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Let's suppose that there's a container that is filled with gas and liquid as seen in the picture, and that the surface areas $S$ and $2S$. My book says that when the container gets rotated so that surface areas get reversed, the pressure force on the bottom decreases. Can you explain how this is true?

First, since the volume of gas is invariant under this rotation, $P_{\text{gas}}$ remains invariant. It's also clear to me that since the height of the liquid rises, pressure $P_{\text{liquid}}$ at the bottom also rises from Pascal law $\Delta P = \Delta h\rho g$. If the following assertion is true

All liquids can be treated as rigid bodies, as long as they are contained in a closed system and do not sustain shear-stress.

Then just like in solids, liquids will exert a force $\vec{F} = G_{\text{gas}}+G_{\text{liquid}}$ on the surface, where $G_{\text{gas}}, G_{\text{liquid}}$ denote the weight of the liquid and the gas. The shape of the container seems to influence whether $\vec{F}<G_{\text{gas}}+G_{\text{liquid}}$ or $\vec{F}>G_{\text{gas}}+G_{\text{liquid}}$. But I would be so happy to hear what you're thinking.

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  • $\begingroup$ By "pressure force" do you mean pressure $\times$ force? Which surface are we calculating the "pressure force" for? Also, you introduced $G_{liquid}$ without defining it. Why is "moment of inertia" being discussed when that is usually used in the context of angular acceleration? $\endgroup$
    – KDP
    Commented Jan 28 at 5:52
  • $\begingroup$ The moment of inertia of an incompressible liquid in a closed system does not change only if the liquid fills the closed system. This is not the case in your example. Morover, as already noticed, the moment of inertia doesn't play any role in the pressure in static problems. The get rotated in the text shouldn't be interpreted as "during the rotation" but "after" the rotation. $\endgroup$ Commented Jan 28 at 7:00
  • $\begingroup$ Pressure is Force/Area, so why are you multiplying the weight (force) by the area in your last comment? $\endgroup$
    – PM 2Ring
    Commented Jan 28 at 12:41
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    $\begingroup$ @PM2Ring Sorry you're right, surface area $S$ shouldn't appear in the above equation. $\endgroup$
    – user390010
    Commented Jan 28 at 12:54

2 Answers 2

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The pressure at the bottom increases with increasing hight , that is true , but it does not increase enough to compensate for the smaller S in F=p*S since the hight does not double.

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Let "pressure force" be the force acting on the bottom of the container due the the hydrostatic pressure of the liquid.

Let the height of the container be H and the initial height of the liquid be $h_1 = H/2$ initially.

The initial pressure force is then $ h_1 \rho g * 2S = (H/2) \rho g * 2S = HS \rho g$

We know when we turn the container upside down the area being acted on will be reduced, but the height of the liquid will be increased, so we need to calculate the new height of the liquid to determine the new pressure on the bottom.

The area of the interface is a linear function of the height of the liquid and is initially $S_{int} = S + S*h_1/H =1.5 S$

For convenience, assume the depth into the page is equal to unity so that we can use the formula for the area of a trapezoid to calculate the volume, for convenience.

The initial volume of the liquid is then:

$V_1 = h_1 (1.5S +2S)/2 = 3/4 \ S H$

The final volume of the liquid is:

$V_2 = h_2 (S + S h_2/H +S)/2 = h_2 S/2 + h_2^2 S/(2H) +h_2 S/2$

The volume of the liquid does not change, so $V_1 = V_2$ and we can equate the above two equations:

$3/4 \ S H = h_2 S/2 + h_2^2 S/(2H) +h_2 S/2$ and rearrange and simplify:

$h_2^2+2*h*H -3*H^2/2 =0$

This is a quadratic equation and the positive root is $h_2 \approx 0.581 H$

The final pressure force on the bottom of the container is now:

$ h_2 \rho g * S = (0.581 H) \rho g * S = 0.581 HS \rho g$

This is lower than the initial $HS \rho g$

The volume and therefore pressure of the gas in the container does not change so it does not affect the above calculation.

Reply to additional request by Keane in the comments:

Consider a completely filled sealed container placed on some weighing scales. From the outset, common sense tells us the container will weigh the same when placed upside down.

At any point in a liquid, the hydrostatic pressure acts in all directions, but the force caused by it has a vector that is orthogonal to the surface it is acting on. In a rectangular container, the hydrostatic pressure is also acting outwards on the side of the vessel. If the sides are flexible, you will see them bulge outwards due to this sideways pressure. Since this force is directed sideways it does not contribute anything to the weight we see indicated on the scales.

If the rectangular vessel has a surface area of 1.5S top and bottom the force acting on the bottom will be proportional to the 1.5S.

If we modify the completely filled container so that the volume is the same but walls are tapered inward, so that the top surface is S and the bottom surface is 2S, the force acting on the horizontal bottom surface is proportional to 2 S, yet the scales are still indicating a weight proportional to 1.5S. Why don't the scales indicate 2S? The reason is that the hydrostatic pressure on the sides has a force vector that such that the normal to the side surface is slightly upward and the upward component of the force is equal to -0.5S giving a total downward force of 1.5S on the weighing scales.

When this slanted vessel is turned upside down, the force on the bottom is proportional to S, but the forces on the side now have a normal that is pointing partly downward and the vertical component of this sideways force contributes an additional 0.5S so that total downward force indicated on the scales is still 1.5S.

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  • $\begingroup$ This makes sense to me, although I still have a doubt in my mind: is it true that the fluids will exert a force $\vec{F} = \text{G}_{\text{gas}}+\text{G}_{\text{liquid}}$ on the surface area? But depending on the shape of container, $\vec{F}>\text{G}_{\text{gas}}+\text{G}_{\text{liquid}}$ (initial) or $\vec{F}<\text{G}_{\text{gas}}+\text{G}_{\text{liquid}}$ (after rotation) $\endgroup$
    – user390010
    Commented Jan 28 at 17:03
  • $\begingroup$ For a given density liquid, the force per unit area on the bottom of the container depends only on the height of the liquid and is independent of the shape of the container. It is just that inverting the container alters the height of the liquid in your example. $\endgroup$
    – KDP
    Commented Jan 28 at 17:11
  • $\begingroup$ Yes, force per unit area is simply pressure itself, and it is independent of the shape of the container. But if we consider pressure force (pressure times surface area), can't the fluids in the container be treated as though they're solids (since they're in equilibrium) and the pressure force is immediately sum of their weights? But shape of the container should influence whether this force is felt greater/smaller on the surface (that's what my textbook says) $\endgroup$
    – user390010
    Commented Jan 28 at 17:14
  • $\begingroup$ The problem with that is the weight remains the same when the liquid or solid block is inverted and the pressure force on the bottom would only depend on area. You also have to remember there is hydrostatic pressure acting on the slanted sides of the container, that contributes an upward component initially and a downward component when inverted. $\endgroup$
    – KDP
    Commented Jan 28 at 17:21
  • $\begingroup$ Thank you. Is it also possible for you to talk briefly about this too, in your answer? The reason why I insisted on looking at weights is because in my textbook, it says that if we consider a rectangular container, then the pressure force equals the weight of the liquid. For containers with slanted sides curved inwards and outwards, it is greater and smaller respectively. $\endgroup$
    – user390010
    Commented Jan 28 at 17:47

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