3 Solving Two Vectors in Perpendicular Axis

It has now come time to add two vectors within the second dimension. Utilizing the basic formulas from previous chapters we can calculate for vector addition. For this scenario we will be calculating within the 𝑥, 𝑦 plane for the simplicity of the reader. Our goal is to calculate the resultant length of the two vectors, and the resultant angle.

If you happen to be currently holding a calculator, I have a few things I wish to communicate. For one, you are a nerd, and I admire your ambition. Second, if you are indeed calculating alongside me, please ensure to keep negative and positive values. The thing that sets vectors apart from scalar values is that vectors have a direction, represented by either a positive or a negative sign in front of the value. These signs indicate where on the calculated axis your line is on. A negative value for your 𝑥 component will mean that your values 𝑥 position will be to the left of the origin^[1], and a positive 𝑥 value will mean the opposite, that it is to the right of the origin.

Our givens we will be working with are:

[latex]\vec{f_1}=23kg[/latex]

[latex]\theta_{1xy}={37}^\circ[/latex]

[latex]\vec{f_2}=39kg[/latex]

[latex]\theta_{2xy}={277}^\circ[/latex]

We cannot simply combine these two values. We need to find the coordinates for the endpoint of each of these lines, and then combine them either in a matrix or by simply adding the values. A matrix will typically be used in more complex problems such as >3D problems, things like CNC toolpath graphing, or Lorentz’s equations. For this reason, we will just be adding the values together.

The first step to solve for our resultant is to solve for the separate directions of the force. We can use sine and cosine to solve for our [latex]f_{1x}[/latex], [latex]f_{1y}[/latex], [latex]f_{2x}[/latex] and [latex]f_{2y}[/latex] by using the exact same method previously discussed. Our values are now:

[latex]f_{1x}=18.36kg[/latex]

[latex]f_{1y}=13.84kg[/latex]

[latex]f_{2x}=4.75kg[/latex]

[latex]f_{2y}=-38.70kg[/latex]

Now that we have these values, we combine them to find the sum of directions. Doing this will give us the 𝑥 and 𝑦 values of our resultant. From here we can plainly use Pythagorean theorem to solve for our resultant.

[latex]{\vec{R}}_{xy}=\sqrt{{(18.36+4.75)}^2{+(13.84-38.70)}^2}[/latex]

[latex]{\vec{R}}_{xy}=33.94kg[/latex]

Now we must calculate the resultant angle. To solve for this, we can either use angles in standard form^[2] or use angles in radians. I do not care to use radians, they are filled with too many unnecessary fractions, therefore we will be utilizing degrees to solve for our resultant angle.

[latex]\theta_{\vec{R}xy}={tan}^{-1}\left(\frac{f_{1y}+f_{2y}}{f_{1x}+f_{2x}}\right)[/latex]

Essentially what we are doing with our 𝑦 and 𝑥 values is breaking down our hypotenuses, also known as our magnitude, into their most basic forms and combining the like terms to solve the equation. When we combine our 𝑦 terms, we are literally combining the like terms of the two vectors, the same is for the 𝑥 values. By plugging our values into the presented equation we end up with our resultant angle as [latex]\theta_{\vec{R}xy}=-{47.08}^\circ[/latex].

For the final step the angle must be converted into its standard form. This is the final step in solving for a resultant. In order to do this, we need to take the angular values from each quadrant and combine it with our resultant angle. We do this because we have calculated the angle relative to its quadrant. Because we always work relative to the 𝑥 axis, we need to calculate accordingly. To solve in quadrant one, we need to add the calculated angle to zero, because positive 𝑥 is our starting point. In quadrant two, we subtract the calculated angle from 180. For quadrant three add the calculated angle to 180, and for quadrant four we subtract the calculated angle from 360. So,

[latex]Q1=0+calculated\ angle[/latex]

[latex]Q2=180-calculated\ angle[/latex]

[latex]Q3=180+calculated\ angle[/latex]

[latex]Q4=360-calculated\ angle[/latex]

Therefore, our angle in standard form is equal to [latex]\theta_{\vec{R}xy}=\ 360-{47.08}^\circ[/latex] which equals to [latex]\theta_{\vec{R}xy}=312.{92}^\circ[/latex].