# Pascal's Laws

## Pascal's laws relates to pressures in fluids

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Pascal's Laws relates to pressures in fluids - in liquid or gaseous state:

- 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 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 as

*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*

*p = pressure in small and large cylinder*

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*

*p = pressure in small and large cylinder*

(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

- 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/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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