Hydraulic Force vs. Pascal's Law

Pascal's Laws relates to pressures in incompressible fluids - liquids.

• if the weight of a fluid is neglected the pressure throughout an enclosed volume will be the same
• the static pressure in a fluid acts equally in all directions
• the static pressure acts at right angles to any surface in contact with the fluid

Example - Pressure in a Hydraulic Cylinder

The pressure of 2000 Pa in an hydraulic cylinder acts equally on all surfaces. The force on a piston with area 0.1 m² can be calculated

F = p A                              (1)

where

F = force (N)

p = pressure (Pa, N/m²)

A = area (m²)

or with values

F = (2000 Pa) (0.1 m²)

= 200 (N)

Example - Force in a Hydraulic Jack

The pressure acting on both pistons in a hydraulic jack is equal.

The force equation for the small cylinder:

F _s = p A _s (2)

where

F _s = force acting on the piston in the small cylinder (N)

A _s = area of small cylinder (m²)

p = pressure in small and large cylinder  (Pa, N/m²)

The force equation for the large cylinder:

F _l = p A _l (2b)

where

F _l = force acting on the piston in the large cylinder (N)

A _l = area of large cylinder (m²)

p = pressure in small and large cylinder (Pa, N/m²)

(2) and (2b) can be combined to

F _s / A _s = F _l / A _l (2c)

or

F _s = F _l A _s / A _l (2d)

The equation indicates that the effort force required in the small cylinder to lift a load on the large cylinder depends on the area ratio between the small and the large cylinder - the effort force can be reduced by reducing the small cylinder area compared to the large cylinder area.

Related Mobile Apps from The Engineering ToolBox

• Hydraulic Force Calculator App

- free apps for offline use on mobile devices.

A Hydraulic Jack Lifting a Car

The back end (half the weight) of a car of mass 2000 kg is lifted by an hydraulic jack where the A _s / A _l ratio is 0.1 (the area of the large cylinder is 10 times the area of the small cylinder).

The force - weight - acting on the large cylinder can be calculated with Newton's Second Law:

F _l = m a

where

m = mass (kg)

a = acceleration of gravity (m/s²)

or

F _l = 1/2 (2000 kg) (9.81 m/s²)

= 9810 (N)

The force acting on the small cylinder in the jack can be calculated with (2d)

F _s = (9810 N) 0.1

= 981 (N)