I'm struggling a bit on a homework problem where a cylindrical rod of radius r is floating in equilibrium in water. The bottom 4 cm of the rod has a density of 2000 kg/m^3 while the other 6 cm of the rod has a density of 900 kg/m^3. The question asks what the depth of the bottom of the rod is. Would I be correct in using the following equation to solve for the amount V of water displaced by the rod? (900 kg/m^3)(9.8 m/s^2)(0.06 m) - (2000 kg/m^3)(9.8 m/s^2)(0.04 m) + (1000 kg/m^3)(V) = 0 Am I on the right path here? If not, can someone point me in the right direction?