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 2nd 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

Example 14.5: Archimedes Principle & “Bath Legend”

• In Air: Newton’s 2nd Law gives:

∑Fy = T1 – Fg = 0. Result:

T1 = Fg = 7.84 N = mcrowng; mcrown = 0.8 kg

• In Water: Newton’s 2nd Law gives:

∑F = B + T2 – Fg = 0, or T2 = Fg – B = 6.84 N

Newton’s 3rd Law gives: T2 = “weight” in water.

From above, B = Fg – T2.

• Archimedes’ Principle says B = ρwatergV

Above numbers give: B = 7.84 – 6.84 = 1.0 N So, ρwatergV = 1.0 N. Note: ρwater = 1000 kg/m3.

Solve for V & get V = 1.02 10-4 m3

• Find the material of the crown from

ρcrown = mcrown/V = 7.84 103 kg/m3

Density of gold = 19.3 103 kg/m3 . 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 = ρswVswg

miceg = ρiceViceg

Newton: ∑Fy= 0 = B - miceg

ρswVsw = ρiceVice

(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/m3

• Newton’s 2nd Law: W´= ∑Fy = W – B = mag.

B = Buoyant force on rock.

• Archimedes’ Principle: B = ρwaterVg.

Combine (g cancels out!): mr - ρwaterV = ma. Algebra:

V = (mr - ma)/ρwater = [(9.28 – 6.18)/1000] = 3.1 10-4 m3 Definition of density in

terms of mass & volume gives:

ρr = (mr/V) = 2.99 103 kg/m3

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 = (mHe + m)g, Note: mHe = ρHeV

Archimedes: B = 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

B

Example: (Variation on previous example)

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? Newton: ∑Fy= 0 0 = B - mHeg - mballoon g - mcargogArchimedes: B = ρairVgAlso: mHe = ρHeV, ρair = 1.29 kg/m3, ρHe = 0.179 kg/m3

0 = ρairV - ρHeV - mballoon - mcargo

mcargo = 918 kg