1. Archimedes Principle & “Bath Legend”

2. 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, ∑F y = 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 Example 14.5: Archimedes Principle & “Bath Legend”

3. In Air: Newton’s 2 nd Law gives: ∑F y = T 1 – F g = 0. Result: T 1 = F g = 7.84 N = m crown g; m crown = 0.8 kg In Water: Newton’s 2 nd Law gives: ∑F = B + T 2 – F g = 0, or T 2 = F g – B = 6.84 N Newton’s 3 rd Law gives: T 2 = “weight” in water. From above, B = F g – T 2. Archimedes’ Principle says B = ρ water gV Above numbers give: B = 7.84 – 6.84 = 1.0 N So, ρ water gV = 1.0 N. Note: ρ water = 1000 kg/m 3. Solve for V & get V = 1.02  10 -4 m 3 Find the material of the crown from ρ crown = m crown /V = 7.84  10 3 kg/m 3 Density of gold = 19.3  10 3 kg/m 3. So crown is NOT gold!! (Density is near that of lead!)

4. Floating Iceberg! ρ ice /ρ water = 0.917, ρ sw /ρ water = 1.03 What fraction f a of iceberg is ABOVE water’s surface? Ice volume  V ice Volume submerged  V sw Volume visible  V = V ice - V sw Archimedes: B = ρ sw V sw g m ice g = ρ ice V ice g Newton: ∑F y = 0 = B - m ice g  ρ sw V sw = ρ ice V ice (V sw /V ice ) = (ρ ice /ρ sw ) = 0.917/1.03 = 0.89 f a = (V/V ice ) = 1 - (V sw /V ice ) = 0.11 (11%!)

5. Example: Moon Rock in Water In air, a moon rock weighs W = m r g = 90.9 N. So it’s mass is m r = 9.28 kg. In water it’s “Apparent weight” is W´ = m a g = 60.56 N. So, it’s “apparent mass” is m a = 6.18 kg. Find the density ρ r of the moon rock. ρ water = 1000 kg/m 3 Newton’s 2 nd Law: W´= ∑F y = W – B = m a g. B = Buoyant force on rock. Archimedes’ Principle: B = ρ water Vg. Combine (g cancels out!): m r - ρ water V = m a. Algebra: V = (m r - m a )/ρ water = [(9.28 – 6.18)/1000] = 3.1  10 -4 m 3 Definition of density in terms of mass & volume gives: ρ r = (m r /V) = 2.99  10 3 kg/m 3

6. 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: ∑F y = 0  B = W He + W load B = (m He + m)g, Note: m He = ρ He V Archimedes: B = ρ air Vg  ρ air Vg = (ρ He V + m)g  V = m/(ρ air - ρ He ) Table: ρ air = 1.29 kg/m 3, ρ He = 0.18 kg/m 3  V = 160 m 3 B

7. Example: (Variation on previous example) Spherical He balloon. r = 7.35 m. V = (4πr 3 /3) = 1663 m 3 m balloon = 930 kg. What cargo mass m cargo can balloon lift? Newton:∑F y = 0 0 = B - m He g - m balloon g - m cargo g Archimedes: B = ρ air Vg Also: m He = ρ He V, ρ air = 1.29 kg/m 3, ρ He = 0.179 kg/m 3  0 = ρ air V - ρ He V - m balloon - m cargo  m cargo = 918 kg