PHYSICS 231 Lecture 21: Buoyancy and Fluid Motion

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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 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

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

Pressures at same heights are the same

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