Archimedes' Principle - Submerged object in a fluid

In summary: Given its mass, you can use the density of iron to get its volume. Subtract that from the volume of the block to get the volume of the wood.
  • #1
somemonkey
1
0

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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  • #2
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).
 

Related to Archimedes' Principle - Submerged object in a fluid

What is Archimedes' Principle?

Archimedes' Principle is a fundamental law of physics that states that the buoyant force exerted on an object immersed in a fluid is equal to the weight of the fluid that the object displaces.

How does Archimedes' Principle work?

Archimedes' Principle works by comparing the weight of an object in air to its weight when submerged in a fluid. The difference in weight is equal to the weight of the fluid that the object has displaced, which is also known as the buoyant force.

What is the relationship between an object's density and its buoyancy?

The relationship between an object's density and its buoyancy is inverse. An object with a higher density than the fluid it is submerged in will sink, while an object with a lower density will float. This is because the weight of the fluid that the object displaces is greater than the weight of the object itself, causing it to float.

How does the shape of an object affect its buoyancy?

The shape of an object does not affect its buoyancy, as long as the object's volume remains the same. This is because the buoyant force is dependent on the weight of the fluid displaced, which is determined by the volume of the object, not its shape.

What is the practical application of Archimedes' Principle?

Archimedes' Principle has many practical applications, including determining the buoyancy of ships and submarines, designing hot air balloons and blimps, and understanding the behavior of objects in different fluids, such as oil and water.

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