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Eureka!. Specific Gravity – The REST of the story…. AKA “Owl” Archimedes saved his life by taking a bath!. Hiero of Syracuse (really!)…. challenged Archimedes. to determine if his new crown was made of pure gold.

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Specific gravity the rest of the story

Eureka!

Specific Gravity – The REST of the story…

AKA

“Owl” Archimedes saved his life by taking a bath!


Hiero of syracuse really

Hiero of Syracuse (really!)…

challenged Archimedes

to determine if his new crown was made of pure gold.



What is going on

F LESS when weighed in water!g

FT

What is going on?

FT

=

Fg



No so

F LESS when weighed in water!g

FT

No! So…

FT

>

Fg


Therefore

F LESS when weighed in water!g

F?

FT

Therefore…

F?

FT

+

=

Fg


This new force is called

F LESS when weighed in water!g

F?

FT

This new force is called…

Oh, BOY!

F?

FT

+

=

Fg


The buoyant force

F LESS when weighed in water!g

F?

FT

the BUOYANT force!

Oh, BOY!

F?

FT

+

=

Fg


The buoyant force1

F LESS when weighed in water!g

FB

FT

the BUOYANT force!

Oh, BOY!

FB

FT

+

=

Fg


What causes the buoyant force

F LESS when weighed in water!g

FB

FT

Hmm...

What causes the BUOYANT force?

FB

FT

+

=

Fg




Because h 2 h 1 the force up is greater than the force down

F LESS when weighed in water!B = |Fup| - |Fdown|

Because h2 > h1, the force up is greater than the force down.

= PbotAbot - PtopAtop

= ρgh2Abot – ρgh1Atop

{ Abot = Atop }

= ρgA(h)

h1

= ρgV

h2

h = h2 - h1


Since the fluid was at rest

|F LESS when weighed in water!B|= Fgfluid

Since the fluid was at rest…

= ρfluidgV

= mfluidg

Remember that:

h1

{ ρ = m/V }

h2

h = h2 - h1

{ m = ρV }


So f b f g fluid

So | F LESS when weighed in water!B |= | Fgfluid |

Oh, BOY!

h1

h2

h = h2 - h1


So f b f g fluid1

Is this LESS when weighed in water!

always true?

So | FB |= | Fgfluid |

h1

h2

h = h2 - h1


Let s consider a still fluid

Let’s consider a still fluid: LESS when weighed in water!

The fluid was there already, and if we encase it in a plastic bag that has the same density as the fluid, it will continue to be there.


F b f g fluid

| F LESS when weighed in water!B |= | Fgfluid |

Oh, BOY!

So the buoyant force on an object equals the weight of the fluid displaced, regardless of the shape of the object.



In all cases f net f g object f b

In all cases, negative):Fnet = Fgobject + FB


If f net is positive then the object is positively buoyant and it will rise upward

F negative):net = Fgobject + FB

If Fnet is positive, then the object is positively buoyant and it will rise upward.

FB

Fg


If f net is negative then the object is negatively buoyant and it will sink downward

F negative):net = Fgobject + FB

If Fnet is negative, then the object is negatively buoyant and it will sink downward.

FB

Fg


If F negative):net is zero, then the object is neutrally buoyant and it will neither rise nor sink while it is under the surface.

Fnet = Fgobject + FB

FB

Fnet = 0 N

Fg


Please note f b was always

ALWAYS negative):

upward

Please note, FBwas ALWAYS______________

FB

FB

Fg

Fg

FB

Fg


float negative):

(i.e. The object would displace enough fluid such that Fnet = ________.

Please note also, if Fnetwere positive, the object would simply __________ if it were not held below the surface.

0 N


What does all of this have to do with sg

F negative):g

FB

FT

Hmm...

What does all of this have to do with SG?

FB

FT

+

=

Fg


Remember that sg o f

F negative):g

FB

FT

Oh, yeah...

Remember that SG = ρo/ρf

FB

FT

+

=

Fg


So sg o f

SG = negative):mo/Vo/mf/Vf

SG = mo/mf

So… SG = ρo/ρf

SG = mog/mfg

SG = Fgo/Fgf

SG = Fgo/|FB|

SG = Fgo/ΔFgo


Eureka! negative):

So… Archimedes realized that an object’s specific gravity (relative to the fluid used) equaled the weight of the object out of the water divided by the change of weight that occurred when the object was submerged in the fluid!

Oh, BOY!

SG = Fg/ΔFg


SG = negative):Fgo/ΔFgo

SG = Fgo/|Fgo– Fgo’|

SG = ρoVo/ρfVf

Note:

SG = ρo/ρf

If the fluid involved is water at 4ºC, then this ratio is the specific gravity for the object, not just a relative SG.


For floating objects objects sink enough that

|F negative):B| = |Fg|

ρo = 0.2 g/mL

For floating objects, objects sink enough that:

ρf.Vdisplaced.g=ρo.Vo.g

Vdisplaced/Vo = ρo/ρf

So 80% of the ball is above the water.


Hydrometers follow

ρ negative):1/ρ2 = V2/V1

Hydrometers follow:

ρ1/ρ2 = A2h2/A1h1

ρ1/ρ2 = h2/h1

h2/h1

Ex. .8/1= .8 /1

If placed into a denser fluid: .8/2= x /1


Hydrometers follow1

ρ negative):1/ρ2 = V2/V1

Hydrometers follow:

x = 0.4

ρ1/ρ2 = A2h2/A1h1

ρ1/ρ2 = h2/h1

Ex. .8/1= .8 /1

.8/2= x /1


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