# Pascal's Law and Hydraulic Force

## Pascal's laws relates to pressures in fluids

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)*

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