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This isn't a HW question. I've already choose the correct answer. I'm just using it for illustration.

I may refer to the top and bottom of the sphere as point P and R whereas to the right and left as point Q and S.

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I imagine the force(P) to be caused by the fact that the top of the sphere is carrying a whole "column" of water. Meanwhile the tip sides of the sphere aren't carrying anything, they are parallel to the water. I also imagine the pressure caused on either side to be due to the constant and random movements of water molecules and their collisions with the sides thus exerting a roughly equal forces on either sides.

What I don't understand is the pressure at R. I know from my textbook that the pressure at point R is equal to the pressure on P plus the pressure caused by a column of water of the height of the sphere but the bottom isn't carrying any more water than the top. It's just carrying the pressure on P plus the pressure caused by the sphere weight.

Secondly, how could the force vector (R) be drawn facing upward! Every force I could imagine to put a significant pressure on the bottom is (P) and the sphere weight, which are all facing downwards. The only force that could act in that direction could only be caused by the random movements of molecules just like points Q and S leading point R to be equal to them too.

I know there's something seriously wrong with my understanding of liquid pressure, but I can't put my hands on it. I've read some of the answers on questions a bit similar to mine but I couldn't really understand most of them.

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  • $\begingroup$ Is this an AP Physics B problem? $\endgroup$ Jan 16, 2022 at 2:58
  • $\begingroup$ No, I'm not an American student. This is a question from GCSE AS physics. $\endgroup$
    – Manar
    Jan 16, 2022 at 11:36

4 Answers 4

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In a fluid, pressure acts equally in all directions. This observation is attributed to Pascal who discovered it. So even though R does not have a column of fluid directly above it, the pressure is uniform with horizontal position at the depth of R because it is transmitted horizontally to R from positions at that depth that do have columns of fluid above them.

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  • $\begingroup$ Shouldn't that pressure, the pressure that's caused at the depth of R because it is transmitted horizontally to R from positions at that depth that do have columns of fluid above them, be pointing downwards though? $\endgroup$
    – Manar
    Jan 16, 2022 at 15:36
  • $\begingroup$ Like I (and Pascal) said, at a given depth, it acts equally in all directions, both horizontal and vertical. $\endgroup$ Jan 16, 2022 at 17:12
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There are a few key concepts that apply to this problem.

The increase in pressure at a depth $h$ below the surface, for a liquid whose density if $\rho$, is equal to $\rho g h$. This pressure applies to all points in the fluid that exist at the given depth because pressure is transmitted equally in all directions at that depth (as pointed out by other posters).

Given that the bottom of the sphere is at depth $h$, it will experience the same pressure as all other points at that depth.

For objects submerged in a liquid, the differential force (from the local pressure) on each differential area of the object is always applied normal to that differential surface. This means that at the bottom of the sphere, the force from the pressure at depth $h$ is pointing up.

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Imagining the microscopic collisions of water molecules with the surface of the sphere, as you describe, is a nice way to think about the pressure. It should be clear from this perspective that the force on the sphere at $R$ is upward—after all, the water molecules there are striking the sphere from below.

What’s a bit subtler, when thinking in terms of molecular collisions, is the other part of your question: why is the force at $R$ greater than at $Q$ or $S$? The molecules are moving at the same average speed everywhere (as long as the temperature is uniform) so shouldn’t they apply the same force everywhere? The reason this isn’t the case is that as you go deeper below the surface, the water molecules are packed ever so slightly closer together (so as to remain in equilibrium with the combined effect of the individually tiny forces that gravity exerts on all the molecules above) so they all repel each other a bit more. Imagine the free body diagram of a water molecule as it collides with the sphere. The surface of the the sphere exerts a force on the molecule; if the molecule was part of a gas, rather than a liquid, this would be the only important force. But in a liquid, there are also forces from the neighboring molecules, on the average in the opposite direction (because there are only neighbors on one side). The net force on the molecule is the sum of all the forces, so in order to bounce away an incoming molecule, a greater applied force from the sphere is required when the molecule’s neighbors are closer, and therefore repelling each other more.

As an aside, it is not really correct to speak of the force at $P$ or $Q$. Pressure is force per unit area, and a point has no area. Better would be to say the force on a 1 square cm patch, or whatever, at each location.

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  • $\begingroup$ One last question, I wonder why we don't speak of the weight of the sphere when we try to get the pressure at the bottom of the sphere. We always consider it if the sphere is on, for example, the ground. $\endgroup$
    – Manar
    Jan 16, 2022 at 21:16
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Indeed pressure in a fluid, as a function of column height, can be counter-intuitive.

Let's look at pressure gradient in the atmosphere first. There is a presssure gradient from the ground level to a height where the atmosphere is so thin it is negligable.


Altitude

Today's astronomical observatories are constructed at locations where the altitude is 4 to 5 kilometers, such as on the highest mountain tops of the Hawaiian islands.

Astronomers have to acclimatize in order to work at an altitude like that. The acclimatization period allows the body to adapt to the lower air pressure (higher amount of red blood cells per unit of volume of blood.) Without acclimatization people are prone to getting altitude sickness.

This illustrates that with height above ground level there is a rapid decline in air density.

Air is compressible, and at ground level the air must have sufficient pressure to carry the entire column of air above it.

So we have that the case of atmospheric pressure is quite intuitive. Gradient in density and gradient in pressure going hand in hand.


In the case of a fluid column we have that the fluid is hardly compressible. So it's harder to get a feel that that there is a significant pressure gradient all the same.

In a column of fluid the pressure (as a function of height) is definitely not uniform. There is a gradient in pressure, consistent with the weight of the fluid column above it.


A fluid

As you mention: a fluid (be it a gas or a liquid) exerts a pressure by way of the constituent molecules bumping against the surface of the object that is immersed in the fluid.

The molecules of a gas (at 1 atmosphere of pressure) have a lot of room to travel.

Water does not compress much under pressure, but it does compress a little.

Water at 1 atmosphere of pressure will have its molecules bumping against the surface of an immersed object at a rate such that the pressure exerted is 1 atmosphere of pressure.

Pressurizing the water reduces how much wiggle room there is for the thermal motion of the water molecules.

Imagine you are pacing back and forth in a room, using the walls of the room to reverse your direction of motion. Pacing from wall to wall is like the life of an air molecule.

Imagine you are a molecule of liquid: you can no longer do any pacing; you are down to a vibration in place. Now increase density: less wiggle room, so the amplitude of your vibration is smaller. But you still have the same velocity from bump to bump, so the amount of bumps per unit of time is larger. This higher frequency of bumps is the higher pressure.

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  • $\begingroup$ I've a question. In this experiment youtu.be/SvnuqkUj1Lo (the video name in case you don't want to click on the link: Air Pressure And The Bottle by George Mehler YouTube channel). When he closes the bottle ,again after opening, the stream of water stops. Why is that though? Shouldn't the air that entered and stayed in the bottle even after closing the lid be as dense and thus as fast as the air outside, creating the same pressure?? $\endgroup$
    – Manar
    Jan 16, 2022 at 22:27

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