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EAS1600
Lab 09
“Archimedes' Principle and Isostasy”
Objectives
In this lab you will investigate the relationships between mass, volume, density, and
floatation in order to understand basic physical principles that determine topographic (heights
above sea level) and physiographic (depths below sea level) features of planet Earth. These
principles also determine the depths of mountain roots below the continents.
After completion of this exercise you should be able to:



determine the density of an object using different methods;
understand how density determines the elevation of the floating objects above the fluid
level;
apply these principles to the Earth’s asthenosphere and crust.
Theoretical background
Density, , is a measure of how tightly matter is packed in a unit volume of an object.
Every day we handle materials with different densities and have an intuitive idea of how to
distinguish very dense substances, like metals, from less dense objects, like wood.
Mathematically, density is defined as the ratio between the mass (m) of an object to its volume
(V):

m
V
(1)
Density is usually expressed in units of kg/m 3 or g/cm3. Water has a density of
approximately 1 g/cm3, while densities of gold and air (at normal atmospheric pressure) are about
19.3 g/cm3 and 0.0012 g/cm3 correspondingly. Objects that are less dense than water will float
whereas objects that are denser than water will sink.
On the Earth’s surface the difference between densities of different kinds of rocks has very
important physical consequences. Both oceanic and continental rocks “float” on top of the
mantle. Why? Crustal rocks (  2.7 - 3.0 g/cm3) are less dense than the upper mantle (  3.3
g/cm3) and therefore they "float" in the mantle. Crustal rocks float in the mantle similar to the way
icebergs float in the ocean. The iceberg’s density is lower, but close to the density of water, so it
floats with a significant part submerged.
Archimedes' principle explains floatation: a body completely or partially submerged in a
fluid is pushed up by a buoyant force (uplift) equal to the weight of the fluid displaced by the
body. This is simply a balance of forces: the force exerted by the object (weight) on the fluid is
opposite and equal to the uplift force or the weight of the displaced fluid.
Let’s consider the details of the floatation physics. On the surface of the Earth, the force of
gravity that is acting on the object is called an object's "weight" (W). The force is equal to mass
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times the acceleration or F = ma and on the Earth the acceleration is
gravitational acceleration - “gravity” (g). Therefore we have W =F= mg.
Using the equation (1) we can replace m with V:
W = Vg,
due to Earth's
(2)
where g = 9.8 m/s2. Equating downward gravitational force acting on the object and upward
buoyant forces we will get:
o Vo g = f Vd g ,
(3)
o Vo = f Vd ,
(4)
mo= md .
(5)
Here o is the density of the object, Vo is the volume of the object, mo is the mass of the object, f
is the density of the fluid, and Vd and md are the volume and mass of the fluid displaced by the
object.
From equation (4), the displacement volume of the floating object is proportional to the
ratio of the densities of the fluid and object:
 o Vd

 f Vo
(6)
From the equation (6) we see, that if density of the object is lower then density of the fluid o <
f , then displaced volume is less then total volume of the object: Vd < Vo and the object will float.
If o > f, then the object sinks and Vd = Vo (it can not displace more than its total volume).
Since the displaced volume (Vd) must equal the volume of the submerged portion of the object
(Vs), then for a floating object, the ratio of densities also equals to the ratio of the volume below
water to the total volume:
o Vs

 f Vo
(7)
From this equation follows the "iceberg" rule: The density of ice is 0.9 g/cm3 while the
density of water is 1.0 g/cm3, and therefore we understand that the submerged portion of an
iceberg is about 90% of its total volume - only 10% is above water and visible.
This way the Archimedes' Principle explains why continental and oceanic crusts sit on top
of the mantle at a height determined by their relative densities and thicknesses. The special case
of application of Archimedes' Principle to the Geology and Earth science is known as the
principle of isostasy: the thicker, less dense and more buoyant crust (the continents) is
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topographically higher than ocean crust that is thinner and denser. It also explains why highelevation mountains have roots that extend deep into the upper mantle.
Online activities
http://serc.carleton.edu/NAGTWorkshops/visualization/collections/IsosGrav.html
http://highered.mcgraw-hill.com/sites/0072402466/student_view0/chapter17/animations_and_movies.html#
http://www.geo.cornell.edu/hawaii/220/PRI/isostasy.html
http://nsdl.org/resource/2200/20110312193900657T
- animations that illustrates the Isostasy principle and rebound.
Safety precautions
-
Be careful when handling glass beakers and mixing rods.
Clean all spills immediately.
Keep all tanks and beakers with water as far away as possible from computers and any
equipment that is connected to a power grid.
Equipment and procedures in the lab
Part 1
1. Write down your block set number in the space above Table 1.
2. Weigh each block (note: make sure your blocks are dry before you proceed). Tare the
electronic scale to get a reading of 0.0 g when nothing is being weighed. Place one block on
the scale and record the mass (mo) into Table 1. Repeat for each block.
3. Use the ruler to measure the length (l), width (w), and height (h) of each of the blocks and
record these values into Table 1. Note: metric measurements only!
4. Calculate and record into Table 1 the volume, Vo , of each block from the direct
measurements:
Vo = length  width  height
5. Calculate and record the density of each block (o) based on mo and Vo using equation 1.
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Part 2
1. Fill the tank with water to a level that causes slight overflow through the overflow tube.
Let the water level equilibrate until there is no overflow.
2. Place an empty 500 ml beaker under the outflow tube.
3. Hold the block horizontally (so that the scale attached to the block is facing you) and
carefully lower it into the tank. It is best to lower one side in first and then the other.
Avoid trapping air bubbles beneath the block or water being splashed on the top of the
block. Allow the block to stabilize and the water level to decrease until there is no
overflow. The volume of the water that was displaced will need to be measured.
Figure 1. Photograph of the experimental setup.
4. Record the volume of displaced water Vfloat into Table 2 (to do this, transfer the displaced
water from the beaker into a dry tall graduated cylinder and record the volume of the
displaced water (1 ml = 1 cm3)).
5. Look at the side of the block and record the total thickness (height) of the block (h), the
thickness above the surface of the water (ha), and the thickness submerged (hs), using the
small scale attached to the block. Note that each division is 1 mm; record your
measurements to the nearest mm and later convert this to cm.
6. Replace the emptied and dried beaker under the outflow tube.
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7. Using the provided rods, carefully and slowly push the block below the surface of the
water until it is completely submerged (the easiest way is to pin it against the bottom).
8. Record in Table 2 the volume of displaced water and Vsub (to do this, transfer the
displaced water from the beaker into a dry tall graduated cylinder and record the volume
of the water).
9. Remove the block from water and dry it. Refill the tank and let the water level equilibrate
until there is no overflow.
10. Repeat steps 2 to 9 for blocks B and C.
Calculations
1. Calculate the total volume of each block (Vtotal) by adding together the two displacement
volumes VFloat and VSub. Place this value into Table 2:
Vtotal= VFloat + VSub.
2. Enter the mass of water displaced from the tank by each floating block into Table 2:
mDisp = VFloat f
where f = 1.00 g/cm3
(for fresh water)
3. Determine the density of each block (denoted B1) using the volume (VTotal) and mass (mDisp)
of displaced water, and record the values on Table 2.
4. Now determine the density (denoted B2) of each block using the ratio of submerged and total
volumes. Since the blocks have a regular shape, and the volume of a block is simply the
product of the length of the sides, equations 6 and 7 can be rewritten as:
 o hs

f
h
where hs/h is the "percentage submerged" or the fraction of the object's volume that is under
the water surface (i.e. h = ha + hs where h is the thickness of the block and hs is the thickness
submerged and ha is the thickness above water). Therefore,
B2 
 f hs
h
Enter your value into Table 2.
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Name ________________________________________________________ Lab section ________
Questions.
Question 1. Which block floated lowest in the water? Choose one from the following and enter
the letter that best corresponds to your answer into Clicker. (2 pts).
A. Block A
B. Block B
C. Block C
Question 2. Which block floated highest out of the water? Choose one from the following and
enter the letter that best corresponds to your answer into Clicker. (2 pts).
A. Block A
B. Block B
C. Block C
Question 3. A flat iceberg is floating in a fresh-water gulf with 9% of it above the surface of the
water. What is the density of the iceberg? Assume water density is 1 g/cm3. Enter the answer in
g/cm3 into your Clicker. (2 pts.)
Question 4. Assume that the total thickness of the iceberg from the above question is 120 m.
How many meters above the water is the flat top of the iceberg? Enter the answer in meters into
Clicker. (2 pts.)
Question 5. A man standing on the top of the iceberg described above wants to lower the
elevation of the iceberg so he can tow it under a low river bridge. He starts chipping away
horizontal layers at the exposed top surface. If he manages to chop off 1-meter thick layer of ice
from the top, what will be the new elevation, i.e. how high above the water will the top of the
iceberg be? Enter the answer in meters into your Clicker. (3 pts.)
Question 6. The same iceberg (use numbers from questions 3 and 4) floats out into the open
ocean and eventually stabilizes with only 88.0% of its total mass submerged. Calculate the
density of the saltwater. Enter the answer in g/cm3 into your Clicker, with at least 3 numbers after
decimal point . (3 pts.)
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Name ________________________________________________________ Lab section ________
Part 3
1. Place the 3 blocks one on top of each other, with the densest block on the bottom and the less
dense on top, ensuring that the scales on the side of each block are adjacent to each other.
Wrap 2-3 rubber bands around all three blocks to hold them in place.
2. Refill the tank with water. Let the water level equilibrate until there is no overflow. Place an
empty dry 500 ml beaker under the outflow tube.
3. Carefully lower the stack of blocks into the full tank, allow them to stabilize, and allow the
water level to come to equilibrium with its overflow. You may need to adjust the blocks and
the rubber band to ensure that the overall mass floats level with the surface of the water, i.e.
that one end isn't lower than the other. Make sure there are no bubbles trapped
underneath and there is no water on the top.
4. Record the displaced water volume (VFloat) into Table 3.
5. Record the total thickness of the blocks (the 3-blocks total height – (h)), the thickness above
the surface of the water (ha), and the thickness submerged (hs), using the small scales attached
to the blocks.
6. Take the blocks out of the water. Remove the top (less dense) layer and re-secure the other
two blocks with the rubber bands.
7. Repeat steps 2 through 5 with the top block removed.
8. Dry off the blocks with a paper towel.
Calculations
1. Using the ratio of submerged and total volumes, determine the bulk (average) density of the
assemblage of three blocks and enter the results into Table 3.
2. Repeat the above calculations for the case when the top block is removed, and enter the results
into Table 3.
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Name ________________________________________________________ Lab section ________
Questions
In the Southeastern United States, the surface of the continental lithosphere is constantly
worn away by erosion. One hundred fifty million years ago, the thickness h of the continental
crust was 60 km. Its density increased with depth and can be described by the following function:
 L (d )  (0.0033 g  cm 3 km1 )d  2.6 g  cm 3 ,
where L(d) is the density of the lithosphere in g/cm3 at depth d in km. Plot the density of the
continental crust as a function of depth on the graph given below, from depth 0 to 60 km.
(4 points)
Question 7. What was the average density (L average ) of the crust one hundred million years
ago? (Hint: Use the graph to estimate it!) Enter the answer in g/cm3 into your Clicker. (2 pts.)
Question 8. What was the elevation of the continental crust with respect to the mantle (mantle
density M = 3.3 g cm-3) one hundred fifty million years ago, when the thickness h of the
continental crust was 60 km? (Hint:consider continental crust floating in the mantle) Enter the
answer in km into your Clicker. (3 pts.)
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Name ________________________________________________________ Lab section ________
Question 9. The top 30 km of the continental crust have been worn away by erosion during the
last 150 million years. Calculate today’s elevation of the crust above the mantle using the average
density of the remaining crust. (Hint: During erosion of the top 30 km of crust, the densities of
the underlying layers remain unchanged) Enter the answer in km into your Clicker. (3 pts.)
Question 10. What is the difference in elevation of the crust one hundred fifty million years ago
and today? Enter the answer in km into your Clicker. (2 pts.)
Question 11. Neglecting sea level changes, what is the difference in elevation with respect to sea
level? Assume oceanic crust thickness did not change in last 150 million years. Enter the answer
in km into your Clicker. (2 pts.)
Problems for individual work
1. Describe how you could estimate your body density (which relates to average body fat %) with a
water tank and scales. (3pts)
2. In the Earth system, what do the blocks from the experiment (section C) represent? (2pts)
3. In the Earth system, what does the water from the experiment (section C) represent? (2pts)
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Name ________________________________________________________ Lab section ________
4. What happened when the block was removed from the top of the pile (experiment in section
C)? Qualitatively describe the vertical displacement after the assembly is "unroofed" and
explain why this happens. (3 pts.)
Conclusions
Based on what you have learned in this lab, explain what you think would happen to the surface
elevation of Greenland (which is currently covered by ice up to one mile thick), if all the ice
melted as a result of global warming? (5 pts.)
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Name ________________________________________________________ Lab section ________
Set # ____ (this is the number on your
blocks)
Table 1. Section A (10 points)
Block
m0
l
(g)
(cm)
A
w
(cm)
h
(cm)
Vo
(cm3)
o
(g/cm3)
B
C
Vo = l∙ w∙ h; o = m0/(Vo)
Table 2. Section B (20 points)
Block VFloat mDisp
VSub
cm3
g
cm3
A
VTotal
cm3
h
cm
ha
cm
hs
cm
B1
3
g/cm
B2
g/cm3
B
C
VFloat is the volume of Water displaced by the floating block.
mDisp is the Mass of the water Displaced by the floating block.
VSub is the volume of the additional water displaced when the block is submerged.
VTotal is the sum of VFloat and VSub.
B1 is density calculated in question B3 ; B2 is density calculated in question B4.
Table 3. Section C (15 points)
Blocks
Block
labels
3 blocks
VFloat
cm3
h
cm
ha
cm
hs
cm
bulk
g/cm3
2 most dense blocks
The most dense block
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