Video Transcript
In the given figure, if the forces that act on a light rod 𝐴𝐵 are equivalent to a couple, find the moment of this couple.
And then we have a diagram showing a rod 𝐴𝐵 with a midpoint at 𝐶 and three forces acting on it. So, with all this in mind, let’s remind ourselves what it means for a system of forces to be equivalent to a couple. If this system of forces is equivalent to a couple, we say that the net force is equal to zero. In other words, the vector sum of the forces is equal to zero. Now, if we take the upwards direction to be positive here, we can see this is indeed the case. The sum of the forces is two plus three. And then we subtract five because it’s acting in the opposite direction. And that is equal to zero, or zero newtons.
Now, as indicated in the question, we know that this system of forces will generate a rotational moment. The question wants us to find the net moment generated. Now, we aren’t actually given an axis of rotation. So, let’s take point 𝐶, which we said is the midpoint of the rod, to be our reference point. Now, we’re going to use the following formula for the magnitude of our moment. If we’re looking at a moment generated by a force of magnitude 𝐹, the magnitude of that moment is 𝐹 times 𝑑, where 𝑑 is the perpendicular distance between the reference point and the line of action of the force.
We also know that when we work with moments, we need to specify a positive direction. It’s generally accepted that, unless otherwise specified, we take the counterclockwise moment to be positive. So, let’s think about the moment generated by each of our forces. We’ll begin with the five-newton force. This force acts at a distance of zero centimeters from 𝐶, and it has a magnitude of five newtons. This means that the moment of the force acting at point 𝐶 is five times zero, which is zero, or zero newton-centimeters.
Let’s now repeat this process with the force acting at point 𝐴. This is a two-newton force, and it acts at a distance of eight centimeters from the reference point. It is, however, trying to turn the rod in the clockwise direction. And we said that that direction is negative when it comes to the rotational moment. So, the magnitude of the moment is two times eight, but the actual moment is negative two times eight. That’s negative 16 newton-centimeters.
We’ll repeat this one further time for the moment of the force at 𝐵. This is a three-newton force, and it acts at a point eight centimeters away from the reference point. This time, this force is attempting to rotate the rod in a counterclockwise direction, in the positive direction. So, the moment is three times eight, which is 24 newton-centimeters.
In order to find the moment of the couple, we find the sum of all of the respective moments. In other words, the net moment 𝑚 of the system is zero plus negative 16 plus 24, which is equal to eight. And so the moment of this couple is eight newton-centimeters.
