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PHYSICS 231
Lecture 23: Buoyancy and fluid motion
Remco Zegers
Walk-in hour: Thursday 11:30-13:30 am
Helproom
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Previously...
P0
Pdepth=h =Pdepth=0+ gh
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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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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Unstable system
This system is not stable; the pressure at
different heights is not 0.
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The pressure at A and B is the same
if the fluid column is not moving
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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 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
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T=w-B=6*1*9.8
=58.8 N 9
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
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
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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
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
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
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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Last lecture...
P0
y
Pdepth=h =Pdepth=0+ gh
h
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=1m and y=3m what is X?
P0 B
y
h
Use Bernoulli’s law
A
PA+½vA2+gyA= PB+½vB2+gyB
At A: PA=P0 vA=? yA=y=3
At B: Pb=P0 vB=0 yB=y+h=4
P0+½vA2+g3=P0+g4
vA=(g/2)=2.2 m/s
x1
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vA
3m
0
Each water element of mass m has the same
velocity vA. Let’s look at one element m.
vA=(g/2)=2.2 m/s
In the horizontal direction:
x(t)=x0+v0xt+½at2=2.2t
x1
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
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Fluid flow
P1+½v12+gy1= P2+½v22+gy2
P1+½v12
= P2+½v22
(v22-v12)=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 to
be 30 m/s, what is v1? (=1E+03 kg/m3)
900-v12=2*(2.0E+5)/(1E+03) v1=22.3 m/s
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