Pascal's law transmits pressure equally through a confined fluid, letting a hydraulic press multiply force: F2 = F1·(d2/d1)². Drag the sliders below to change the input force and the two piston diameters, and watch the fluid pressure and output force respond in real time.
This hydraulic press hands you a bargain with fine print: a modest shove on a narrow piston becomes an enormous lifting force on a wide one, yet the multiplier is fixed entirely by the geometry you dial in, and not one joule of energy is invented along the way. Set Input force F1, then size the two pistons with the Small piston diameter (d1) and Large piston diameter (d2) sliders and watch the readouts reprice that push in real time.
Behind the scene is Pascal's principle: pressure applied to a confined, incompressible fluid is transmitted undiminished, so the fluid pressure P is identical at both pistons, P = F1/A1 = F2/A2. The areas obey A = π·(d/2)², which is why the output force F2 follows F2 = F1·(A2/A1) = F1·(d2/d1)². Because area scales with diameter squared, the force climbs with the diameter ratio squared — make d2 five times d1 and F2 leaps to twenty-five times F1, enough to hoist the mass shown by m = F2/g with g = 9.81.
The catch lives in that same geometry. Equal pressure on unequal areas produces unequal forces, but the incompressible fluid enforces A1·s1 = A2·s2, so the wide piston creeps up as little as its force is large. You buy force by spending distance, exactly the trade a lever makes — work in stays roughly work out. Put numbers to the exchange with the Pascal's law calculator, or take the controls of another machine in our library of physics simulations.
Pascal's law says that pressure applied to a confined, incompressible fluid is transmitted equally throughout it. So the pressure is the same on both pistons of a hydraulic press: P = F1/A1 = F2/A2.
The output force is F2 = F1·(A2/A1) = F1·(d2/d1)². Because a piston's area grows as the square of its diameter, a large piston five times the diameter of the small one gives twenty-five times the force.
No. The large piston moves proportionally less than the small one (A1·s1 = A2·s2), so the work put in is about equal to the work got out — you trade distance for force, exactly like a lever.
The pressure is the same everywhere in the fluid; it is the different piston areas that turn that shared pressure into different forces, since force = pressure × area. A bigger piston, same pressure, means a bigger force.