Download Demonstrating Archimedes` Principle

Demonstrating Archimedes’ Principle
Archimedes’ Principle states that a floating body will displace its own weight of liquid. In
this first exercise we will test this idea by comparing the masses of different blocks of
wood to the mass of water that each displaces. Figure 1 shows the experimental set-up.
1.
Determine the masses of one block of pine and one block of oak to the nearest
1/100 of a gram (0.01 g). The pine is the light-colored wood and the oak is the dark
wood. Record the masses on the answer sheet (page 11).
2.
Weigh an aluminum pie plate. Set the 1000 mL beaker in the pie plate on top of
several rubber stoppers. Fill the beaker until the water is just at the lip. Top it off
with a squeeze bottle containing a small amount of Calgon (soap breaks the
water’s surface tension).
NOTE: You want the beaker as close as possible to overflowing without
spilling, and you want the pie plate to be absolutely dry. Line the plate
with a couple of paper towels while filling the beaker in case of spills.
3.
Slowly lower the pine block into the water until it is just floating. The runoff should
flow down the side of the beaker and into the pan. When it stops flowing, use a 25
mL plastic pipette and bulb to remove a bit of water from the beaker to prevent
spillage when you lift the beaker.
4.
Carefully remove the beaker and the rubber stoppers from the pie plate. Weigh the
pie plate with the water. Subtract the weight of the pie plate from the total, and
record the weight of the water on the answer sheet.
5.
Using a funnel, pour the runoff from the pie plate into a 25 mL graduated cylinder.
There will be more than 25 mL. Repeat steps 2-4 with the oak block and record
both of the water volumes on the answer sheet.
NOTE: We are using a 25 mL graduated cylinder here because it has
finer measurement divisions (greater precision). There will be more than
25 mL of water in the pan, so you will need to fill the cylinder almost to
the top, record the volume, empty it, and then pour the rest of the water
in, adding the second volume to the first. Do not fill the graduated
cylinder higher than the 25 mL limit and estimate the volume or you are
throwing out the precision this device provides.
6.
Since water has a density of 1.0 grams/cm3, and since 1 cm3 = 1 mL, the volume of
water in mL should be equal to the mass of the water in grams. Also the volume of
water in milliliters should be equal to the mass of the pine block, in grams. Is this
true?
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Figure 1.
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Set-up for determining the density/buoyancy relationship.
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NOTE: The density of water can vary slightly with temperature (as
you will learn in subsequent lab exercises). The small amount of
soap used here can also affect the density, but there is enough
error introduced here due to water sticking to the sides of the
beaker, to the rubber stoppers and to the pie plate, that these slight
density variations can be safely ignored. What other sources of
error might be involved in this experiment?
Figure 2.
7.
Significant dimensions for determining the
volume of a wood block with cylindrical hole.
Measure the dimensions of each of the blocks and calculate the total volume of
each of the blocks. Record your measurements on the answer sheet.
Volume of a solid, square block:
Vblock = T x L2
(1)
*since the blocks are square (L = W), L x W = L2
Remember that the blocks have a cylindrical hole in the center that reduces the
total mass. To account for this measure the diameter of the hole and calculate the
missing mass in the cylinder:
Volume of a cylinder:
Vhole = π x R2 x T
(2)
Vb - h = T x (L2 – π x R2)
(3)
π = 3.1416; R = D/2
Volume of block without hole:
8.
Now calculate the density for each of the blocks:
Density of block:
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ρ = mass/volume
(4)
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B.
Modeling the Continental and Oceanic Lithosphere
Geologists divide the upper Earth into a chemically distinct crust and mantle. The crust
comes in two flavors: continental crust varies in thickness, averaging about 35 km, and
has a heterogeneous composition which, on average, resembles granite (silicate-based
rocks rich in sodium, potassium and aluminum); oceanic crust is much thinner, typically
8-10 km thick, and is composed of basalt (a silicate-based rock higher in iron,
magnesium and calcium).
The mantle is a homogeneous, mostly solid mass made of a rock called peridotite (a
silicate rock dominated by iron and magnesium). Only rarely does this rock get hot
enough to melt, but the region known as the asthenosphere is close enough to its
melting temperature that it may be up to 1% molten. As a result the asthenosphere,
while not a fluid, displays fluid-like behavior under most conditions (geophysicists refer
to this silly putty-like behavior as “plastic”).
The region of the mantle located above the asthenosphere is completely solid. The
combination of the solid crust and this solid, upper mantle is known as the lithosphere. In
the rare case when we are referring only to this upper solid layer of the mantle (separate
from the crust), it is called the “mantle lithosphere” (Figure 3).
In this exercise we will create a simple physical model of the upper layers of the Earth
using our wood blocks and water in an aquarium to represent the various layers of rock:
Figure 3.
1.
Aquarium model of the Earth’s lithosphere and asthenosphere.
Fill an aquarium with about 10 cm of water and place the ring stand in the
aquarium (Figure 3).
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2.
Slide a small pine block down the ring stand. Place the large pine block with the
rectangular cutout over the ring stand so that it fits around the small pine block.
This is a model of the continental lithosphere with no topography. Float the thin oak
block next to the pine block – this represents the oceanic lithosphere.
3.
Measure the water depth beneath the pine block and beneath the oak block using
a metric ruler. Enter the data in the space provided on the answer sheet.
NOTE: Measuring water depth with a ruler here is much easier if
you put the ruler in the water next to the wood block and put your
head down so that you are looking at the ruler from “underwater.”
4.
Calculate the volume and weight of the water column under each block, assuming
a wood block and a water column with horizontal dimensions of 1 cm x 1 cm and
whose thickness is the water depth under the block (Figure 4). We can assume the
density of fresh water to be 1.0 g/cm3. Enter these on the answer sheet.
5.
Assuming wood blocks of the same thicknesses as measured but with horizontal
dimensions of 1 cm x 1 cm, calculate the mass of each of the wood blocks using
the wood densities you determined in part A (answer sheet, p. 11).
6.
Add up the wood and water masses under each column and enter them on the
answer sheet.
7.
Measure the depth of the water with no overlying block of wood and determine the
mass of this water column, assuming the same horizontal dimensions (Figure 4).
Figure 4.
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According to Archimedes, the total pressure
acting on a square centimeter (or any fixed
area) at a fixed depth should be constant.
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8.
Put another small pine block over the ring stand to represent a mountain (Figure 5
below). Note how it displaces part of the lower "continental crust" in the water.
9.
Again assuming blocks of 1 cm x 1 cm, measure and record the depth of water
under this "mountain." Calculate and record the mass of this water column.
10. Calculate the volume and mass of the pair of pine blocks, using the measured
thickness and horizontal dimensions of 1 cm x 1 cm. Calculate the mass of the pair
of blocks using the previously determined density. Add this mass to the mass of the
underlying water column (Figure 6).
Figure 5.
If the lithosphere becomes thicker it should
increase in depth as well as in elevation.
+
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Figure 6.
If the lithosphere becomes thicker it should
increase in depth as well as in elevation.
C.
Density and Specific Gravity of Rocks
In creating our simple wood/water model of the upper layers of the Earth we have made
a couple of key assumptions: that the asthenosphere does behave as a fluid over
relatively short time scales; that the asthenosphere is denser than the lithosphere,
otherwise the lithosphere would not float (since the mantle lithosphere is approximately
the same density as the asthenosphere, it is the lighter crust that decreases the overall
density of the lithosphere).
NOTE: Much of what we know about the rheology of the
asthenosphere comes from seismologists: We know, for instance,
that the asthenosphere can transmit seismic shear waves (Swaves), indicating that it is not a pure fluid; we know that no
earthquakes originate in the asthenosphere, indicating that it is too
soft (plastic) to break under stress; and we know that all seismic
body waves slow down significantly in this region, indicating that
the rock there is closer to its melting point than adjacent layers
(asthenosphere means “weak sphere,” and it is also known as the
“low velocity layer”).
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In this portion of the lab we will consider how well our model works in the real world of
rocks. Unlike cut wooden blocks, which can be measured with a ruler, rocks have odd
shapes that make volume measurements very difficult. We can get around this problem
by using Archimedes’ Principle: “an immersed object displaces water equal to its own
volume.”
To determine density of a rock we must know its mass and its volume. We could
determine volume using the method we used in part A: immerse the rock in a full beaker
of water and measure the water that spills out. This is complicated and messy however.
But let’s assume that we have a beaker of water full to the brim with 100 grams (100 mL)
of water. If you drop in a rock that is 1 cm3 (= 1 mL) and a mass of 3 grams (ρ = 3 g/cm3)
then 1cm3 of water will spill out. The net gain in mass is 2 grams (+3 grams of rock, -1
gram of water). We conclude that even though the rock will not float, the water is
exerting a buoyant force on the rock that negates a weight equivalent to the mass of the
water displaced (this is what Archimedes did with the King’s crown).
Therefore, the difference between the weight of the rock in air and its weight underwater
is the weight of the water displaced, which is equivalent to its volume (since 1.0 gram of
water = 1.0 mL of water), which is equal to the volume of the rock. This is the numerical
gram equivalent of the volume, not the volume itself, as the unit is grams rather than cm3
or mL. Dividing this equivalent volume by the density of water yields the true volume of
the rock:
(5)
Alternatively we can simply divide the weight in air by the difference between weight in
air and weight in water, yielding the specific gravity of the sample:
(6)
Since the units here are grams divided by grams, specific gravity is a unitless quantity
whose value is the numerical equivalent of density. Specific gravity is sometimes called
“relative density”, and can be thought of as the ratio of rock density to the density of
water.
Measuring the specific gravity of rocks
Each student will be given three rocks: granite, representing the continental crustal
rocks; basalt, representing oceanic crust; and peridotite, representing the mantle rocks
in which they are "floating."
1.
Using the experimental set up shown below in Figure 7, determine the mass of
each rock on the top tray of the scale (weight in air). Record the results on the
answer sheet.
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2.
Determine the mass of each rock sample on the hanging tray below the scale
(weight in water). Depth does not matter so long as the rock is completely
submerged. Be sure that no part of the tray is touching the sides of the beaker.
Again, record the results on the answer sheet.
3.
Determine the specific gravity of each rock using equation 6. Record the results on
the answer sheet.
Figure 7.
Lab set-up for determining specific gravity of rocks, or any odd-shaped
substances denser than water.
D.
Determination of total mass exerted at a fixed depth within the
asthenosphere
Using the specific gravity calculations from the previous exercise and Figure 8 below,
calculate the total mass a column of rock exerts at 150 km depth beneath the
mountains, the continents and the ocean. Make the following assumptions:
1.
Assume that the water in the ocean has a density of 1.03 g/cm3.
2.
Assume that the continental crust has the density of granite, oceanic crust has the
density of basalt, and the lithosphere and the asthenosphere each have the density
of peridotite.
3.
Since both layers have the same density, we will treat the mantle lithosphere and
the asthenosphere as a single layer of mantle.
4.
We are determining the masses of three columns with varying layer thicknesses
and densities, but we will assume each to have horizontal dimensions of 1 cm x 1
cm.
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5.
Record your data and calculations on the answer sheet. Pay attention to units and
convert where necessary.
Figure 8.
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Simple three-layer model of the Earth’s upper layers.
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Name
Lab Section
Date
.
MS20 Laboratory:
Density, Specific Gravity, Archimedes and Isostasy
Answer sheet: record all data in the appropriate metric units (centimeters, grams, etc.).
Remember to use significant figure rules and to indicate appropriate units (if the scale
reads 13.4 g, your answer is not 13.4, but 13.4 g (or 13.4 grams).
A.
Demonstrating Archimedes’ Principle
Pine Block
Oak Block
Mass of block <grams>
(Wt of water in pan – dry weight of pan)
Weight of water in pan <grams>
Displaced volume <cm3 or mL>
(volume of water in pan)
Block thickness <cm>
Length of side <cm>
Vblock (equation 1) <cm3>
Rhole <cm>
(½ diameter)
Vhole (equation 2) <cm3>
Vb - h (equation 3) <cm3>
ρwood (equation 4) <g/cm3>
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Archimedes' Principle states that a floating body will displace its own weight of fluid. Are
the masses of the blocks the same as the masses of water each displaces? If there is a
difference, what are the likely sources of error?
B.
Modeling the Continental and Oceanic Lithosphere
Pine Block
Oak Block
Depth of water <cm>
Volume of water <cm3 or mL>
Mass of water <g>
Mass of block <g>(1cm x 1cm x t)
Combined masses <g>
(water column + wood block)
Depth of water/no blocks <cm>
Mass of water/no blocks <g>
Again, according to Archimedes' Principle the pressure (or the total weight) acting on the
bottom of the tank (or at some depth in the asthenosphere) should not change as more
floating masses are added; i.e., the combined masses of each of the wood blocks and
the water columns beneath them should be the same as the total mass of the open
water. Is this true here? Again, try to account for likely sources of error.
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Figure 5 shows our model of the earth's crust using two different types of wood to
represent the two different types of crust. Assume you shaved some wood off the top of
the pine block (representing continental erosion). What would happen to the pine block?
Explain.
Depth of water under two pine blocks <cm>
Volume of water column <cm3 or mL>
Mass of water column <g>
Total mass of two pine blocks <g>
Combined masses of blocks and water column <g>
Is the total mass acting on the bottom of the aquarium approximately the same as in the
previous calculations? Again, try to account for likely sources of error?
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Gravity and time, aided by various processes of physical and chemical weathering,
removes rock from higher continental elevations and transports it to lower elevations,
ultimately to the ocean floor. For every meter of rock removed from a mountain range,
would you expect the elevation to decrease by one meter? Explain.
What happens to the oceanic crust as water, ice and wind continuously deliver and
deposit continental sediments? Explain.
C.
Density and Specific Gravity of Rocks
GRANITE
BASALT
PERIDOTITE
Mass in air
Mass in water
Massair - Masswater
Rock volume (equation 5)
Rock density (equation 4)
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Can you think of another way to measure the volume of the rock specimens?
D.
Determination of
asthenosphere
total
mass
exerted
at
a
fixed
depth
within
the
weight under ocean crust
5.0 x 105 cm
A. Depth of water
x
10 x 105 cm
B. Depth of ocean crust
x
135 x 105 cm
C. Depth of mantle
x
x
1.0 cm
Length (L)
x
1.0 cm
Width (W)
=
x
1.0 cm
Length (L)
x
1.0 cm
Width (W)
=
x
1.0 cm
Length (L)
x
1.0 cm
Width (W)
=
Total weight under ocean (A + B + C)
=
Water density
Basalt density
Peridotite density
weight under continental crust
30 x 105 cm
D. Depth of crust
x
120 x 105 cm
E. Depth of mantle
x
x
1.0 cm
Length (L)
x
1.0 cm
Width (W)
=
x
1.0 cm
Length (L)
x
1.0 cm
Width (W)
=
Total weight under continent (D + E)
=
Granite density
Peridotite density
weight under mountains
55 x 105 cm
F. Depth of crust
x
100 x 105 cm
G. Depth of mantle
x
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x
1.0 cm
Length (L)
x
1.0 cm
Width (W)
=
x
1.0 cm
Length (L)
x
1.0 cm
Width (W)
=
Total weight under continental mountains (F + G)
=
Granite density
Peridotite density
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During ice ages slightly cooler temperatures at mid-to-high latitudes cause less snow to
melt in spring and summer than accumulates in winter. Over many years this ice piles up
to form continental glaciers that can exceed 2 kilometers in thickness. Glacial ice has a
density of about 0.9 g/cm3, just below that of water, and about 1/3 that of granite. What
would be the effect of a 2000 meter thick ice sheet on the continent? What would
happen when the ice melted?
The total weight under the continents should precisely equal the weight under the ocean
basin. The calculations, although close, don't really match. What does this tell us about
our simple three-rock model? Explain.
The last ice age ended in northern Europe and in North America approximately 10,000
years ago, yet there is excellent evidence that the Scandinavian peninsula (which
includes Norway, Sweden and Finland) is still uplifting rapidly. From this information,
what can we conclude about the physical properties of the asthenosphere?
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