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Relationship between forces in a hydraulic system

We can derive a relationship between the forces in the simple hydraulic system shown in [link] by applying Pascal’s principle. Note first that the two pistons in the system are at the same height, and so there will be no difference in pressure due to a difference in depth. Now the pressure due to F 1 size 12{F rSub { size 8{1} } } {} acting on area A 1 size 12{A rSub { size 8{1} } } {} is simply P 1 = F 1 A 1 size 12{P rSub { size 8{1} } = { {F rSub { size 8{1} } } over {A rSub { size 8{1} } } } } {} , as defined by P = F A size 12{P= { {F} over {A} } } {} . According to Pascal’s principle, this pressure is transmitted undiminished throughout the fluid and to all walls of the container. Thus, a pressure P 2 size 12{P rSub { size 8{2} } } {} is felt at the other piston that is equal to P 1 size 12{P rSub { size 8{1} } } {} . That is P 1 = P 2 size 12{P rSub { size 8{1} } =P rSub { size 8{2} } } {} .

But since P 2 = F 2 A 2 size 12{P rSub { size 8{2} } = { {F rSub { size 8{2} } } over {A rSub { size 8{2} } } } } {} , we see that F 1 A 1 = F 2 A 2 size 12{ { {F rSub { size 8{1} } } over {A rSub { size 8{1} } } } = { {F rSub { size 8{2} } } over {A rSub { size 8{2} } } } } {} .

This equation relates the ratios of force to area in any hydraulic system, providing the pistons are at the same vertical height and that friction in the system is negligible. Hydraulic systems can increase or decrease the force applied to them. To make the force larger, the pressure is applied to a larger area. For example, if a 100-N force is applied to the left cylinder in [link] and the right one has an area five times greater, then the force out is 500 N. Hydraulic systems are analogous to simple levers, but they have the advantage that pressure can be sent through tortuously curved lines to several places at once.

Calculating force of slave cylinders: pascal puts on the brakes

Consider the automobile hydraulic system shown in [link] .

When the driver applies force on the brake pedal the master cylinder transmits the same pressure to the slave cylinders but results in a larger force due to the larger area of the slave cylinders.
Hydraulic brakes use Pascal’s principle. The driver exerts a force of 100 N on the brake pedal. This force is increased by the simple lever and again by the hydraulic system. Each of the identical slave cylinders receives the same pressure and, therefore, creates the same force output F 2 size 12{F rSub { size 8{2} } } {} . The circular cross-sectional areas of the master and slave cylinders are represented by A 1 size 12{A rSub { size 8{1} } } {} and A 2 size 12{A rSub { size 8{2} } } {} , respectively

A force of 100 N is applied to the brake pedal, which acts on the cylinder—called the master—through a lever. A force of 500 N is exerted on the master cylinder. (The reader can verify that the force is 500 N using techniques of statics from Applications of Statics, Including Problem-Solving Strategies .) Pressure created in the master cylinder is transmitted to four so-called slave cylinders. The master cylinder has a diameter of 0.500 cm, and each slave cylinder has a diameter of 2.50 cm. Calculate the force F 2 size 12{F rSub { size 8{2} } } {} created at each of the slave cylinders.

Strategy

We are given the force F 1 size 12{F rSub { size 8{1} } } {} that is applied to the master cylinder. The cross-sectional areas A 1 size 12{A rSub { size 8{1} } } {} and A 2 size 12{A rSub { size 8{2} } } {} can be calculated from their given diameters. Then F 1 A 1 = F 2 A 2 size 12{ { {F rSub { size 8{1} } } over {A rSub { size 8{1} } } } = { {F rSub { size 8{2} } } over {A rSub { size 8{2} } } } } {} can be used to find the force F 2 size 12{F rSub { size 8{2} } } {} . Manipulate this algebraically to get F 2 size 12{F rSub { size 8{2} } } {} on one side and substitute known values:

Solution

Pascal’s principle applied to hydraulic systems is given by F 1 A 1 = F 2 A 2 size 12{ { {F rSub { size 8{1} } } over {A rSub { size 8{1} } } } = { {F rSub { size 8{2} } } over {A rSub { size 8{2} } } } } {} :

F 2 = A 2 A 1 F 1 = πr 2 2 πr 1 2 F 1 = 1.25 cm 2 0.250 cm 2 × 500 N = 1 . 25 × 10 4 N . size 12{F rSub { size 8{2} } = { {A rSub { size 8{2} } } over {A rSub { size 8{1} } } } F rSub { size 8{1} } = { {πr rSub { size 8{2} } rSup { size 8{2} } } over {πr rSub { size 8{1} } rSup { size 8{2} } } } F rSub { size 8{1} } = { { left (1 "." "25"`"cm" right ) rSup { size 8{2} } } over { left (0 "." "250"`"cm" right ) rSup { size 8{2} } } } times "500"`N=1 "." "25" times "10" rSup { size 8{4} } `N} {}

Discussion

This value is the force exerted by each of the four slave cylinders. Note that we can add as many slave cylinders as we wish. If each has a 2.50-cm diameter, each will exert 1 . 25 × 10 4 N . size 12{1 "." "25" times "10" rSup { size 8{4} } `N "." } {}

A simple hydraulic system, such as a simple machine, can increase force but cannot do more work than done on it. Work is force times distance moved, and the slave cylinder moves through a smaller distance than the master cylinder. Furthermore, the more slaves added, the smaller the distance each 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. The movement of the legs of a spider is achieved partly by hydraulics. Using hydraulics, a jumping spider can create a force that makes it capable of jumping 25 times its length!

Making connections: conservation of energy

Conservation of energy applied to a hydraulic system tells us that the system cannot do more work than is done on it. Work transfers energy, and so the work output cannot exceed the work input. Power brakes and other similar hydraulic systems use pumps to supply extra energy when needed.

Section 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 slave cylinder. Explain how this will affect the force produced at the slave cylinder.

Problems&Exercises

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

2.55 × 10 7 Pa ; or 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 the slave cylinder? The master cylinder has a 2.00-cm diameter and the slave has a 24.0-cm diameter.

A crass host pours the remnants of several bottles of wine into a jug after a party. He then inserts a cork with a 2.00-cm diameter into the bottle, placing it in direct contact with the wine. He is amazed when he 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 × 10 3 N size 12{5 "." "76" times "10" rSup { size 8{3} } `N} {} 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 slave cylinder 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 to friction.

(a) Verify that work input equals work output for a hydraulic system assuming no losses 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. (b) 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?

(a) V = d i A i = d o A o d o = d i A i A o . size 12{ V=d rSub { size 8{i} } A rSub { size 8{i} } =d rSub { size 8{o} } A rSub { size 8{o} } drarrow d rSub { size 8{o} } =d rSub { size 8{i} } left ( { {A rSub { size 8{i} } } over {A rSub { size 8{o} } } } right ) "." } {}

Now, using equation:

F 1 A 1 = F 2 A 2 F o = F i A o A i . size 12{ { {F rSub { size 8{1} } } over {A rSub { size 8{1} } } } = { {F rSub { size 8{2} } } over {A rSub { size 8{2} } } } drarrow F rSub { size 8{o} } =F rSub { size 8{i} } left ( { {A rSub { size 8{o} } } over {A rSub { size 8{i} } } } right ) "." } {}

Finally,

W o = F o d o = F i A o A i d i A i A o = F i d i = W i . size 12{W rSub { size 8{o} } =F rSub { size 8{o} } d rSub { size 8{o} } = left ( { {F rSub { size 8{i} } A rSub { size 8{o} } } over {A rSub { size 8{i} } } } right ) left ( { {d rSub { size 8{i} } A rSub { size 8{i} } } over {A rSub { size 8{o} } } } right )=F rSub { size 8{i} } d rSub { size 8{i} } =W rSub { size 8{i} } } {}

In other words, the work output equals the work input.

(b) If the system is not moving, friction would not play a role. With friction, we know there are losses, so that W out = W in W f size 12{W rSub { size 8{"out"} } =W rSub { size 8{"in"} } - W rSub { size 8{f} } } {} ; therefore, the work output is less than the work input. In other words, with friction, you need to push harder on the input piston than was calculated for the nonfriction case.

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Source:  OpenStax, College physics ii. OpenStax CNX. Nov 29, 2012 Download for free at http://legacy.cnx.org/content/col11458/1.2
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