A 950-kg cylindrical can buoy floats vertically in sea-water. The diameter of the buoy is 0.900 m. Calculate the additional distance the buoy will sink when an 80.0-kg man stands on top of it.

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Hey, everyone in this problem, we have a hollow metallic container of mass 400 kg with a square base floating in water. The length of a base is 1 m, a 70 kg person holding luggage of mass, 20 kg jumps into the container and we're asked what extra height of the container will be submerged. So let's imagine a free body diagram of this situation. And we have our container and it has some weight acting downwards and then we get a buoyant force from the water acting upwards. Hm This is related to Archimedes's principle and we're gonna say that up is the positive Y direction. Now, in equilibrium condition, recall that the sum of the forces is going to be equal to zero. We're talking about forces in the Y direction. And so are some of forces in the Y direction is going to be equal to zero. Now, what forces do we have in the wide direction? Well, we have this buoyant force FB and we have this weight acting downwards. So we have minus W and that's equal to zero. This tells us that our buoyant force is going to be equal to our weight W and we're looking for information about the height. OK. So let's try to break these two forces down to get some sort of equation or relationship with the height H. So recall that the buoyant force is given by the density of the fluid. In this case, we are in water. So the density of water, the gravitational acceleration G times the volume of the container that is inside the water. And the weight recall is equal to mg mass times gravitational acceleration. So we can divide both sides by G. OK. We have the density of the water times the volume. Well, this is a square base and so the volume is going to be B squared times the height H and that's equal to M. So let's consider the two cases. OK. Initially, we have a mass, we'll call it M one of 400 kg. OK? We have the container that just has 400 kg in it after the person jumps in. OK. We have a, the mass is M two. This is gonna be the mass of the container 400 kg plus the mass of the person that jumps in 70 kg plus the mass of the luggage, they're holding 20 kg. And so the mass M two is going to be equal to 490 kg. And from our equation with our forces, OK. Using the equilibrium condition for our forces, we have this relationship between the mass M and the height H OK. So we want to find this difference in height related to mass. So let's look at the height that we have in each situation. So we have the density of water times B squared H one is equal to M one. If we wanna isolate for H one, the height of the container submerged in this initial situation, we know that H one is equal to M one row W times B squared. This is going to be equal to 400 kg divided by the density of water. Maybe we can look this up in a table in our textbook 1000 kg per meter. Cued times B square. We're told that the base is 1 m. So we have 1 m squared for the base squared of our container. And this is going to give us an initial height H one of 0.4 m. OK? And so what this tells us is that initially when we have just the weight of the container or just the mass of the container of 400 kg, that that container is submerged 0.4 m. Let's do the same when we have a mass of 490 kg. OK. We have that row W times B squared H two is equal to M two. We can isolate for H two just like we did. For H one, we have M two over row WB squared. This is gonna be 490 kg divided by 1000 kilograms per meter, cubed times 1 m squared. And this gives us a height H two of 0.49 m. And so what this tells us is that after the person jumps in with the suitcase, the container is now going to be submerged 0.49 m. Ok. Now looking back at the question, the question is asking us what the extra height submerged is. Ok? And so the extra height is just going to be the difference between these two heights, ok? It's going to be the height that we're at after the person has jumped in minus the initial height that we were at before that person jumped in. This is gonna be 0.49 m minus 0.4 m. And we get an extra height of 0.09 m. And so the extra height that the container is going to be submerged because of the person jumping into that container with luggage is going to be 0.09 m. And if we look at our answer choices, we see that that corresponds with answer choice. E thanks everyone for watching. I hope this video helped see you in the next one.

A 950-kg cylindrical can buoy floats vertically in sea-water. The diameter of the buoy is 0.900 m. Calculate the additional distance the buoy will sink when an 80.0-kg man stands on top of it.

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Key Concepts

Buoyancy

Weight and Density

Volume Displacement