Question Video: Finding the Moment of a Couple Equivalent to Three Forces Acting on the Sides of a Regular Hexagon Mathematics

๐ด๐ต๐ถ๐ท๐ธ๐ is a regular hexagon of side length 8 cm, and forces of magnitudes 2, 13, and 11 newtons are acting at ๐ด๐ต, ๐ถ๐, and ๐ธ๐ท respectively. If the system is equivalent to a couple, determine the magnitude of the moment of the forces.

04:07

Video Transcript

๐ด๐ต๐ถ๐ท๐ธ๐ is a regular hexagon of side length eight centimeters, and forces of magnitudes two, 13, and 11 newtons are acting at ๐ด๐ต, ๐ถ๐, and ๐ธ๐ท respectively. If the system is equivalent to a couple, determine the magnitude of the moment of the forces.

Okay, so letโs say that this is our regular hexagon. And weโre told that between vertices ๐ด and ๐ต, thereโs a force of two newtons acting; then between vertices ๐ถ and ๐, a force of 13 newtons; and lastly from ๐ธ to ๐ท, a force of 11 newtons. As we begin solving for the magnitude of the moment due to these three forces, letโs recall the equation for the moment of a single force. Itโs equal to the product of that force times the perpendicular distance between where the force is applied and an axis of rotation. For our regular hexagon, that axis of rotation will be located in the center of this shape.

This point lies along the line from ๐ถ to ๐. And thatโs important when it comes to the overall moment of these forces. Looking again at this equation, we would say that for the force from ๐ถ to ๐, the 13-newton force, its perpendicular distance from the axis of rotation is zero. This means that this force doesnโt contribute to the overall moment of these three forces. To find our answer then, we really only need to consider the two-newton and the 11-newton force acting on these sides.

Sketching in the lines of action of these forces, the perpendicular distance between these lines of action and the center of our shape is the same. Weโll call this distance ๐. And weโll need to solve for this value in order to calculate our overall moment. To start solving for ๐, letโs observe that if we begin at this line and then move all the way around our center point, weโll have gone through an angular displacement of 360 degrees. Knowing that, if we sketched dashed lines from our center point to two adjacent vertices, then we can see that because this angle applies to exactly one side of our hexagon and there are six sides, itโs angular extent must be 360 over six degrees.

But then, and remember our goal here is to solve for ๐, we can go even further in dividing up this shape. In this pink right triangle, and weโve drawn an expanded view of it here, the height of the triangle is ๐. And this angular extent must be 360 degrees divided by 12. And that equals 30 degrees. Along with the measure of this angle, we can also solve for the length of this side of our triangle. Since the length of each side of our hexagon is given as eight centimeters, this particular side length of our triangle must be four centimeters, half of that.

We can now write down a relationship between this angle, this distance, and our unknown distance ๐. And we can say that the tan of 30 degrees equals four divided by ๐. Rearranging for ๐, we find itโs equal to four divided by the tan of 30 degrees. Since the tan of 30 degrees equals the square root of three over three, we can say that ๐ equals 12 over the square root of three or, equivalently, if we multiply top and bottom by the square root of three, four times that square root.

Now that weโve solved for ๐, which applies to both our two- and our 11-newton forces, weโre ready to start our overall calculation for the magnitude of the moment involved. As we do that, letโs set up the convention that a moment in the counterclockwise direction around our axis of rotation is positive. This means the moment created by our two-newton force, which will tend to create a counterclockwise rotation about the center of our shape, is positive, while that created by our 11-newton force will be the opposite direction and therefore negative.

We can write then that the total moment acting about the center of our shape is two times four root three minus 11 times four root three. This equals negative nine times four root three or negative 36 times the square root of three. Our question asks us to solve for the magnitude of the moment of these forces. So as our final answer, weโll say that the magnitude of this moment, the absolute value of the moment, is 36 times the square root of three newton centimeters.