Triple Physics Topic 8 - Hydraulics

As a sixth former (more years ago than I care to remember), I was involved in a Technology project to design and build a car for use in towns and cities.  This was long before the Smart Car or the Toyota IQ were conceived.  Our car was conceived as the "Three Metre Car for the Two Metre Man".  We used our Headmaster (who was well over two metres tall) as a model.  Even he could sit in it with some comfort.  The car finally emerged as this.


Instead of a normal clutch and gearbox, we opted for a hydraulic transmission system, an early day attempt at a constant velocity transmission system.  The rear-mounted 700 cm3 engine drove a variable output pump which was connected to two hydraulic traction motors in the wheel-hubs.   The gear-lever operated the output of the pump - a higher output was equivalent to a higher gear.



The pump can just be seen at the lower left.  The main filter sits above the pump.  Under the rear seat was a 20 litre reservoir for the hydraulic oil.



Although the car worked, it was not very successful.  The pump (which was the gearbox) sounded like a band-saw cutting through sheet steel.  It widdled hydraulic oil all over the place.  It was almost impossible to get the transmission into top gear.  The transmission was heavy, making the car decidedly heavy.  It had a prodigious appetite for petrol - not very good even in those days.  Its performance was none too good either.


Eventually the system was replaced with a conventional clutch and gearbox.  But it was not a bad attempt for a number of teenage would-be engineers.  The car itself was able to carry a large load and was not uncomfortable, either. 


I demonstrated it one afternoon to some school governors and other very important people.  They were thrilled.  Not so delighted was the Deputy Headmaster, as the wretched vehicle had widdled a number of fine lines of hydraulic oil across the car park.  Today's Health and Safety freaks would have had a fit!


Hydraulic systems such as the one described here are usually found on dump-trucks and excavators.  Continuously Variable Transmissions for cars are based on a rubber belt placed between two cones.



So how does a hydraulic system work?  Let us look at the key concepts.



Hydraulic Systems

All hydraulic systems transmit force through a liquid.  Liquids cannot be compressed like gases.  So if 1 cm3 of liquid leaves the pump, 1 cm3 liquid must go into the motor.  If it doesn't, it means that there is a leak somewhere.   A simple hydraulic system is shown below.



The motor can be a simple cylinder with a piston.  It is called a motor because it moves.  In this case, the motor is a ram, a cylinder in which there is a piston that moves when oil is pumped in.


The system acts as a force multiplier.  A small force is used to make a big force.


A key principle of working with hydraulic systems is that the pressure is the same throughout.  The pressure in the motor is the same as the pressure in the pump.  Also the hydraulic pressure acts equally in all directions.




The hydraulic pressure is defined as:

the force per unit area


In other words, how many newtons there are acting over every square metre.  The formula is:


Pressure = force area

In Physics code:


p = F



In triangle form:


The units for pressure are newtons per square metre (N m-2) or Pascals (Pa).


1 Pa = 1 N m-2


If you are using Pascals, you must convert square centimetres to square metres:


1 cm2 = 1 10-4 m2


The same relationship for pressure applies to gases as well.


Worked example

A hydraulic system is shown below

The force applied to the pump is 5 N and the area is 10 cm2.  The ram that acts as the motor has an area of 250 cm2.  Calculate:

(a) The pressure in Pascals in the system;

(b) The force from the ram;

(c) The factor by which the force is multiplied.


(a) Pressure = force area = 5 N (10 10-4) = 5000 Pa

(b) Force = pressure area = 5000 (250 10-4) = 125 N

(c) Factor by which force is multiplied = 125 5 = 25 times.

Notice that the area of the ram is 25 times bigger than the area of the pump.


As with any force multiplier, the pump has to go a lot further than the motor.  In this case, the travel of the pump would be 25 times further than the travel of the ram.  This is because the pump has to travel 25 times the distance in the direction of the force to do the same job of work that is done by the ram (25 times the force).



Examples of Hydraulic Systems

The picture below shows a bottle jack being used to lift a car.

Image from diy tools.


Notice that the car is on axle stands for safety.  If you jack up a car, you must use axle stands.  If not, the jack may slip and the car will fall onto you, leading to serious injury.  A bottle jack is shown in this diagram.



I have left the release valve out of this diagram for the sake of clarity.  There is a channel that runs between the high pressure side and the low pressure side.  This is normally blocked with a pointed screw that is undone when you want to lower the jack.


Ball valves stop the oil going back to the reservoir.


We are going to look at how the jack works by doing quantitative calculations (i.e. using numbers).  Each question will lead on to the next part of the study.  By a series of numerical calculations, we will see how much force a mechanic needs to raise a 10 tonne load (e.g. the back axle of a lorry).


The picture shows some numerical dimensions that we will use.



Formula for area of a circle is:

A = pr2

Remember to convert square cm into square metres.  Make sure you know the difference between the radius and the diameter!


Question 1

Work out the area of the ram



Question 2

The jack has a maximum load of 10 tonnes (10 000 kg).  What is the weight that acts on the ram?



Question 3

Calculate the pressure acting on the oil under the ram



Question 4

What is the pressure in the pump?  Explain your answer.



Question 5

Calculate the force needed to operate the pump.



Question 6

Calculate the factor by which the forces are multiplied.



This factor is the mechanical advantage of the jack.  It is also the ratio of the stroke of the pump to the stroke of the ram


Question 7

What is the ratio of the stroke of the pump to the stroke of the ram?



Now we use moments to work out the action of the lever that operates the pump.  The pivot is right at the end of the lever.  You may want to review moments in the previous topic.


Question 8

How many strokes are needed to raise the load by 5 cm?



Question 9

Calculate the moment from the pump (assume that the handle is perpendicular to the shaft of the pump)



Question 10

What is the moment at the end of the handle?  Explain your answer.



Question 11

Calculate the force that the mechanic needs to push down on the handle.



Question 12

Calculate the factor by which the force from the mechanic is multiplied to lift the 10 tonne load.



From this calculation, you have found out the extent to which a lever combined with a hydraulic pump acts as a very effective force multiplier.



Other examples of hydraulic systems include:


Pneumatic systems work on the same principle as hydraulic systems.  A compressor fills a reservoir with compressed air, which can drive a variety of tools.  The advantages of this are:

However pneumatic systems cannot transfer such large forces, as air is easily compressed.