Metallic coin over wooden block in glass of water - buoyancy

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CaptCoonoor
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Homework Statement


So there is some water inside a container; the height of water inside the container is l. I placed a wooden block on the water and it's floating to some height x, on top of the block is a metallic coin (see the diagram below).

What will happen if I throw that coin inside the water? Will the Height x and l change? Can we derive an expression for this change? I tried doing it, but failed.

cdtEi.png

Homework Equations


Fb=gρhA

The Attempt at a Solution


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Here is what I've tried Let's assume that our wooden block has mass m1 Volume v1height h1, height underneath the water as y and let's assume mass of Metallic coin as m2 Now total mass on block will be equal to (in which buoyant force is actin upon) : m = m2 + m1 and let's also say that total height h1 = x + y

Principle of buoyancy of a liquid is Give as : Fb=gρhA
Simplifying :
Fb=g(m/v1)(x+y)A ... {1}

Now, archimedes principle is given as F=gρhA
F=g(ρf - ρb)hA // ρb is Density of the Body

And then I am not sure what to do next .. Can anyone help me out?
 
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Discussion

haruspex said:
Where ρ, h and A are what?
A - Area
ρ - density
h -
CaptCoonoor said:
h1 = x + y
 
haruspex said:
Sure, but of what?
Area of Wooden Block, ρb --> Density of Body
ρf --> Density of Liquid (water)
h is of Wooden block too
 
CaptCoonoor said:
Area of Wooden Block, ρb --> Density of Body
ρf --> Density of Liquid (water)
OK, but you previously posted
CaptCoonoor said:
Principle of buoyancy of a liquid is Give as : Fb=gρhA
Simplifying :
Fb=g(m/v1)(x+y)A ... {1}
If ρ is the density of the liquid, why do you have m/v1 there (where you have defined m = m1+m2)?
CaptCoonoor said:
h is of Wooden block too
If h is the total height of the block, then Ah would be the volume of the block. If you multiply that by the density of the liquid, what mass will that give you?
 
Let me try to explain again:
CaptCoonoor said:
Principle of buoyancy of a liquid is Give as : Fb=gρhA
Simplifying :
Fb=g(m/v1)(x+y)A ... {1}
This is wrong. The buoyant force does not depend in any way on the mass or density of the object nor on any part of its volume that is not immersed. It depends on the density of the fluid and the volume of the object that is submerged (and gravity, of course).
CaptCoonoor said:
Now, archimedes principle is given as F=gρhA
The principle of buoyancy is Archimedes' principle. It applies independently of whether the set-up is static and of whether the object is fully or partly immersed. Your other equation should come from the fact that the block+coin float.
 
haruspex said:
Let me try to explain again:

This is wrong. The buoyant force does not depend in any way on the mass or density of the object nor on any part of its volume that is not immersed. It depends on the density of the fluid and the volume of the object that is submerged (and gravity, of course).

The principle of buoyancy is Archimedes' principle. It applies independently of whether the set-up is static and of whether the object is fully or partly immersed. Your other equation should come from the fact that the block+coin float.
Didnt Realize it, Thanks .. Gonna rederive it