Hydraulic Force vs. Pascal's Law
Pascal's law and the hydraulic force acting 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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