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Buoyancy: Example

by Jared Rovny
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    00:01 In case there’s a little abstract, we have an example here that might make it a little more clear.

    00:06 We’re asking, if we know the specific gravity of ice is approximately 0.9, what fraction of an iceberg would really be under water? So, let’s see if we can use this principle.

    00:16 If you’d like, you can do this example yourself first by using the slides that we just went through and see if you can find how much of an iceberg would really be underwater given what we just said.

    00:26 If you’ve tried this example, hopefully, it went something like this. You have that the amount of the object is below the water, so you have some sort of an iceberg here.

    00:38 What we’re saying is that this amount that’s below the water, this will be equal to the volume of the displaced water. What we’ll require is that the mass of the water times g, which is the buoyant force pushing upwards on our object will be equal to the force of gravity acting downwards on our object. So, the buoyant force is equal to the mass times gravitational acceleration of the water, the displaced water which is equal to the density times the volume of water.

    01:12 So, this is the mass times the gravitational acceleration. This has to be equal to Fg because our object, our iceberg in this case is floating in equilibrium. We know the force of gravity on our object. So, this is the mass of the iceberg times the gravitational acceleration, g.

    01:32 Here is our condition right here that these have to be equal to each other. So, the density of water times the volume of the water times g has to be equal to the mass of the ice which is the density of the ice times the volume of the ice times g. Canceling the g’s, we see exactly the condition that we just discovered. The volume of the ice divided by the volume of the water has got to be equal to the density of water over the density of ice. If we rewrite this side by flipping both sides of the equation saying 1 over each side. So, we could rewrite it in this way or rewritten it this way instead. We can have the volume of the water over the volume of the ice.

    02:17 In other words, the ratio of the amount that’s submerged over the total volume of the ice cube has got to be equal to the density of the ice divided by the density of water. You might remember, this is exactly the definition of the specific gravity of the ice which we were given in this problem to be 0.9. So now, we know that the ratio of the volume of the water, in other words, the volume down here, the volume that’s displaced to the volume of the entire ice cube, the entire iceberg here has got to be 0.9. So in other words, 90% of an iceberg really is under water.

    03:00 This is an example of how you could solve a specific gravity problem or a buoyancy problem by again comparing the mass of the water displaced to the mass of your object as a whole. Again, always remembering that you can always rewrite the mass of an object as its density times its volume.

    03:20 Here’s a second example using a classic idea of Archimedes principle which asks, if a king gives Archimedes a crown and asks him to find out whether it’s actually made of gold, he can weigh the crown to be 10 kilograms and then submerge it in water to find that its weight lessens in water because of the buoyancy force. So, the weight lessens from a particular value, 93 newtons from 98 newtons. The question is if we know the experimental specific gravity for gold is somewhere between 19 and 20, is it possible that we can confirm that this crown might be made of gold? Give this problem a shot on your own using the definitions that we’ve just introduced. Then we’ll try it here as well. In this situation, what we have is quite straightforward. You have an object, some crown. I don’t really know how to draw a crown.

    04:13 That’s alright. But we have an object and then we submerge it in water. So, if we put this same object under water (I should stop trying to draw a crown), then what’s going to happen is its weight is going to be lessened by the buoyant force. From this problem, what we know is that the weight of the crown was initially 98 and that when the buoyant force is alleviating it, the weight is now only 93 newtons. This means that the buoyant force must be 5 newtons upwards.

    04:46 So, we’ll keep this in mind as we go through the rest of our problem. What we would like to know is what the density of our object is. The density of our object will be equal to the mass of the object divided by the volume of the object. Now, the mass of the object is not so bad because we’re giving that in the problem. Mass equals 10 kilograms. The question is can we somehow find the volume of our object just based on the information that we’ve got in this problem. Let's keep this in mind as we go. Let’s see if we can use this buoyant force equation to figure out what the volume of our object is since the crown is completely submerged.

    05:26 Well, we have that the buoyant force is equal to the weight of the water, which is the mass of the water times g. We also know that the mass of the water, one more time, can be written as its density times the volume and that this will be the buoyant force of the water.

    05:44 So now, we know what the volume of the displaced water is because we can solve this equation by saying the volume of the water equals the buoyant force divided by the density of water times g.

    05:55 Everything you see on the right hand side of this equation are things that we know.

    05:59 We know the buoyant force from the givens in the problem that’s up here. We know the density of water.

    06:04 We know the gravitational acceleration. The key thing is that this volume of water that’s displaced will be equal to the volume of the crown because the crown is completely submerged in the water.

    06:14 So, this is something we can use in order to find the density of our crown. So, let’s do that.

    06:18 The density of the crown is its mass divided by its volume, where we found the volume from here, the buoyant force divided by the density of water times g. We can just rewrite this very quickly and see that this is mass of the crown times the density of water times g divided by the buoyant force which we found to be 5 newtons. Now, one thing we might want to do here since we have the density of the crown on this left hand side and we have the density of the water on the right hand side is divide both sides by the density of the water because that way, we can find this quantity which is the specific gravity of our crown. We know from the problem that for gold, this should be between 19 and 20. So, let’s see what happens if we just write this.

    07:04 Remember, we’ve divided out by this density of water. So, we now have mg divided by the buoyant force. Writing this out, this will be approximately equal to the mass of the crown which is 10 kilograms times the gravitational acceleration which we'll approximate and that’s reason for this here, that's 10 meters per second squared divided by the buoyant force which we saw to be 5 newtons. Fortunately, we have kilograms meters per second squared in the numerator and newtons in the denominator. So, the units cancel as we expect.

    07:35 10 times 10 is 100 then we have 5 where 100 divided by 5 is 20. So, this is the specific gravity of our crown.

    07:45 If you need it to be anywhere between 19 and 20 where we've slightly overestimated by using 10 instead of 9.8, this is a great estimate for the density of gold. So in this case, whoever made this crown gets off scot-free because we can very easily analyze what the density of the object that we were given is by submerging it in water and therefore finding the volume of the object using Archimedes principle for buoyancy.


    About the Lecture

    The lecture Buoyancy: Example by Jared Rovny is from the course Fluids.


    Author of lecture Buoyancy: Example

     Jared Rovny

    Jared Rovny


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