Homework Help: Fluid mechanics : Barometer - slight confusion

1. Jun 16, 2010

ludic

What would happen to the mercury in the column if a barometer is accelerated upwards?

In static condition, acceleration is g.

When the system moves upward with an acceleration a, effective acceleration = g + a.

The air above the mercury in the trough will press it with higher force (i.e the normal reaction force at this interface increases), so the pressure at this interface will be higher than atm. P . To make the pressure at the base of the column equal to this increased pressure, mercury rises in the column.

Am I not right in thinking that on being accelerated upward the mercury would rise in the column?
Please do point out where I am amiss if I am wrong.

2. Jun 16, 2010

kuruman

You are saying that the weight of the air outside the column will increase and that is correct. However, would the weight of the mercury inside the column also not increase?

3. Jun 16, 2010

ludic

yes..... now i come to think of it, it will...... so, will the height of mercury in the column remain same ?

4. Jun 16, 2010

kuruman

What do you think?

5. Jun 16, 2010

ludic

I do think so. ... because:

Mg / A = h(rho)g ---------------- static condition
M(g+a) / A = h'(rho)(g+a) ------------ accelerating condition.

ergo, h' = h

where rho is the density of mercury and M is the mass of the section of atmosphere above the trough of the barometer, and A is the CSA of the trough.

Last edited: Jun 16, 2010
6. Jun 16, 2010

zzzoak

In static condition
$$P=\rho gh$$
where
P is atmospheric pressure.

7. Jun 16, 2010

ludic

so, am I right ? Will the height remain same ?

8. Jun 17, 2010

kuruman

Isn't that what you proved in posting #5?

9. Jun 17, 2010

ludic

Please once say that it will. I need to cite this thread somewhere.... please...

10. Jun 17, 2010

zzzoak

My opinion is that when it is moved up the weight of Hg is increased
$$P=\rho (g+a)h_1$$

11. Jun 17, 2010

ludic

Please be so kind and patient to clarify this doubt:

I understand the part where mercury in column becomes heavier and goes down because of increased effective acceleration

What I don't understand is why - if we call the pressure of atmosphere it's weight upon unit surface area being considered - would the effective weight of atmosphere, and hence its pressure not increase on the barometer trough due to this same increased effective aceleration, and force the mercury in the column back up....... effectively keeping the height of mercury in the column same as when the barometer was static.

please tell me what is the flaw in the above reasoning..............

I am thinking that maybe what I reason considers the atmosphere above the trough static, but in actuality, it won't have any increased pressure because the air will shift sideways as the barometer ascends.