In Pressure and Depth you found $P = P_0 + \rho g h$ at every point of a still liquid. Now press on the liquid's surface with a piston, raising $P_0$. What happens to the pressure down below?
Why it works
Every point's pressure is $P_0$ plus its own $\rho g h$. Pushing on the piston raises $P_0$ by some amount $\Delta P$, and a liquid barely compresses, so the depths and the $\rho g h$ terms stay the same. Then every point's pressure rises by exactly $\Delta P$. That is Pascal's principle: a pressure change in an enclosed fluid reaches every part of it, undiminished.
Key idea
In a hydraulic system the pressure is the same on both pistons: $\dfrac{F_1}{A_1} = \dfrac{F_2}{A_2}$, so $F_2 = F_1 \dfrac{A_2}{A_1}$. The liquid's volume is fixed, so $A_1 d_1 = A_2 d_2$: the big piston moves a shorter distance. Work in equals work out, $F_1 d_1 = F_2 d_2$.
Worked examples
Worked example: A car lift
A car lift's large piston has an area of $0.50\,\text{m}^2$ and its small piston $0.010\,\text{m}^2$. What force on the small piston holds up a $1500\,\text{kg}$ car?
- The large piston must push up with the car's weight: $F_2 = mg = (1500\,\text{kg})(9.8\,\text{m/s}^2) = 14{,}700\,\text{N}$.
- Equal pressures: $F_1 = F_2 \dfrac{A_1}{A_2} = (14{,}700\,\text{N}) \dfrac{0.010\,\text{m}^2}{0.50\,\text{m}^2}$.
- $F_1 = 294\,\text{N}$, about the weight of a $30\,\text{kg}$ child.
Answer: $294\,\text{N}$
Worked example: Where the work goes
In the same lift you push the small piston down $25\,\text{cm}$. How far does the car rise, and how do the two works compare?
- Volume moved: $A_1 d_1 = A_2 d_2$, so $d_2 = d_1 \dfrac{A_1}{A_2} = (25\,\text{cm})\dfrac{0.010}{0.50} = 0.50\,\text{cm}$.
- Work in: $F_1 d_1 = (294\,\text{N})(0.25\,\text{m}) = 73.5\,\text{J}$.
- Work out: $F_2 d_2 = (14{,}700\,\text{N})(0.0050\,\text{m}) = 73.5\,\text{J}$. Equal, as work and energy conservation demand.
Answer: $0.50\,\text{cm}$; both works are $73.5\,\text{J}$
Worked example: Round pistons
A press has round pistons with diameters $2.0\,\text{cm}$ and $10\,\text{cm}$. By what factor does it multiply force?
- The force grows by the area ratio, and a circle's area is $\pi d^2/4$.
- $\dfrac{A_2}{A_1} = \left(\dfrac{d_2}{d_1}\right)^2 = \left(\dfrac{10}{2.0}\right)^2 = 25$.
- Five times the diameter gives twenty-five times the force.
Answer: $25$
Common mistake
Thinking a hydraulic lift gives you energy for free. It multiplies force, not energy: to lift the car $1\,\text{cm}$ with $\tfrac{1}{50}$ of its weight, you must push $50\,\text{cm}$. Check any machine by comparing $F d$ on the way in and on the way out; without friction they are equal, and with friction the output is less.
Car brakes, excavator arms and barber chairs all use Pascal's principle: a liquid in a closed line carries a pressure change from where it is made to where it is needed.