Question Video: Finding the Moment of the Couple Resulting from Forces Acting on an Equilateral Triangle | Nagwa Question Video: Finding the Moment of the Couple Resulting from Forces Acting on an Equilateral Triangle | Nagwa

# Question Video: Finding the Moment of the Couple Resulting from Forces Acting on an Equilateral Triangle Mathematics • Third Year of Secondary School

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In a triangle 𝐴𝐵𝐶, 𝐴𝐵 = 𝐵𝐶 = 32 cm, and 𝑚∠𝐵 = 120°. Forces of magnitudes 2, 2, and 2√3 newtons are acting at 𝐴𝐵, 𝐵𝐶, and 𝐶𝐴, respectively. If the system is equivalent to a couple, find the magnitude of its moment considering the positive direction is 𝐴𝐵𝐶.

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### Video Transcript

In a triangle 𝐴𝐵𝐶, 𝐴𝐵 equals 𝐵𝐶 equals 32 centimeters, and the measure of angle 𝐵 is 120 degrees. Forces of magnitudes two, two, and two root three newtons are acting at 𝐴𝐵, 𝐵𝐶, and 𝐶𝐴, respectively. If the system is equivalent to a couple, find the magnitude of its moment considering the positive direction is 𝐴𝐵𝐶.

Okay, so let’s say that this is our triangle 𝐴𝐵𝐶, and we’re told here that the measure of angle 𝐵 is 120 degrees. And we also know that the lengths of the two sides 𝐴𝐵 and 𝐵𝐶 are the same. That means we’re working with an isosceles triangle, and therefore this interior angle is the same as this one. If we call that angle 𝜃, we can say that two times 𝜃 plus 120 degrees equals 180 degrees. If we solve this equation for 𝜃, we find it equals 30 degrees. Knowing then about the geometry of this triangle, we’re also told about forces that act along its sides. There’s a two-newton force acting from 𝐴 to 𝐵, an identical two-newton force from 𝐵 to 𝐶, and then from 𝐶 to 𝐴, a force of two root three newtons.

We’re told that this system of forces is equivalent to a couple. That means that the total force on this triangle is zero. However, there is a moment that’s created about its center. And in fact, it’s the magnitude of that moment that we want to solve for. To start doing that, let’s clear a bit of space. And to start analyzing the forces involved here, we’ll set up a coordinate system. We’ll say that corner 𝐴 in our triangle is at the origin of this coordinate frame, and the 𝑦-axis moves vertically upward and the 𝑥 horizontally to the right.

Now, as we’ve seen, there are three forces that are acting on this triangle. One of them acts purely in the 𝑥-direction. But the other two forces, these two newton forces on the other sides, can be divided up into their vertical and horizontal components. We’re saying here that this two-newton force effectively originates at this point, at the origin, while this other two-newton force effectively originates at point 𝐶. It’s by choosing these two points of origin, we could call them 𝐴 and 𝐶 in our triangle, that we can model all the forces involved as a couple.

When we go to calculate the components of these two two-newton forces, we can recall that in a right triangle, where one of the other interior angles is called 𝜃, the sine of this angle is given by the ratio of the opposite side length to the hypotenuse length, while the cosine equals the adjacent side length over the hypotenuse. If we consider then all the horizontal forces acting on our triangle, first, there’s this component of our one two-newton force. That component equals two times the cos of 30 degrees. Secondly, there’s this component of the other two-newton force. That’s also equal to two times the cos of 30 degrees.

Lastly, though, there’s this force acting from point 𝐶 to point 𝐴 in our triangle. Based on the way we’ve set up our 𝑥-axis, this force will be negative. If we recall that the cos of 30 degrees is equal to the square root of three over two, then we get two root three over two plus two root three over two minus two root three. And as we expect, this equals zero. The important thing for our purposes, though, is that all of these horizontal force components are along the same line of action. This means that since they sum to zero, they apply no moment about the center of our triangle. In other words, the horizontal force components don’t contribute to that moment.

Next, let’s consider the vertical components of the forces involved here. First, we see this vertical component of what we can call our first two-newton force. That will be equal to two times the sin of 30 degrees. And then there’s our second vertical component over here. That equals negative two times the sin of 30 degrees. So as we expect, the net force in the vertical direction is zero. But note that these two forces don’t act along the same line of action. Therefore, they will apply a moment to our triangle, and its magnitude will depend on the perpendicular distance between this dashed line that goes through the center of our triangle and the lines of action of our two vertical forces. We can call that perpendicular distance 𝑑. And we see it’s equal to one-half the length of the longest side of our triangle.

Recalling that each of the shorter sides has a length of 32 centimeters, we can say that the distance 𝑑 equals 32 times the cos of 30 degrees. That’s 32 times the square root of three over two or simply 16 root three. We’re now ready to calculate the magnitude of the moment created by our vertical forces. Each of those two forces has a magnitude of two newtons times the sin of 30 degrees. And we multiply this by the perpendicular distance between the lines of action of those forces and the axis of rotation of our shape. Altogether, this moment equals 32 times the square root of three. And as for the units, our forces have units of newtons and our distances have units of centimeters. The magnitude of the moment created by this system of forces equals 32 times the square root of three newton centimeters.

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