# Intuition problem about an accelerating barometer

• Kaushik
In summary: Yes, that is correct. The first equation (7) is correct because it takes into account the effective gravity (g+a) due to the acceleration of the elevator. The second equation (3) is incorrect because it only considers the change in gravity (g-a) and does not take into account the effect of the pseudo force. Therefore, (7) is the correct equation because it takes into account both the change in gravity and the effect of the pseudo force. This explains why the water level should rise (x_0 > h) instead of decrease (x_0 < h) when the elevator accelerates upwards.
Kaushik
Consider a barometer kept in an elevator. When the elevator accelerates upwards, there is a pseudo force along with gravitational force acting on the liquid in the barometer. Due to which pressure on the liquid should increase, right? If pressure on the liquid increases then to balance this, the water level should rise and not decrease. Isn't?
But my book states that the water level decreases. I am not able to understand why that is the case.

When the elevator accelerates downwards the pseudo force is upwards and hence the air pressure inside the elevator decreases and the level of liquid should also drop. What is wrong in my intuition here?

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Kaushik said:
there is a pseudo force along with gravitational force acting on the liquid in the barometer. Due to which air pressure should increase, right?
A force acting on the liquid in the barometer is increasing the air pressure?

Kaushik
A.T. said:
A force acting on the liquid in the barometer is increasing the air pressure?
Oops! I meant 'increase the pressure on the liquid'. As the pressure on the interface increases the pressure of the point in the same level (under the inverted tube) also increases. Due to this increase, won't the water level rise?

Is it done like this:
Let the external pressure be ##P_{ext}## before the elevator accelerates and ##x_0## be the new height.
Then when it starts to accelerate,
##P_{ext} + ρax_0 = ρgx_0 \tag1##
But,
##P_{ext} = ρgh \tag2##
Comparing the above,
##gh = x_0(g-a) \tag3##
## \implies x_0 > h \tag4##Or is it done like this:
##P_{ext} = ρ(g_{eff})x_0 = ρ(g+a)x_0 \tag5##
But,
##P_{ext} = ρgh \tag6##
Comparing the above ##2##,
##gh = (g+a)x_0 \tag7##
##\implies x_0 < h \tag8##

Which one is correct and why?

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Kaushik said:
Oops! I meant 'increase the pressure on the liquid'.
The pseudo force is acting on the entire volume of the liquid, not at some interface as pressure.

Kaushik
A.T. said:
The pseudo force is acting on the entire volume of the liquid, not at some interface as pressure.
If it is acting on the entire volume, then won't the air pressure increase?

Kaushik said:
If it is acting on the entire volume, then won't the air pressure increase?
If the elevator is airtight then the air pressure at the bottom will increase. But the pseudo force is acting on both: the liquid and the air.

Kaushik
A.T. said:
If the elevator is airtight then the air pressure at the bottom will increase. But its acting on both: the liquid and the air.
So the air pressure at the bottom will increase because more molecules are pressing against it as compared to other heights. So then the position of the barometer in the elevator also matters?

Kaushik said:
So then the position of the barometer in the elevator also matters?
If the elevator is airtight it has some effect. But I'm not sure if your book wants you to consider that, or just the change in effective gravity.

russ_watters and Kaushik
A.T. said:
If the elevator is airtight it has some effect. But I'm not sure if your book wants you to consider that, or just the change in effective gravity.
My book never mentioned about the difference in air pressure with height in an accelerating elevator. As of now, I think we can neglect its effect.

Kaushik said:
##gh = (g+a)x_0 \tag7##
##\implies x_0 > h \tag8##
Are you sure?

Kaushik
Kaushik said:
My book never mentioned about the difference in air pressure with height in an accelerating elevator. As of now, I think we can neglect its effect.
Yes. This effect is typically neglected. Similarly, we normally neglect the pressure difference between the front and rear of an accelerating car. Except when carrying helium balloons to a birthday party.

And yes, position matters. The pressure at the top of the sealed elevator will decrease when the elevator accelerates upward. The pressure at the bottom will increase.

Kaushik
A.T. said:
Are you sure?
According to my perception(as of now), this seems right!
## \frac{g+a}{g} > 1 \implies x_0 >h##
If this is wrong, where am I going wrong and why?

Kaushik said:
Is it done like this:
Let the external pressure be ##P_{ext}## before the elevator accelerates and ##x_0## be the new height.
Then when it starts to accelerate,
##P_{ext} + ρax_0 = ρgx_0 \tag1##
But,
##P_{ext} = ρgh \tag2##
Comparing the above,
##gh = x_0(g-a) \tag3##
## \implies x_0 < h \tag4##Or is it done like this:
##P_{ext} = ρ(g_{eff})x_0 = ρ(g+a)x_0 \tag5##
But,
##P_{ext} = ρgh \tag6##
Comparing the above ##2##,
##gh = (g+a)x_0 \tag7##
##\implies x_0 > h \tag8##

Which one is correct and why?
I know that both are completely opposite. It would be nice if you point out the error. Thanks!

Kaushik said:
According to my perception...

Kaushik
Oops!
##(4)## was actually ##x_0 > h## and ##(8)## is ##x_0 < h##. Is it correct now?

Kaushik said:
##(8)## is ##x_0 < h##. Is it correct now?
Yes.

Kaushik
A.T. said:
Yes.
Ok so now, which one is correct? 1st attempt or the 2nd attempt?
Kaushik said:
Is it done like this:
Let the external pressure be ##P_{ext}## before the elevator accelerates and ##x_0## be the new height.
Then when it starts to accelerate,
##P_{ext} + ρax_0 = ρgx_0 \tag1##
But,
##P_{ext} = ρgh \tag2##
Comparing the above,
##gh = x_0(g-a) \tag3##
## \implies x_0 < h \tag4##Or is it done like this:
##P_{ext} = ρ(g_{eff})x_0 = ρ(g+a)x_0 \tag5##
But,
##P_{ext} = ρgh \tag6##
Comparing the above ##2##,
##gh = (g+a)x_0 \tag7##
##\implies x_0 > h \tag8##

Which one is correct and why?

EDIT: I feel like the second attempt is wrong because the two ##P_{ext}## i equated are actually not equal. Am I correct?

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Kaushik said:
Ok so now, which one is correct? 1st attempt or the 2nd attempt?
2nd

Kaushik
A.T. said:
2nd
Kaushik said:
EDIT: I feel like the second attempt is wrong because the two PextPextP_{ext} i equated are actually not equal. Am I correct?

The external pressure before acceleration is actually lesser than the external pressure after acceleration. So how can I equate both of them?

Kaushik said:
Is it done like this:
Let the external pressure be ##P_{ext}## before the elevator accelerates and ##x_0## be the new height.
Then when it starts to accelerate,
##P_{ext} + ρax_0 = ρgx_0 \tag1##
You will need to justify this starting point better. In words, what are you equating to what? And if ##x_0## is the new height, what was the old height?

Kaushik
Kaushik said:
The external pressure before acceleration is actually lesser than the external pressure after acceleration. So how can I equate both of them?
See post #9.

jbriggs444 said:
You will need to justify this starting point better. In words, what are you equating to what? And if ##x_0## is the new height, what was the old height?
The following is what I am thinking:
##h## is the old height.
## P_{ext} = ρgh ## (before accelerating)
Now, as it starts to accelerate the pressure increases.
It increases by ##ρax_0## due to upward acceleration.
New external pressure is ##P_{ext} + ρax_0##
This new external pressure should be equal to the weight of the liquid column above it.
##P_{ext} + ρax_0 = ρgx_0##

Where am I going wrong?

Kaushik said:
The following is what I am thinking:
##h## is the old height.
## P_{ext} = ρgh ## (before accelerating)
Now, as it starts to accelerate the pressure increases.
So you are contemplating a column of uniform density air above the elevator which is h units in height. Say about ten miles high.

And you imagine that the elevator is sealed to the sides of the [implausibly tall] shaft like a piston in a cylinder.

And you imagine that the column of air is sufficiently incompressible that it rises up the shaft along with the elevator as soon as the acceleration begins.

Is that what you are imagining?

Kaushik said:
Where am I going wrong?
Hard to say. Confusing the surrounding air with the liquid? Failing to decide if the book wants you to consider the change in air pressure, and provides enough information to actually do it?

Kaushik
A.T. said:
Hard to say. Confusing the surrounding air with the liquid? Failing to decide if the book wants you to consider the change in air pressure, and provides enough information to actually do it?
I think I am getting confused in forming the equations. Like the 'cause and effect' is where I'm getting confused!

Kaushik said:
I think I am getting confused in forming the equations. Like the 'cause and effect' is where I'm getting confused!
The intent in the book is that the elevator is vented to the shaft and that the shaft is ventilated to the outside. The pressure inside the elevator is always going to be the same as the pressure outside the elevator and is not affected by the elevator's acceleration.

You are not given the elevator's speed or direction of motion. There is no intent that any Bernoulli, air-ram or altitude effects be accounted for.

jbriggs444 said:
So you are contemplating a column of uniform density air above the elevator which is h units in height. Say about ten miles high.
The air pressure against the surface of the mercury would only be the same as the pressure from the mercury column if you were to accelerate a total column of air up to space at the same rate as the elevator. Alternatively you could add mass to the centre of the Earth so the local g would be the same for air and mercury. The 'weights of air' and mercury would track.

Kaushik said:
I think I am getting confused in forming the equations. Like the 'cause and effect' is where I'm getting confused!
'Cause and effect' is irrelevant for the equations.

Kaushik and sophiecentaur
A.T. said:
'Cause and effect' is irrelevant for the equations.
Yeah. That problem is common until you find that the approach that ignores cause and effect actually works. In another context, amplifier feedback is mind blowing until you realize that the steady state can be described and solved very easily.

I think it is pretty straight forward... there would be more pressure at the bottom of the elevator due to the acceleration of the elevator creating a higher pressure zone near the bottom, and the mercury (barometer liquid) would also have more effective weight, so 14.7psi (the benchmark for ambient at 29.92") would equate to a lower (position) value on the barometer. say the acceleration was double that of gravity, then 15" of mercury might equate to 14.7psi. (for example)

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zanick said:
then 15" of mercury might equate to 14.7psi. (for example)
A shorter column of mercury could certainly do the job if the elevator is rising.
The point is that the air pressure in the elevator could balance the mercury if the elevator cab was tall enough to increase the effective weight of air inside it.
I have also considered another experiment. If a sealed piston were at the top of the cab, with neutral density (a weighted helium balloon, for instance) could also maintain the same ambient pressure as the cab went up or down. (A bit sketchy on the details though, I'm afraid.)

sophiecentaur said:
A shorter column of mercury could certainly do the job if the elevator is rising.
The point is that the air pressure in the elevator could balance the mercury if the elevator cab was tall enough to increase the effective weight of air inside it.
I have also considered another experiment. If a sealed piston were at the top of the cab, with neutral density (a weighted helium balloon, for instance) could also maintain the same ambient pressure as the cab went up or down. (A bit sketchy on the details though, I'm afraid.)
why would the elevators height be a factor ? elevator's are no air tight as you can feel your ears pop as you go up or down in one. the air pressure in the cab is the weight of the air above you to the edge of space. It would make no sense to seal it, and expand it to the edge of space when that is a condition you already have. ;) that's where the 14.7psi comes from. (and slightly less as you arrive at your 100th floor destination) . or do you mean , if you wanted to double the pressure in the elevator, that was sealed, it would have to be extended to space AND accelerated at 1g. vs accelerating 1g of the elevator sealed where the weight of the air would not change as much due to much less air mass to have its weigh increased causing a smaller pressure gradient rise?

zanick said:
why would the elevators height be a factor ? elevator's are no air tight as you can feel your ears pop as you go up or down in one. the air pressure in the cab is the weight of the air above you to the edge of space. It would make no sense to seal it, and expand it to the edge of space when that is a condition you already have. ;) that's where the 14.7psi comes from. (and slightly less as you arrive at your 100th floor destination) . or do you mean , if you wanted to double the pressure in the elevator, that was sealed, it would have to be extended to space AND accelerated at 1g. vs accelerating 1g of the elevator sealed where the weight of the air would not change as much due to much less air mass to have its weigh increased causing a smaller pressure gradient rise?
I maybe didn't describe the thought very well. It's easy to increase the 'weight' of the mercury column by accelerating it upwards. It's much harder to increase the 'weight' of the air above it. One way would be to use a very high cab, containing all the significant volume of the air above. Accelerating it upwards would increase the weights of mercury and air columns and produce the same reading for pressure. OR you could replace the 100km column of air with a piston which would produce an equivalent increase in internal pressure as the cab accelerates upwards. Both experiments are particularly 'thought-only' types and a bit futile but - hell why not?

PS it wouldn't be necessary too move far, vertically. The acceleration could be over a very small distance so the change in ambient pressure with altitude need not be relevant.

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## 1. What is an accelerating barometer?

An accelerating barometer is a device used to measure changes in atmospheric pressure. It typically consists of a glass tube filled with mercury or another liquid, and a vacuum at the top. As the atmospheric pressure changes, the liquid in the tube will rise or fall, indicating the pressure.

## 2. How does an accelerating barometer work?

An accelerating barometer works based on the principle of hydrostatic equilibrium. As the atmospheric pressure changes, the weight of the air above the liquid in the tube changes, causing the liquid to rise or fall. This change in height is then measured and used to determine the atmospheric pressure.

## 3. What is the intuition problem with an accelerating barometer?

The intuition problem with an accelerating barometer arises when trying to understand why the liquid in the tube rises or falls in response to changes in atmospheric pressure. It can be difficult to intuitively grasp the concept of hydrostatic equilibrium and how it relates to changes in pressure.

## 4. How can the intuition problem be resolved?

The intuition problem can be resolved by understanding the scientific principles behind the functioning of an accelerating barometer. This includes understanding the concept of hydrostatic equilibrium and how it is affected by changes in atmospheric pressure. Additionally, visual aids and demonstrations can help to illustrate the concept and make it easier to understand.

## 5. Why is an accelerating barometer important?

An accelerating barometer is important because it allows us to accurately measure changes in atmospheric pressure, which is a crucial factor in weather forecasting. It is also used in various industries such as aviation, agriculture, and construction to monitor changes in pressure and make necessary adjustments. Without an accurate way to measure atmospheric pressure, our understanding and predictions of weather patterns would be significantly hindered.

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