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I am currently in high school and I came across a physics problem that asked to find whether the height of the mercury in a mercury barometer would change when it is kept in an elevator that accelerating upwards with a magnitude of 'a'.

Here is how I approached the problem(demonstrated in the image file attached):

1)In the frame of the elevator, the acceleration due to gravity would be (g+a) downwards

2)I assumed the Mass of the air column above the liquid mercury exposed to the atmosphere to be 'M'

3)I assumed the area of the mercury in the container exposed to the atmosphere to be 'A'.

4)I assumed that a vacuum exists in the space between the tube and the mercury.

  1. I assumed the density of mercury to be 'rho'

After you read my attempt at solving the problem, you see that at the end I get an equation for 'h' is not a function of either g or a. So that must imply that the height of the mercury inside the elevator accelerating upwards must be the same as in the case where the elevator was not accelerating.

But after checking various solutions on the internet, which propose that the height of the mercury changes, I am confused and not able to determine what I assumed wrong or where I made a mistake.

I would appreciate if someone could clarify my conceptual doubt.

(P.S I do not know how to write mathematical equations in PhysicsStackExchange so please excuse any distastefulness in my handwriting or illustration skills)

enter image description here

enter image description here

Qmechanic
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Archit Chhajed
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  • Your mistake is when you assume that the atmospheric pressure in the lift (your $P_A$) is changed by the acceleration of the lift - see https://physics.stackexchange.com/questions/101524. – gandalf61 Feb 16 '21 at 15:37
  • @gandalf61 I visited the link you sent but the link does not explain why the pressure within the elevator would remain the same. It's something that maybe I haven't learnt or conceptualized yet, I would greatly appreciate it if you would elaborate more in a reply to me explaining why the atmospheric pressure inside the lift would remain the same. Because from what I understand, me assuming that atmospheric pressure changes in the elevator should give me a contradiction at the end that tells me that my implicit assumption was wrong.Moreover, I never took atmospheric pressure into consideration – Archit Chhajed Feb 16 '21 at 18:16
  • @gandalf61 I only worked out the equation for a general case for where the acceleration is upwards for the elevator. I never took the atmospheric pressure when the lift is not accelerating into consideration. As you can see I only considered the mass of the air column above the mercury in the container that is exposed to the atmosphere and the area of the container, and at the end, (g+a) cancels out. I don't know where exactly is my mistake. I greatly appreciate you providing your time and effort to help me out with this. :) – Archit Chhajed Feb 16 '21 at 18:21
  • I have posted a more complete answer below. – gandalf61 Feb 16 '21 at 18:55
  • To learn how to write equations here, please see What notation and symbols are commonly used here. Also see https://math.meta.stackexchange.com/q/1773/207316 & https://www.onemathematicalcat.org/MathJaxDocumentation/TeXSyntax.htm – PM 2Ring Feb 16 '21 at 21:02
  • @PM2Ring Thank you for providing me the link to learn how to write mathematical equations on PhysicsStackExhange, but I have difficulty in following the instructions, I would be grateful if you could link me a video tutorial on how to use the required language to use mathematical symbols. :) – Archit Chhajed Feb 17 '21 at 08:51
  • Sorry, Archit, I don't keep track of video tutorials, I prefer written material. And sorry if I overloaded you with too much information. The "MathJax basic tutorial and quick reference" can be rather intimidating, (and slow to load). There's a current discussion on Math SE meta about a MathJax tutorial for first time users, but no solid plans, as yet. – PM 2Ring Feb 18 '21 at 12:24
  • You don't need to learn a huge amount of info to start using MathJax, you can do most of what you need, once you understand a few basic principles. And you can easily see the MathJax code for any equation displayed on these sites: right-click (or long-press on a touch screen) and a menu will pop up. Select Show Math As/TeX Commands. – PM 2Ring Feb 18 '21 at 12:30

2 Answers2

1

Your expression for $P_B$ is correct, but your expression for $P_A$ is incorrect. When you write

$\displaystyle P_A=\frac{M(g+a)}{A}$

you are implicitly assuming that the whole mass $M$ of the atmosphere above the open area of mercury (all the way to the top of the atmosphere) is being accelerated. This is not true. Only the small amount of air in the lift is being accelerated, and its mass is negligible compared to the much larger mass $M$. You should have

$\displaystyle P_A=\frac{Mg}{A}$

On the other hand, all of the mercury is being accelerated, so we do have to include the acceleration when calculating the pressure of the column of mercury, as you have done in your expression for $P_B$:

$\displaystyle P_B = \rho (g+a)h$

Equating $P_A$ and $P_B$ now gives

$ \displaystyle \Rightarrow \frac {Mg}{A} = \rho(g+a)h \\ \displaystyle \Rightarrow h = \frac{Mg}{A\rho (g+a)}$

We can also write $\frac{Mg}{A}$ as atmospheric pressure$P_{atm}$, so we have

$\displaystyle h=\frac{P_{atm}}{\rho (g+a)}$

As you can see, the height $h$ of the mercury column now depends on the acceleration $a$, and it decreases as $a$ increases.

gandalf61
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  • Nice explanation why atmospheric pressure on the barometer has neglible increase. – John Darby Feb 16 '21 at 21:03
  • You mentioned that only a small mass of the air is accelerated. But how is that possible, from what I understand is that the elevator is completely enclosed and encloses everything inside it, so when the elevator accelerates upwards, the air inside it must also accelerate, So in the frame of the elevator the weight of the air above the mercury exposed to the atmosphere should be M(g+a). I wish you could clarify why this reasoning is wrong and provide a more detailed explanation. – Archit Chhajed Feb 17 '21 at 08:41
  • Moreover, I did not propose to approximate or neglect any variable. I solved the equations as I did in my original post for a general case. If the elevator was accelerating with a very high value of a, it would be equivalent to an elevator not accelerating and being on an entirely different planet with a much higher value of g. (Assuming that the density of the air in the elevator remains the same). – Archit Chhajed Feb 17 '21 at 08:47
  • (Continued Paragraph from my last reply)And in this scenario, we would expect the pressure to change inside the elevator and it does, since P=(rho)(g+a)h, since the acceleration changes, but the height of mercury inside the tube still remains the same. I don't see any loopholes in this explanation. – Archit Chhajed Feb 17 '21 at 08:57
  • @JohnDarby I humbly request you to weigh in on my explanation as well. :) – Archit Chhajed Feb 17 '21 at 08:58
  • @ArchitChhajed The mass $M$ of air that contributes to the atmospheric pressure is the entire mass of a column of air right to the top of the atmosphere. It is not just the air in the lift. But only the air in the lift (which has a negligible mass) is accelerated. This is why the acceleration of the lift does not change the atmospheric pressure within the lift. On the other hand, all of the mercury is being accelerated, so the pressure of a column of mercury with height $h$ does change due to the acceleration of the lift. This is why there is an asymmetry between $P_A$ and $P_B$. – gandalf61 Feb 17 '21 at 10:56
  • @ArchitChhajed I have edited my answer to try to make this point clearer. – gandalf61 Feb 17 '21 at 11:05
  • The acceleration has a negligible effect on pressure, but it is not zero. A very high acceleration would squeeze most of the air together at the bottom of the elevator - a "mini-atmosphere" of sorts - in which case the answer would depend not only on the acceleration, but also on where in the elevator the measurement was done. This would be a fairly atypical elevator, mind. – Kristoffer Sjöö Feb 17 '21 at 11:36
  • @Archit Chhajed I added my answer. This is a good question. – John Darby Feb 17 '21 at 13:30
1

This is response to a comment from @Archit Chhajed requesting my answer.

Any object of mass $m$ in the elevator experiences an inertial (fictitious) force $-ma$ downward where $a$ is the acceleration of the elevator upward. The total force on each object is $-m(g + a)$ downward where $g$ is the acceleration of gravity. The weight of each object is increased from $mg$ to $m(g + a)$. There are three objects in the elevator to be considered: the air in the elevator (not the total atmosphere), the mercury exposed to the air in the elevator, and the mercury in the tube. The increase in the pressure from the air in the elevator on the mecury exposed to the air is very small due to the small mass of air; use the ideal gas law to calculate the small mass: $PV = nRT$ where $P$ is pressure, $V$ volume of the elevator, $n$ moles of air (convert this to mass of air), $R$ the universal gas constant, and $T$ temperature. So the increase in the pressure from the air in the elevator is very small. Both the mercury exposed to the air and the mercury in the tube increase in weight. (It is as if with the elevator at rest the mercury is replaced by a denser liquid, except for the accelerating elevator the density is not increased, but the effective acceleration of gravity is increased from $g$ to $g + a$). As others (e.g. @gandalf16) have explained, the increase in weight of the mercury in the tube decreases the height of the mercury in the tube, neglecting the small change in air pressure. That is, the air pressure pushes a heavier liquid less-far up the tube. The level of the mercury outside the tube increases as the level of mercury inside the tube decreases. (I have neglected the small vapor pressure of mercury in the open space at the top of the tube; that is, I have assumed a vacuum in the open space at the top of the closed-ended tube.)

Hope this helps.

John Darby
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