moving fluid

47. Air moving past an airfoil:
Fast-moving air
Slow-moving air

48. Air moving past an airfoil:
Lift
Fast-moving air; low pressure
Slow-moving air; high pressure

49. Frisbee:
fast air
slow air

50. Spinning ball through air:

51. Other examples: umbrellas
on a windy day; standing
next to a train track as a
train passes; toilets on a
windy day

52. Bernoulli’s
Equation
P1 + ½ ρ v12 + ρ g h1 = P2 + ½ ρ v22 + ρ g h2
ex: A frisbee of area m2 has air flowing under it at 15.0 m/s
and over it at 15.8 m/s. Find the lift force in newtons.
Lift comes from a difference in pressure, P2 - P1
P1 + ½ ρ v12 + ρ g h1 = P2 + ½ ρ v22 + ρ g h2
h1 (height of air at bottom of frisbee) = h2 (height at top)
P1 + ½ ρ v12 = P2 + ½ ρ v22

53. P1 + ½ ρ v12 = P2 + ½ ρ v22
Solve for ( P2 - P1 )
- P P1
½ ρ v12 = P2 - P1 + ½ ρ v22
- ½ ρ v ½ ρ v22
½ ρ v ½ ρ v22 = P2 - P1
P2 - P1 = ½ ρ v ½ ρ v22
= ½ ρ ( v v22 )
ρair = kg/m3
= ½ ( 1.29 kg/m3 )( [15.0 m/s]2 - [15.8 m/s]2 )
P2 - P1 = N/m2 F
P =
F = P A
= ( 15.9 N/m2 )( m2 ) A
F = N

54. ex: Water is pumped from the ground 35 m up to the top of a water
tower. At the bottom, the pipe is 1.5 m in diameter and the water
flows at 0.45 m/s under a pressure of 4.8 atm. The pipe tapers to a
diameter of 0.50 m at the top. Find the flow rate and the pressure
at the top.
P1 + ½ ρ v12 + ρ g h1 = P2 + ½ ρ v22 + ρ g h2
A1 v1 = A2 v2
A1 = π ( 0.75 m )2 = m2
A2 = π ( 0.25 m )2 = m2
v1 = m/s
v2 = ?
P1 = 4.8 atm
N/m2
atm
P1 = N/m2
P2 = ?

55. ex: Water is pumped from the ground 35 m up to the top of a water
tower. At the bottom, the pipe is 1.5 m in diameter and the water
flows at 0.45 m/s under a pressure of 4.8 atm. The pipe tapers to a
diameter of 0.50 m at the top. Find the flow rate and the pressure
at the top.
A1 = m2
A2 = m2
v1 = m/s
A1 v1 = A2 v2
A1 v1
v2 = A2
( m2 )( 0.45 m/s ) = m2
v2 = 4.0 m/s

56. ex: Water is pumped from the ground 35 m up to the top of a water
tower. At the bottom, the pipe is 1.5 m in diameter and the water
flows at 0.45 m/s under a pressure of 4.8 atm. The pipe tapers to a
diameter of 0.50 m at the top. Find the flow rate and the pressure
at the top.
P1 + ½ ρ v12 + ρ g h1 = P2 + ½ ρ v22 + ρ g h2
v1 = m/s
v2 = 4.0 m/s
P1 = N/m2
P2 = ?
Let h1 = 0
h2 = 35 m

57. v1 = m/s
v2 = 4.0 m/s
h1 = 0
P1 = N/m2
P2 = ?
h2 = 35 m
P1 + ½ ρ v12 + ρ g h1 = P2 + ½ ρ v22 + ρ g h2
- ½ ρ v22 - ρ g h2
- ½ ρ v22 - ρ g h2
P2 = P1 + ½ ρ v12 - ½ ρ v ρ g h2
ρ = kg/m3
½ ρ v12 = ½ ( 1000 kg/m3 )( 0.45 m/s )2 = N/m2
½ ρ v22 = ½ ( 1000 kg/m3 )( 4.0 m/s )2 = N/m2
ρ g h2 = ( 1000 kg/m3 )( 9.8 m/s2 )( 35.0 m ) = N/m2
P2 = ( ) + ( ) - ( ) - ( )
P2 = N/m2
atm
P2 = atm
N/m2