Pascal's law — pressure transmits equally: Worked examples

The scenario matrix

Pascal problems live in a small grid. Two knobs turn: the area ratio $k = A_{2} / A_{1}$ and what they ask for (output force, input force, distance, pressure change, energy). Everything below fills one cell of this table.

Cell	Situation	What's special	Example
A	$k > 1$ (big output piston)	force multiplied	Ex 1
B	Distance trade-off	small piston travels far	Ex 2
C	Energy check	work in $=$ work out	Ex 3
D	$k < 1$ (reversed)	force reduced, distance gained	Ex 4
E	$k = 1$ (equal pistons)	degenerate — nothing multiplied	Ex 5
F	Pure $Δ p$ transmission	no pistons, just depth	Ex 6
G	Word problem (car jack)	real numbers, unit care	Ex 7
H	Exam twist (piston has weight)	extra $ρ g h$ + weight bias	Ex 8
I	Limiting case ( $A_{1} \to 0$ )	infinite advantage myth	Ex 9

Every numeric answer below is machine-checked in the verify block.

The one picture that runs the whole page

Figure — Pascal's law — pressure transmits equally

Look at the figure. The red small piston (area $A_{1}$ ) is pushed down with force $F_{1}$ . The confined fluid carries the same pressure change $Δ p = F_{1} / A_{1}$ to the large piston (area $A_{2}$ ), which pushes back with $F_{2} = Δ p A_{2}$ . Because the fluid is incompressible, the volume $A_{1} d_{1}$ that leaves the small side must equal the volume $A_{2} d_{2}$ that arrives at the big side — that single fact runs cells B, C, D, and I.

Formula The four equations we will reuse

$Pascal: equal p \frac{F _{1}}{A _{1}} = \frac{F _{2}}{A _{2}}$ Why this one? Pascal says the applied pressure is identical everywhere, so force-over-area on the small side equals force-over-area on the big side. It answers "what is shared between the pistons?" — the pressure. $incompressible A_{1} d_{1} = A_{2} d_{2}$ Why this one? The fluid can't be squashed, so the volume pushed out of one cylinder ( $A_{1} d_{1}$ ) must reappear in the other ( $A_{2} d_{2}$ ). It answers "if one piston moves, how far does the other move?" $force mult. F_{2} = F_{1} \frac{A _{2}}{A _{1}}$ Why this one? Rearranging the first equation for $F_{2}$ gives the output force directly. It answers "how much force do I get out?" $energy conserved F_{1} d_{1} = F_{2} d_{2}$ Why this one? Multiply force by distance on each side and the area ratios cancel — proving work in equals work out. It answers "is this a free-energy machine?" (No.)

Circular areas: $A = π r^{2}$ , so $\frac{A _{2}}{A _{1}} = (\frac{r _{2}}{r _{1}})^{2}$ . Why squared? Area grows with the square of radius — double the radius, quadruple the area. That is the entire source of the "huge boost".

Cell A — big output piston ( $k > 1$ )

In the figure, the red thin input piston is being pushed with $F_{1}$ ; the wide black output piston delivers the multiplied $F_{2}$ . Watch how the same $Δ p$ spreads across a far bigger black face.

Cell B — the distance you pay

The figure marks the two travel distances: the red double-arrow on the small side ( $d_{1} = 50 cm$ ) is huge, the black one on the big side ( $d_{2} = 2 cm$ ) is tiny. That visual mismatch is the price of the force boost.

Cell C — energy is conserved (no free lunch)

The figure draws work as area: the red input bar is tall-and-thin (small force, long distance), the black output bar is short-and-wide (big force, short distance). The two rectangles have the same area — that equal area is the $100 J$ conserved.

Cell D — reversed press ( $k < 1$ )

Here the red arrow sits on the big piston — this time we push the wide side. The figure shows the small black piston delivering the reduced output force. Everything from Cell A runs backwards.

Cell E — degenerate case, equal pistons ( $k = 1$ )

The figure shows two pistons of identical width. Input (red) and output (black) arrows are the same length — no multiplication, no reduction. This is the exact boundary between Cell A and Cell D.

Cell F — pure transmission, no pistons

The figure shows a sealed tank; the red top piston is squeezed by $Δ p = 6000 Pa$ . Follow the label at the bottom: the same $6000 Pa$ arrives there, untouched by the $h = 4 m$ depth marked on the left.

Worked example Ex 6 — pressing a sealed tank

A sealed tank of water, $4 m$ deep, has a piston on top. You raise the surface pressure by $Δ p = 6000 Pa$ . By how much does the pressure at the very bottom rise? (Take $ρ = 1000 kg/m^{3}$ , $g = 9.8$ .)

Forecast: deeper water already carries more pressure. Does the extra push weaken by the time it reaches the bottom?

Absolute bottom pressure: $p_{bottom} = p_{surface} + ρ g h$ (from hydrostatics). Here plain $p$ means the total pressure at the bottom, $p_{surface}$ is the total pressure at the top. Why this step? This splits the total bottom pressure into a surface part (which we change) and a depth part $ρ g h$ (which we don't touch).
Raise the surface's total pressure by the applied change $Δ p$ : new bottom $= (p_{surface} + Δ p) + ρ g h = p_{bottom,old} + Δ p$ . Why this step? The $ρ g h$ term is identical before and after (same fluid, same depth), so it cancels — only the applied change $Δ p$ survives.
Therefore $Δ p_{bottom} = 6000 Pa$ — the depth is irrelevant. Note: the absolute $p_{bottom}$ is much larger; it is only the change that equals $6000 Pa$ . Why this step? Subtracting the old bottom pressure from the new one leaves only $Δ p$ ; the fixed $ρ g h$ baseline drops out of the difference, so the change cannot depend on depth.

Verify: The unchanged baseline $ρ g h = 1000 \times 9.8 \times 4 = 39200 Pa$ appears in both old and new bottom pressure, so it cannot affect the change. $Δ p_{bottom} = 6000 Pa$ exactly. ✓ Transmission is undiminished — Pascal in its purest form.

Cell G — real-world word problem

The figure sets up the jack: the red narrow handle piston, the wide lifting piston with a car block on top labelled $8820 N$ , and the output force $6250 N$ . Reading it, you can see the output is shorter than the load — the one-stroke failure.

Worked example Ex 7 — the car jack

A hydraulic jack has a pump handle piston of diameter $1.2 cm$ and a lifting piston of diameter $6 cm$ . A person pushes the handle piston with $F_{1} = 250 N$ . (a) What weight can it lift? (b) The car mass is $900 kg$ — can it lift it?

Forecast: ratio of diameters is $5$ . Enough for a $900 kg$ car?

Use diameters directly: $\frac{A _{2}}{A _{1}} = (\frac{d _{2}}{d _{1}})^{2} = (\frac{6}{1.2})^{2} = 25$ . Why this step? Since $A = π (D /2)^{2}$ , the $π$ and the $\frac{1}{4}$ cancel in the ratio — you may use diameters, no need to halve.
Output force: $F_{2} = 250 \times 25 = 6250 N$ . Why this step? Pascal press formula.
Weight the car needs lifted: $W = m g = 900 \times 9.8 = 8820 N$ . Why this step? Weight is mass times gravity — this is the load to beat.
Compare: $6250 N < 8820 N$ → cannot lift it in one press. Why this step? Output force must at least equal the load. Real jacks use a ratchet: many pump strokes, each transferring a slug of fluid, so the effective advantage is far higher over many strokes.

Verify: $F_{2} = 6250 N$ , load $= 8820 N$ , and $6250 < 8820$ is true → single stroke fails. ✓ Units: N vs N, consistent.

Cell H — exam twist: the piston has weight

The figure shows the load and the platform's own weight stacked on the big piston, plus a red input arrow of $\approx 30 N$ and a height marker $h = 0.5 m$ between the two piston levels. Those two extras — grey platform weight and the height gap — are exactly what push the input above the naive value.

Worked example Ex 8 — heavy platform on the big side

Small piston area $A_{1} = 5 cm^{2}$ , large piston area $A_{2} = 250 cm^{2}$ . The large piston itself (platform) weighs $W_{p} = 400 N$ and sits $h = 0.5 m$ higher than the small one. You want to just start lifting a load $L = 1000 N$ on the big piston. What input force $F_{1}$ is needed? (fluid $ρ = 800 kg/m^{3}$ oil, $g = 9.8$ .)

Forecast: naive answer ignores platform weight and height. Guess low, then watch the corrections push it up.

Here plain $p_{1}, p_{2}$ are the absolute fluid pressures just under each piston (total pressure sitting there), not applied changes.

Total downward force the big side presents $= L + W_{p} = 1000 + 400 = 1400 N$ , over area $A_{2} = 250 cm^{2} = 0.025 m^{2}$ . Why this step? The fluid pressure under the big piston must hold the load and the platform's own weight.
Required pressure just under big piston: $p_{2} = \frac{1400}{0.025} = 56000 Pa$ . Why this step? Pressure $=$ force/area; this is the total pressure the fluid must supply there.
The small piston is lower by $h = 0.5 m$ , so by hydrostatics its fluid pressure is higher by $ρ g h$ : $p_{1} = p_{2} + ρ g h = 56000 + 800 \times 9.8 \times 0.5 = 56000 + 3920 = 59920 Pa$ . Why this step? Pascal transmits the applied change equally, but the two pistons sit at different heights, so their absolute pressures differ by the baseline $ρ g h$ — you must add it back.
Input force: $F_{1} = p_{1} A_{1} = 59920 \times 5 \times 1 0^{- 4} = 29.96 N$ . Why this step? Force $=$ pressure $\times$ area on the small piston.

Verify: $F_{1} \approx 29.96 N$ . Sanity: a naive "ignore weight & height" answer would be $\frac{L}{A _{2}} \cdot A_{1} = \frac{1000}{0.025} \times 5 \times 1 0^{- 4} = 20 N$ — the platform weight and the height difference correctly raised the requirement to $\approx 30 N$ . ✓

Cell I — limiting case ( $A_{1} \to 0$ )

The figure shows the input piston shrunk to a red sliver next to a wide black output piston, with the input stroke drawn stretching far off the top — a visual reminder that as $A_{1}$ shrinks, the distance the little piston must travel blows up just as fast as the force does.

Worked example Ex 9 — infinite advantage?

As the input piston shrinks, $A_{1} \to 0$ , the ratio $A_{2} / A_{1} \to \infty$ . Does a whisper of force lift anything, for free?

Forecast: the multiplier blows up — is this a free-energy machine?

Force out: $F_{2} = F_{1} \frac{A _{2}}{A _{1}} \to \infty$ as $A_{1} \to 0$ (for fixed $F_{1}$ ). Why this step? The ratio diverges, so any fixed input force gives unbounded output force — mathematically true.
Distance in: $d_{1} = d_{2} \frac{A _{2}}{A _{1}} \to \infty$ too. Why this step? Volume conservation forces the tiny piston to sweep the same volume with a smaller face, so its travel distance grows by the very same diverging ratio — an infinite stroke to move the big piston a finite amount.
Energy: $W = F_{1} d_{1}$ stays finite and equal to $F_{2} d_{2}$ ; the infinities cancel. Why this step? $F_{2} d_{2} = (F_{1} \cdot k) (d_{2})$ while $F_{1} d_{1} = F_{1} (d_{2} \cdot k)$ — both equal $F_{1} k d_{2}$ , so the diverging $k$ appears identically on both sides and never creates extra energy.

Verify: Take $k = 1000$ , $F_{1} = 10 N$ , $d_{2} = 0.01 m$ : $F_{2} = 10000 N$ , $d_{1} = 10 m$ . $W_{1} = 10 \times 10 = 100 J$ , $W_{2} = 10000 \times 0.01 = 100 J$ . ✓ Equal. The limit is not free energy — it demands an impractical stroke length. Reality also caps it: no fluid is perfectly incompressible and pistons have finite size.

The whole matrix on one map

The decision tree above is also drawn as a figure so you can follow it even if Mermaid does not render. Start at "equal pressure", split on the area ratio $k$ for force questions, drop to volume conservation for distance/energy questions, and add $ρ g h$ only when the two pistons sit at different heights.

Recall Which cell am I in?

First question: are they asking force, or distance, or energy? ::: Force → use $F_{2} = F_{1} A_{2} / A_{1}$ (cells A/D/E). Distance → $A_{1} d_{1} = A_{2} d_{2}$ (cell B). Energy → $F_{1} d_{1} = F_{2} d_{2}$ (cell C). Big output piston multiplies force by which number? ::: $A_{2} / A_{1} = (r_{2} / r_{1})^{2}$ — the squared radius ratio. Pistons at different heights — what must you add? ::: The $ρ g h$ baseline difference (cell H), on top of the transmitted change. Does $A_{1} \to 0$ give free energy? ::: No — work stays finite; the tiny piston needs infinite travel (cell I). If a question only raises the surface pressure and asks the change deep down? ::: It transmits undiminished — $Δ p$ is the same at every depth (cell F); track the change, not the absolute pressure.

Connections

Parent: Pascal's law — the principle these examples exercise.
Pressure — force per unit area — $p = F / A$ , used in every cell.
Hydraulic systems — brakes, lifts, jacks — cells A, D, G in the real world.
Incompressibility & continuity equation — the $A_{1} d_{1} = A_{2} d_{2}$ behind cells B, C, I.
Conservation of energy in machines — why cell C and cell I are not free lunches.
Hydrostatic pressure — p = p0 + ρgh — the $ρ g h$ correction in cells F and H.