1. The problem statement, all variables and given/known data A rectangular wooden block of weight W floats with exactly one-half of its volume below the waterline. Masses are stacked on top of the block until the top of the block is level with the waterline. This requires 20g of mass. What is the mass of the wooden block? 2. Relevant equations Buoyant Force: F_b = (p_f)(V_f)(g) = w_o = (p_o)(V_o)(g) Mass Density: m = pV 3. The attempt at a solution Since the block is initially half submerged, its density is one-half of the water: p_o = 0.50g/cm^3 I'm assuming that the density of the block and the density of the fluid don't change. Without the volume, I don't see how to use the formulas... My only guess at the answer is 40g because it would require 20g to overcome the initial buoyant force (where the block floats halfway submerged)?
No need to guess. Use Archimedes' principle and equilibrium. (1) The buoyant force equals the weight of the displaced fluid--the volume of that fluid is just half the volume of the block. BF = weight of block. (2) After adding weight, what happened to the buoyant force? (How does it compare to the previous buoyant force?) That new buoyant force BF' must equal the total weight of block plus added weights.
Sorry, this is all new to me.. If the wood block is floating, then the buoyant force is equal to the weight of the block. But since the wood block with the added weight is now submerged under the water, then the weight of the block must be greater than the buoyant force? Or does the buoyant force change to accommodate the new weight?
The buoyant force definitely changes as weight is added--that's why more water is displaced (it sinks lower). In the first case, the buoyant force just has to support the weight of the block; in the second case it must support the block plus the added weight. Hint: By what factor does the buoyant force increase? (Compare the amount of water displaced.)