**6.4 ****
Hydraulic systems -
mechanical devices for the transmission of forces in liquids **
**and
****
calculations**

Hydraulics is how we can use fluids (e.g. air or oil) to
transfer force to achieve some useful work e.g. operating machinery like car
jacks, car brakes or JCB digging machines.

As we have said above, pressure in fluids is caused by particle
collisions with themselves and the surface of a container.

**These collisions cause a net force at right angles to all
surfaces the fluid is in contact with**. Hence the equation

**pressure = force ÷ area,
P = F / A**

**P**, pascals (**Pa**);
**F**, force in newtons (**N**); **A**, area on which
force acts in square metres (**m**^{2})

A force of **1 N**
acting on **1 m**^{2} creates a pressure of **1 Pa**

(I've repeated these
details because there are more calculations coming up!)

However, there is an important new idea to understand the
applications of hydraulics.

**
The pressure in a fluid is transmitted equally in all
directions.**

**AND, for the purposes of calculations, you
consider the force involved to act at right-angles to the surface.**

Since liquids are effectively
**incompressible**, **if a force
is applied to any point in an enclosed fluid system, the net force is transmitted
to any other point in the fluid (gas or liquid)**. This is the basis of all**
hydraulic systems**.

**It is no good using a gas in a
hydraulic system**, because a gas is readily compressed, affecting the
action of the pistons in a hydraulic system - see diagram below.

A lot of the force and energy of the
compression action is used to compress the gas and so reducing the
transmitted force-pressure.

**The
principles and calculations involving hydraulic systems.**

Diagram
of the basic principles of a hydraulic system

The diagram above illustrates the idea of using
**a hydraulic system to apply a small force to produce a large force** - an
example of a force multiplier system to effect a mechanical advantage (remember
moments and lever
systems!)

The formula for pressure is **P = F/A** (already
dealt with in detail on this page).

Therefore: P1 = F1_{input}/A1 and output
P2 = F2_{output}/A2 (where = pressure, A = area, F = force)

BUT, P1 = P2 because the pressure at any given
moment in time is the same throughout the hydraulic system.

Therefore: F1_{input}/A1 = F2_{output}/A2, rearranging gives
the output force F2_{output} = F1_{input} x A2/A1

So the ratio of the two piston areas (A2/A1)
gives you the **force multiplying effect** (as long as A2 > A1).

The **mechanical advantage** can be simply defined as:

**output force / input force = F2 / F1**

Note: In hydraulic systems, piston 1 is in
the **master cylinder** and piston 2 is in a **slave cylinder** (can be
several of the latter e.g. in a car braking system of a system for raising a car
in a garage).

**
Some examples of how to do hydraulic system calculations**

**Q3.1**
In a simple hydraulic system piston 1 has a cross-section area of 0.000050 m^{2},
and piston 2 has a cross-section area of 0.00025.

If a force of 35 N is applied to piston 1,

(a) What is the force multiplying ratio?

(b) What output force is generated by piston
2?

Worked out ANSWERS to the hydraulic system calculation
questions

Q3.2 A medium
sized car typically weighs 1000 kg. (gravity force 9.8 N/kg)

A hydraulic system to jack the car up above
an inspection pit has four slave cylinders whose pistons have a cross
section area of 0.01 m^{2}. The cross-section area of the master
cylinder piston is 0.00125 m^{2}.

(a) What is the weight of the car?

(b) What is the minimum force that must be
applied to the master cylinder to raise the car upwards?

Worked out ANSWERS to the hydraulic system
calculation questions

**Q3.3**
Solve the following problem with a little help from the diagram on the right of
a simple hydraulic system of two pistons and cylinders connected together.

The cross-section area of piston A1 is
0.000400 m^{2}.

The cross-section area of piston A2 is
0.00800 m^{2}.

(a) If a force of 40 N (F1) is applied to
piston 1, calculate the pressure created in the fluid.

(b) Calculate the force F2 created by the
force of 40 N on piston A1.

(c) What is the multiplying or mechanical
advantage factor in this hydraulic system?

(d) In order to create a force of 2000 N
from cylinder 2, what should the cross-section area of piston 2 be if the
force applied to piston 1 is still 40 N?

Worked out ANSWERS to the hydraulic system calculation
questions

Index physics Forces notes 6. Forces & pressure in
fluids, calculations

**
Keywords, phrases and learning objectives for forces involving **
**pressure situations**

Be able to describe and explain the principles of hydraulic systems
and how they produce a mechanical advantage by increasing an applied
force by transmission of the force through an enclosed liquid
system.

Be able to define the mechanical advantage of a
given hydraulic system e.g. in terms of the input and output force
ratio..

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Index physics Forces notes 6. Forces & pressure in fluids, calculations