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..


1 Answer 1

  1. The buoyancy force is not fundamentally a single force, but the resultant of forces due to fluid pressure acting on all parts of the surface of the submerged body.

  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...

  • $\begingroup$ The first point of your answer is where I had an error, thank you $\endgroup$ Commented May 17, 2020 at 17:23
  • $\begingroup$ I $thought$ that was the problem. Glad it's sorted. $\endgroup$ Commented May 17, 2020 at 17:25
  • $\begingroup$ It's funny because in my language the word for buoyancy force is "Üleslükkejõud" which literally means "upwards pushing force" and I find it to be bad design that the thing called "upwards pushing force" is in fact NOT simply the force that pushes something upwards. $\endgroup$ Commented May 17, 2020 at 17:33
  • $\begingroup$ In the UK we usually use the term π‘’π‘π‘‘β„Žπ‘Ÿπ‘’π‘ π‘‘ (or π΄π‘Ÿπ‘β„Žπ‘–π‘šπ‘’π‘‘eπ‘Žπ‘› π‘’π‘π‘‘β„Žπ‘’π‘ π‘‘) rather than π‘π‘’π‘œπ‘¦π‘Žπ‘›π‘π‘¦ π‘“π‘œπ‘Ÿπ‘π‘’. So the same problem arises as with the Estonian term. But then if you're swimming you probably 𝑒π‘₯π‘π‘’π‘Ÿπ‘–π‘’π‘›π‘π‘’ the upthrust as if it were a single force; you're not usually aware of pushes on different parts of your body – or are you? $\endgroup$ Commented May 17, 2020 at 17:53
  • $\begingroup$ @PhilipWood There is always someone who always damages my answers and those who give me the answers like you. :-( Always thanks and I hope to compensate for my last answer. $\endgroup$
    – Sebastiano
    Commented May 18, 2020 at 21:49

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