Question Video: Analysis of Two Coplanar Forces Acting at a Point to Produce a Resultant Mathematics

Two forces of magnitudes 6๐น and 12๐น act at a point. The magnitude of their resultant is 9๐น. Find the measure of the angle between them, giving your answer to the nearest minute.

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Video Transcript

Two forces of magnitudes six ๐น and 12๐น act at a point. The magnitude of their resultant is nine ๐น. Find the measure of the angle between them, giving your answer to the nearest minute.

Letโ€™s begin by imagining what these two forces might look like. They act at some point ๐‘ƒ. So letโ€™s suppose the force of magnitude six ๐น acts in the horizontal direction as shown. The force of magnitude 12๐น acts at the same point, but it must act in a different direction. We might assume it acts approximately in the direction shown. Weโ€™re trying to find the measure of the angle between these two forces. Letโ€™s call that ๐œƒ. Now, in order to find the value of ๐œƒ, we need to use the fact that the magnitude of the resultant of these two forces is nine ๐น.

Remember, the resultant of the two forces is simply their sum. But that doesnโ€™t mean we add six ๐น and 12๐น. These are their magnitudes. So weโ€™re going to instead redraw our diagram as a triangle of forces. To do so, we begin with the six-๐น force as before. We then model the force of magnitude 12๐น so that it starts at the terminal point of the force of magnitude six ๐น. Since the resultant is the sum of the forces, the resultant can be modeled as starting at the initial point of the six-๐น force and finishing at the terminal point of the force of magnitude 12 newtons as shown.

We now notice that this is a non-right triangle and we know the lengths, which are the magnitudes of each force of all three sides. We can therefore find the measure of the angle between the six-๐น force and the 12๐น force in this triangle. Letโ€™s call that ๐›ผ. We can then use information about angles in parallel lines to find the value of ๐œƒ. So how do we find the value of ๐›ผ? Well, letโ€™s label our triangle as shown. Since we know all three sides and weโ€™re trying to find an angle, we can use the law of cosines. We can write that as cos ๐ด equals ๐‘ squared plus ๐‘ squared minus ๐‘Ž squared over two ๐‘๐‘. Since capital ๐ด is angle ๐›ผ, the left-hand side is simply cos ๐›ผ.

Then, we substitute everything we know about each side in our triangle. On the right-hand side, we then get 12๐น squared plus six ๐น squared minus nine ๐น squared over two times 12๐น times six ๐น. This becomes 144๐น squared plus 36๐น squared minus 81๐น squared over 144๐น squared, which simplifies even further to 99๐น squared over 144๐น squared. And of course, ๐น is part of the magnitude of our force, so we know it cannot be equal to zero. This means weโ€™re able to divide both the numerator and denominator of our fraction by ๐น squared. So the cosine of our angle ๐›ผ must be equal to 99 over 144.

To find the measure of angle ๐›ผ then, letโ€™s take the inverse or arccos of both sides. The inverse cosine of 99 over 144 is 46.567 and so on. And of course, thatโ€™s in degrees. Now, we wonโ€™t round this just yet because weโ€™re looking to find ๐œƒ, not ๐›ผ. Now, if we go back to our earlier diagrams and compare them both, we see that angle ๐œƒ and angle ๐›ผ can be linked by a pair of parallel lines. In fact, theyโ€™re supplementary angles. This means that ๐œƒ plus ๐›ผ must be equal to 180 degrees. So ๐œƒ is 180 minus ๐›ผ. And of course, weโ€™ve calculated ๐›ผ. So ๐œƒ must be equal to 180 minus that exact value of 46.567 and so on. That gives us 133.432.

Now the question asks us to give our answer to the nearest minute. To do so, we need to extract the decimal part of our answer and multiply it by 60. 0.432 and so on multiplied by 60 is 25.95. Now, of course, correct to the nearest whole number, that will be 26. And so we found the measure of the angle between our two forces. Correct to the nearest minute, itโ€™s 133 degrees and 26 minutes.

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