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Course: Class 9 Physics (India) > Unit 3
Lesson 6: Density and condition for floatingCalculating mass of displaced liquid
Let's solve 2 problems in finding the mass of displaced water. Created by Mahesh Shenoy.
Want to join the conversation?
- a block of wood weighing 2.0N is held under water by a string attached to the bottom ofa container.The tension in the string is 0.5N.Determine the density of the wood.(4 votes)
- At7:42, couldn't the buoyant force be greater than the force of gravity ?
Yes, the stone is at rest, but, to move we need to apply a force. The stone could be at rest because gravity is pulling it and it has nothing to push itself off of !(I'm not claiming I'm right, just asking a doubt.)
We need friction in order to move and for the stone to move to move in the upward direction, it would need a force. However, since there is no vertical force on the stone,it wouldn't move upward. So, the buoyant force could be greater ! Can somebody explain where I am wrong with my line of thought ??(4 votes)- if buoyant force was greater, the object would possibly completely float in the water, like if a fully blown air balloon was placed in water. So, the object could still move upward (although I am fairly certain that I am right, I still recommend that someone with better knowledge of the subject)(1 vote)
- i love your explanation but answer of first question is wrong it should be 40 g(2 votes)
- What about for a partially submersed object? Is it the same as mass of liquid displaced = mass of object because the force of gravitation is equal to the buoyant force?(1 vote)
- the mass of fluid displaced is equal to the volume*density of the submerged portion of the object. So, if three-fourths of an object's volume is submerged, the mass of water displaced= three-fourths of the volume of the object * density of water(2 votes)
- How can the mass of the displaced water be equal to the weight displaced water? (because weight = g*mass)(1 vote)
Video transcript
- [Instructor] Let's
solve a couple of problems to see how to calculate the amount of liquid displaced when something is floating on top of it and when something sinks into it. Here's the first one. An object of mass 30 grams
and volume 40 cm cube is dropped in water. Calculate the mass of displaced water, given the density of water
is one gram per cm cubed. Let's first try and draw the situation and then see how to solve it. So we're given an object of mass 30 grams. So let's draw an object. So let's say here is our object, imagine it's a stone. We know it's mass and we know it's volume, we'll write that down in a while, but it's dropped in water. So let's say we have
a container with water and it's given that we take this stone and we'd drop it in water. We're asked to calculate
the mass of displaced water. What does that mean? Well, whenever you drop any
object inside any liquid, or any gas for that matter, then as that object gets submerged, some of that liquid has to move away to make space for that object, right? So that liquid that goes away is what we call the displaced
water or the displaced liquid. And of course, it's
that liquid that comes, moves over here and
raises the level of water, and that's why we knew when
you submerged something, the level of water rises because the water gets displaced. And we're asked to calculate what the mass of that displeased water is. That's what we need to figure out. Okay. So let's quickly get write
on what is given to us. So what is given? We are given the mass of the object. So let's write that down
over here somewhere. We know the mass of
the object is 30 grams, we are given the volume of the object. So we know the volume that has given us 40 cm cubed, we are also given the density of water. So density, now instead of writing the word density, I usually like to use this symbol rho, which is just to save space. So anyways, density of water that is given to us as
one gram per cm cube, and we need to calculate what the mass of the displaced water is. So I'm just gonna call
that as Mw, w for water. That's what you need to figure out. Not the mass of the entire water, but how much water got displaced, that's what we need to figure out. Now, how do we do this? How do we figure that out? Well, first of all that value depends on whether our object
is floating or sinking, right? I mean, think about it, if our object were to float like this, let's say somewhat this way, then because only a small
amount of object is submerged, only a small amount of water
would get displaced, right? In that case, we'll have a smaller value. On the other hand if that
same stone where to sink, now, the entire body got submerged. That means that a lot of water must
have gotten displaced, right? So this value would be more. So I guess the first thing we need to do is to calculate or figure out whether this object is
going to float or sink, and then think about how
to calculate this, right? So that's what we do first. Now, how do we do that? How do we figure this out, whether this is going to float or sink? Well, they've seen this before. The condition for floating is that the density of the
object, again rho means density, so the density of the
object should be smaller than the density of the fluid, in this case, the density of the water. So if this stone has
smaller density than water, it will float, if it has larger density
than water, it will sink. So first, and by the way, how do
we calculate density? We calculate density as
mass divide by its volume. And this is something that we talked a lot in previous videos. So if you need more clarity on why these things are there or you need more clarity on this, great idea to go back and watch those videos on densities
and condition for floating. So anyways, the first step
for us is to calculate what the density of the
object is, our stone, and figure it out whether it
is going to float or sink. In fact, can you try this first, before we do it over here. We know the mass of the stone, we know the volume of the stone, so can you go ahead and
calculate what the density of the stone is, and check whether it's
going to float or sink. Go ahead, give it a try. Okay, let's do it. So the density of our object is going to be, of our object, is going to be its mass
divided by its volume. So it's 30 grams divided by 40 cm cubed, that will be the zero's cancel, 3 by 4. 3 by 4 is just 0.75, 0.75 grams per cm cubed. The density of our object that means is 0.75 grams per cm cube, density of water is one gram per cm cube. So object has a smaller
density than water, which means the condition
of a floating is satisfied, and that means our
object is going to float. So here's a stone, that is going to float. So it might be somewhat like this. So if I put the stone in water, it's gonna look somewhat this way, it's gonna be floating there, okay? Now that we know it's gonna float, next is how do we calculate the
mass of the displaced water? How do I do that? So in this example, in this case, this much water is going
to get displaced, right? That much water is
going to move up, right? Let me just take that much
water and put it at the side. So here is the liquid that got displaced. It's this much amount of
liquid that actually moves up, but I'm putting it to the
right over here so we can see, and we need to calculate
what this mass is. What is the mass of this water? How do we do that? How do we figure this out? Well, we can use Archimedes'
Principle to figure this out. So Archimedes' Principle says that whenever an object
is submerged in liquid, there's an upward force acting on it, let me use this color, upward force acting on it
called the buoyant force, and that force equals the
weight of this liquid, this displaced liquid. So whatever is the weight
of this displaced liquid, whatever it's weight is, that weight equals the buoyant force. Which means if I displace more liquid, it'll have more weight and so the buoyant force will increase. That's what's Archimedes Principle, okay? Now, how does that help? Well, think about the
forces on this stone. It's being pushed up by
the weight of this liquid, the buoyant force, but it's
also being pushed down, it's also being pushed
down by its own weight by, due to gravity. So this is the weight of the stone and this is the buoyant force, which equals the weight
of the displaced liquid. Now, we know that our was
stone is floating, right? If the stone is floating
and just staying there, it's stationary over there, what can we say about these two forces? A, they have to be equal
to each other, right? Because if this force
was larger than this, the store would be rising up. If this force was larger than this, the stone would be sinking. So the two forces must be exactly equal, which means the weight of this stone should equal the weight
of this displaced liquid. So displaced liquid has the
same weight as the stone, or the mass of that liquid should be the same as that of the stone. Because (indistinct) weight
is just mass times G. So if the weights are the same, their masses must be the same as well. So from this, we can say, the liquid should have the
same mass as the stone. And you know this mass
of the stone is 30 grams and so from this, we can
say the mass of that liquid or the mass of that
water, that is this water, has to be 30 grams. So whenever objects are floating, this means that the displaced liquid or the displaced fluid
should have the same mass as that of the object. Because the object is stationary, the two weights have to
be equal to each other. Well, what if the object is sinking? Well, let's do one more problem for that. So here's the second problem, and the only difference in
the second problem is that the mass of the stone and the volume of the stone has changed. Everything else is the same. It's dropped in water, we need to calculate the
mass of the displaced water. Since the mass and the volume
of the stone has changed, we have a different stone
altogether, so that's it. Here is our new stone. And again, if we calculate its density, that's if we quickly go ahead and do that, which I'm pretty sure you can do, so I'm just directly writing it, if we directly go ahead and do that, we'll this time, we'll
get it to be 15 grams (indistinct), five grams
per centimeter cube. That means this time, the object has a larger
density than water, it's denser than water. So this condition is not met, and so the object will sink. So if I put that object inside the water, it's going to sink. And as a result, it's going to displace
that much amount of water. So whatever is the amount
of water present over here, that will move away, and again, this is the displaced water. Okay? Now, can I do the same
thing as I did before? Can we say that the stone
is being pushed up with a buoyant force, which equals the weight of the displaced liquid, and it's being pushed
down by its own weight and the two forces are balanced, so this should equal this, so that means the two
weights should be equal and so their masses should be equal, can I do the same thing as before? No, I can't do that. Can you think why? Pause the video, this is super important, okay? Can you think why now I can say that their
weights must be the same? Think about this. Okay, well, in the previous case, the object was floating, it was at rest. And that's why the forces were balanced, that the buoyant force
was balancing the weight and that's why the two
had the same weight. But this time object is sinking. If it is sinking, it's accelerating down, the forces are not balanced. Does that make sense? It's sinking, therefore the buoyant force is smaller than the weight of the object, otherwise it would never sing,
it would just stay there. Therefore, this is smaller than this, and so the weight of the
displaced liquid will be smaller than the weight of the stone. So the mass of the liquid will be smaller than
the mass of the stone. Mass of the stone is equal to 50 grams, so most of the liquid won't be 50 grams. This time displaced water won't have 50 grams, it'll be less than that. Does that make sense? Okay, so how do we calculate this time? What do we do? Because the forces are not
equal, I can't equate them, so this time what to do? Well, this time, let me write that down
over here, this time because our object is
completely submerged, we know that the volume
of the displaced liquid should be equal to the
volume of the stone, right? Think about it, because the opposite is
completely submerged, water has to move away to make space for our entire object. And so this much, whatever
is the volume of that stone, that much water must have gone
away to make space for that. So I know for sure now that the volume, I don't
know the mass of the liquid, but I know the volume
of that displaced liquid volume of that water displaced, that should be the same as
the volume of the stone, which is 10 cm cubed. Okay? And now I know the volume, I know the density of the water. can I calculate the mass? Yes, we can. Again, good idea to pause the video and see
if you can try yourself, all right? So density is mass divided by volume. So mass will be density
times volume, right? I'm pretty sure you can do that. So mass will be density of water, mass of the water times volume of the water, and that would equal now, density of water is given
as one gram per cm cubed, times room of water, we just found out this is the displace water, okay? And so the cm cubed cancels
out and we get 10 grams as the onset. And so the amount of liquid displaced, the mass of liquid displaced, we see is 10 grams, smaller than the mass of the stone. That's why the rate is smaller, buoyant force is smaller, and that's why it's sinking down. So long story short, when things are floating, we can say that the weight
equals the buoyant force, which is the same as the
weight of the displaced liquid. And so that means the two
which should be equal, and so when things are floating, the mass of the displaced liquid should equal the mass of the object, okay? On the other hand, what happens
when things are sinking? When things are sinking, the weight of the object is
more than the buoyant force. So this weight is more than this weight, but because it's completely immersed, this time we can say that their volumes must be the same. So this time we say the
volume of the displaced liquid equals the volume of the object. But of course, don't think
of these as formulae, we can easily get confused, instead, always use logic
to arrive at these steps.