Sect 10 7 BuoyancyArchimedes Principle Experimental facts Objects

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.

• Buoyant Force: Occurs because the pressure in a fluid increases with depth! P = ρg h (fluid at REST!!)

Archimedes Principle The total (upward) buoyant force FB on an object of volume V completely or partially submerged in a fluid with density ρF: FB = ρFVg (1) ρFV m. F 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 F B = m Fg (2) FB = weight of fluid displaced by the object! (1) or (2) Archimedes Principle Proved for cylinder. Can show valid for any shape

Archimedes Principle

• 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 & “Bath Legend”

• Archimedes Principle: Valid for floating objects F B = m Fg = ρFVdispl g (m. F = mass of fluid displaced, Vdispl = volume displaced) W = m O g = ρO VO g (m. O = 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 = ρsw. Vdisplg miceg = ρice. VOg ∑Fy= 0 = FB - miceg ρsw. Vdispl = ρice. VO (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 = ρwater. Vg. Combine (g cancels out!): mr - ρwater. V = ma. Algebra & use definition of ρ: V = (mr - ma)/ρwater. ρ = (mr/V) = 2. 99 103 kg/m 3

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 = (m. He + m)g , Note: m. He = ρHe. V Archimedes: FB = 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

Prob. 25: (Variation on example 10 -10) Spherical He balloon. r = 7. 35 m. V = (4πr 3/3) = 1663 m 3 mballoon = 930 kg. What cargo mass mcargo can balloon lift? ∑Fy= 0 0 = Fbouy - m. Heg - mballoon g - mcargog Archimedes: Fbouy = ρ air. Vg Also: m. He = ρHe. V, ρ air = 1. 29 kg/m 3, ρHe = 0. 179 kg/m 3 0 = ρair. V - ρHe. V - mballoon - mcargo = 918 kg