Video Transcript
𝐴𝐵𝐶𝐷𝐻𝑂 is a regular hexagon with sides of length five centimeters. Two forces of the same magnitude 13 newtons are acting along 𝐶𝐵 and 𝑂𝐻, respectively. Determine the magnitude of the moment of the couple.
We will begin by sketching the regular hexagon as shown. We are told that it has side length five centimeters. There are two forces of magnitude 13 newtons acting along 𝐶𝐵 and 𝑂𝐻, respectively. These two forces form a couple, as they are a pair of parallel forces with equal magnitude and opposite direction which do not lie on the same line of action. We are asked to calculate the magnitude of the moment of this couple.
This is equal to 𝐹𝑑 sin 𝜃, where 𝐹 is the magnitude of the forces — in this question, 13 newtons — 𝑑 is the distance between their points of action — in this case, the line segment 𝑂𝐶 — and 𝜃 is the angle that the forces make with this line segment. If we let the center of the hexagon be point 𝑃, we can use our properties of hexagons to help determine the length of 𝑂𝐶 and the angle 𝜃. The line segment 𝐴𝐵 is parallel and equal in length to the line segment 𝑃𝐶. This means that 𝑃𝐶 is equal to five centimeters. Likewise, the line segment 𝐻𝐷 is the same length and parallel to the line segment 𝑂𝑃. This is also equal to five centimeters, meaning that the line segment 𝑂𝐶 is equal to 10 centimeters.
Triangle 𝑂𝑃𝐻 is therefore equilateral. And we know that the angles in an equilateral triangle equal 60 degrees. The measure of angle 𝜃 is therefore equal to 60 degrees. And since the two forces form a couple, this is also true of the angle 𝐵𝐶𝑃. We now have values of 𝐹, 𝑑, and 𝜃. The magnitude of the moment of the couple is equal to 13 multiplied by 10 multiplied by the sin of 60 degrees. Since the sin of 60 degrees is root three over two, this simplifies to 130 multiplied by root three over two, which in turn is equal to 65 root three.
The magnitude of the moment of the couple is therefore equal to 65 root three newton-centimeters.
