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Understanding Archimedes' Principle Through Classic Examples

Archimedes' Principle explains buoyancy, demonstrated through historical tales like Archimedes and the crown, as well as modern examples including moon rocks in water and helium balloons. The king sought proof if the crown was pure gold by calculating its weight in air and water, revealing its true material through density comparisons. The principle also applies to iceberg visibility and the lifting capacity of helium balloons in air. Understanding these principles of physics is crucial for grasping fluid dynamics and material properties.

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Understanding Archimedes' Principle Through Classic Examples

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  1. Archimedes Principle & “Bath Legend”

  2. Example 14.5: Archimedes Principle & “Bath Legend” • Archimedes was (supposedly) asked by the King, “Is the crown made of pure gold?”. To answer, he weighed the crown in air & completely submerged in water. See figure. He compared the weights, used Archimedes’ Principle & found the answer. • Weight in air = 7.84 N • Weight in water (submerged)= 6.84 N • Use Newton’s 2nd Law, ∑Fy= 0in both cases, do some algebra & find that the buoyant force B will equal the apparent “weight loss” • Difference in scale readings will be the buoyant force

  3. In Air: Newton’s 2nd Law gives: ∑Fy = T1 – Fg = 0.Result: T1 = Fg = 7.84 N = mcrowng; mcrown = 0.8 kg • In Water: Newton’s 2nd Law gives: ∑F = B + T2 – Fg = 0, or T2 = Fg – B = 6.84 N Newton’s 3rd Law gives: T2 = “weight” in water. From above, B = Fg – T2. • Archimedes’ Principle says B = ρwatergV Above numbers give: B = 7.84 – 6.84 = 1.0 N So, ρwatergV = 1.0 N. Note: ρwater = 1000 kg/m3. Solve for V & get V = 1.02  10-4 m3 • Find the material of the crown from ρcrown = mcrown/V = 7.84  103 kg/m3 Density of gold = 19.3  103 kg/m3 . So crown is NOT gold!! (Density is near that of lead!)

  4. Floating Iceberg! ρice/ρwater= 0.917, ρsw/ρwater= 1.03 What fraction fa of iceberg is ABOVE water’s surface? Ice volume Vice Volume submerged Vsw Volume visible  V = Vice -Vsw Archimedes:B = ρswVswg miceg = ρiceViceg Newton:∑Fy= 0 = B - miceg  ρswVsw = ρiceVice (Vsw/Vice) = (ρice/ρsw)= 0.917/1.03 = 0.89 fa = (V/Vice) = 1 - (Vsw/Vice) = 0.11 (11%!)

  5. Example: Moon Rock in Water • In air, a moon rock weighs W = mrg = 90.9 N. So it’s mass is mr = 9.28 kg.In water it’s “Apparent weight” is W´ = mag = 60.56 N.So, it’s “apparent mass” is ma = 6.18 kg.Find the density ρr of the moon rock. ρwater = 1000 kg/m3 • Newton’s 2nd Law:W´= ∑Fy = W – B = mag. B=Buoyant force on rock. • Archimedes’ Principle:B= ρwaterVg. Combine (g cancels out!):mr - ρwaterV = ma.Algebra: V = (mr - ma)/ρwater = [(9.28 – 6.18)/1000] = 3.1  10-4 m3Definition of density in terms of mass & volume gives: ρr= (mr/V) = 2.99  103 kg/m3

  6. Example: Helium Balloon B • Air is a fluid  There is a 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? Newton:∑Fy= 0 B = WHe + Wload B = (mHe + m)g, Note:mHe = ρHeV Archimedes:B = ρairVg  ρairVg = (ρHeV + m)g  V = m/(ρair - ρHe) Table: ρair = 1.29 kg/m3 , ρHe = 0.18 kg/m3  V = 160 m3

  7. Example:(Variation on previous example) Spherical He balloon. r = 7.35 m. V = (4πr3/3) = 1663 m3 mballoon = 930 kg. What cargo mass mcargo can balloon lift? Newton:∑Fy= 0 0 = B- mHeg - mballoon g - mcargog Archimedes:B = 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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