Clicker Question
Room Frequency BA
A force of 200 lbs is applied as FIN. Assume that AIN is 1 cm2 and
AOUT is 20 cm2. What’s POUT?
A) 200 lb/cm2 B) 400 lb/cm2
C) 2000 lb/cm2 D) 4000 lb/cm2
By Pascal’s Principle, the applied pressure is the same every in the
fluid (at the same depth).
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Announcements
• CAPA assignment #13 is due on Friday at 10 pm.
• This week in Section: Assignment #6
• Start reading Chapter 11 on Vibrations and Waves
• Prof. Nagle will have his regular office hours this
afternoon at 2-3pm (in his office).
• Prof. Uzdensky’s office hours Tuesday 11am are
canceled this week
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Pressure as a function of depth
Ph  Patm  Water hg
Pressure only depends on the depth; at
a given depth the pressure is the same.
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Forces on a Small Box of Water
Consider again a small imaginary cubic box at depth h
We argued last time that in static fluids,
the net force on the box must be zero.
Again we conclude that the side forces
all cancel and only the top and bottom
forces are left.
Fnet = Ftop + Fbottom = 0
Using our depth formula we can find that
h
Δy
Fnet   Water g(h  y)A  Water ghA  mWater g  0
The pressure difference between top and bottom
Water gyA  mWater g
is just the weight of the water in the box
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Forces on a Small Box of Gold
Now suppose the cubic box is a block of gold at depth h
The fluid outside the block is the undisturbed, so the
pressures are not changed. The net force however is no
longer necessarily zero!!!
Again we conclude that the side forces
all cancel and only the top and bottom
forces are left.
Fnet = Ftop + Fbottom
Using our depth formula we can find that
h
Δy
Fnet   Water g(h  y)A  Water ghA  mGold g
The pressure difference between top and bottom
Water gyA  mWater g
is just the weight of the water in the box
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Buoyant Force
So we conclude that the net force is
Fnet  Water gyA  mGold g  mwater g  mGold g
The fluid exerts an upward force of mwaterg
on the gold cube. This is known as the
buoyant force.
For some general fluid, we can write
FBuoyancy  Fluid gyA  mFluid g
You can show that this is true for any shape
volume V!
FBuoyancy  Fluid gV  mFluid g
h
Δy
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Archimedes Principle
A solid body, either partially or totally submerged in a fluid, will
experience an upward buoyant force equal to the weight of the
displaced fluid.
FBuoyancy  Fluid gV  mFluid g
Let’s check with a demonstration!
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Clicker Question
Room Frequency BA
Two identical bricks are held under water in a bucket. One of the
bricks is lower in the bucket than the other. The upward buoyant
force on the lower brick is…..
The weight of
displaced fluid does
not depend on depth
Fbuoy  m fluid g   fluidVg
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Room Frequency BA
Clicker Question
A solid piece of plastic of volume V and
density ρplastic would ordinarily float in water,
but it is held under water by a string tied to the
bottom of bucket as shown. (The density of
water is ρwater.)
FB
What is the buoyant force on the plastic?
A)
B)
C)
D)
E)
Zero
ρplastic V
ρwater V
ρwater V g
ρplastic V g
Fbuoy  m fluid g   fluidVg
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Motion of Submerged objects
Fbuoyant
When will an object sink or float?
Fbuoy = mfluidg
Fg
Δy
= mobjectg
Fnet  mFluid g  mObject g
Fg
Object will rise:
Object will sink:
Object will be in equilibrium:
Fbuoy > Fg
Fbuoy < Fg
Fbuoy = Fg
(mfluid > mobject)
(mfluid < mobject)
(mfluid = mobject)
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Motion of Submerged objects
Since Fbuoy and Fg both depend on the
same volume V we can also write
Object will rise:
Fbuoyant
ρfluid > ρobject
Δy
Object will sink: ρfluid < ρobject
Object will be in equilibrium:
ρfluid = ρobject
Fg
Diet vs Regular Pepsi!
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Example. A block of copper (Cu) with mass 400 g and density
ρCU = 8.9 g/cm3 is suspended by a string while under water. How
does the tension in the string compare with the weight of the
copper block?
0  Fnet   F  FT  FB  Fg
 FT  m fluid g  mCU g
FT  g(mCU  m fluid )  g(0.4kg  (0.4 / 8.9)kg)  g(0.355kg)  0.345N
Fg  (0.4kg)g  0.392N
FT 0.345N

 88%
Fg 0.392N
The buoyant force helps support the weight of the block.
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Room Frequency BA
Clicker Question
A carefully-made sphere, when placed
under water, remains at rest, in equilibrium
as shown above. How does the magnitude
of the upward buoyant force FB compare to
the weight mg of the sphere?
FB
mg
A) FB > mg
A) FB = mg
A) FB < mg
Net force = 0
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Clicker Question
Room Frequency BA
A helium-filled balloon of volume V can
carry a total mass M (M includes the mass
of the rubber balloon but not the mass of
the Helium inside).
What is the correct expression for the
buoyant force FB on the balloon?
A)
B)
C)
D)
ρair V g
Weight of displaced
fluid
ρhelium V g
Mg
(ρair– ρhelium) V g
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Room Frequency BA
Clicker Question
A helium-filled balloon of volume V can
carry a total mass M (M includes the mass
of the rubber balloon but not the mass of
the Helium inside).
What is the correct expression for the net
force on the balloon?
A)
B)
C)
D)
ρair V g -Mg
(ρair+ ρhelium) V g - Mg
ρhelium V g - Mg
(ρair– ρhelium) V g - Mg
Helium has weight!
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Floating objects
Floating Object: Displaces a mass of water equal to its mass.
FB  mg
 fluidVdisplaced g  objectVobject g
mass of fluid displaced =
mass of the object
or
Vdisplaced object

Vobject
 fluid
Vdispl
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Room Frequency BA
Clicker Question
What fraction of a piece of Aluminum will be
submerged if it is placed in a tank of Mercury?
(ρAl = 2.7 x 103 kg/m3, ρHg = 13.6 x 103 kg/m3)
A) 20%
B) 40%
C) 60%
D) 80%
Vdisplaced
Vdisplaced
Vobject
object

 fluid
 object 
 2.7x10 3 

Vobject  
V
 0.2Vobject

3  object
 13.6x10 
  fluid 
The volume of the Mercury displaced is the
amount of the Aluminum that will be submerged.
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Waterfall Death Trap
Every year several people drown near/underneath waterfalls!
Why?
Foamy, frothy water is much less dense than regular water!!!
Hence the force of buoyancy is much less! Humans rapidly sink
deep under the surface.
Also “explains” mysterious disappearance of ships in the
Bermuda Triangle: Seabed releases of large quantities of gas
(“avalanches” at the edge of the continental shelf) just under the
ship. The ship immediately sinks below the surface.
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