FORCE 2 -
FORCE 1 -
10 in.2 = 10 in.3). Since liquid is practically
incompressible, it must go somewhere. What happens
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
= 2 in.3). To accommodate these 2 cubic inches of
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
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
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
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
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
indicated in pounds per square inch, or pressure.
Volume and Distance Factors
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.
To find pressure, use
Assume that the increase in pressure forces the
input piston downward 1 inch. This results in the
To find area, use
displacement of 10 cubic inches (in.3) of fluid (1 in. ×