Bouyant force with water and in equilibrium

bdh2991
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Homework Statement


A cube of wood having an edge dimension of 20.9 cm and a density of 651 kg/m3 floats on water.

what is the distance between the top of the cube to the water level?


Homework Equations



F = density*g*Volume

The Attempt at a Solution



I believe i have the right concept down in which i did the sum of the forces so it would be:

-(density of cube*g*Vtotal) + (density of water*g*Vdisplaced) = 0

I would like someone to confirm this for me and i also wanted to ask...couldn't i just use mg for the force of the cube downward and then add the buoyant force of water to it? (if not then why not?)

for example: -mcube*g + (density of water*g*Vdisplaced)
 
on Phys.org
nvm...i complete forgot that d = m/v, therefor the force due to gravity is the same for mg or density*g*volume...sorry about that
 
Hint: check out Archimedes principle.
 

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