Question Video: Finding Missing Forces When the Forces Are Equivalent to a Couple Mathematics

In the given figure, if the forces that act on a light rod 𝐴𝐵 are equivalent to a couple and the moment of this couple is equal to 17 N⋅cm, find 𝐹₁ and 𝐹₂.

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Video Transcript

In the given figure, if the forces that act on a light rod 𝐴𝐵 are equivalent to a couple and the moment of this couple is equal to 17 newton centimeters, find 𝐹 sub one and 𝐹 sub two.

Let’s begin by reminding ourselves what we mean when we say that the forces are equivalent to a couple. Now, a couple consists of a pair of parallel forces that are equal in magnitude, opposite in sense and do not share a line of action. This means that the resultant of the forces are equal to zero. And as such, a couple does not produce a translation; it only produces a rotation. So this means that the resultant of a couple is just a pure moment.

In this case, we’re told that the moment of the couple is equal to 17 newton centimeters. And the most important fact that’s going to help us answer this question, in fact, is that the moment of the couple is independent of the reference location. In practice, this means that we can take the sum of the moments about point 𝐴 or the sum of the moments about point 𝐶. Or, indeed, we can even calculate the sum of the moments about point 𝐷 or point 𝐵. And we should always get a value of 17 newton centimeters.

Now, this is really useful because, in fact, if we take moments about either point 𝐷 or point 𝐵, that will allow us to calculate the value of either force 𝐹 sub one or 𝐹 sub two. Let’s take moments about 𝐷. This will allow us to find the value of 𝐹 sub one. And the reason this happens is because to calculate the moment of a force, we multiply the magnitude of that force by the perpendicular distance from the line of action of the force to the pivot. Now, of course, force 𝐹 sub two acts at a distance of zero units from 𝐷. So even if we attempted to calculate the moment, we know it would be zero for force 𝐹 sub two.

Let’s begin then by looking at the moment of the force acting at point 𝐴. This has a magnitude of five newtons, and it’s looking to move the object 𝐴𝐵 in a counterclockwise direction. So the moment is going to be positive. The distance away from point 𝐷 is one plus two. So the moment is five times one plus two or five times three newton centimeters. We’ll evaluate this when we’ve calculated the sum of all of our moments. Let’s next consider the moment of the force that acts at point 𝐶 . This force is trying to rotate the object in a clockwise direction. And so, its moment is going to be negative. It has a magnitude of two newtons and it acts at a point two centimeters away from 𝐷. So its moment is negative two times two.

Finally, let’s consider the moment of the force acting at point 𝐵. This is trying to rotate the object in a counterclockwise direction, and so the moment is going to be positive. And it has a magnitude of 𝐹 sub one and acts at a distance of one centimeter away from 𝐷. So the moment is 𝐹 sub one times one. Now, of course, this is the sum of all of the moments about point 𝐷. And we know that the sum of the moments will be independent of the reference location. So the sum of these is equal to 17. Simplifying the left-hand side and we get 15 minus four plus 𝐹 sub one, which simplifies further to 11 plus 𝐹 sub one equals 17.

We’ll solve for 𝐹 sub one by subtracting 11 from both sides. And that gives us 𝐹 sub one is 17 minus 11 or 𝐹 sub one is equal to six. So we’ve calculated the value of 𝐹 sub one. We could repeat this process but take moments about point 𝐷 to find the value of 𝐹 sub two. Alternatively, we said earlier that the sum of the forces acting on our body must be equal to zero. So let’s calculate the sum of the forces and we’ll take upwards to be positive. In the upward direction, we have two newtons and 𝐹 sub one, and in the downwards direction, we have five and 𝐹 sub two. So the sum of our forces is two plus 𝐹 sub one minus five minus 𝐹 sub two. And we know that that is equal to zero.

In fact, we’ve already calculated the value of 𝐹 sub one as well; it’s six. So our equation becomes two plus six minus five minus 𝐹 sub two equals zero. That simplifies to three minus 𝐹 sub two equals zero. So adding 𝐹 sub two to both sides of this equation and we get 𝐹 sub two equals three. We’re working in newtons, so we’re done. The value of 𝐹 sub one and 𝐹 sub two are six newtons and three newtons.

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