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Question Video: Analysis of the Resolved Components of a Force Mathematics

A force of magnitude 41 N acts due south. It is resolved into two components. Find the magnitudes of 𝐅₁ and 𝐅₂. Give your answer to two decimal places.

03:55

Video Transcript

A force of magnitude 41 newtons acts due south. It is resolved into two components as shown on the diagram. Find the magnitudes of 𝐅 sub one and 𝐅 sub two. Give your answer to two decimal places.

Let’s begin by adding our force of 41 newtons to the diagram. It acts due south, so it’s this one here. That force is itself then resolved into two parts or components. Those are 𝐅 sub one and 𝐅 sub two. Now, you might notice that these two components are given in vector form. But actually, we’re trying to find the magnitudes of these. The magnitude of the vector is essentially its length. And so we want to find these two dimensions here. And what we might try to do is add right-angled triangles to our diagram.

But it’s not particularly easy to do so here since the forces 𝐅 sub one and 𝐅 sub two are not acting at right angles to one another. So instead, we draw something called the parallelogram of forces. We’re drawing lines parallel to 𝐅 sub one and 𝐅 sub two, therefore creating a parallelogram. The force of 41 newtons acting due south then makes up the diagonal of this parallelogram. And so we can then use angle facts to complete some of our missing measurements.

We know that alternate angles are equal, so we can add a 45-degree angle here and a 60-degree angle here. We also know that angles in a triangle add up to 180 degrees. So by subtracting 45 and 60 from 180, we find the missing two angles in our triangles. Let’s enlarge the triangle on the left-hand side, linking the force of 41 newtons and 𝐅 sub two. We have a non-right-angled triangle for which we know the measure of all three angles and one of the sides. This means that to find the measure of a second side, we can use the law of sines.

Labeling our triangle as shown, we see we’re going to link π‘Ž over sin 𝐴 with 𝑏 over sin 𝐡. Substituting what we know about our triangle into the formula, and we see that the magnitude of 𝐅 sub two divided by sin of 60 is equal to 41 over sin of 75. To solve this equation to find the magnitude of 𝐅 sub two β€” notice I’ve written that with bars β€” we multiply by sin of 60. And so the magnitude of 𝐅 sub two is 41 over sin of 75 times sin of 60, and that’s equal to 36.759 and so on. Correct two decimal places then, the magnitude of 𝐅 sub two is 36.76 newtons.

We’re going to repeat this process to find the magnitude of 𝐅 sub one. Now, to do this, we could repeat the process of redrawing the triangle on the right-hand side. However, we know that this side, the side labeled 𝐴𝐡, on the triangle we drew is parallel to the side that represents the vector component 𝐅 sub one. In fact, since this is a parallelogram, it’s also of equal length. And so if we calculate the length of line segment 𝐴𝐡, that will tell us the magnitude of 𝐅 sub one. This time, we’re linking 𝑏 and sin 𝐡 with 𝑐 and sin 𝐢. And so our equation is 41 over sin 75 equals the magnitude of 𝐅 sub one over sin of 45.

To solve for 𝐅 sub one, we’re going to multiply both sides by sin of 45 degrees. And so the magnitude of 𝐅 sub one is 41 over sin of 75 times sin of 45, which is 30.014 and so on. Correct to two decimal places, that’s 30.01. And so the magnitude of 𝐅 sub one is 30.01 newtons and the magnitude of 𝐅 sub two is 36.76 newtons.

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