Register to reply

Calculate new water height in container after adding floating object

by AmoLago
Tags: adding, container, floating, height, object, water
Share this thread:
Aug31-12, 10:03 AM
P: 1
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 in to 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.

Attached Thumbnails
Phys.Org News Partner Physics news on
Physical constant is constant even in strong gravitational fields
Physicists provide new insights into the world of quantum materials
Nuclear spins control current in plastic LED: Step toward quantum computing, spintronic memory, better displays
Sep1-12, 02:06 AM
Sci Advisor
HW Helper
P: 9,938
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 realised 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.

Register to reply

Related Discussions
Usiing angles and height to calculate height/altitude of object Differential Geometry 0
How to calculate the height of something by throwing an object off it? General Math 4
Object floating on the surface of water Classical Physics 2
Lenses: If given object height and focal length, is it possible to calculate... Introductory Physics Homework 4