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