PHY 2311

PHYSICS 231Lecture 21: Buoyancy and Fluid

Motion

Remco ZegersWalk-in hour: Tue 4-5 pm

Helproom

PHY 2312

Previously

Solids: LA

FL

LL

AFY

0

0/

/ Young’s modulus

xA

Fh

hx

AFS

/

/Shear modulus

pressureP

VV

P

VV

AFB

00 //

/ Bulk modulusAlso fluids

P=F/A (N/m2=Pa) Fpressure-difference=PA=M/V (kg/m3)

General:

Pascal’s principle: a change in pressure applied to a fluid that is enclosed is transmitted to the wholefluid and all the walls of the container that hold the fluid.

PHY 2313

Pascal’s principle

Pascal’s principle: a change in pressure applied to a fluid that is enclosed is transmitted to the wholefluid and all the walls of the container that hold the fluid.

HOLDS FOR A FLUID FULLY ENCLOSED ONLY

PHY 2314

Pascal’s principle

In other words then before: a change in pressure applied to a fluid that is enclosed in transmitted to the wholefluid and all the walls of the container that hold the fluid.

P=F1/A1=F2/A2

If A2>>A1 thenF2>>F1.

So, if we apply a smallforce F1, we can exerta very large Force F2.

Hydraulic press demo

PHY 2315

Pressure vs Depth

Horizontal direction:P1=F1/A P2=F2/A F1=F2 (no net force)So, P1=P2

Vertical direction:Ftop=PatmAFbottom=PbottomA-Mg=PbottomA-gAh

Since the column of water is not moving:Ftop-Fbottom=0PatmA=PbottomA-gAhPbottom=Patm+ gh

PHY 2316

Pressure and Depth:

Pdepth=h =Pdepth=0+ gh

Where:Pdepth=h: the pressure at depth hPdepth=0: the pressure at depth 0=density of the liquidg=9.81 m/s2

h=depthPdepth=0=Patmospheric=1.013x105 Pa = 1 atm =760 TorrFrom Pascal’s principle: If P0 changes then the pressures at all depths changes with the same value.

PHY 2317

A submarine

A submarine is built in such a way that it can stand pressuresof up to 3x106 Pa (approx 30 times the atmospheric pressure). How deep can it go?

PHY 2318

Does the shape of the container matter?

NO!!

PHY 2319

Pressure measurement.

The open-tube manometer.The pressure at A and B isthe same:P=P0+ghso 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 23110

Pressure Measurement: the mercury barometer

P0= mercurygh

mercury=13.6E+03 kg/m3

mercury,specific=13.6

PHY 23111

Pressures at same heights are the same

P0

P=P0+gh

h

P0

P=P0+gh

h

P=P0+gh

h

PHY 23112

Buoyant force: B

htop

hbottom

Ptop =P0+ wghtop

Pbottom =P0+ wghbottom

p = wg(htop-hbottom)

F/A = wghF = wghA=gVB =wgV=Mwaterg Fg=w=MobjgIf the object is not moving:B=Fg so: wgV=Mobjg

P0

-

Archimedes (287 BC) principle: the magnitude of the buoyantforce is equal to the weight of the fluid displaced by the object

PHY 23113

Comparing densities

B =fluidgV Buoyant force

w =Mobjectg=objectgV

Stationary: B=wobject= fluid

If object> fluid the object goes down!If object< fluid the object goes up!

PHY 23114

A floating object

h

A

B

w

w=Mobjectg=objectVobjectg

B=weight of the fluid displaced by the object =Mwater,displacedg = waterVdisplacedg = waterhAg h: height of the object under water!

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

h= objectVobject/(waterA) only useable if part of the objectis above the water!!

PHY 23115

question?? N

10N of water insidethin hollow sphere

A hollow sphere with negligible weightif not filled, is filled with water andhung from a scale. The weight is 10 N.It is then submerged. What is theweight read from the scale?

a) 0 Nb) 5 Nc) 10 Nd) 20 Ne) impossible to tell

PHY 23116

An example?? N

1 kg of water insidethin hollow sphere

A)?? N

7 kg iron sphere ofthe same dimensionas in A)

B)

Two weights of equal size and shape, but different mass aresubmerged in water. What are the weights read out?

PHY 23117

Another one

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

equation of continuity

A1,1 A2,2

v1v21

2

the mass flowing into area 1 (M1) must be the same as themass flowing into area 2 (M2), else mass would accumulate in the pipe).M1= M2 1A1x1= 2A2x2 (M=V=Ax)1A1v1t =2A2v2t (x=vt)1A1v1 =2A2v2

if is constant (liquid is incompressible) A1v1 =A2v2

x1

x2

PHY 23119

Bernoulli’s equation

W1=F1x1=P1A1 x1=P1VW2=-F2x2=-P2A2 x2=-P2VNet Work=P1V-P2V

m: transported fluid mass

same

KE=½mv22-½mv1

2 & PE=mgy2-mgy1

Wfluid= KE+ PEP1V-P2V=½mv2

2-½mv12+ mgy2-mgy1 use =M/V and div. By V

P1-P2=½v22-½v1

2+ gy2- gy1

P1+½v12+gy1= P2+½v2

2+gy2

P+½v2+gy=constantP: pressure ½v2:kinetic Energy per unit volume

gy: potential energy per unit volume

Another conservation Law

PHY 23120

Moving cansP0

P0

Top view Before air is blown in betweenthe cans, P0=P1; the cans remainat rest and the air in between the cans is at rest (0 velocity)P1+½v1

2+gy1= PoP1

When air is blown in between thecans, the velocity is not equal to 0.P2+½v2

2 (ignore y)

case:1: no blowing2: blowing

PHY 23121

Applications of Bernoulli’s law: moving a cart

No spin, no movementVair