FORCE 2 -

FORCE 1 -

10 in.2 = 10 in.3). Since liquid is practically

200 LBS.

20 LBS.

2

incompressible, it must go somewhere. What happens

INPUT

OUTPUT

PISTON 1

PISTON 2

is that the displaced fluid moves the output piston

20 SQ. IN.

20 SQ. IN.

upward. Since the area of the output piston is also 10

square inches, the piston moves 1 inch upward to

accommodate the 10 cubic inches of fluid. The pistons

are of equal areas, so they will move equal distances,

PRESSURE 10 LBS.

PER SQ. INCH

though in opposite directions.

If you apply this reasoning to the system shown

in figure 8-3, and if piston 1 is pushed down 1 inch,

only 2 cubic inches of fluid is displaced (1 in. 2 in.2

1"

1/10"

= 2 in.3). To accommodate these 2 cubic inches of

ASf08003

fluid, piston 2 will have to move only one-tenth of an

inch, because its area is 10 times that of piston 1.

creates 10 psi (20 lb 2 in.2) in the fluid. Although

This leads to the second basic rule for two pistons in

2

this applied force is much smaller than the applied

the same fluid power system, which is the distances

force shown in figure 8-2, the pressure is the same

moved are inversely proportional to their areas.

because the force is concentrated on a relatively small

To understand how Pascal's law is applied to

area.

hydraulics, you must make a distinction between the

This pressure of 10 psi acts on all parts of the fluid

terms *force *and *pressure*. Force may be defined as the

container, including the bottom of piston 2. Therefore,

push or pull exerted against the total area of a particular

the upward force on piston 2 is 10 pounds for each of its

surface and is expressed in pounds. In figure 8-3, the

20 square inches of area, or 200 psi (10 lb 20 in.2). In

2

force exerted on piston 1 is 20 pounds. Pressure, on the

this case, the original force has been multiplied tenfold

other hand, is the amount of push or pull on a unit area

using the same pressure in the fluid as before. In any

of the surface acted upon. In hydraulics, the unit area

system with these dimensions, the ratio of output force

used is the square inch, and pressure is expressed in

to input force is always 10 to 1, regardless of the value

pounds per square inch (psi). In figure 8-3, 20 pounds

of the applied force. For example, if the applied force

is exerted on 2 square inches, so pressure is 10 psi (20

lb 2 in.2). It is important for you to remember that

of piston 1 is 50 pounds, the pressure in the system is

2

increased to 25 psi. This will support a resistant force

"pressure is the amount of force acting upon 1 square

of 500 lb on piston 2.

inch of area."

The system works the same way in reverse. In

In figure 8-3, the effort exerted on piston 1 and the

figure 8-3, consider piston 2 as the input and piston 1 as

work accomplished by piston 2 are indicated in

pounds, and they are both referred to as force. Since the

the output. In that case, the output force will be

confined liquid that transmits this force acts on all

one-tenth the input force. Sometimes such results are

sides of the container, the result of these forces is

desired.

indicated in pounds per square inch, or pressure.

To show the relationship among force, pressure,

and area, use the formula F = PA. F represents force (in

In the systems shown in views A and B of figure

pounds), P represents pressure (in pounds per square

8-2, the pistons have areas of 10 square inches. Since

inch), and A represents area (in square inches). When

the areas of the input and output pistons are equal, a

any two of these factors are known, you can use this

force of 100 pounds on the input piston will support a

formula to find the unknown.

resistant force of 100 pounds on the output piston. At

To find force, use

F = PA

this point the pressure of the fluid is 10 psi. A slight

force in excess of 100 pounds on the input piston,

however, will slightly increase the pressure of the fluid.

F

To find pressure, use

P=

This will work to overcome the resistance force.

A

Assume that the increase in pressure forces the

F

input piston downward 1 inch. This results in the

To find area, use

A=

displacement of 10 cubic inches (in.3) of fluid (1 in.

P

2

8-3