Post on 02-Feb-2016
description
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