submerged log

Hi,I have a buoyancy question where I can work out a few things with formulas

q)I dont know how to find out the additional part of the log submerged  after placing a 20kg block on the wood
I used Archimedes principles to find buoyancy force

originally the log dimensions 0.2 X  0.1m X 2m
lies in water with density 1000 kg / m^3

the part of the log submerged in the water has dimensions 0.2 X  0.03m X 2m
 then placing 20kg mass has how much effect on submerging the log
I have tried recalculating the density, Fb but I am getting stuck with extra mass

q)how much weight do I need to fully submerge the log (min mass needed)?
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Paul SauvéRetiredCommented:
Quite simple - in order to sink the log, the density (mass/volume) of the log and the added weight must be greater or equal to that of water.

Since you know total volume of the log and the volume of the log in the water, you can calculate the calculate the volume not under water.

Next, the added volume 20kg mass must be taken into consideration.

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jagguyAuthor Commented:
I add 20kg mass which is 20000cm^3 and converted to m^3 and the addition mass is correct
Paul SauvéRetiredCommented:
>>I add 20kg mass which is 20000cm^3 and converted to m^3 and the addition mass is correct

this calculatuon  gives a density equal to that of water.

you have to decrease the volume of this mass in order to to increase the density!
jagguyAuthor Commented:
yes 20kg is 20kg without any buoyancy factor applied to this weight.
it could be 20kg of lead
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