Archimedes' Principle - Submerged object in a fluid

AI Thread Summary
The discussion centers on a physics problem involving Archimedes' Principle and the behavior of a wooden block with an iron ball attached. For part (a), the submerged volume of wood is determined by the weight of the iron ball when it is not submerged. In part (b), it is suggested that the submerged volume of wood will decrease when the block is inverted, as the iron ball displaces its own volume of water. However, participants note that the problem lacks necessary information, such as the mass of the iron ball, making it impossible to solve parts (a) and (c) accurately. The conversation emphasizes the importance of using mass and volume correctly in buoyancy calculations.
somemonkey
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Homework Statement


A 1.25kg wooden block has an iron ball glued to one side. (a) if the blocks floats in water with the iron ball "high and dry," what is the volume of wood that is submerged? (b) If the block is now inverted, so that the iron ball is completely immersed, does the volume of wood that is submerged increase, decrease or remain the same? (c) Calculate the volume of wood that is submerged when the block is in the inverted position.


Homework Equations


Volume submerged = V of solid (density of solid/density of fluid)
Density of water = 1000
Density of iron = 7860
Density of wood = 1220

The Attempt at a Solution


So, for part a, I was thinking that since the iron ball isn't submerged, the amount of water is displaces is equal to its weight.
Therefore, the volume of wood that is submerged can be found with (Vsub/Vsolid) = ((density of wood + density of iron ball)/(density of water)), where (Vsub/Vsolid) is the answer.

For part b and c, I was thinking that the volume of wood will decrease, as when the iron ball is submerged, it displaces a volume of water equal to its own volume.
I know that volume of a sphere is V = (4/3)(pi)(r^3), however I don't know exactly how I would plug it into the equation: (Vsub/Vsolid) = ((density of wood + density of iron ball)/(density of water))

thanks in advance
 
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somemonkey said:
Density of wood = 1220
Wood that dense is not going to float, with or without an iron ball.
somemonkey said:
(density of wood + density of iron ball)/(density of water)
It makes no sense to add densities. You can masses and you can add volumes.
somemonkey said:
(Vsub/Vsolid) is the answer.
Being a ratio of volumes, that will give a dimensionless number, not a volume.
somemonkey said:
For part b and c, I was thinking that the volume of wood will decrease, as when the iron ball is submerged, it displaces a volume of water equal to its own volume.
Yes.
somemonkey said:
volume of a sphere is
The shape of the iron mass is irrelevant if you are not given its radius. Unfortunately post #1 does not give its mass either, making it impossible to answer a) and c).
 
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