Pascal's principle applies to fluids. It says

This stems from the fact that liquids, unlike gases, are virtually incompressible. The atoms and molecules of liquids are very close together, and compression would entail strong electron-cloud repulsion between them. That's the same kind of repulsion that keeps your dinner plate on top of the table rather than falling right through it, even though solid materials are mostly the empty space between electron clouds and nuclei.

One way to look at Pascal's principle is shown in this diagram. A fluid is enclosed in a U-shaped tube with a moveable piston in either end. If a force, F, is applied to the left side, that force will be transmitted through the fluid to the piston on the other side. We could measure the upward force on that piston, and it would equal the downward force on the left-side piston.

In this figure, we've set the force to 1 Newton, and the cross-sectional area of the tube to 1 m^{2} for convenience. The resulting pressure is 1 Pa (Pascal).

Now imagine that we change the geometry of the tube & system just a bit, keeping Pascal's principle in mind. In the figure below, the area of the left-side tube is 1, but the area of the right side tube is a factor of 100 larger.

If pressure is Force divided by Area, $P = F/A,$ then we can rearrange to calculate the force:

$$P = \frac{F}{A} \: \color{#E90F89}{\longrightarrow} \: F = A \cdot P$$

Now if we take the pressure to be 1 Pa, the force applied to the right-side piston is

$$F = 1 \, Pa \cdot 100 \, m^2 = 100 \, N.$$

Notice that by changing the tube size, we created a force multiplier. We don't get something for nothing, though, because we have to push a lot of fluid from the narrow tube into the wider tube in order to get its piston to rise by a given amount. This is completely analogous to the ideas behind simple machines like the inclined plane or the lever.

Pressure applied to any part of an enclosed liquid will be applied equally in any direction in the fluid.

Changes in cross-sectional area of a fluid reservoir or conduit can act as a force multiplier, like a simple machine.

The relationship between the forces applied to apertures of area A_{1} and A_{2} is

$$\frac{F_1}{A_1} = \frac{F_2}{A_2}$$

A simple hydraulic jack, like a car jack, is a good example of how Pascal's law is put into practice. Using such jacks, one can even lift *buildings*.

The hydraulic jack consists of a lever that pushes a piston into a narrow tube of some fluid, usually an oil. The lever allows a human to apply a lot of force to the fluid reservoir. The size of the fluid tube is expanded below a moveable piston that can be pushed upward to raise a mass. A reservoir of low-pressure oil and a couple of valves complete the jack so that it can be extended its full length, and sometimes further if the piston can telescope.

To full the pump, the handle is raised. The high-pressure (lower) valve shuts off, maintaining the fluid pressure under the lifting piston. The filling valve opens, as it allows fluid movement from right to left (in the figure) only. Fluid from the low-pressure reservoir enters the pump and is ready to be pushed into the high-pressure side.

When the lever is pushed downward, the one-way filling valve closes, and the lower valve (which allows flow only in the right → left direction (in this figure). Because the diameter of the pumping piston is smaller than the diameter of the lifting piston, the force is multiplied and the rising jack can lift large masses.

A 2000 Kg car is to be lifted by a hydraulic jack. The diameter of the lifting piston is 5.5 cm and the diameter of the pumping piston is 1 cm. If 100 N of force is applied to the pumping piston, how much force will be available to lift the car?

**Solution**

$$ \begin{align} A_1 &= \pi(0.5)^2 = 0.785 \, cm^2. \; \text{and}\\[5pt] A_2 &= \pi(2.25)^2 = 15.9 \, cm^2 \end{align}$$

The force to hold the piston in place is given by

$$\frac{F_1}{A_1} = \frac{F_2}{A_2} \: \color{#E90F89}{\longrightarrow} \: F_2 = \frac{A_2 F_1}{A_1}$$

So

$$ \begin{align} F_2 &= \frac{(15.9 \, cm^2)(100 \, N)}{0.785 \, cm^2} \\[5pt] &= 2025 \, N \end{align}$$

So the jack should provide plenty of force to lift the car, particularly when we consider that we generally only have to *tip* the car a bit to change a tire.

This photo shows a part of the arm of an excavator or backhoe, a machine for digging. Most of the motion of machines like these are acheived using hydraulic cylinders and pistons. The arm shown has to bend at the hinge at the top. Notice the black lines, through which hydraulic fluid is pumped. These are flexible and can bend with the bending of the arm. Because the force mechanism is fluid, we can use lines (tubes) like this to direct force in any direction we want.

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