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In this Physics tutorial, you will learn:
Tutorial ID | Title | Tutorial | Video Tutorial | Revision Notes | Revision Questions | |
---|---|---|---|---|---|---|
3.2 | Position, Reference Point |
In the previous Physics Tutorial, "Types of Motion", we described the motion as "the phenomenon in which an object changes its location over time." But, is it always easy to know whether an object is moving or not?
Consider the following examples:
Is it easy to detect any motion in the above examples? Why?
Which action would help you find whether the abovementioned objects are moving or not? Why?
Let's try to explain the concept of reference point by providing answers to the questions posed in the "Introduction" section.
1. The downloading bar seems unmoveable. At first sight, it seems the downloading process has stopped. You can check this by putting the cursor at the end of the orange bar, which shows the downloading progress. After a while, you can check whether the orange bar has moved on the right of the cursor or not. If yes, the downloading process is active. Therefore, the cursor acts as a reference point for you as it helps understanding whether the downloading progress bar is moving or not.
Look at the figure below:
2. The bright star is slightly on the left of pyramid in the first figure. After a while, it "shifts" on the right of the pyramid. In this way, the pyramid acts as a reference point, which helps us to understand whether the stars are in the same place as before or they "have moved".
Look at the figure:
3. In the figure below, the star in the wall helps us understand that the clock is working properly as the minute hand has moved in respect to the star. Therefore, it (the star) acts as a reference point.
In all the above examples, it was clearly outlined the necessity and importance of a reference point from which we start making any measurement or estimation in order to detect any change in object's location.
Thus, by definition, "A fixed point with respect to which a body changes its location is called Reference Point or Origin."
When dealing with calculations in Kinematics, reference point is usually the origin of the coordinate system, no matter whether the system is 1, 2 or 3 dimensional. Reference point (origin) is usually denoted by the number 0 or the letter O.
We can show the location of an object by a finger, or by drawing a small dot in the place where the object lies. If the object is voluminous, we usually put a small dot at the object's centre to show its location as in the figure below.
It is obvious that location does not imply the use of any reference point or coordinate. Therefore, no numerical values are involved when dealing with the location of an object.
On the other hand, Position is a physical quantity that shows how far an object from the origin (reference point) is. Position not only has a magnitude (numerical value) but it also has a direction. It is not the same thing if we say, "the object is 6m on the left of the reference point" and "the object is 6m on the right of the reference point" although the distance from the origin (reference point) is the same in both cases (6 m).
Thus, there are two men in the figure; each of them is at 6m from the tree, which in this case acts as a reference point or origin. Therefore, it is obvious it is not enough knowing only the distance from the reference point but we must know the direction as well. Only then, we can exactly determine the position of a given object.
As we discussed in Physics Tutorial 2.1 "Vectors and Scalars in Physics", a quantity for which direction information is required is known as "vector quantity." This is the case for the position of an object.
By definition, "Position is a vector quantity that shows how far an object is from the origin in a given direction."
Position can be positive, negative or zero. It can be positive when the location of the object is at the positive part of the position axis. For example, the position of the green bicycle shown below is positive as its location is at (+4) m. As for the position of the blue bicycle, it is negative because its location is at (-1) m. (Remember, for voluminous and irregularly shaped objects we consider the actual location at the centre of the object). Position can be zero when the object is located at the reference point.
It is known that in a system of coordinates we can assign a letter to each direction available. Thus, if there is only one direction available (1-D) as shown in the above figure, we denote the axis by the letter x and the object's position by the vector x⃗ or Ox)⃗. We can write
and
When two directions of motion (2-D) are available, we can express the position of an object using a pair or coordinates (one for each direction). We must use two letters (usually x and y) to label the directions. Therefore, we need to know both coordinates to determine the location of an object in 2-D. Look at the figure below:
From the above graph, we can see that the object A is 4m on the right and 2m above the reference point. Therefore, we say the position of the object A is at (4m, 2m) and by this, we understand that the position of object A is represented through the vector
and the object A is
away from the origin in the direction of the vector OA⃗.
Likewise, we can use the same approaches in 3-D (in space) as well. We have another direction added in this case. Usually, it is denoted by the letter z. Therefore, we must write all three coordinates to determine the position of an object. Look at the graph below.
The object A is in the 3 dimensional space. It is diverted 6m from the origin according the x-direction, 5 m from the origin in the y-direction and 6m from the origin in the z-direction, all in the positive direction. This means the components of the vector OA⃗ which represents geometrically the linear distance from the origin, are OAx = 6m, OAy = 5m and OAz = 6m respectively.
From the concept of vector's magnitude, we know that
Therefore, substituting the values, we obtain for the magnitude of the vector OA⃗
Thus, the object A is nearly 10m away from the origin (reference point) in the direction of the vector OA⃗.
Write the position of the objects A, B and C shown in the figure. How far are they from the origin?
The object A is in the xOy plane. It has no z-coordinate, so we need to know only two coordinates to show its position.
From the figure, we can see that Ax = 4m, Ay = 3m (and Az = 0m). Therefore, the position OA⃗ of the object A is
The distance from the origin of the Object A is found by calculating the magnitude of the vector OA⃗. Therefore, using the known procedure explained in the article "Vectors and Scalars" for calculating the magnitude of a vector, we can write
Substituting the values, we obtain for the magnitude of the vector OA⃗
This means the point A is 5m away from the origin in the direction of the vector OA⃗.
The same procedure is used for the other two objects. Thus, for the object B we have only one coordinate Bz = 5m as it lies on the z-axis only. It is not necessary to calculate the magnitude of the vector OB⃗ as it is clear that |OB⃗|= 5m.
As for the object C, we can see from the figure that it contains all three coordinates. Thus, Cx = 3m, Cy = 6m and Cz = 4m. Therefore, the magnitude of the vector OC⃗ which represents the position of the object C (its linear distance from the origin), is
Substituting the values, we obtain for the magnitude of the vector (OC)⃗
Therefore, the object C is 7.81m away from the origin in the direction of the vector OC⃗.
The figure below shows the position vectors for the three objects.
Remark! The 1 and 2 dimensional motions are special cases of the 3 dimensional motion. We can either write 0 in the place of the missing coordinates or simply represent the position in as many coordinates as given. We can illustrate this aspect using the position of the vector B. This position can be mathematically represented in three ways:
In one dimension (according z only)
In two dimensions (according x and z, or y and z)
In three dimensions (according x, y and z)
All these three presentations show the same thing: the vector OB⃗. Hence, they are all equivalent.
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