Vertical manometer

Consider a manometer filled with water. Given a pressure $p_0$ at height $y=h$, The formula for the pressure $p$ at $y=0$ is $$ p = p_0+\rho g h $$ One way to derive this is to consider the slug of liquid/gas above the fluid at $y=0$, which has mass $\rho h A$. The pressure force at $y=0$ is the static pressure plus the pressure needed to balance this force.

Tilted manometer

For a manometer tilted at angle $\theta$, the change in height is instead $l\sin\theta$, so the pressure is $$ p=p_0+\rho g l \sin\theta $$ where $l$ is the length of manometer going off at angle $\theta$ from $y=0$.

I was trying to derive this formula by considering the "slug of mass" above the fluid at $y=0$. This mass is $$ m=\rho l A $$ The gravitational force acting at the center of this slug of fluid/gas is $$ W = \rho g l A $$ This gravitational force can be decomposed into a force acting axially along the manometer and adding to the fluid pressure, and a force perpendicular to the lower manometer surface.

$$ W_{axial} = \rho g l A \sin\theta $$ $$ W_{perp} = \rho g l A \cos\theta $$


My question is, what do we do with the force component acting perpendicular to the manometer? It's a fluid force that acts on the lower, inner surface of the manometer, which makes me think it should somehow add to the pressure. But for a given height, the pressure in a static fluid is the same everywhere.

Somewhere, my reasoning is breaking down. I'd appreciate some help to understand where I'm going wrong.

manometer drawing


1 Answer 1


If the left had end was open then the liquid surface would not be inclined as in your diagram it would be horizontal.

Thus your pressure $P_0$ being the same across a cross-section of the tube is not correct.

The isobars in your tube are horizontal and then the situation is no different from having liquid in a container whose sides are not vertical and whose base is not horizontal.

  • $\begingroup$ I see your point that the isobars should be horizontal. That pretty much negates the my argument! What would be the correct way to construct this "slug of fluid" argument? It would depend on "what is the mass of fluid between the original isobar and the new isobar"? $\endgroup$
    – nwsteg
    Commented Jul 2, 2023 at 1:00
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    $\begingroup$ The answer to the post Pressure at base of 3 different dam designs might be of use? $\endgroup$
    – Farcher
    Commented Jul 2, 2023 at 8:04
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    $\begingroup$ It does help. I just thought of another way to show it. Start from the top position $y=h$. Consider a small column of water that goes straight down a small distance until it just barely hits the slanted manometer wall. Use the normal hydrostatic equation to get the pressure at this new isobar. Continue this process until you reach the desired depth. The vertical distance will be $l\sin\theta$. $\endgroup$
    – nwsteg
    Commented Jul 2, 2023 at 13:57
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    $\begingroup$ @nwsteg Yes, you are using the idea of jumping between the horizontal isobars. $\endgroup$
    – Farcher
    Commented Jul 2, 2023 at 14:27

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