Video: Resultant of Coplanar Forces

In this video, we will learn how to find the resultant of a group of forces acting on a point.

17:31

Video Transcript

In this video, we’ll learn how to find the resultant of a group of forces acting on a point. Now, we might recall that there are many ways in which forces can be manipulated. And it’s often easier to work with a large complicated system of forces by reducing it to a number of smaller problems. We call this process the resolution of forces or force systems. And it’s one way to simplify what may otherwise seem to be an impossible system of forces acting on a body.

In this video, we’re primarily interested in coplanar force systems. Now, when we talk about this type of force system, we’re talking about a system in which the forces act in just one plane. Now, those forces might be parallel or perpendicular to one another, or even acting at an angle. Now, in those cases, we’ll need to use trigonometry to resolve them into their various components, usually in terms of the horizontal and vertical components. Then, once we’ve done that and defined a positive direction of course, we’re able to find the resultant of our forces. And that’s just the sum of the separate forces.

So we’re going to begin by just looking at an example of how this works. And we’re going to consider how to apply right angle trigonometry to a system of forces given in vector notation.

The resultant of forces 𝐅 sub one equals negative four 𝐢 plus two 𝐣 newtons, 𝐅 sub two equals five 𝐢 minus seven 𝐣 newtons, and 𝐅 sub three equals two 𝐢 plus nine 𝐣 newtons makes an angle 𝜃 with the positive 𝑥-axis. Determine 𝑅, the magnitude of the resultant, and the value of tan of 𝜃.

We’re given three vector forces in this question. These vector forces are given in terms of 𝐢- and 𝐣-components. 𝐢 and 𝐣 are unit vectors, and they act perpendicular to one another. And we generally say that the unit vector 𝐢 acts in the positive 𝑥-direction, whilst the unit vector 𝐣 acts in a positive 𝑦-direction.

We’re actually being asked to calculate some information about the resultant of these forces. And we recall of course that the resultant of a number of forces is simply the vector sum of those forces. So we’re going to add together the vector 𝐅 sub one, 𝐅 sub two, and 𝐅 sub three. We can therefore say that the resultant of our three forces is going to be a single vector in terms of 𝐢 and 𝐣. We calculate this vector by adding negative four 𝐢 plus two 𝐣, five 𝐢 minus seven 𝐣, and two 𝐢 plus nine 𝐣.

Now, of course, to add a number of vectors, we simply add their individual components. So we’ll begin by looking at the 𝐢-components. That’s negative four 𝐢 plus five 𝐢 plus two 𝐢. Negative four plus five plus two is three. So we know the 𝐢-component of our resultant is three. Then we add the 𝐣-components, that’s two, negative seven, and nine, giving us a 𝐣-component of four for our resultant vector. And of course this is in newtons still. And so that’s the resultant of our forces.

But in fact, we were looking to find 𝑅, the magnitude of the resultant. And we might recall that, as a direct result of the Pythagorean theorem, the magnitude of a vector whose 𝐢-component is 𝑎 and whose 𝐣-component is 𝑏 is the square root of the sum of the squares of the individual components. So it’s the square root of 𝑎 squared plus 𝑏 squared. And this, therefore, means that the magnitude of our resultant will be the square root of three squared plus four squared. Well, three squared plus four squared is 25. So we’re finding the positive square root of 25, which is simply five. The resultant vector is in newtons, and so the magnitude of this vector must also be in newtons. And we can, therefore, say that the magnitude of the resultant is five newtons.

We still need to find the value of tan of 𝜃, where 𝜃 is the angle that our resultant makes with the positive 𝑥-axis. Let’s sketch this out and just see what it might look like. Here is our resultant vector. It’s three 𝐢 plus four 𝐣. Since the horizontal component, the 𝐢-component, is three units and the vertical component 𝐣 is four units, we can add a right-angled triangle with the measurements three and four units as shown. Of course, we’ve worked out the resultant, and that’s equal to five newtons. 𝜃 is this angle here. It’s the angle that our resultant, the yellow line, makes with the positive 𝑥-axis.

We have a right-angled triangle for which we actually know three of its sides. And so to find the value of the included angle, we’re going to use right angle trigonometry. We then see that the side adjacent to the included angle is three units and the side opposite to it is four units. We can therefore use the tan ratio to work out the value of 𝜃. Since tan of 𝜃 is opposite over adjacent, here tan of 𝜃 must be equal to four-thirds. And in fact, we’re finished.

We could, if we were required to, work out the value of 𝜃 by finding the inverse or arc tan of both sides of our equation. But we were actually asked to find the value of tan of 𝜃, and so we’re done. The value of 𝑅 is five newtons, and tan of 𝜃 is equal to four-thirds.

So that’s great. We’ve seen how to find the resultant of coplanar forces given in vector form. And we’ve also seen how to calculate bits of information from that. But what do we do if we’re given a diagram and forces acting at an angle? In our next example, we’ll find out.

A body has a force of 10 newtons acting on it horizontally, 25 newtons acting on it vertically upward, and five newtons acting on it at an angle of 45 degrees to the horizontal, as shown in the figure. What is the magnitude of the single resultant force acting on the body and at what angle to the horizontal does it act? Give your answers correct to one decimal place.

As specified in the question, we have three forces we’re interested in. We’ve got 25 newtons acting vertically upwards, 10 newtons acting to the right horizontally, and five newtons acting at an angle. The question is asking us to find some information about the single resultant force acting on the body.

There is a little bit of a problem though. We have coplanar forces. They’re all acting in one plane. And we have two forces perpendicular to one another. But then we have this force acting at an angle. We need to take into account that that force acting at an angle will have components that act horizontally and components that act vertically. So we’re going to split that five-newton force into its horizontal and vertical components. By doing so, we’ll be able to find the resultant of the horizontal components and the resultant of the vertical components. And that’s because the resultant force of two or more forces is the vector sum of those forces.

To resolve our five-newton force into its horizontal and vertical components, we’re going to drop in a right-angled triangle as shown. We’ll call the horizontal component of this force 𝑥, and the vertical component we’ll call 𝑦. The magnitude of this force is five newtons. So that’s actually the length of the hypotenuse in this right-angled triangle. Then we can use trigonometric convention to label our triangle with respect to the 45-degree angle. The side 𝑥 is the side adjacent to our included angle. The side labeled 𝑦 is the side opposite it. And we already saw that we knew the hypotenuse is five newtons.

To calculate the horizontal component 𝑥, we’re going to link the adjacent and hypotenuse by using the cosine ratio. And substituting what we know about our triangle into this formula, we get cos of 45 degrees equals 𝑥 divided by five. To solve for 𝑥, we simply multiply both sides of our equation by five. And we find that 𝑥 is five times cos of 45 degrees. Now, in fact, we know cos of 45 degrees is root two over two. So 𝑥 is five root two over two, and in fact that’s in newtons.

We need to repeat this process to find the value of 𝑦. This time, we use the sine ratio since that links the opposite side with the hypotenuse. And substituting what we know into the formula, we get sin of 45 equals 𝑦 over five. Once again, we multiply both sides of this equation by five. And we find 𝑦 is equal to sin of 45 degrees, and that’s five root two over two newtons.

Now, it’s very interesting that 𝑥 and 𝑦 are the same. And if we go back to our right-angled triangle, we see that we could’ve saved ourselves a little bit of time. It’s a right-angled triangle for which one of the angles is 45 degrees. This means that the third angle, the angle at the top of our triangle here, must also be 45 degrees. And we, therefore, have an isosceles triangle, and 𝑥 and 𝑦 must in fact be the same. And so we know both 𝑥 and 𝑦 are five root two over two.

Our next job is to find the vector sum of all of our forces. To do so though, we’re going to need to define a positive direction. As is convention on the 𝑥𝑦-plane, it makes sense that horizontally to the right would be positive in the 𝑥-direction and vertically upwards would be positive in the 𝑦-direction.

Let’s begin by considering the sum of the forces acting horizontally. We have that 10-newton force, and then we have the horizontal component of our five-newton force. Their sum is simply 10 plus five root two over two. Notice that since the five-newton force is acting to the right and up, we can infer that the horizontal component must be acting in the positive direction. Let’s repeat this process in the vertical direction. We have the 25-newton force and the vertical component of the five-newton force. Once again, this vertical component must be acting upwards, and so it’s positive. And so the sum of the forces in this direction is 25 plus five root two over two. And whilst not entirely necessary, we can choose to write this using vector notation. We see the resultant vector 𝐑 in newtons is 10 plus five root two over two 𝐢 plus 25 plus five root two over two 𝐣.

Now that we know the resultant, we need to work out the magnitude of that resultant and the angle to the horizontal. If a direction isn’t given for the angle, we do assume that we’re taking the angle from the positive horizontal, in other words, the positive 𝑥-axis. Now, remember, the magnitude of a vector is the square root of the sum of the squares of its components. So the magnitude of our force is the square root of 10 plus five root two over two squared plus 25 plus five root two over two squared. That gives us 31.58, which correct to one decimal place is 31.6 newtons. And so we have the magnitude of the resultant. And what about the angle to the horizontal?

Let’s sketch the resultant vector 𝐑. We now know both the horizontal and vertical components of this vector. And so we can use right angle trigonometry to find the angle that the resultant makes with the positive horizontal. Let’s call that 𝜃. Relative to our included angle, we know the opposite and the adjacent sides. And so we use the tan ratio. We find that tan of 𝜃 is 25 plus five root two over two divided by 10 plus five root two over two. And then if we evaluate that fraction, we get 2.108 and so on. To find the value of 𝜃, we simply find the inverse or arc tan of both sides of this equation. So 𝜃 is the inverse tan of 2.108 and so on, which is 64.623. Correct to one decimal place, that’s 64.6. The magnitude of our single resultant force then is 31.6 newtons, and the angle to the horizontal it makes is 64.6 degrees.

At this stage, it’s really important to note that whilst the process for finding the magnitude of the resultant is generally the same, the process for finding the angle it makes with the positive horizontal axis isn’t. If both the horizontal and the vertical components of our resultant are positive, we can simply apply the fact that our angle 𝜃 will be the inverse tan of the 𝐣-component divided by the 𝐢-component. If this is not the case though, it’s really sensible to draw out a diagram and just see where our resultant force lies. We will be able to use the tan ratio, but we might need to do some adding or subtracting from 180.

In our final example, we’ll see how knowing the resultant can help us to find missing values.

Coplanar forces of magnitudes 𝐹 newtons, eight root three newtons, root three newtons, and nine root three newtons act on a particle, as shown in the diagram. Given that the magnitude of their resultant is nine root three newtons, determine the value of 𝐹.

We’re given information about the magnitude of the resultant of four coplanar forces, where of course the resultant is the vector sum of those forces. The problem is, one of those forces is 𝐹, and that’s what we’re trying to find. So what we’re going to do is resolve each of our forces into perpendicular components. And then we’re going to find the vector sum of these and then find the magnitude and equate that to nine root three. So let’s define the direction in which the 𝐹 force is acting to be the positive horizontal direction. And then perpendicular to this and upwards is the positive vertical direction.

Let’s begin by resolving our forces then in a horizontal direction. We know 𝐹 is acting in this direction. But actually, we need to find the horizontal component of the eight root three, nine root three, and root three forces. We add a right-angled triangle to our eight root three force. We know that the horizontal component will be acting to the right and the vertical component this will be acting upwards. And then we can use right angle trigonometry to see that the adjacent side, and that’s the horizontal component, is eight root three cos 60 and the opposite side, the vertical component, is eight root three sin 60.

Let’s repeat this for the root three force. Since angles on a straight line add up to 180 degrees, the included angle this time is 60. And then we see that the side opposite to this, that’s the vertical component acting upwards, is root three sin 60. And the adjacent, which is the horizontal component acting to the left, is root three cos 60. There’s one more triangle that we’re interested in. Finally, we have the nine root three force. We could add a triangle to the left of this, but there’s an awful lot going on there already. So we add a triangle to the right as shown.

We know that this angle here is 60 degrees. So the included angle must be 30. And once again, we resolve it into its horizontal and vertical components. We get nine root three cos of 30 for the vertical component and nine root three sin of 30 for the horizontal component.

Now that we’ve resolved all of these forces, let’s find the horizontal sum. We have 𝐹 acting in a positive direction plus eight root three cos of 60. That’s the component of the eight root three force that acts in the horizontal direction. And then we subtract root three cos of 60 since that’s acting to the left, and nine root three sin of 30. By using the fact that both sin of 30 and cos of 60 are one-half, we simplify this to 𝐹 plus eight root three times one-half minus root three times one-half minus nine root three times one-half. And all that rather complicated working simplifies to just 𝐹 minus the square root of three.

Let’s do the same in the vertical direction. We know that we have eight root three sin 60 acting upwards, so in the positive direction, and root three sin 60. But in the opposite direction, we have nine root three cos of 30. So we’re going to subtract this from our sum. This time, we use the fact that sin of 60 and cos of 30 are root three over two. And so we get eight root three times root three over two plus root three times root three over two minus nine root three times root three over two, which is in fact equal to zero. And that’s really useful because we can actually work out the magnitude of the resultant really easily.

Since our force is only acting in one direction, the magnitude of our resultant must be 𝐹 minus root three. But we were told that the magnitude of the resultant is nine root three. So we can set up a simple equation for 𝐹. 𝐹 minus the square root of three is equal to nine root three. And we’ll solve for 𝐹 by adding root three to both sides. And so we see that 𝐹 is 10 root three or 10 root three newtons.

We’ll now recap the key points from this lesson. In this lesson, we learned that coplanar forces are forces which act in just one plane. And the resultant of a set of coplanar forces is the vector sum of all of those forces. We also saw that if the forces were given in vector form, this was really straightforward. But if they’re not, we need to split them into perpendicular directions, usually 𝑥 and 𝑦. And that allows us to solve problems involving the resultant. Of course, when we do this, we mustn’t forget to define a positive direction.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.