A coordinate system provides a reference frame to describe a system we want to analyze. Although there are many different coordinate systems that can be used, selecting an appropriate one will make a problem easier to solve.
In statics, we use orthogonal coordinate systems, where orthogonal means “perpendicular.” The advantage of orthogonal coordinate systems is that the coordinate directions are all perpendicular and, therefore, independent. The intersection of the coordinate axes is called the origin, and measurements are made from there. Both points and vectors are described with a set of numbers called the coordinates. For points in space, the coordinates specify the distance you must travel in each of the coordinate directions to get from the origin to the point in question. Together, the coordinates can be thought of as specifying a position vector, a vector from the origin directly to the point. The position vector gives the magnitude and direction needed to travel directly from the origin to the point.
In the case of force vectors, the coordinates are the scalar components of the force in each of the coordinate directions. These components locate the tip of the vector and they can be interpreted as the fraction of the total force which acts in each of the coordinate directions.
Three coordinate directions are needed to map our real three-dimensional world but in this section we will start with two, simpler, two-dimensional orthogonal systems: rectangular and polar coordinates, and the tools to convert from one to the other.
The most important coordinate system is the Cartesian system, which was named after the French mathematician René Descartes. In two dimensions the coordinate axes are straight lines rotated 90° apart named \(x\text{,}\) and \(y\text{.}\)
In most cases, the \(x\) axis is horizontal and points to the right, and the \(y\) axis points vertically upward, however, we are free to rotate or translate this entire coordinate system if we like. It is usually mathematically advantageous to establish the origin at a convenient point to make measurements from, and to align one of the coordinate axes with a major feature of the problem.
The polar coordinate system is an alternate orthogonal system which is useful in some situations. In this system, a point is specified by giving its distance from the origin \(r\text{,}\) and \(\theta\text{,}\) an angle measured counter-clockwise from a reference direction – usually the positive \(x\) axis.Polar coordinates are useful if we know more about distances and angles from a central point. Examples of systems that would use polar coordinates would be tower cranes to build skyscrapers, robotic arm motion, and radar systems.
You should be able to translate points from one coordinate system to the other whenever necessary. The relation between \((x,y\)) coordinates and \((r;\theta)\) coordinates are illustrated in the diagram and right-triangle trigonometry is all that is needed to convert from one representation to the other.
Luckily, it is quite easy to translate points between the rectangular \((x, y)\) and polar \((r; \theta)\) coordinate systems using right-triangle trigonometry. As you move point \(P\) around the figure below, you can see how the two coordinate types are related.
Take care when using the inverse tangent function on your calculator. Calculator angles are always in the first or fourth quadrant, and you may need to add or subtract 180° to the calculator angle to locate the point in the correct quadrant.
Rectangular To Polar for forces (Given: rectangular components).
If you are working with forces rather than distances, the process is exactly the same but triangle is labeled differently. The hypotenuse of the triangle is the magnitude of the vector, and sides of the right triangle are the scalar components of the force, so for vector \(\vec{A}\)
You must be careful here and use some common sense. The \(\ang{-30}\) angle your calculator gives you in this problem is incorrect because point \(P\) is in the second quadrant, but your calculator doesn’t know this. It can’t tell whether the argument of \(\tan^{-1}(-0.577)\) is negative because the \(x\) was negative or because the \(y\) was negative, so it must make an assumption and in this case it is wrong.
The arctan function on calculators will always return values in the first and fourth quadrant. If, by inspection of the \(x\) and the \(y\) coordinates, you see that the point is in the second or third quadrant, you must add or subtract \(\ang{180}\) to the calculator’s answer.
So in this problem, \(\theta\) is really \(\ang{-30} + \ang{180}\text{.}\) After making this adjustment, the location of \(P\) in polar coordinates is:
Most scientific calculators include handy polar-to-rectangular and rectangular-to-polar functions that can save you time and help you avoid errors. Perhaps you should google your calculator model 1
google.com
to find out if yours does and learn how to use it?
Given: The magnitude of force \(\vec{F} = \N{200}\text{,}\) and from the diagram we see that the direction of \(\vec{F}\) is \(\ang{30}\) counter-clockwise from the negative \(x\) axis.
If you would prefer not to apply the negative signs by hand, you can convert the \(\ang{30}\) to an angle measured from the positive \(x\) axis and let your calculator takes care of the signs. You may use either \(\theta = \ang{30} \pm \ang{180}\text{.}\)
Although this approach is mathematically correct, experience has shown that it can lead to errors and we recommend that when you work with right triangles, use angles between zero and \(\ang{90}\text{,}\) and apply signs manually as required by the physical situation.