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Sect. 10-7: Buoyancy/Archimedes Principle

Sect. 10-7: Buoyancy/Archimedes Principle. Experimental facts: Objects submerged (or partially submerged) in a fluid APPEAR to “weigh” less than in air. When placed in a fluid, many objects float! Both are examples of BUOYANCY.

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Sect. 10-7: Buoyancy/Archimedes Principle

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  1. Sect. 10-7: Buoyancy/Archimedes Principle • Experimental facts: • Objects submerged (or partially submerged) in a fluid APPEARto “weigh” less than in air. • When placed in a fluid, many objects float! • Both are examples of BUOYANCY.

  2. Buoyant Force:Occurs becausethe pressure in a fluid increases with depth! P = ρg h (fluid atREST!!)

  3. Archimedes Principle  The total (upward) buoyant force FBon an objectof volume V completely or partially submerged in a fluid with density ρF: FB = ρFVg (1) ρFV  mF  Mass of fluid which would take up same volume as object, if object were not there. (Mass of fluid that used to be where object is!)  Upward buoyant force FB = mFg (2) FB = weight of fluid displaced by the object! (1) or (2)  Archimedes Principle Proved for cylinder. Can show valid for any shape

  4. Archimedes Principle

  5. Object, mass m in a fluid. Vertical forces are buoyant force, FB& weight, W = mg • “Apparent weight” = net downward force: W´ ∑Fy = W - FB < W Object appears “lighter”!

  6. Archimedes Principle & “Bath Legend”

  7. Archimedes Principle: Valid for floating objects FB = mFg = ρFVdispl g (mF = mass of fluid displaced, Vdispl = volume displaced) W = mOg = ρOVOg (mO = mass of object, VO = volume of object) Equilibrium:  ∑Fy = 0 = FB - W 

  8. Archimedes Principle: Floating objects Equilibrium:  ∑Fy = 0 = FB -W  FB = W or ρFVdispl g = ρOVOg  f = (Vdispl/V) = (ρO/ρF) (1) f  Fraction of volume of floating object which is submerged. Note: If fluid is water, right side of (1) is specific gravity of object!

  9. Example: Floating log (a) Fully submerged: FB > W  ∑Fy = FB -W = ma (It moves up!) (b) Floating: FB = W or ρFVg = ρOVg  ∑Fy = FB -W = 0 (Equilibrium: It floats!)

  10. Prob. 33: Floating Iceberg! (SG)ice= 0.917 (ρice/ρwater), (SG)sw= 1.025  (ρsw/ρwater) What fraction fa of iceberg is ABOVE water’s surface? Iceberg volume VO Volume submerged Vdispl Volume visible  V = VO -Vdispl Archimedes: FB = ρswVdisplg miceg = ρiceVOg ∑Fy= 0 = FB - miceg  ρswVdispl = ρiceVO (Vdispl/VO)= (ρice/ρsw)= [(SG)ice/(SG)sw] = 0.917/1.025 = 0.89 fa = (V/VO) = 1 - (Vdispl/VO) = 0.11 (11%!)

  11. Example 10-9: Hyrdometer (ρO/ρF)= (Vdispl/V)

  12. Prob. 22:Moon Rock in water Moon rock mass mr = 9.28 kg. Volume V is unknown.  Weight W = mrg = 90.9 N Put rock in water & find “apparent weight” W´ = mag  “apparent mass” ma = 6.18 kg  W´ = 60.56 N. Density of rock = ρ (mr/V) =? W´ ∑Fy = W - FB = mag . FB = Buoyant force on rock. Archimedes: FB = ρwaterVg. Combine (g cancels out!): mr - ρwaterV = ma . Algebra & use definition of ρ: V = (mr - ma)/ρwater.ρ= (mr/V) = 2.99 103 kg/m3

  13. Example 10-10:Helium Balloon • Air is a fluid  Buoyant force on objects in it. Some float in air. • What volume V of He is needed to lift a load of m=180 kg? ∑Fy=0 FB = WHe + Wload FB = (mHe + m)g , Note: mHe = ρHeV Archimedes: FB = ρairVg  ρairVg = (ρHeV + m)g  V = m/(ρair - ρHe) Table: ρair = 1.29 kg/m3 , ρHe = 0.18 kg/m3  V = 160 m3

  14. Prob. 25:(Variation on example 10-10) Spherical He balloon. r = 7.35 m. V = (4πr3/3) = 1663 m3 mballoon = 930 kg. What cargo mass mcargo can balloon lift? ∑Fy= 0 0 = Fbouy - mHeg - mballoon g - mcargog Archimedes: Fbouy = ρairVg Also: mHe = ρHeV, ρair = 1.29 kg/m3, ρHe = 0.179 kg/m3  0 = ρairV - ρHeV - mballoon - mcargo  mcargo = 918 kg

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