Video Transcript
If 𝐴𝐵𝐶𝐷𝐸𝑂 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 𝐴𝐵, 𝐵𝐶, 𝐶𝑂, 𝐸𝐷, and 𝐶𝐴, respectively, and the system is equivalent
to a couple, find its moment norm.
Okay, so let’s say that this is a
regular hexagon with vertices 𝐴, 𝐵, 𝐶, 𝐷, 𝐸, 𝑂. As far as forces acting on this
hexagon, we’re told that there’s a force that acts along the line from 𝐴 to 𝐵. This is a 20-newton force. Then, along the line from 𝐵 to 𝐶,
there’s another 20-newton force. From 𝐶 to 𝑂, along this dashed
line, there’s a 13-newton force. Then from 𝐸 to 𝐷, in this
direction, there’s another 13-newton force. And then from 𝐶 to 𝐴, along this
dashed line, there’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 𝐶
to 𝑂 lies along that center point. This matters because when we go to
calculate the moment due to a force, it’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 𝐶 to 𝑂, that perpendicular distance 𝑑 is zero. The line of action of this force
passes through the axis of rotation. And therefore, it doesn’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 we’ll want to solve for in our shape. We’ve called one of those 𝑑 one
here and the other 𝑑 two. These distances are, on the one
hand, the perpendicular distance between one of the sides of our hexagon and the
center point, that’s 𝑑 one, and then the perpendicular distance between this line
from 𝐶 to 𝐴 at our center point; that’s 𝑑 two. 𝑑 one and 𝑑 two are the values
we’ll substitute in for 𝑑 in this equation to compute our overall moment. So let’s solve for them, and we’ll
start with 𝑑 one.
If we start at the dashed line
representing 𝑑 one and we move all the way around our center point until we reach
that line again, we’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 𝑑 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, we’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 𝑑 one. The tan of our angle of 30 degrees
equals three divided by 𝑑 one or equivalently 𝑑 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 𝑑 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 𝑑 two. Once again, we can sketch in a
right triangle so that 𝑑 two equals the triangle’s height. Now back on our original sketch,
note that if we extend this side of our right triangle, it will reach vertex 𝐵 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 𝑑 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 we’re considering
now has the same length as the hypotenuse of our earlier triangle that we used to
solve for 𝑑 one.
If we consider that triangle once
more, where here we haven’t drawn the two triangles to scale, then we can use the
fact that this side length is three and this side length 𝑑 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
𝑑 two, that hypotenuse equals six.
We can now use our 60-degree angle
and our hypotenuse length to solve for 𝑑 two. The cos of this angle of 60 degrees
equals 𝑑 two over six. Recalling that the cos of 60
degrees itself equals one-half, we can say that this equation implies that 𝑑 two
equals six times one-half or three. Now that we have values for 𝑑 one
and 𝑑 two, we’re ready to apply this equation to all of our forces. As we do, let’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.
Here’s what we can write then. The overall moment around the
center of our hexagon equals 20 times 𝑑 one plus 20 times 𝑑 one minus 13 times 𝑑
one minus 20 root three times 𝑑 two. If we then substitute in our values
for 𝑑 one and 𝑑 two into this expression, we find we can actually factor out three
root three from all of these terms. When we calculate what’s inside the
parentheses, we see that positive 20 cancels out with negative 20. And we’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.
