The Liquid Phase Help (page 4)

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By — McGraw-Hill Professional
Updated on Sep 5, 2011

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. 10-4).

Basic States of Matter The Liquid Phase Pascal’s Law For Incompressible Liquids

Fig. 10-4 . Pascal’s law for confined, incompressible liquids. The forces are directly proportional to the areas of the pistons.

Because the pipes have unequal diameters, the surface areas of the pistons are different. One of the pistons has area A 1 (in meters squared), and the other has area A 2 . Suppose that you push downward on piston number 1 (the one whose area is A1) with a force F 1 (in newtons). How much upward force F 2 is produced at piston number 2 (the one whose area is A 2 )? 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. 10-4, piston number 2 is smaller than piston number 1, so the force F 2 is proportionately less than the force F 1 . Mathematically, the following equations both hold:

F 1 / F 2 = A 1 / A 2

A 1 / F 2 = A 2 / F 1

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.

Pascal's Law Practice Problem


Suppose that the areas of the pistons shown in Fig. 10-4 are A 1 = 12.00 cm 2 and A 2 = 15.00 cm 2 . (This does not seem to agree with the illustration, where piston number 2 looks smaller than piston number 1, but forget about that while we solve this problem.) If you press down on piston number 1 with a force of 10.00 N, how much upward force will result at piston number 2?


At first, you might think that we have to convert the areas of the pistons to meters squared in order to solve this problem. In this case, however, it is sufficient to find the ratio of the areas of the pistons because both areas are given to us in the same units:

A 1 / A 2 = 12.00 cm 2 /15.00 cm 2

= 0.8000

Thus we know that F 1 / F 2 = 0.8000. We are given F 1 = 10.00 N, so it is easy to solve for F 2 :

10.00/ F 2 = 0.8000

1/ F 2 = 0.08000

F 2 = 1/0.08000 = 12.50 N

We are entitled to four significant figures throughout this calculation because all the input data were provided to this degree of precision.

Practice problems of these concepts at: Basic States Of Matter Practice Test

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