2. 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.

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

5. Archimedes Principle  The total (upward) buoyant force F B on an object of volume V completely or partially submerged in a fluid with density ρ F : F B = ρ F Vg (1) ρ F V  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 F g (2) F B = weight of fluid displaced by the object! (1) or (2)  Archimedes Principle Proved for cylinder. Can show valid for any shape

7. Archimedes Principle

8. Object, mass m in a fluid. Vertical forces are buoyant force, F B & weight, W = mg “Apparent weight” = net downward force: W´  ∑F y = W - F B < W  Object appears “lighter”!

9. Archimedes Principle & “Bath Legend”

10. Archimedes Principle: Valid for floating objects F B = m F g = ρ F V displ g (m F = mass of fluid displaced, V displ = volume displaced) W = m O g = ρ O V O g (m O = mass of object, V O = volume of object) Equilibrium:  ∑F y = 0 = F B - W 

11. Archimedes Principle: Floating objects Equilibrium:  ∑F y = 0 = F B -W  F B = W or ρ F V displ g = ρ O V O g  f = (V displ /V) = (ρ O /ρ F ) (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!

12. Example: Floating log (a) Fully submerged: F B > W  ∑F y = F B -W = ma (It moves up!) (b) Floating: F B = W or ρ F Vg = ρ O Vg  ∑F y = F B -W = 0 (Equilibrium: It floats!)

13. Prob. 33: Floating Iceberg! (SG) ice = 0.917  (ρ ice /ρ water ), (SG) sw = 1.025  (ρ sw /ρ water ) What fraction f a of iceberg is ABOVE water’s surface? Iceberg volume  V O Volume submerged  V displ Volume visible  V = V O - V displ Archimedes: F B = ρ sw V displ g m ice g = ρ ice V O g ∑F y = 0 = F B - m ice g  ρ sw V displ = ρ ice V O (V displ /V O )= (ρ ice /ρ sw ) = [(SG) ice /(SG) sw ] = 0.917/1.025 = 0.89 f a = (V/V O ) = 1 - (V displ /V O ) = 0.11 (11%!)

14. Example 10-9: Hyrdometer (ρ O /ρ F )= (V displ /V)

15. Prob. 22: Moon Rock in water Moon rock mass m r = 9.28 kg. Volume V is unknown.  Weight W = m r g = 90.9 N Put rock in water & find “apparent weight” W´ = m a g  “apparent mass” m a = 6.18 kg  W´ = 60.56 N. Density of rock = ρ  (m r /V) = ? W´  ∑F y = W - F B = m a g. F B = Buoyant force on rock. Archimedes: F B = ρ water Vg. Combine (g cancels out!): m r - ρ water V = m a. Algebra & use definition of ρ: V = (m r - m a )/ρ water. ρ = (m r /V) = 2.99  10 3 kg/m 3

16. 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? ∑F y =0 F B = W He + W load F B = (m He + m)g, Note: m He = ρ He V Archimedes: F 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

17. Prob. 25: (Variation on example 10-10) 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? ∑F y = 0 0 = F bouy - m He g - m balloon g - m cargo g Archimedes: F bouy = 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