Video Transcript
The measure of the angle between two forces is 120 degrees, and the magnitude of their resultant is 79 newtons. Find their magnitudes, given that they have a difference of 51 newtons.
Let’s begin by defining each of our forces. Let’s define the forces to be vector quantities 𝐅 sub one and 𝐅 sub two with magnitudes 𝐹 sub one and 𝐹 sub two, respectively. Then we’ll define their resultant to be the vector 𝐑. And since the resultant of two forces is found by calculating their sum, the vector 𝐑 is the sum of vector 𝐅 sub one and vector 𝐅 sub two. Let’s suppose then that vector 𝐅 sub one acts in the positive 𝑥-direction as shown. We’re told that the angle between the forces is 120 degrees. So we can model force 𝐅 sub two as acting in the direction shown. If we now model the initial part of 𝐅 sub two as acting from the end of 𝐅 sub one, then we can add the resultant force to the diagram as shown.
So, let’s look at this as a triangle of forces. The length of each side represents the magnitude of each vector. So, the base of this triangle is 𝐹 sub one, and one of its other sides is 𝐹 sub two. Since the magnitude of the resultant is 79, the third side in this triangle is 79 newtons. Now, there is an extra calculation we can do. We can find the angle, let’s call that 𝜃, between two sides in our triangle. Looking at our original force diagram, we see that we constructed two sides representing force 𝐹 sub two. By very definition, these absolutely must be parallel. This means the angle 𝜃 is supplementary to the angle of 120 degrees. That is, they sum to 180. That means 𝜃 must be equal to 180 minus 120, which is 60 degrees. So, we have a triangle of forces with an included angle of 60 degrees and one side of 79 newtons.
We can use the law of cosines to link all of these. Defining the angle of 60 degrees to be angle capital 𝐴 and the side opposite to be lowercase 𝑎, the law of cosines tells us that 𝑎 squared equals 𝑏 squared plus 𝑐 squared minus two 𝑏𝑐 cos 𝐴. We then substitute everything we know about our triangle into this formula. We get 79 squared equals 𝐹 sub one squared plus 𝐹 sub two squared minus two 𝐹 sub one 𝐹 sub two cos 60. Now, 79 squared is 6241, whilst cos 60 is one-half. So, this simplifies to 6241 equals 𝐹 sub one squared plus 𝐹 sub two squared minus 𝐹 sub one times 𝐹 sub two.
Now we’ll have a problem because we have an equation involving two unknowns. So, we need to use one final fact, and that is that the magnitude of the forces have a difference of 51 newtons. Now we don’t know which is the greater force. So, we make an assumption. We’ll assume that 𝐹 sub one is, so 𝐹 sub one minus 𝐹 sub two equals 51. We can then rearrange and say that that means that 𝐹 sub one equals 51 plus 𝐹 sub two. With this in mind, we can now substitute 51 plus 𝐹 sub two for each instance of 𝐹 sub one in our previous equation. This will result in a quadratic equation purely in 𝐹 sub two, which we can then solve. The right-hand side will become 51 plus 𝐹 sub two squared plus 𝐹 sub two squared minus 51 plus 𝐹 sub two times 𝐹 sub two. And then we distribute our parentheses as shown.
We simplify first by noticing that 𝐹 sub two squared minus 𝐹 sub two squared is zero. And so we get the equation 6241 equals 2601 plus 51𝐹 sub two plus 𝐹 sub two squared. Finally, let’s subtract 6241 from both sides, and we’re now ready to solve this equation. We can factor the right-hand side to get 𝐹 sub two minus 40 times 𝐹 sub two plus 91. Then, setting each expression inside each pair of parentheses to zero and solving, we find 𝐹 sub two equals 40 or 𝐹 sub two equals 91. Remember though, 𝐹 sub two is the magnitude of this force, so it must be positive. And so we choose 𝐹 sub two equals 40.
With that in mind, we can now find the value of 𝐹 sub one. We go back to our earlier equation 𝐹 sub one equals 51 plus 𝐹 sub two. Since 𝐹 sub two is 40, 𝐹 sub one is 51 plus 40, which is equal to 91. And so we found the magnitudes of the two forces in question. They are 40 newtons and 91 newtons.
