Understanding the slanted manometer - what happens to the tube normal force?

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.

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

Question

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.

Thus your pressure $$P_0$$ being the same across a cross-section of the tube is not correct.
• 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 $l\sin\theta$. Jul 2, 2023 at 13:57