Pascal's law

Push on one spot of a liquid trapped inside a closed container and, almost magically, every wall of that container feels the extra push equally — instantly, in every direction. That single fact is Pascal's law, and it is the quiet principle behind why a small force on a small piston can lift a car on a large piston. This article stays inside the ICSE Class 9 "Fluids" chapter: transmission of pressure in an enclosed liquid, the relation $F_1/A_1 = F_2/A_2$ , and naming (not dissecting) hydraulic machines. We will state the law precisely, see why it differs from ordinary depth-pressure, work through two-piston numericals fully, and flag the traps examiners love to set.

The core insight

Pressure applied at any point of an enclosed liquid is transmitted equally and undiminished in all directions, so $F_1/A_1 = F_2/A_2$ links the forces on any two pistons connected to that liquid.

The core idea

Pascal's law states that pressure applied at any point of an enclosed liquid is transmitted equally and undiminished throughout the liquid, and acts perpendicularly on every surface it touches. Push anywhere on the liquid and the increase in pressure does not stay local — it spreads through the whole enclosed body of liquid at once, with the same intensity everywhere. The most useful consequence for numericals: if an enclosed liquid connects two pistons of cross-sectional areas $A_1$ and $A_2$ , the pressure must be identical at both, because it is transmitted undiminished between them.

p_1 = p_2 \quad\Rightarrow\quad \frac{F_1}{A_1} = \frac{F_2}{A_2}

where: $F_1$ is the force applied on the piston of area $A_1$ , and $F_2$ is the force appearing on the piston of area $A_2$ ; both forces act perpendicular to their respective piston faces, and pressure is measured in pascals ( $1\ \text{Pa} = 1\ \text{N/m}^2$ ). Because the pressure ( $F/A$ ) is forced to stay equal while the areas $A_1$ and $A_2$ can be very different, a small force on a small piston can balance — or produce — a much larger force on a larger piston. This is the idea (never the internal mechanism) behind the hydraulic press, hydraulic jack, and hydraulic brakes; the ICSE syllabus asks only that you name these as applications of Pascal's law. It is worth separating Pascal's law sharply from the liquid-column formula $p = h\rho g$ . That formula tells you how pressure changes with depth inside an open liquid — a fact about the liquid sitting still under gravity. Pascal's law instead describes what happens when an extra pressure is deliberately applied to a liquid sealed inside a container: that extra pressure adds on top of any depth-pressure already present and is passed on equally to every part of the enclosed liquid and every wall it touches. The word "enclosed" is doing real work here — an open liquid does not transmit an applied push evenly the way a sealed system of connected cylinders does.

Derivation AFrom equal transmitted pressure to F1/A1 = F2/A2

The goal is to show, starting from nothing more than the statement of Pascal's law and the definition of pressure, that a force applied on one piston of an enclosed hydraulic system is related to the force appearing on a second piston purely through the ratio of their areas — the single relation that every two-piston numerical in this chapter rests on.

Step 1. Write the definition of pressure at each piston. A force $F_1$ pressing on a piston of area $A_1$ creates a pressure $p_1 = F_1/A_1$ at that piston; a force $F_2$ on a piston of area $A_2$ creates a pressure $p_2 = F_2/A_2$ at that piston. Both pistons are in contact with the same enclosed liquid.

Step 2. Apply Pascal's law. Because the liquid is enclosed and pressure applied anywhere in it is transmitted equally and undiminished throughout, the pressure the liquid exerts at the first piston must equal the pressure it exerts at the second piston: $p_1 = p_2$ . This is not an assumption about the pistons — it is a direct statement of the law.

Step 3. Substitute the two definitions from Step 1 into the equality of Step 2. Since $p_1 = F_1/A_1$ and $p_2 = F_2/A_2$ , and $p_1 = p_2$ , it follows immediately that $F_1/A_1 = F_2/A_2$ . Rearranged, $F_2 = F_1 \times (A_2/A_1)$ , which shows explicitly that the force is scaled up (or down) by the ratio of the areas.

Step 4. Read the physical meaning. Nothing here says the forces are equal — only the pressures are. If $A_2$ is much larger than $A_1$ , then $F_2$ is much larger than $F_1$ , even though the underlying pressure never changed. This is exactly the relation used to solve every hydraulic-piston numerical in this chapter.

Worked example

An enclosed liquid in a container has two pistons. A force of 40 N is applied on the smaller piston of area $0.02\ \text{m}^2$ . Find the transmitted pressure and the force produced on a larger piston of area $0.5\ \text{m}^2$ .

Step 1. Find the pressure at the smaller piston using $p = F_1/A_1$ : $p = 40/0.02 = 2000\ \text{Pa}$ .

Step 2. Apply Pascal's law: this pressure is transmitted equally and undiminished throughout the enclosed liquid, so the pressure at the larger piston is also $2000\ \text{Pa}$ .

Step 3. Find the force on the larger piston using $F_2 = p \times A_2$ : $F_2 = 2000 \times 0.5 = 1000\ \text{N}$ .

A force of only 40 N on the small piston produces a force of 1000 N on the large piston — the pressure stayed equal, but the forces differ because the areas differ.

Competitive stretch

In a closed system, piston A has area $0.01\ \text{m}^2$ and piston B has area $0.08\ \text{m}^2$ . A force of 25 N is applied to piston A. What force on piston B keeps the system in equilibrium?

Step 1. For equilibrium, Pascal's law requires the pressures on both pistons to be equal: $F_A/A_A = F_B/A_B$ .

Step 2. Substitute the known values: $25/0.01 = F_B/0.08$ , so $2500 = F_B/0.08$ .

Step 3. Solve for $F_B$ : $F_B = 2500 \times 0.08 = 200\ \text{N}$ . A force of 200 N on piston B balances a force of only 25 N on piston A, since $F_B$ must be proportionally larger to match the same pressure spread over a larger area.

Common traps

Thinking a larger force means a larger pressure. In these problems, pressure stays equal on both pistons; only the forces differ, because the areas differ.
Trying to explain the internal working of a hydraulic press, jack, or brake. The syllabus asks only that you name these as applications of Pascal's law, not describe valves or pistons in detail.
Confusing Pascal's law with $p = h\rho g$ . Pascal's law is about transmission of an applied pressure through an enclosed liquid; $p = h\rho g$ is about pressure varying with depth in an open liquid at rest.
Forgetting the liquid must be enclosed. In an open liquid, pressure still varies with depth, and a push at one point is not transmitted evenly to every part of the liquid.

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Why can a small force on a small piston lift something heavy on a large piston, doesn't that break the rules of physics?

No rule is broken because it is the pressure, not the force, that stays equal. Since $F_1/A_1 = F_2/A_2$ , if $A_2$ is much larger than $A_1$ , then $F_2$ must be proportionally larger than $F_1$ to keep the pressure the same on both sides. You are trading a small force over a small area for a large force over a large area — the pressure link between them is what makes it possible.

Do I need to know how a hydraulic press or hydraulic brake actually works inside, with all the valves and pipes?

No. The ICSE Class 9 syllabus explicitly excludes the internal working of hydraulic machines. You only need to know that the hydraulic press, hydraulic jack, and hydraulic brakes are named applications of Pascal's law, and be able to apply $F_1/A_1 = F_2/A_2$ to numericals about them. Describing valves or fluid pathways is beyond what is asked.

How is Pascal's law different from the pressure formula p equals h rho g that I learned for liquids at rest?

$p = h\rho g$ describes how pressure changes with depth inside a liquid simply due to the weight of liquid above a point — it applies whether or not the liquid is enclosed. Pascal's law describes something different: what happens when you actively apply an extra pressure to a liquid sealed inside a container — that extra pressure is passed on equally and undiminished to every part of the liquid and every wall of the container, on top of whatever depth-pressure already exists.

Pascal's law

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads