I have an elevator accelerating upwards at an acceleration of $a$. A mercury barometer measures the atmoshperic pressure at this point. Will the reading be less than, more than or equal to $76 \ cm$ of $Hg$?

I remembered from my class that the pressure in a liquid in an nonintertial frame accelerating upwards at $a$ is $\rho h (g + a)$. So, my first thought was it would be less that $76cm$ of $Hg$. But, then I realized that as you are accelerating upwards, you are hitting the air molecules faster and more often. So, technically, the atmoshperic pressure in our frame of reference should increase. But, then again I thought that air was compressible, so maybe that would skew off the results. What's the correct answer?


The height of the mercury in the barometer would be less.

$P = \rho gh$, without acceleration

$h = \frac{P}{\rho g}$, without acceleration

$P = \rho (g + a)h$ with acceleration

$h = \frac{P}{\rho (g +a)}$ with acceleration

The actual atmospheric presure would be the same in either case, but the barometer would report a lower pressure in units of cm or mm of mercury.

If the elevator and pool of mercury was exposed instead of a normal enclosed elevator, then movement relative to the air would create an additional pressure. The pressure from wind is very small at the speed of an elevator, and can be calculated by:

$P_{wind} = \frac{1}{2}\rho_{air} v^2$, where $v$ is the velocity of the wind (elevator in this case).

$\rho_{air} =1.25 kg/m^3$

So at a speed of $1 m/s$ , the additional pressure is only $0.6Pa$ compared to atmospheric pressure of $101325 Pa$!

  • $\begingroup$ How about the change change in air density from air beyond smashed into the bottom of the elevator? $\endgroup$ – aPhysicist Aug 19 '14 at 17:51
  • $\begingroup$ In this are both the $P$, i.e. before and after acceleration same? If yes, how? Won't there be change in air pressure inside the elevator as there is pseudo force acting on the molecules inside the elevator $\endgroup$ – Kaushik Nov 12 '19 at 6:00

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