Question Video: Finding the Moment of a Couple Equivalent to a System of Five Forces Acting on a Regular Hexagon | Nagwa Question Video: Finding the Moment of a Couple Equivalent to a System of Five Forces Acting on a Regular Hexagon | Nagwa

# Question Video: Finding the Moment of a Couple Equivalent to a System of Five Forces Acting on a Regular Hexagon Mathematics • Third Year of Secondary School

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If ๐ด๐ต๐ถ๐ท๐ธ๐ is a regular hexagon having a side length of 6 cm, where forces of magnitudes 20 N, 20 N, 13 N, 13 N, and 20โ3 newtons are acting along ๐ด๐ต, ๐ต๐ถ, ๐ถ๐, ๐ธ๐ท, and ๐ถ๐ด, respectively, and the system is equivalent to a couple, find its moment norm.

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### Video Transcript

If ๐ด๐ต๐ถ๐ท๐ธ๐ is a regular hexagon having a side length of six centimeters, where forces of magnitudes 20 newtons, 20 newtons, 13 newtons, 13 newtons, and 20 root three newtons are acting along ๐ด๐ต, ๐ต๐ถ, ๐ถ๐, ๐ธ๐ท, and ๐ถ๐ด, respectively, and the system is equivalent to a couple, find its moment norm.

Okay, so letโs say that this is a regular hexagon with vertices ๐ด, ๐ต, ๐ถ, ๐ท, ๐ธ, ๐. As far as forces acting on this hexagon, weโre told that thereโs a force that acts along the line from ๐ด to ๐ต. This is a 20-newton force. Then, along the line from ๐ต to ๐ถ, thereโs another 20-newton force. From ๐ถ to ๐, along this dashed line, thereโs a 13-newton force. Then from ๐ธ to ๐ท, in this direction, thereโs another 13-newton force. And then from ๐ถ to ๐ด, along this dashed line, thereโs a force of 20 root three newtons.

Considering all these forces as a system, we want to find the moment norm, in other words, the absolute value of the moment these forces create around the center of our hexagon. That center lies at a point right around here. And importantly, the line from ๐ถ to ๐ lies along that center point. This matters because when we go to calculate the moment due to a force, itโs equal to that force times the perpendicular distance between where the force is applied and the axis of rotation. In the case of this 13-newton force from ๐ถ to ๐, that perpendicular distance ๐ is zero. The line of action of this force passes through the axis of rotation. And therefore, it doesnโt contribute to the moment around the center point. All the other forces, though, do contribute to this moment because none of them have a line of action passing through the center.

To compute our overall moment, there are two distances that weโll want to solve for in our shape. Weโve called one of those ๐ one here and the other ๐ two. These distances are, on the one hand, the perpendicular distance between one of the sides of our hexagon and the center point, thatโs ๐ one, and then the perpendicular distance between this line from ๐ถ to ๐ด at our center point; thatโs ๐ two. ๐ one and ๐ two are the values weโll substitute in for ๐ in this equation to compute our overall moment. So letโs solve for them, and weโll start with ๐ one.

If we start at the dashed line representing ๐ one and we move all the way around our center point until we reach that line again, weโll have moved through an angular displacement of 360 degrees. If we divide that angle into six equal parts, then one of those parts would cover an entire side of our shape. The measure of this angle then is 360 over six degrees. And then, if we divide this angle in half so it becomes 360 over 12 degrees, then we have a right triangle whose height is the distance we want to solve for ๐ one. And in this triangle, we know the measure of this angle here, 360 degrees over 12 or 30 degrees.

Not only this, we can also figure out the length of this side. Remember, weโre told that each one of the sides of our hexagon has a length of six centimeters. This side then of our right triangle will be half that, three centimeters. Knowing all this, we can now write a relationship between this known angle, this known side length, and our unknown side length ๐ one. The tan of our angle of 30 degrees equals three divided by ๐ one or equivalently ๐ one equals three over the tan of 30 degrees. Since the tan of 30 equals the square root of three over three, we can simplify our expression for ๐ one and show that it equals three times the square root of three.

Knowing this, we can now move on to calculating this other distance ๐ two. Once again, we can sketch in a right triangle so that ๐ two equals the triangleโs height. Now back on our original sketch, note that if we extend this side of our right triangle, it will reach vertex ๐ต of our hexagon. That tells us that this angle or this angle here in our larger sketch is equal to 360 degrees divided by six or 60 degrees. To solve for ๐ two, we need to know the length of one more of the sides of this right triangle. We can solve for the length of the hypotenuse by recognizing that the hypotenuse here of the triangle weโre considering now has the same length as the hypotenuse of our earlier triangle that we used to solve for ๐ one.

If we consider that triangle once more, where here we havenโt drawn the two triangles to scale, then we can use the fact that this side length is three and this side length ๐ one is three root three along with the Pythagorean theorem to say that the length of this hypotenuse is the square root of three squared plus three root three quantity squared. This equals the square root of 36 or six. As we mentioned, the length of this hypotenuse is the same as the length of this one. So in the triangle whose height is ๐ two, that hypotenuse equals six.

We can now use our 60-degree angle and our hypotenuse length to solve for ๐ two. The cos of this angle of 60 degrees equals ๐ two over six. Recalling that the cos of 60 degrees itself equals one-half, we can say that this equation implies that ๐ two equals six times one-half or three. Now that we have values for ๐ one and ๐ two, weโre ready to apply this equation to all of our forces. As we do, letโs set up the convention that any moment in the counterclockwise direction is positive. Therefore, a moment in the clockwise direction will be negative. Looking at the four forces for which we will calculate moments, we see that the moments created by our two 20-newton forces will be positive, while those created by our 13-newton force and our 20 root three newton force will be negative.

Hereโs what we can write then. The overall moment around the center of our hexagon equals 20 times ๐ one plus 20 times ๐ one minus 13 times ๐ one minus 20 root three times ๐ two. If we then substitute in our values for ๐ one and ๐ two into this expression, we find we can actually factor out three root three from all of these terms. When we calculate whatโs inside the parentheses, we see that positive 20 cancels out with negative 20. And weโre left with three root three times seven or 21 times the square root of three. This is a positive value, and we know it has units of newtons times centimeters. The moment norm of these forces then equals 21 times the square root of three newton centimeters.

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