13.11 - Pascal’s principle

Pascal's principle: Pressure in a confined fluid

is transmitted unchanged to all parts of the fluid

and to the containing walls.

If you jump down on one side of a waterbed, a person sitting on the other side will get a

jolt up. This illustrates Pascal’s principle: A variation in pressure in the enclosed fluid is

being transmitted unchanged throughout the fluid.

In the illustration to the right, you see a five-kilogram mass “balancing” a 50-kilogram

mass. How does something that seems so counterintuitive í a small mass balancing a

large mass í occur?

First, Pascal’s principle asserts that the pressure exerted by the weight of the first

mass on the fluid is transmitted to the second mass unchanged. The two masses

“balance” because the surface area of the fluid under the 50-kilogram mass is 10 times

larger than the surface area under the five-kilogram mass. The pressure is the same

under both masses, but since the surface area is 10 times larger under the more

massive object, the upward force, the pressure times the area, is 10 times greater than

it is under the less massive object, so the system is in equilibrium. Although we focus

on the pressures supporting the weights, according to Pascal’s principle the pressure

acts in all directions at every point in the fluid, so it presses on the walls of the hydraulic

system as well.

In some ways, the equilibrium in such a system is similar to a small mass located at the

end of a long lever arm balancing a large mass located close to the lever’s pivot point.

In fact, this similarity has caused systems like those shown to the right to be called

hydraulic levers. You may be familiar with them from the “racks” in car repair shops:

they are the gleaming metal pistons that lift cars.

Illustrations of hydraulic levers are shown to the right. In each case the pressure on

both pistons must be the same in order to conform to Pascal’s principle, but the force

will depend on the areas of the pistons. In the example problem you are asked to

calculate the downward force needed on the left piston to lift the small automobile on

the right. It turns out that this force is about 20 newtons (only 4½ pounds)!

Do you get “something for nothing” when you raise a heavy car with such a small force?

No, the amount of work you do equals the work the piston on the right does on the car.

You apply less force, but through a greater displacement. If you press your piston down

a meter in the scenario shown in Example 1, the car will rise only about 1.6 millimeters,

as you can verify by considering the incompressible volume of water displaced on each

side.

Pascal’s principle

An enclosed fluid transmits pressure:

·unchanged, and

·in all directions

P 1 = P 2

F 1 /A 1 = F 2 /A 2

P = pressure

F = force

A = surface area

How much force must be exerted

on the left-hand piston to lift the

automobile on the right?

P 1 = P 2

F 1 /A 1 = F 2 /A 2

F 1 = mgA 1 /A 2

(^260) Copyright 2000-2007 Kinetic Books Co. Chapter 13