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## Archimedes Principle

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### 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 ... – PowerPoint PPT presentation

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Title: Archimedes Principle

1
Archimedes Principle Bath Legend
2
Example 14.5 Archimedes Principle Bath Legend
• Archimedes was (supposedly) asked by the King,
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 Newtons 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

3
• In Air Newtons 2nd Law gives
• ?Fy T1 Fg 0. Result
• T1 Fg 7.84 N mcrowng mcrown 0.8 kg
• In Water Newtons 2nd Law gives
• ?F B T2 Fg 0, or T2 Fg B 6.84 N
• Newtons 3rd Law gives T2 weight in
water.
• From above, B Fg T2.
• Archimedes Principle says B ?watergV
• Above numbers give B 7.84 6.84 1.0 N
• So, ?watergV 1.0 N. Note ?water 1000 kg/m3.
• Solve for V get V 1.02 ? 10-4 m3
• Find the material of the crown from
• ?crown 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!)

4
Floating Iceberg!
• ?ice/?water 0.917, ?sw/?water 1.03
• What fraction fa of iceberg is ABOVE
• waters surface? Ice volume ? Vice
• Volume submerged ? Vsw
• Volume visible ? V Vice - Vsw
• Archimedes B ?swVswg
• miceg ?iceViceg
• Newton ?Fy 0 B - miceg
• ? ?swVsw ?iceVice
• (Vsw/Vice) (?ice/?sw) 0.917/1.03 0.89
• fa (V/Vice) 1 - (Vsw/Vice) 0.11 (11!)

5
Example Moon Rock in Water
• In air, a moon rock weighs W mrg 90.9 N. So
its mass is mr 9.28 kg. In water its
Apparent weight is
• W mag 60.56 N. So, its apparent mass is
ma 6.18 kg. Find the density ?r of the moon
rock. ?water 1000 kg/m3
• Newtons 2nd Law W ?Fy W B mag.
• B Buoyant force on rock.
• Archimedes Principle B ?waterVg.
• Combine (g cancels out!) mr - ?waterV ma.
Algebra
• V (mr - ma)/?water (9.28 6.18)/1000
3.1 ? 10-4 m3 Definition of density in terms of
mass volume gives
• ?r (mr/V) 2.99 ? 103 kg/m3

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 ?Fy 0 ? B WHe Wload
• B (mHe m)g, Note mHe ?HeV
• Archimedes B ?airVg
• ? ?airVg (?HeV m)g
• ? V m/(? air - ? He)
• Table ?air 1.29 kg/m3 , ?He 0.18 kg/m3
• ? V 160 m3

B
7
Example (Variation on previous example)
Spherical He balloon. r 7.35 m. V (4pr3/3)
1663 m3 mballoon 930 kg. What cargo mass
mcargo can balloon lift? Newton ?Fy 0 0 B -
mHeg - mballoon g - mcargog Archimedes B
?airVg Also mHe ?HeV, ?air 1.29 kg/m3, ?He
0.179 kg/m3 ? 0 ?airV - ?HeV - mballoon -
mcargo ? mcargo 918 kg