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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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**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.**Buoyant Force:Occurs becausethe pressure in a fluid**increases with depth! P = ρg h (fluid atREST!!)**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**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”!**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 **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!**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!)**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%!)**Example 10-9: Hyrdometer**(ρO/ρF)= (Vdispl/V)**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**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**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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