Pascal's law — pressure transmits equally: Exercises

Before we start, here is the whole toolkit on one card so no symbol appears unexplained.

Formula The five tools (each earned in the parent note)

Pressure $p = \frac{F}{A}$ — force $F$ (in newtons, N) pressed straight into an area $A$ (in square metres, $m^{2}$ ). See Pressure — force per unit area.
Pascal transmission — a pressure change $Δ p$ applied to enclosed incompressible fluid reaches every point equally. See the parent note.
Same pressure across pistons $\frac{F _{1}}{A _{1}} = \frac{F _{2}}{A _{2}} \Rightarrow F_{2} = F_{1} \frac{A _{2}}{A _{1}}$ .
Volume conservation (incompressible): $A_{1} d_{1} = A_{2} d_{2}$ — fluid pushed out of one cylinder ( $A_{1} d_{1}$ ) appears in the other. See Incompressibility & continuity equation.
Hydrostatic add-on $p = p_{0} + ρ g h$ — extra pressure from a column of fluid of density $ρ$ , height $h$ , with $g \approx 9.8 m/s^{2}$ . See Hydrostatic pressure — p = p0 + ρgh.

Here $A$ = area, $d$ = distance a piston moves, $ρ$ (Greek "rho") = density in $kg/m^{3}$ .

Figure — Pascal's law — pressure transmits equally

Look at the figure: the small piston (left, lavender) has area $A_{1}$ ; you push it with force $F_{1}$ and it sinks a long way $d_{1}$ . The large piston (right, coral) has area $A_{2}$ ; it rises with force $F_{2}$ but only a short way $d_{2}$ . The mint fluid between them carries the same pressure to both. Keep this picture in mind for every problem.

Level 1 — Recognition

Recall Solution L1.1

The pressure change is equal at both pistons — that is exactly Pascal's law. The force is bigger on the large piston, and the distance moved is smaller on the large piston. So neither force nor distance is equal; only pressure.

Recall Solution L1.2

By exactly $3000 Pa$ . Write $p = p_{0} + ρ g h$ . Raising $p_{0}$ by $Δ p$ gives $p_{new} = (p_{0} + Δ p) + ρ g h$ . The $ρ g h$ term (which depends on depth) is untouched, so the rise is $Δ p$ at every depth — including $4 m$ . The depth is a distractor.

Level 2 — Application

Recall Solution L2.1

Area scales with radius squared, so we never need $π$ — it cancels: $\frac{A _{2}}{A _{1}} = \frac{π r _{2}^{2}}{π r _{1}^{2}} = (\frac{r _{2}}{r _{1}})^{2} = (\frac{30}{3})^{2} = 1 0^{2} = 100.$ Why squared? Look at the figure: doubling the radius quadruples the disc's area, so force multiplies by area ratio, not radius ratio. $F_{2} = F_{1} \frac{A _{2}}{A _{1}} = 200 \times 100 = 20, 000 N .$

Recall Solution L2.2

Fluid is incompressible, so volume out = volume in: $A_{1} d_{1} = A_{2} d_{2}$ . $d_{1} = d_{2} \frac{A _{2}}{A _{1}} = 2 cm \times 100 = 200 cm = 2 m .$ The small piston travels a huge distance to move the big one a tiny bit — that's the price of the force boost.

Recall Solution L2.3

Areas are given directly, so no squaring is needed here: $F_{2} = F_{1} \frac{A _{2}}{A _{1}} = 150 \times \frac{20}{5} = 150 \times 4 = 600 N .$

Level 3 — Analysis

Recall Solution L3.1

Work = force $\times$ distance moved in the force's direction.

Input work: $W_{1} = F_{1} d_{1} = 200 \times 2 = 400 J$ .
Output work: $W_{2} = F_{2} d_{2} = 20, 000 \times 0.02 = 400 J$ . They are equal: $W_{1} = W_{2} = 400 J$ . The press multiplies force by 100 but shrinks distance by 100, so their product — the energy — is unchanged. See Conservation of energy in machines.

Recall Solution L3.2

Pressure at the large piston = pressure from your push plus the weight of the oil column of height $h$ standing above it (from Hydrostatic pressure — p = p0 + ρgh): $p_{2} = \frac{F _{1}}{A _{1}} + ρ g h .$ Step 1 — applied pressure: $\frac{F _{1}}{A _{1}} = \frac{300}{4 \times 1 0 ^{- 4}} = 750, 000 Pa$ . Step 2 — hydrostatic add-on: $ρ g h = 850 \times 9.8 \times 1.5 = 12, 495 Pa$ . Step 3 — total at big piston: $p_{2} = 750, 000 + 12, 495 = 762, 495 Pa$ . Step 4 — force: $F_{2} = p_{2} A_{2} = p_{2} \times 50 A_{1} = 762, 495 \times 50 \times 4 \times 1 0^{- 4}$ $F_{2} = 762, 495 \times 0.02 = 15, 249.9 N \approx 1.52 \times 1 0^{4} N .$ The pure-Pascal estimate would be $300 \times 50 = 15, 000 N$ ; the oil column adds about $250 N$ .

Level 4 — Synthesis

Recall Solution L4.1

(a) Required output force = car's weight: $F_{2} = m g = 1500 \times 9.8 = 14, 700 N$ . Area ratio: $\frac{A _{2}}{A _{1}} = (\frac{r _{2}}{r _{1}})^{2} = (\frac{25}{2.5})^{2} = 1 0^{2} = 100$ . $F_{1} = F_{2} \frac{A _{1}}{A _{2}} = 14, 700 \times \frac{1}{100} = 147 N .$ (b) Volume conserved: $A_{1} d_{1} = A_{2} d_{2} \Rightarrow d_{1} = d_{2} \frac{A _{2}}{A _{1}} = 0.5 \times 100 = 50 m$ . (c) Input work $W_{1} = F_{1} d_{1} = 147 \times 50 = 7350 J$ . Gravitational energy gained by car $= m g d_{2} = 14, 700 \times 0.5 = 7350 J$ . ✓ They match: every joule you push in reappears as the car's height energy. No free lunch — you just pushed a small piston a long $50 m$ .

Recall Solution L4.2

Step 1 — bottom pressure before pressing: $p_{before} = p_{0} + ρ g h = 101, 325 + 1000 \times 9.8 \times 3 = 101, 325 + 29, 400 = 130, 725 Pa$ . Step 2 — Pascal adds the applied change undiminished: $p_{after} = p_{before} + Δ p = 130, 725 + 20, 000 = 150, 725 Pa$ . Step 3 — force on hatch: $F = p_{after} A = 150, 725 \times 0.02 = 3014.5 N$ .

Level 5 — Mastery

Recall Solution L5.1

(a) Same height ⇒ same pressure $p = F_{1} / A_{1} = 50/ (1 \times 1 0^{- 4}) = 500, 000 Pa$ everywhere. $F_{2} = p A_{2} = 500, 000 \times 4 \times 1 0^{- 4} = 200 N, F_{3} = p A_{3} = 500, 000 \times 9 \times 1 0^{- 4} = 450 N .$ (b) Fluid is incompressible: the volume the small piston pushes in equals the total volume the others take up: $A_{1} d_{1} = A_{2} d_{2} + A_{3} d_{3} .$ Divide by $A_{1} = 1 cm^{2}$ (so ratios $A_{2} / A_{1} = 4$ , $A_{3} / A_{1} = 9$ ): $d_{1} = \frac{A _{2}}{A _{1}} d_{2} + \frac{A _{3}}{A _{1}} d_{3} = 4 d_{2} + 9 d_{3} . ■$ (c) Real fluids and seals cause friction, and any trapped air is compressible, so some input work is lost as heat / absorbed by compression — the output forces fall a little short and transmission isn't perfectly instant. (See the "any fluid" mistake in the parent note and Conservation of energy in machines.)

Recall One-line self-test

Force question ::: use $F_{2} = F_{1} A_{2} / A_{1}$ (pressures equal at equal height). Distance question ::: use $A_{1} d_{1} = A_{2} d_{2}$ (volume conserved). Different heights ::: add $ρ g h$ for the column between the pistons. Does it ever give free energy? ::: No — $F_{1} d_{1} = F_{2} d_{2}$ always.

Connections

Parent: Pascal's law — the theory these drills test.
Pressure — force per unit area — the $p = F / A$ used throughout.
Incompressibility & continuity equation — the $A_{1} d_{1} = A_{2} d_{2}$ volume rule.
Hydrostatic pressure — p = p0 + ρgh — the $ρ g h$ term in L3.2 and L4.2.
Hydraulic systems — brakes, lifts, jacks — L2.3 brake and L4.1 lift.
Conservation of energy in machines — the "no free lunch" checks in L3.1, L4.1, L5.1.