Visual walkthrough — Manometers, barometers

This page is the picture-by-picture version of the parent note Manometers & Barometers. Read it slowly; each figure is the argument.

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Step 1 — What is pressure? (a picture of pushing)

WHAT. Look at the figure: the same fist-sized push spread over a big plate barely dents it, but concentrated on a pin-tip it punctures. Same force $F$, different area $A$ — different pressure.

WHY define it as force per area? Because a fluid never pushes as "a total force"; it pushes on whatever surface you dip in. To compare a thin tube with a wide dish we must talk per unit area, and — crucial for later — this is exactly why the tube's width will drop out.

Recall Check yourself

If you double the area but keep the same force, what happens to pressure? ::: It halves — pressure is force divided by area.

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Step 2 — Why does going deeper mean more pressure?

WHAT. The figure marks two depths in a still tank. The deeper red dot has a taller column of water above it than the upper teal dot.

WHY it must increase. The fluid is not moving. So the fluid above any patch is being held up by the fluid below it. The only thing that can hold up weight is a push from underneath — pressure. More weight above ⇒ more push needed below ⇒ more pressure with depth. (See Hydrostatic Pressure.)

PICTURE. Notice the arrows: gravity (plum, down) on the column, and the support pressure (orange, up) from beneath. They must balance because nothing accelerates.

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Step 3 — Nail down the amount: a force balance on one cylinder

We now turn "more" into an exact formula.

WHAT. Cut out an imaginary cylinder of fluid: cross-section area $A$, its top at depth $h_1$, its bottom at depth $h_2$ (deeper, so $h_2>h_1$). Three vertical forces act, drawn in the figure:

• Top face pushed down by fluid above: force $=P_{1}A$.
• Bottom face pushed up by fluid below: force $=P_{2}A$.
• Weight of the cylinder itself pulling down: $W=\rho A(h_2-h_1)g$.

WHY a cylinder and why balance? Because the fluid is static (equilibrium), the up-pushes and down-pulls on this chunk must sum to zero: $\underset{\text{up}}{\underbrace{P_{2}A}}=\underset{\text{down}}{\underbrace{P_{1}A}}+\underset{\text{weight down}}{\underbrace{\rho A(h_2-h_1)g}}$

Now divide every term by $A$. WHY we may: pressure is per area, so area was never really the point — it cancels cleanly: $\boxed{P_2=P_1+\rho g(h_2-h_1)}$

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Step 4 — Same fluid, same level, same pressure

This tiny fact is the hinge of every U-tube.

WHAT. In the figure, a bent tube of one connected fluid has two points $L$ and $R$ drawn on the same horizontal dashed line.

WHY they are equal. In the law $\Delta P=\rho gh$, the height difference $h$ between $L$ and $R$ is zero — they sit at the same level. Zero height ⇒ zero pressure difference ⇒ $P_L=P_R$.

PICTURE the alternative: if $P_L$ were bigger, the extra push would shove fluid sideways from $L$ toward $R$ — but the fluid is still. So the pressures must match. This is Pascal's Law at work.

Recall Why does the horizontal bend not matter?

Because $\Delta P=\rho gh$ uses only vertical $h$; travelling sideways adds no height, so no pressure change. ::: Correct — only up/down changes pressure.

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Step 5 — Build the manometer and pick the reference line

WHAT. The gas pushes on the liquid in the left arm, driving it down and lifting the right arm up. The two free surfaces end up at different heights; their vertical gap is $h$.

WHY pick a reference line. From Step 4, a horizontal line drawn through the fluid gives equal pressure on both arms. We choose the dashed line at the lower free surface (here the left/gas arm). Everything below that is the same connected liquid, so we can equate the two sides there.

PICTURE. In the figure: dashed reference line through the left surface; on the right arm, a liquid column of height $h$ stands above that same line.

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Step 6 — Walk down each arm to the reference line

WHAT — left (gas) arm. Start at the gas, which presses $P_{\text{gas}}$ directly on the left liquid surface — and that surface is our reference line. So the pressure at the line from the left: $P_{\text{left}}=P_{\text{gas}}$

WHAT — right (open) arm. Start at the open top: atmosphere presses $P_{\text{atm}}$ on the right surface. To reach the reference line we go down by the height $h$, adding weight of that column (Step 3, "down adds"): $P_{\text{right}}=P_{\text{atm}}+\rho gh$

Term by term: $P_{\text{atm}}$ is the air's push on top; $\rho gh$ is the extra push from the $h$-tall slug of liquid sitting above the line ($\rho$ its density, $g$ gravity, $h$ its vertical height).

WHY set them equal. Both describe the pressure at the same reference line in the same fluid, so by Step 4: $P_{\text{left}}=P_{\text{right}}⟹P_{\text{gas}}=P_{\text{atm}}+\rho gh$

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Step 7 — The other case: gas below atmospheric (sign flip)

Never leave a case unshown. What if the atmosphere is the stronger one?

WHAT. Now the gas arm is the higher surface. Put the reference line at the lower (open-arm) surface.

• Right/open arm at the line: $P_{\text{right}}=P_{\text{atm}}$ (open surface is the line).
• Left/gas arm: from gas surface go down by $h$ to reach the line: $P_{\text{left}}=P_{\text{gas}}+\rho gh$.

Equate: $P_{\text{gas}}+\rho gh=P_{\text{atm}}⟹\boxed{P_{\text{gas}}=P_{\text{atm}}-\rho gh}$

WHY the minus. The gas is too weak to hold its side down, so the atmosphere pushes the open side lower and the gas side rises. A rising gas side ⇒ subtract $\rho gh$.

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Step 8 — Gauge vs absolute (what the number means)

WHAT. The figure stacks a bar from $0$ up: first $P_{\text{atm}}$, then the manometer's extra $\rho gh$ on top.

WHY it matters. The manometer only ever gives you the difference $\rho gh$ between the gas and the atmosphere — that difference is called gauge pressure: $P_{\text{gauge}}=P_{\text{abs}}-P_{\text{atm}}=\rho gh$ To get the true absolute pressure you must add back the atmosphere: $P_{\text{abs}}=P_{\text{atm}}+\rho gh.$ (A barometer differs because its top is a vacuum, so it reads absolute directly — see the parent note.)

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The one-picture summary

This single figure compresses all eight steps: weight of a fluid column ⇒ pressure grows with depth ($\Delta P=\rho gh$) ⇒ same level = same pressure ⇒ equate the two arms at a reference line ⇒ read the height gap as $P_{\text{gas}}=P_{\text{atm}}\pm \rho gh$.

Recall Feynman retelling — say it to a friend

Fluid is heavy, so the deeper you go the more stuff is stacked on you and the harder it presses — that "harder press" is pressure, and it grows by exactly the weight of the fluid above per unit area, which works out to $\rho gh$ for a height $h$. Now bend a tube of this liquid into a U. Because the liquid is connected and still, any level line has the same pressure on both sides — if it didn't, liquid would slosh across until it did. Put gas on one side and open air on the other. Whichever push is stronger shoves its side of the liquid down and lifts the other side up, leaving a height gap $h$. Walk down each arm to a shared line, adding $\rho gh$ every time you descend, and set the two sides equal. Out pops $P_{\text{gas}}=P_{\text{atm}}\pm \rho gh$: plus when the open side rides higher (strong gas), minus when the gas side rides higher (weak gas), and dead level when they're equal. The manometer never tells you the whole pressure — only how much it beats or trails the air, the gauge pressure — so add the atmosphere back if you want the absolute value.

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Connections

• Hydrostatic Pressure — Step 3's $\rho gh$ is proven here.
• Pascal's Law — the "same level, same pressure" of Step 4.
• Atmospheric Pressure — the $P_{\text{atm}}$ the open arm feels.
• Density and Specific Gravity — why $\rho$ (mercury vs water) sets the height.
• Buoyancy and Archimedes Principle — also built on depth-dependent pressure.
• Bernoulli's Equation — extends pressure to moving fluids.