Archimedes Principle Bath Legend Example 14 5 Archimedes

Archimedes Principle & “Bath Legend”

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 2 nd Law, ∑Fy = 0 in 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

• In Air: Newton’s 2 nd Law gives: ∑Fy = T 1 – Fg = 0. Result: T 1 = Fg = 7. 84 N = mcrowng; mcrown = 0. 8 kg • In Water: Newton’s 2 nd Law gives: ∑F = B + T 2 – Fg = 0, or T 2 = Fg – B = 6. 84 N Newton’s 3 rd Law gives: T 2 = “weight” in water. From above, B = Fg – T 2. • Archimedes’ Principle says B = ρwaterg. V Above numbers give: B = 7. 84 – 6. 84 = 1. 0 N So, ρwaterg. V = 1. 0 N. Note: ρwater = 1000 kg/m 3. Solve for V & get V = 1. 02 10 -4 m 3 • Find the material of the crown from ρcrown = mcrown/V = 7. 84 103 kg/m 3 Density of gold = 19. 3 103 kg/m 3. So crown is NOT gold!! (Density is near that of lead!)

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 = ρsw. Vswg miceg = ρice. Viceg Newton: ∑Fy= 0 = B - miceg ρsw. Vsw = ρice. Vice (Vsw/Vice) = (ρice/ρsw) = 0. 917/1. 03 = 0. 89 fa = (V/Vice) = 1 - (Vsw/Vice) = 0. 11 (11%!)

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/m 3 • Newton’s 2 nd Law: W´= ∑Fy = W – B = mag. B = Buoyant force on rock. • Archimedes’ Principle: B = ρwater. Vg. Combine (g cancels out!): mr - ρwater. V = ma. Algebra: V = (mr - ma)/ρwater = [(9. 28 – 6. 18)/1000] = 3. 1 10 -4 m 3 Definition of density in terms of mass & volume gives: ρr = (mr/V) = 2. 99 103 kg/m 3

Example: Helium Balloon • 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 = (m. He + m)g, Note: m. He = ρHe. V Archimedes: B = ρair. Vg = (ρHe. V + m)g V = m/(ρ air - ρ He) Table: ρair = 1. 29 kg/m 3 , ρHe = 0. 18 kg/m 3 V = 160 m 3 B

Example: (Variation on previous example) Spherical He balloon. r = 7. 35 m. V = (4πr 3/3) = 1663 m 3 mballoon = 930 kg. What cargo mass mcargo can balloon lift? Newton: ∑Fy= 0 0 = B - m. Heg - mballoon g - mcargog Archimedes: B = ρair. Vg Also: m. He = ρHe. V, ρair = 1. 29 kg/m 3, ρHe = 0. 179 kg/m 3 0 = ρair. V - ρHe. V - mballoon - mcargo = 918 kg