Pascal's law — pressure transmits equally: Foundations

Before you can believe that idea, you must be fluent with a handful of symbols and pictures. The parent note tosses around $p$ , $A$ , $F$ , $ρ$ , $g$ , $h$ , $Δ p$ , $θ$ , $d A$ , wedge geometry, and "isotropic." This page builds each one from zero, in the order they depend on each other, so nothing is ever used before it's earned.

1. Force — a push, measured

Why does the topic need it? Because the whole payoff of Pascal's law — "a small push lifts a car" — is a statement about forces. We must be able to say "this force is 100 N pushing down" precisely.

2. Piston — the sliding lid that presses the fluid

Why introduce it now? Because the very next idea (area) needs something concrete whose area matters. A piston is a circle, so its area is $π r^{2}$ — and the two pistons' different areas are the whole trick of the lift.

3. Area — how much surface a push lands on

The face of a circular piston of radius $r$ has area $A = π r^{2} .$

Figure — Pascal's law — pressure transmits equally

Figure s01 — Two circles of radius 1 and 2 drawn to scale; the larger one covers four times the area even though its radius is only doubled. This is why $A = π r^{2}$ makes big pistons enormously stronger.

Why does the topic need it? Because pressure is force spread over an area, and the lift's magic is that the big piston has a much larger area. Radius-squared is where the number 100 in the worked example comes from.

4. Pressure — force shared out over area

Now we combine the two arrows above.

Why "force per area" and not just "force"? Because a fluid doesn't care about your total push — it responds to how concentrated that push is. Two different-sized pistons feeling the same pressure produce different forces. That single fact is the hydraulic press. See Pressure — force per unit area.

5. The hydraulic press formula $F_{2} = F_{1} (A_{2} / A_{1})$

We now have exactly what we need to see why a tiny push lifts a heavy load — the destination of the whole topic. Hold two pistons joined by confined fluid: a small one (area $A_{1}$ ) you push with force $F_{1}$ , and a big one (area $A_{2}$ ) carrying the load with force $F_{2}$ .

Here $F_{1}, A_{1}$ label the small (input) piston and $F_{2}, A_{2}$ label the big (output) piston. Every symbol in this formula was defined above; the equal-pressure step is the one thing still owed, and §10 supplies it.

6. Density — how tightly packed the matter is

Why does the topic need it? Because a taller column of fluid weighs more, and how much more depends on how heavy the fluid is per volume — that's $ρ$ . It appears next, in the depth term.

7. Gravity $g$ and depth $h$ — the weight of the fluid above

Figure s02 — A water tank with the surface marked $h = 0$ ; a yellow arrow measures depth $h$ downward, and pink arrows at the bottom show the weight of the column above pressing down. Deeper means a taller, heavier column, hence more pressure.

Why does the topic need $g$ and $h$ ? Together with $ρ$ they measure the pressure added by the fluid's own weight — the part of the pressure that Pascal's law leaves untouched when you press extra on the top.

8. Hydrostatic pressure — deriving $p = p_{0} + ρ g h$

We don't just quote this formula — we build it by stacking the weight of the fluid, slab by slab.

Figure s03 — A single thin slab of fluid at depth $y$ : blue arrows show pressure $p (y)$ pushing down on the top and $p (y + d y)$ pushing up on the bottom, a pink arrow shows the slab's weight $ρ A g d y$ pulling down. Balancing these three gives $d p / d y = ρ g$ .

Why derive it? Because the transmission argument works by noticing that when you raise $p_{0}$ , the stacked-up $ρ g h$ piece does not change. See Hydrostatic pressure — p = p0 + ρgh.

9. The change symbol $Δ p$ — "how much it went up"

10. Isotropic pressure & the tiny wedge — $θ$ , $d A$ , $p_{x} = p_{y} = p_{n}$

Above (§5) we assumed the pressure from the small piston reaches the big one undiminished and equal. The reason it can is that pressure at a point is the same in every direction. The word for that is isotropic (Greek: "same in all turns"). Here is the proof.

Figure s04 — A tiny triangular wedge of still fluid with angle $θ$ ; blue arrows push normal to each face ( $p_{x}$ into the vertical face, $p_{y}$ into the horizontal face, $p_{n}$ into the slant). The slant push splits into horizontal and vertical shares that exactly balance the other two, forcing $p_{x} = p_{y} = p_{n}$ .

Formula Writing out the balance (this is the WHY)

Geometry of the wedge (from the triangle's shape): $d A_{x} = d A_{n} sin θ, d A_{y} = d A_{n} cos θ .$ Horizontal ( $x$ ) balance. The push on the vertical face is $p_{x} d A_{x}$ . Only the horizontal component of the slant push opposes it; the slant push is $p_{n} d A_{n}$ and its horizontal share is $p_{n} d A_{n} sin θ$ . Equal and opposite: $p_{x} d A_{x} = p_{n} d A_{n} sin θ .$ Substitute $d A_{x} = d A_{n} sin θ$ : $p_{x} (d A_{n} sin θ) = p_{n} d A_{n} sin θ .$ The whole factor $d A_{n} sin θ$ appears on both sides, so we divide it out — that cancellation is why the areas and the angle vanish, leaving $p_{x} = p_{n} .$ Vertical ( $y$ ) balance. Likewise the vertical component of the slant push is $p_{n} d A_{n} cos θ$ : $p_{y} d A_{y} = p_{n} d A_{n} cos θ \Rightarrow p_{y} (d A_{n} cos θ) = p_{n} d A_{n} cos θ \Rightarrow p_{y} = p_{n} .$ Therefore $p_{x} = p_{y} = p_{n}$ : pressure at the point is the same in every direction.

Why does this let §5 work? Because pressure doesn't "prefer" a direction, the pressure you create by pushing the small piston presses equally on the fluid and on the far piston — which is exactly what "transmitted undiminished" needs.

What is a piston and why does its shape matter?

A sliding disc sealing fluid in a cylinder; it's a circle, so its area is

π r^{2}

What does the symbol

r

mean and its units?

The radius of the circular piston face — centre-to-edge distance, in metres.

Do I know the area of a circle and why it grows so fast?

A = π r^{2}

; area scales with radius squared, so doubling

r

quadruples

A

Can I write pressure three ways?

p = F / A

F = p A

A = F / p

; unit is the pascal,

1 Pa = 1 N / m^{2}

Can I derive the press formula?

Equal pressure

p_{1} = p_{2}

gives

F_{1} / A_{1} = F_{2} / A_{2}

, so

F_{2} = F_{1} A_{2} / A_{1}

What is the difference between mass and weight?

Mass in kg (amount of stuff); weight in N (force

= m g

);

1000

kg of water weighs

\approx 9.8

kN.

What do

y

and

p_{0}

mean in the hydrostatic derivation?

y

is the depth coordinate (down from the surface,

y = 0

at top);

p_{0}

is the pressure at that surface.

Why is

d p / d y = + ρ g

positive?

Because

y

is measured downward, and going deeper raises the pressure, so the slope is positive.

Can I derive

p = p_{0} + ρ g h

from a slab?

Balance weight vs pressure difference to get

d p / d y = ρ g

, then integrate from

0

h

What does

Δ p

(and a subscript) mean, and its key property?

The increase

p_{new} - p_{old}

at the labelled spot;

Δ p

is the same at every depth.

What do

p_{x}

p_{y}

p_{n}

denote?

The pressures pushing normal to the vertical face, the horizontal face, and the slanted face of the wedge.

Why is fluid pressure isotropic at a point?

A still tiny wedge has zero net force; dividing out

d A_{n} sin θ

forces

p_{x} = p_{y} = p_{n}

What does "incompressible" give us?

Volume can't be squashed, so

A_{1} d_{1} = A_{2} d_{2}

and the push transmits undiminished.

Why is the press not free energy?

F_{1} d_{1} = F_{2} d_{2}

— you trade force for distance; work is conserved.

Foundations — Pascal's law — pressure transmits equally