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Specific Gravity – The REST of the story… PowerPoint Presentation
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Specific Gravity – The REST of the story…

Specific Gravity – The REST of the story…

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Specific Gravity – The REST of the story…

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  1. Eureka! Specific Gravity – The REST of the story… AKA “Owl” Archimedes saved his life by taking a bath!

  2. Hiero of Syracuse (really!)… challenged Archimedes to determine if his new crown was made of pure gold.

  3. Archimedes realized (eventually) that the crown would weigh LESS when weighed in water!

  4. Fg FT What is going on? FT = Fg

  5. Has the gravity stopped pulling as hard on the crown? Fg

  6. Fg FT No! So… FT > Fg

  7. Fg F? FT Therefore… F? FT + = Fg

  8. Fg F? FT This new force is called… Oh, BOY! F? FT + = Fg

  9. Fg F? FT the BUOYANT force! Oh, BOY! F? FT + = Fg

  10. Fg FB FT the BUOYANT force! Oh, BOY! FB FT + = Fg

  11. Fg FB FT Hmm... What causes the BUOYANT force? FB FT + = Fg

  12. To keep this simple, let’s look at a cylinder in water.

  13. Does the fluid exert any force on the cylinder?

  14. FB = |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

  15. |FB|= Fgfluid Since the fluid was at rest… = ρfluidgV = mfluidg Remember that: h1 { ρ = m/V } h2 h = h2 - h1 { m = ρV }

  16. So | FB |= | Fgfluid | Oh, BOY! h1 h2 h = h2 - h1

  17. Is this always true? So | FB |= | Fgfluid | h1 h2 h = h2 - h1

  18. Let’s consider a still fluid: 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.

  19. | FB |= | Fgfluid | Oh, BOY! So the buoyant force on an object equals the weight of the fluid displaced, regardless of the shape of the object.

  20. Now let’s look at some situations (and assume that down is negative):

  21. In all cases, Fnet = Fgobject + FB

  22. Fnet = Fgobject + FB If Fnet is positive, then the object is positively buoyant and it will rise upward. FB Fg

  23. Fnet = Fgobject + FB If Fnet is negative, then the object is negatively buoyant and it will sink downward. FB Fg

  24. If Fnet 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

  25. ALWAYS upward Please note, FBwas ALWAYS______________ FB FB Fg Fg FB Fg

  26. float (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

  27. Fg FB FT Hmm... What does all of this have to do with SG? FB FT + = Fg

  28. Fg FB FT Oh, yeah... Remember that SG = ρo/ρf FB FT + = Fg

  29. SG = mo/Vo/mf/Vf SG = mo/mf So… SG = ρo/ρf SG = mog/mfg SG = Fgo/Fgf SG = Fgo/|FB| SG = Fgo/ΔFgo

  30. Eureka! 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

  31. SG = 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.

  32. |FB| = |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.

  33. ρ1/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

  34. ρ1/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