# 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 m ^{ 2 } * can be calculated

* F = p A (1) *

* where *

* F = force (N) *

* p = pressure (Pa, N/m ^{ 2 } ) *

* A = area (m ^{ 2 } ) *

or with values

* F = (2000 Pa) (0.1 m ^{ 2 } ) *

* = 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 (m ^{ 2 } ) *

* p = pressure in small and large cylinder (Pa, N/m ^{ 2 } ) *

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 (m ^{ 2 } ) *

* p = pressure in small and large cylinder (Pa, N/m ^{ 2 } ) *

* (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.

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### 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/s ^{ 2 } ) *

or

* F _{ l } = 1/2 (2000 kg) (9.81 m/s ^{ 2 } ) *

* = 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) *

## Related Topics

### • Fluid Mechanics

The study of fluids - liquids and gases. Involving velocity, pressure, density and temperature as functions of space and time.

### • Gases and Compressed Air

Air, LNG, LPG and other common gas properties, pipeline capacities, sizing of relief valves.

## Related Documents

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Calculate hydraulic cylinder force.

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Introduction to pressure - online pressure units converter.