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 [tex]\Sigma[/tex] F = ma = 0 (it's floating) [tex]\rho[/tex] = F / A Here's what I tried doing. [tex]\Sigma[/tex] 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.... [tex]\rho[/tex] * V_l * g - [tex]\rho[/tex] * V_w = [tex]\rho[/tex] * V_w * H I'm not sure if this is near the correct approach.