P = F , or P = 20,000 lb = 2,000 psi

You might compare this to the way you manipulate

10 in.2

A

Ohm's law to find voltage, current, and resistance.

Now, we use this formula to confirm the size of the

To generate the 2,000 psi needed to balance the

pressure developed by piston 1 in the system shown in

10-ton load, a force of 500 pounds must be applied to

figure 8-3.

the top of piston 1.

To find pressure, insert the known values of force

2

(20 lb) and area (2 in.2) into the formula F = PA.

= 500 lb

F = PA, or F = 2,000 psi × 0.25 in.

2

Transposing the formula, we have:

The 500 pounds of force applied to piston 1 will

equal the resistant pressure of 2,000 psi and support or

P = F , or P = 20 lb = 10 psi

2 in.2

A

balance the load. To lift the load, however, requires

s o m e t h i n g i n ex c e s s o f 5 0 0 p o u n d s o f f o r c e .

This pressure is applied to the confined liquid

Application of the force can be mechanical, such as by

and, in turn, to all sides of the container, including

the lever shown in figure 8-4, or by motor, or by any

the 20-square-inch area of piston 2. Note that the

other means that will generate the force required by the

upward force exerted by the liquid on piston 2 (F =

10 psi × 20 in.2 = 200 lb) is the same as the downward

system. For complete and detailed descriptions of the

2

force (200 lb) exerted by the piston. Thus, the

c l a s s e s o f l eve r s , c o n s u l t *B a s i c M a ch i n e s *,

20-pound force applied to piston 1 supports a force

NAVEDTRA 14037.

of 200 pounds acting against piston 2.

You should remember that the force exerted on

piston 1 (20 lb) only creates a flow of liquid. and that

the resistance force (200 lb) is required to create

pressure. With no resistance against piston 2, any force

exerted on piston 1 will develop only that pressure

resulting from the friction of liquid flow and the weight

of piston 2.

For a more realistic application of multiplication

of forces, consider the 10-ton hydraulic jack system

shown in figure 8-4. Assume that piston 2 is the

lifting piston, with an area of 10 square inches, and

that piston 1 is the pump, with an area of 0.25 square

inch. Next, assume that a 10-ton (20,000 pounds)

load is placed on piston 2. To support this force, a

pressure of 2,000 psi must be exerted against the

bottom of piston 2.

HANDLE

PUMP

10-TON LOAD

500 LB

PISTON 2

(20,000 LB)

PISTON 1

10 SQ. IN.

0.25 SQ. IN.

CYLINDER 2

CYLINDER 1

ASf08004

8-4