# PHYSICS 231 Lecture 21: Buoyancy and Fluid Motion - PowerPoint PPT Presentation

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PHYSICS 231 Lecture 21: Buoyancy and Fluid Motion. Remco Zegers Walk-in hour: Tue 4-5 pm Helproom. Previously. Young’s modulus. Solids:. Shear modulus. Bulk modulus Also fluids. General:. P=F/A (N/m 2 =Pa) F pressure-difference = PA =M/V (kg/m 3 ).

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PHYSICS 231 Lecture 21: Buoyancy and Fluid Motion

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## PHYSICS 231Lecture 21: Buoyancy and Fluid Motion

Remco Zegers

Walk-in hour: Tue 4-5 pm

Helproom

PHY 231

### Previously

Young’s modulus

Solids:

Shear modulus

Bulk modulus

Also fluids

General:

P=F/A (N/m2=Pa) Fpressure-difference=PA

=M/V (kg/m3)

Pascal’s principle: a change in pressure applied

to a fluid that is enclosed is transmitted to the whole

fluid and all the walls of the container that hold the fluid.

PHY 231

### Pascal’s principle

Pascal’s principle: a change in pressure applied

to a fluid that is enclosed is transmitted to the whole

fluid and all the walls of the container that hold the fluid.

HOLDS FOR A FLUID FULLY ENCLOSED ONLY

PHY 231

### Pascal’s principle

In other words then before: a change in pressure applied

to a fluid that is enclosed in transmitted to the whole

fluid and all the walls of the container that hold the fluid.

P=F1/A1=F2/A2

If A2>>A1 then

F2>>F1.

So, if we apply a small

force F1, we can exert

a very large Force F2.

Hydraulic press demo

PHY 231

### Pressure vs Depth

Horizontal direction:

P1=F1/A P2=F2/A

F1=F2 (no net force)

So, P1=P2

Vertical direction:

Ftop=PatmA

Fbottom=PbottomA-Mg=PbottomA-gAh

Since the column of water is not moving:

Ftop-Fbottom=0

PatmA=PbottomA-gAh

Pbottom=Patm+ gh

PHY 231

### Pressure and Depth:

Pdepth=h =Pdepth=0+ gh

Where:

Pdepth=h: the pressure at depth h

Pdepth=0: the pressure at depth 0

=density of the liquid

g=9.81 m/s2

h=depth

Pdepth=0=Patmospheric=1.013x105 Pa = 1 atm =760 Torr

From Pascal’s principle: If P0 changes then the pressures

at all depths changes with the same value.

PHY 231

### A submarine

A submarine is built in such a way that it can stand pressures

of up to 3x106 Pa (approx 30 times the atmospheric

pressure). How deep can it go?

PHY 231

NO!!

PHY 231

### Pressure measurement.

The open-tube manometer.

The pressure at A and B is

the same:

P=P0+gh

so h=(P-P0)/(g)

If the pressure P=1.01 atm, what

is h? (the liquid is water)

h=(1.01-1)*(1.0E+05)/(1.0E+03*9.81)=

=0.1 m

PHY 231

### Pressure Measurement: the mercury barometer

P0= mercurygh

mercury=13.6E+03 kg/m3

mercury,specific=13.6

PHY 231

P0

P0

h

h

h

P=P0+gh

P=P0+gh

P=P0+gh

PHY 231

### Buoyant force: B

P0

Ptop=P0+ wghtop

Pbottom=P0+ wghbottom

p= wg(htop-hbottom)

F/A= wgh

F= wghA=gV

B=wgV=Mwaterg

Fg=w=Mobjg

If the object is not moving:

B=Fg so: wgV=Mobjg

-

htop

hbottom

Archimedes (287 BC) principle: the magnitude of the buoyant

force is equal to the weight of the fluid displaced by the object

PHY 231

### Comparing densities

B=fluidgV Buoyant force

w=Mobjectg=objectgV

Stationary: B=w

object= fluid

If object> fluid the object goes down!

If object< fluid the object goes up!

PHY 231

### A floating object

A

w=Mobjectg=objectVobjectg

B=weight of the fluid displaced by

the object

=Mwater,displacedg

= waterVdisplacedg

= waterhAg

h: height of the object under water!

h

B

w

The object is floating, so there is no net force (B=w):

objectVobject= waterVdisplaced

h= objectVobject/(waterA) only useable if part of the object

is above the water!!

PHY 231

?? N

question

A hollow sphere with negligible weight

if not filled, is filled with water and

hung from a scale. The weight is 10 N.

It is then submerged. What is the

10N of water inside

thin hollow sphere

• 0 N

• 5 N

• 10 N

• 20 N

• impossible to tell

PHY 231

?? N

A)

### An example

?? N

B)

7 kg iron sphere of

the same dimension

as in A)

1 kg of water inside

thin hollow sphere

Two weights of equal size and shape, but different mass are

submerged in water. What are the weights read out?

PHY 231

### Another one

An air mattress 2m long 0.5m wide and 0.08m thick and has

a mass of 2.0 kg. A) How deep will it sink in water? B) How

much weight can you put on top of the mattress before it

sinks? water=1.0E+03 kg/m3

PHY 231

### equation of continuity

x2

x1

2

v2

v1

1

A1,1

A2,2

the mass flowing into area 1 (M1) must be the same as the

mass flowing into area 2 (M2), else mass would accumulate

in the pipe).

M1= M2

1A1x1= 2A2x2 (M=V=Ax)

1A1v1t=2A2v2t (x=vt)

1A1v1 =2A2v2

if  is constant (liquid is incompressible) A1v1 =A2v2

### Bernoulli’s equation

same

W1=F1x1=P1A1 x1=P1V

W2=-F2x2=-P2A2 x2=-P2V

Net Work=P1V-P2V

m: transported fluid mass

KE=½mv22-½mv12 & PE=mgy2-mgy1

Wfluid= KE+ PE

P1V-P2V=½mv22-½mv12+ mgy2-mgy1 use =M/V and div. By V

P1-P2=½v22-½v12+ gy2- gy1

P1+½v12+gy1= P2+½v22+gy2

P+½v2+gy=constant

P: pressure ½v2:kinetic Energy per unit volume

gy: potential energy per unit volume

Another conservation Law

PHY 231

When air is blown in between the

cans, the velocity is not equal to 0.

P2+½v22 (ignore y)

### Moving cans

P0

Before air is blown in between

the cans, P0=P1; the cans remain

at rest and the air in between

the cans is at rest (0 velocity)

P1+½v12+gy1= Po

Top view

P1

case:

1: no blowing

2: blowing

P0

PHY 231

### Applications of Bernoulli’s law: moving a cart

No spin, no movement

Vair

PHY 231