Lecture 19Lecture 19Fluids: density, pressure,

Pascal’s principle and Buoyancy.Water tower

Pascal’s vases

Hydraulic press

Barometer

What is a fluid?

Fluids are “substances that flow”…. “substances that take the shape of the container”

Atoms and molecules must be free to move .. No long range correlation between positions (e.g., not a crystal).

Gas or liquid… or granular materials (like sand)

In a fluid:

Density, pressure

Density:

Units:

Pascal Pa = 1 N/m2

psi = lb/in2

atmosphere 1 atm = 1.013 × 105 Pabar 1 bar = 105 Pa

Ex: Pure water: 1000 kg/m3

Vm

Pressure: Fp

A

Particles are always moving

i.e., exerting (perpendicular) forces on surfaces

F

i.e., hitting surfaces

Surface of area A

Atmospheric pressure

DEMO: Piston and weight

The atmosphere of Earth is a fluid, so every object in air is subject to some pressure.

At the surface of the Earth, the pressure is

patm ~ 1.013 x 105 Pa = 1 atm

Area of a hand ~ 200 cm2 = 0.02 m2

atm ~ 2000 N on your hand due to air!F p A

Vacuum gun

Sealed tube, air pumped out

Ping-pong ball

What happens if we punch a little hole on one side?

DEMO: Vacuum gun

Length of tube L ~ 3 mMass of ball m ~ 3 gRadius of tube R ~ 2 cm

W = ΔKE

FL ∼ (patmR2)L =12

mv 2− 0

v ∼ √2patmR2L

m∼300 km/s

Pressure vs. depth

Fbottom

Ftop

mg

Imaginary box of fluid with density ρ with bases of area A

and height h

bottom topp p gh

m Ah

hFor floating object, net force must be zero!

bottom topF F mg

DEMO: Plastic tube

with cover

Example: How deep under water is p = 2 atm ?

h =pbottom−ptop

ρfluidg=

1.01×105 Pa(103 kg/m3) (9.81 m/s2)

= 10.3 m

(i.e. 1 atm is produced by a 10.3 m high column of water)

Called guage pressure

Pbottom/top =Fbottom/top

A

Fluid in an open container

Pressure is the same at a given depth, independent of the shape of the container. p(y)

y

Fluid level is the same everywhere in a connected container (assuming no surface forces)

•A

•B

If liquid height was higher above A than above B

If liquid height was higher above A than above B

pA > pB pA > pB Net force

Net force

Net flow

Net flow

This is not equilibrium!This is not equilibrium!

DEMO: Pascal’s vases

ACT: U tube

Two liquids Y and G separated by a thin, light piston (so they cannot mix) are placed in a U-shaped container. What can you say about their densities?

A. ρG < ρY

B. ρG = ρY

C. ρG > ρY

YG

•A

•B

ACT: U tube

Two liquids Y and G separated by a thin, light piston (so they cannot mix) are placed in a U-shaped container. What can you say about their densities?

A. ρG < ρY

B. ρG = ρY

C. ρG > ρY

YG

•A

•B

Pressure at A and B must be the same:

Y 1 G 2 atm G 3 atmgh gh p gh p

h1

h2

h3

Y 1 G 3 2h h h

DEMO: U-tube with

water and kerosene

Since h1< h3 − h2 ⇒ ρy> ρG

Water towers

Water towers are a common sight in the Midwest… because it’s so flat!

h

house atm waterp p hg

So physics sucks, but how much?

Your physics professor sucks on a long tube that rises out of a bucket of water. He can get the liquid to rise 5.5 m (vertically). What is the pressure in His mouth at this moment?

A. 1 atm

A. 0.67 atm

B. 0.57 atm

C. 0.46 atm

D. 0 atm

h

x A

x B

So physics sucks, but how much?

Your physics professor sucks on a long tube that rises out of a bucket of water. He can get the liquid to rise 5.5 m (vertically). What is the pressure in His mouth at this moment?

A. 1 atm

A. 0.67 atm

B. 0.57 atm

C. 0.46 atm

D. 0 atm

DEMO: Sucking

through a hose

water atmmouth

atm watermouth

5 3 3 2 10 Pa 10 kg/m 9.8 m/s 5.5 m 46100 Pa 0.46 atm

p gh pp p gh

h

x A

x B

Pascal’s principle

Any change in the pressure applied to an enclosed fluid is transmitted to every portion of the fluid and to the walls

of the containing vessel.

Pascal’s Principle is most often applied to incompressible fluids (liquids):

Increasing p at any depth (including the surface) gives the same increase in p at any other depth

Hydraulic chamber

F2 can be very large…

No energy is lost:

F1

A1

=F2

A2

F2 =A2

A2

F1

W = F1d1=(F2

A1

A2)(d2

A2

A1)=F2d2

Incompressible fluid: A1d1 = A2 d2

ACT: Hydraulic chambers

In each case, a block of mass M is placed on the piston of the large cylinder, resulting in a difference di between the liquid levels. If A2 = 2A1, then:

A1 A10

A2 A10

M

MdB

dA

A. dA < dB

B. dA = dB

C. dA > dB

ACT: Hydraulic chambers

In each case, a block of mass M is placed on the piston of the large cylinder, resulting in a difference di between the liquid levels. If A2 = 2A1, then:

A1 A10

A2 A10

M

MdB

dA

A. dA < dB

B. dA = dB

C. dA > dB

Pressure depends only on the height of the water column above it.

Measuring pressure with fluids

Barometer Measures absolute pressure Top of tube evacuated (p = 0) Bottom of tube submerged into pool of mercury

open to sample (p) Pressure dependence on depth:

vacuum

p=0

h atmosphere

p=p0

Sample at p

Sample at p

hh

Vacuum p = 0

Vacuum p = 0

pp patmpatm

∆h∆h

patm

h

p

Manometer Measures gauge pressure: pressure relative to

atmospheric pressure. Pressure dependence on depth: Δ h =

p−patm

ρHg g

A unit for pressure760 mm Hg = 760 torr = 1 atm

h =p

ρHgg

ACT: Side tube

A sort of barometer is set up with a tube that has a side tube with a tight fitting stopper. What happens when the stopper is removed?

vacuum

stopper

A. Water spurts out of the side tube.

B. Air flows in through the side tube.

C. Nothing, the system was in equilibrium and remains in equilibrium.

ACT: Side tube

A sort of barometer is set up with a tube that has a side tube with a tight fitting stopper. What happens when the stopper is removed?

vacuum

stopper

A. Water spurts out of the side tube.

B. Air flows in through the side tube.

C. Nothing, the system was in equilibrium and remains in equilibrium.

DEMO: Side tube

Buoyancy and the Archimedes’ principle

ybottom

ytop

hA

A box of base A and height h is submerged in a liquid of density ρ.

bottom topAp Ap

atm bottom atm topA p gy A p gy

A hg

Archimedes’ principle: The liquid exerts a net force upward called buoyant force whose magnitude is equal to the weight of the displaced liquid.

direction upVg

topbottomF F F

Ftop

Fbottom

Net force by liquid:

In-class example: Hollow sphere

A hollow sphere of iron (ρFe = 7800 kg/m3) has a mass of 5 kg. What is the maximum diameter necessary for this sphere to be completely submerged in water? (ρwater = 1000 kg/m3)

A. It will always be submerged.

B. 0.11 m

C. 0.21 m

D. 0.42 m

E. It will always only float.

In-class example: Hollow sphere

A hollow sphere of iron (ρFe = 7800 kg/m3) has a mass of 5 kg. What is the maximum diameter necessary for this sphere to be completely submerged in water? (ρwater = 1000 kg/m3)

A. It will always be submerged.

B. 0.11 m

C. 0.21 m

D. 0.42 m

E. It will always only float.

FB

mg

The sphere sinks if FB< mg

maxMaximum diameter 2 0.21 mR

ρwater43

πR3 g < mg ⇒ R <3√

3m4πρwater

= 0.106 m

What if the sphere is solid?!

Density rule

A hollow sphere of iron (ρFe = 7800 kg/m3) has a mass of 5 kg. What is the maximum diameter necessary for this sphere to be fully submerged in water? (ρwater = 1000 kg/m3) Answer: R = 0.106 m.

And what is the average density of this sphere?

3

sphere water33

5 kg 1000 kg/m4 4 0.106 m3 3

m

R

An object of density ρobject placed in a fluid of density ρfluid

• sinks if ρobject > ρfluid

• is in equilibrium anywhere in the fluid if ρobject = ρfluid

• floats if ρobject < ρfluid (will not be completely submerged)

DEMO: Frozen helium

balloon

This is why you cannot sink in the Dead Sea

(ρDead Sea water = 1240 kg/m3 , ρhuman body = 1062 kg/m3 ) !

Buoyancy (in the Dead Sea)

ACT: Styrofoam and lead

A piece of lead is glued to a slab of Styrofoam. When placed in water, they float as shown.

What happens if you turn the system upside down?

A

Pb

styrofoam

Pb

styrofoam

Pb

styrofoam

B

C. It sinks.

ACT: Styrofoam and lead

A piece of lead is glued to a slab of Styrofoam. When placed in water, they float as shown.

What happens if you turn the system upside down?

A

The displaced volume in both cases must be the same (volume of water whose weight is equal to the weight of the lead+Styrofoam system)

Pb

styrofoam

Pb

styrofoam

Pb

styrofoam

B

C. It sinks.

1 2

ACT: Floating wood

Two cups have the same level of water. One of the two cups has a wooden block floating in it. Which cup weighs more?

A. Cup 1

B. Cup 2

C. They weigh the same.

1 2

ACT: Floating wood

Two cups have the same level of water. One of the two cups has a wooden block floating in it. Which cup weighs more?

A. Cup 1

B. Cup 2

C. They weigh the same.

The weight of the wood is equal to the weight of the missing liquid (= “displaced liquid”) in 2.

Cup 2 has less water than cup 1.

DEMO: Bucket of water

with wooden block

ACT: Aluminum and lead

Two blocks of aluminum and lead with identical sizes are suspended from the ceiling with strings of different lengths and placed inside a bucket of water as shown. In which case is the buoyant force greater?

A. Al

B. Pb

C. It’s the same for both

Al

Pb

ceiling

ACT: Aluminum and lead

Two blocks of aluminum and lead with identical sizes are suspended from the ceiling with strings of different lengths and placed inside a bucket of water as shown. In which case is the buoyant force greater?

A. Al

B. Pb

C. It’s the same for both

AlThe displaced volume (= volume of the block) is the same in both cases.

Depth or object density do not play any role.Pb

ceiling

The different weight is compensated with a different tension in the strings.

ACT: Wooden brick

When a uniform wooden brick (1 m x 1 m x 2 m) is placed horizontally on water, it is partially submerged and the height of the brick above the water surface is 0.5 m. If the brick was placed vertically, the height of brick above the water would be:

A. 0.5 m

B. 1.0 m

C. 1.5 m.

0.5 m

ACT: Wooden brick

When a uniform wooden brick (1 m x 1 m x 2 m) is placed horizontally on water, it is partially submerged and the height of the brick above the water surface is 0.5 m. If the brick was placed vertically, the height of brick above the water would be:

A. 0.5 m

B. 1.0 m

C. 1.5 m.

0.5 m

The displaced volume in both cases needs to be the same (Because the weight of the wood did not change) : half of the volume of the brick.

Same volume