Three forces, measured in newtons, are acting along the sides of an equilateral triangle 𝐴𝐵𝐶 as shown in the figure. Given that the triangle has a side length of seven centimeters, determine the algebraic sum of the moments of the forces about the midpoint of 𝐴𝐵 rounded to two decimal places.
Okay, so taking a look at our figure, we see this equilateral triangle 𝐴𝐵𝐶. We also notice that there are forces of different magnitudes acting along each side length. What we want to do is calculate the total sum of the moments created by these forces about the midpoint of line segment 𝐴𝐵. That line segment is here, and we can say that the midpoint of that line is right at this pink dot. What we’re going to do then is calculate the moment of all three of these forces about this point and then add those moments together.
We can start out by recalling that the moment created by a force is equal to the component of that force that is perpendicular to a distance between where the force is applied and an axis of rotation. In our situation, that axis of rotation is the midpoint of 𝐴𝐵. Knowing then that we’ll apply this relationship to each of our three forces, let’s take note of the fact that each side of our equilateral triangle is seven centimeters long. We’ll call this side length 𝑠, and then we’ll clear some space at the top of our screen, so we can start calculating the moments due to each of our three forces.
We can call these moments 𝑀 300, 𝑀 100, and 𝑀 150, indicating the forces that cause them. And the total moment overall, we’ll call that simply 𝑀, is the sum of these three. As we’ve seen, the moment created by a force depends on the distance between the line of action of that force and the axis of rotation. If we consider our 300-newton force, its line of action looks like this, and we see this goes right through the midpoint of 𝐴𝐵. Therefore, for the moment created by our 300-newton force, 𝑑 is zero. That is, there’s no perpendicular distance between where this force is applied and the axis of rotation. Therefore, the moment created by this force about our particular axis of rotation is zero.
Knowing that, let’s move on to calculate the moment created by our 100-newton force. This force’s line of action looks like this. If we say that the distance 𝑑 between our axis of rotation and the perpendicular application of our force is shown here, then that means we want to know what component of our 100-newton force acts along this blue line, which is perpendicular to the orange line. If we can figure that out, then that will be 𝐹 sub perpendicular by which we multiply 𝑑 to solve for the moment 𝑀. To solve for that force component, we’ll want to know this angle right here. Let’s call that angle 𝜃.
At this point, we can recall that we’re working with an equilateral triangle. That tells us all the interior angles of our triangle are the same, and therefore they must each be 60 degrees. This fact helps us solve for 𝜃 because we see now that 60 degrees plus 𝜃 must equal 90 degrees. This is so because our blue line here is at 90 degrees to our orange line. And we can see that that angle was made up of 60 degrees and 𝜃. This implies that 𝜃 equals 30 degrees. And so now, we can say that for this particular force, the perpendicular component of that force is equal to 100, that’s 100 newtons, multiplied by the cos of 30 degrees.
When we do take the cos of 30 degrees and multiply it by our 100-newton force, we’ll get this component of that 100-newton force. Note that if we had taken the sine instead of cosine, we would have the force component parallel to 𝑑, rather than perpendicular to it. In other words, we’ll get 𝐹 perpendicular.
According to our moment equation, the next thing to do is to multiply this perpendicular component of our force by 𝑑. From our sketch, we can see that 𝑑 is one-half the length of 𝐴𝐵. The length of 𝐴𝐵 equals the length of one side of our triangle, seven centimeters. Leaving out units then, 𝑑 is equal to seven-halves. So then, this product equals the moment created by our 100-newton force. And we can note that the cos of 30 degrees is equal to the square root of three over two. This expression then equals 100 over four times seven times root three or equally 175 times the square root of three.
What we’ve calculated here is the magnitude of the moment created by our 100-newton force about our point of interest. Note, though, that according to our chart, moments have a sign associated with them. A moment that tends to create a counterclockwise rotation is positive. We see further that the moment created by this force would tend to be clockwise about our point of interest, in other words, negative. This means when we plug in our value for 𝑀 sub 100, we’ll use negative 175 times root three. Doing this takes into account the sign of our moment. Okay, so much for the moment created by our 100-newton force. Now for that created by the 150-newton force.
Once again, we’ll look at the force’s line of action, and we’d like to figure out what component of this force is perpendicular to this distance 𝑑. Once again, we draw a line that’s at 90 degrees to 𝑑. And now we’d like to know this angle, what we’ll call 𝜙. If we extend our orange line, then notice that the angle between these two lines here is the same as this interior angle of our triangle. In other words, it’s 60 degrees. And once again, we have this angle of 60 degrees added to our unknown angle equaling 90 degrees. 60 degrees plus 𝜙 equals 90 degrees. And so 𝜙, just like 𝜃, is 30 degrees.
The magnitude of our force that’s perpendicular to 𝑑 then is 150 times the cos of 30 degrees. And to solve for the magnitude of the moment created by this force, we multiply it by 𝑑. Again, that’s seven over two. Recalling that the cos of 30 degrees equals the square root of three over two, this expression equals 525 over two times the square root of three.
The moment created by this force, just like that created by the 100-newton force, also tends to create a clockwise rotation about our axis of rotation. By our sign convention, this also then is a negative moment, and we’ll substitute this value in for 𝑀 150. We now have an expression we can evaluate for the total moment 𝑀 created by all three of these forces. Entering this expression on our calculator and rounding our result to two decimal places, we get negative 757.77. Recall that our units are newtons, the units of our forces, multiplied by centimeters, the units of distance. Our final answer then is that the net moment created by these forces about the midpoint of 𝐴𝐵 is negative 757.77 newton centimeters.