sect. 10-7: buoyancy/archimedes principle experimental facts: –objects submerged (or partially...

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 mF 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

FB = mFg (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

FB = mFg

= ρFVdispl g

(mF = mass of fluid

displaced, Vdispl =

volume displaced)

W = mOg = ρOVOg

(mO = 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 = ρswVdisplg

miceg = ρiceVOg

∑Fy= 0 = FB - miceg ρswVdispl = ρiceVO

(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 = ρwaterVg. Combine (g cancels out!):mr - ρwaterV = ma . Algebra & use definition of ρ: V = (mr - ma)/ρwater. ρ = (mr/V) = 2.99 103 kg/m3

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 = (mHe + m)g , Note: mHe = ρHeV

Archimedes: FB = ρairVg

ρairVg = (ρHeV + m)g

V = m/(ρ air - ρ He)

Table: ρair = 1.29 kg/m3 , ρHe = 0.18 kg/m3

V = 160 m3

Prob. 25: (Variation on example 10-10)

mballoong

Fbuoy

mcargog

mHeg

Spherical He balloon. r = 7.35 m. V = (4πr3/3) = 1663 m3

mballoon = 930 kg. What cargo mass mcargo can balloon lift? ∑Fy= 0 0 = Fbouy - mHeg - mballoon g - mcargogArchimedes: Fbouy = ρ airVgAlso: mHe = ρHeV, ρ air = 1.29 kg/m3, ρHe = 0.179 kg/m3

0 = ρairV - ρHeV - mballoon - mcargo

mcargo = 918 kg