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To begin with I am working in a high school classroom where the students are working on the applications of vectors. The beginning of the lesson is about calculating direction and magnitude of vectors and follows with an example question that states:

Two ropes are tied to a wagon. a child pulls one with a force of 20 pounds, while another child pulls the other with a force of 30 pounds. If the angle between the two ropes is 28 degrees, how much force must be exerted by a third child, standing behind the wagon, to keep the wagon from moving? Hint: assume the wagon is at the origin and one rope runs along the positive x-axis. Find the resultant forces on the wagon from the ropes.

Now I know that students can solve this by thinking about it in terms of a parallelogram and applying the law of cosines and quadratic equation.

They can also use the magnitude formula $V = \sqrt {x^2+y^2}$ where child 3 is approximately $(47.67, 9.39)$.

Personally it has been a while since I have worked with vectors and can't really think of other ways to approach this although I know there probably are. Is there other methods that would be appropriate to show high school students? This is an advanced math course where they have just finished law of sines and cosines, and end goal of vectors is mostly just airplane problems. Students are not expected to have any background knowledge in physics.

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    $\begingroup$ You can actually find this in the text book Contemporary Precalculus (5th Edition). I don't know if an image would help but link There should be no second angle needed. The wagon is not intended to be moving. The third child is exerting enough force that the wagon doesn't move in any direction. Imagery wise I think of siblings unwilling to share, although I can't tell you why the wagon would have two ropes. $\endgroup$ – Alex Rost Nov 12 '18 at 19:26
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    $\begingroup$ A possible third technique is graphical addition, with a ruler and protractor. You pick a scale, e.g., 1 lb->5 mm, and draw the two vectors tip to tail, then measure the resultant. $\endgroup$ – Ben Crowell Nov 12 '18 at 21:41
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    $\begingroup$ Just take the other two forces and divide them into x and y components using trig. (Only have to do it for the one not on the x axis.) The counteracting force is equal in magnitude but opposite in sign to the the sum of the x components and the sum of y components. [You need to make an assumption on which force is left or right, but this does not affect the problem result if you state the answer with respect to one of the forces.] $\endgroup$ – guest Nov 12 '18 at 21:45
  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – quid Nov 15 '18 at 22:46
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Physically, the problem is to find a third force $\vec F_3$ that balances the two given forces $\vec F_1$ and $\vec F_2$. (This is Newton's Second Law $\vec F_{net}=m\vec a$, where $\vec F_{net}=\vec F_1+\vec F_2+\vec F_3$ and $\vec a=\vec 0$ for equilibrium.) So, $\vec F_3= -(\vec F_1+\vec F_2)$.
In Physics classes, this is an opportunity to practice vector addition.

In your post, you describe using the parallelogram rule [tails-together] to find $(\vec F_1+\vec F_2)$ along the diagonal of the parallelogram. Then $\vec F_3$ is the vector opposite to $(\vec F_1+\vec F_2)$.

As a geometry problem, $\vec F_1+\vec F_2+\vec F_3=\vec 0$ means that the vectors added head to tail form a closed polygon (here, a triangle). You are given have two legs and an exterior angle. Find the missing leg (magnitude and direction)... using trigonometry.

In Physics classes, we usually don't use the closed polygon method.
Instead, we usually draw a Free Body Diagram with the force tails at the origin, then use components:
$$F_{1,x}+F_{2,x}+F_{3,x}=0$$ $$F_{1,y}+F_{2,y}+F_{3,y}=0$$ and try to find unknowns $F_{3,x}$ and $F_{3,y}$.

(I get $\vec F_3=(( -46.488)\hat x+(-14.084)\hat y) \rm{\ lbs}$, then, if needed, find magnitude and direction.)

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I hope this approach isn't too elementary for your question. I taught myself trig during the summer between my junior and senior year in high school, and developed the following approach. For the "airplane problems" you described, I started with a 4-column table:

    r  |   angle  |   x = r cos(theta)  |   y = r sin(theta)
=============================================================
  3 m  |   45 deg |                     |
2.5 m  |  127 deg |                     |  

For the vectors that were given, I worked left-to-right to decompose them into x and y projections:

   r  |   angle  |   x = r cos(theta)  |   y = r sin(theta)
=============================================================
   3m |  45 deg  |  2.12 m             |  2.12 m
2.5 m |  127 deg |  -1.5 m             |  1.99 m

This worked well for any number of vectors (not just two).

Then in the last row I worked backwards from right-to-left:

  • Sum the x and y components separately
  • Let Pythagoras calculate the distance
  • Use the inverse tangent formula to determine the angle
           r  |   angle  |   x = r cos(theta)  |   y = r sin(theta)
     =========|==========|=====================|===================
           3m |  45 deg  |  2.12 m             |  2.12 m
        2.5 m |  127 deg |  -1.5 m             |  1.99 m
     =========|==========|=====================|===================
        4.16 m| 81.4 deg |   0.62 m            |  4.11 m

At first, I paid attention to the quadrant that a vector was in so that I could determine the signs of the x and y projections. But then I realized with delight that my calculator would do that for me. Neither did I need to worry whether I was adding or subtracting vectors. The angle took care of the correct signs (so that some components were negative), so I could just add the columns of positive and negative numbers.

The WWII-era trig book I read would use different axes as the 0 degree reference line. It took examples from navigation, surveying, artillery, and several others I no longer remember. So I couldn't always use the exact same formulas to decompose a vector into its x and y components. That's when I started to write the formulas in the header row of the table: I thought through it (and maybe had to diagram it) just once. But then I could rapidly apply the same formula to all of the rows.

It was all very cut-and-dried. And systematic enough that I still remember the approach today, more than 30 years later. That might be it's greatest strength.

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