### Key Questions

- Why are orthogonal coordinate systems useful?
- How do you transform between polar and Cartesian coordinates?

- Why are orthogonal coordinate systems useful?
- How do you transform between polar and Cartesian coordinates?

A coordinate system gives us a frame of reference to describe a system which we would like to analyze. In statics we normally use orthogonal coordinate systems, where orthogonal means “perpendicular.” In an orthogonal coordinate system the coordinate direction are perpendicular to each other and thereby 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.

Points are specified as an ordered pair of coordinate values separated by a *comma* and enclosed in parentheses, \(P = (x,y)\text{.}\)

Similarly, forces and other vectors will be specified with an ordered pair of scalar components enclosed by angle brackets,

\begin{equation*}
\vec{F} = \langle F_x, F_y \rangle
\end{equation*}

.

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.

In this text, points in polar coordinates will be specified as an ordered pair of values separated by a *semicolon* and enclosed in parentheses

\begin{equation*}
P = (r\ ; \theta)
\end{equation*}

.

Angles can be measured in either radians or degrees, so be sure to include a degree sign on angle \(\theta\) if that is what you intend.

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.

\begin{align}
r \amp = \sqrt{x^2 + y^2}\tag{2.3.1}\\
\theta \amp = \tan^{-1}{\left(\frac{y}{x}\right)}\tag{2.3.2}\\
P \amp = (r \; ; \theta)\tag{2.3.3}
\end{align}

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.

\begin{align}
x \amp = r \cos \theta\tag{2.3.4}\\
y \amp = r \sin \theta\tag{2.3.5}\\
P \amp = (x,y)\tag{2.3.6}
\end{align}

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 \(\vec{A}\)

\begin{align}
A \amp = \sqrt{A_x^2 + A_y^2}\tag{2.3.7}\\
\theta \amp = \tan^{-1}{\left(\frac{A_y}{A_x}\right)}\tag{2.3.8}\\
\vec{A} \amp = (A\; ; \theta)\tag{2.3.9}
\end{align}

\begin{align}
A_x \amp = A \cos \theta\tag{2.3.10}\\
A_y \amp = A \sin \theta\tag{2.3.11}\\
\vec{A} \amp = \langle A_x, A_y \rangle = A \langle \cos \theta, \sin \theta \rangle\tag{2.3.12}
\end{align}

Express point \(P = (-8.66, 5)\) in polar coordinates.

Answer.
Solution 1.
Solution 2.

\(P = (10\; ; 150°)\)

Given: \(x = -8.66\text{,}\) \(y = 5\)

\begin{align*}
r \amp = \sqrt{x^2 +y^2} \amp \theta \amp =\tan^{-1}
\left(\frac{y}{x}\right)\\
\amp = \sqrt{(-8.66)^2 + (5)^2 } \amp \amp =\tan^{-1}
\left(\frac{5}{-8.66}\right)\\
\amp = 10 \amp \amp = \tan^{-1} (- 0.577)\\
\amp \amp \amp = -\ang{30}
\end{align*}

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:

\begin{gather*}
P = (10; \ang{150})
\end{gather*}

Most scientific calculators include handy polar-to-rectangular and rectangular-to-polar functions which can save you time and help you avoid errors. Perhaps you should google your calculator model^{ 1 } to find out if yours does and learn how to use it?

Express \(\N{200}\) force \(\vec{F}\) as a pair of scalar components.

Answer.
Solution 1.
Solution 2.

\begin{gather*}
\vec{F} = \langle \N{-173.2}, \N{-100} \rangle
\end{gather*}

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.

Letting \(\theta = \ang{30}\) we can find the components of \(\vec{F}\) with right triangle trigonometry.

\begin{align*}
F_x \amp = F \cos \theta \amp F_y \amp = F \sin \theta\\
\amp = \N{200} \cos \ang{30} \amp \amp = \N{200} \sin \ang{30}\\
\amp = \N{173.2} \amp \amp = \N{100}
\end{align*}

Since the force points down and to the left into the third quadrant, these values are actually negative, and the signs must be applied manually.

After making this adjustment, the location of \(\vec{F}\) expressed in rectangular coordinates is:

\begin{equation*}
\vec{F} = \langle \N{-173.2}, \N{-100} \rangle
\end{equation*}

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 take care of the signs. You may use either \(\theta = \ang{30} \pm
\ang{180}\text{.}\)

For \(\theta = \ang{-150}\)

\begin{align*}
F_x \amp = F \cos \theta \amp F_y \amp = F \sin \theta\\
\amp = \N{200} \cos (\ang{-150}) \amp \amp = \N{200} \sin( \ang{-150})\\
\amp = \N{-173.2} \amp \amp = \N{-100} \\\\
\vec{F} \amp = \langle \N{-173.2}, \N{-100} \rangle
\end{align*}

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.

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