Goal: To understand that
objects whose density is less than the fluid they are placed in will float.
Previously Discussed: The
concepts of mass and density.
Materials:
One, unopened, 12 fl. oz. can of Coke
One, unopened, 12 fl. oz. can of Diet Coke
Water-filled fish tank (or other clear vessel
holding sufficient water)
39 g sugar in a display vial
1 g aspartame
(from a sweetener packet) in a display vial
(Note: There is only 3.5 mg of aspartame in a 1-gram sweetener
packet, the remainder is filler to give the contents some bulk)
Activity:
Break up the class
in groups of 3 or 4 and have them discuss the question:
- Why
does a battleship float?
After 5 minutes
end discussions and list their ideas on the board.
- If any ideas use motion of the ship or waves pushing on
the ship as a reason, as other students in the class what they think of this
idea. If no one else bring it up, point out to them that a ship or boat at
anchor in quiet water floats.
Discuss the remaining ideas and at the end of the discussion ask if
anyone in the class wants to add any more ideas as to why a battleship (or any
boat or object) floats.
We cant float a battleship, but we can try it with unopened cans of
Coke and Diet Coke.
Weigh the unopened can of Coke (~421 g)
Weigh the unopened
can of Diet Coke (~403 g)
Provide them with the volume of the unopened
cans (~414 cm3)
Have them calculate the average density of each
unopened can
- Unopened Coke can ~1.02 g/cm3
- Unopened Diet Coke can ~0.97 g/cm3
Inform the class that water has a density of
1.00 g/cm3
Point out the relationship in densities of the three items (i.e.,
Diet Coke < water < Coke).
In their groups, have the class decide on whether the unopened Coke
can will float or sink and whether the unopened Diet Coke can will float or
sink.
After 5 minutes have the groups state their decisions and ask them
the basis for those decisions.
Put the cans in the water tank/container (the Coke can sinks and the Diet Coke
can just barely floats).
Ask the class if any conclusions can be drawn from what we are seeing and what
we know about the two cans.
If students do not come to the realization that the Diet Coke
can floats because its density is less than water and the Coke can sinks because
its density is greater than water, guide them to it. Also point out that with
any fluid (liquid or gas), objects that have an average density less than the
fluid will float and those whose average density is greater than the fluid will
sink (show them a picture of a density column - see Attachment A).
Now point out the following:
- The aluminum in each can has a
density of ~2.70 g/cm3
- The liquid Coke has a density of
~1.15 g/cm3
- The liquid Diet Coke has a density
of ~1.10 g/cm3
So the can is denser than water and both soda liquids are denser than water, so
why did the Diet Coke float?
Have a class discussion on this question, whose answer is the gas bubble. See
if anyone realizes it and if not guide them towards the answer (i.e., What
happens when you open a can/bottle of soda? When you pour soda, what happens in
the glass? etc.)
Now circle back to the original question - What
makes a battleship float?
The answer is that the interior of a battleship (or any boat for that matter) is
basically air, this allows the average density of the battleship to be much less
than water.
Assessment:
The assessment is via the class response to the discussions
taking place. The central concept is that, to float, the average density of the
object has to be less than the fluid on which it is floating.
IF TIME PERMITS:
Invariably, you will get the question of why the mass of
the full soda cans differ (i.e., the densities of the soda liquids differ) and
the answer is the sweetener used:
In a 12 fl. oz. can of Coke, there are 39 g of
sugar.
In a 12 fl. oz. can of Diet Coke, there are 182
mg (0.182 g) of aspartame.
Show the 39 g sugar vial and the 182 mg
aspartame vial.
Show the other two vials explaining that in a 1 g aspartame packet there is only
3.5 mg (0.0035 g) of aspartame and the rest is filler. The purpose behind this
filler is so that the consumer can see that they are putting something into
their beverage. The aspartame is uniformly blended into the filler.
Aspartame is about 200 times sweeter than sugar
(i.e., 39 g/0.182 g is ≈214).
Finally, if someone points out that there was not a 39 g difference in the two
cans (i.e., more like an ~18 g difference), point out the difference in
ingredients:
- High fructose corn syrup and/or
sucrose (i.e., the 39 g of sugar)
- Diet Coke also has:
- Aspartame, potassium benzoate
and citric acid
So the addition of the potassium benzoate and
citric acid diminish the difference to ~18 g.
PART 2:
WHAT FORCES DETERMINE DENSITY IN A FLUID?
Goal: To understand the forces
acting on an object placed in a fluid and how that determines if it will float
or sink.
Materials:
One, unopened, 12 fl. oz. can of Coke
Water-filled fish tank (or other clear vessel
holding sufficient water)
Spring scale (in newtons)
String
Activity:
From the previous work we know that an unopened Coke can will sink.
Weighing the can in air (use the string to make a sling to hold the can), it
weighs about 4.1 N.
Put the can completely in the water and the scale now shows a weight of about
0.1 N.
Break up the class
in groups of 3 or 4 and have them discuss the question:
- What
happened to the rest of the weight?
After 5 minutes
end discussions and list their ideas on the board.
Ask the class if any of them remember playing with something that floats on the
water (i.e., beach ball, rubber ducky, etc.) and if they tried to push it
completely under the water, they felt something pushing back. Point out to them
that the force that was pushing back is the buoyant force and is the force that
also makes the unopened Coke can weigh less in water. Further inform them that
we are now going to figure out where that buoyant force comes from.
But to understand about buoyant force, first define fluids (i.e., both water and
air are fluids, and a fluid is defined as a substance that can easily change
shape), second when a fluid is not in motion, the pressure at any point in a
fluid is transmitted equally in all directions, third that pressure is force
divided by area (P = F/A), and forth to understand what pressure is, discuss air
pressure (PAIR).
Explanation of PAIR:
Hold your hand out in front of you, palm
up. You are holding up a weight equivalent to that of a washing machine. How
can this be? You are surrounded by a fluid that presses down on you all the
time. This fluid is the mixture of gases that makes up Earth's atmosphere. The
pressure exerted by the air is usually referred to as air pressure, or
atmospheric pressure.
Air exerts pressure because it has mass.
You may forget that air has mass, but each cubic meter of air around you has a
mass of about 1 kilogram. The force of gravity on this mass, produces air
pressure. The pressure from the weight of air in the atmosphere is great
because the atmosphere is over 100 kilometers high.
Think about a square measuring one
centimeter by one centimeter on the palm of your hand. (draw a hand with a tall
narrow square bar above it) At sea level, the air is pushing against that small
square with a force of 10.13 newtons (N). Thus, air pressure at sea level is
about 10.13 N/cm2. The total surface area of your hand is probably
about 100 square centimeters. So the total force due to the air pressure on
your hand, at sea level, is about 1,000 newtons.
How could your hand possibly support the
weight of the atmosphere when you dont feel a thing? In a fluid that is not
moving, pressure at a given point is exerted equally in all directions. Air is
pushing down on the palm of your hand with 10.13 N/cm2 of pressure.
It is also pushing up on the back of your hand with the same 10.13 N/cm2
of pressure. These two pressures balance each other exactly.
So why aren't you crushed even though the
air pressure outside your body is so great? The reason again has to do with a
balance of pressures. Pressure inside your body balances the air pressure
outside your body. But where does the pressure inside your body come from? It
comes from fluids within your body. Some parts of your body, such as your
lungs, sinus cavities, and your inner ear, contain air. Other parts of your
body, such as your cells and your blood, contain liquids.
Therefore, PAIR = 10.13 N/cm2, but for now, well still
call it PAIR for now.
Ask the class to discuss in their groups the following: If they were in water
(i.e., at a pool or at a beach) and they dove under the water, what the pressure
on them would be due to what?
After about five minutes, have the groups present their ideas. Foster
cross-discussion and provide guidance and examples to get them to (The answer is
that the pressure is due not only to the column of water above them but also to
the column of air above that column of water - guide them to this answer.)
Draw on the board a container of water with a cube (i.e., edge dimension a and
mass m), with the cube held in place somewhere underneath the water (i.e., top
of cube at a depth of d beneath the surface).
In their groups, have them discuss what are the forces acting on the cube
(ignoring whatever force is holding it in place) and in what direction.
After about 5 minutes, bring the class together and list each groups
conclusions.
While listing the conclusions, also discuss amongst the class these
conclusions. In the end the guided discussion should lead to the following
forces:
- An downward force due to water and air pressure on the cube top (FW(d)).
- A downward force due to gravity (FG = m x g or mg).
- An upward force due to water and air pressure on the cube bottom (FW(a+d)).
- A sideway force due to water pressure on the side (FW(s)).
One force for each side, with each force pushing inward.
At this point draw a cube on the board and label the forces acting on the cube.
The forces FW(a+d) and FW(a) are due to water pressure at
some depth z (Pz) and air pressure (PAIR).
The relationship between force and pressure is that Pressure = Force/Area.
So the forces due to pressure above can be rewritten as:
- FW(d) = a2 x PW(d) or a2PW(d)
- FW(a+d) = a2 x PW(a+d) = a2PW(a+d)
- FW(s) = a2 x PW(s) = a2PW(s)
So lets look at the water pressure at some depth z. PW(z) = PAIR
+ Pz. Where Pz is the pressure due only to the water at
depth z.
Form a square column measuring b by b at the base and extends z tall
(i.e., from depth z to the surface of the water). The mass of this column is
the density of water (rW)
times volume so mz = rW
x z x b x b or b2zrW.
The force that this column applies at the base is mass times the acceleration
due to gravity (i.e., Fz = mzg or b2zgrW).
Now pressure at the base is the force divided by area at the base (i.e., Pz = b2zgrW)/b2
or zgrW). Therefore PW(z)
= zgrW + PAIR.
First look at the side forces. As previously discussed, a fluid, at a given
point has the same pressure in all directions and since opposite sides of the
cube have the force (and thus the pressure) pointing inwards, the PAIR
from each side cancels out. While the force (and thus the pressure) of the
water increases as the depth increases, at any given depth, the force (and
pressure) are the same in all directions so this force/pressure also cancels
out. As an example, ask them if they have ever put an object into still/quiet
water and whether it floats or sinks does it move around on the surface or on
the bottom. The answer is no. the side forces/pressures are in equilibrium and
have no effect on the movement of the object
Buoyant force (FB) is defined as the sum of the upward force due to
the fluid minus the downward force due to the fluid. From before, FW(a+d)
is the upward force and FW(d) is the downward force.
Therefore:
- FB = FW(a+d) - FW(d)
= a3grW
+ a2dgrW + a2PAIR
- (a2dgrW + a2PAIR)
= a3grW
+ a2dgrW + a2PAIR
- a2dgrW - a2PAIR
= a3grW+ a2dgrW + a2PAIR
- a2dgrW - a2PAIR
= a3grW
Point out to the class that the buoyant force (FB) is due to the
density of the fluid (in this case water) not the density of the object.
Look at the difference (D) between FB and FG (i.e., D = FB
- FG).
Ask:
- If D > 0 (positive)? -- (float - positively buoyant)
- If D < 0 (negative)? -- (sink - negatively buoyant)
- If D = 0 -- (will stay where you put it -- neutrally buoyant)
Back to the Coke can, where did the weight go? (guided discussion -- D is
negative so the can will sink. When the can was in the air, the scale was
measuring only FG, but when the can is in the water it is now
measuring FG - FB)
Assessment:
Have the class look at the following two containers (draw
it on the board):
Container
1
Container
2
<----WATER
LEVEL---->
C
A
B
D
E
Two identical containers.
The fluid in both containers is water (r
= 1.0 g/cm3).
Both containers are filled to the same level.
All five blocks are cubes with edge dimension a.
The cubes are held in place in the water (i.e., thin rods, wires, threads, etc.)
Their relative positions in the water are to scale, That is:
- A and B are at the same level.
- D and E are at the same level.
- C is at the least depth, A and B are at the middle depth, and D and
E are the
deepest.
The blocks have the following masses:
- A = 1.0 gram
- B = 5.0 gram
- C = 0.5 gram
- D = 1.0 gram
- E = 2.0 gram
Questions:
1. Is the buoyant force of Block A <, >, or = Block B?
2. Is the buoyant force of Block C <, >, or = Block D?
3. Is the buoyant force of Block C <, >, or = Block E?
4. Is the buoyant force of Block D <, >, or = Block E?
5. Can any comparison be made of the buoyant force on the blocks in
Container 1 versus the blocks in Container 2?
6. If the edge dimension a is one centimeter, and the five blocks are
released from where they are held, what would happen to each of the blocks?
ATTACHMENT A
DENSITY COLUMN
(From:
Science Explorer Motion, Forces, and Energy,
published by Prentice Hall)
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