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# Buoyant force example problems

A couple of problems involving Archimedes' principle and buoyant forces. Created by Sal Khan.

Video transcript

Let's say that I have some
object, and when it's outside of water, its weight
is 10 newtons. When I submerge it in water-- I
put it on a weighing machine in water-- its weight
is 2 newtons. What must be going on here? The water must be exerting some
type of upward force to counteract at least
8 newtons of the object's original weight. That difference is the
buoyant force. So the way to think about is
that once you put the object in the water-- it could be a
cube, or it could be anything. We know that we have a downward
weight that is 10 newtons, but we know that once
it's in the water, the net weight is 2 newtons, so there
must be some force acting upwards on the object
of 8 newtons. That's the buoyant force that
we learned about in the previous video, in the video
about Archimedes' principle. This is the buoyant force. So the buoyant force is equal
to 10 minus 2 is equal to 8. That's how much the water's
pushing up. And what does that
also equal to? That equals the weight of the
water displaced, so 8 newtons is equal to weight of
water displaced. What is the weight of
the water displaced? That's the volume of the water
displaced times the density of water times gravity. What is the volume of
water displaced? It's just the volume of water,
divide 8 newtons by the density of water,
which is 1,000 kilograms per meter cubed. A newton is 1 kilogram meter
per second squared. Then, what's gravity? It's 9.8 meters per
second squared. If we look at all the units,
they actually do turn out with you just ending up having
just meters cubed, but let's do the math. We get 8 divided by 1,000
divided by 9.8 is equal to 8.2 times 10 to the negative 4. V equals 8.2 times 10 to the
minus 4 cubic meters. Just knowing the difference in
the weight of an object-- the difference when I put
it in water-- I can figure out the volume. This could be a fun game
to do next time your friends come over. Weigh yourself outside of water,
then get some type of spring or waterproof weighing
machine, put it at the bottom of your pool, stand on it, and
figure out what your weight is, assuming that you're dense
enough to go all the way into the water. You could figure out somehow
your weight in water, and then you would know your volume. There's other ways. You could just figure out how
much the surface of the water increases, and take
that water away. This was interesting. Just knowing how much the
buoyant force of the water was or how much lighter we are when
the object goes into the water, we can figure out the
volume of the object. This might seem like a very
small volume, but just keep in mind in a meter cubed, you
have 27 square feet. If we multiply that number
times 27, it equals 0.02 square feet roughly. In 0.02 square feet, how many--
in a square foot, there's actually-- 12 to the
third power times 12 times 12 is equal to 1,728 times 0.02. So this is actually
34 square inches. The object isn't as small as you
may have thought it to be. It's actually maybe a little bit
bigger than 3 inches by 3 inches by 3 inches, so it's
a reasonably sized object. Anyway, let's do another
problem. Let's say I have some balsa
wood, and I know that the density of balsa wood is 130
kilograms per meter cubed. I have some big cube of balsa
wood, and what I want to know is if I put that-- let
me draw the water. I have some big cube of balsa
wood, which I'll do in brown. So I have a big cube of balsa
wood and the water should go on top of it, just so
that you see it's submerged in the water. I want to know what percentage
of the cube goes below the surface of the water? Interesting question. So how do we do that? For the object to be at rest,
for this big cube to be at rest, there must be zero net
forces on this object. In that situation, the buoyant
force must completely equal the weight or the force
of gravity. What's the force of gravity
going to be? The force of gravity is just the
weight of the object, and that's the volume of the balsa
wood times the density of the balsa wood times gravity. What's the buoyant force? The buoyant force is equal to
the volume of the displaced water, but that's also the
volume of the displaced water and it's the volume of the cube
that's been submerged. The part of the cube that's
submerged, that's volume. That's also equal
to the amount of volume of water displaced. We could say that's the volume
of the block submerged, which is the same thing, remember,
as the volume of the water displaced times the density
of water times gravity. Remember, this is density
of water. So remember, the buoyant force
is just equal to the weight of the water displaced and that's
just the volume of the water displaced times the density
of water times gravity. Of course, the volume of the
water displaced is the exact same thing as the volume
of the block that's actually submerged. Since the block is stationary,
it's not accelerating upwards or downwards, we know that
these two quantities must equal each other. So V, the volume of the wood,
the entire volume, not just the amount that's submerged,
times the density of the wood times gravity must equal the
volume of the wood submerged, which is equal to the volume of
the water displaced times the density of water
times gravity. We have the acceleration
of gravity. We have that on both sides,
so we can cross it out. Let me switch colors to
ease the monotony. What happens if we divide both
sides by the volume of the balsa wood? I'm just rearranging
this equation. I think you'll figure it out. We divide both sides by that,
and you get the volume submerged divided by the volume
of the balsa wood-- I just divided both sides by VB
and switched sides-- is equal to the density of the balsa
wood divided by the density of water. Does that make sense? I just did a couple of quick
algebraic operations, but hopefully that got rid of
the g, and that should make sense to you. Now we're ready to solve
our problem. My original question is
what percentage of the object is submerged? That's exactly this number. If we say this is the volume
submerged over the total volume, this is the
percent submerged. That equals the density of
balsa wood, which is 130 kilograms per meter cubed,
divided by the density of water, which is 1,000 kilograms
per meter cubed, so 130 divided by 1,000 is 0.13. Vs over VB is equal to
0.13, which is the same thing as 13%. So, exactly 13% percent of this
balsa wood block will be submerged in the water. That's pretty neat to me. It actually didn't have
to be a block. It could have been shaped
like a horse. I'll see you in the
next video.