Lesson Video: Equivalent System of Forces to a Couple | Nagwa Lesson Video: Equivalent System of Forces to a Couple | Nagwa

Lesson Video: Equivalent System of Forces to a Couple Mathematics

In this video, we will learn how to identify the conditions for a system of coplanar forces to be equivalent to a couple and find its moment.

17:21

Video Transcript

In this video, we’ll learn how to identify the conditions for a system of coplanar forces to be equivalent to a couple, and we’ll find its moment. As we get started, let’s remind ourselves what some of these terms mean.

First, a couple is a pair of forces that act in equal and opposite directions and do not lie along the same line of action. So, for example, if each of these two arrows represents a force vector, we can see because these arrows are the same length, these vectors must have the same magnitude. They also point in opposite directions. And the lines of action of these two forces don’t overlap. Therefore, they form a couple. Another thing we can say about these forces is that they’re coplanar, that is, they lie in the same plane, in this case the plane of our screen.

Now see that we had a very thin, effectively massless rod positioned like this so that the forces of our couple push the rod in opposite directions. In this case, the couple would create what’s called a moment about the center of the rod. We can call this moment 𝑀 sub c since it’s caused by a couple. And in general, the moment due to a couple is equal to two times one of the forces in that couple multiplied by the perpendicular distance between where that force is applied and the axis of rotation. In our specific case, that distance 𝑑 would be marked out here in the dashed line in pink. It’s half the perpendicular distance between the lines of action of our two forces.

If we say further that the magnitude of each one of these forces is 𝐹, then just as our general equation tells us, the magnitude of the moment created by this couple would be two times 𝐹 times 𝑑. Now in this lesson, we’re considering systems of forces which, when we combine them altogether, are equivalent to a couple. As an example of this, on our opening screen we had four people pushing on a merry-go-round. If we looked at the merry-go-round top down, it would look like this with those force vectors. This is a system of forces. And from the perspective of the moment that it creates around the center of the merry-go-round, we can say that this system of forces is equivalent to a couple like this.

As we study this topic then, there are two basic questions we’ll be asking. First, given a system of forces, we want to know if that system is equivalent to a couple. And if not, we want to determine what force or forces would make it that way. To see how we work with these questions in practice, let’s look now at an example.

In the figure shown, 𝐴𝐵𝐶𝐷 is a right-angled trapezoid at 𝐴, where 𝐴𝐵 equals 12 centimeters, 𝐵𝐶 equals 32 centimeters, and 𝐴𝐷 equals 16 centimeters. The shown forces are measured in newtons and are represented completely by the sides of the trapezoid, where the magnitude of the forces is proportional to their corresponding side lengths. If the system of forces is equivalent to a couple, find 𝐹 one, 𝐹 two, and 𝐹 three.

All right, so here we see this trapezoid with corners 𝐴, 𝐵, 𝐶, and 𝐷. And we also see the three unknown forces 𝐹 one, 𝐹 two, and 𝐹 three we want to solve for. Each of these forces acts along a side of the trapezoid along with this known force of 30 newtons. Our problem statement tells us that the side length 𝐴𝐵 is 12 centimeters, 𝐵𝐶 is 32 centimeters, and side length 𝐴𝐷 is 16. We’re told further that the magnitude of the forces acting on this trapezoid is proportional to their corresponding side lengths.

This means, for example, that since side length 𝐵𝐶 is twice as long as side length 𝐴𝐷, then the magnitude of the force 𝐹 three must be twice as great as the magnitude of 𝐹 one. This same proportionality relationship applies to the other sides. And we’re also told that this system of four forces is equivalent to a couple. That means we can equate these four forces to two equal and opposite forces that act along different lines of action. Knowing all this, it’s the unknown forces 𝐹 one, 𝐹 two, and 𝐹 three we want to solve for.

To begin doing that, let’s clear some space on screen and let’s think about what it means that these four forces are equivalent to a couple. If we draw the lines of action of these four forces, they effectively trace out the sides of this trapezoid. Because these four forces are equivalent to a couple, that means we can pick two points of intersection of these lines of action. And so long as those points lie along all four lines of action, then we can say that a force couple effectively originates from those points.

Here’s what we mean by that. Say we picked the corners of the trapezoid 𝐵 and 𝐷. At point 𝐵, the lines of action of forces 𝐹 two and 𝐹 three meet. And point 𝐷, we see, is an intersection of the lines of action of our 30-newton force and 𝐹 one. In that sense then, we’ve accounted for all four forces. And so if we sketch in these forces as though they originate at these two corners of our trapezoid, they would look something like this.

So we’re modeling all four forces as though they start at these two corners of our trapezoid. Note that we could just as well have picked the corners 𝐴 and 𝐶 since those two corners intersect all four lines of action. But either way, we can analyze these four forces as a couple. All this means that the net force acting at point 𝐷 is equal in magnitude but opposite in direction to the net force acting at point 𝐵. And actually since forces in what we could call the vertical direction are independent of forces in the horizontal direction, we can also say that the total vertical force at 𝐷 is equal and opposite that at 𝐵, and likewise for the horizontal components.

This means there are two force balance equations we can write out, one for the vertical direction and one for the horizontal. If we decide that forces to the right and forces pointing upward are positive, then when it comes to the vertical forces acting at points 𝐵 and 𝐷, we can write that the vertical component of our 30-newton force, we’ll call it 30 sub v, minus the force 𝐹 two acting at point 𝐵 equals zero. Rearranging this equation slightly, we see then that if we can solve for that vertical component of our 30-newton force, we’ll know 𝐹 two.

Coming back to our sketch, we see that this 30-newton force is effectively the hypotenuse of a right triangle. If we think of this triangle though not in terms of forces but in terms of distances, we know that this side is equal to 12 centimeters, this side is 32 minus 16 or 16 centimeters. And that means the hypotenuse is equal to the square root of 12 squared plus 16 squared, or 20 centimeters. Knowing these side lengths is helpful because if we call this interior angle of our right triangle 𝜃, then we can say that the vertical component of our 30-newton force is equal to 30 times the sin of 𝜃.

We recall that given a right triangle where another interior angle is 𝜃, the sin of that angle is equal to the ratio of the opposite side length to the hypotenuse length. This means that in our right triangle of interest, the sin of 𝜃 is equal to the ratio of 12 divided by 20. 30 times 12 over 20 equals 18. And so we now know the magnitude of force 𝐹 two. And we know that this is in units of newtons. We’ll record this result and now move on to solving for the two remaining unknown forces 𝐹 one and 𝐹 three. Both of these forces, we see, are directed horizontally.

For our force balance equation, we can write that 𝐹 three minus 𝐹 one minus the horizontal component of the 30-newton force, 30 times the cos of 𝜃, all equal zero. Coming back to our example right triangle, we can say that the cos of this angle 𝜃 is equal to the adjacent side length divided by the hypotenuse. In the triangle formed from our trapezoid, this ratio is equal to 16 divided by 20. Even knowing this though, we can see that here we have one equation but two unknowns. We’ll need to bring in some other information then to solve for 𝐹 one and 𝐹 three.

At this point, we can recall that the forces involved in this scenario are proportional to one another according to their corresponding side lengths. This means, for example, that the ratio of side length 𝐴𝐷, 16 centimeters, to the side length 𝐴𝐵 is equal to the ratio of 𝐹 one to 𝐹 two, the forces on those respective sides. This is great because it means that 𝐹 one equals 16 over 12 times 𝐹 two, which we know. So 𝐹 one then equals 16 over 12 times 18 newtons, or 24 newtons.

Recalling this proportionality method then, there are now two ways we could go about solving for 𝐹 three. We could use the same method we used to solve for 𝐹 one. Or, now that we know 𝐹 one, we could use this force balance equation here. Since we’ve already written this equation out, let’s continue on with it. Adding 𝐹 one and 30 times 16 over 20 to both sides, we get that 𝐹 three equals 𝐹 one plus 30 times 16 over 20. And knowing that 𝐹 one without units is 24 and that 16 over 20 equals four-fifths, we find that 𝐹 three is equal to 48 with units of newtons.

We find then that the three forces 𝐹 one, 𝐹 two, and 𝐹 three acting on the sides of this trapezoid have magnitudes of 24, 18, and 48 newtons, respectively. These magnitudes mean that all four forces involved with the trapezoid act as a couple.

Let’s now look at an example where we calculate the moment created by a system of forces.

𝐴𝐵𝐶𝐷 is a rectangle, in which 𝐴𝐵 equals 45 centimeters, 𝐵𝐶 equals 55 centimeters, and 𝐷𝐸 equals 28 centimeters. Forces of magnitudes 225, 275, 265, and 135 newtons act along 𝐴𝐵, 𝐵𝐶, 𝐶𝐸, and 𝐸𝐴 respectively. If the system of forces is equivalent to a couple, determine the magnitude of the moment of the forces.

Okay, so here we have this rectangle 𝐴𝐵𝐶𝐷. And we’re told that side length 𝐴𝐵 equals 45 centimeters, 𝐵𝐶 is 55 centimeters, and that side length 𝐷𝐸 is 28 centimeters. We’re Also told the magnitude of four forces acting on this rectangle and that this system of forces is equivalent to a couple. This means we can effectively replace the four forces with two equal and opposite forces that don’t lie along the same line of action. Knowing all this, we want to determine the magnitude of the moment of these four forces.

To start doing that, let’s clear a bit of space on screen. And let’s consider what it means that these four forces are equivalent to a couple. It means if we were to draw the lines of action of these four forces, then if we were to pick two intersection points of these lines so that these points included all four lines of action, then we could effectively model all four forces as though they originated from those two points. So for example, if we pick the points 𝐶 and 𝐴 in our rectangle, then from point 𝐴 we can consider our 225-newton force to be acting down and our 135-newton force to be acting to the right, likewise with point 𝐶 where we say our 275-newton force acts to the left and our 265-newton force acts up and to the right. Because our system of forces is equivalent to a couple, we can say that the total force acting at point 𝐴 is equal and opposite that total force acting at point 𝐶.

And actually we can go further than this. Since forces can be separated into independent vertical and horizontal components, we can say that the net horizontal force at point 𝐴 is equal and opposite that at point 𝐶, and similarly for the net vertical force at these two points. As a side note, we didn’t need to use the points 𝐴 and 𝐶 as a model for the origins of our four forces. Taken together, these two points meet our condition of overlapping all four lines of action, but, looking back to our rectangle, so did the points, say, 𝐸 and 𝐵. Either of these pairs of points would work for analyzing this scenario.

But anyway, knowing that the total forces at 𝐶 and 𝐴 form a couple, our goal is to solve for the moment created by that couple. We can recall that the moment created by a couple of forces is equal to two times the force component perpendicular to the distance between where the force is applied and where the axis of rotation is located. The idea here is that each of the forces in the couple contributes equally to the moment overall. That’s why if we solve for this perpendicular component of one of those two forces and then multiply it by the distance 𝑑, we need only multiply that result by two to solve for the overall moment.

Using 𝐴 and 𝐶 as the points of origin of the two forces in our couple, we can think of those forces as acting on a line like this that joins at the two points. So the moment we’re solving for acts about this midpoint of that line. In our rectangle, that line and midpoint would look like this. So here’s the idea. If we can find the net force that’s perpendicular to this line acting either at point 𝐴 or point 𝐶, then we will have solved for 𝐹 perpendicular in this equation. We could choose either point 𝐶 or point 𝐴 at which to solve for this force.

And since the forces at point 𝐴 act in what we could call the purely vertical and purely horizontal directions, let’s choose that point. Our goal then, is to solve for the components of the 225- and 135-newton forces that are perpendicular to this dashed orange line. Adding those together will give us 𝐹 perpendicular here. Looking at the right triangle created by our 225-newton force, we can call this interior angle of that triangle 𝜃 sub one. We’re interested in this angle because the sin of 𝜃 sub one times 225 equals the perpendicular component of this force.

Now, if we go back to our original diagram, this angle in that diagram is also equal to 𝜃 sub one. And we see that indeed this is an interior angle in this right triangle drawn in orange. The shorter two sides of the triangle have lengths of 45 and 55 centimeters, respectively. And by the Pythagorean theorem, we can say then that the hypotenuse has a length of the square root of 45 squared plus 55 squared. That’s equal to the square root of 5050.

At this point, we can recall that given a right triangle where 𝜃 is another interior angle of that triangle, then the sin of 𝜃 equals the ratio of the opposite side length to the hypotenuse. And that means when it comes to our angle of interest, the sin of 𝜃 sub one, this is equal to 55 divided by the square root of 5050. For 𝜃 sub one, that’s the opposite side length to the hypotenuse length ratio. So then we now have an expression for the perpendicular component of this 225-newton force. And we can start to keep a tally of this using a variable we’ll call 𝐹 perpendicular. This is exactly that variable we see in our equation for 𝑀 sub c. And we now know that it’s equal to this term plus a second term we’ll figure out soon.

That second term is equal to this component of our 135-newton force. To solve for it, we can use a similar approach. Let’s say that this angle interior in this right triangle is 𝜃 sub two. On our rectangle, the angle 𝜃 sub two would look like this. And notice that this is the same as this angle here in our triangle. Once more, to calculate the component of this force perpendicular to the dashed line, we’ll want to use the sin of this angle. We want to calculate 135 times the sin of 𝜃 sub two. The sin of this angle equals the opposite side length, 45 centimeters, divided by the hypotenuse.

We now know the component of the 135-newton force that’s perpendicular to our dashed orange line. This then is the second and final term in our equation for 𝐹 perpendicular. All right, so to find 𝑀 sub c, the only thing that remains for us now is to solve for this distance 𝑑 and then multiply it by 𝐹 perpendicular and two. Looking at our original diagram, that distance is half the distance from point 𝐶 to point 𝐴. In other words, it’s this distance here. We see right away that that equals one-half, our hypotenuse length. In other words, 𝑑 is equal to the square root of 5050 over two.

Knowing all this, we’re now ready to calculate 𝑀 sub c. Plugging in the values we calculated for 𝐹 perpendicular and 𝑑, notice that this leading factor of two will cancel out with this denominator of two and that also this denominator of the square root of 5050 that appears in both terms in 𝐹 perpendicular will be multiplied by that same value and therefore become equal to one. We find then that 𝑀 sub c equals 225 times 55 plus 135 times 45. And this equals 18450. We recall that the units of our forces are newtons and the units of our distances in this example are centimeters. The moment then created by this system of forces, which is equivalent to a couple, equals 18450 newton centimeters.

Let’s finish up our lesson now by summarizing a few key points. In this lesson, we’ve seen that a system of forces may be equivalent to a couple. If that’s so, then the moment created by that system of forces equals the moment of a couple. We’ve also seen that forces in a system may act along a single line or along different axes, for example, the sides of a shape. And lastly, we saw that when a system of forces is equivalent to a couple, the forces in that system can be modeled as acting from two different points.

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