11.5

Pascal’s Principle

As we have seen, the pressure in a fluid increases with depth, due to the weight of the fluid above the point of interest. A completely enclosed fluid may be subjected to an additional pressure by the application of an external force. For example, Figure 11-15a shows two interconnected cylindrical chambers. The chambers have different diameters and, together with the connecting tube, are completely filled with a liquid. The larger chamber is sealed at the top with a cap, while the smaller one is fitted with a movable piston. We now follow the approach outlined earlier in Figure 11-5 and begin by asking what determines the pressure P1 at a point immediately beneath the piston. According to the definition of pressure, it is the magnitude F1 of the external force divided by the area A1 of the piston: . If it is necessary to know the pressure P2 at any deeper place in the liquid, we just add to the value of P1 the increment , which takes into account the depth h below the piston: . The important feature here is this: The pressure P1 adds to the pressure due to the depth of the liquid at any point, whether that point is in the smaller chamber, the connecting tube, or the larger chamber. Therefore, if the applied pressure P1 is increased or decreased, the pressure at any other point within the confined liquid changes correspondingly. This behavior is described by Pascal’s principle.

Figure 11-15 (a) An external force is applied to the piston on the left. As a result, a force is exerted on the cap on the chamber on the right. (b) The familiar hydraulic car lift.

| Pascal’s Principle
| Any change in the pressure applied to a completely enclosed fluid is transmitted undiminished to all parts of the fluid and the enclosing walls.

The usefulness of the arrangement in Figure 11-15a becomes apparent when we calculate the force F2 applied by the liquid to the cap on the right side. The area of the cap is A2 and the pressure there is P2. As long as the tops of the left and right chambers are at the same level, the pressure increment is zero, so that the relation becomes . Consequently, , and

| (11.5)

If area A2 is larger than area A1, a large force can be applied to the cap on the right chamber, starting with a smaller force on the left. Depending on the ratio of the areas A2/A1, the force can be large indeed, as in the familiar hydraulic car lift shown in part b. In this device the force is not applied to a cap that seals the larger chamber, but, rather, to a movable plunger that lifts a car. Examples 7 and 8 and 7 and 8 deal with a hydraulic car lift.

| Example 7 | | | A Car Lift

In the hydraulic car lift shown in Figure 11-15b, the input piston on the left has a radius of and a negligible weight. The output plunger on the right has a radius of . The combined weight of the car and the plunger is 20 500 N. Since the output force has a magnitude of , it supports the car. Suppose that the bottom surfaces of the piston and plunger are at the same level, so that in Figure 11-15b. What is the magnitude F1 of the input force that is needed so that ?
Reasoning When the bottom surfaces of the piston and plunger are at the same level, as in Figure 11-15a, Equation 11.5 applies, and we can use it to determine F1.
Solution According to Equation 11.5, we have

Using for the circular areas of the piston and plunger, we find that

Problem solving insight Note that the relation , which results from Pascal’s principle, applies only when the points 1 and 2 lie at the same depth () in the fluid.

A N A L Y Z I N G M U L T I P L E - C O N C E P T P R O B L E M S

| Example 8 | | | A Car Lift Revisited

The data are the same as in Example 7. Suppose now, however, that the bottom surfaces of the piston and plunger are at different levels, such that in Figure 11-15b. The car lift uses hydraulic oil that has a density of . What is the magnitude F1 of the input force that is now needed to produce an output force having a magnitude of ?

The physics of a hydraulic car lift.

Reasoning Since the bottom surfaces of the piston and plunger are at different levels, Equation 11.5 no longer applies. Our approach will be based on the definition of pressure (Equation 11.3) and the relation between pressure and depth given in Equation 11.4. It is via Equation 11.4 that we will take into account the fact that the bottom of the output plunger is below the bottom of the input piston.
Knowns and Unknowns We have the following data:

Description | Symbol | Value | Comment
Radius of input piston | r1 | 0.0120 m |
Radius of output piston | r2 | 0.150 m |
Magnitude of output force | F2 | 20 500 N |
Density of hydraulic oil |  |  |
Level difference between input | h | 1.10 m | See Figure 11-15b.
piston and output plunger |  |  |
Unknown Variable |  |  |
Magnitude of input force | F1 | ? |

Modeling the Problem

| Step 1 Force and Pressure
| The input force F1 is determined by the pressure P1 acting on the bottom surface of the input piston, which has an area A1. According to Equation 11.3, the input force is , where because the piston has a circular cross section with a radius r1. With this expression for the area A1, the input force can be written as in Equation 1 at the right. We have a value for r1, but P1 is unknown, so we turn to Step 2 in order to obtain it.

| (1)

| Step 2 Pressure and Depth in a Static Fluid
| In Figure 11-15b, the bottom surface of the plunger at point B is at the same level as point A, which is at a depth h beneath the input piston. Equation 11.4 applies, so that

This result for P1 can be substituted into Equation 1, as shown at the right. Now it is necessary to have a value for P2, the pressure at the bottom surface of the output plunger, which we obtain in Step 3.

| Step 3 Pressure and Force
| By using Equation 11.3, we can relate the unknown pressure P2 to the given output force F2 and the area A2 of the bottom surface of the output plunger: . The area A2 is a circle with a radius r2, so . Thus, the pressure P2 is

At the right, this result is substituted into Equation 2.

Solution Combining the results of each step algebraically, we find that
Thus, we find that the necessary input force is

The answer here is less than that in Example 7 because the weight of the 1.10-m column of hydraulic oil provides some of the input force to support the car and plunger.
Related Homework: Problem 33

In a device such as a hydraulic car lift, the same amount of work is done by both the input and output forces in the absence of friction. The larger output force moves through a smaller distance, while the smaller input force moves through a larger distance. The work, being the product of the magnitude of the force and the distance, is the same in either case since mechanical energy is conserved.
An enormous variety of clever devices use hydraulic fluids, just as the car lift does. In Figure 11-16 for instance, the fluids multiply a small input force into the large output force required to rotate a truck into a position where it can be easily emptied.

Figure 11-16 At this energy plant, trucks carrying wood waste are emptied with the aid of a system that uses a hydraulic fluid to generate a large output force, starting with a small input force. (Creatas/PictureQuest)

Copyright © 2007 John Wiley & Sons, Inc. All rights reserved.