Calculate new water height in container after adding floating object

In summary, the conversation discusses the calculation of the new fluid height within a container after adding a floating solid, including the principles of displacement and buoyancy. The formula for determining the new fluid height is also mentioned, and clarification is provided regarding the concept of displacement and the role of Archimedes' principle.
  • #1
AmoLago
1
0
Hi all,

I've been looking at some buoyancy problems and one continues to vex me, how to calculate the new fluid height within a container after a floating solid is added to it.

I understand that for a completely submerged solid Archimedes comes into play, volume of the solid equals volume of the displaced fluid. Volume of the fluid divided by the surface area of the fluid in the container added to the original fluid height gives me the new fluid height... hwith object = (Vsolid/Aliquid)+hstart ...easy.

Searching around I find that a floating solid displaces it's own weight in fluid. This sounds great, however, I also find that the percent of the solid submerged is equal to the difference between the fluid density and the solid density. So I understand that and object with average density of 0.4g/cm3 will have 40% of it's body submerged in water.

But, what I don't get is that in the case of the water surface area in the container being very close to the surface area of the solid, only one of the above can be true. Either it displaces it's weight in fluid and sinks below the surface, or it remains partially submerged and it doesn't replace it's own weight in fluid.

So I assume there must be a formula to combine the two to work out how high the the fluid will be. Could someone please tell me what it is, or at least tell me what I've misunderstood?

I've attached a picture to help demonstrate what I'm on about.

Cheers
Amo
 

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  • #2
I think you misunderstand what's meant by displacement here.
The 'displaced' volume is exactly the volume of solid that's below the (final) waterline. If a rectangular block of density 40% that of water floats level, 40% of it will be submerged.

Btw, you've connected Archimedes with the wrong principle. That the submerged volume is the displaced volume would have been obvious. What Archimedes realized is that the buoyancy (upward force) is equal to the weight of the displaced fluid - and that is true whether or not the object floats.
 

What is the formula for calculating the new water height after adding a floating object to a container?

The formula for calculating the new water height is: Final Water Height = Original Water Height + (Volume of Floating Object / Area of Container).

What units should be used for the volume of the floating object and the area of the container?

The units for the volume of the floating object and the area of the container should be consistent. It is recommended to use the metric system, such as centimeters or meters, for more accurate calculations.

Can this formula be used for any shape of container and floating object?

Yes, as long as the volume of the floating object and the area of the container are accurately measured, this formula can be used for any shape of container and floating object.

What if the floating object is partially submerged in the water?

In this case, the volume of the floating object should be adjusted to account for the submerged portion. For example, if half of the floating object is submerged, its volume should be halved before plugging it into the formula.

Are there any factors that could affect the accuracy of this calculation?

Yes, there are a few factors that could affect the accuracy of this calculation, such as the density of the floating object, the buoyancy force acting on the object, and any other objects in the container that may affect the water level. These factors should be taken into consideration for a more precise calculation.

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