If I have a body (for simplicity a cube with side d) on a fluid (with density ρ) with its top at a depth h, the force acting on top of it is ma = (h*d*d)*ρ*(g); where g is the acceleration of gravity, (d*d*h) is the volume of the column of liquid on top of the cube and that times ρ is the mass of the column. (Assume no atmospheric pressure).
Now, I get everything up to that point. What I don't get is the force acting on the bottom of the cube. Shouldn't the force being exerted by the column of liquid bellow the cube on the cube be equal and opposite to the weight of the column of liquid on top of the cube plus the weight of the cube? But the derivation I saw said that the force acting on the bottom is equal to ((h + d)*d*d)*ρ*(g); where ((h + d)*d*d) is the volume of the column of liquid on top of the cube plus the volume of the cube. Isn't that assuming that the density of the cube is the same as the density of the liquid?
I don't get how you can get to Archimedes principle without assuming that the net force on the cube (buoyant force), is equal to the volume of liquid displaced by the cube in the first place.
Now, I get everything up to that point. What I don't get is the force acting on the bottom of the cube. Shouldn't the force being exerted by the column of liquid bellow the cube on the cube be equal and opposite to the weight of the column of liquid on top of the cube plus the weight of the cube? But the derivation I saw said that the force acting on the bottom is equal to ((h + d)*d*d)*ρ*(g); where ((h + d)*d*d) is the volume of the column of liquid on top of the cube plus the volume of the cube. Isn't that assuming that the density of the cube is the same as the density of the liquid?
I don't get how you can get to Archimedes principle without assuming that the net force on the cube (buoyant force), is equal to the volume of liquid displaced by the cube in the first place.