Hydrostatic pressure at a point inside a water tank that is accelerating

[songoku]

Homework Statement

A cube-shaped tank is filled with water and attached firmly to a cart moving with an acceleration of 20 cm/s^2. Find the pressure at a point which is 10 cm deep and 10 cm from the wall of the tank.

Relevant Equations

Hydrostatic pressure

I draw this diagram:

The formula for hydrostatic pressure is: $P=\rho g h$ so I just plug everything
$$P=1000 \times 9.8 \times 0.1=980 Pa$$

Will the acceleration of the cart affect the hydrostatic pressure?

Another thing that came to my mind was there would be extra force coming from normal force between water and left and right side of the tank and since the tank is accelerated to the right, normal force from the left part of the tank is bigger than the right part causing net force to the right.

But I am not sure whether I need to consider this to answer this question and how to proceed.

Thanks

 

[erobz]

Are you sure you are not supposed to measure $10 \, \rm{cm}$ deep as from the point on the wall where the water touches it on the far left? I believe this problem is leading you to find the slope of the water due to the acceleration of the cart. The depth will not be $10 \, \rm{cm}$ from surface in that case, you will have to determine it. Like you say, the problem is trivial otherwise.

I set up Newtons Second Law in a non-inertial frame, the coordinate $x {}^+\leftarrow$ is fixed to the front of the cart. The entire cart is accelerating to the right at $a \rightarrow^-$. Look a thin vertical cross-section of length $dx$, height $h(x)$ and mass $dm$.

 

[kuruman]

In the non-inertial frame of the cart there is an apparent acceleration of gravity vector $$\mathbf{g}_{\text{app}}=\mathbf{g}+(\mathbf{-a}).$$The surface of the liquid is perpendicular to the apparent acceleration vector. To find the pressure, you need to find the perpendicular distance $d$ from point A to the surface, then use $$p=\rho~g_{\text{app}}~d.$$You can read more about apparent acceleration of gravity here if you wish.

 

[Orodruin]

find the pressure, you need to find the perpendicular distance d from point A to the surface, then use p=ρ gapp d.

You don’t, actually. If the situation is as depicted it is something of a trap to make you think you need to do all that. However, in an incompatible fluid in a constant gravitational field $\vec g_{\rm app}$ the pressure difference between two points separated by $\vec d$ is $\rho \vec d\cdot \vec g_{\rm app}$. If you pick your reference point as OP did, then $\vec d \perp \vec a$ and $\vec d \parallel \vec g$. Therefore the pressure difference is $\rho g d$, where $d$ is 10 cm and $g$ the regular gravitational acceleration.

Alternatively you can reach the same conclusion by looking at the similar triangles for the distances and accelerations.

 

[kuruman]

You don’t, actually.

Yes, I see that now. I fell into the trap thinking of equal pressure planes below the surface.

 

[Orodruin]

Yes, I see that now. I fell into the trap thinking of equal pressure planes below the surface.

It is a typical problem where knowing too much easily leads you astray whereas knowing not quite as much may give you the right answer but for the wrong reason.

 

[kuruman]

It is a typical problem where knowing too much easily leads you astray whereas knowing not quite as much may give you the right answer but for the wrong reason.

On the other hand, if you try too hard to think outside the box you might fall off a cliff.

 

[Lnewqban]

@songoku , this two problems are similar:
https://www.physicsforums.com/threads/acceleration-of-u-pipe-containing-water.999370/

 

[erobz]

On the other hand, if you try too hard to think outside the box you might fall off a cliff.

Did I fall off a cliff? I got that the slope was simply $\frac{a}{g}$?

 

[songoku]

I am really sorry for late reply

You don’t, actually. If the situation is as depicted it is something of a trap to make you think you need to do all that. However, in an incompatible fluid in a constant gravitational field $\vec g_{\rm app}$ the pressure difference between two points separated by $\vec d$ is $\rho \vec d\cdot \vec g_{\rm app}$. If you pick your reference point as OP did, then $\vec d \perp \vec a$ and $\vec d \parallel \vec g$. Therefore the pressure difference is $\rho g d$, where $d$ is 10 cm and $g$ the regular gravitational acceleration.

Alternatively you can reach the same conclusion by looking at the similar triangles for the distances and accelerations.

Which similar triangles for distances and accelerations?

@songoku , this two problems are similar:
https://www.physicsforums.com/threads/acceleration-of-u-pipe-containing-water.999370/

I also think they are similar but I can not connect the dots

Thanks

 

[Lnewqban]

I also think they are similar but I can not connect the dots

Thanks

Is the shown sketch coming from the problem or you draw it yourself?

Regarding the question:
Find the pressure at a point which is 10 cm deep and 10 cm from the wall of the tank.
What is the meaning of “10 cm deep”?
From which wall of the tank?
What angle is formed between the surface and the horizon for that acceleration?

 

[haruspex]

Homework Statement:: A cube-shaped tank is filled with water

So perhaps the cube has a lid, not open as you've drawn it.

 

[haruspex]

an incompatible fluid

Incompatible with being compressed, right?

 

[erobz]

So perhaps the cube has a lid, not open as you've drawn it.

That would resolve the issue of which wall the measurement is to be taken from. EDIT. Never mind, I believe the issue of which wall is still an issue either way.

 

[songoku]

Is the shown sketch coming from the problem or you draw it yourself?

I draw it myself, there is no picture from the original question

Regarding the question:
Find the pressure at a point which is 10 cm deep and 10 cm from the wall of the tank.
What is the meaning of “10 cm deep”?

I interpret it as 10 cm from the surface of water which is now slanted because of the acceleration

From which wall of the tank?

I don't know so I just take one of the side

What angle is formed between the surface and the horizon for that acceleration?

Not given. Do we need to find the angle?

So perhaps the cube has a lid, not open as you've drawn it.

Maybe but does it affect the analysis and calculation?

Thanks

 

[erobz]

Maybe but does it affect the analysis and calculation?

Thanks

If it has a lid and is completely filled, I believe there is a positive pressure gradient opposite the direction of acceleration. So the pressure 10 cm down and 10 cm over will be different front to back. I think that is the same in either case.

 

[jbriggs444]

I interpret it as 10 cm from the surface of water which is now slanted because of the acceleration

10 cm down from what point on the slanted surface?

 

[songoku]

If it has a lid and is completely filled, I believe there is a positive pressure gradient opposite the direction of acceleration. So the pressure 10 cm down and 10 cm over will be different front to back. I think that is the same in either case.

I will search about pressure gradient first then response back because I don't know what it is

10 cm down from what point on the slanted surface?

Like what I draw in OP

Thanks

 

[Lnewqban]

... I interpret it as 10 cm from the surface of water which is now slanted because of the acceleration

Should the measurement of that distance be perfectly vertical or perpendicular to the slanted surface?

... I don't know so I just take one of the sides.

I don't see how that horizontal distance is relevant, as each molecule of the fluid is subjected to the same vertical (gravity) and horizontal accelerations.

... Not given. Do we need to find the angle?

A vectorial addition of both accelerations would result in a total acceleration vector having certain angle respect to a vertical line.

Do you know which value of pressure is the correct answer for this problem?

 

[Orodruin]

A vectorial addition of both accelerations would result in a total acceleration vector having certain angle respect to a vertical line.

Which, with the OP’s interpretation of the problem (with the 10 cm depth being measured from the surface right above the point) is not something you need to compute.

 

[haruspex]

Maybe but does it affect the analysis and calculation?

It removes any doubt re the meaning of depth here. It would also mean the top of the water above the point is not necessarily at atmospheric pressure.

 

[Lnewqban]

,

Which, with the OP’s interpretation of the problem (with the 10 cm depth being measured from the surface right above the point) is not something you need to compute.

The angle, or the a and g vector's summation?

 

[jbriggs444]

The angle, or the a and g vector's summation?

Neither is relevant. You have a reference pressure at the surface directly above the test point. You have a known gradient of pressure along a vertical axis. What more do you need?

 

[erobz]

I think it's either what you have already solved... a question which basically tries to trip you up

or

The tank is completely filled as @haruspex interprets, and you have to do some analysis.

The other possible solutions seem to leave out too much information about where specifically this test point is relative to the walls.

 

[Lnewqban]

Neither is relevant. You have a reference pressure at the surface directly above the test point. You have a known gradient of pressure along a vertical axis. What more do you need?

I believe that the gradient of pressure is not known, unless calculated.

 

[Orodruin]

I believe that the gradient of pressure is not known, unless calculated.

The point is, you only need the component of the pressure gradient in the vertical direction, which makes the acceleration irrelevant.

 

[erobz]

I believe that the gradient of pressure is not known, unless calculated.

I believe in the case where it is an open top tank as drawn by the OP, the pressure at a depth (vertically) $h$ from any point on the surface is just $P = \rho g h$.

The pressures from the acceleration and the hydrostatic pressure must be additive to produce the slope of the water; If they didn't add together then why would it slope? If we draw a horizontal line at some depth $h$ as a function of $x$ the distance from the leading wall of the tank, then the pressure at any point on that line is given by: $$ P = \rho g ( h + \frac{a}{g}x )$$ But if we draw a line parallel to the surface at depth $h$ from the leading edge the pressure at every point on that line is just: $$ P = \rho g h $$

 

[erobz]

The point is, you only need the component of the pressure gradient in the vertical direction, which makes the acceleration irrelevant.

Unless it's to be interpreted as @haruspex points out?

 

[Orodruin]

Unless it's to be interpreted as @haruspex points out?

Yes, I am talking about the interpretation depicted in the OP.

 

[Lnewqban]

The point is, you only need the component of the pressure gradient in the vertical direction, which makes the acceleration irrelevant.

I am probably wrong, as other members seem to imply, but I am totally missing that point.
I would love to see my error and to avoid misleading @songoku

The way I see it:
These two graphical situations are comparable to this problem in principle, if some relation can be established among static pressure inside the fluid in equilibrium and the total reaction of the represented bodies, also in equilibrium, at the support point.

The value of the unique static pressure at certain point inside the fluid, which is reacting to the horizontal acceleration of its container, can't be the traditional ρgh (pressure gradient in the vertical direction).

That point is supporting the force (each mass x common diagonal acceleration) of all the molecules that are aligned with the resultant acceleration vector and located between that point and the surface.

 

[jbriggs444]

can't be the traditional ρgh (pressure gradient in the vertical direction).

Note that $\rho g h$ is not the pressure gradient. It is the pressure. More correctly, it is the pressure difference between the test point and a reference pressure at a point on the surface directly above.

A "gradient" is a rate of change. It is a vector. Its direction is the direction in which pressure increases most rapidly with displacement. Its magnitude is the rate at which pressure increases with displacement in that direction. In the drawing that you provide (one gee of vertical gravity and one gee of horizontal acceleration), the direction of the gradient is 45 degrees down and to the left. Its magnitude is $\sqrt{2}\rho g$. As you clearly agree.

But we do not much care how rapidly pressure is increasing in a direction 45 degrees down and to the left. Our reference point is directly above our test point. We care about how rapidly pressure is increasing in the purely vertical direction. We want the vertical component of the pressure gradient.

If you wanted, you could calculate that vertical component as $\sqrt{2}\rho g \cos 45 \text{ degrees}$ That simplifies to $\rho g$.

Or you could just use the obvious fact that the vertical component is $\rho g$. Just like the horizontal component is $a g$ (with $a = g$ in the case you portray).

That point is supporting the force (each mass x common diagonal acceleration) of all the molecules that are aligned with the resultant acceleration vector and located between that point and the surface.

View attachment 315891

Right. So if you have a column of water $\frac{h}{\sqrt{2}} \text{meters}$ in diagonal extent so that our test point is $\frac{h}{\sqrt{2}}\text{ meters}$ below a reference point at the surface in a direction 45 degrees up and to the right. It will be at a pressure delta of $\sqrt{2} \rho g h\frac{1}{\sqrt{2}}$ above surface pressure.

Equivalently, it is $h \text{ meters}$ below a reference point on the surface directly above and is at a pressure delta of $\rho g h$ above surface pressure.

Both calculations yield the same result. But one is simpler than the other.

 

[kuruman]

Here is an explanation in the accelerated frame. First refer to the picture on the left below. The apparent acceleration of gravity is the vector sum of the horizontal $(\mathbf{-a})$ and the vertical acceleration of gravity $\mathbf{g}$. The fluid surface (blue line) will be perpendicular to the apparent acceleration of gravity. In the non-inertial frame the hydrostatic pressure will be $p=\rho~g_{\text{app}}~d$ where $d$ is the perpendicular distance to the surface from the point of interest. That was my approach in post #3 where I suggested that one needs to find $d$ and $g_{\text{app}}$. However, that is unnecessary extra work.

Refer to the picture on the right where the perpendicular distance to the surface is shown as $d$. Note (what I missed at first until @Orodruin pointed it out) that from the picture on the left $$g = g_{\text{app}} \cos \theta.$$
Then the hydrostatic pressure at perpendicular distance $d$ from the surface is $$p=\rho~g_{\text{app}}~d=\rho~\frac{g}{\cos\theta}~d=\rho~g~h.$$On edit: The same argument stated differently. Planes parallel to the surface are planes of constant hydrostatic pressure. So when you go from point A to point B, the pressure change will be the same as when go from A directly to C because there is no change in pressure when you go along the surface from B to C.

On second edit: If the picture on the right were rotated counterclockwise by $\theta$, so that the apparent gravity is aligned from the top of the screen to the bottom, the proposition that the pressure difference between A and B is the same as between A and C would be perfectly acceptable to "inside the box" thinkers.

 

[Orodruin]

Incompatible with being compressed, right?

Indeed. I blame auto complete.

 

[Lnewqban]

Note that $\rho g h$ is not the pressure gradient. It is the pressure. More correctly, it is the pressure difference between the test point and a reference pressure at a point on the surface directly above.
...
Both calculations yield the same result. But one is simpler than the other.

Thank you much for the detailed explanation, excellent as usual.

Yes, I referred to the natural rate of pressure change in the vertical direction as pressure gradient, being the measuring point at one singular point of it.

It is just a simply trigonometric thing: what we gain in acceleration in that direction, we gain in number of molecules in a direction 45 degrees down and to the left.
Sorry about wasting yours, @erobz, @kuruman time, I should have noted that dependence.

Respectfully, I still prefer to see what actually happens in the direction it happens.

 

[Lnewqban]

@songoku , this two problems are similar:
https://www.physicsforums.com/threads/acceleration-of-u-pipe-containing-water.999370/

I also think they are similar but I can not connect the dots

@songoku
I believe that this link could help you connect those missing dots:
https://www.me.psu.edu/cimbala/Learning/Fluid/Rigid_body/rigid_body.htm