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PHYSICS 231
Lecture 24: Buoyancy and Fluid Motion
Remco Zegers
Walk-in hour: Monday 9:15-10:15 am
Helproom
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Solids:
General:
Previously
FL0
F/A
Young’s modulus
Y

L / L0 AL
F/A
Fh
Shear modulus
S

x / h Ax
F / A
P
Bulk modulus
B

V / V0
V / V0 Also fluids
P  pressure
P=F/A (N/m2=Pa)
=M/V (kg/m3)
Fpressure-difference=PA
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.
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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
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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.
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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?
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Does the shape of the container matter?
NO!!
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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
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Pressure Measurement: the mercury
barometer
P0= mercurygh
mercury=13.6E+03 kg/m3
mercury,specific=13.6
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Pressures at same heights are the same
P0
P0
h
P=P0+gh
h
h
P=P0+gh
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P=P0+gh
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P0
htop
hbottom
Buoyant force: B
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
Archimedes (287 BC) principle: the magnitude of the buoyant
force is equal to the weight of the fluid displaced by the object
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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!
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A floating object
A
B
w
h
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 object
is above the water!!
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A)
?? N
An example
B)
?? N
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?
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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
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equation of continuity
x2
x1
v1
1
A1,1
2
v2
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
W1=F1x1=P1A1 x1=P1V
W2=-F2x2=-P2A2 x2=-P2V
Net Work=P1V-P2V
same
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 Another conservation Law
P+½v2+gy=constant
P: pressure ½v2:kinetic Energy per unit volume
gy: potential energy per unit volume
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P0
Moving cans
Top view
P1
case:
1: no blowing
2: blowing
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
When air is blown in between the
cans, the velocity is not equal to 0.
P2+½v22 (ignore y)
Bernoulli’s law:
P1+½v12+gy1= P2+½v22+gy2
P0=P2+½v22 so P2=P0-½v22
So P2<P0
Because of the pressure difference
left and right of each can, they move inward
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Applications of Bernoulli’s law: moving a cart
No spin, no movement
Vair
V2=Vair-v
P2
Spin and movement
P1 V1=Vair+v
Near the surface of the rotating cylinder: V1>V2
P1+½v12= P2+½v22
P1-P2= ½[(vair-v)2-(vair+ v) 2]
P2>P1 so move to the left
P1-P2= ½(v22- v12)
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hole in a tank
P0
y
h
Pdepth=h =Pdepth=0+ gh
If h=1m & y=3m what is x?
Assume that the holes are small
and the water level doesn’t drop
noticeably.
x
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If h=1 m and y=3 m what is X?
P0 B
y
h
A
x1
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