Wednesday 4 April 2012
Class meeting
Topics
Textbook sections
Ponderables
Mini-labs (deliv.)
Lab
Demonstrations
Mini-lectures
PHYS 116 SCALE-UP
34
Fluids. Archimedes’ principle
14-7
Crystal growth, Archimedes and the crown of gold
Weight in air and water
Polar bear on an ice floe
Other
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Demo: Weight in air and water (Archimedes’ principle)
o Spring scale, block, large beaker
Mini-lecture: polar bear on an ice floe, PhET simulation, videos
Ponderable: Crystal growth, Archimedes and the crown of gold
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Wednesday 4 April 2012
PHYS 116 SCALE-UP
Consider the forces on a piece of fluid, Fbottom  Ftop  mg if the fluid is at equilibrium. We call
Fb  Fbottom  Ftop the buoyant force, and it results because the pressure at the bottom of the piece of
fluid is larger than the pressure at the top so there is a net upward force from the surrounding fluid.
Here the buoyant force just balances the weight of the fluid in our piece, so the net force on the piece
(buoyant force upward minus gravitational force downward) is zero. But if we replace the piece of fluid
by an identical volume of material with a density smaller than that of water (e.g. ice, which has a density
of 917 kg/m3 compared to water at 998 kg/m3), then the gravitational force on the ice would be smaller
than the buoyant force on it, so it would accelerate upward. Similarly if we replace the piece of fluid
with iron (7900 kg/m3) the net force (buoyant minus gravitational) would be downward and the iron
would accelerate downward. However, the acceleration would not be as large as if the iron were in air,
because the gravitational force is partially offset by the buoyant force.
Going back to Fb  mg   fluidVg , note that the buoyant force on whatever occupies the volume of
the piece of fluid is equal to the weight of that volume of fluid. Another way to say this is that the
buoyant force on an object is equal to the weight of the fluid that it displaces. An object will be in
equilibrium (buoyant force equals gravitational force) if the weight of the fluid it displaces is equal to the
weight of the object. If the object is less dense than the fluid, so that the volume of fluid that has the
same mass as the object is smaller than the object itself, then the object will float on the surface with
part of its volume not immersed (and therefore not displacing fluid). Consider an iceberg. The density
of seawater water = 1030 kg/m3 is larger than that of ice ice = 917 kg/m3. If the iceberg has volume V
then it has mass M   iceViceberg . The volume of water that has the same mass as the iceberg is
Vwater 
M
 water

 ice
V
= 0.89Viceberg. Thus an iceberg will float with 11% of its volume sticking
 water iceberg
out of the water.
What if there is a polar bear standing on the iceberg? Then the equilibrium equation becomes
Fb  Mg  mbear g  0 , so a larger buoyant force is needed to balance the additional weight of the bear
(who feels no buoyant force because he is not in the water). To increase the buoyant force the iceberg
will float lower in the water so that more of it is submerged and it displaces a larger mass of water.
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Wednesday 4 April 2012
PHYS 116 SCALE-UP
CRYSTAL GROWTH
One method of growing crystals of a substance from solution is to dissolve the substance in two solvents
that are of different densities and are immiscible in one another, so that they separate in the beaker
(like oil and water). One of the solvents can dissolve more of the substance than the other. As the
temperature of the liquids is lowered, the crystal will grow at the interface between the two solvents.
You have gotten a summer job in a crystal growth laboratory (you are very lucky—there are not so many
such labs in the world these days!), and have been given the task of growing a crystal of an organic
compound. You find two immiscible solvents with densities of 1.10 g/cm3 and 0.86 g/cm3 and succeed
in growing a crystal that is in the form of a tiny rectangular block. Looking closely at the beaker, you
notice that the crystal is suspended such that one-fifth of its height is below the interface between the
two solvents and the rest is above the interface. The density of the crystal you are trying to grow is 0.95
g/cm3. Have you succeeded in growing the substance you want, or has the chemistry gone awry?
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Wednesday 4 April 2012
PHYS 116 SCALE-UP
ARCHIMEDES AND THE CROWN OF GOLD
According to legend, the following challenge led Archimedes to the discovery of his famous principle:
Hieron, king of Syracuse, had given the royal goldsmith a quantity of pure gold from which to make a
crown. The king was suspicious that the goldsmith may have substituted base metal for some of the
gold in the crown and profited from the difference in value. Archimedes was ordered to determine
whether the crown was in fact made of pure gold, with the condition that only a nondestructive test
would be allowed. Rather than answer the problem in the politically most expedient way (or perhaps
extract a bribe from the goldsmith), Archimedes thought about the problem scientifically. A method
equivalent to what Archimedes is said to have done (compare the volume of water displaced by the
crown to the volume displaced by the same weight of pure gold) is to compare the actual weight of the
crown, Wactual , and the apparent weight of the crown Wapparent when it is submerged in water as shown
below.
Suppose you are Archimedes and you find that the apparent weight of the crown is 4.50 N and the
actual weight is 5.00 N (or whatever units were used in Greece in the 3rd century BCE). What do you tell
the king?
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