### Video Transcript

𝐴𝐵𝐶𝐷 is a square having a side length of 50 centimeters. Forces of magnitudes 30, 60, 160, and 10 newtons are acting at 𝐴𝐵, 𝐵𝐶, 𝐶𝐷, and 𝐷𝐴, respectively, while two forces of magnitudes 40 root two and 90 root two newtons are acting at 𝐴𝐶 and 𝐷𝐵, respectively. If the system is equivalent to a couple, find its moment considering the positive direction is 𝐷𝐶𝐵𝐴.

All right, so here is our square. And we’re told each side length is 50 centimeters. And there are also these four forces each of which acts along the side of the square. Along 𝐴𝐵, there’s a force of 30 newtons; along 𝐵𝐶, a force of 60; from 𝐶 to 𝐷, a force of 160 newtons; and then a force of 10 newtons from 𝐷 to 𝐴. Then even going beyond this, if we define the diagonals 𝐴𝐶 and 𝐷𝐵, forces of 40 root two and 90 root two newtons act along these lines, respectively. We have then six total forces acting on our square. And we’re told that this system of forces is equivalent to a couple. That means that while the net force on the square is zero, the net moment is not.

And it’s exactly that moment that we want to calculate. And we’ll consider the positive direction of rotation to be in the direction from 𝐷𝐶𝐵 to 𝐴. This tells us that counterclockwise rotation is positive and clockwise must be negative. As we get started, we can recall that the moment created by a single force is equal to that force times the perpendicular distance between where the force is applied and the axis of rotation. In the case of our square, that rotation axis will point into and out of the screen located here at the center. For all six of our forces then, we’ll want to calculate 𝐹 perpendicular times 𝑑 and then sum them all together to get the overall moment.

Let’s clear some space to work. And the first forces we’ll consider here are the forces that act along the diagonals of our square. Since they’re directed this way, that means the lines of action of these forces pass through our axis of rotation. In terms of the moment created by these forces then, 𝑑 in this equation would be zero for them. That’s the perpendicular distance between where the forces are applied and where the axis of rotation is located. As we calculate overall moment then, we can effectively disregard these two forces along the diagonals. They don’t contribute to the overall moment, so we won’t include them in our calculation.

In the case of the other four forces though, if we sketch in their lines of action, we see that there is a nonzero perpendicular distance between these lines and the center. That means each one will contribute to the overall moment about that point. So let’s start adding up the contributions to the overall moment from each of these forces.

We’ll start with the 30-newton force acting from 𝐴 to 𝐵. This force would tend to create a clockwise rotation about the center. Therefore, this force will contribute a negative moment. And the distance between the line of action of this force and the center is 50 over two, or 25 centimeters. And in fact, that’s the perpendicular distance between all four forces on the sides of our square and the center of the square.

Knowing that, let’s move on to considering the contribution of our 60-newton force. This also will create a negative moment. And as we saw, the distance value we’ll use is 25 centimeters. So we’ll move on to the 160-newton force, also creating a negative moment. And lastly, we’ll include the 10-newton force, which again tends to create a clockwise rotation about the center.

When we add together these four terms, we get a result of negative 6500. The units here are newtons times centimeters. And this is our final answer. All six of these forces together create a moment of negative 6500 newton centimeters.