PHYSICS 231 Lecture 23: Buoyancy and fluid motion

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PHYSICS 231Lecture 23 Buoyancy and fluid motion

• Remco Zegers
• Walk-in hour Thursday 1130-1330 am
• Helproom

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Previously...
P0
Pdepthh Pdepth0 ?gh
Pascals 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
h
h
PP0?gh
PP0?gh
PP0?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 sameif the fluid
column is not moving
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Buoyant force B
P0
Ptop P0 ?wghtop Pbottom P0 ?wghbottom ?p
?wg(htop-hbottom) F/A ?wg?h F
?wg?hA?gV B ?wgVMwaterg FgwMobjg If the
object is not moving BFg so ?wgVMobjg
-
htop
hbottom
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 Bw ?object ?fluid If ?objectgt
?fluid the object goes down! If ?objectlt ?fluid
the object goes up!
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A floating object
A
wMobjectg?objectVobjectg Bweight of the fluid
displaced by the object Mwater,displacedg
?waterVdisplacedg ?waterhAg h height of
the object under water!
h
B
w
The object is floating, so there is no net force
(Bw) ?objectVobject ?waterVdisplaced h
?objectVobject/(?waterA) only useable if part of
the object is above the water!!
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An example
?? N
A)
?? N
B)
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?
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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?
?water1.0E03 kg/m3
A) h ?objectVobject/(?waterA)
hMobject/(1.0E0320.5)2.0/1.0E032.0E-03m2mm
B) if the objects sinks the mattress is just
completely submerged hthickness of
mattress. 0.08(Mweight2.0)/(1.0E0320.5)
So Mweight78 kg
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Bernoullis equation
same
W1F1?x1P1A1 ?x1P1V W2-F2?x2-P2A2
?x2-P2V Net WorkP1V-P2V m transported fluid
mass
?KE½mv22-½mv12 ?PEmgy2-mgy1 Wfluid ?KE
?PE P1V-P2V½mv22-½mv12 mgy2-mgy1 use ?M/V
and div. By V P1-P2½?v22-½?v12 ?gy2-
?gy1 P1½?v12?gy1 P2½?v22?gy2
P½?v2?gyconstant P pressure ½?v2kinetic
Energy per unit volume ?gy potential energy
per unit volume
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Moving cans
P0
Before air is blown in between the cans, P0P1
the cans remain at rest and the air in between
the cans is at rest (0 velocity) P1½?v12?gy1
Po
Top view
P1
Bernoullis law P1½?v12?gy1
P2½?v22?gy2 P0P2½?v22 so P2P0-½?v22 So P2ltP0
Because of the pressure difference left and
right of each can, they move inward
P0
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Last lecture...
P0
Pdepthh Pdepth0 ?gh
h
y
If h1m y3m what is x? Assume that the holes
are small and the water level doesnt
drop noticeably.
x
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If h1m and y3m what is X?
P0
B
Use Bernoullis law
h
PA½?vA2?gyA PB½?vB2?gyB At A PAP0 vA?
yAy3 At B PbP0 vB0 yByh4 P0½?vA2?g3P0
?g4 vA?(g/2)2.2 m/s
y
A
x1
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Each water element of mass m has the
same velocity vA. Lets look at one element
m. vA?(g/2)2.2 m/s
vA
In the horizontal direction x(t)x0v0xt½at22.2
t In the vertical direction y(t)y0v0yt½at23-
0.5gt2 0 when the water hits the ground,
so t0.78 s so x(0.78)2.20.781.72 m
3m
0
x1
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Fluid flow
P1½?v12?gy1 P2½?v22?gy2 P1½?v12
P2½?v22 (v22-v12)2(P1-P2)/?
If P14.0105 Pa, P22.0105 Pa and by counting
the amount of water coming from the right v2 is
found to be 30 m/s, what is v1? (?1E03
kg/m3) 900-v122(2.0E5)/(1E03) v122.3 m/s