Lesson Video: Triangle of Forces

Video Transcript

In this lesson, we’ll learn how to solve problems about the equilibrium of a particle under the action of three forces using the triangle of forces method. By this stage, you’ll probably have worked with systems of forces fairly extensively using the Pythagorean theorem and right-angled trigonometry to find the magnitude and direction of these forces and also to split them into their component parts. You might recall that a particle is in equilibrium if the vector sum of these forces, in other words, their resultant, is equal to zero. We’re now interested in what happens when there are exactly three forces acting at a point.

The key point to remember here is that when three coplanar forces, in other words, three forces that lie in the same two-dimensional plane, acting at a point are in equilibrium, they can be represented in magnitude and direction by the adjacent sides of a triangle taken in order. With that process complete, we can then use facts about triangles such as trigonometry and similar triangles to solve problems involving these forces. For instance, let’s imagine we have three forces 𝐹 sub one, 𝐹 sub two, and 𝐹 sub three acting at a point as shown. We can represent these alternatively using a triangle, remembering to take them in order and consider the direction of each force. We begin by adding the force 𝐹 sub one as the first side in our triangle.

Then we move on to force 𝐹 sub two. This begins at the terminal or endpoint of 𝐹 sub one and travels in the same direction. Finally, we repeat this process with the third force, 𝐹 sub three. This time, that begins at the terminal point of 𝐹 sub two and travels in the same direction as in our first diagram. And there we have it, a triangle of forces. We’re now going to see how we can solve problems using this triangle of forces method.

Three coplanar forces 𝐹 sub one, 𝐹 sub two, and 𝐹 sub three are acting on a body in equilibrium. Their triangle of forces forms a right triangle as shown. Given that 𝐹 sub one is equal to five newtons and 𝐹 sub two is equal to 13 newtons, find the magnitude of 𝐹 sub three.

Remember, when three coplanar forces acting at a point are in equilibrium, they can be represented in magnitude and direction by the adjacent sides of a triangle taken in order. We actually have the triangle of forces drawn for us, and we know the magnitude of two of the forces. 𝐹 sub one is equal to five newtons and 𝐹 sub two is equal to 13 newtons. This triangle now represents the relative magnitude of each of our forces. And since it forms a right triangle, we can find the magnitude of the third force by using the Pythagorean theorem. This tells us that, in a right triangle, the sum of the squares of the two shorter sides is equal to the square of the hypotenuse. If we let the hypotenuse be equal to 𝑐, then we say that 𝑎 squared plus 𝑏 squared equals 𝑐 squared.

In this case, the longest side in our triangle is the side represented by the 13-newton force. And so, using the magnitudes we’ve been given and letting the magnitude of 𝐹 sub three be equal to 𝑏 newtons, we can say that five squared plus 𝑏 squared equals 13 squared. That is, 25 plus 𝑏 squared equals 169. And subtracting 25 from both sides of this equation, we find 𝑏 squared is equal to 144. To solve for 𝑏, we simply need to find the square root of both sides of this equation. Now, usually we would find both the positive and negative square root of 144. But since this represents a magnitude, we know it absolutely must be positive. And so 𝑏 is equal to the square root of 144, which is 12. Given that 𝐹 sub one is five newtons and 𝐹 sub two is 13 newtons then, we can say the magnitude of 𝐹 sub three is 12 newtons.

In our next example, we’ll look at how to take three coplanar forces acting at a point and form a triangle of forces.

In the figure, three forces of magnitudes 𝐹 sub one, 𝐹 sub two, and 𝐹 sub three newtons meet at a point. The lines of action of the forces are parallel to the sides of the right triangle. Given that the system is in equilibrium, find the ratio of 𝐹 sub one to 𝐹 sub two to 𝐹 sub three.

We know that when three coplanar forces acting at a point are in equilibrium, they can be represented in magnitude and direction by the adjacent sides of a triangle taken in order. So we’re going to begin by representing the three forces in our question using a triangle. We’re going to take these in order, so let’s begin with force 𝐹 sub one. Then, 𝐹 sub two is perpendicular to 𝐹 sub one. So we can add that force to our diagram, noting that we must start at the terminal point of 𝐹 sub one. Then, we add 𝐹 sub three starting at the terminal point of 𝐹 sub two to complete our triangle. This is a right triangle since we said that 𝐹 sub one and 𝐹 sub two are perpendicular to one another.

We might also notice that the force 𝐹 sub one is parallel to the side in our original triangle measuring 87 centimeters. 𝐹 sub two is colinear to the side measuring 208.8 centimeters. And then there’s a shared side represented by this 𝐹 sub three force. Since this is the case, we can say that the two triangles, that is, the force triangle and the one whose dimensions we know, must be similar. They’re proportional to one another. We can therefore say that the magnitudes of the forces in our triangle of forces will be directly proportional to the lengths of the respective sides in that original triangle.

And so to find the ratio of 𝐹 sub one to 𝐹 sub two to 𝐹 sub three, we’re going to find the ratio of the lengths of the sides in this triangle. Let’s find the length of the third side then. We’ll label it 𝑥 centimeters. Since this is a right triangle, we can use the Pythagorean theorem to find the length of 𝑥. For a right triangle whose longest side is 𝑐 units, the Pythagorean theorem says that 𝑎 squared plus 𝑏 squared equals 𝑐 squared. In this case, our hypotenuse is 𝑥 centimeters. So the Pythagorean theorem gives us 87 squared plus 208.8 squared equals 𝑥 squared. Evaluating the left-hand side of this equation and we find that that’s equivalent to 51166.44. To find the value of 𝑥 then, we find the square root of both sides. The square root of 51166.44 is 226.2. And so the length of the third side in our triangle is 226.2 centimeters.

Remember though, we’re trying to find the ratio of the forces 𝐹 sub one to 𝐹 sub two to 𝐹 sub three. And we said that that will be the same as the ratio of the relevant sides. Listing these in the relevant order and we find the ratio of 𝐹 sub one to 𝐹 sub two to 𝐹 sub three to be equivalent to 87 to 208.8 to 226.2. Dividing each of these numbers by a rather unusual shared factor, that’s 17.4, and we get five to 12 to 13. Alternatively, if we had calculated 87 divided by 226.2 and 208.8 divided by 226.2, we would have found five thirteenths and twelve thirteenths, respectively. The ratio of 𝐹 sub one to 𝐹 sub two to 𝐹 sub three is five to 12 to 13.

In this example, we had some diagram to go from. In our next example, we’ll need to draw the triangle of forces completely.

A body is under the effect of three forces of magnitudes 𝐹 sub one, 𝐹 sub two, and 36 newtons, acting in the directions of line segments 𝐴𝐵, 𝐵𝐶, and 𝐴𝐶, respectively, where triangle 𝐴𝐵𝐶 is a triangle such that 𝐴𝐵 equals four centimeters, 𝐵𝐶 equals six centimeters, and 𝐴𝐶 equals six centimeters. Given that the system is in equilibrium, find 𝐹 sub one and 𝐹 sub two.

We know that when three coplanar forces acting at a point are in equilibrium, they can be represented in magnitude and direction by the adjacent sides of a triangle taken in order. So we’re going to represent the three forces given using a triangle. But we’re told that they act in the directions of the various sides of triangle 𝐴𝐵𝐶. So we’ll sketch triangle 𝐴𝐵𝐶 first. Triangle 𝐴𝐵𝐶 looks a little something like this, and we notice that the sides 𝐴𝐶 and 𝐵𝐶 are both six centimeters in length. So it’s actually an isosceles triangle.

We’ll now use this triangle to sketch a triangle of forces representing 𝐹 sub one, 𝐹 sub two, and the 36-newton force. The force with magnitude 𝐹 sub one newtons acts in the direction of line segment 𝐴𝐵. Then the force with magnitude 𝐹 sub two acts in the direction of line segment 𝐵𝐶. Notice that the force with magnitude 𝐹 sub two begins at the terminal point of our previous force. And so we have to begin our third force at the terminal point of 𝐹 sub two. But we were told that this 36-newton force acts in the direction of the line segment 𝐴𝐶, not the line segment 𝐶𝐴. However, since we know that the magnitude of this force is 36 newtons, we can label it as shown. If we were considering the direction of the force, we would need to consider that this would be the negative direction of our original force. But for magnitudes which just represent size, this is absolutely fine.

We’re now ready to compare our triangles. Since each of our forces acts in the same direction as each side in our triangle 𝐴𝐵𝐶, the two triangles must in fact be similar. And so we can say that the magnitudes of each of our forces must be directly proportional to the lengths of the sides in triangle 𝐴𝐵𝐶. So we can find force 𝐹 sub two really easily. We know that the sides 𝐴𝐶 and 𝐵𝐶 are equal in length. So this force and this force must be equal in magnitude. And so 𝐹 sub two must be equal to 36 newtons. And then we have two different ways that we can calculate the magnitude 𝐹 sub one.

One way is to say that the ratio of the line segment 𝐴𝐵 to the ratio of line segment 𝐴𝐶 will be equal to the ratio of the magnitude 𝐹 sub one to the magnitude 36 newtons. In other words, four divided by six will give us the same outcome as 𝐹 sub one divided by 36. And whilst we could simplify the fraction four-sixths, it doesn’t make a lot of sense to do this because we’re going to multiply both sides of this equation by 36. Then we spot that 36 and six have a common factor of six. So 𝐹 sub one will be equal to six times four over one, which is simply equal to 24. And so 𝐹 sub one is 24 newtons. It’s worth noting at this point that we could’ve used scale factor to calculate the value of 𝐹 sub one.

Since the two triangles are similar, we can deduce that one is an enlargement or a dilation of the other. And thus, the scale factor for enlargement would be 36, that’s one of the dimensions on our force triangle, divided by six, the corresponding dimension on triangle 𝐴𝐵𝐶. 36 divided by six is six. And so we can transform any measurement on our triangle 𝐴𝐵𝐶 onto the dimensions of our force triangle by multiplying by six. This means 𝐹 sub one would be equal to four times six, which is once again 24. 𝐹 sub one is 24 newtons and 𝐹 sub two is 36 newtons.

In our final example, we’ll look at how to use a similar technique to solve a problem involving tension.

A uniform rod of length 50 centimeters and weight 143 newtons is freely suspended at its ends from the ceiling by means of two perpendicular strings attached to the same point on the ceiling. Given that the length of one of the strings is 30 centimeters, determine the tension in each string.

We’re going to begin by drawing a free-body diagram of this scenario. Let’s define the ends of our rod to be 𝐴 and 𝐵. And so we have our rod suspended from two perpendicular strings, one of which measures 30 centimeters. Now, in fact, we can work out the measurement of the second piece of string. It’s 40 centimeters. And we can calculate that using the Pythagorean theorem or just recognizing that we have a multiple of a Pythagorean triple — three, four, five. Since the rod is uniform, we can say that the downwards force of its weight acts at a point exactly halfway along the rod. Now, since the rod exerts a downwards force on the pieces of string, there will be an opposite reaction force of tension. Let’s call that 𝑇 sub 𝐴 for the tensional force in the first bit of string and 𝑇 sub 𝐵 for the tensional force in the second.

We have our free-body diagram, but there’s still an awful lot going on here. So how do we simplify it further? Well, we know that when three coplanar forces acting at a point are in equilibrium, they can be represented in magnitude and direction by the adjacent sides of a triangle taken in order, since the systems in equilibrium will lay the force vectors end to end to make a triangle. 𝑇 sub 𝐴 and 𝑇 sub 𝐵 will be perpendicular to one another, since the pieces of string along which they run are also perpendicular to one another. But in fact, with a little bit of inspection, we can see that these are similar triangles. We’ll define the angle between the rod and the 30-centimeter piece of string to be 𝑥 degrees. This is equal to the angle between the 143-newton force and the tensional force at 𝐵.

This isn’t hugely intuitive, but we can convince ourselves that this is true by adding a line parallel to the 143-newton force on our first diagram and then using the fact that angles in a triangle sum to 180. Since the angles in our force triangle and the angles in the triangle represented by the lengths of each item are equal, we know these triangles are similar. And so we can use scale factor or ratios to find the magnitudes of 𝑇 sub 𝐴 and 𝑇 sub 𝐵. Let’s consider the ratio of 𝑇 sub 𝐴 to 143. This must be equal to the ratio of the 40-centimeter length to the 50-centimeter length. And we chose these pairs of sides in both our diagrams because we’re interested in the pairs of sides that are adjacent to the angle measuring 90 minus 𝑥 degrees.

Multiplying both sides of this equation by 143 and we find that 𝑇 sub 𝐴 is 40 over 50 or four-fifths times 143, which is 114.4. So the magnitude of 𝑇 sub 𝐴 is 114.4 newtons. In a similar way, the ratio of 𝑇 sub 𝐵 to 143 newtons will be equal to the ratio of the 30-centimeter length to the 50-centimeter length. To solve for 𝑇 sub 𝐵 once again, we multiply by 143, which gives us 85.8 or 85.8 newtons. The tensional forces in each string are 85.8 newtons and 114.4 newtons.

We’ll now recap the key points from this lesson. We’ve seen that when three coplanar forces acting at a point are in equilibrium, they can be represented in magnitude and direction by the adjacent sides of a triangle taken in order. Once we have this triangle of forces, we can solve problems by using similarity, the Pythagorean theorem, and even trigonometry.