Video: Moment of a Couple | Nagwa Video: Moment of a Couple | Nagwa

# Video: Moment of a Couple

In this video, we will learn how to calculate the moment of a couple of two forces and the resultant of two or more couples.

16:43

### Video Transcript

In this video, weβre talking about the moment of a couple. This means something a little bit different from what we see on screen. But as weβll learn, it does involve vectors that point in opposite directions. Letβs jump right in.

We can recall that the mathematical definition of a moment involves a force, weβve called it π, as well as a point. If our force is exerted a perpendicular distance π away from this point weβve picked, then we say that about this point it exerts a moment π. And itβs helpful to recall that in this equation the force weβre talking about is explicitly perpendicular to the distance between where the force is applied and the point of rotation. So, this variable π, we say, is the moment of the force π about this point we picked. But now letβs expand this idea to talk about the moment of a couple.

A couple is a pair of forces that are equal in magnitude but point in opposite directions. And importantly, for a pair of forces like π one and π two to form a couple, they canβt act on the same line of action. For example, if we had two forces acting like this, π π΄ and π π΅, even though they had the same magnitude and act in opposite directions, theyβre not a couple because they do lie along the same line of action. So, coming back to our point here, if we were to apply a force couple about this point, then out a distance π to the right of our point, we would apply this force, weβve called it π prime, which is equal in magnitude but opposite in direction to π.

π prime and π form a force couple, and we can see what effect this would have on the moment π about our point. Whenever we have a couple of forces acting about a point, the moment created by one of the forces is equal to the moment created by the other. And this equality is not just in magnitude but also in direction or sign. For example, notice that the force π creates a counterclockwise moment about our point that would be positive, and the force π prime creates a moment in the same direction. This will always be true when we work with couples because recall that force couples always act in opposite directions. The total moment then created by our couple is simply two times the moment created by either one of the forces in the couple.

Knowing all this, letβs get some practice with these ideas now through a few examples.

In the figure below, π΄π΅πΆπ· is a rectangle in which π΄π΅ equals five centimeters and π΅πΆ equals four centimeters. The forces that are shown in the figure are in newtons, and the system is in equilibrium. Find the value of πΉ one plus πΉ two.

Okay, looking at our figure, we see this rectangle with corners, π΄, π΅, πΆ, π·. And that each of the four sides has a force acting on it. On the left side, a force of magnitude πΉ two acts upward, and on the right a force of that same magnitude acts downward. And then on the top, a force of magnitude πΉ one acts to the left, and on the bottom a force of magnitude 35 newtons acts to the right.

A key piece of information weβre given is that our system is in equilibrium. This means that even though these four forces are acting on it, it doesnβt translate and it also doesnβt rotate. The forces acting on the four sides of our rectangle effectively cancel one another out. This tells us that these four forces comprise two force couples. The first couple is force πΉ one and this 35-newton force. And the second is made of these two forces of magnitude πΉ two.

The fact that these pairs of forces do form couples tells us that the magnitude of force πΉ one must equal 35 newtons. This is because the forces in a couple have the same magnitude but act in opposite directions. So, we can say right away that πΉ one equals 35 newtons, which means that now to answer our question, we just need to solve for πΉ two.

If we consider the point at the very center of our rectangle to be the axis of rotation for these two force couples, then we can see that the moment about this point created by our πΉ two force couple would point in a clockwise direction. Weβll call that π two. And then notice that the moment created by our other force couple acts in opposition to this. And that actually makes sense as we recall that our system is an equilibrium. That means these two moments must cancel one another out, and therefore they have to point in opposite directions.

If we think just in terms of magnitudes, then we can write that π one is equal in magnitude to π two. And now letβs recall that, in general, the moment created by a couple of forces, weβll call it π sub c, is equal to two times the perpendicular component of one of those forces multiplied by the distance away from the axis of rotation.

Going back to our sketch, we could say, for example, that the distance between the force πΉ one at the top of our rectangle and the axis of rotation is π one and the distance between πΉ two and that center point is π two, meaning that we can write our moment balance equation like this. We see that the factors of two cancel from each side of our equation.

And now, since itβs πΉ two that we want to solve for, letβs rearrange so that that is the subject of our equation. If we divide both sides by π two, that cancels out on the right. And we can see that πΉ two equals πΉ one multiplied by this distance ratio π one to π two. We know the force πΉ one. And letβs recall that weβre given the side lengths of this rectangle. π΄π΅ is equal to five centimeters, which, we can see is equal to two times π two and π΅πΆ is four centimeters or two times π one. These relationships tell us that π two equals five half centimeters and π one equals two centimeters.

Substituting in these values and leaving out the units of centimeters, we see that πΉ two equals 35 newtons times two over five-halves, which simplifies to four-fifths times 35 newtons or 28 newtons. So, since πΉ two equals 28 newtons and πΉ one equals 35 newtons, the sum of these must be 63 newtons. Thatβs the sum of the magnitudes of the forces in our two couples.

Letβs now look at an example where our forces are not at 90 degrees to the axis of our system.

In the figure below, πΉ one equals three newtons, and πΉ one and πΉ two form a couple. Find the algebraic measure of the moment of that couple.

Looking at our figure, we see the forces πΉ two and πΉ one, which are acting at 45 degrees to the axis of this system. Weβre told that these forces form a couple, and we see that they act at a distance of seven root two centimeters from one another. We want to calculate the moment that this couple of forces produces, and that moment will be calculated about the midpoint of this black line.

Based on the directions of πΉ one and πΉ two, we can see that they will create a counterclockwise moment about this point. By convention, moments in this direction are considered positive. So now, to answer our question, we need to calculate the magnitude of this moment. Letβs recall that the moment created by a couple of forces, we can call it π sub c, is equal to two times the perpendicular component of the magnitude of one of those forces in the couple multiplied by the distance between where that force is applied and the axis of rotation.

In our case, that distance would be the length of this pink line here. Whether weβre talking about force πΉ one or πΉ two, that distance is the same. And since πΉ one and πΉ two form a couple, we can work with just one of them and use that in our calculation for π sub c. Letβs choose to work with the force πΉ sub two. Since this force forms a couple with πΉ sub one, we know that its magnitude must be three newtons. And what we want to do is multiply the component of this force perpendicular to our black line by the distance, weβll call it π, between where the force is applied and our rotation point.

As for that perpendicular component of our force πΉ two, because this angle here is marked as 45 degrees, this one here must be the same. Thatβs because we know that they add up to 90 degrees. So, if we call this component of πΉ sub two πΉ sub two perpendicular, then we can say that itβs equal to three newtons times the cos of 45 degrees. The cos of 45 degrees equals the square root of two over two. So, the component of πΉ two perpendicular to the axis of our system is three root two over two newtons. This is the value weβll use in our equation to calculate the moment of our couple of forces.

π sub c equals two times the perpendicular component of πΉ sub two multiplied by π, where π is this distance. And we see that distance is seven root two over two centimeters. With all these values plugged in to calculate π sub c, we find that all of our factors of two and root two cancel one another out. So, we end up with three times seven or 21 newton centimeters. This is the moment of this couple of forces.

Letβs now look at an example where our forces are given as vectors.

Given that two forces π one equals negative π’ plus two π£ and π two are acting at two points π΄: two, two and π΅: negative two, negative two, respectively, to form a couple, find the perpendicular distance between the two forces.

Okay, to get started here, letβs plot the two points π΄ and π΅ where these forces π one and π two are acting. Here, we have our π₯- and π¦-axes marked out in arbitrary units. And we know that the point π΄ is here; thatβs at two, two. And the point π΅ is located here at negative two, negative two. Weβre told that at point π΄ a force π one is acting. And based on its components, we can say that π one acts along this line, passing through the point one, four. When it comes to plotting the force π two, weβre told that this force, along with π one, forms a couple. And we can recall that forces in a couple have the same magnitude but act in opposite directions.

So, starting at point π΅ and moving in the opposite direction to the force π one, π two would look like this where it passes through the point negative one, negative four. Our question then says, what is the perpendicular distance between these two forces? What this is saying is that if we draw the lines of action of these two forces, hereβs the line of action of π two, and hereβs the line of action of π one, then we want to solve for the perpendicular distance between these lines. We could measure that distance here or here or here or anywhere along these lines so long as that distance is perpendicular to these two parallel lines of action.

Now, since these two lines of action are lines, letβs give them names according to that property. Letβs call the line of action from force π one π¦ one and that of force π two π¦ two. And we can recall that in an π₯π¦-plane, the general formula for a line is that the π¦-coordinate is equal to the slope of the line or gradient multiplied by the π₯-coordinate added to the π¦-axis intercept of the line.

So, for π¦ one and π¦ two then, the lines of action of our two forces, letβs solve for the equations of those lines. First, letβs consider the slope or gradient of the lines. And thankfully because theyβre parallel, they have the same slope. As we consider again the components of the force π one, this tells us that for every increase of one in our π₯-value, our π¦-value correspondingly decreases by two. Therefore, the slope of this line is negative two. And as we mentioned, this is a match for the slope of π¦ two since the lines are parallel.

And now, letβs consider where these two lines intercept the π¦-axis. Starting with line π¦ two, if we follow this line of action, we see that it intercepts the π¦-axis at negative six. So, we now have the complete equation for the line π¦ two. For line π¦ one, this intersects the π¦-axis off of the graph that weβve drawn. But knowing the points that this line passes through, both point π΄ as well as the point one, four, we can see that when π₯ is equal to zero, π¦ must be equal to positive six.

So, now, we have the equations of our two parallel lines and we want to solve for the perpendicular distance between them. To start doing that, we can write the equation of a third line, weβll call it π¦ perpendicular that is perpendicular to π¦ one and π¦ two. The key property of a line that is perpendicular to another line, say π¦ two, is that the slope or gradient of either one of the lines is the negative inverse of the slope of the other.

In our case then, since the slope of our two lines of action is negative two, that means the slope of our perpendicular line will be negative the inverse of this. In other words, it will be positive one-half. And as we mentioned earlier, it doesnβt matter where the π¦-intercept of our perpendicular line is because whether itβs here or here or here or anywhere else, that distance we want to calculate will be the same. So, we can leave π¦ perpendicular as it is.

And now what we want to do is find where π¦ perpendicular intercepts π¦ one and where it intercepts π¦ two. Letβs begin with where π¦ perpendicular and π¦ one meet. To find where this is, weβll set them equal to one another. So, negative two π₯ plus six equals one-half π₯. And if we add two π₯ to both sides of this equation, we get the result that five-halves π₯ is equal to six, which implies that π₯ is equal to twelve-fifths. This value weβve solved for is the π₯-coordinate of the point where π¦ one is equal to π¦ perpendicular.

To find the corresponding π¦-coordinate of that point, we just need to substitute this value of π₯ in for π₯ in either our π¦ perpendicular or π¦ one equations. If we do that here, then we find that π¦ is equal to one-half times twelve-fifths or six-fifths. If we plot this point on our sketch, it would be somewhere around here. As weβve seen, its coordinates are twelve-fifths and six-fifths. The line that passes through this point and is perpendicular to both of our lines of action will also intersect the line π¦ two.

Now, we want to solve for where that point of intersection is. Similar to earlier, weβll set π¦ two equal to π¦ perpendicular, which means that negative two π₯ minus six equals one-half π₯. And now adding two π₯ to both sides, we find that five-halves π₯ equals negative six, telling us that this π₯-value of intersection is negative twelve-fifths. Substituting this value in for π₯ here, we find the corresponding π¦-value is negative six-fifths.

So, now, we have our two points of intersection between our lines of action and a line perpendicular to them both. To calculate the distance π separating our two lines of action, we just need to calculate the distance between these two points. In general, the distance between two points in the π₯π¦-plane is equal to the square root of the sum of the squares of the difference between the two π₯-coordinates Ξπ₯ and the difference between the two π¦-coordinates Ξπ¦.

In our case, Ξπ₯ is equal to the π₯-coordinate of one point twelve-fifths minus the π₯-coordinate of the other point negative twelve-fifths, giving us twenty-four fifths. And likewise, the change in the π¦-coordinate, Ξπ¦, equals six-fifths minus negative six-fifths or twelve-fifths. Writing these values off to the side, we can say that π equals the square root of twenty-four fifths squared plus twelve-fifths squared. We can factor a one-fifth out from under the square root. And if we square 24 and 12 and add those results together, we get 720.

And it turns out that 720 can be written as 144 multiplied by five. And 144 we know is equal to 12 squared, which means we can factor 12 out from the square root and get a simplified result of twelve-fifths times the square root of five. Recalling that this is a distance that weβve calculated, we can give our final answer as twelve-fifths times the square root of five length units. This is the perpendicular distance between the lines of action of our two forces.

Letβs now finish up this lesson by summarizing a few key points. In this lesson, weβve seen that a couple is a pair of equal and opposite forces acting along different lines of action. Along with this, we saw that a couple creates a moment. That moment is equal to two times the perpendicular component of one of the forces in the couple multiplied by the distance between where that force is applied and the axis of rotation. This summarizes the moment of a couple.

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