PHY 2311

PHYSICS 231Lecture 23: Buoyancy and fluid

motion

Remco ZegersWalk-in hour: Thursday 11:30-13:30 am

Helproom

PHY 2312

Previously...

Pdepth=h =Pdepth=0+ gh

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.

P0

PHY 2313

Pressures at same heights are the same

P0

P=P0+gh

h

P0

P=P0+gh

h

P=P0+gh

h

PHY 2314

Unstable system

This system is not stable; the pressure atdifferent heights is not 0.

PHY 2315

The pressure at A and B is the sameif the fluid column is not moving

PHY 2316

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 2317

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 2318

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 2319

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 weight read out?

B= waterVdisplacedg w= sphereVsphereg A) B= waterVsphereg w= waterVsphereg so B=w and 0 N is read out!

B) B= waterVsphereg=Mwater sphere w= ironVsphereg=Miron sphereg=7Mwater sphereg

T=w-B=6*1*9.8 =58.8 N

PHY 23110

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

A) h= objectVobject/(waterA) h=Mobject/(1.0E+03*2*0.5)=2.0/1.0E+03=2.0E-03m=2mm

B) if the objects sinks the mattress is just completely submerged: h=thickness of mattress. 0.08=(Mweight+2.0)/(1.0E+03*2*0.5) So Mweight=78 kg

PHY 23111

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

PHY 23112

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

Bernoulli’s law: P1+½v1

2+gy1= P2+½v22+gy2

P0=P2+½v22 so P2=P0-½v2

2

So P2<P0 Because of the pressure differenceleft and right of each can, they move inward

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

2

PHY 23113

Last lecture...

Pdepth=h =Pdepth=0+ ghh

x

y If h=1m & y=3m what is x?Assume that the holes are smalland the water level doesn’t dropnoticeably.

P0

PHY 23114

h

x1

y

If h=1m and y=3m what is X?

Use Bernoulli’s law

PA+½vA2+gyA= PB+½vB

2+gyB

At A: PA=P0 vA=? yA=y=3At B: Pb=P0 vB=0 yB=y+h=4P0+½vA

2+g3=P0+g4

vA=(g/2)=2.2 m/s

A

P0 B

PHY 23115

Each water element of mass m has the samevelocity vA. Let’s look at one element m.vA=(g/2)=2.2 m/s

x1

3m

vA

In the horizontal direction:x(t)=x0+v0xt+½at2=2.2t

In the vertical direction:y(t)=y0+v0yt+½at2=3-0.5gt2

= 0 when the water hits the ground, so t=0.78 s

so x(0.78)=2.2*0.78=1.72 m

0

PHY 23116

Fluid flow

P1+½v12+gy1= P2+½v2

2+gy2

P1+½v12 = P2+½v2

2

(v22-v1

2)=2(P1-P2)/

If P1=4.0*105 Pa, P2=2.0*105 Pa and by counting the amount of water coming from the right v2 is found tobe 30 m/s, what is v1? (=1E+03 kg/m3)900-v1

2=2*(2.0E+5)/(1E+03) v1=22.3 m/s