# Calculating the pressure on the bottom of an object submerged in a fluid, how come it only depends on three variables?

If I know the height of the object and how much liquid is above it, why is the formula for calculating the pressure acting on the bottom of the object still height times density times the gravitational constant?

As far as I can make sense of it if the object is in the liquid then the forces acting on the bottom of the object are two:

1) one is coming from underneath and would be equal to the buoyancy force

2) the other one is coming from the top with the atmosphere, the liquid and the object itself pressing down on it.

In the first case I would simply divide the buoyant force with the area of the bottom to get the pressure, why is this wrong?

In the second case it doesn't make sense to me to simply use the formula, because the object itself has a different weight and density than the liquid, so how can we not take that into account when calculating the force that is acting on it? Yet all my teachers and everyone keeps assuring me to just do G * height *density without any explanation of the logic behind it..

2. The pressure on the bottom of the object is greater than the pressure on the top, because the bottom is deeper down. The pressure at depth $$h$$ in a fluid of density $$\rho$$ is given by $$p=h\rho g +\text{atmos press}$$
3. Now take the case of a cuboidal body with four of its faces vertical, the top (horizontal) face (area $$A$$) at depth $$h_T$$ and its bottom side (area $$A$$) at depth $$h_B$$. The buoyancy force, $$F_\text {buoy}$$, is therefore $$F_\text {buoy}=\text{upward force on bottom – downward force on top}$$ So $$F_\text{buoy}=\text{pressure on bottom} \times A\ – \text{pressure on top} \times A$$ So $$F_\text{buoy}=h_B \rho g A +\text{atmos press} \times A\ – (h_T \rho g A +\text{atmos press} \times A)$$ So $$F_\text{buoy}=(h_B-h_T)A \rho g =\text{volume of body}\times \text{density of fluid} \times g$$ So $$F_\text{buoy}=\text{weight of fluid displaced by body}$$ There is a neat way of showing that the same result applies for a body of any shape...
• I $thought$ that was the problem. Glad it's sorted. May 17, 2020 at 17:25