Question

A homogeneous log of relative density 0.81 is 3 metres long, 0.5 metres square cross-section, and is floating in fresh water. Find the displacement of the log, and the distance between the centres of gravity and buoyancy.

Answer 1

I understand the 0.5m square cross-section to mean that the side of the square cross-section is 0.5m, making its area 0.25m². Volume of the log=3×0.25m³=0.75m³. Relative density is based on the density of freshwater (1000kg/m³), so the density of the log is 810kg/m³, and its mass is 0.75×810=607.5kg. It displaces 607.5kg of water (which counterbalances its weight). This I understand to be the displacement. Density of water=1000=mass/volume, so 607.5kg has a volume of 607.5/1000=0.6075m³.

The base area of the log is 3×0.5=1.5m², therefore the water level is 0.6075/1.5=0.405m above the base of the log. The centre of buoyancy (COB) is the centre of gravity (COG) of the displacement volume which is half this distance=0.2025m.

The COG of the log (when not immersed in water) is half its height, that is, 0.5/2=0.25m. The distance between the COG and COB=0.25-0.2025=0.0475m=4.75cm.

CORRECTION

I discovered a typing error in one of the calculations and I was using the wrong criteria for measuring the COB. These errors are now corrected.

Comments

Thanks for your correction. I have guessed this solution since I am not familiar with the term "centre of buoyancy". I was about to get more details on what centre of buoyancy means when I saw your comment. I will revise my solution when I have more details. I've also assumed that the centre of gravity refers to the log when it is out of water, rather than when in water. Nevertheless I hope my answer has been a little helpful.