PASCAL’S LAW FOR INCOMPRESSIBLE LIQUIDS
 

Imagine a watertight, rigid container. Suppose that there are two pipes of unequal diameters running upward out of this container. Imagine that you fill the container with an incompressible liquid such as water so that the container is completely full and the water rises partway up into the pipes. Suppose that you place pistons in the pipes so that they make perfect water seals, and then you leave the pistons to rest on the water surface (Fig. 1).
Because the pipes have unequal diameters, the surface areas of the pistons are different. One of the pistons has area A₁ (in meters squared), and the other has area A₂. Suppose that you push downward on piston number 1 (the one whose area is A₁) with a force F₁ (in newtons). How much upward force F₂ is produced at piston number 2 (the one whose area is A₂)? Pascal’s law provides the answer: The forces are directly proportional to the areas of the piston faces in terms of their contact with the liquid. In the example shown by Fig. 1, piston number 2 is smaller than piston number 1, so the force F₂ is proportionately less than the force F₁. Mathematically, the following equations both hold:

Fig. 1. Pascal’s law for confined, incompressible liquids. The forces are directly proportional to the areas of the pistons.

F₁/F₂ = A₁/A₂
A₁F₂ = A₂F₁
When using either of these equations, we must be consistent with units throughout the calculations. In addition, the top equation is meaningful only as long as the force exerted is nonzero.