6 Solving Many Vectors in the 𝑥 and 𝑦 Axis

Now that we understand the basics of my first two equations, we can try them out. Allow us to run a few questions to deepen your understanding. For the purpose of solving all of the following questions we will be using the new equations and concepts that we have learned.

For this chapter’s question, we will be solving for three vectors utilizing my summed equation. Throughout the chapter I will explain in detail how to effectively use my equations. Our angles can be stated relative to something, such as [latex]30°[/latex] West of North, but for my system all angles will be in standard position relative to positive [latex]x[/latex], moving counterclockwise. Our givens will be as follows:

[latex]\vec{f_1}=10[/latex]

[latex]\vec{f_2}=12[/latex]

[latex]\vec{f_3}=\ 8[/latex]

[latex]\theta_{1xy}={40}^\circ[/latex]

[latex]\theta_{2xy}={176}^\circ[/latex]

[latex]\theta_{3xy}={300}^\circ[/latex]

Based on these, we can now decide on the equation we choose to solve for the addition of the vector magnitudes. The equation we will be using is the most recent summed formula from chapter five. It is important to note that we are choosing this formula due to the fact that we have more than two vectors we are solving for. The use of summations will be a lifesaver for our eyes, and for the amount of paper that needs to be printed. As opposed to using a multitude of formulas to solve for many different amounts of vectors, we can use one summation to do all the work for us.

[latex]{\vec{R}}_{xy}=\sum_{i=1}^{n}\sqrt{\left(f_icos\theta_{ixy}\right)^2+\left(f_isin\theta_{ixy}\right)^2}[/latex]

This is the base equation. We can modify it especially for making it work for the current scenario,

[latex]{\vec{R}}_{xy}=\sqrt{\left(\left(f_1cos\theta_{1xy}\right)+\left(f_2cos\theta_{2xy}\right)+\left(f_3cos\theta_{3xy}\right)\right)^2+\left(\left(f_1sin\theta_{1xy}\right)+\left(f_2sin\theta_{2xy}\right)+\left(f_3sin\theta_{3xy}\right)\right)^2}[/latex]

And now we plug our values into the equation,

[latex]{\vec{R}}_{xy}=\sqrt{\left(\left(10cos40°\right)+\left(12cos176°\right)+\left(8cos300°\right)\right)^2+\left(\left(10sin40°\right)+\left(12sin176°\right)+\left(8sin300°\right)\right)^2}[/latex]

We can solve for the things inside the brackets,

[latex]{\vec{R}}_{xy}=\sqrt{\left(7.660-11.970+4\right)^2+\left(6.427+0.837-6.928\right)^2}[/latex]

And once again we combine the brackets,

[latex]{\vec{R}}_{xy}=\sqrt{\left(-0.31\right)^2+\left(0.336\right)^2}[/latex]

The values we have just calculated are the and values of the resultant. These values can be reused to determine the resultant angle. Our resultant in the and is,

[latex]{\vec{R}}_{xy}=0.4571[/latex]

Excellent. From these values we can now finish the question by solving for the resultant angle. For the equation we will be using, we will choose the equation from the same part as the previously used resultant equation.

[latex]\theta_{\vec{R}xy}=\sum_{i=1}^{n}{{tan}^{-1}\left(\frac{f_isin\theta_{ixy}}{f_icos\theta_{ixy}}\right)}[/latex]

Rearranged:

[latex]\theta_{\vec{R}xy}={tan}^{-1}\left(\frac{\left(f_1sin\theta_{1xy}\right)+\left(f_2sin\theta_{2xy}\right)+\left(f_3sin\theta_{3xy}\right)}{\left(f_1cos\theta_{1xy}\right)+\left(f_2cos\theta_{2xy}\right)+\left(f_3cos\theta_{3xy}\right)}\right)[/latex]

Plugged in is:

[latex]\theta_{\vec{R}xy}={tan}^{-1}\left(\frac{\left({10sin40}^\circ\right)+\left({12sin176}^\circ\right)+\left({8sin300}^\circ\right)}{\left({10cos40}^\circ\right)+\left({12cos176}^\circ\right)+\left({8cos300}^\circ\right)}\right)[/latex]

Now we solve for inside the brackets:

[latex]\theta_{\vec{R}xy}={tan}^{-1}\left(\frac{\left(6.4\right)+\left(0.8\right)+\left(-6.9\right)}{\left(7.6\right)+\left(-11.9\right)+\left(4\right)}\right)[/latex]

This solves as [latex]\theta_{\vec{R}xy}=-{45}^\circ[/latex]. This answer is great, but we can make it better. We need to put this in standard form in order to get the correct answer. To do so we need the quadrant value and combine it with our calculated angle. Due to our [latex]x[/latex] and [latex]y[/latex] values being [latex]-0.3[/latex] and [latex]0.3[/latex] accordingly, we know that our vector resultant is in the second quadrant making our equation [latex]{180}^\circ-{45}^\circ=\theta_{\vec{R}xy}={135}^\circ[/latex]. Of course, this answer is off by a few degrees, subsequently because we are rounding to a single decimal place for visual organization. If you were to plug the formula into your calculator and run it, it would result in a calculated angle of [latex]132.667°[/latex].