Fluids and Elasticity Chapter 15. Density ( ) = mass/volume Rho ( ) – Greek letter for density Units - kg/m 3 Specific Gravity = Density of substance.

Fluids and Elasticity

Chapter 15

Density ()

= mass/volume

• Rho ( ) – Greek letter for density

• Units - kg/m3

• Specific Gravity = Density of substance

Density of water (4oC)

• Unitless ratio

• Ex: Lead has a sp. Gravity of 11.3 (11.3 times denser than water

Ex: 1

Estimate the mass of air in this classroom

Pressure

• Force per unit area

• P = F/A

• Unit - N/m2 (Pascal)

• The larger the area, the less the pressure– Shoeshoes– Elephant feet– Bed of nails

Fluid Pressure

• A fluid exerts the same pressure in all directions at a given depth

• P = Po + gh

• The atmosphere is a fluid

• Po often 1 atm (101.3 kPa)

Pressure: Example 1

A water storage tank is 30 m above the water faucet in a house. Calculate the pressure at the faucet:

We will neglect the atmospheric pressure since it is the same at the tank and at the surface

P = gh = (1000 kg/m3)(9.8 m/s2)(30 m)

P = 29,000 kgm2/m3s2 = 29,000 kg m/s2m2

P = 29,000 N/m2

Pressure: Example 2The Kraken can live at a depth of 200 m. Calculate

the pressure the creature can withstand (neglect atmospheric pressure)

Pressure: Example 2

P = gh = (1000 kg/m3)(g)(200 m)

P = 1.96 X 106 N/m2

Atmospheric Pressure

• 1 atm = 1.013 X 105 N/m2 = 101.3 kPa

• 1 bar = 1 X 105 N/m2 (used by meteorologists)

• Gauge pressure

P = Patm + PG

Absolute pressure atmospheric pressure Gauge pressure

• We usually measure gauge pressure

• Ex: A tire gauge reads 220 kPa, what is the absolute pressure?

P = Patm + PG

P = 101.3 kPa + 220 kPa = 321 ka

Straw Example

You can pick up soda in a straw using your finger. Why doesn’t the soda fall out?

Another Straw Example

What pushes soda up a straw when you drink through it?

Torrecilli’s Work

P = Patm + PG

P = Patm + gh

What is the highest column of water that the atmosphere can support?

P = Patm + gh

0 = 1.013 X 105 N/m2 + (1000kg/m3)(9.8m/s2)(h)

h = 10.3 m

• No vacuum pump can pump more than ~30 feet

Try the same calculation with mercury

P = Patm + gh

0 = 1.013 X 105 N/m2 + (13,600kg/m3)(9.8m/s2)(h)

h = 0.760 m (760 mm)

1 atm = 760 mm Hg (760 torr)

Can an astronaut attach suction cups to the boots of his spacesuit to help him climb around the space shuttle while in space?

Pascal’s Principle

• Pressure applied to a confined fluid increases the pressure the same throughout

Pin = Pout

Fin = Fout

Ain Aout

for pistons:

F = g(A1 + A2)d2

Pascal’s Principle: Ex 1

A hydraulic lift can produce 200 lb of force. How heavy a car can be lifted if the area of the lift is 20 times larger that the input of the Force?

Fin = Fout

Ain Aout

Fout = Fin Aout = (200 lb) (20) = 4000 lbs

Ain 1

Pascal’s Principle: Ex 1

A hydraulic lift has the car rest ona 25 cm pipe. The lift the car, compressed air pushes on a 6.0 cm pipe.

a. Calculate the force needed to lift a 1300 kg car.

b.Calculate how much the air pressure force must be increased to lift the car 2.0 m.

Buoyancy

• Buoyancy – The “lift” provided by water– Objects weight less in water than out

• Caused by pressure differential between top and bottom of an object.

Fbouyant = gV

Derivation of the Buoyancy Formula

Fb = F2 – F1

P = F/AF = PAF = ghA

Fb = gh2A – gh1A

Fb = gA(h2 - h1)

Fb = gV

Archimedes Principle

“The bouyant force on an object equals the weight of fluid displaced by the object”

w’ = weight of an object in water (or any liquid)

w’ = mg - Fb

Buoyancy: Example 1

A 7000-kg ancient statue lies at the bottom of the sea. Its volume is 3.0 m3. How much force is needed to lift it?

Fb = gV

Fb = (1000 kg/m3)(9.8 m/s2)(3.0m3)

Fb = 2.94 X 104 kg-m/s2

Fb = 2.94 X 104 N mg

Fb

w’ = mg - Fb

w’ = (7000 kg)(9.8m/s2) - 2.94 X 104 N

w’ = 3.92 X 104 N

Say, isn’t w’ just the sum of the forces?

Yep.

F = w’mg

Fb

Buoyancy: Example 2

Archimedes tested a crown for the king. Out of water, it masses 14.7 kg. In water, it massed 13.4 kg. Was the crown gold?

w’ = mcrg – Fb

w’ = mcrg – gVcr

(13.4 kg)(g) = (14.7 kg)(g) – (1000 kg/m3)(g)(Vcr)

131 N = 144 N – (9800 kg/ms2)(Vcr)

Vcr = 0.00133 m3

Now we can calculate the density of the crown:

= m/V = 14.7 kg/0.00133 m3

= 11,053 kg/m3

Gold’s density is about 19,000kg/m3. This is much closer to lead.

Example 3

A cube of wood that is 10 cm on a side is held underwater by tying a string to the cube and the bottom on the container. The wood has a density of 700 kg/m3.

a. Draw a free body diagram showing all the forces on the block.

b.Calculate the force of bouyancy

c. Calculate the tension in the string.

Floating• Objects that are less dense than water will float

• Part of the object will be above the water line

• A case of static equilibrium

F = 0

mg

Fb

Floating: Example 1

A 1200 kg log is floating in water. What volume of the log is under water?

F = 0

F = 0 = mg – Fb

mg = Fb

mg = gVlog

Vlog = mg

g mg

Fb

Vlog = m (Hey, the g’s cancel!)

Vlog = 1200 kg = 1.2 m3

1000 kg/m3

Floating: Example 1

A wooden raft has a density of 600 kg/m3, an area of 5.7 m2, and a volume of 0.60 m3. How much of the raft is below water in a freshwater lake?

Let’s first calculate the mass of the raft:

= m/V

m = V = (600 kg/m3)(0.60 m3) = 360 kg

Now we can worry about the raft.

F = 0

F = 0 = mg – Fb

mg = Fb

mg = gVsubmerged

mg = ghsubmergedA

mg = ghsubmergedA

m = hsubmergedA (Hey, the g’s cancelled!)

hsubmerged = m/A

hsubmerged = 360 kg = 0.063 m

(1000 kg/m3)(5.7 m2)

Floating: Example 3

Suppose a continent is floating on the mantle rock. Estimate the height of the continent above the mantle (assume the continent is 35 km thick).

F = 0 = mg – Fb

0 = mcg – mangVc(submerged)

mcg = mangVc(submerged)

mc = manVc(submerged)

We don’t know the mass of the continent

c = mc/Vc(total)

mc =cVc(total)



mc =cVc(total)

manVc(submerged) = cVc(total)

Vc(submerged) = c = (2800 kg/m3) = 0.85

Vc(total) man (3300 kg/m3)

This means that 85% of the continent is submerged, and only 15% is above:

(0.15)(35 km) = 5.25 km

Floating: Ex 4

A block is placed in water and 5.8 cm is submerged. The same block is placed in an unknown liquid and 4.6 cm is submerged. Calculate the density of the unknown liquid. Assume the same face of the block pointed downward in both cases (A).

Fluid Flow

• Laminar Flow – Smooth, streamline flow (laminar means “in layers”)

• Turbulent Flow – erratic flow with “eddies”

• Viscosity – Internal friction of a liquid– High viscosity = slow flow– Viscosity is NOT the same as density

Equation of Continuity

A1v1 = A2v2

A = Area of a pipe

v = velocity of the liquid

Equation of Continuity

v1A1 = v2A2

• Fluid will flow faster through a smaller opening

• Placing your finger over a hose opening.

The term “vA” is the “volume rate of flow”

A = m2

v = m/s

vA = m3/s

Q = vA

Eqn. Of Continuity: Example 1

A garden hose has a radius of 1.00 cm and the water flows at a speed of 0.80 m/s. What will be the velocity if you place your finger over the hose and narrow the radius to 0.10 cm?

A1 = r2 = (3.14)(0.01 m)2 = 3.14 X 10-4 m2

A2 = r2 = (3.14)(0.001 m)2 = 3.14 X 10-6 m2

A1v1 = A2v2

v2 = A1v1

A2

v2 = (3.14 X 10-4 m2)(0.80 m/s) = 80 m/s

(3.14 X 10-6 m2)


A water hose 1.00 cm in radius fills a 20.0-liter bucket in one minute. What is the speed of water in the hose?

A1 = r2 = (3.14)(1 cm)2 = 3.14 cm2

Remember that Av is volume rate of flow.

A2v2=20.0 L 1 min 1000 cm3 = 333 cm3/s

1 min 60 s 1 L

A1v1 = A2v2

v1 = A2v2/A1

v1 = 333 cm3/s = 160 cm/s or 1.60 m/s

3.14 cm2


A sink has an area of about 0.25 m2. The drain has a diameter of 5 cm. If the sink drains at 0.03 m/s, how fast is water flowing down the drain?

Ad =r2 = ()(0.025 m)2 = 1.96 X 10-3 m3

Advd = Asvs

vd = Asvs/Ad=[(0.25 m2)(0.03 m/s)]/(1.96 X 10-3 m3)

vd = 3.82 m/s


The radius of the aorta is about 1.0 cm and blood passes through it at a speed of 30 cm/s. A typical capillary has a radius of about 4 X 10-4 cm and blood flows through it at a rate of 5 X 10-4 m/s. Estimate how many capillaries there are in the human body.

Aava = NAcvc(N is the number of capillaries)

Aa = r2 = (3.14)(0.01 m)2 = 3.14 X 10-4 m2

Ac = r2 = (3.14)(4 X 10-6 cm)2 = 5.0 X 10-11 cm2

N = Aava/ Acvc

N = (3.14 X 10-4 m2)(0.30 m/s) = ~ 4 billion

(5.0 X 10-11 cm2)(5 X 10-4 m/s)


How large must a heating duct be to replenish the air in a room 300 m3 every 15 minutes? Assume air moves through the vent at 3.0 m/s.

Advd = Arvr

Arvr = volume rate of flow:

Arvr = 300 m3 1 min = 0.333 m3/s

15 min 60 s

Ad = Arvr/vd

Ad = 0.333 m3/s = 0.11 m2

3.0 m/s

Bernoulli’s Principle

The velocity and pressure of a fluid are

inversely related.

Why does a shower curtain sometimes “attack” a person taking a shower?

What will happen to closed windows during a tornado? Will they blow in or out?

Applications of Bernoulli’s Principle

1. Airplane wing


2. Prairie Dog Burrows1. Air moves faster (lower pressure) at the top

2. Draws air through the burrow

3. The exact same thing happens with our chimneys


3. Spray Paint Flow of air (low pressure)


4. Dime in a cup

Bernoulli’s Equation

Pt + ½vt2 + gyt = Pb + ½vb

2 + gyb

(note: you often have to use the Eqn. Of Continuity in these situations:)

A1v1 = A2v2

• Often useful when you have both a change in height and area:

Pipe from a water reservoir to a house

Pipe from a house into a sewer pipe

If there is no change in altitude, the equation simplifies:


2 + gyb

Pt + ½vt2 = Pb + ½vb

2

Bernoulli’s Equation: Example 1

A water heater in the basement of a house pumps water through a 4.0 cm pipe at 0.50 m/s and 3.0 atm. What will be the flow speed and pressure through a 2.6 cm spigot on the second floor, 5.0 m above?

3.0 atm1.013 X 105 N/m2 = 3.0 X 105 N/m2

1 atm

Flow speed:

Atvt = Abvb

vt = Abvb/At (Remember A = r2)

vt = r2bvb (Hey, the ’s cancel!)

r2t

vt = r2bvb = (0.02 m)2(0.50m/s) = 1.2 m/s

r2t (0.013 m)2

Now the pressure:


2 + gyb

Pt + ½(1000)(1.2)2 + (1000)(9.8)(5) = 3.0X105 + ½(1000)(0.50)2 + (1000)(9.8)(0)

Pt = 2.5 X 105 N/m2

Bernoulli’s Equation: Example 2

A drunken redneck shoots a hole in the bottom of an aboveground swimming pool. The hole is 1.5 m from the top of the tank. Calculate the speed of the water as it comes out of the hole.

yt = 1.5 m

yb = 0 m

The top of the pool is a much larger area than the hole. We will assume that the vt = 0.


2 + gyb

Pt + gyt = Pb + ½vb2 + gyb

Also, both the top and the hole are open to the atmosphere, so Pt = Pb

Pt + gyt = Pb + ½vb2 + gyb

gyt = ½vb2 + gyb

Set the bottom of the pool as yb = 0.

gyt = ½vb2 + gyb

gyt = ½vb2

vb2 = 2gyt/

vb2 = 2gyt

vb2 = (2)(9.8m/s2)(1.5 m)

vb = 5.42 m/s

Example 3

A hydroelectric dam is 200 m above the power plant. The inlet hose at the top has a diameter of 100 cm, and the outlet hose to the turbine has a diameter of 50 cm.

a. Calculate the speed of the water into the turbine (both are open to the atmosphere)

Elasticity: Hooke’s Law

• Hooke’s Law – usually used with a spring

• Can consider anything to be like a spring

• F = kL (F=kxspring)

• k = proportionality (spring) constant

• Can’t stretch things forever

Elastic region – material will still bounce back

Plastic region – material will not return to original length (but has not broken)

F = kL

This is only linear

in the proportional

region

HOPE I DON’T REACH MY

ELASTIC LIMIT!

Elastic Region: Young’s Modulus(E)

Stress = Force = F

Area A

Strain = Change in length = L

Original length Lo

Y = stress

strain

Y = F/A or F = Y L

L/Lo A Lo

Young’s Modulus: Example 1

A 1.60 m long steel piano wire has a diameter of 0.20 cm. How great is the tension in the wire if it stretches 0.30 cm when tightened?

A = r2 = (0.0010 m)2 = 3.1 X 10-6 m2

F = Y L

A Lo

F = AY L

Lo

F = AY L

Lo

F = (3.1 X 10-6m2)(200 X 1011N/m2)(0.0030 m)

1.60 m

F = 1200 N


A steel support rod of radius 9.5 mm and length 81 cm is stretched by a force of 6.2 X 104 N (about 7 tons).

a. Calculate the stress.

b.Calculate the elongation.

Area = r2 = (0.0095 m2

Stress = Force = 6.2 X 10-4m = 2.2 X 108 N/m2

Area 2.84 X 10-4 m2

F = YL

A Lo

L = FL = (6.2 X 104 N)(0.81m)YA (200 X 109 N/m2)(2.84 X 10-4

m2)

L = 8.84 X 10-4 m = 0.89 mm


A 2.0 m long, 1.0 mm diameter wire is suspended. Hanging a 4.5 kg mass stretches the wire length by 1.00 mm.

a. Calculate Young’s modulus

b.Identify the material from the table.

The Three Types of Stress

Stretching Squeezing Horizontal

Other Modulus’

Shear Modulus – Used for shear stress

Bulk Modulus – Used for even compression on all sides (an object when submerged)

Fracture

• Breaking Point

• Uses– Tensile Strength – Stretching– Compressive Strength – under a load– Shear Stress – Shearing

• Safety Factor – reciprocal that is multiplied by the tensile strength

• Ex: A safety factor of 3 means you will only use 1/3 of the maximum stress

Fracture: Example 1

A concrete column 5 m tall will have to support 1.2 X 105 N (compression). What area must it have to have a safety factor of 6?

Max stress = (1/6)(compressive strength)

Max stress = (1/6)(20 X 106 N/m2)= 3.3X106 N/m2

Stress = F

A

Stress = F

A

Area = F = (1.2 X 105 N) = 3.64 X 10-2 m

Stress 3.3 X 106N/m2

How much will the column compress under the load?

F = EL

A Lo

L = FL = (1.2 X 105 N)(5 m)

EA (20 X 109 N/m2)(3.64 X 10-2 m2)

L = 8.3 X 10-4 m = 0.83 mm

Fracture: Example 2

Spider-man’s webbing has a tensile strength of 600 X 106 N/m2 and he wishes to use a safety factor of 3. What is the diameter of the webbing if the maximum force at the bottom of a swing is 1500 N?

Maximum Stress

(1/3)(600 X 106 N/m2) = 200 X 106 N/m2

Stress = Force

Area

Area = Force = 1500 N = 7.5 X 10-6m2

Stress 200X106N/m2

Area = r2

r = (A/)1/2 = 1.55 X 10-3 m or 1.55 mm

Diameter = 3.10 mm

Concrete

• Concrete is much stronger under compression than tension– Tensile Strength – 2 X 106N/m2

– Compressive Strength – 20 X 106N/m2

• Prestressed concrete – rods or mesh are stretched when the concrete is poured. Released after concrete dries.

• Now under compression

Fluids and Elasticity Chapter 15. Density ( ) = mass/volume Rho ( ) – Greek letter for density Units - kg/m 3 Specific Gravity = Density of substance.

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