### Video Transcript

๐ด๐ต๐ถ๐ท is a square having a side length of 50 centimeters. Forces of magnitudes 30, 60, 160, and 10 newtons are acting at ๐ด๐ต, ๐ต๐ถ, ๐ถ๐ท, and ๐ท๐ด, respectively, while two forces of magnitudes 40 root two and 90 root two newtons are acting at ๐ด๐ถ and ๐ท๐ต, respectively. If the system is equivalent to a couple, find its moment considering the positive direction is ๐ท๐ถ๐ต๐ด.

All right, so here is our square. And weโre 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 ๐ด๐ต, thereโs a force of 30 newtons; along ๐ต๐ถ, a force of 60; from ๐ถ to ๐ท, a force of 160 newtons; and then a force of 10 newtons from ๐ท to ๐ด. Then even going beyond this, if we define the diagonals ๐ด๐ถ and ๐ท๐ต, 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 weโre 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 itโs exactly that moment that we want to calculate. And weโll consider the positive direction of rotation to be in the direction from ๐ท๐ถ๐ต to ๐ด. 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, weโll want to calculate ๐น perpendicular times ๐ and then sum them all together to get the overall moment.

Letโs clear some space to work. And the first forces weโll consider here are the forces that act along the diagonals of our square. Since theyโre 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, ๐ in this equation would be zero for them. Thatโs 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 donโt contribute to the overall moment, so we wonโt 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 letโs start adding up the contributions to the overall moment from each of these forces.

Weโll start with the 30-newton force acting from ๐ด to ๐ต. 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, thatโs the perpendicular distance between all four forces on the sides of our square and the center of the square.

Knowing that, letโs 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 weโll use is 25 centimeters. So weโll move on to the 160-newton force, also creating a negative moment. And lastly, weโll 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.