It will still only displace its own weight in water whatever shape it is.
The shape will contribute or otherwise to stability but not displacement.
Maths questions - buoyancy
- Thread starter englishpremier
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It will still only displace its own weight in water whatever shape it is.
The shape will contribute or otherwise to stability but not displacement.
The foam won't effect buoyancy in normal conditions* but there could be many reasons for doing that:
-It'll stop any moisture building up in the chamber and causing mould/rot which could weaken the structure.
-It'll increase the strength of the chamber walls significantly.
-Even a small amount of water leaking into the chamber over time could effect the stability when it's weight moves with the centre of mavity of the boat
-Presumably there's a chance the chamber could be breached so filling it with foam would negate that?
*other than the weight of the foam increasing the total weight of the boat - which is probably a negligible difference?
while technically true surely the effect in reality is rather negligible, I doubt the boat is noticeable lower in the water just because some chambers are filled with foam?
1)
Buoyancy force is:
Fb = ρf·V·g
Where ρf is the fluid in which the solid is floating.
As both boxes have the same volume they will experience an identical buoyancy force at the same displacement from the surface.
To find the equilibrium position the force of mavity plus the buoyancy force must equal 0:
Fb + (Wc + Wi) = 0 -> Fb = -(1 + Wi) -> Fb = -1 - m·g -> Fb = -1 - ρ·V·g
As the accelerational constant of g is negative...:
Fb = ρ·V·g - 1
...where V and g never change, then as ρ increases, so must Fb. As the density of the foam is greater than the density of the air inside the other cube, then the buoyancy force experienced by the foam containing cube is greater. Since buoyancy is directly proportional to density, volume, and mavity, where density of fluid and mavity never change, then the immersed volume must increase. Thus we can infer since the cube containing the foam has a higher buoyancy force acting on it, it sits lower in the water.
2)
If we make the assumption that neither cube sinks then the immersed volume again is identical at the same displacement from the surface. We can simply assume the cube itself weighs 101kg i.e.
Fb = -(101 + Wi)
Since as the cube sinks only the immersed volume changes, and ρ and g never change (in the same cube) then they can be neglected. Thus we are looking for the change in Fb for each example:
Δa = (ρa - 1) / (ρa - 101) = (1.225-1) / (1.225 - 101) = a change of 99.77%
Δf = (ρf - 1) / (ρf - 101) = (40-1) / (40 - 101) = a change of 36.07%
Therefore the added mass has more of an effect on the air filled box. However, the foam filled box will still sit lower in the water.
Buoyancy force is:
Fb = ρf·V·g
Where ρf is the fluid in which the solid is floating.
As both boxes have the same volume they will experience an identical buoyancy force at the same displacement from the surface.
To find the equilibrium position the force of mavity plus the buoyancy force must equal 0:
Fb + (Wc + Wi) = 0 -> Fb = -(1 + Wi) -> Fb = -1 - m·g -> Fb = -1 - ρ·V·g
As the accelerational constant of g is negative...:
Fb = ρ·V·g - 1
...where V and g never change, then as ρ increases, so must Fb. As the density of the foam is greater than the density of the air inside the other cube, then the buoyancy force experienced by the foam containing cube is greater. Since buoyancy is directly proportional to density, volume, and mavity, where density of fluid and mavity never change, then the immersed volume must increase. Thus we can infer since the cube containing the foam has a higher buoyancy force acting on it, it sits lower in the water.
2)
If we make the assumption that neither cube sinks then the immersed volume again is identical at the same displacement from the surface. We can simply assume the cube itself weighs 101kg i.e.
Fb = -(101 + Wi)
Since as the cube sinks only the immersed volume changes, and ρ and g never change (in the same cube) then they can be neglected. Thus we are looking for the change in Fb for each example:
Δa = (ρa - 1) / (ρa - 101) = (1.225-1) / (1.225 - 101) = a change of 99.77%
Δf = (ρf - 1) / (ρf - 101) = (40-1) / (40 - 101) = a change of 36.07%
Therefore the added mass has more of an effect on the air filled box. However, the foam filled box will still sit lower in the water.
It's interesting when you put it in those terms. Intuitively you'd think if it had a higher buoyancy force acting on it compared to an identical sized object, then it would sit higher in the water. But of course the opposite is true. An object with ~0 buoyancy force acting on it would sit basically on the surface of the water.