Floating Wood: Finding the Ratio Above Water

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A block of wood with a density of 750 kg/m3 is floating in water, which has a density of 1000 kg/m3. To determine how much of the wood is above the water, the principle of buoyancy must be applied, where the weight of the displaced water equals the weight of the wood. The discussion suggests using a symbolic approach to relate the densities and volumes involved. The ratio of the densities indicates that approximately 25% of the wood would be above the water's surface. Understanding the forces acting on the wood is essential for solving the problem correctly.
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A block of wood with a density that is 750 kg/m3 is floating in a tank of water. How much of the wood is above the surface of the water



D=M/V



I've been trying for hours to try to figure this one out. The only info I'm given the the density of a block of wood in the water at 750kg/m^3. I'm thinking that it has something to so with the approx. density of water at 1000kg/m^3, so the ratio is 4:3 so that approx 25% would be above the water, but not at all sure I'm on the right track.
 
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You're on the right track. Now all you need to do is derive it :-)

Why not try a strictly symbolic approach. Let the density of the wood be r1 and that of the water r2. Now suppose the volume of the block is v. What's the weight of the block? How much water needs to be displaced for it to float? ...
 
thank you ...I think I get it!
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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