Lesson Video: Resolution of Forces | Nagwa Lesson Video: Resolution of Forces | Nagwa

# Lesson Video: Resolution of Forces Mathematics

In this video, we will learn how to solve problems about the resolution of a force into two directions.

17:39

### Video Transcript

In this lesson, we’ll learn how to solve problems about the resolution of a force in two directions. We might recall that a force is an interaction that can change the motion of an object. We often describe them as push or pull forces, but there’s much more to them than that. We’re primarily interested in how we can take a single force and resolve it into component — that means part — forces. These components are often at right angles to one another, but occasionally they might not be. And so we’re going to use right angle trigonometry in addition to non-right angle trigonometry to help us to do this. We’re going to begin with a very simple example to see what this process might look like.

Resolve a force of 81 newtons into two perpendicular components, 𝐹 sub one and 𝐹 sub two, as shown in the figure. Give your answer correct to two decimal places.

And then we have a diagram that shows a force of 81 newtons acting at an angle of 54 degrees to the horizontal. We’re also told that 𝐹 sub one and 𝐹 sub two, which are components of our 81-newton force, are perpendicular. This of course means that they meet at an angle of 90 degrees, at a right angle. So how does this help us to calculate the values of 𝐹 sub one and 𝐹 sub two?

Well, a nice method that we have is to add in a right-angled triangle to our diagram. Since the two vertical lines are parallel and 𝐹 sub two is itself a component of the 81-newton force, we can say that the length of the vertical line that we added to our diagram must be equal to the magnitude of 𝐹 sub two. And so since we have a right-angled triangle for which we know the measure of one of its lengths and its angles, we can use right angle trigonometry.

Our included angle is 54 degrees, and our hypotenuse is 81 newtons. The side opposite to the angle measuring 54 degrees is 𝐹 sub two, and the side adjacent to it is 𝐹 sub one. We’re going to begin by calculating the measure of 𝐹 sub one. Since 𝐹 sub one is the length of the adjacent side and we know the length of the hypotenuse, we’re going to use the cosine ratio here. cos of 𝜃 is adjacent over hypotenuse. And substituting everything we know about our triangle into this formula, we get cos of 54 degrees equals 𝐹 sub one over 81. We’ll solve for 𝐹 sub one by multiplying through by 81. And so 𝐹 sub one is 81 times cos of 54, which is 47.610 and so on. Correct two decimal places then, we see that 𝐹 sub one is 47.61 newtons.

Let’s repeat this process for 𝐹 sub two. This time, we’re trying to find the opposite side in our triangle. And so we’re going to use the sine ratio, where sin 𝜃 is opposite over hypotenuse. Now, since at this point we already knew two of the lengths, we could’ve used the Pythagorean theorem. But we’ll get the same answer either way. This time, we get an equation sin of 54 equals 𝐹 sub two over 81. Again, we solve for 𝐹 sub two by multiplying both sides by 81. So 𝐹 sub two is 81 times sin of 54, which is 65.530 and so on. Correct two decimal places then, 𝐹 sub two is 65.53 newtons. And so we’ve resolved the force of 81 newtons acting at an angle of 54 degrees to the horizontal into two perpendicular components.

Now, if we think about it, it does make sense that the measures of 𝐹 one and 𝐹 two are smaller than 81. When we resolve a force, we’re breaking it into its components, into parts. So 𝐹 sub one and 𝐹 sub two are a part of the 81-newton force. And so they must be lower in value.

In our next example, we’ll look at how to resolve components of a force when they are not acting at right angles to one another.

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 𝑎 over sin 𝐴 with 𝑏 over sin 𝐵. 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 — notice I’ve written that with bars — 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 𝐴𝐵, 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 𝐴𝐵, that will tell us the magnitude of 𝐅 sub one. This time, we’re linking 𝑏 and sin 𝐵 with 𝑐 and sin 𝐶. 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.

Now, a key skill is to be able to take the information in the question and construct your own force diagram. Let’s see what that looks like when we’re working with another parallelogram of forces.

The angle between two forces, 𝑎 one and 𝑎 two, is 75 degrees. Their resultant is 2,900 newtons and makes an angle of 45 degrees with 𝑎 one. Find the forces 𝑎 one and 𝑎 two. Give your answers to two decimal places.

Before we do anything, we’re going to begin by drawing a force diagram. Here are our two forces 𝑎 one and 𝑎 two, which meet at an angle of 75 degrees. Then, we have this resultant force of 2,900 newtons that makes an angle of 45 degrees with 𝑎 one. Remember, the resultant force is just the single force that we get by combining a system of forces, so here 𝑎 one and 𝑎 two. Our job is to find the forces 𝑎 sub one and 𝑎 sub two. And so we’re going to add a couple of things to our diagram.

Firstly, we add in lines parallel to those representing the forces 𝑎 one and 𝑎 two. This gives us a parallelogram. And so we know that not only are the opposite sides parallel, but they’re also equal in length. And that’s going to be useful later down the line. Then we use the fact that alternate angles are equal. And this angle must be 45 degrees. Then, by subtracting 45 from 75, we get two angles of 30 degrees. The third angle is found by subtracting 45 and 30 from 180 degrees, since there are 180 degrees in a triangle. And so we see that the third angle in both our triangles is 105 degrees.

Let’s separate out our triangles. And we see that we’re dealing with a pair of identical non-right-angled triangles. We have a length of 2,900 in both triangles. And then we have 𝑎 sub one here. And since we said that opposite sides in a parallelogram are parallel and equal in length, we have 𝑎 sub two over here.

Since we know the measure of all the angles in our triangle and one of its lengths, we can use the law of sines to find the other two. Let’s label our triangle as shown. And to begin with, we’ll work out the value of 𝑎 sub one. This means we want 𝑎 over sin 𝐴 equals 𝑐 over sin 𝐶. Substituting everything we know about our triangle into this formula gives us 𝑎 sub one over sin of 30 degrees equals 2,900 over sin of 105. We’re going to multiply both sides by sin of 30 to find the value of 𝑎 sub one. And so 𝑎 sub one is 2,900 over sin of 105 times sin of 30, which correct to two decimal places is 1,501.15. So 𝑎 sub one is 1,501.15 newtons.

Let’s clear some space and perform the same process to find 𝑎 sub two. This time, we’re going to use 𝑏 over sin 𝐵 equals 𝑐 over sin 𝐶. Substituting into the formula, and we see that we can solve for 𝑎 sub two by multiplying both sides by sin of 45. And 𝑎 sub two then is 2,900 over sin of 105 times sin of 45. And that, correct to two decimal places, is 2,122.95. 𝑎 sub one is 1,501.15 newtons, and 𝑎 sub two is 2,122.95 newtons.

In our final example, we’ll look at how we can resolve the weight of the body that’s resting on an inclined plane into two components in two directions.

A body weighing 72 newtons is placed on a plane that is inclined at 45 degrees to the horizontal. Resolve its weight into two components 𝐹 sub one and 𝐹 sub two, where 𝐹 sub one is the component in the direction of the plane and 𝐹 sub two is the component normal to the plane.

Being able to model bodies resting on a plane is going to be really important when it comes to working with Newton’s laws of motion. So to see how it works, we’re going to begin by drawing a sketch of the body resting on a plane. Here is a body resting on a plane inclined at an angle of 45 degrees. The body has a weight of 72 newtons. And that’s essentially the downwards force that it exerts on the plane due to gravity.

Now, usually, we’d look to add any extra forces to the diagram. But actually, we’re just interested in resolving this 72-newton force into two components, 𝐹 sub one and 𝐹 sub two. We’re actually told that 𝐹 sub one is the component in the direction of the plane. Essentially, it’s the component of the weight that acts parallel to the plane. Then, 𝐹 sub two is the component normal to the plane. It’s the component of our weight that’s perpendicular to the plane. And so we can add these two forces in as shown.

Now, in fact, you might be used to seeing 𝐹 sub one here. This is parallel and of the same length as the line that we’ve drawn on our triangle. And so 𝐹 sub one will be the same value in either position. Now, in fact, 𝐹 sub one and 𝐹 sub two are perpendicular to one another. And so we have a right-angled triangle in which we know the length of the hypotenuse. That’s 72 newtons. This means then that if we can find the measure of one other angle in our triangle, we can use right angle trigonometry to resolve our weight into the components 𝐹 sub one and 𝐹 sub two. In other words, we can find expressions and values for 𝐹 sub one and 𝐹 sub two.

Now, to find another angle then, we’re going to draw a slightly larger right-angled triangle. In this triangle, we see that we have a right angle at the point where the body meets the plane. We also have an angle of 45 degrees. This actually means we have an isosceles right triangle. But we could also work out the other angle by calculating 180 minus 90 minus 45. We see that this angle here is 45 degrees.

Now, we know that the 72 newtons acts vertically downwards. And so this force must be acting at an angle of 90 degrees to the horizontal. This means we can work out this angle here. We have a right-angled triangle with one angle of 90 and another of 45. And so 180 minus 90 minus 45 is 45 degrees. And so we now know one other angle in the right-angled triangle we were interested in. In fact, it will always be the case that the angle at which the plane is inclined to the horizontal will be the same as this angle here. So let’s enlarge the triangle we’re looking at to make it a little bit easier to work out what’s happening.

We now have a right-angled triangle for which we know one of the sides and the measure of one of its angles. We’re going to use right angle trigonometry to find the value of 𝐹 one and 𝐹 two. This 72-newton length is the hypotenuse. Then, 𝐹 sub one is the opposite, and 𝐹 sub two is the adjacent. To find 𝐹 sub one, we’ll use the sine ratio. We know that 𝐹 sub one is the opposite, and we have the hypotenuse. We can therefore say that sin of 45 degrees must be equal to 𝐹 sub one over 72. And to find the value of 𝐹 sub one, we’ll multiply both sides by 72. That gives us that 𝐹 sub one is 72 times sin of 45.

But actually, we can quote that sin of 45 is root two over two. So this becomes 72 root two over two, which is 36 root two, and that’s in newtons. We then have two possible options for calculating the value of 𝐹 sub two. We could use the cosine ratio. If we went down this route, we’d find that 𝐹 sub two is 72 times cos of 45, which is also 36 root two or 36 root two newtons. Alternatively, we might have spotted that we were working with an isosceles right-angled triangle. And so 𝐹 sub one and 𝐹 sub two absolutely had to be the same length. 𝐹 sub one and 𝐹 sub two are 36 root two newtons.

We might even choose to use the Pythagorean theorem to double check our answer. By summing the squares of 36 root two and 36 root two and then finding the square root of that value, we’d then get a value of 72 newtons as expected. Now, we can generalize this idea a little bit, although it is useful to be able to see where these values come from. For a body resting on a plane inclined at an angle of 𝜃 to the horizontal with a weight of 𝑤 newtons, the component of the weight that acts parallel to the plane will always be 𝑤 sin 𝜃, whereas the component of the weight that acts perpendicular to the plane will always be 𝑤 cos 𝜃.

Let’s now recap the key points from this lesson. In this video, we saw that to resolve a force into component parts where those components are perpendicular to one another, we can add right-angled triangles to our diagram and use right angle trigonometry. This process is particularly useful when working with bodies on inclined planes. When this isn’t possible, however, that is, when the components are not at right angles to one another, we use a diagram called a parallelogram of forces. Then, we use angle facts and the sine rule to help us calculate the individual components.