Chapter 11 Sections 1 – 4

Fluids, Density, Pressure

What is a Fluid?

• Before we begin to understand the nature of a Fluid

we must understand the nature of all the states of

matter:

• The 3 primary states of matter:

• Solid: Definite shape and volume.

• Liquid: Takes the shape of its container, yet has

a definite volume.

• Gas: Takes the shape and volume of its container.

• Special "states”

• Plasma - high temperature matter, most common in the

universe, ionized gas, found in stars

A Fluid is:

• By definition, a fluid is any material that is unable to

withstand a static shear stress.

• Unlike an elastic solid which responds to a shear

stress with a recoverable deformation, a fluid

responds with an irrecoverable flow.

HOT AIR BALLOONS use

heated air, which is less dense

than the surrounding air, to

create an upward buoyant

force. According to Archi-

emedes’ Principle, the buoyant

force is equal to the weight of

the air displaced by the

balloon.

Fluids at Rest

Paul E. Tippens

Objectives

• Define and apply the concepts of density and fluid pressure to solve physical problems.

• Define and apply concepts of absolute, gauge, and atmospheric pressures.

Density • The 3 primary states have a distinct density,

which is defined as mass per unit of volume.

• Mass – material or matter in a body or “stuff”.

• Volume – The space an object takes up.

• Two ways to find volume:

• Measure the length, width, and height.

• L x w x h = volume

• Water displacement method – put in water and measure amount of water displaced or pushed out.

Mass Density

2 kg, 4000 cm3

Wood

177 cm3

45.2 kg

; mass m

Densityvolume V

Lead: 11,300 kg/m3

Wood: 500 kg/m3

4000 cm3

Lead

Same volume as the wood

2 kg Lead

Same mass as the wood

Demonstration of Density

• Density of water – 1 g = 1 ml = 1 cm3, so the density of water is 1 g/cm3 or 1000 kg/m3

• Normal wood block #2 – mass = 75 g, v = 3.8 x 3.8 x 8 cm = 115.5 cm3, Density = .65 g/cm3

• “Second “ Wood Block – m = 141 g, v = 4.3 x 4.3 x 7 cm = 129.4 cm3, Density = 1.1 g/ cm3.

• Which will Float in water?

Density of Each Coke

• πr 2h = volume of cylinder, Diameter = 6.5 cm/2 =

3.25 cm, height = 12 cm, so volume = 398.2 cm3.

Minus some of the top and bottom = ~ 380 cm3.

• Diet coke – mass = ~370 g, so Density = .974 g/

cm3.

• Regular coke – mass = ~390 g, so Density = 1.03 g/

cm3.

• From the data, what will happen when each is

put is a beaker of water?

Example 1: The density of steel is 7800 kg/m3. What is the volume of a 4-kg block of steel?

4 kg

3

4 kg;

7800 kg/m

m mV

V

V = 5.13 x 10-4 m3

What is the mass if the volume is 0.046 m3?

m = 359 kg

3m/kg7800

kg4

ρ

mV;

V

mρ

)m 046.0)(kg/m (7800 33 V m

Specific Gravity The relative density r of a material is the ratio of its density to the density of water (1000 kg/m3).

Steel (7800 kg/m3) r = 7.80

Brass (8700 kg/m3) r = 8.70

Wood (500 kg/m3) r = 0.500

Examples:

31000 kg/m

xr

Example Problem

2. A king orders a gold crown having a mass of

0.5 kg. When it arrives from the metal smith,

the volume of the crown is found to be

185 cm3. Is the crown made of gold, which

has a density of 19.3 x 103 kg/m3?

Pressure Pressure is the ratio of a force, F, to the area, A, over which it is applied:

Pressure ; Force F

PArea A

A = 2 cm2

1.5 kg

P = 73,500 N/m2

2

2

m .

)m/s kg)(9.8 .(

A

F P

00020

51

The Unit of Pressure (Pascal):

A pressure of one pascal (1 Pa) is defined as a

force of one newton (1 N) applied to an area of one

square meter (1 m2).

21 Pa = 1 N/mPascal:

In the previous example the pressure was 73,500

N/m2. This should be expressed as:

P = 73,500 Pa

Example Problem

3. A 50 kg woman balances on one heel of a

pair of high heel shoes. If the heel is circular

with radius 0.5 cm, what pressure does she

exert on the floor?

Example Problem

4. What must be the contact area between a

suction cup (completely exhausted) and a

ceiling in order to support the weight of an

80 kg student?

Fluid Pressure A liquid or gas cannot sustain a shearing stress - it is

only restrained by a boundary. Thus, it will exert a

force against and perpendicular to that boundary.

• The force F exerted by a

fluid on the walls of its

container always acts

perpendicular to the walls. Water flow shows perpendicular Flow

Fluid Pressure Fluid exerts forces in many directions. Try to submerse a

rubber ball in water to see that an upward force acts on

the float.

• Fluids exert pressure in

all directions.

F

Pressure vs. Depth in Fluid Pressure = force/area

h

mg Area

• Pressure at any point in a

fluid is directly proportional

to the density of the fluid

and to the depth in the fluid. P = gh

Fluid Pressure:

A

hgAp

A

Vg P So,

Ah V V; m ;A

mg P

Independence of Shape and Area.

Water seeks its own level,

indicating that fluid pressure

is independent of area and

shape of its container.

• At any depth h below the surface of the water

in any column, the pressure P is the same.

The shape and area are not factors.

Properties of Fluid Pressure

• The forces exerted by a fluid on the walls of its container are always perpendicular.

• The fluid pressure is directly proportional to the depth of the fluid and to its density.

• At any particular depth, the fluid pressure is the same in all directions.

• Fluid pressure is independent of the shape or area of its container.

Example 5. A diver is located 20 m below the surface of a lake ( = 1000 kg/m3). What is the pressure due to the water?

h = 1000 kg/m3

DP = gh

The difference in pressure

from the top of the lake to

the diver is:

h = 20 m; g = 9.8 m/s2

ΔP = 196 kPa

20m))(m/s (9.8)(1000kg/m P 23D

Atmospheric Pressure

atm atm h

Mercury

P = 0 One way to measure atmospheric

pressure is to fill a test tube with

mercury, then invert it into a bowl of

mercury.

Density of Hg = 13,600 kg/m3

Patm = ρgh h = 0.760 m

Patm = (13,600 kg/m3)(9.8 m/s2)(0.760 m)

Patm = 101,300 Pa

Absolute Pressure

Absolute Pressure: The sum of the

pressure due to a fluid and the pressure

due to atmosphere.

Gauge Pressure: The difference between

the absolute pressure and the pressure due

to the atmosphere:

Absolute Pressure = Gauge Pressure + 1 atm

h

DP = 196 kPa

1 atm = 101.3 kPa

ΔP = 196 kPa

1 atm = 101.3 kPa

Pabs = 196 kPa + 101.3 kPa

Pabs = 297 kPa

Atmospheric Pressure at sea level = 1.013 X 105 Pa = 101.3 kPa

= 760 mm Hg = 14.7 lb/in2 = 1 atmosphere

Example Problem

6. At what depth in a lake is the absolute

pressure equal to three times atmospheric

pressure?

Example Problem

7. If the fluid in the barometer is water, what

will be the height of the water column in the

vertical tube at atmospheric pressure?

Example Problem

8. The spring of a pressure gauge has a force

constant of 1000 N/m, and the piston has a

diameter of 2 cm. Find the depth in water

for which the spring compresses by 0.5 cm.

Assignment

• pages 350 – 352,

• #3, 6, 11, 14, 16, 21, 27