5.1
Physics 2010
Archimedes' Principle
Experiment 5
The density  of an object is defined as its mass m divided by its volume V.

m
V
(1)
(The Greek letter  , rho, is pronounced "row".) In this experiment, you will determine
the density of some irregularly shaped solids whose masses can be determined easily by
weighing, and whose volumes are hard to measure directly.
The simplest way to find V is to lower the solid into a calibrated beaker of water.
The rise in the level of the water tells you the volume of the object, as shown below.
When you use this method, you will discover that it is not very accurate.
V=volume of solid
A much more accurate method is based on Archimedes' Principle which states:
When a solid of volume V is immersed in a fluid, it experiences a buoyant force FB equal
to the weight mg of fluid which it displaces: that is,
FB  mg  f Vg
(2)
where f is the density of the fluid. If f is known and FB is measured, then from Eqn (2)
one can calculate the volume V of the solid. In this experiment, you will immerse the
solid in water, whose density is known to be
 water  1000
.
gram / cm3  1000
.
kg / liter = 1000 kg / m3 .
To find the buoyant force FB , you will weigh a solid object, first out of the water
and then suspended in the water, as shown below. In the first case, the weight of the
object is supported by the string alone: therefore, the tension T in the supporting string is
T = mg. In the second case, the weight is born by the string and the buoyant force
together, and the tension is T  mg  FB . The difference of these two tensions is the
required force FB.
5.2
reading = map
FB
T = mg - FB
reading = m
T = mg - FB
T = mg
mg
Forces on solid when immersed.
A small complication is that most scales (including those in this lab) are calibrated
in units of mass, not in units of force. The reading of the scale is not the tension in the
string, but the tension divided by g. Therefore, when you weight the block out of the
water and the tension is mg, the reading is just m, the mass of the block. (This is
obviously convenient and is why the scale is calibrated in this way.)
When you weigh the block under water, the tension (or apparent weight) is
T  mg  FB . Thus, the reading on the scale is the apparent mass map,
map 
mg  FB
g
(3)
You can solve (3) for FB to give
FB  ( m  map ) g .
(4)
Combining (2) and (4), we find that
water V g  ( m  map ) g ,
or
V
m  map
 water
.
(5)
Notice how the factors of g cancel.
The expression (5) for V is based on Archimedes' Principle. Thus, by comparing
your value obtained from (5) with your earlier rough measurement of V, you can test
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5.3
Archimedes' principle. However, since the value obtained from (5) is much more
accurate, it is this value that you will use in the rest of the experiment.
Knowing the mass and volume of your block, you can calculate its density. By
comparing your value with the known densities of the metals available in this laboratory,
you can identify what the metal is. You will then repeat this exercise for a second metal
block.
Once you know the volume of a solid, you can use the relation (5) to find the
density of a fluid other than water. In Part II of this experiment, you will immerse both of
your blocks in anti-freeze. Eqn (5) still holds, except that the denominator is now af , the
density of anti-freeze, and m ap is the apparent mass obtained by weighing the block in the
anti-freeze. Solving (5) for af , we find that
af 
m  map
V
.
(6)
Since you can do this with both blocks, you can get two answers for af . Comparing
them gives a sensitive test of the accuracy of your measurements. If you have been
careful, your two values will agree reasonably well to four significant figures.
Procedure
Part I. Densities of Metals
In this experiment it will be convenient to use cgs units (centimeter-gram-second).
1. There is an assortment of irregularly shaped blocks of aluminum, brass, steel, and lead.
Pick two blocks, of different metals, weigh both of them, and record their masses. With
care, you can find the masses within one or two hundredths of a gram on a four-arm
balance, but you must be careful that the balance is zeroed properly; that is, that it reads
zero when nothing is on it. (Your instructor can help you to adjust this if necessary.)
You might weigh one block on two or three different balances to find out how reliable
they are.
2. Put some water into a calibrated beaker (the kind with a handle, which has 5 ml
graduations, 1 ml = 1 milliliter = 1cm3) and lower one of your blocks into the water.
Record the volumes to the nearest 1 ml, without and with the block, and calculate V by
subtraction. The beaker has 5ml divisions, but you can estimate volumes to the nearest
ml. Do this two or three times for the same block, with different amounts of water in the
beaker. State your best estimate for V and its approximate uncertainty.
3. To find the volume of each of your two blocks using Archimedes' principle, you need
to weigh each block while it hangs immersed in water. The four-arm balance has an
adjustable platform on which the beaker of water can rest, as shown in the diagram
below. (Only the beakers without handles will fit). Put this platform above the pan of the
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5.4
balance, and place the beaker on it. Suspend the first block from the appropriate hook,
making sure that it is totally immersed. Be sure that the block is not touching the side of
the beaker and that the beaker is not rubbing against the bars that support the balance pan.
Also, check that no bubbles are adhering to the block, especially in the hole for the string.
Adjust the length of the supporting string and the amount of water, if necessary. Now
find the apparent mass map of the first block immersed in water. Use Eqn (5) and your
measurements of m and map to calculate the volume V of your first block. Repeat this
determination of map and V for the second block.
Comment in your lab book about whether the two values of V for the first block
obtained in steps 2 and 3 agree.
4-arm balance
platform
pan
m
V
of each block. Given the following know approximate densities, identify the composition
of your blocks. Is this identification consistent with the appearance of the blocks? (Brass
is yellowish-gold, lead is a dull gray and soft, aluminum is silvery and light, and steel is
silvery-gray and considerably heavier than aluminum.)
4. Using your more accurate values of V for the two blocks, calculate the density  
metal
aluminum
brass
steel
lead
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density (gram/cm3)
2.7
8.5
7.8
11.3
5.5
(Note: These densities are approximate. Both brass and steel are alloys or mixtures of
various pure metals, and their densities vary according to their composition. Although
aluminum and lead are, in principle, pure metals, the samples in the laboratory almost
certainly contain impurities which change their densities slightly.)
Part II. Density of Anti-Freeze
Using the known volumes of your blocks, you can now find the density of the
anti-freeze available in the bottles near the sink. Caution! Anti-freeze is poisonous. It is
also moderately expensive. When you have finished, pour it back into the bottle, not
down the drain. Never pour anything toxic down a drain. It all eventually goes into the
drinking water supply and purification is never perfect!
Empty your beaker of water, dry it, and fill it with the anti-freeze. Now find the
apparent masses of your two blocks when immersed in the anti-freeze. Use Eqn (6) and
your data to get two values for the density af of the anti-freeze.
Comment on how well your two values of af agree. If you have been careful, the
percent discrepancy may be as small as 0.1%.
Questions
1. Suppose there was a tiny bubble of air clinging to the block when you weighed it in
water. If the volume of the bubble was Vbub , then the volume V computed from Eqn (5)
would have been Vblock  Vbub . If Vbub were 1mm3, what would have been your computed
density of your first block? Note that 1mm = 0.1 cm, so 1mm3 = 0.001 cm3. What is the
percent change in computed density? Is it a significant effect?
2. In part I, you weighed the blocks out of the water to find their masses. In truth, they
were immersed in a fluid - namely the air - and there was a small buoyant force on them.
Therefore, the values that you took to be their true masses were really their apparent
masses in air. Using the known volume V of your first block, calculate how big a
difference this makes to your measurement of m. The density of air is about 1 kg/m3 or
10-3 gram/cm3 in Boulder. Is the effect significant?
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5.6
Physics 2010
Pre Lab Questions
Experiment 5
1. What is Archimedes Principle?
2. A rock is thrown in a beaker of antifreeze and it sinks to the bottom. Is the buoyant
force on the rock greater than, less than, or equal to the weight of the rock? Explain
your answer.
3. A chunk of metal is held stationary, suspended by a string and the tension in the string
is 19.6 N. What is the mass of the chunk of metal?
4. If the metal in question 3 is brass, use the known density of brass (given in the lab
writeup) to find the volume of the chunk of metal.
5. A mass m is weighed while immersed in water and suspended by a string attached to a
balance. The balance shows the apparent mass map of the mass when it under water.
How is the volume V of the rock related to the density of water, the mass of the rock,
and the apparent mass read on the balance? (Also, what is the density of water?)
6. Dry Lab.
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