# Why sections are irrelevant in a mercury barometer

My understanding of a mercury barometer is that the force exerted by air is equal to the force exerted by mercury in the emerged portion of the tube. The forces here are weight of air and weight of mercury.

But in the picture above, there are more air molecules involved in the left side experience than on the right side one. Still the balance is not changed and the same height of mercury is actually balanced by a variable weight of air.

I don't understand why air-mercury interface area (noted S1) and tube section area (S2) don't influence the height of the mercury column.

I suspect the answer is: "because the weight by unit of area (definition of the pressure) is the same in both cases". So what I don't understand is actually why pressures are taken into account rather than weights.

I'm looking for a didactic explanation.

The problem with your reasoning is that each small volume of water on the surface of the lower water level feels forces of $\rho g h dA$ and $P_{air}dA$ (Where $dA$ is the area of that small volume). You need to notice that the water are not a rigid body.... So if the water are not moving each volume element should have a force balance on it, and the force balance is not necessarily true for the "body of water".

• It is worth noting that the barometric fluid is mercury and not water as described in this answer. Nov 29, 2019 at 15:41

Equation of continuity: $S_1 v_1 = S_2 v_2$

Because the fluid is at rest it is $v_1=v_2=0$

Bernoulli equation: $\rho g h_1 + p_1 = \rho g h_2 + p_2$

This equation does not depend on the section area. It can pe easily derived by considering a volume $V$ with specific weight $\rho g$. It is $V = S_i\Delta h$ with $i =1,2$ for both sectional areas with a height difference of the fluid volume $\Delta h = h_2-h_1$. Then conpute which pressure difference $\Delta p$ one obtains: $\Delta p = \frac{F_i}{S_i} = \frac{\rho g S_i\Delta h}{S_i}$.

The end result is independent on the $S_i$.