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12.3: Pascal’s Law

Difficulty Level: At Grade Created by: CK-12
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The student will:

  • Understand and be able to solve problems using Pascal’s Principle.


  • Pascal’s Law: Increasing the pressure of fluid anywhere in a system increases the pressure everywhere in the system.


Pressure is defined as force divided by area, but this does not explain in itself how pressure transfers. Blaise Pascal, after whom the metric unit of pressure is named, also clarified a useful physical principle that is now named after him. Pascal's Law states that any confined incompressible fluid under pressure will transmit pressure equally throughout the system.  In other words, increasing the pressure of the fluid anywhere increases the pressure everywhere. 

If you have ever been to an auto repair shop, you’ve probably seen cars raised high enough above the ground so that the mechanics can perform their repairs. The device that raises the car is called a hydraulic lift (Figure below).  A hydraulic lift can create a very large force to lift the car with only a small force. 

How does it work?  The diagram in the Figure above shows a cross-section of a hydraulic lift with square pistons. The pressure applied at the narrow piston of surface \begin{align*}A\end{align*}A is transmitted to the bottom of the wide piston of surface area \begin{align*}A' = 9A\end{align*}A=9A. If, for example, the pressure is \begin{align*}P\end{align*}P at the narrow piston is \begin{align*}1000 ~ \frac{\text{N}}{\text{m}^2}\end{align*}1000 Nm2, the bottom of the wide piston will have the same pressure applied to it, but over nine times the area. The total force on the wide piston will be nine times greater, for a total of \begin{align*}F' = PA' = P (9A) = 9000 ~ \text{N}\end{align*}F=PA=P(9A)=9000 N.

The force is multiplied in proportion to the ratio \begin{align*}\frac{A_{wide}}{A_{narrow}}\end{align*}AwideAnarrow.

More formally, we can write \begin{align*}P_{narrow} = P_{wide} \rightarrow \frac{F_{narrow}}{A_{narrow}} = \frac{F_{wide}}{A_{wide}} \rightarrow \frac{A_{wide}}{A_{narrow}} = \frac{F_{wide}}{F_{narrow}}\end{align*}Pnarrow=PwideFnarrowAnarrow=FwideAwideAwideAnarrow=FwideFnarrow; it is often the case that “narrow” is replaced with “in” and “wide” is replaced with “out.”

It may seem that we “are getting more out of the system than we’re putting into it.” After all, in the example above we only needed to input 1,000 N in order to output 9,000 N.

As you may suspect, this is not the case.

If we wish to raise the car, work must be done. And energy conservation tells us that we can never get more energy out of a system than we put into a system. In fact, because of friction, we always need to put more energy in than we get out.

If we wish to raise a 9,000 Newton car by 10 centimeters, it requires 900 joules of energy.

\begin{align*}W = Fx = (9,000 ~ \text{N})(0.10 ~ \text{m}) = 900 ~ \text{J}\end{align*}W=Fx=(9,000 N)(0.10 m)=900 J

Therefore, we will have to move the narrow piston nine times farther than the distance that the wide piston rises. By the Law of Energy Conservation (assuming no energy is transformed into heat by friction):

\begin{align*}E_i = E_f ~~ \rightarrow ~~ & W_{input} = W_{output} \\ & (1,000 ~ \text{N})(0.90 ~ \text{m}) = (9,000 ~ \text{N})(0.10 ~ \text{m})\end{align*}Ei=Ef    Winput=Woutput(1,000 N)(0.90 m)=(9,000 N)(0.10 m)

Interactive Simulation

Check Your Understanding

The input force of a hydraulic lift is 800 Newtons and the output force is 16,000 Newtons. What is the ratio \begin{align*}\frac{A_{out}}{A_{in}}\end{align*}AoutAin?

Answer: \begin{align*}\frac{A_{out}}{A_{in}} = \frac{F_{out}}{F_{in}} = \frac{16,000 ~ \text{N}}{800 ~ \text{N}} = 20\end{align*}AoutAin=FoutFin=16,000 N800 N=20.

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Date Created:
Mar 11, 2013
Last Modified:
Sep 20, 2016
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