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FHSST Physics/Waves/Definition

=What are waves?= Waves are phenomena that everyone experiences constantly; water waves, sound waves, light waves, human waves when the home team scores... the list goes on. When asked what makes a wave a wave, the most common responses would probably be that a wave is something that moves, or propagates, or perhaps that it is something that repeats over and over again. These properties do capture the essential qualities of waves. Now we must determine these properties quantitatively, and discover what governs their behavior.

Generally, a wave is defined as any phenomenon which can be modeled by a function of the form $$f(\bar k\bar r-\omega t)$$ where the $$r$$-vector represents a position in space, and $$t$$ represents a time, and the $$k$$-vector and omega are both constants. Don't be intimidated by the vectors in the argument - most of our time at first will be spent on one-dimensional waves. If the wave is in only one spatial dimension 'x', for instance a wave travelling on a taut string, it can be written simply as $$f(kx-wt)$$.

Any function of this form "propagates" along the $$\bar k$$ direction over time. As time increases, the argument of the function increases; over time the form of the function effectively advances through space. Try coming up with functions of this form, and plot them at time $$t=0$$, then plot them again at a later time. This progressive property will become obvious. Try to figure out the velocity with which your function advances! (we will study this later) The negative sign in front of the time term causes the wave to propagate in the direction defined as positive (if that seems confusing, try plotting more functions over time, and examine the results). If you replace the negative with a positive (or instead consider a negative value of omega), the wave will propagate in the negative direction.

A very special and important case of a wave is the mathematical function $$f(\bar k\bar r-\omega t)=\sin(\bar k\bar r-\omega t)$$, or in one dimension, $$f(kx-\omega t)=\sin(kx-\omega t)$$. This is a sinusoidal wave – it oscillates up and down infinitely in both directions, and moves as time progresses. I mentioned that waves have the quality of repeating over and over, the quality of periodicity. However, many functions of the form mentioned above do not seem to repeat. However, you will find that ALL waves can be decomposed into a sum of many of these simple, infinitely repeating waves when you learn about Fourier transformations.

More than any other concept, physicists are finding that waves characterize the structure of the universe at every scale imaginable. As you learn about the physics of waves in everyday life, keep an open mind towards finding waves and wave behavior everywhere you turn.

Let's consider a very well-known case of a wave phenomenon: water waves. Waves in water consist of moving peaks and troughs. A peak is a place where the water rises higher than when the water is still and a trough is a place where the water sinks lower than when the water is still.

So waves have peaks and troughs. This could be our first property for waves. The following diagram shows the peaks and troughs on a wave.

In physics we try to be as quantitative as possible. If we look very carefully we notice that the height of the peaks above the level of the still water is the same as the depth of the troughs below the level of the still water.

Waves are repetitions of physical quantity in a periodic manner, carrying energy in the process. The water waves, for example, can be visualized to repeat any of the physical quantities like "peaks", "troughs", "potential energy" or "kinetic energy". Even, we can visualize water waves as the motion of disturbance (energy). It is the energy aspect of waves that is central to the understanding of different types of waves, many of which are not visible.

Looking closely at the water wave, we can recognize that crests and troughs basically represent of extreme potential and kinetic energies in addition to representing rise and fall of water from the still level. At the peak, energy is only potential, whereas energy is only kinetic at the trough. Similarly, propagation of electromagnetic wave is associated with repetitions of magnetic and electric field in space with certain periodicity. As existence of electrical or magnetic fields does not require any medium, electromagnetic waves can move even in the absence of any medium.

Characteristics of Waves: Amplitude
We use symbols agreed upon by convention to label the characteristic quantities of the waves. The characteristic height of a peak and depth of a trough is called the amplitude of the wave. The vertical distance between the bottom of the trough and the top of the peak is twice the amplitude. To put it simply, the amplitude is the distance of the wave from the medium, to the crest or trough

Worked Example 1

Question: (NOTE TO SELF: Make this a more exciting question) The height of the wave from the medium is 2m. What is the distance from the peak to the trough. What

The amplitude is 2m. (Read above paragraph to know why). The distance from the peak to trough is 4m.
 * Answer

Characteristics of Waves : Wavelength
Look a little closer at the peaks and the troughs. The distance between two adjacent (next to each other) peaks is the same no matter which two adjacent peaks you choose. So there is a fixed distance between the peaks.

Similarly, you'll notice that the distance between two adjacent troughs is the same no matter which two troughs you look at. But, more importantly, its is the same as the distance between the peaks. This distance which is a characteristic of the wave is called the wavelength.

Waves have a characteristic wavelength. The symbol for the wavelength is the Greek letter lambda, $$\lambda$$.

The wavelength is the distance between any two adjacent points which are in phase. Two points in phase are separate by an integer (0,1,2,3,...) number of complete wave cycles. They don't have to be peaks or trough but they must be separated by a complete number of waves.

Characteristics of Waves : Period
Now imagine you are sitting next to a pond and you watch the waves going past you. First one peak, then a trough and then another peak. If you measure the time between two adjacent peaks you'll find that it is the same. Now if you measure the time between two adjacent troughs you'll find that its always the same, no matter which two adjacent troughs you pick. The time you have been measuring is the time for one wavelength to pass by. We call this time the period and it is a characteristic of the wave.

The period of the wave is denoted with the symbol $$T$$.

Characteristics of Waves : Frequency
There is another way of characterising the time interval of a wave. We timed how long it takes for one wavelength to go past. We could also turn this around and say how many waves go by in 1 second.

We can easily determine this number, which we call the frequency and denote f. To determine the frequency, how many waves go by in 1s, we work out what fraction of a waves goes by in 1 second by dividing 1 second by the time it takes T. If a wave takes 1/2 a second to go by then in 1 second two waves must go by. $$\frac{1}{\frac{1}{2}} =2$$. The unit of frequency is the Hz or s−1.

Waves have a characteristic frequency.

generally, the frequency of a wave is the number of crests that pass by per unit time.

Characteristics of Waves : Speed
Now if you are watching a wave go by you will notice that they move at a constant velocity. Thinking back to rectilinear motion you will be able to remember that we know how to work out how fast something moves. The speed is the distance you travel divided by the time you take to travel that distance. This is excellent because we know that the waves travel a distance $$\lambda$$ in a time T. This means that we can determine the speed.

There are a number of relationships involving the various characteristic quantities of waves. A simple example of how this would be useful is how to determine the velocity when you have the frequency and the wavelength. We can take the above equation and substitute the relationship between frequency and period to produce an equation for speed of the form

Is this correct? Remember a simple first check is to check the units! On the right hand side we have speed which has units ms−1. On the left hand side we have frequency which is measured in s−1 multiplied by wavelength which is measure in m. On the left hand side we have ms−1 which is exactly what we want.

Speed of a wave through strings
The speed of a wave traveling along a vibrating string (v) is directly proportional to the square root of the tension (T) over the linear density (μ):

v=\sqrt{\frac{T}{\mu}} \, $$

μ is equal to the mass of the string divided by the length of the string.

$$ \mu={\frac{M}{L}} $$

title: FHSST Physics/Waves/Definition


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FHSST Physics/Waves/Types

= Two Types of Waves =

We agreed that a wave was a moving set of peaks and troughs and we used water as an example. Moving peaks and troughs, with all the characteristics we described, in any medium constitute a wave. It is possible to have waves where the peaks and troughs are perpendicular to the direction of motion, like in the case of water waves. These waves are called transverse waves.

There are two additional types of waves. The first is called longitudinal waves and have the peaks and troughs in the same direction as the wave is moving. The question is how do we construct such a wave?

An example of a longitudinal wave is pressure waves moving through a gas. The peaks in this wave are places where the pressure reaches a peak and the troughs are places where the pressure is a minimum.

In the picture below we show the random placement of the gas molecules in a tube. The piston at the end moves into the tube with a repetitive motion. Before the first piston stroke the pressure is the same throughout the tube.

When the piston moves in it compresses the gas molecules together at the end of the tube. If the piston stopped moving the gas molecules would all bang into each other and the pressure would increase in the tube.

When the piston moves out again before the molecules have time to bang around then the increase in pressure moves down the tube like a pulse (single peak and trough, a single wave cycle).

As this repeats we get waves of increased and decreased pressure moving down the tubes. We can describe these pulses of increased pressure (peaks in the pressure) and decreased pressure (troughs of pressure) by a sine or cosine graph.

The second additional type of wave is the torsional wave. The peaks and troughs rotate around the direction of motion. In simpler terms, a "twisting motion" is transmitted through the medium. Of the two wave types, this is the hardest one to describe and visualize.

There are a number of examples of each type of wave. Not all can be seen with the naked eye but all can be detected.

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

title: FHSST Physics/Waves/Types


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FHSST Physics/Waves/Properties

= Properties of Waves =

We have discussed some of the simple properties of waves that we need to know. These have just been describing the characteristics that waves have. Now we can progress onto some more interesting and, perhaps, less intuitive properties of waves.

Properties of Waves : Reflection
When waves strike a barrier they are reflected. This means that waves bounce off things. Sound waves bounce off walls, light waves bounce off mirrors, radar waves bounce off planes and how bats can fly at night and avoid things as small as telephone wires. etc. The property of reflection is a very important and useful one.

(NOTE TO SELF: Get an essay by an air traffic controller on radar) (NOTE TO SELF: Get an essay by on sonar usage for fishing or for submarines)

When waves are reflected, the process of reflection has certain properties. If a wave hits an obstacle at a right angle to the surface (NOTE TO SELF: diagrams needed) then the wave is reflected directly backwards.

If the wave strikes the obstacle at some other angle then it is not reflected directly backwards. The angle that the waves arrives at is the same as the angle that the reflected waves leaves at. The angle that waves arrives at or is incident at equals the angle the waves leaves at or is reflected at. Angle of incidence equals angle of reflection

In the optics chapter you will learn that light is a wave. This means that all the properties we have just learnt apply to light as well. Its very easy to demonstrate reflection of light with a mirror. You can also easily show that angle of incidence equals angle of reflection.

If you look directly into you see yourself ....

'''Need to mention that the incident wave, normal to the surface and the reflected wave all lie in the same plane. The same also holds for refraction at a surface.'''

Phase shift of reflected wave
When a wave is reflected from a more dense medium, it undergoes a phase shift. That means that the peaks and troughs are swapped around.

The easiest way to demonstrate this is to tie a piece of string to something. Stretch the string out flat and then flick the string once so a pulse moves down the string. When the pulse (a single peak in a wave) hits the barrier that the string is tied to, it will be reflected. The reflected wave will look like a trough instead of a peak. This is because the pulse had undergone a phase change. The fixed end is like reflection off a more dense medium.

If the end of the string was not fixed, i.e. it could move up and down then the wave would still be reflected but it would not undergo a phase shift.

Properties of Waves : Refraction
Sometimes waves move from one medium to another. The medium is the substance that is carrying the waves. In our first example this was the water. When the medium properties change it can affect the wave.

Let us start with the simple case of a water wave moving from one depth to another. The speed of the wave depends on the depth. If the wave moves directly from the one medium to the other than we should look closely at the boundary. When a peak arrives at the boundary and moves across it must remain a peak on the other side of the boundary. This means that the peaks pass by at the same time intervals on either side of the boundary. The period and frequency remain the same! But we said the speed of the wave changes, which means that the distance it travels in one time interval is different i.e. the wavelength has changed.

Going from one medium to another the period or frequency does not change only the wavelength can change.

Now if we consider a water wave moving at an angle of incidence not 90 degrees towards a change in medium then we immediately know that not the whole wave will arrive at once. So if a part of the wave arrives and slows down while the rest is still moving faster before it arrives the angle of the wavefront is going to change. This is known as refraction. When a wave bends or changes its direction when it goes from one medium to the next.

If it slows down it turns towards the perpendicular.

If the wave speeds up in the new medium it turns away from the perpendicular to the medium surface.

When you look at a stick that emerges from water it looks like it is bent. This is because the light from below the surface of the water bends when it leaves the water. Your eyes project the light back in a straight line and so the object looks like it is a different place.

Properties of Waves : Interference
If two waves meet interesting things can happen. Waves are basically collective motion of particles. So when two waves meet they both try to impose their collective motion on the particles. This can have quite different results.

If two identical (same wavelength, amplitude and frequency) waves are both trying to form a peak then they are able to achieve the sum of their efforts. The resulting motion will be a peak which has a height which is the sum of the heights of the two waves. If two waves are both trying to form a trough in the same place then a deeper trough is formed, the depth of which is the sum of the depths of the two waves. Now in this case the two waves have been trying to do the same thing and so add together constructively. This is called constructive interference.

If one wave is trying to form a peak and the other is trying to form a trough then they are competing to do different things. In this case they can cancel out. The height of the peak less the depth of the trough will be the resulting effect. If the depth of the trough is the same as the height of the peak nothing will happen. If the height of the peak is bigger than the depth of the trough a smaller peak will appear and if the trough is deeper then a less deep trough will appear. This is destructive interference.

Properties of Waves : Standing Waves
When two waves move in opposite directions, through each other, constructive interference happens. If the two waves have the same frequency and wavelength then a specific type of constructive interference can occur: standing waves can form.

Standing waves are disturbances which don't appear to move, they stand in the same place. Lets demonstrate exactly how this comes about. Imagine a long string with waves being sent down it from either end. The waves from both ends have the same amplitude, wavelength and frequency as you can see in the picture below:

To stop from getting confused between the two waves we'll draw the wave from the left with a dashed line and the one from the right with a solid line. As the waves move closer together when they touch both waves have an amplitude of zero:

If we wait for a short time the ends of the two waves move past each other and the waves overlap. Now we know what happens when two waves overlap, we add them together to get the resulting wave.

Now we know what happens when two waves overlap, we add them together to get the resulting wave. In this picture we show the two waves as dotted lines and the sum of the two in the overlap region is shown as a solid line:

The important thing to note in this case is that there are some points where the two waves always destructively interfere to zero. If we let the two waves move a little further we get the picture below:

Again we have to add the two waves together in the overlap region to see what the sum of the waves looks like.

In this case the two waves have moved half a cycle past each other but because they are out of phase they cancel out completely. The point at 0 will always be zero as the two waves move past each other.

When the waves have moved past each other so that they are overlapping for a large region the situation looks like a wave oscillating in place. If we focus on the range -4, 4 once the waves have moved over the whole region. To make it clearer the arrows at the top of the picture show peaks where maximum positive constructive interference is taking place. The arrows at the bottom of the picture show places where maximum negative interference is taking place.

As time goes by the peaks become smaller and the troughs become shallower but they do not move.

For an instant the entire region will look completely flat.

The various points continue their motion in the same manner.

Eventually the picture looks like the complete reflection through the x-axis of what we started with:

Then all the points begin to move back. Each point on the line is oscillating up and down with a different amplitude.

If we superimpose the two cases where the peaks where at a maximum and the case where the same waves where at a minimum we can see the lines that the points oscillate between. We call this the envelope of the standing wave as it contains all the oscillations of the individual points. A node is a place where the two waves cancel out completely as two waves destructively interfere in the same place. An anti-node is a place where the two waves constructively interfere.

To make the concept of the envelope clearer let us draw arrows describing the motion of points along the line.

Every point in the medium containing a standing wave oscillates up and down and the amplitude of the oscillations depends on the location of the point. It is convenient to draw the envelope for the oscillations to describe the motion. We cannot draw the up and down arrows for every single point!

Reflection from a fixed end
If waves are reflected from a fixed end, for example tying the end of a rope to a pole and then sending waves down it. The fixed end will always be a node. Remember: Waves reflected from a fixed end undergo a phase shift.

The wavelength, amplitude and speed of the wave cannot affect this, the fixed end is always a node.

Reflection from an open end
If waves are reflected from end, which is free to move, it is an anti-node. For example tying the end of a rope to a ring, which can move up and down, around the pole. Remember: The waves sent down the string are reflected but do not suffer a phase shift.

Wavelengths of standing waves with fixed and open ends

Beats
If the waves that are interfering are not identical then the waves form a modulated pattern with a changing amplitude. The peaks in amplitude are called beats. If you consider two sound waves interfering then you hear sudden beats in loudness or intensity of the sound.

The simplest illustration is to draw two different waves and then add them together. You can do this mathematically and draw them yourself to see the pattern that occurs.

Here is wave 1:

Now we add this to another wave, wave 2:

When the two waves are added (drawn in coloured dashed lines) you can see the resulting wave pattern:

To make things clearer the resulting wave without the dashed lines is drawn below. Notice that the peaks are the same distance apart but the amplitude changes. If you look at the peaks they are modulated i.e. the peak amplitudes seem to oscillate with another wave pattern. This is what we mean by modulation.

The maximum amplitude that the new wave gets to is the sum of the two waves just like for constructive interference. Where the waves reach a maximum it is constructive interference.

The smallest amplitude is just the difference between the amplitudes of the two waves, exactly like in destructive interference.

The beats have a frequency which is the difference between the frequency of the two waves that were added. This means that the beat frequency is given by

Properties of Waves : Diffraction
One of the most interesting, and also very useful, properties of waves is diffraction. When a wave strikes a barrier with a hole, only part of the wave can move through the hole. If the hole is similar in size to the wavelength of the wave diffractions occurs. The waves that comes through the hole no longer looks like a straight wave front. It bends around the edges of the hole. If the hole is small enough it acts like a point source of circular waves.

This bending around the edges of the hole is called diffraction. To illustrate this behaviour we start with Huygen's principle.

Huygen's Principle
Huygen's principle states that each point on a wavefront acts like a point source or circular waves. The waves emitted from each point interfere to form another wavefront on which each point forms a point source. A long straight line of points emitting waves of the same frequency leads to a straight wave front moving away.

To understand what this means lets think about a whole lot of peaks moving in the same direction. Each line represents a peak of a wave.

If we choose three points on the next wave front in the direction of motion and make each of them emit waves isotropically (i.e. the same in all directions) we will get the sketch below:

What we have drawn is the situation if those three points on the wave front were to emit waves of the same frequency as the moving wave fronts. Huygens principle says that every point on the wave front emits waves isotropically and that these waves interfere to form the next wave front.

To see if this is possible we make more points emit waves isotropically to get the sketch below:

You can see that the lines from the circles (the peaks) start to overlap in straight lines. To make this clear we redraw the sketch with dashed lines showing the wavefronts which would form. Our wavefronts are not perfectly straight lines because we didn't draw circles from every point. If we had it would be hard to see clearly what is going on.

Huygen's principle is a method of analysis applied to problems of wave propagation. It recognizes that each point of an advancing wave front is in fact the center of a fresh disturbance and the source of a new train of waves and that the advancing wave as a whole may be regarded as the sum of all the secondary waves arising from points in the medium already traversed. This view of wave propagation helps better understand a variety of wave phenomena, such as diffraction.

Wavefronts Moving Through an Opening
Now if we allow the wavefront to impinge on a barrier with a hole in it, then only the points on the wavefront that move into the hole can continue emitting forward moving waves - but because a lot of the wavefront have been removed the points on the edges of the hole emit waves that bend round the edges.

The wave front that impinges (strikes) the wall cannot continue moving forward. Only the points moving into the gap can. If you employ Huygens' principle you can see the effect is that the wavefronts are no longer straight lines.

Riaan Note: still cant find this image, have to get it from the pdf

For example, if two rooms are connected by an open doorway and a sound is produced in a remote corner of one of them, a person in the other room will hear the sound as if it originated at the doorway. As far as the second room is concerned, the vibrating air in the doorway is the source of the sound. The same is true of light passing the edge of an obstacle, but this is not as easily observed because of the short wavelength of visible light.

This means that when waves move through small holes they appear to bend around the sides because there aren't enough points on the wavefront to form another straight wavefront. This is bending round the sides we call diffraction.

Properties of Waves : Dispersion
Dispersion is a property of waves where the speed of the wave through a medium depends on the wavelength. So if two waves enter the same dispersive medium and have different wavelengths they will have different speeds in that medium even if they both entered with the same speed.

title: FHSST Physics/Waves/Properties


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FHSST Physics/Waves/Sound Applications

= Practical Applications of Waves: Sound Waves =

Doppler Effect
The Doppler Effect is an interesting phenomenon that occurs when an object producing sound is moved relatively to the listener.

Consider the following: When a car blaring its horn is behind you, the pitch is higher as it is approaching, and becomes lower as it is moving away. This is only noticeable if the object is moving at a fairly high speed, although it is still theoretically present at any speed.

When an object is moving away from the listener, the sound waves are stretched over a further distance meaning they happen less often. The wavelength ends up being greater so the frequency is less and the pitch is lower. When an object is moving towards the listener, the waves are compressed over a small distance making a very small wavelength and therefore a large frequency and high pitch. Since the pitch of the sound depends on the frequency of the waves, the pitch increases when the object is moving towards the listener.

f' is the observed frequency, f is the actual frequency, v is the speed of sound ($$ v=336+0.6T $$) T is temperature in degrees Celsius, $$v_0$$ is the speed of the observer, and $$v_s$$ is the speed of the source. If the observer is approaching the source, use the top operator (the +) in the numerator, and if the source is approaching the observer, use the top operator (the -) in the denominator. If the observer is moving away from the source, use the bottom operator (the -) in the numerator, and if the source is moving away from the observer, use the bottom operator (the +) in the denominator.

Example problems
A. An ambulance, which is emitting a 40 Hz siren, is moving at a speed of 30 m/s towards a stationary observer. The speed of sound in this case is 339 m/s.

$$f' = 40( \frac{339 + 0}{339 - 30} )$$

B. An M551 Sheridan, moving at 10 m/s is following a Renault FT-17 which is moving in the same direction at 5 m/s and emitting a 30 Hz tone. The speed of sound in this case is 342 m/s.

$$f' = 30( \frac{342 + 10}{342 + 5} )$$

Ultra-Sound
still to be completed

Ultrasound is sound that has too high a frequency for humans to hear. Some other animals can hear ultrasound though. Dog whistles are an example of ultrasound. We can't hear the sound, but dogs can. Audible sound is in the frequency range between 20 Hz and 20000 Hz. Anything above that is ultrasound, and anything below that is called infrasonic.

Ultrasound also has medical applications. It can be used to generate images with a sonogram. Ultrasound is commonly used to look at fetuses in the womb.

title: FHSST Physics/Waves/Sound Applications


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FHSST Physics/Waves/Electromagnetic Applications

= Practical Applications of Waves: Electromagnetic Waves =

In physics, wave-particle duality holds that light and matter simultaneously exhibit properties of waves and of particles. This concept is a consequence of quantum mechanics.

In 1905, Einstein reconciled Huygens' view with that of Newton. He explained the photoelectric effect (an effect in which light did not seem to act as a wave) by postulating the existence of photons, quanta of energy with particulate qualities. Einstein postulated that the frequency of light, $$f$$, is related to the energy, $$E$$, of its photons:

where $$h$$ is Planck's constant ($$6.626\times10^{-34}Js$$).

In 1924, De Broglie claimed that all matter has a wave-like nature. He related wavelength $$\lambda$$ and momentum p:

This is a generalization of Einstein's equation above, since the momentum of a photon is given by

where $$c$$ is the speed of light in vacuum, and $$f = \frac{c}{\lambda}$$.

De Broglie's formula was confirmed three years later by guiding a beam of electrons (which have rest mass) through a crystalline grid and observing the predicted interference patterns. Similar experiments have since been conducted with neutrons and protons. Authors of similar recent experiments with atoms and molecules claim that these larger particles also act like waves. This is still a controversial subject because these experimenters have assumed arguments of wave-particle duality and have assumed the validity of de Broglie's equation in their argument.

The Planck constant h is extremely small and that explains why we don't perceive a wave-like quality of everyday objects: their wavelengths are exceedingly small. The fact that matter can have very short wavelengths is exploited in electron microscopy.

In quantum mechanics, the wave-particle duality is explained as follows: every system and particle is described by state functions which encode the probability distributions of all measurable variables. The position of the particle is one such variable. Before an observation is made the position of the particle is described in terms of probability waves which can interfere with each other.


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FHSST Physics/Waves/Equations and Quantities

= Important Equations and Quantities =

Frequency:

Speed:

$$\begin{matrix}v &= &f\lambda\\&=& \frac{\lambda}{T}\end{matrix}$$


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FHSST Physics/Vectors

=Vectors= NOTE TO SELF: SW: initially this chapter had a very mathematical approach. I have toned this down and tried to present in a logical way the techniques of vector manipulation after first exploring the mathematical properties of vectors. Most of the PGCE comments revolved around the omission of the graphical techniques of vector addition (i.e. scale diagrams), incline questions and equilibrium of forces. I have addressed the first two equilibrium of forces and the triangle law of three forces in equilibrium I think would be better off in the Forces Chapter. Also the Forces chapter should include some examples of incline planes. Inclines are introduced here merely as an example of components in action!

title: FHSST Physics/Vectors


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FHSST Physics/Vectors/PGCE Comments

= PGCE Comments =
 * Use a+b = c to illustrate that a = c-b under subtraction.
 * Summarise properties of vectors at end.
 * Mention that vector multiplication exists in two forms- neither introduced at school, but used implicitly in work and motor rules.

title: FHSST Physics/Vectors/PGCE Comments


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FHSST Physics/Vectors/TO DO LIST

= `TO DO' LIST =
 * Please mention that although a rotation about an axis can be represented as a vector which statisfies your definition, the addition (composition) of two successive rotations about distinct axes is not obtained by simply adding the corresponding vectors. A better definition is: something which has direction and magnitude and adds by the parallelogram rule. (Alex)
 * Mention that the direction of a vector is measured from its tail.
 * Also include explanations of the ways in which to specify direction (e.g. bearing, compass points, angle from vertical, etc.)
 * Include exam-type questions involving flowing rivers or aeroplanes in wind. These questions link displacement, velocity and time and also test the learners understanding of vectors combining to give a combined result (i.e. resultant).
 * Mention that displacement at a point is the directed line from the start to that point.
 * Rewrite section on velocity to make clear the distinction between average and instantaneous rates of change (velocity and speed). The $$\Delta$$'s in the equations imply we are calculating average quantities. Mention that we take the limit of a small time interval to give instantaneous quantities. Perhaps the example of a parabola with average gradient and gradient of tangent can be used as an illustration. Else defer until chapter on Graphs and Equations of Motion. Instantaneous velocity: reading on the speedometer in a direction tangent to the path. Instantaneous speed is magnitude of instantaneous velocity but average speed is not equal to magnitude of average velocity. Average speed and average velocity are the total distance and resultant displacement over the time interval related to that part of the path. The example of the circular track uses these definitions and is an important illustration of the differences. Instantaneous calculated at a certain instant in time while average is calculated over an interval.
 * Include relative velocity.
 * Address PGCE comments above and comments in the text.

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FHSST Physics/Vectors/Introduction

Introduction
A vector is 'something' that has both magnitude and direction. "'Thing'? What sorts of 'thing'?" Any piece of information which contains a magnitude and a related direction can be a vector. A vector should tell you how much and which way.

Consider a man driving his car east along a highway at 100 km/h. What we have given here is a vector — the car's velocity. The car is moving at 100 km/h (this is the magnitude) and we know where it is going — east (this is the direction). Thus, we know the speed and direction of the car. These two quantities, a magnitude and a direction, form a vector we call velocity.

In physics, magnitudes often have directions associated with them. If you push something it is not very useful knowing just how hard you pushed. A direction is needed too. Directions are extremely important, especially when dealing with situations more complicated than simple pushes and pulls.

Different people like to write vectors in different ways. Any way of writing a vector so that it has both magnitude and direction is valid.

Are vectors physics? No, vectors themselves are not physics. Physics is just a description of the world around us. To describe something we need to use a language. The most common language used to describe physics is mathematics. Vectors form a very important part of the mathematical description of physics, so much so that it is absolutely essential to master the use of vectors.

Mathematical representation
Numerous notations are commonly used to denote vectors. In this text, vectors will be denoted by symbols capped with an arrow. As an example, $$\overrightarrow{s}$$,  $$\overrightarrow{v}$$ and $$\overrightarrow{F}$$are all vectors (they have both magnitude and  direction). Sometimes just the magnitude of a vector is required. In this case, the arrow is omitted. In other words, F denotes the magnitude of vector $$\overrightarrow{F}$$. $$|\overrightarrow{F}|$$ is another way of representing the size of a vector.

Graphical representation
Graphically vectors are drawn as arrows. An arrow has both a magnitude (how long it is) and a direction (the direction in which it points). For this reason, arrows are vectors.

In order to draw a vector accurately we must specify a scale and include a reference direction in the diagram. A scale allows us to translate the length of the arrow into the vector's magnitude. For instance if one chose a scale of 1cm = 2N (1cm represents 2N), a force of magnitude 20N would be represented as an arrow 10cm long. A reference direction may be a line representing a horizontal surface or the points of a compass.

Worked Example 2: Drawing vectors
Question: Using a scale of $$1cm=2m.s^{-1}$$ represent the following velocities:

a) $$6m.s^{-1}$$ north

b) $$16m.s^{-1}$$ east

Answer:

Introdução à Física/Grandezas vetoriais

title: FHSST Physics/Vectors/Introduction


text:
Scouting/BSA/Archaeology Merit Badge

Requirement 1
Tell what archaeology is and explain how it differs from anthropology, geology, paleontology, treasure hunting, and history.

Archaeology is the study of past human life by examining buildings and objects found in the ground; geology is the science that deals with the dynamics and physical history of the earth; anthropology is the science that deals with the origins, physical and cultural development, biological characteristics, and social customs and beliefs of humankind; paleontology is the science of the forms of life existing in former geologic periods, as represented by their fossils; treasure hunting is seeking relics of history for the sake of finding them rather than scientific goals or pursuit of knowledge; history is the branch of knowledge dealing with past events.

Requirement 2
Describe each of the following steps of the archaeological process: site location, development of a research design, historical research, site excavation, artifact identification and examination, interpretation, preservation, and information sharing.

Site location: First, the archaeologists try to locate sites. There are many ways to find sites. Archaeologists can locate sites by looking for places where there is a possibility that ancient cultures might have survived. A likely idea is near a river, but not close enough for houses to be affected by floods. An unlikely idea is in the center of a desert, far way from a source of water. Some sites are found before a construction. Before a building construction, archaeologists are informed of the building. If they find artifacts, construction plans might have to be changed. Construction might also be delayed. Site excavation: 2nd, before the excavation the archaeologists separate the field into several areas. After that, they begin to protect the site. They forbid access to other people. At the end, they begin to excavate. Artifact identification and examination: The assistance of specialists and the recourse to laboratories often are essential to date to determine materials or organizations, to study the human remainders. Interpretation: After the study of the new data, the researcher proposes several assumptions on the object find. This aspect determining of research often calls upon a gathering of specialized knowledge joined together in data bases. Preservation: the preservation is of course out of question of preserving all the sites concealing of the vestiges. It’s imperative to preserve the sites major and necessary to study the others before any destruction. It is also advisable to ensure the conservation of the vestiges put at the day: they are intended to enter public collections (museums), where they are then accessible to the researchers. Information sharing: The diffusion of the results of the excavations and the archaeological studies constitutes an essential stage of the work of the archaeologist. The ministry for the Culture and the Communication took this route by the means of a policy of edition and exposures in order to sensitize the public with archaeological research. The regional services of archaeology publish each year a regional scientific assessment.

Requirement 3
Describe at least two ways in which archaeologists determine the age of sites, structures, or artifacts. Explain what relative dating is.

1) The dating with carbon-14 is a method of measurement of the age of an organic subtance (time passed since the death). The absolute age limits which can be measured is approximately 50 000 years.  The dating with the carbon-14 is a powerful tool thanks to which them archaeologists dates from the events before indatable, in particular those which are old of more than 6000 years (prehistoric). 2)the archaeologists can date object with the style. For example in Greece, people black, background red (-620 ; -530) and people red, background black (from -530).

Requirement 4
Do TWO of the following:

 * A. Learn about three archaeological sites located outside the United States.

 * B. Learn about three archaeological sites located within the United States.

 * C. Visit an archaeological site and learn about it.

 * For EACH site you research for options a, b, or c, point it out on a map and explain how it was discovered. Describe some of the information about the past that has been found at each site. Explain how the information gained from the study of these sites answers questions that archaeologists are asking and how the information may be important to modern people. Compare the relative ages of the sites you research.

Requirement 5
Choose ONE of the sites you picked for Requirement 4 and give a short presentation about your findings to a Cub Scout pack, your Scout troop, your school class, or another group.

Requirement 6
Do the following:
 * A. Explain why it is important to protect archaeological sites.

It is important to protect archaeological sites. The archaeological sites are unique. Many of the sites are broken. They need protection. The sites have been made by our ancestors. It’s important to protect these sites.

 * B. Explain what people should do if they think they have found an artifact.

When a person finds old objects, he must call to the City Hall. The City Hall can call to an archaeologist. Many sites are found during a construction of a road, buildings, landscaping, etc... All states have laws to protect sites.

 * C. Describe the ways in which you can be a protector of the past.

To protect archaeological sites, you must obey the law of protection of archaeological objects. don’t do excavation alone

Requirement 7
Do ONE of the following:

 * A. Make a list of items you would include in a time capsule. Discuss with your merit badge counselor what archaeologists a thousand years from now might learn about you and the culture in which you live based on the contents of your capsule.

Archaeologists will understand what the world was like with the photos, the map, and the books. They will understand what some people of the world think with the religious text. * photos of all the world * Bible and other religious texts * current map of the world

 * B. Make a list of the trash your family throws out during one week. Discuss with your counselor what archaeologists might learn about you and your family if they found your trash a thousand years from now.

Requirement 8
Do ONE of the following:

 * A. Under the supervision of a qualified archaeologist, spend at least eight hours helping to excavate an archaeological site.

 * B. Under the supervision of a qualified archaeologist, spend at least eight hours in an archaeological laboratory helping to prepare artifacts for analysis, storage, or display.

 * C. If you are unable to work in the field or in a laboratory under the supervision of a qualified archaeologist, you may substitute a mock dig. To find out how to make a mock dig, talk with a professional archaeologist, trained avocational archaeologist, museum school instructor, junior high or high school science teacher, adviser from a local archaeology society, or other qualified instructor. Plan what you will bury in your artificial site to show use of your "site" during two time periods.

Requirement 9
Under the supervision of a qualified archaeologist or instructor, do ONE of the following:
 * A. Help prepare an archaeological exhibit for display in a museum, visitor center, school, or other public area.

 * B. Use the methods of experimental archaeology to re-create an item or to practice a skill from the past. Write a brief report explaining the experiment and its results.

Requirement 10
Do ONE of the following:

 * A. Research American Indians who live or once lived in your area. Find out about traditional lifeways, dwellings, clothing styles, arts and crafts, and methods of food gathering, preparation, and storage. Describe what you would expect to find at an archaeological site for these people.

 * B. Research settlers or soldiers who were in your area at least one hundred years ago. Find out about the houses or forts, ways of life, clothing styles, arts and crafts, and dietary habits of the early settlers, farmers, ranchers, soldiers, or townspeople who once lived in the area where your community now stands. Describe what you would expect to find at an archaeological site for these people.

Requirement 11
Identify three career opportunities in archaeology. Pick one and explain how to prepare for such a career. Discuss with your counselor what education and training are required, and explain why this profession might interest you. Egyptologist studies old Egypt. An archaeologist need to work Latin and Greek. He goes an archaeologist's school and to choose option Egyptologist. Archaeologist under water: Sub-aqueous archaeology is research and the study of the vestiges in order to know the human activities of last and is practiced in interior water: artificial often closed: wells, cisterns, underground conduits drowned, natural:  lakes, rivers, ponds, peat bogs, marshes, karst, and ground water. The aquatic environment is very favorable to archaeology: water protects from the destructive human actions, the organic and mineral elements safe from oxygen, the light and the biological organizations are preserved often perfectly. A archaeologist under water need do a school of archaeology an to practice swimming under water. Restorer restores objects like ceramic and glass found in an archaeological excavation. They need to do history of art and after to work in “conservation and restoration of the object”

title: Scouting/BSA/Archaeology Merit Badge


text:
FHSST Physics/Vectors/Examples

= Some Examples of Vectors =

Displacement
Imagine you walked from your house to the shops along a winding path through the veld. Your route is shown in blue in Figure 3.1. Your sister also walked from the house to the shops, but she decided to walk along the pavements. Her path is shown in red and consisted of two straight stretches, one after the other.

Although you took very different routes, both you and your sister walked from the house to the shops. The overall effect was the same! Clearly the shortest path from your house to the shops is along the straight line between these two points. The length of this line and the direction from the start point (the house) to the end point (the shops) forms a very special vector known as  displacement. Displacement is assigned the symbol $$\overrightarrow{s}$$

OR

(NOTE TO SELF: choose one of the above)

In this example both you and your sister had the same displacement. This is shown as the black arrow in Figure 3.1. Remember displacement is not concerned with the actual path taken. It is only concerned with your start and end points. It tells you the length of the straight-line path between your start and end points and the direction from start to finish. The distance travelled is the length of the path followed and is a scalar (just a number). Note that the magnitude of the displacement need not be the same as the distance travelled. In this case the magnitude of your displacement would be considerably less than the actual length of the path you followed through the veld!

Velocity
The terms rate of change and with respect to are ones we will use often and it is important that you understand what they mean. Velocity describes how much displacement changes for a certain change in time.

We usually denote a change in something with the symbol $$\Delta$$ (the Greek letter Delta). You have probably seen this before in maths — the gradient of a straight line is $$\frac{\Delta y}{\Delta x}$$. The gradient is just how much y changes for a certain change in x. In other words it is just the rate of change of y with respect to x. This means that velocity must be

$$\begin{matrix}\overrightarrow{v}=\frac{\Delta \overrightarrow{s}}{\Delta t} =\frac{\overrightarrow{s}_{final}-\overrightarrow{s}_{initial}}{t_{final}-t_{initial}}\end{matrix}$$

(NOTE TO SELF: This is actually average velocity. For instantaneous $$\Delta$$'s change to differentials. Explain that if $$\Delta$$ is large then we have average velocity else for infinitesimal time interval instantaneous!)

What then is speed? Speed is how quickly something is moving. How is it different from velocity? Speed is not a vector. It does not tell you which direction something is moving, only how fast. Speed is the magnitude of the velocity vector (NOTE TO SELF: instantaneous speed is the magnitude of the instantaneous velocity.... not true of averages!).

Consider the following example to test your understanding of the differences between velocity and speed.

Worked Example 3: Speed and Velocity
Question: A man runs around a circular track of radius 100m. It takes him 120s to complete a revolution of the track. If he runs at constant speed, calculate:

 * 1) his speed,
 * 2) his instantaneous velocity at point A,
 * 3) his instantaneous velocity at point B,
 * 4) his average velocity between points A and B,
 * 5) his average velocity during a revolution.

Answer:
 * 1. To determine the man's speed, we need to know the distance he travels and how long it takes. We know it takes $$120 s$$ to complete one revolution of the track.   What distance is one revolution of the track? We know the track is a circle and we know its radius, so we can determine the perimeter or distance around the circle. We start with the equation for the circumference of a circle:

$$\begin{matrix}C & =& 2\pi r \\ & = & 2\pi (100m) \\& = & 628.3\;m.\end{matrix}$$

 * 2. Now that we have distance and time, we can determine speed. We know that speed is distance covered per unit time. If we divide the distance covered by the time it took, we will know how much distance was covered for every unit of time.

$$\begin{matrix} v & = &\frac{Distance\ travelled}{time\ taken} \\ & = & \frac{628.3m}{120s} \\ & = & 5.23\ m.s^{-1} \end{matrix}$$

 * 3. Consider point A in the diagram:

We know which way the man is running around the track, and we know his speed. His velocity at point A will be his speed (the magnitude of the velocity) plus his direction of motion (the direction of his velocity). He is moving at the instant that he arrives at A, as indicated in the diagram below.

His velocity vector will be $$5.23\ m.s^{-1}$$ West.

 * 4. Consider point B in the diagram:

 * We know which way the man is running around the track, and we know his speed. His velocity at point B will be his speed (the magnitude of the velocity) plus his direction of motion (the direction of his velocity). He is moving at the instant that he arrives at B, as indicated in the diagram below.

 * His velocity vector will be $$5.23\ m.s^{-1}$$ South.

 * 4. So, now, what is the man's average velocity between Point A and Point B?

As he runs around the circle, he changes direction constantly. (Imagine a series of vector arrows pointing out from the circle, one for each step he takes.) If you add up all these directions and find the average it turns out to be ...Right. South west. And, notice that if you just looked for the average between his velocity at Point A and at Point B, that comes out south west, too. So his average velocity between Point A and Point B is $$5.23\ m.s^{-1}$$ south west.

 * 5. Now we need to calculate his average velocity over a complete revolution. The definition of average velocity is given earlier and requires that you know the total displacement and the total time. The total displacement for a revolution is given by the vector from the initial point to the final point. If the man runs in a circle, then he ends where he started. This means the vector from his initial point to his final point has zero length. A calculation of his average velocity follows:

$$\begin{matrix} \overrightarrow{v}&=&\frac{\Delta\overrightarrow{s}}{\Delta t} \\ &=& \frac{0m}{120s} \\ &=& 0\ m.s^{-1} \end{matrix}$$

Acceleration
Acceleration is also a vector. Remember that velocity was the rate of change of displacement with respect to time so we expect the velocity and acceleration equations to look very similar. In fact:

(NOTE TO SELF: average and instantaneous distinction again! expand further — what does it mean?)

Acceleration will become very important later when we consider forces.

Force
Imagine that you and your friend are pushing a cardboard box kept on a smooth floor. Both of you are equally strong. Can you tell me in which direction the box will move ? Probably not. Because I have not told you in which direction each of you are pushing the box. If both of you push it towards north, the box would move northwards. If you push it towards north and you friend pushes it towards east, it would move north-eastwards. If you two push it in opposite directions, it wouldn't move at all !

Thus in dealing with force applied on any object, it is equally important to take into account the direction of the force, as the magnitude. This is the case with all vectors.

title: FHSST Physics/Vectors/Examples


text:
FHSST Physics/Vectors/Mathematical Properties

= Mathematical Properties of Vectors =

Vectors are mathematical objects and we will use them to describe physics in the language of mathematics. However, first we need to understand the mathematical properties of vectors (e.g. how they add and subtract).

We will now use arrows representing displacements to illustrate the properties of vectors. Remember that displacement is just one example of a vector. We could just as well have decided to use forces to illustrate the properties of vectors.

Addition of Vectors
If we define a displacement vector as 2 steps in the forward direction and another as 3 steps in the forward direction then adding them together would mean moving a total of 5 steps in the forward direction. Graphically, this can be seen by first following the first vector two steps forward, and then following the second one three steps forward:

We add the second vector at the end of the first vector, since this is where we now are after the first vector has acted. The vector from the tail of the first vector (the starting point) to the head of the last (the end point) is then the sum of the vectors. This is the tail-to-head method of vector addition.

The order in which you add vectors does not matter. In the example above, if you decided to first go 3 steps forward and then another 2 steps forward, the end result would still be 5 steps forward.

The final answer when adding vectors is called the resultant.

In other words, the individual vectors can be replaced by the resultant — the overall effect is the same. If vectors $$\overrightarrow{a}$$ and $$\overrightarrow{b}$$ have a resultant $$\overrightarrow{R}$$, this can be represented mathematically as, $$\begin{matrix}\overrightarrow{R} &=& \overrightarrow{a} + \overrightarrow{b}.\end{matrix}$$

Let us consider some more examples of vector addition using displacements. The arrows tell you how far to move and in what direction. Arrows to the right correspond to steps forward, while arrows to the left correspond to steps backward. Look at all of the examples below and check them.

Let us test the first one. It says one step forward and then another step forward is the same as an arrow twice as long — two steps forward.

It is possible that you end up back where you started. In this case the net result of what you have done is that you have gone nowhere (your start and end points are at the same place). In this case, your resultant displacement is a vector with length zero units. We use the symbol $$\overrightarrow{0}$$ to denote such a vector:

Check the following examples in the same way. Arrows up the page can be seen as steps left and arrows down the page as steps right.

Try a couple to convince yourself!

It is important to realise that the directions aren't special — forward and backwards or left and right are treated in the same way. The same is true of any set of parallel directions:

In the above examples the separate displacements were parallel to one another. However the same tail-to-head technique of vector addition can be applied to vectors in any direction.

Now you have discovered one use for vectors; describing resultant displacement — how far and in what direction you have travelled after a series of movements.

Although vector addition here has been demonstrated with displacements, all vectors behave in exactly the same way. Thus, if given a number of forces are acting on a body, you can use the same method to determine the resultant force acting on the body. We will return to vector addition in more detail later.

Subtraction of Vectors
What does it mean to subtract a vector? Well this is really simple: if we have 5 apples and we subtract 3 apples, we have only 2 apples left. Now lets work in steps — if we take 5 steps forward, and then subtract 3 steps forward, we are left with only two steps forward:

What have we done? You originally took 5 steps forward but then you took 3 steps back. That backward displacement would be represented by an arrow pointing to the left (backwards) with length 3. The net result of adding these two vectors is 2 steps forward:

Thus, subtracting a vector from another is the same as adding a vector in the opposite direction (i.e. subtracting 3 steps forwards is the same as adding 3 steps backwards).

This suggests that in this problem, arrows to the right are positive, and arrows to the left are negative. More generally, vectors in opposite directions differ in sign (i.e. if we define up as positive, then vectors acting down are negative). Thus, changing the sign of a vector simply reverses its direction:

In mathematical form, subtracting $$\overrightarrow{a}$$ from $$\overrightarrow{b}$$ gives a new vector $$\overrightarrow{c}$$

$$\begin{matrix}\overrightarrow{c} &=& \overrightarrow{b} - \overrightarrow{a}\\&=& \overrightarrow{b} + (-\overrightarrow{a})\end{matrix}$$

This clearly shows that subtracting vector $$\overrightarrow{a}$$ from $$\overrightarrow{b}$$ is the same as adding $$(-\overrightarrow{a})$$ to $$\overrightarrow{b}$$. Look at the following examples of vector subtraction.

Scalar Multiplication
What happens when you multiply a vector by a scalar (an ordinary number)?

Going back to normal multiplication we know that $$2 \times 2$$ is just 2 groups of 2 added together to give 4. We can adopt a similar approach to understand how vector multiplication works.


text:
FHSST Physics/Vectors/Addition

= Techniques of Vector Addition =

Now that you have been acquainted with the mathematical properties of vectors, we return to vector addition in more detail. There are a number of techniques of vector addition. These techniques fall into two main categories- graphical and algebraic techniques.

Graphical Techniques
Graphical techniques involve drawing accurate scale diagrams to denote individual vectors and their resultants. We next discuss the two primary graphical techniques, the tail-to-head technique and the parallelogram method.

The Tail-to-head Method
In describing the mathematical properties of vectors we used displacements and the tail-to-head graphical method of vector addition as an illustration. In the tail-to-head method of vector addition the following strategy is followed:

 * Choose a scale and include a reference direction.
 * Choose any of the vectors to be summed and draw it as an arrow in the correct direction and of the correct length- remember to put an arrowhead on the end to denote its direction.
 * Take the next vector and draw it as an arrow starting from the arrowhead of the first vector in the correct direction and of the correct length.
 * Continue until you have drawn each vector- each time starting from the head of the previous vector. In this way, the vectors to be added are drawn one after the other tail-to-head.
 * The resultant is then the vector drawn from the tail of the first vector to the head of the last. Its magnitude can be determined from the length of its arrow using the scale. Its direction too can be determined from the scale diagram.

Worked Example 4 Tail-to-Head Graphical Addition I
Question: A ship leaves harbour H and sails 6km north to port A. From here the ship travels 12km east to port B, before sailing 5.5km south-west to port C. Determine the ship's resultant displacement using the tail-to-head technique of vector addition.

Answer:

Now, we are faced with a practical issue: in this problem the displacements are too large to draw them their actual length! Drawing a 2km long arrow would require a very big book. Just like cartographers (people who draw maps), we have to choose a scale. The choice of scale depends on the actual question- you should choose a scale such that your vector diagram fits the page. Before choosing a scale one should always draw a rough sketch of the problem. In a rough sketch one is interested in the approximate shape of the vector diagram.

Step 1 :

Let us draw a rough sketch of the situation

In a rough sketch one should include all of the information given in the problem. All of the magnitudes of the displacements are shown and a compass has been included as a reference direction.

Step 2 :

Next we choose a scale for our vector diagram. It is clear from the rough sketch that choosing a scale where 1cm represents 1km (scale: 1cm = 1km) would be a good choice in this problem )- the diagram will then take up a good fraction of an A4 page. We now start the accurate construction.

Step 3 :

Construction Step 1: Starting at the harbour H we draw the first vector 6cm long in the direction north (remember in the diagram 1cm represents 1km):

Construction Step 2: Since the ship is now at port A we draw the second vector 12cm long starting from this point in the direction east:

Construction Step 3: Since the ship is now at port B we draw the third vector 5.5cm long starting from this point in the direction south-west. A protractor is required to measure the angle of 45o.

Construction Step 4: As a final step we draw the resultant displacement from the starting point (the harbour H) to the end point (port C). We use a ruler to measure the length of this arrow and a protractor to determine its direction

Step 4 :

We now use the scale to convert the length of the resultant in the scale diagram to the actual displacement in the problem. Since we have chosen a scale of 1cm = 1km in this problem the resultant has a magnitude of 8.38  km. The direction can be specified in terms of the angle measured either as 75.4o east of north or on a bearing of 75.4o.

Step 5 :

Now we can quote the final answer: The resultant displacement of the ship is 8.38  km on a bearing of 75.4o!

Worked Example 5 Tail-to-Head Graphical Addition II
Question: A man walks 40  m East, then 30   m North.

a) What was the total distance he walked?

b) What is his resultant displacement?

Answer:

Step 1 :

What distance did the man travel? In the first part of his journey he traveled 40  m and in the second part he traveled 30   m. This gives us a total distance traveled of $$40+30=70\ m$$

Step 2 :

What is his resultant displacement? The man's resultant displacement is the vector from where he started to where he ended. It is the sum of his two separate displacements. We will use the tail-to-head method of accurate construction to find this vector. Firstly, we draw a rough sketch:

Step 3 :

Next we choose a scale suitable for the problem. A scale of 1cm represents 5m (1cm = 5m) is a good choice here. Now we can begin the process of construction.

Step 4 :

We draw the first displacement as an arrow 8cm long (according to the scale $$8cm = 8\times5m = 40m$$) in the direction east:

Step 5 :

Starting from the head of the first vector we draw the second displacement as an arrow 6cm long (according to the scale $$6cm = 6\times5m = 30m$$) in the direction north:

Step 6 :

Now we connect the starting point to the end point and measure the length and direction of this arrow (the resultant)

Step 7 :

Finally we use the scale to convert the length of the resultant in the scale diagram to the actual magnitude of the resultant displacement. According to the chosen scale 1cm = 5m. Therefore 10cm represents 50m. The resultant displacement is then 50m 36.9o north of east.

The Parallelogram Method
When needing to find the resultant of two vectors another graphical technique can be applied- the parallelogram method. The following strategy is employed:

 * Choose a scale and a reference direction.
 * Choose either of the vectors to be added and draw it as an arrow of the correct length in the correct direction.
 * Draw the second vector as an arrow of the correct length in the correct direction from the tail of the first vector .
 * Complete the parallelogram formed by these two vectors.
 * The resultant is then the diagonal of the parallelogram. Its magnitude can be determined from the length of its arrow using the scale. Its direction too can be determined from the scale diagram.

Worked Example 6
Parallelogram Method of Graphical Addition I

Question: A force of F1 = 5 N is applied to a block in a horizontal direction. A second force F2 = 4 N is applied to the object at an angle of 30° above the horizontal.

Determine the resultant force acting on the block using the parallelogram method of accurate construction.

Answer:

Step 1 :

Firstly we make a rough sketch of the vector diagram:

Step 2 :

Now we choose a suitable scale. In this problem a scale of 1 cm = 0.5 N would be appropriate, since then the vector diagram would take up a reasonable fraction of the page. We can now begin the accurate scale diagram.

Step 3 :

Let us draw F1 first. According to the scale it has length 10 cm:

Step 4 :

Next we draw F2. According to the scale it has length 8 cm. We make use of a protractor to draw this vector at 30° to the horizontal:

Step 5 :

Next we complete the parallelogram and draw the diagonal:

 RIAAN NOTE: Image missing img155.png PDF page 51

Step 6 :

Finally we use the scale to convert the measured length into the actual magnitude. Since 1 cm = 0.5 N, 17.4 cm represents 8.7 N. Therefore the resultant force is 8.7 N at 13.3° above the horizontal.

The parallelogram method is restricted to the addition of just two vectors. However, it is arguably the most intuitive way of adding two forces acting at a point.

Vectors in a Straight Line
Whenever you are faced with adding vectors acting in a straight line (i.e. some directed left and some right, or some acting up and others down) you can use a very simple algebraic technique:

 * Choose a positive direction. As an example, for situations involving displacements in the directions west and east, you might choose west as your positive direction. In that case, displacements east are negative.
 * Next simply add (or subtract) the vectors with the appropriate signs.
 * As a final step the direction of the resultant should be included in words (positive answers are in the positive direction, while negative resultants are in the negative direction).

Let us consider a couple of examples.

Worked Example 7
Adding vectors algebraically I

Question: A tennis ball is rolled towards a wall which is 10m away to the right. If after striking the wall the ball rolls a further 2.5m along the ground to the left, calculate algebraically the ball's resultant displacement.

(NOTE TO SELF: PGCE suggest a `more real looking' diagram, followed by a diagram one would draw to solve the problem (like our existing one with the positive direction shown as an arrow))

Answer:

Step 1 :

Let us draw a picture of the situation:

Step 2 :

We know that the resultant displacement of the ball ( $$\overrightarrow{s}_{resultant}$$) is equal to the sum of the ball's separate displacements ($$\overrightarrow{s}_1$$ and $$\overrightarrow{s}_2$$ ):

$$\begin{matrix}\overrightarrow{s}_{resultant} & = & \overrightarrow{s}_{1} + \overrightarrow{s}_{2}\end{matrix}$$

Since the motion of the ball is in a straight line (i.e. the ball moves left and right), we can use the method of algebraic addition just explained.

Step 3 :

First we choose a positive direction. Let's make to the right the positive direction. This means that to the left becomes the negative direction.

Step 4 :

With right positive:

$$\begin{matrix}\overrightarrow{s}_{1} & = & +10.0m \\& and & \\\overrightarrow{s}_{2} & = & -2.5m\end{matrix}$$

Step 5 :

Next we simply add the two displacements to give the resultant:

$$\begin{matrix}\overrightarrow{s}_{resultant} & = & (+10m) + (-2.5m) \\& = & (+7.5)m\end{matrix}$$

Step 6 :

Finally, in this case right means positive so:

$$\begin{matrix}\overrightarrow{s}_{resultant} & = & 7.5m \rm {\ to \ the \ right}\end{matrix}$$

Let us consider an example of vector subtraction.

Worked Example 8
Subtracting vectors algebraically I

Question: Suppose that a tennis ball is thrown horizontally towards a wall at 3m.s−1 to the right. After striking the wall, the ball returns to the thrower at 2m.s−1. Determine the change in velocity of the ball.

Answer:

Step 1 :

Remember that velocity is a vector. The change in the velocity of the ball is equal to the difference between the ball's initial and final velocities:

$$\begin{matrix}\Delta\overrightarrow{v} & = & \overrightarrow{v}_{final} -\overrightarrow{v}_{initial}\end{matrix}$$

Since the ball moves along a straight line (i.e. left and right), we can use the algebraic technique of vector subtraction just discussed.

Step 2 :

Let's make to the right the positive direction. This means that to the left becomes the negative direction.

Step 3 :

With right positive:

$$\begin{matrix}\overrightarrow{v}_{initial} & = & +3m.s^{-1} \\& and & \\\overrightarrow{v}_{final} & = & -2m.s^{-1}\end{matrix}$$

Step 4 :

Thus, the change in velocity of the ball is:

$$\begin{matrix}\Delta\overrightarrow{v} & = & (-2m.s^{-1}) - (+3m.s^{-1}) \\& = & (-5)m.s^{-1}\end{matrix}$$

Remember that in this case right means positive so:

$$\begin{matrix}\Delta\overrightarrow{v} & = & 5m.s^{-1} \rm \textbf{\ to \ the \ \emph{left}}\end{matrix}$$

Remember that the technique of addition and subtraction just discussed can only be applied to vectors acting along a straight line.

A More General Algebraic technique
In worked example 3 the tail to head method of accurate construction was used to determine the resultant displacement of a man who travelled first east and then north. However, the man's resultant can be calculated without drawing an accurate scale diagram. Let us revisit this example.

Worked Example 9
An Algebraic solution to Worked Example 3

Question: A man walks 40 m East, then 30  m North.

 * 1) Calculate the man's resultant displacement.

Answer:

Step 1 :

As before, the rough sketch looks as follows:

Step 2 :

Note that the triangle formed by his separate displacement vectors and his resultant displacement vector is a right-angle triangle. We can thus use Pythogoras' theorem to determine the length of the resultant. If the length of the resultant vector is called s then:

$$\begin{matrix}s^2&=&(40m)^2 + (30m)^2\\s^2&=&2500m^2\\s&=&50m\\\end{matrix}$$

Step 3 : Now we have the length of the resultant displacement vector but not yet its direction. To determine its direction we calculate the angle $$\alpha$$ between the resultant displacement vector and East.

We can do this using simple trigonometry:

$$\begin{matrix}\tan \alpha &=& \frac{opposite}{adjacent}\\\tan \alpha &=& \frac{30}{40}\\\alpha& =& \arctan (0.75) \\\alpha &=& 36.9^o\\\end{matrix}$$

Step 4 :

Our final answer is then:

 * Resultant Displacement: 50 m at 36.9o North of East

This is exactly the same answer we arrived at after drawing a scale diagram!

In the previous example we were able to use simple trigonometry to calculate a man's resultant displacement. This was possible since the man's directions of motion were perpendicular (north and east). Algebraic techniques, however, are not limited to cases where the vectors to be combined are along the same straight line or at right angles to one another. The following example illustrates this.

Worked Example 10
Further example of vector addition by calculation

Question: A man walks from point A to point B which is 12km away on a bearing of 45o. From point B the man walks a further 8km east to point C. Calculate the man's resultant displacement.

Answer:

Step 1 : Let us begin by drawing a rough sketch of the situation

RIAAN NOTE: Image on page 56 is missing

$$B\hat{A}F = 45^o$$ since the man walks initially on a bearing of 45o.Then, $$A\hat{B}G = B\hat{A}F = 45^o$$ (alternate angles parallel lines). Both of these angles are included in the rough sketch.

Step 2 :

Now let us calculate the length of the resultant (AC). Since we know both the lengths of $$AB$$ and $$BC$$ and the included angle $$A\hatC$$, we can use the cosine rule:

$$\begin{matrix}AC^2 &=& AB^2 +BC^2 -2\cdot AB\cdot BC\cos(A\hat{B}C)\\&=& (12)^2 + (8)^2 -2\cdot (12)(8)\cos(135^o)\\&=& 343.8\\AC &=& 18.5\ km\end{matrix}$$

Step 3 :

Next we use the sine rule to determine the angle $$\theta$$:

$$\begin{matrix}\frac{\sin\theta}{8}&=&\frac{\sin 135^0}{18.5}\\\sin\theta &=& \frac{8\times\sin 135^o}{18.5}\\\theta &=& \arcsin(0.3058)\\\theta &=& 17.8^o\end{matrix}$$

Thus, $$F\hat{A}C=62.8^o$$

Step 4 :

Our final answer is then:

 * Resultant Displacement: 18.5km on a bearing of 62.8o

title: FHSST Physics/Vectors/Addition


text:
FHSST Physics/Vectors/Components

= Components of Vectors =

In the discussion of vector addition we saw that a number of vectors acting together can be combined to give a single vector (the resultant). In much the same way a single vector can be broken down into a number of vectors which when added give that original vector. These vectors which sum to the original are called components of the original vector. The process of breaking a vector into its components is called resolving into components.

While summing a given set of vectors gives just one answer (the resultant), a single vector can be resolved into infinitely many sets of components. In the diagrams below the same black vector is resolved into different pairs of components. These components are shown in red. When added together the red vectors give the original black vector (i.e. the original vector is the resultant of its components).

In practice it is most useful to resolve a vector into components which are at right angles to one another.

Worked Example 11
Resolving a vector into components

Question: A motorist undergoes a displacement of 250km in a direction 30o north of east. Resolve this displacement into components in the directions north ($$\overrightarrow{s}_N$$ and east ($$\overrightarrow{s}_E$$).

Answer:

Step 1 :

Firstly let us draw a rough sketch of the original vector

Step 2 :

Next we resolve the displacement into its components north and east. Since these directions are orthogonal to one another, the components form a right-angled triangle with the original displacement as its hypotenuse:

Notice how the two components acting together give the original vector as their resultant.

Step 3 :

Now we can use trigonometry to calculate the magnitudes of the components of the original displacement:

$$\begin{matrix}s_N &=& 250\sin30^o\\&=& 125\ km\end{matrix}$$

and

$$\begin{matrix}s_E &=& 250\cos30^o\\&=& 216.5\ km\end{matrix}$$

Remember sN and sE are the magnitudes of the components- they are in the directions north and east respectively.

Block on an incline
As a further example of components let us consider a block of mass m placed on a frictionless surface inclined at some angle $$\theta$$ to the horizontal. The block will obviously slide down the incline, but what causes this motion?

The forces acting on the block are its weight mg and the normal force N exerted by the surface on the object. These two forces are shown in the diagram below.

Now the object's weight can be resolved into components parallel and perpendicular to the inclined surface. These components are shown as red arrows in the diagram above and are at right angles to each other. The components have been drawn acting from the same point. Applying the parallelogram method, the two components of the block's weight sum to the weight vector.

To find the components in terms of the weight we can use trigonometry:

$$\begin{matrix}W_{\|} &=& mg\sin\theta\\W_{\perp} &=& mg\cos\theta\end{matrix}$$

The component of the weight perpendicular to the slope W$\perp$ exactly balances the normal force N exerted by the surface. The parallel component, however, $$W_{\|}$$ is unbalanced and causes the block to slide down the slope.

Vector addition using components
In Figure 3.3 two vectors are added in a slightly different way to the methods discussed so far. It might look a little like we are making more work for ourselves, but in the long run things will be easier and we will be less likely to go wrong.

In Figure 3.3 the primary vectors we are adding are represented by solid lines and are the same vectors as those added in Figure 3.2 using the less complicated looking method.

Each vector can be broken down into a component in the x-direction and one in the y-direction. These components are two vectors which when added give you the original vector as the resultant. Look at the red vector in figure 3.3. If you add up the two red dotted ones in the x-direction and y-direction you get the same vector. For all three vectors we have shown their respective components as dotted lines in the same colour.

But if we look carefully, addition of the x components of the two original vectors gives the x component of the resultant. The same applies to the y components. So if we just added all the components together we would get the same answer! This is another important property of vectors.

Worked Example 12
Adding Vectors Using Components

Question: Lets work through the example shown in Figure 3.3 to determine the resultant.

Answer:

Step 1 :

The first thing we must realise is that the order that we add the vectors does not matter. Therefore, we can work through the vectors to be added in any order.

Step 2 :

Let us start with the bottom vector. If you are told that this vector has a length of 5.385 units and an angle of 21.8o to the horizontal then we can find its components. We do this by using known trigonometric ratios. First we find the vertical or y component:

$$\begin{matrix}\sin \theta & = & \frac{y}{\mbox{hypotenuse}} \\\sin (21.8) & = &\frac{y}{5.385}\\y & = & 5.385 \sin (21.8)\\y & = & 2\end{matrix}$$

Secondly we find the horizontal or x component:

$$\begin{matrix}\cos \theta & = & \frac{x}{\mbox{hypotenuse}} \\\cos (21.8) & = &\frac{x}{5.385}\\x & = & 5.385 \cos (21.8)\\x & = & 5\end{matrix}$$

We now know the lengths of the sides of the triangle for which our vector is the hypotenuse. If you look at these sides we can assign them directions given by the dotted arrows. Then our original red vector is just the sum of the two dotted vectors (its components). When we try to find the final answer we can just add all the dotted vectors because they would add up to the two vectors we want to add.

Step 3 :

Now we move on to considering the second vector. The green vector has a length of 5 units and a direction of 53.13 degrees to the horizontal so we can find its components.

$$\begin{matrix}\sin \theta & = & \frac{y}{\mbox{hypotenuse}} \\\sin (53.13) & = &\frac{y}{5}\\y & = & 5 \sin (53.13)\\y & = & 4\end{matrix}$$

$$\begin{matrix}\cos \theta & = & \frac{x}{\mbox{hypotenuse}} \\\cos (53.13) & = &\frac{x}{5}\\x & = & 5 \cos (53.13)\\x & = & 3\end{matrix}$$

Step 4 :

Now we have all the components. If we add all the x-components then we will have the x-component of the resultant vector. Similarly if we add all the y-components then we will have the y-component of the resultant vector.

The x-components of the two vectors are 5 units right and then 3 units right. This gives us a final x-component of 8 units right.

The y-components of the two vectors are 2 units up and then 4 units up. This gives us a final y-component of 6 units up.

Step 5 :

Now that we have the components of the resultant, we can use Pythagoras' theorem to determine the length of the resultant. Let us call the length of the hypotenuse l and we can calculate its value

$$\begin{matrix}l^2&=&(6)^2 + (8)^2\\l^2&=&100\\l&=&10.\\\end{matrix}$$

The resultant has length of 10 units so all we have to do is calculate its direction. We can specify the direction as the angle the vectors makes with a known direction. To do this you only need to visualize the vector as starting at the origin of a coordinate system. We have drawn this explicitly below and the angle we will calculate is labeled $$\alpha$$.

Using our known trigonometric ratios we can calculate the value of $$\alpha$$

$$\begin{matrix}\tan \alpha & = & \frac{6}{8} \\\alpha & = & \arctan \frac{6}{8}\\\alpha & = & 36.8^o.\end{matrix}$$

Step 6 :

Our final answer is a resultant of 10 units at 36.8o to the positive x-axis.

title: FHSST Physics/Vectors/Components


text:
FHSST Physics/Vectors/Importance

= Do I really need to learn about vectors? Are they really useful? =

Vectors are essential to do physics. Absolutely essential. This is an important warning. If something is essential we had better stop for a moment and make sure we understand it properly.

title: FHSST Physics/Vectors/Importance


text:
FHSST Physics/Vectors/Summary

= Summary of Important Quantities, Equations and Concepts =

Vector: A vector is a measurement which has both magnitude and direction. Displacement: Displacement is a vector with direction pointing from some initial (starting) point to some final (end) point and whose magnitude is the straight-line distance from the starting point to the final point.

Distance: The distance traveled is the length of your actual path.

Velocity: Velocity is the rate of change of displacement with respect to time.

Acceleration: Acceleration is the rate of change of velocity with respect to time.

Resultant: The resultant of a number of vectors is the single vector whose effect is the same as the individual vectors acting together

title: FHSST Physics/Vectors/Summary


text: Template:FHSST Physics/Forces

title: Template:FHSST Physics/Forces


text:
FHSST Physics/Forces

= Forces =

title: FHSST Physics/Forces


text:
FHSST Physics/Forces/TO DO LIST

= `TO DO' LIST =

 * introduce concept of a system for use in momentum- in NIII for instance, also concept of isolated system and external forces
 * add incline plane examples
 * generally `bulk-up' the Newton's Laws sections

title: FHSST Physics/Forces/TO DO LIST


text:
FHSST Physics/Forces/Definition

= What is a force? =

The simplest way to describe force is to say that it is a `push' or a `pull'. The push or pull on an object may cause either deformation or may change the state of motion of the object under consideration. The harder you 'push' or 'pull', the more force you are applying.

If we leave aside the deformation aspects, then force can be considered to produce change in the state of the motion of the object i.e. velocity. We have, though, experienced in real life that a 'push' or 'pull' does not always manifest in the change of motion. The reason is simple. A change in the state of motion requires a net force. For example, if the force is great enough to overcome friction the object being pushed or pulled will move. So long as the forces on an object are balanced (i.e. net force is zero), the state of motion described by "velocity" will remain same.

In fact, the acceleration of a body is directly proportional to the net force acting on it. The word net is important- forces are vectors and what matters in any situation is the vector sum of all the forces acting on an object.

The unit of force is the newton (symbol N)

 * Force was first described by Archimedes. Archimedes of Syracuse (circa 287 BC - 212 BC), was a Greek mathematician, astronomer, philosopher, physicist and engineer. He was killed by a Roman soldier during the sack of the city, despite orders from the Roman general, Marcellus, that he was not to be harmed.

EQUILIBRIUM OF FORCES. DEFINITION-The Equilibrium of forces is the single fore required to produce an equilibrium When only three forces act on an object this closed figure is a triangle. This leads to the triangles law for three forces in equilibrium.

title: FHSST Physics/Forces/Definition


text:
FHSST Physics/Forces/Diagrams

= Force diagrams =

The resultant force acting on an object is the vector sum of the set of forces acting on that one object. It is very important to remember that we consider all the forces that act on the object under consideration - not the forces that the object might, in turn, apply on other objects.

The easiest way to determine this resultant force is to construct what we call a force diagram. In a force diagram we represent the object by a point and draw all the force vectors connected to that point as arrows. Remember from the Vectors chapter that we use the length of the arrow to indicate the vector's magnitude and the direction of the arrow to show which direction it acts in.

The second step is to rearrange the force vectors so that it is easy to add them together and find the resultant force.

Let us consider an example to get started:

Two people push on a box from opposite sides with a force of 5 N.

When we draw the force diagram we represent the box by a dot. The two forces are represented by arrows, with their tails on the dot.

See how the arrows point in opposite directions and have the same magnitude (length). This means that they cancel out and there is no net force acting on the object.

This result can be obtained algebraically too, since the two forces act along the same line. Firstly we choose a positive direction and then add the two vectors taking their directions into account.

Considering direction towards right as the positive direction

$$\begin{matrix}F_{res} &=& (+5\mbox{ N})+(-5\mbox{ N})\\&=& 0N\end{matrix}$$

As you work with more complex force diagrams, in which the forces do not exactly balance, you may notice that sometimes you get a negative answer (e.g. -2 N). What does this mean? Does it mean that we have something which is opposite of the force? No, all it means is that the force acts in the opposite direction to the one that you chose to be positive. You can choose the positive direction to be any way you want, but once you have chosen it you must stick with it.

Once a force diagram has been drawn the techniques of vector addition introduced in the previous chapter can be implemented. Depending on the situation you might choose to use a graphical technique such as the tail-to-head method or the parallelogram method, or else an algebraic approach to determine the resultant. Since force is a vector, all of these methods apply!

Always remember to check your signs

Worked Example 13 Single Force on a block
Question: A block on a frictionless flat surface weighs 100 N. A 75 N force is applied to the block towards the right. What is the net force (or resultant force) on the block?

Answer:

Step 1 : Firstly let us draw a force diagram for the block:

RIAAN Note image on page 68 is missing

Be careful not to forget the two forces perpendicular to the surface. Every object with mass is attracted to the centre of the earth with a force (the object's weight). However, if this were the only force acting on the block in the vertical direction then the block would fall through the table to the ground. This does not happen because the table exerts an upward force (the normal force) which exactly balances the object's weight.

Step 2 :

Thus, the only unbalanced force is the applied force. This applied force is then the resultant force acting on the block.

title: FHSST Physics/Forces/Diagrams


text:
FHSST Physics/Forces/Equilibrium of Forces

= Equilibrium of Forces =

At the beginning of this chapter it was mentioned that resultant forces cause objects to accelerate. If an object is stationary or moving at constant velocity then either:

 * no forces are acting on the object, or
 * the forces acting on that object are exactly balanced.

A resultant force would cause a stationary object to start moving or an object moving with a given velocity to speed up or slow down or change direction such that the velocity of the object changes.

In other words, for stationary objects or objects moving with constant velocity, the resultant force acting on the object is zero. The object is said to be in equilibrium.

If a resultant force acts on an object then that object can be brought into equilibrium by applying an additional force that exactly balances this resultant. Such a force is called the equilibrant and is equal in magnitude but opposite in direction to the original resultant force acting on the object.

Definition: The equilibrant of any number of forces is the single force required to produce equilibrium.

As an example of an object in equilibrium, consider an object held stationary by two ropes in the arrangement below:

Let us draw a force diagram for the object. In the force diagram the object is drawn as a dot and all forces acting on the object are drawn in the correct directions starting from that dot. In this case, three forces are acting on the object.

Each rope exerts a force on the object in the direction of the rope away from the object.

Since the object has mass, it is attracted towards the centre of the earth. This weight is represented in the force diagram as $$\overrightarrow{W}$$.

Since the object is stationary, the resultant force acting on the object is zero. In other words the three force vectors drawn tail-to-head form a closed triangle:

[In general, when drawn tail-to-head the forces acting on an object in equilibrium form a closed figure with the head of the last vector joining up with the tail of the first vector. When only three forces act on an object this closed figure is a triangle. This leads to the triangle law for three forces in equilibrium:

Triangle Law for Three Forces in Equilibrium: Three forces in equilibrium can be represented in magnitude and direction by the three sides of a triangle taken in order.


text:
User talk:Mreult~enwikibooks

Hi Mreult! Before you put much effort into putting your book here, did you think about putting it to Wikisource: http://wikisource.org/wiki/Author:Isaac_Newton ? If you plan to put the original book here, it definitely should go to Wikisource. See also the comment on Talk:Opticks --Andreas 14:15, 25 Mar 2005 (UTC)

It seems you intended to copy the entire original book here. According to What Wikibooks is not that should go on Wikisource, so you can now find the book at Opticks. If you were intending to annotate this, however, we can leave the pages here (or undelete them if they are already deleted by the time you read this). If you were just intending to convert the book to wiki format, please feel free to continue work at the new Wikisource location. Thank you for your time. :) GarrettTalk 12:14, 1 August 2005 (UTC)

Thanks for everyone who warned me about this. I haven't visit here for a very long time and the book seems to vanished. However, thank you again. --Mreult 09:16, 10 March 2006 (UTC)

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title: User talk:Mreult~enwikibooks