A Thermodynamic (supposed to be easy) Problem

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To float a steel tank weighing 10,000 kg at a depth of 10 meters in the ocean, the total weight must equal the weight of the water displaced. The tank's cross-sectional area is 3 m² and its height is 16 m, giving it a volume of 30 m³. The density of seawater is approximately 997 kg/m³, leading to a displaced water weight of about 29,910 kg. Therefore, to achieve buoyancy, 19,910 kg of concrete must be added to the tank. This calculation confirms the necessary weight for the tank to float at the desired depth.
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Homework Statement



A steel tank of cross sectional area 3 m2 and height 16 m weighs 10,000 kg and is open at the top. We want to float it in the ocean so that it is positioned 10 m straight down by pouring concrete into its bottom. How much concrete should we use?


Homework Equations





The Attempt at a Solution



I have thought about Potential gravity. I have looked that Waters density is 997 kg/m3. I got this container's ensity (It is 208.33) but I just don't know how to solve it! :) I don't remember much about buoyancy and Bouyancy might be the thing needed here!

Answer is 19910 kg...Please help! :)
 
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¿you need to have 10 meters down water and 6 up water?

Then you have to put inside the tank same weight as the weight of water evacuated. (Arquimides)

Volume of the tank inside water (section X 10 m) ---> Volume of water evacuated ----> density of water -----> mass of water evacuated = 10000kg of the tank + x kg of concrete

¿It can be?
 
30 m^3 of tank ------> 30 m^3 of water ----- x 997kg/m^3 -----> 29910 kg

Then your tank have to weight 29910; 10000 of the tank and 19910 of the concrete
 
Awesome! THanks sooo much! :)
 
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