# Archimedes principle problem? *1 fluid, 1 solid*

I'm having trouble arriving at a finalized answer dealing with the sum of the forces with this problem dealing with Archimedes principle (buoyancy is weight of the displacement).

Here's the question:

A cube of wood whose edge is 12 mm is floating in a liquid in a glass with one of its faces parallel to the liquid surface. The density of wood is 762 kg/m^3, that of liquid is 1296 kg/m^3. How far (h) below the liquid surface is the bottom face of the cube?

Relevant equations

$$\Sigma$$ F = ma = 0 (it's floating)

$$\rho$$ = F / A

Here's what I tried doing.

$$\Sigma$$ F_y = B - F_l - F_w = 0 , with B as Buoyancy, F_l is the liquids force, and the F_w is the forceo f the wood . I ended up using Volume and density to plug in for Forces obtaining this....

$$\rho$$ * V_l * g - $$\rho$$ * V_w = $$\rho$$ * V_w * H

I'm not sure if this is near the correct approach.

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Doc Al
Mentor
The buoyant force is the force of the liquid on the wood. Only two forces act on the wood: its weight and the buoyant force.

Hint: What volume of liquid must be displaced to support the wood? How does that compare to the volume of the wood?

Oh, so it's similar to the case of the fully sugmerged object, only with different densities given and g is negligible.