Video Transcript
In this video, we’re talking about
the buoyant force, which is the force that pushes up on an object whenever it’s in a
fluid. We’re going to learn what causes
this force. And we’ll also see how to calculate
mathematically whether an object will float or sink in a given liquid.
To get started on this topic, say
that we have a basin. And then we fill that basin with
water to a particular level, that is, to a particular depth. And then imagine that we have a
cube of wood and we toss that cube into the basin of water. Now, if our cube is light enough,
we know what will happen when we do this. Instead of sinking to the bottom,
our cube will float. But as we look closely, we notice
that that’s not the only thing that goes on here.
Notice what’s happened to the level
of water in the basin. It’s gone up from its original
level. If we draw in a dashed line where
the water level was at first, then we can see that by dropping this cube into the
water, we’ve raised the average water level. More on that in a moment. But for now, let’s notice that
something very interesting is happening here.
If we were to take a second
identical wooden cube to our first cube and just drop that cube to the ground,
nothing would stop it. We know the cube would fall to the
ground. But here, in the case of our cube
in the basin of water, the cube hasn’t fallen all the way to the bottom. Something is holding it up. That something is called the
buoyant force. And we know it needs to be acting
upward on our cube with a force equal in magnitude to the weight of this cube
downward. That’s the requirement for our cube
to float rather than sinking to the bottom of the basin.
If we think about what could be
causing the buoyant force. As we look at our cube in the
basin, we see the only thing the cube is in contact with is the water. It’s not touching the sides of the
basin or the bottom or any other object. So somehow, the fluid in this basin
must be exerting this buoyant upward force. If we were to pick a certain point
within our basin of water. Say that point right there. We know that thanks to the weight
of the water on itself, there’s pressure exerted at that point. If we were to use blue arrows to
represent the pressure, we know that pressure acts downward at that point. But then it also acts to the side
and upward and really in every direction equally from that location. This fact that pressure at a point
within a fluid is equal in all directions is known as Pascal’s principle.
It’s this idea of Pascal’s
principle that explains to us how there can be an upward-acting force on our
floating cube. It’s because the pressure within a
fluid doesn’t only act downward, but it also acts upward or to the sides or in any
direction. So when it comes to the buoyant
force in general, this upward-acting force on objects in fluids, we can say that
this force is due to the pressure that’s exerted by the fluid that the object is
in. That pressure acts on the object in
the fluid and pushes up on it, hence the buoyant force.
Okay, so that’s what causes this
upward-acting force. But then we wonder how do we know
how big this force is. In other words, what determines
just how much pressure is exerted on an object in a fluid? How do we know whether that object
will float or sink? To answer that question, we’ll come
back to what we noticed earlier, that the height of water in our basin changed when
we put this wooden cube in it. When we dropped the cube in the
basin, the water level went up by this much, which means that this cross section of
water that we’re seeing here wasn’t there originally. This is water that’s been moved out
of the way by the cube.
This water, which we could say was
originally here in the basin before the cube was dropped in, was pushed out of the
way when the cube landed. The fancy word for this is it was
displaced. It’s this amount of water, the
water that was displaced when we dropped our cube into the basin that contributes to
the pressure exerted upward on the cube by the fluid. In particular, it’s the weight of
this amount of water, this amount of displaced fluid, that equals the upward buoyant
force on a cube. We could think of it this way.
Say that we have a measuring scale
and say that we were able to put on one side of the scale all of the water that was
displaced by dropping our wooden cube into the water basin. So all of this displaced water here
that we’ve marked out in our sketch is collected on one side of the scale. And then on the other side of the
scale, we put the object that we dropped into our fluid, in this case, the wooden
block. Now, if the amount of fluid that
our object displaced has the same weight as the object we dropped in the fluid, then
our scale will balance out like we’re seeing here. And the object will float in the
fluid. And then if we push the wood block
down so it was completely underwater, say with our hand. In that case, the amount of fluid
pushed out of the way would outweigh the object, because now we’re displacing
slightly more fluid. So again, the block would
float.
As a side note, when an object does
float, that means it’s displacing its weight in fluid. So a partly submerged floating
object like our block displaces less fluid than if the block was fully underwater
while it floated. And then, of course, there’s a
third possibility, which is that our object actually outweighs the amount of fluid
that it displaces. In that instance, our scale would
look something like this. And the result would be that this
object in our fluid would sink. We can see that this third
possibility is not happening in our case because this wooden block is actually
floating in the water.
But the key point here is that the
two things on either side of this scale, so to speak, the things that determine
whether an object floats or sinks in a liquid are, on the one hand, the object
itself and, on the other, the amount of fluid that that object pushes out of the way
when it’s in the fluid. It’s the weight of that amount of
pushed out of the way or displaced fluid that determines the buoyant force, the
upward force acting on the object.
Now let’s consider a special case
of a floating object. Say that we drop our wood cube into
this water basin. And this time, the cube still
floats. But the top of the cube is exactly
level with the level of the water in the basin. In other words, the cube is
completely submerged. It’s completely underwater, except
for the very top surface. But it’s not sinking. It still floats. Now, if we were to once again
consider the amount of water that was pushed out of the way or displaced by dropping
our cube in the water and then put that amount on our scale along with the
block. We would see what we saw in the
very first instance that the weight of the water displaced by the cube is exactly
equal to the weight of the cube.
Now here’s something interesting
because our cube is completely submerged underwater. That means that the amount of water
it displaces is equal to the volume of the cube itself. After all, all of the cube is
underwater, just barely. This means that really, the
displaced water on our scale would take up exactly as much space as the wooden
cube. They would be the same size. So let’s think about this. If the water takes up exactly as
much space as the wooden block and based on the fact that our scale is balanced,
they must have the same weight. Then that means the displaced water
and the wooden block have the same volume and they have the same mass. Now it’s at this point that we can
recall the definition for the density symbolized using the Greek letter 𝜌 of an
object.
The density of a material is equal
to the mass of that material divided by the space it takes up, its volume. But we’ve just said that both the
mass and the volume of the wooden block and the displaced water are equal when the
wooden block is completely submerged under the water but still floating. And if they have the same mass and
volume, then they must also have the same density. So when the top of our wooden block
is exactly on level with the top of the water in the basin, we can say that the
density of the block — we’ll call it 𝜌 sub b — is equal to the density of the water
— we’ll call it 𝜌 sub w.
But then, what if these two
densities weren’t equal to one another? Well, there are two possibilities
for that. In the one case, the density of the
block could be greater than the density of the water. In that case, when we tossed the
block in the water, the buoyant force created by the water wouldn’t be enough to
hold up the block. So the block would sink to the
bottom of the basin. The other possibility is that the
density of the water is greater than the density of the block. In that scenario, what we saw
earlier takes place. The block not only floats, but part
of it floats above the level of the water. So in answer to our question of how
do we know whether an object will sink or float when we drop it in a certain fluid,
we see that the answer comes down to the density of the object compared to the
density of that fluid.
What we’re finding then is that
it’s the differences in density between the object and the fluid the object is in
that determines whether the object floats or whether it sinks. And not only this, but we see we
can now extend our definition of the buoyant force. This force, we know, is due to the
pressure exerted by a fluid on an object. And the buoyant force is equal to
the weight of the fluid the object displaces or pushes out of the way. So going over here to our mass of
water that’s displaced by the object, the weight of that mass of water we can call
it capital 𝑊 is equal to the mass of the water times the acceleration due to
gravity.
Whenever we use the weight of a
displaced fluid to calculate the buoyant force, we need to be careful. That’s because of this term here,
the mass in this equation. It’s important to realize that this
mass, in the case of a buoyant force calculation, is not equal to the overall mass
of our object, but rather it’s equal to the mass of the fluid it displaces. So if we call the mass of fluid
that’s displaced by an object 𝑚 sub f, we can say that 𝑚 sub f multiplied by 𝑔 is
equal to the buoyant force — we’ll call it 𝐹 sub B — created by that displaced
fluid. The important point to remember is
not to confuse the mass of the fluid displaced with the mass of the object that’s
displacing it. That all may sound a bit confusing,
but let’s look at an example which should help clarify these ideas.
A cube-shaped object with sides
that are 130 centimeters long has a density of 950 kilograms per cubic meter. The cube is placed in a body of
water. The water has a density of 1000
kilograms per cubic meter. What is the volume of the object in
cubic meters? What is the mass of the object? Answer to the nearest kilogram.
Alright, so in this scenario, we
have this cube-shaped object. And then if we mark out the height
and width and depth of the cube, we know that because it’s a cube, these three
quantities are all equal to one another. And we’re told in the problem
statement that they’re equal to 130 centimeters. Our first question asks, what is
the volume of this object in cubic meters? That volume, which we can call 𝑉,
is equal to the total amount of space that this cube takes up. In other words, it’s equal to the
height of the cube multiplied by its width multiplied by its depth. And we know that each one of those
values is 130 centimeters.
So if we took that amount 130
centimeters and then cubed it, that would be equal to the volume of the cube but in
units of cubic centimeters rather than cubic meters like our question asks. So before we multiply out the
right-hand side of this expression, let’s convert 130 centimeters into whatever that
distance is in meters. To do that, let’s recall that 100
centimeters is equal to one meter. Which means that to do this
conversion, we’ll take the decimal place in our current value of 130 centimeters,
and we’ll move it two spots to the left.
By doing that, we convert to a unit
of meters and our value for that distance is 1.30. When we go ahead and cube this
value, we find a result of 2.197 cubic meters. That’s how much space this cubic
object takes up. Knowing this, we can move on to our
second question, what is the mass of the object?
In the problem statement, we’re
told about this object’s density, the value of that. And we can recall that density —
symbolized using the Greek letter 𝜌 — is equal to the mass of an object divided by
how much space it takes up, its volume. It’s not density we want to solve
for for this object, but rather the mass. So to do that, let’s multiply both
sides of the equation by the volume 𝑉. When we do, that term cancels out
on the right-hand side and we find that the mass of an object is equal to its
density multiplied by its volume.
Now, if we call the mass of our
cube 𝑚 sub c, then we know that that’s equal to the density of the cube. We’ll call it 𝜌 sub c multiplied
by the cube’s volume 𝑉. 𝜌 sub c is equal to 950 kilograms
per cubic meter and 𝑉 the cube’s volume we just solved for earlier. So the cube’s mass is equal to the
product of these two values. And before we multiply them, notice
what happens to the units. We have kilograms per cubic meter
multiplied by cubic meters, so that unit of cubic meters cancels out and we’re left
with units of kilograms. That’s perfect because we’re
calculating a mass after all. And to the nearest kilogram, this
number comes out to be 2087. That’s the total mass of our
cube-shaped object. We can now write that result off to
the side. And then, we’ll move on to the last
two questions in our scenario.
The first of those two questions
says this: How many cubic meters of water have a mass equal to that of the
object?
At this point, we can recall from
our problem statement that our cubic object is placed in a body of water and we’re
given the density of that water. Now, when we talk about this
question of how many cubic meters of water have a mass equal to the mass of the
object, what we’re saying is what is the volume of water, whose mass will be equal
to the mass of our cubic object. Though we don’t yet know what that
volume is, we do know a condition on that volume. The mass of that volume of water —
we’ll call that mass 𝑚 sub w — must be equal to the mass of our cubic object, 𝑚
sub c.
That’s what it means when we say
that this amount of water, however much it is, has a mass equal to the mass of our
object. And at this point, we can again
recall the definition of density. That density is equal to an
object’s mass divided by its volume. And as we saw earlier, this means
that the density of an object times its volume is equal to its mass. This means that we can rewrite our
equation here for the mass of the water and the mass of the cubic object. We can replace these terms with the
densities and volumes of the water and cubic object, respectively.
Here’s what that would look
like. We can replace the mass of the
water with the density of the water — we’ll call it 𝜌 sub w — times the volume of
the water, 𝑉 sub w. And then, we’ll replace the mass of
the cubic object 𝑚 sub c with the density of the cubic object 𝜌 sub c multiplied
by its volume 𝑉. And now that we have this second
equation, we see that it’s 𝑉 sub w the volume of water that we want to solve
for. That’s how many cubic meters of
water have a mass equal to that of the object. To go about doing that, we’ll
rearrange this equation by dividing both sides by the density of water 𝜌 sub w. That cancels that term on the
left-hand side.
So the volume of water we’re after
is equal to the density of our cubic object divided by the density of water
multiplied by the volume of our cubic object. And this whole equation was made
possible by the fact that we’re setting the mass of our water equal to the mass of
our object. And we’re asking the question just
how much water in volume is that? Well, there’s good news. The density of our cubic object is
given to us in the problem statement. And so is the density of water. It’s 1000 kilograms per cubic
meter. We’re not given the volume of our
cubic object. But remember we solved for that
earlier. So it is a known quantity.
When we plug in for all these
values, here’s what we get. Notice that in our numerator, the
units of cubic meters cancel out. And along with that, the units of
kilograms cancel from numerator and denominator. When all the dust settles, we’ll be
left with units of cubic meters, units of volume. This is a good sign to us that
we’re on the right track with our calculation. And when we compute this value, we
find it’s 2.087 cubic meters. That’s the volume of water that
would have a mass equal to the mass of the cubic object. Now, let’s move on to the last
question in this exercise.
How far below the water’s surface
must the base of the object be in order to displace a massive water equal to that of
the object? Answer to the nearest
centimeter.
Okay, so now, we’re finally facing
the fact that this cube is not sitting off by itself, but rather is placed in a body
of water. So we can say that this line
represents the water level, and we want to know how far below that surface must the
base of our cubic object be so that the object displaces a massive water equal to
the mass of the object. Now, before we go any further,
let’s give a name to this distance. Let’s call this distance 𝑑. And here’s what we’re saying with
this distance, with our cube submerged at distance 𝑑 under the surface of the
water, some fraction of that cube’s volume is pushing water out of the way. It’s displacing that water. And the idea is that that mass of
water displaced is equal to the total mass of the cube. Let’s clear some space on screen to
get into this a bit more.
What we have is a cube with this
total volume, but only part of that volume is submerged underwater. It’s only that part of the volume,
which is underwater, which displaces or pushes water out of the way. Because the cube is floating, that
means that this amount of water displaced is equal to the total mass of the
cube. That’s what it means for the weight
force of the object pulling it down to be balanced out by the buoyant force, pushing
it up. For these two forces to be equal,
here’s what needs to be true. The total mass of the cube 𝑚 sub c
needs to be equal to the mass of the displaced water. We’ll call it 𝑚 sub dw.
We know the total cube mass from
earlier. And using our modified density
equation, we can say that the mass of the displaced water from the cube is equal to
the volume of the cube underwater multiplied by the density of water. And the volume of the cube
underwater is equal to 𝑑, what we want to solve for, times the width of the cube
times its depth.
As a next step then, we can
rearrange this equation to solve for lowercase 𝑑. We’ll do it by dividing both sides
by the width of the cube times its depth times the density of water. Doing so cancels these terms out on
the right-hand side. When we plug in for all these
values on the left, using a distance of 1.30 meters for both 𝑊 and 𝑑, and then
calculate this term. We find a result of 1.23 meters or
123 centimeters. That’s how far below the surface
the bottom edge of this cube is.
Let’s summarize now what we’ve
learned in this lesson. First off, we learned that buoyant
force is caused by pressure within a fluid. Buoyant force is equal to the
weight of fluid displaced by an object. And lastly, we saw that the
relative densities of an object and fluid determine whether the object sinks or
floats in that fluid.
