Two forces 𝐹 sub one equals 39 newtons and 𝐹 sub two equals 39 root two newtons act at a point. The measure of the angle between them is 45 degrees. Determine 𝑅, the magnitude of their resultant, and find 𝜃, the measure of the angle between their resultant and 𝐹 sub one. Give your answer to the nearest minute.
Let’s begin by sketching a diagram showing the two forces 𝐹 sub one and 𝐹 sub two acting at a point. Let’s suppose 𝐹 sub one acts in the direction of the positive 𝑥-axis as shown. Then, we can construct 𝐹 sub two at an angle of 45 degrees. We then have their resultant force. If we think about each of these forces as being vector quantities, the resultant vector 𝐑 is the sum of the vector 𝐅 sub one and the vector 𝐅 sub two as shown. And what this means is we can construct a triangle of forces using 𝐹 sub one, 𝐹 sub two, and 𝑅. Then, since we now have a pair of parallel lines, we can find the angle between 𝐹 sub one and 𝐹 sub two — let’s call that 𝜃 — by recognizing that 𝜃 and 45 degrees are supplementary. They sum to 180 degrees. So 𝜃 is 180 minus 45, which is 135. It’s 135 degrees.
If we think about the sides of the triangle as simply representing the magnitude of each force, we see we can use non-right triangle trigonometry to find the magnitude of 𝑅. In this case, the law of cosines can be used. Labeling the side that we are trying to find lowercase 𝑎, we find that 𝑎 squared equals 𝑏 squared plus 𝑐 squared minus two 𝑏𝑐 cos 𝐴. So 𝑅 squared is equal to 39 root two squared plus 39 squared minus two times 39 root two times 39 cos of 135. 39 root two squared plus 39 squared is 4563. Then, cos of 135 is negative root two over two. So the second part simplifies to negative 3042 root two times negative root two over two. This means 𝑅 squared is 4563 plus 3042, which is 7605.
To find the value of 𝑅, which represents a magnitude so must be positive, we take the square root. The square root of 7605 is 39 root five. So the value of 𝑅, the magnitude of the resultant, is 39 root five newtons.
The next part of this question wants us to find 𝜃, the measure of the angle between the resultant and 𝐹 sub one. Not to be confused with the earlier value of 𝜃 we found, this is the measure of angle 𝐵 in the triangle we’ve just drawn. This time, we can use the law of sines to help us establish the value of 𝜃. We’re going to use sin 𝐵 over 𝑏 equals sin 𝐴 over 𝑎. We’ve just established the value of 𝑎 to be 39 root five. So we get sin 𝜃 over 39 root two equals sin 135 over 39 root five. Next, we multiply by 39 root two. When we do, the 39s cancel, and we find sin 𝜃 is root two sin 135 over root five.
In order to ensure that we’re working with exact values, let’s simply take the inverse or arcsine of both sides of this equation to find the value of 𝜃. When we do, we find 𝜃 is 26.565 and so on. To give this to the nearest minute, we’re going to take the decimal part and multiply by 60. When we do, we get 33.9 and so on. Correct to the nearest whole number then, that’s 34. And so 𝜃 is 26 degrees and 34 minutes. So the value of 𝑅, the magnitude of the resultant, is 39 root five newtons. And the angle between the resultant and 𝐹 sub one, 𝜃, is 26 degrees and 34 minutes.