Bouyant force with water and in equilibrium

AI Thread Summary
A cube of wood with a density of 651 kg/m³ and an edge length of 20.9 cm is floating on water, prompting a calculation of the distance from the top of the cube to the water level. The participant correctly applies the equilibrium condition, stating that the sum of forces must equal zero, using the equation involving the density of the cube and the buoyant force. They also consider using the mass of the cube to represent the downward force but realize that it relates to the same gravitational force as the density and volume approach. The discussion emphasizes the importance of Archimedes' principle in understanding buoyant force. The participant seeks confirmation of their understanding and calculations regarding buoyancy.
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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)
 
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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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