So basically in a video I was taught that, for a submerged cylinder the forces acting on the bottom curved surface is equal to the weight of water directly above + imaginary water weight, if someone could explain me this thing, I would be grateful.enter image description here


2 Answers 2


I think the word "imaginary" is not helping here.

The force on the bottom of the cylinder is the weight of the round bottom column water above it, period.

And that weight can be calculated by breaking the volume of the round bottom column into a rectangular and a half circle. Exactly as you have sketched.

To show what you mean in another way possibly easier to envision, let's assume the cylinder was floating on the surface of the water so its center was level with the surface of the water.

Then the force applied to the bottom of the cylinder would be the sum of infinitesimally small columns of water. Let's say the cylinder with radius R starts at the origin.

  • R radius of the cylinder The force on the bottom is the sum of infinitesimally small columns of water with the width dx of the function

$$Y=\sqrt {R^2-X^2}$$

$$F_{bottom}=\Sigma Y *dx$$

$$F_{bottom}= \int_0^{2R} \sqrt(R^2 - x^2) dx $$

And we already know that integral is equal to $ \ \pi*R^2/2$, which is the are of half circle.

So if we move the cylinder dawn to the depth of D. the force on the bottom will be

$$F_{bottom}=2RD+ \pi R^2/2$$

The force on the top again is exactly as you have shown

$$F_{top}=2RD-\pi R^2/2$$

And buoyancy is $$F_{bottom}- F_{top}= \pi R^2 \quad = area \ of \ cylinder$$

For simplicity, I assumed the axial length of the cylinder as one!

  • $\begingroup$ but the water is imaginary, since the cylinder is not full of water. $\endgroup$
    – user253751
    Apr 4, 2023 at 13:01
  • $\begingroup$ The hydrostatic pressure of water at any depth is equal to the pressure of the column of water at that depth. this pressure is applied equally in all directions including downward,upward, and sideways. if you have an object it applies to any point on that object. proportional to the depth of that point. The pressure to the bottom surface of the cylinder is from the water around it. $\endgroup$
    – kamran
    Apr 4, 2023 at 15:08
  • $\begingroup$ and this column is imaginary, since there isn't actually a column of water going up from that point to the surface $\endgroup$
    – user253751
    Apr 4, 2023 at 15:14
  • $\begingroup$ @user253751, it's real, would not be there if the water was not there. this is not a chat blog. it ends here! $\endgroup$
    – kamran
    Apr 4, 2023 at 16:50
  • $\begingroup$ the cylinder is not full of water $\endgroup$
    – user253751
    Apr 4, 2023 at 16:51

Just imagine there is water instead of cylinder. In such a case, the water is just sitting there in equilibrium (not moving up or down). You could still do the same force equilibrium with the water cylinder: force from pressure acting on the bottom part will have to be equal to the force from pressure acting on the top part combined wit the cylinder weight. Then you can go back to cylinder from other material than water and see, that the only thing changing in the equilibrium is the weight of the cylinder relative to water, which determines if the resulting force acting on the cylinder will point up or down.


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