Buouyancy force and Archimedes' principle

AI Thread Summary
The discussion centers on calculating how much more a fishing boat will submerge when loaded with 3.0 m³ of fish, given the fish's density of 0.90 kg/dm³ and using the principle of buoyancy. Participants emphasize the need to equate the weight of the fish with the weight of the displaced water to find the additional submerged volume. The density of water is assumed to be 1 kg/dm³ for the calculations. The correct approach involves using the equation for buoyancy, balancing the mass of the fish against the mass of the displaced water. The focus remains on determining the additional displacement caused by the added weight of the fish.
lin1430
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How much more will the volume of a fishingboat go under water, if I load the boat with 3.0m^3 fish with the density 0.90kg/dm^3?

Fish : 3.0m^3
Density of fish: 0.90kg/dm^3?

Homework Equations



Archimedes principle: density * volume * g[/B]

The Attempt at a Solution



Tried setting upward force = downwards force (mg)

substituting m for ro*V[/B]

changing V to A*h, then finding h. But it is obviously wrong.
 
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Unfortunately it's not possible to comprehend your attempt to solve the problem. Please try to write down all of your steps, explaining your used symbols.

The principle of Archimedes states that the buoyancy force acting on a body corresponds to the weight (force) of the displaced fluid. My first guess, after reading your post, would be that you mixed up the density of the water and the fish...
 
This was my attempt
 

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lin1430 said:
This was my attempt

I think it would be easier to calculate the additionally displaced volume of water, when the fish is loaded on the boat. You can calculate that with the equation below your drawing, all you need is the values for the densities of fish and water and the volume of fisch you want to load. The data of the fish is provided as you wrote in your first post. Do you have any density given for the water? If not, I propose to use 1 kg/dm^3. But with these three values you are able to calculate the displaced volume by the weight of the fish. I don't know if this is already the demanded answer or if you have to calculate how much the boat will immerge (if the geometry of the boat is given).
 
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stockzahn said:
I think it would be easier to calculate the additionally displaced volume of water, when the fish is loaded on the boat. You can calculate that with the equation below your drawing, all you need is the values for the densities of fish and water and the volume of fisch you want to load. The data of the fish is provided as you wrote in your first post. Do you have any density given for the water? If not, I propose to use 1 kg/dm^3. But with these three values you are able to calculate the displaced volume by the weight of the fish. I don't know if this is already the demanded answer or if you have to calculate how much the boat will immerge (if the geometry of the boat is given).

Sounds good. The density of water is to be 1kg/dm^3. I do not have to calculate how much the boat wil immerge, just how much additional change will happen. But how would I proceed to calculate the displaced volume by the equation under my drawing?
 
lin1430 said:
Sounds good. The density of water is to be 1kg/dm^3. I do not have to calculate how much the boat wil immerge, just how much additional change will happen. But how would I proceed to calculate the displaced volume by the equation under my drawing?

As you wrote in your first post, balance of forces must be fulfilled after loading the fish. The additional weight of the fish must be compensated by additional displaced water. Just equal the mass of the loaded fish (##\rho_FV_F##) and the mass of the displaced water (##\rho_WV_W##); the gravitational acceleration(s) ##g## cancels each other on both sides of the equation. You stick with one unknown (##V_W##) for which you have one equation to solve for it.
 
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