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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,
    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 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
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