Is it possible to determine the vertical centre of mass without knowing the mass of an object?

I live in a floathouse. This is a house floating on water on top of two pontoons. I can determine all length measurements but can not for obvious reasons weigh the house. I have seen on tv when they use weights to tilt a boat and use that combined with distance and area measurements to determine the centre of buoyancy but I can not find any information online how to do this.

Do you know?

  • $\begingroup$ the center of mass and the center of buoyancy are two different things. engineeringtoolbox.com/centre-gravity-buoyancy-d_1286.html . $\endgroup$
    – anna v
    Jan 9, 2018 at 6:52
  • $\begingroup$ Can you measure the volume of water displaced by the house? (i.e. the volume of the submerged portion of the house) $\endgroup$
    – The Photon
    Jan 9, 2018 at 17:15
  • $\begingroup$ @annav right sorry I am just figuring this out. I believe we might need to figure out both to determine the stability but I feel like we should be able to since we can determine the lifting power of the pontoons and we know the shape $\endgroup$ Jan 10, 2018 at 17:08
  • $\begingroup$ @ThePhoton Yes I can determine that, I know the shape of the pontoons and I know the type of foam they are filled with also. $\endgroup$ Jan 10, 2018 at 17:08
  • $\begingroup$ Then you can get the weight of the boat, because the weight of the boat is equal to the weight of the water it displaces. $\endgroup$
    – The Photon
    Jan 10, 2018 at 17:33

1 Answer 1


Assuming you know the exact shape, you do know the mass, as it is equal to the density of water multiplied by the volume displaced. I will assume you mean that you don't know the distribution of the mass.

You know (or can calculate) the horizontal location of the center of mass--it is vertically aligned with the center of buoyancy, which you get from the centroid of the submerged portion (if it was not so aligned, there would be a torque which would tilt the house until alignment was achieved). So imagine a vertical line passing through this position, and know that the CG of the house lies on that line.

Add a known amount of weight (let's make it equal to the weight of the house--this is not necessary but makes the arithmetic a little simpler) in a known spot and let the house come to equilibrium. Then calculate the new center of buoyancy (it's different now because the submerged portion is different). The center of gravity of the house+weight is now vertically aligned with this center of buoyancy.

The center of gravity of the house alone still lies along the line you determined before, now no longer vertical, as it tilted over with the house when you added the weight. The CG of the house+weight lies at the midpoint of the line connecting the weight with the CG of the house.

There is only one spot along the tilted line containing the CG of the house that satisfies this requirement: the midpoint of the line connecting it to the weight lies directly above the new center of buoyancy. This spot is the CG of the house.

  • $\begingroup$ can you elaborate more about the mass? I can determine the volume of the structure that is under water but I don't think that would equate directly to the mass as the house is floating. The pontoons have foam underneath them which lift the house more. $\endgroup$ Jan 10, 2018 at 17:11
  • $\begingroup$ It is Archimedes' principle: the weight of the floating object is equal to the weight of the displaced fluid. You need to know the volume of everything that is below the water surface, including the foam. Multiply that by the density of water, and you have the mass of the house. $\endgroup$
    – Ben51
    Jan 10, 2018 at 17:13
  • $\begingroup$ I can determine the horizontal centre of gravity of the structure as it is evenly balanced so it would be in the centre of the barge the house is on but I do not know the centre of gravity of the house itself and I do not know how high above or below the water the vertical centre of gravity is. The foam under the house is not evenly distributed, it is distributed so that the house sits flat on the water but the back of the house is heavier as this is where the appliances and septic are. There is additional foam in the back. $\endgroup$ Jan 10, 2018 at 17:15
  • $\begingroup$ So the "lifting power" of the foam doesn't come into play? If there was a different kind of foam used the house would sit lower. To me this doesn't make sense because then in that case using your explanation the house would be heavier, but it wouldn't actually be, it would just be lifted less. $\endgroup$ Jan 10, 2018 at 17:16
  • $\begingroup$ You must include the foam in the definition of what constitutes the "house". If you use heavier foam, yes, the house will sit lower, and more volume will be displaced, and you will get a larger value for the mass of the house. Which is correct. You added mass by switching to heavier foam. $\endgroup$
    – Ben51
    Jan 10, 2018 at 17:20

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