Archimedes' Principle - Submerged object in a fluid

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SUMMARY

The discussion centers on Archimedes' Principle as applied to a 1.25 kg wooden block with an iron ball attached. For part (a), the submerged volume of wood is determined by the weight of the block and the density of water, leading to the conclusion that the volume submerged is a function of the total weight divided by the fluid density. In part (b), when the block is inverted, the volume of wood submerged decreases because the iron ball displaces its own volume of water. The calculations require precise values for the iron ball's mass and radius to solve parts (a) and (c) accurately.

PREREQUISITES
  • Understanding of Archimedes' Principle
  • Knowledge of density and its calculations
  • Familiarity with volume calculations, particularly for spheres
  • Basic algebra for solving equations
NEXT STEPS
  • Calculate the submerged volume using the formula: Vsub = (mass of object)/(density of fluid)
  • Explore the implications of density variations in buoyancy scenarios
  • Learn how to calculate the volume of a sphere using V = (4/3)(pi)(r^3)
  • Investigate the effects of different materials on buoyancy and stability in fluids
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Students studying physics, particularly those focusing on fluid mechanics and buoyancy, as well as educators looking for practical examples of Archimedes' Principle in action.

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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