{"id":413,"date":"2026-07-04T22:18:14","date_gmt":"2026-07-04T22:18:14","guid":{"rendered":"https:\/\/physicsfundamentalsinfo.com\/blog\/?p=413"},"modified":"2026-08-24T13:04:15","modified_gmt":"2026-08-24T13:04:15","slug":"pascals-law-hydraulics","status":"publish","type":"post","link":"https:\/\/physicsfundamentalsinfo.com\/blog\/fluids\/pascals-law-hydraulics\/","title":{"rendered":"Pascal’s Law & Hydraulics"},"content":{"rendered":"\n
Definition<\/div>

\nPascal’s law states that a pressure change applied to a confined, incompressible fluid is transmitted undiminished to every point in the fluid and to the walls of its container. Because pressure equals force divided by area (P1<\/sub> = P2<\/sub>, so F1<\/sub>\/A1<\/sub> = F2<\/sub>\/A2<\/sub>), a small force on a small piston can produce a much larger force on a larger piston.\n<\/p><\/div>\n\n

Watch a mechanic slide a trolley jack under a 1,500 kg car, pump the handle a few times with one hand, and lift the whole thing clear of the ground. No motor, no gears \u2014 just a sealed body of oil and two pistons of different sizes.<\/p>\n\n

That everyday miracle runs on Pascal’s law. The same trick stops your car when you brush the brake pedal, swings the arm of a 20-tonne excavator, and tips the dentist’s chair back \u2014 all from a 17th-century insight into how liquids pass on a push.<\/p>\n\n

What Is Pascal’s Law?<\/h2>\n\n

Take a plastic bottle filled to the brim with water, seal it, and squeeze it in the middle. Prick a pinhole anywhere \u2014 top, bottom, the far side \u2014 and water jets out with the same urgency. Your squeeze did not stay where your fingers were; the whole body of water felt it at once.<\/p>\n\n

Made precise, that observation is Pascal’s law: a change in pressure applied to an enclosed, incompressible fluid at rest is transmitted undiminished to every point in the fluid and to the walls of its container.<\/strong> The transmitted pressure pushes at right angles to every surface it meets. You will also see it called Pascal’s principle, or the principle of transmission of fluid-pressure<\/a>.<\/p>\n\n

Two words in that statement do the heavy lifting. Enclosed<\/em>: the fluid must be confined, or the pressure simply escapes. Incompressible<\/em>: liquids barely squash, so none of the push is soaked up along the way.<\/p>\n\n

Who discovered Pascal’s law?<\/h3>\n\n

Blaise Pascal (1623\u20131662), the French mathematician and physicist, established the law in 1653 in his treatise on the equilibrium of liquids, published in 1663 after his death. The SI unit of pressure \u2014 the pascal (Pa), one newton per square metre \u2014 is named in his honour.<\/p>\n\n

The story goes that in 1646 Pascal burst a sturdy wooden barrel by pouring water into a long, thin vertical tube fixed to its lid. A modest amount of water, raised high, created enough pressure to split the staves \u2014 an early hint that pressure depends on far more than the sheer quantity of liquid.<\/p>\n\n

\n\n \"Portrait\n\n
Blaise Pascal (1623\u20131662) formulated the law of pressure transmission that now bears his name.<\/figcaption>\n\n<\/figure>\n\n

The Pascal’s Law Formula<\/h2>\n\n

Everything starts from the definition of pressure \u2014 force spread over area:<\/p>\n\n

P = F \/ A<\/div>\n\n

In a confined fluid, Pascal’s law makes the transmitted pressure the same at the input piston and the output piston. Setting those two pressures equal gives the working equation of every hydraulic machine:<\/p>\n\n

F1<\/sub> \/ A1<\/sub> = F2<\/sub> \/ A2<\/sub><\/div>\n\n

Rearranged for the output force:<\/p>\n\n

F2<\/sub> = F1<\/sub> \u00d7 (A2<\/sub> \/ A1<\/sub>)<\/div>\n\n

One more relation completes the set. The liquid’s volume is fixed, so whatever volume the small piston sweeps out, the large piston must receive:<\/p>\n\n

A1<\/sub> \u00d7 d1<\/sub> = A2<\/sub> \u00d7 d2<\/sub><\/div>\n\n
\n\n\n\n\n
Symbol<\/th>\nMeaning<\/th>\nSI unit<\/th>\n<\/tr>\n<\/thead>\n
P<\/td>Pressure in the fluid<\/td>pascal (Pa = N\/m2<\/sup>)<\/td><\/tr>\n
F1<\/sub><\/td>Input force on the small piston<\/td>newton (N)<\/td><\/tr>\n
A1<\/sub><\/td>Area of the small piston<\/td>square metre (m2<\/sup>)<\/td><\/tr>\n
F2<\/sub><\/td>Output force on the large piston<\/td>newton (N)<\/td><\/tr>\n
A2<\/sub><\/td>Area of the large piston<\/td>square metre (m2<\/sup>)<\/td><\/tr>\n
d1<\/sub>, d2<\/sub><\/td>Distances the pistons move<\/td>metre (m)<\/td><\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n\n

Want to skip the algebra? Plug your numbers straight into our Pascal’s Law Calculator<\/a> \u2014 it solves for any variable and shows every step.<\/p> Area grows with the square<\/em> of diameter (A = \u03c0d2<\/sup>\/4), so a piston 5 times wider has 25 times the area. Forget to square it and every answer comes out 5 times too small.\n\n

How Pascal’s Law Works<\/h2>\n\n

Why should a push travel so faithfully? Picture the liquid as a crowd of molecules packed shoulder to shoulder. Lean on one edge of the crowd and, because nobody can shrink or step aside, the shove passes from neighbour to neighbour until every wall feels it.<\/p>\n\n

Liquids really are that stiff. Squeezing water down by just 1% takes roughly 22 MPa \u2014 about 220 times atmospheric pressure. For any hydraulic machine you will ever meet, the volume simply does not change, so none of your push is wasted compressing the fluid.<\/p>\n\n

Pressure has no favourite direction, either. At any point it pushes equally every way, and on every wall it pushes at right angles to the surface \u2014 the fluid shoving back on whatever contains it, exactly as Newton’s laws of motion<\/a> demand. The change spreads at the speed of sound in the liquid \u2014 around 1.4\u20131.5 km\/s in water and typical hydraulic oils \u2014 which is why hydraulic controls feel instant.<\/p>\n\n

\"Pascal's<\/figure>\n

An applied pressure change \u0394P reaches every point of a confined fluid, pushing at right angles to every wall.<\/p>\n\n

Step by step inside a hydraulic system<\/h3>\n\n
    \n
  1. You push the small piston with force F1<\/sub> acting over area A1<\/sub>.<\/li>\n
  2. The pressure everywhere in the fluid rises by P = F1<\/sub> \/ A1<\/sub>.<\/li>\n
  3. That extra pressure arrives, undiminished, at the large piston.<\/li>\n
  4. Pressure times area gives the output force: F2<\/sub> = P \u00d7 A2<\/sub>.<\/li>\n
  5. Because A2<\/sub> is larger than A1<\/sub>, F2<\/sub> is larger than F1<\/sub> \u2014 in exact proportion.<\/li>\n<\/ol>\n\n

    Try it yourself below. Drag the force and piston sizes and watch the pressure, output force and distance trade-off respond.<\/p>\n\n

    <\/span>Pascal's Law Lab<\/span><\/div>
    \n