# 14.3 Pascal's principle and hydraulics  (Page 3/6)

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## Strategy

We are given the force ${F}_{1}$ applied to the pedal cylinder. The cross-sectional areas ${A}_{1}$ and ${A}_{2}$ can be calculated from their given diameters. Then we can use the following relationship to find the force ${F}_{2}$ :

$\frac{{F}_{1}}{{A}_{1}}=\frac{{F}_{2}}{{A}_{2}}.$

Manipulate this algebraically to get ${F}_{2}$ on one side and substitute known values.

## Solution

Pascal’s principle applied to hydraulic systems is given by $\frac{{F}_{1}}{{A}_{1}}=\frac{{F}_{2}}{{A}_{2}}:$

$\begin{array}{}\\ \hfill {F}_{2}& =\frac{{A}_{2}}{{A}_{1}}{F}_{1}=\frac{\pi {r}_{2}^{2}}{\pi {r}_{1}^{2}}{F}_{1}\hfill \\ & =\frac{{\left(1.25\phantom{\rule{0.2em}{0ex}}\text{cm}\right)}^{2}}{{\left(0.250\phantom{\rule{0.2em}{0ex}}\text{cm}\right)}^{2}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}500\phantom{\rule{0.2em}{0ex}}\text{N}=1.25\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{\text{10}}^{4}\phantom{\rule{0.2em}{0ex}}\text{N}\text{.}\hfill \end{array}$

## Significance

This value is the force exerted by each of the four wheel cylinders. Note that we can add as many wheel cylinders as we wish. If each has a 2.50-cm diameter, each will exert $1.25\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{4}\phantom{\rule{0.2em}{0ex}}\text{N}\text{.}$ A simple hydraulic system, as an example of a simple machine, can increase force but cannot do more work than is done on it. Work is force times distance moved, and the wheel cylinder moves through a smaller distance than the pedal cylinder. Furthermore, the more wheels added, the smaller the distance each one moves. Many hydraulic systems—such as power brakes and those in bulldozers—have a motorized pump that actually does most of the work in the system.

Check Your Understanding Would a hydraulic press still operate properly if a gas is used instead of a liquid?

Yes, it would still work, but since a gas is compressible, it would not operate as efficiently. When the force is applied, the gas would first compress and warm. Hence, the air in the brake lines must be bled out in order for the brakes to work properly.

## Summary

• Pressure is force per unit area.
• A change in pressure applied to an enclosed fluid is transmitted undiminished to all portions of the fluid and to the walls of its container.
• A hydraulic system is an enclosed fluid system used to exert forces.

## Conceptual questions

Suppose the master cylinder in a hydraulic system is at a greater height than the cylinder it is controlling. Explain how this will affect the force produced at the cylinder that is being controlled.

## Problems

How much pressure is transmitted in the hydraulic system considered in [link] ? Express your answer in atmospheres.

251 atm

What force must be exerted on the master cylinder of a hydraulic lift to support the weight of a 2000-kg car (a large car) resting on a second cylinder? The master cylinder has a 2.00-cm diameter and the second cylinder has a 24.0-cm diameter.

A host pours the remnants of several bottles of wine into a jug after a party. The host then inserts a cork with a 2.00-cm diameter into the bottle, placing it in direct contact with the wine. The host is amazed when the host pounds the cork into place and the bottom of the jug (with a 14.0-cm diameter) breaks away. Calculate the extra force exerted against the bottom if he pounded the cork with a 120-N force.

$5.76\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{3}\phantom{\rule{0.2em}{0ex}}\text{N}\phantom{\rule{0.2em}{0ex}}\text{extra force}$

A certain hydraulic system is designed to exert a force 100 times as large as the one put into it. (a) What must be the ratio of the area of the cylinder that is being controlled to the area of the master cylinder? (b) What must be the ratio of their diameters? (c) By what factor is the distance through which the output force moves reduced relative to the distance through which the input force moves? Assume no losses due to friction.

Verify that work input equals work output for a hydraulic system assuming no losses due to friction. Do this by showing that the distance the output force moves is reduced by the same factor that the output force is increased. Assume the volume of the fluid is constant. What effect would friction within the fluid and between components in the system have on the output force? How would this depend on whether or not the fluid is moving?

If the system is not moving, the friction would not play a role. With friction, we know there are losses, so that ${W}_{\text{o}}={W}_{\text{i}}-{W}_{\text{f}};$ therefore, the work output is less than the work input. In other words, to account for friction, you would need to push harder on the input piston than was calculated.

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