What percentage of a boat is above water when it has a density of 0.75 g/cc?

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    Buoyancy
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A boat with a density of 0.75 g/cc floats in water, which has a density of 1 g/cc, resulting in 25% of the boat being above water. The buoyancy force is equal to the weight of the water displaced, which can be calculated using the formula Fg = Fb, where Fb is the buoyant force. The mass of water displaced is determined by multiplying the density of water by the volume submerged. A Free Body Diagram can help visualize the forces acting on the boat, confirming the relationship between weight and buoyancy. The discussion emphasizes the importance of understanding buoyancy principles to solve such problems accurately.
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Homework Statement


A boat with a density of 0.75 g/cc is floating in water. What percentage of the boat is floating above the surface of the water?


Homework Equations


I don't know any general formulas for buoyancy. I was hoping someone here could provide one.


The Attempt at a Solution


I understand that the density of water is 1 g/cc, so is it really as easy as understanding 1-0.75, or is that simply a coincidence. I know the answer is 25% and expected a rather low number due to the rather low density, but what is the exact procedure for solving this problem?
 
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Every cc of boat in the water receives 1cc weight of water worth of pushing up.
So 1g/cc of flotation, if the boat weights 0.75g/cc then you get 0.25g/cc of overall flotation.
 
That pretty much does it.
It would be clearer if you started with Fg = Fb
where the buoyancy force Fb = (mass of water displaced)*g.
The mass displaced is the density of water times the volume of water displaced.
 
Draw a block partly submerged in the water, and use that at as a Free Body Diagram. Put the forces on it, a bouyant force and a weight force, and then set up the statics problem. Then work it out as Delphi51 says.
 
Thanks everybody. That was my line of thinking, but I wanted to be sure I wasn't mistaken.
 
As Archimedes is supposed to have said in his bath:
"Eureka"
 
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