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I am currently trying to work through why an object with smaller density than the fluid it is in floats upwards. I know that the object experiences pressure in all directions because it has moved water. However already here i start to get confused:

Why doesn't the newly moved water simply increase the height of water in the water container? Is it because to do so, the moved water would have to move all the water molecules above it for that to happen, and therefore often decides to not do so? Does that too explain why the pressure gets larger the deeper we get? because more water molecules have to be moved to increase the height of the water?

Now I'm learning about this from KhanAcademy, and when they had a cube placed in water at given depth $h$, they say that the pressure it experiences is: $$\frac {h\cdot\text{density of water}\cdot g}{\text{area of cube}}$$ Basically, the weight of the water above the cube, divided by the cube area. But when they've previously said that the pressure is NOT caused by the amount of water above our object, this confuses me greatly. And while it DOES make sense for the top of our cube, since the pressure is $F/a$, and the force it experiences IS the weight of the water above it, the same formula is also used to find the pressure at the bottom of the cube, and on the sides of the cubes. And in that case it makes no sense.

So why do we find the pressure with this formula when it seems contradictory with the reason our object experiences pressure?

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Consider a very small cube of water of side length $\Delta h$, density $\rho$ with the gravitational field strength $g$ in equilibrium with surrounding water.

The weight of that cube of water is $\Delta h^3\rho g$.

As the cube of water is in equilibrium then the net force on it must be zero.

Consider vertical forces.
On the top face there is a force $f$ downwards, the weight of the cube is downwards and there is a force $F$ on the bottom face of the cube upwards.

Using Newton's second law with up as positive gives $F - \Delta h^3\rho g - f=0 \Rightarrow \dfrac {F}{\Delta h^2} - \dfrac{f}{\Delta h^2} = \Delta h \rho g$.

Define a quantity called pressure as the force per unit normal area.

$\dfrac {F}{\Delta h^2} - \dfrac{f}{\Delta h^2}$ is the difference in pressure across the cube $\Delta P = \Delta h\rho g$

That formula would still be the same if the cube of water was surrounded by a vessel made out of glass with a square base of side $\Delta h$ with an open top and above the water there was a vacuum - ignore the practicalities of doing this.
The vacuum would exert no force on the top of the cube of water and the base of the glass vessel would exert an upward force on the the base of the cube of water equal to the weight of the cube.
The pressure difference between the top and bottom of the cube of water would still be $\Delta P = \Delta h \rho g$.

You can then extend this to lots of cubes of water piled on top of one another.

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  • $\begingroup$ im still slightly confused: what force excatly are you reffering to with small "f"? Is that the weight above our cube of water? so is: "f" = hpg? $\endgroup$ – Buster Bie Mar 18 '17 at 21:18
  • $\begingroup$ @BusterBie $f$ is a force which is exerted on the top face of the cube of water . It could be due to the weight of water (and air) above the cube of water , so the water and air above the cube of water exerts a force on the top of the cube of water. In turn the top face of the cube of water exerts an equal and opposite force on the water and air above it. $f\ne h\rho g$ as the dimensions on either side are not the same. $\endgroup$ – Farcher Mar 19 '17 at 6:06
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It makes perfect sense that pressure is still increasing with height at the bottom and sides of the cube.

Pressure acts equally in all directions, not just up/down. It also is reasonable to consider it goes up as you have more water above you. This even applies when the body isn't fully submerged, because the pressure still acts on the bottom and sides. That pressure still changes with height.

The only reason pressure varies with height is because gravity is acting on it, so there is more than just internal forces of the liquid causing pressure. If there were no potential energy associated with height (i.e. somewhere with simulated or real free-fall) then pressure shouldn't vary with height.

I'm not sure if this answers your question though, it's somewhat unclear to me what was confusing you.

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  • $\begingroup$ Whats really confusing me is what excatly is causing the pressure to go inwards on our object in all directions? The only explanation ive gotton, (the one with the water being moved) doesn't seem to make sense with all the other theory about pressure. $\endgroup$ – Buster Bie Mar 17 '17 at 11:01
  • $\begingroup$ @BusterBie When you push down on the water, it doesn't just try to push itself directly downwards. It's a fluid, so it is able to change it's shape. The increase in pressure will make it try and go any direction it can, which means it pushes out in every direction. Since it's in a container that can't change shape, pressure increases in all directions (instead of pressure staying the same and the fluid spreading out, this is what happens when there are no walls to contain the pressure, it will spread over the surface since nothing resists it's pushing, and no pressure is built up. $\endgroup$ – JMac Mar 17 '17 at 11:06
  • $\begingroup$ so would that mean that the following is true: imgur.com/a/xaWBo $\endgroup$ – Buster Bie Mar 17 '17 at 13:21
  • $\begingroup$ @BusterBie Yes, that's a pretty good way to think about it. Imagine the container were weak (like sides made of something flexible), the sides could bulge out, which would limit the pressure built up. This increase in area would also make the height of the water level lower (which ties in with the pressure being lower). If the sides were able to burst, the water would want to spread out so that there's no water on top of it, which would make all the water at atmospheric pressure (this is like spilling a glass of water on a flat surface, it spreads everywhere). $\endgroup$ – JMac Mar 17 '17 at 13:25
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Assuming a ideal (non viscous and non compressible),

The pressure $P$ at a depth $h$ of fluid with density $\rho$ on a planet with gravitational acceleration $g$ and atmospheric pressure $P_0$ assuming the upper surface (open end) of fluid is kept open to its atmosphere is

$$P=P_0 + \rho gh$$

JMac's answer covers the rest of your questions.

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