Lesson Video: Moment of a Couple Mathematics

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.

14:52

Video Transcript

In this video, we’ll learn how to calculate the moment of a couple of two forces and the resultant of two or more couples. We define a couple as a system of two forces of equal magnitudes and opposite directions which act in different lines of action. And a couple produces a rotation due to the moments of the forces in that couple.

Let’s consider a pair of equal magnitude forces acting on a rod perpendicularly to the length of that rod. Since the magnitude of these forces are equal and they act in opposite directions and in different lines of action, this is a couple. Let’s explore the moment produced by the couple at the midpoint between these two lines of action, which we’ve labeled as point 𝐴.

The forces 𝐹 one and 𝐹 two produce a clockwise rotation. It’s typical to take counterclockwise to be positive, so they’re producing a rotation in the negative direction. The moment is the product of the force and the perpendicular distance from the pivot to the line of action of the force. The force 𝐹 sub one is acting at a distance 𝑑 over two units from 𝐴. It acts perpendicular to the line that joins point 𝐴 to the line of action. And so the moment of force 𝐹 one about point 𝐴 is negative 𝐹 one times 𝑑 over two, where 𝐹 one here is simply the magnitude of the force. We can simplify this a little and write it as negative 𝐹 sub one 𝑑 over two. And of course it’s negative because it’s acting in that clockwise direction.

In a similar way, we define the magnitude of force 𝐹 sub two as simply 𝐹 sub two. And the moment of this force about point 𝐴 is negative 𝐹 sub two 𝑑 over two. Now, remember, we said that the magnitude of these two forces was equal. So this is equivalent to saying that the moment of the second force is negative 𝐹 sub one 𝑑 over two about 𝐴. The sum then of the moments about 𝐴 is negative 𝐹 sub one 𝑑 over two minus 𝐹 sub two 𝑑 over two. But since negative 𝐹 sub 𝑑 over two is equal to negative 𝐹 sub one 𝑑 over two, we can equivalently write this as negative two 𝐹 sub one 𝑑 over two or negative two 𝐹 sub two 𝑑 over two. We can then simplify by dividing through by two to give us either negative 𝐹 sub one 𝑑 or negative 𝐹 sub two 𝑑.

So what happens if we consider the moment about point 𝐡, the endpoint of our rod? Well, if we define the length of the rod to be 𝑙 units, the moment about point 𝐡 by force 𝐹 sub one is negative 𝑙𝐹 sub one. Similarly, the moment of force two is 𝑙 minus 𝑑 times 𝐹 sub two. Now, this is positive because this is trying to rotate the object in a counterclockwise direction. Now, of course, we said that the magnitude of force 𝐹 sub one and 𝐹 sub two is equal. So we can write this as 𝑙 minus 𝑑 times 𝐹 sub one. Then, the sum of the moments about 𝐡 is negative 𝑙𝐹 sub one plus 𝑙 minus 𝑑 𝐹 sub two, which we’ve just seen can be equivalently written as 𝑙 minus 𝑑 𝐹 sub one.

Distributing our parentheses, and we see that this simplifies to negative 𝐹 sub one 𝑑 or equivalently negative 𝐹 sub two 𝑑. And here is where it gets a little bit clever. We notice that this is the same as the net moment about 𝐴. In fact, this will be the same about any point on the rod as the moment due to a couple is in fact independent of the point about which the moments due to the couple are taken.

Let’s generalize this. The moment of a couple is the sum of the moments of the two forces of that couple about a point in space. The magnitude of the moment of a couple is in fact the product of the magnitude of one of the forces and the distance between them. We define this as π‘š equals 𝐹 times 𝑙, where 𝐹 is the magnitude of the force and 𝑙 is called the arm of the couple.

Now here it’s worth reminding ourselves what we mean by the arm of the couple. The arm is the length between a joint axis and the line of force acting on that joint when that force acts perpendicularly. Finally, the moment of a couple is independent of the point about which we take the moments of the two forces. So with all this in mind, let’s look at an example of how to calculate the length of the moment arm.

If the norm of the moment of a couple is 750 newton meters and the magnitude of one of its two forces is 50 newtons, determine the length of the moment arm.

It’s worth recalling what we mean by the norm of the moment. The norm of the moment is simply the magnitude of that moment. And so we’re given information about the norm of the moment and the magnitude of one of its two forces. And so we can link these by using the formula that determines the magnitude of the moment of a couple. It’s the product of the magnitude of either of the forces in the couple and 𝑙, the length of the moment arm. In this case, we’re told that the norm of the moment or its magnitude is 750 newton meters. And the magnitude of one of its two forces is 50 newtons. We can therefore substitute everything we know about the moment of this couple into the formula, and we get 750 equals 50 times 𝑙.

To solve for 𝑙, we divide both sides of this equation by 50. So 𝑙 is 750 divided by 50, or 15. And since we’re dividing newton meters by meters, the units for our length are simply meters. And so the length of the moment arm in this case is 15 meters.

Now, the forces don’t necessarily act perpendicularly to the line which connects the points from which they act. In these cases though, there is a technique that will help us determine the moment due to the couple. We saw that 𝑙 is the arm of the couple. And it’s the length between a joint axis and the line of force acting on that joint when the force acts perpendicularly. Now that’s the important part. This definition allows us to consider the components of each force which act perpendicular to the line that joins the points at which they act. If we define the distance between the points of the lines of action of each force as 𝑑, we can say that the component of the force that acts perpendicular to this length is the magnitude of 𝐹 times sin πœƒ. And so we’re now able to say that the magnitude of the moment due to the couple is equal to the magnitude of either force 𝐹 times sin πœƒ times 𝑑.

In our next example, we’ll demonstrate an application of this.

In the figure below, 𝐹 sub one is equal to three newtons and 𝐹 sub one and 𝐹 sub two form a couple. Find the algebraic measure of the moment of that couple.

In order to find the algebraic measure of the moment of the couple, we begin by recalling that if the forces form a couple, then their magnitudes must be equal. If we define 𝐹 sub one and 𝐹 sub two to be their respective magnitudes, then we can say that 𝐹 sub one equals 𝐹 sub two, which equals three newtons. Then, to find the magnitude of the moment of the couple, we find the product of the magnitude of one of the forces times sin πœƒ and 𝑑, where 𝑑 is the distance between the line of action of each force. And πœƒ is the angle that each force makes with the line connecting the points that force one and force two act from.

We see from our diagram that πœƒ here is equal to 45 degrees. And 𝑑, the distance between points 𝐴 and 𝐡 here, which are the points where 𝐹 one and 𝐹 two act, is equal to seven root two centimeters. And so the magnitude of the moment of this couple is three times sin 45 degrees times seven root two. Now, of course, sin of 45 degrees is an exact value that we should know by heart. It’s root two over two. But of course, the square root of two divided by two times the square root of two is two over two or simply one. And so the magnitude of the moment of the couple is three times seven, which is 21 or 21 newton centimeters. Since the forces are trying to move the body in a counterclockwise direction, we know that the algebraic measure of the moment is going to be positive. So the answer is 21 newton centimeters.

Now, so far, we’ve considered a couple of formulae. But an alternative way of calculating the moment of a couple is to use the cross product. The moment of a force 𝐹 about a point can be found using the cross product. If we take the vector 𝐫 to be the position vector from the point about which the moment’s being taken to any point on the line of action of the force, the moment of the force is the cross product of 𝐫 and 𝐅. If we consider a couple formed of two forces 𝐹 sub one and 𝐹 sub two that act at 𝐴 and 𝐡, respectively, the moment of the couple is then given by the cross product of the vector 𝐁𝐀 and the vector 𝐅 sub one, which is equivalent to finding the cross product of the vector 𝐀𝐁 and the vector 𝐅 sub two.

Essentially, in the equation for the moment of the couple, the position vector 𝐫 is substituted by the vector between the points of applications of the forces. We can think of this as taking moments about 𝐡 in the first case and moments about 𝐴 in the second. So with this in mind, let’s look at how to find the perpendicular distance between two force vectors that form a couple.

Given that two forces 𝐅 sub one equals negative 𝐒 plus two 𝐣 and 𝐅 sub two are acting at two points two, two and negative two, negative two, respectively, to form a couple, find the perpendicular distance between the two forces.

We’ve been given a vector expression for one of the forces in our couple. Now, of course, for the forces to form a couple, their magnitudes must be equal. They will also be acting in opposite directions. So we could, if we’re required, find the vector 𝐅 sub two. They’ll sum to zero, so 𝐅 sub two will be the negative of vector 𝐅 sub one. That’s the negative of negative 𝐒 plus two 𝐣, which is 𝐒 minus two 𝐣.

Now, in fact, it’s worth noting that we don’t actually need to work this out to be able to answer the question. But it would allow us to check our answer at the end. Then we know that for a couple formed of two forces 𝐅 sub one and 𝐅 sub two that act at 𝐴 and 𝐡, respectively, the moment of the couple is either given by the cross product of the vector 𝐁𝐀 and the vector 𝐅 sub one or equivalently the cross product of the vector 𝐀𝐁 and the vector 𝐅 sub two.

Now, we’re told that the force 𝐅 sub one acts at the point two, two and the vector 𝐅 sub two acts at the point negative two, negative two. So we define 𝐴 and 𝐡 as shown. Then we recall that the vector 𝐁𝐀 can be found by subtracting the vector 𝐎𝐁 from the vector πŽπ€. Since the vector πŽπ€ is the position vector of 𝐴, it’s simply two 𝐒 plus two 𝐣. And similarly, the vector 𝐎𝐁, which is the position vector for 𝐡, is negative two 𝐒 minus two 𝐣. Distributing the parentheses, and we find that the component for 𝐒 is two minus negative two, which is four. And similarly, the 𝐣-component is four as well. So the vector 𝐁𝐀 is four 𝐒 plus four 𝐣.

And so we can say that the cross product of the vector 𝐁𝐀 and the vector 𝐅 sub one is the cross product of the vector four 𝐒 plus four 𝐣 plus zero 𝐀 and the vector negative 𝐒 plus two 𝐣 plus zero 𝐀. And of course, we sometimes think about this in terms of the determinant of a two-by-two matrix. We get four times two minus four times negative one 𝐀. That’s eight plus four 𝐀, which is 12𝐀.

Now we want to find the perpendicular distance between the two forces, in other words the length of the moment arm. And so we use the formula that the magnitude of the moment of a couple is equal to the magnitude of one of the forces times the length of the moment arm. Now, in fact, since the moment is acting in simply one direction, its magnitude is equal to 12. We take the magnitude of either force. And to do so, we find the square root of the sum of the squares of each component. Specifically, the magnitude of force 𝐅 sub one is the square root of negative one squared plus two squared, which is equal to root five. And so substituting what we know about the moment of our couple into the formula, and we get 12 is equal to the square root of five times 𝑙, which we can solve for 𝑙 by dividing through by root five.

Finally, we’ll rationalize the denominator to simplify this. And to do so, we multiply both the numerator and denominator of the fraction by the square root of five, which gives us 12 root five over five. We can therefore say that the perpendicular distance between the two forces is 12 root five over five length units.

We’ll now recap some of the key concepts from this lesson. In this video, we learned that a couple is a system of two forces of equal magnitudes and opposite directions which act in different lines of action. The moment of a couple is the sum of the moments of the two forces of that couple about a point in space. And the magnitude of the moment is equal to the magnitude of one of the two forces and the distance between them, or the arm of the couple.

We learned that the moment of a couple is a constant independent of the point about which we take the moments of the two forces. And if the forces are not acting in a direction perpendicular to the line which connects the points that they act from, we can calculate the magnitude of the moment of the couple by finding the product of the magnitude of one of the forces times sin πœƒ times 𝑑, where 𝑑 is the perpendicular distance between the points that the forces act from. And πœƒ is the angle between the line of action of either force and this line.

Finally, for a couple formed of two forces 𝐹 sub one acting at point 𝐴 and 𝐹 sub two acting at point 𝐡, the moment of the couple could be calculated using the cross product. We say that the moment is equal to the cross product of the vector 𝐁𝐀 and the vector 𝐅 sub one, or equivalently the cross product of the vector 𝐀𝐁 and the vector 𝐅 sub two.

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