### Video Transcript

In this video, we’re going to learn
about buoyancy. We’ll learn what buoyancy is and
how to calculate the buoyancy of various objects.

To get started, imagine that one
day you’re out fishing on your boat on a lake. Just as you think you might be
getting a nibble on the line, a wave approaches the boat from the opposite side and
sends it in a motion. Caught off balance, you tip over
the side of the boat and fall in.

Over the next few moments, you see
some of the things you fell in with come to the surface, your fishing line, for
example. But other items seem to have
disappeared, your sunglasses, your phone, and the sandals you were wearing. To understand why some of these
items float to the surface and some end up on the bottom of the lake, we’ll want to
know a bit about buoyancy.

Buoyancy is the tendency of an
object to experience an upward acting force when the object is in a fluid. It could be a liquid or a gas. For example, if we had a container
full of water and put a table-tennis ball on the surface of that water, the ball
would experience an upward acting force called the buoyant force, able to balance
out the weight force of the ball.

On the other hand, if we took a
nugget of gold and dropped it into the water, that gold would sink all the way to
the bottom and rest on its base. Even though it doesn’t float, the
gold also experiences a buoyant force. And there’s a principle called
Archimedes’ principle that connects these two cases.

This principle says that the upward
buoyant force on an object, whether it’s a floating object like a ping-pong ball or
a submerged object like our nugget, that upward buoyant force is equal to the weight
of fluid that the object displaces. We can expand this principle to
consider specifically the two cases where, first, an object floats such as our
ping-pong ball and second when an object sinks.

When an object floats in a fluid,
it displaces its own weight in that fluid. If we were to consider a zoomed-in
view of our floating table-tennis ball, we would see that the ball isn’t entirely on
the surface of the water. There’s actually a very small
fraction of the ball’s volume that falls below the equilibrium level of the water in
the container. If we could somehow take that
volume of water that the ball replaces and put that amount on a scale, with the
table-tennis ball itself on the other side of the scale, we would find that these
two masses are equal. Their weight comes out the
same.

The second corollary of Archimedes’
principle is that if an object doesn’t float but sinks, then when submerged, it
displaces its volume in the weight of the fluid it’s in. So, now if we put our gold nugget
on one side of a scale and then on the other side put the amount of water which
takes up the same volume as the nugget, when we released the scale, we would find an
imbalance. Though the volumes are the same,
that amount of gold weighs more than that amount of water.

Reflecting on Archimedes’
principle, we come to see that buoyancy is really about relative density. The reason that the table-tennis
ball floats is because, on average, it’s less dense than the fluid it’s in, whereas
the gold nugget sinks because, on average, it’s more dense than water.

We can symbolize density using the
Greek letter 𝜌. And we recall that the density of
an object is equal to the mass of that object divided by the volume it takes up. Here’s a question about buoyancy
and density. Say that I have a tank of
water. And into that tank I drop a board
of solid wood. What happens to the board? Well, it flows, right? Now say I take a bar of solid metal
and I throw that into the tank. The metal will clearly sink. If that’s true, then how is it
possible that boats made of metal are able to float?

It took quite a long time for
people to discover that something like this was possible. And it is possible. And the reason it’s possible has to
do with average density. If my metal hull boat is filled
with air, then the average density of all the air contained inside along with the
metal may actually be less dense than the fluid that it’s in, so that the boat
floats. Keeping this idea of average object
density in mind, let’s try a few examples involving buoyancy.

A chunk of iron with a mass in air
of 390.0 grams is found to have an apparent mass of 350.5 grams when completely
submerged in an unknown liquid. Assume that the iron has a density
of 7.86 times 10 to the third kilograms per meter cubed. What mass of fluid does the iron
replace? What is the volume of the iron? Calculate the fluid’s density.

In the statement, we’re told the
mass of the iron in air, 390.0 grams. We’ll name that mass 𝑚. We’re told that when the iron is
submerged completely in an unknown liquid, it has an apparent mass of 350.5
grams. We’ll name that 𝑚 sub 𝑎. Knowing that the density, 𝜌, of
iron is 7.86 times 10 to the third kilograms per cubic meter, we want to solve for
three things. First, the mass of fluid that the
iron replaces. We’ll name that 𝑚 sub 𝑓. Second, we want to know the volume
of the iron. We’ll name it 𝑉. And finally, we wanna solve for
this unknown fluid’s density. We’ll name that 𝜌 sub 𝑓.

To start in on our solution, let’s
consider the difference between the apparent mass and the true mass of the iron. Say that we have our chunk of iron
and we suspend it by an effectively weightless chord from a scale. The scale reads out a mass. And we’ve called that mass in air
𝑚. Then, imagine we take the same
setup but we submerge the chunk of iron in our unknown fluid. When we do that, the scale records
the value we’ve named 𝑚 sub 𝑎, the apparent mass of the iron.

By a corollary of Archimedes’
principle, the mass of the fluid displaced by the iron must be equal to the
difference between the true mass, we can call it the mass in air of the iron, and
its apparent mass. Written as an equation, we can see
that 𝑚 sub 𝑓 is equal to the mass of the iron in air minus the apparent mass, or
the measured mass when it’s in the fluid. That’s 390.0 grams minus 350.5
grams, or 39.5 grams. That’s the mass of fluid displaced
by the iron.

Next, we wanna solve for the volume
𝑉 of this chunk of iron. We recall that the density of an
object, 𝜌, is equal to its mass divided by its volume. Rearranging that expression for our
case, we can write that the volume of the iron is equal to the mass of the sample
divided by its density. We’ll use the mass of the sample
measured in air. We can call that the true mass of
the iron. Substituting in our values for that
mass of the iron in units of kilograms and the iron’s density, when we calculate
this fraction, we find it’s 49.6 centimeters cubed. That’s the volume taken up by this
chunk of iron.

Finally, we want to solve for the
density of the unknown fluid that the iron is submerged in. To figure this out, we can again
rely on our expression for density. The density of the fluid is equal
to the fluid’s mass divided by its volume, which is the same as the volume of the
submerged iron. Plugging in for these two values,
when we calculate this fraction, we find it’s 0.796 grams per centimeter cubed. That’s the density of the fluid the
iron is submerged in.

Now let’s look at an example
involving the relative densities of a fluid and a floating object.

A woman floats in fresh water of
density 1.00 times 10 to the third kilograms per cubic meter. 4.00 percent of her volume is above
the water’s surface. What is the density of the
woman? What percent of the woman’s volume
would be above water if she floated in sea water of density 1.03 times 10 to the
third kilograms per meter cubed?

In part one, we want to solve for
the density of the woman. We can call that 𝜌 sub 𝑤. And in part two, we want to solve
for the percent of the woman’s volume that would be above water if she was in sea
water. We’ll call that percent capital
𝑃. We’re told the density of fresh
water as well as salt water. And we can name those densities 𝜌
sub 𝑓 and 𝜌 sub 𝑠, respectively.

In the initial example, where the
woman is in fresh water, we’re told that 4.00 percent of her volume is above that
water when she floats. This means that, compared to the
density of fresh water, the density of the woman, 𝜌 sub 𝑤, is four percent
less. Written as an equation, we can say
that 𝜌 sub 𝑤 equals 𝜌 sub 𝑓 minus four percent of 𝜌 sub 𝑓, 0.04 times that
density. This is equal to 0.96 times 𝜌 sub
𝑓, or 0.96 times 1.00 times 10 to the third kilograms per cubic meter. This product equals 960 kilograms
per cubic meter. That’s the density of the
woman.

In part two, we imagine that the
woman moves from fresh water to salt water. And now we want to know what
percentage of her volume is above that level of the water. We’ll figure this out using the
density of salt water, 𝜌 sub 𝑠, and the density of the woman from part one, 𝜌 sub
𝑤. Here’s one way to think of
this. Imagine that 𝜌 sub 𝑤 and 𝜌 sub
𝑠 were the same, so that this fraction is equal to one. In that case, 100 percent of the
woman would be submerged and zero percent would be above water.

So, to solve for 𝑃, the percent
above water, we’ll want to subtract this ratio from 100 percent. And then, more than that, we’ll
want to turn this ratio from a decimal into a percentage value. And we’ll do that by multiplying it
by 100. Now, we have an expression, which
when we calculate it, we’ll solve for the percentage of the woman’s volume that is
above the level of the saltwater.

When we plug in the densities for
the woman and for salt water and calculate this expression, we find it’s equal, to
three significant figures, to 6.80 percent. That’s the percentage of the
woman’s volume that would be above water level if she was floating in sea water.

Let’s summarize what we’ve learned
so far about buoyancy. We’ve seen that buoyancy describes
the upward force that acts on objects in fluid. Archimedes’ principle tells us that
a floating object displaces its weight and a submerged object displaces its volume
in the fluid it’s floating or submerged in. And we’ve seen that buoyancy comes
down to relative average densities between the object that may float or sink and the
fluid that it may float or sink in.