Question Video: Computing the Net Moment of Multiple Couples | Nagwa Question Video: Computing the Net Moment of Multiple Couples | Nagwa

# Question Video: Computing the Net Moment of Multiple Couples Mathematics • Third Year of Secondary School

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𝐴𝐵𝐶𝐷 is a rectangle, in which 𝐴𝐵 = 14 cm and 𝐵𝐶 = 24 cm. Given that 𝑋 and 𝑌 are the midpoints of 𝐴𝐵 and 𝐶𝐷, respectively, and forces of magnitudes 153 N, 199 N, 153 N, 199 N, 73 N, and 73 N are acting as shown in the figure, determine the moment of the resultant couple rounded to two decimal places.

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### Video Transcript

𝐴𝐵𝐶𝐷 is a rectangle, in which 𝐴𝐵 equals 14 centimeters and 𝐵𝐶 equals 24 centimeters. Given that 𝑋 and 𝑌 are the midpoints of 𝐴𝐵 and 𝐶𝐷, respectively, and forces of magnitudes 153 newtons, 199 newtons, 153 newtons, 199 newtons, 73 newtons, and 73 newtons are acting as shown in the figure, determine the moment of the resultant couple rounded to two decimal places.

Looking at our figure, we see these six forces which act as three couples. There’s the 153-newton force couple, the 199-newton force couple, and the 73-newton force couple. Knowing all this along with the fact that side length 𝐴𝐵 equals 14 centimeters and side length 𝐵𝐶 equals 24 centimeters, we want to solve for the moment of the overall or resultant couple from these six forces. As we start doing that, we can clear some space on screen and begin by noting that, in general, if we have some force 𝐹 where the line of action of that force is some perpendicular distance 𝑑 away from an axis of rotation, then the moment 𝑀 about that axis created by this force is equal to 𝐹 times 𝑑.

Sometimes, in this expression, we’ll see a symbol like this, indicating the perpendicular relationship between 𝐹 and 𝑑. This is the moment created by a single force. But then, if we imagine having another force identical in magnitude but opposite in sign to 𝐹 which also acts a perpendicular distance 𝑑 away from our axis of rotation, then we say that together these two forces form a couple. And in that case, the moment about our axis, we’ll call it 𝑀 sub c, equals two times 𝐹 perpendicular times 𝑑. This tells us that each of the forces in a couple contribute equally to the overall moment. This means that we can solve for the overall moment by solving for the moment of either of the forces in a couple and then multiplying it by two. We’ll make use of that fact down here in our scenario.

As we said, there are three force couples here, each one creating a moment. Because of the way these couples are arranged, we can say that the axis of rotation for all three of them is the geometric center of this rectangle. The overall moment created by these three couples together, we’ll call that moment 𝑀, is equal to the moment created by our 153-newton force couple plus that created by our 199-newton force couple plus the moment due to our 73-newton force couple.

Let’s start in on solving for 𝑀, first by looking at this moment. If we sketch in the line of action of one of our 153-newton forces, then we can see that the perpendicular distance between this line and our axis of rotation is one-half the side length 𝐵𝐶. That is, it’s 12 centimeters. So leaving out units, if we take 153 and multiply it by 12, that will give us one-half of the magnitude of the total moment created about our axis of rotation by this couple of forces. To account for the entire couple then, we’ll multiply this result by two.

The last thing we need to figure out for this couple’s moment is whether it is positive or negative. We see that positive moments tend to create counterclockwise rotation. Based on the direction of the arrows of these forces, we see that, indeed, that’s the direction of this moment. So then 𝑀 153 c is equal to two times 153 times 12. Knowing this, let’s now look at our next couple’s moment.

If we sketch in the line of action of one of our 199-newton forces, we can see that the perpendicular distance between this line of action and our axis of rotation is one-half the length of line segment 𝐴𝐵. That is, it’s seven centimeters. For this couple’s moment then, we’ll say 199 times seven is multiplied by two to give the overall moment magnitude. We see that for our axis of rotation, these 199-newton forces tend to create a clockwise rotation. By our sign convention then, this moment is negative.

This then brings us to our final couple moment. Considering the 73-newton forces, if we sketch in the line of action of one of these forces, what we want to do is solve for this perpendicular distance, the distance from that line to our axis. We can note that the length of this line here, whatever it is, must be one-half of the length of this line here. And note that this longer line, we’ll call it 𝑑, can be considered part of a right triangle. This right triangle in orange is similar to another one in our sketch. This pink triangle is similar to the orange one.

Here’s how we know that. Whatever this angle is right here, it must be the case that that angle plus this angle in our orange triangle add up to 90 degrees. That’s because together they form one of the corners of a rectangle. And that means whatever this small angle in our pink triangle is must be the same as this small angle in our orange one. And therefore, this angle in our pink triangle must be equal in size to this one we’ve just marked out in our orange. All this to say, these two triangles are similar, which will help us solve for the distance 𝑑.

Clearing some space to look more closely, we know that this horizontal side length of our pink triangle is 24 centimeters. Then, for the vertical side length, we’re told that 𝑋 is the midpoint of 𝐴 and 𝐵, which tells us that this side of the triangle must be half of 14. So then, we have a right triangle where the two shorter side lengths are seven and 24. By the Pythagorean theorem, we can say that the hypotenuse of this pink triangle equals the square root of seven squared plus 24 squared. That equals exactly 25.

Knowing this, let’s consider that the hypotenuse of our orange triangle is the distance from 𝐶 to 𝑌 in our rectangle. Since 𝑌 is the midpoint of 𝐷𝐶, this distance must also be seven centimeters. And so now, here is what we can do. We can set up our ratio between the side lengths of our pink triangle and those of our orange. Specifically, we can say that the hypotenuse of our larger triangle, that’s 25, divided by the hypotenuse of our smaller triangle, that’s seven, equals the ratio of our pink triangle’s larger leg, that’s 24, divided by our orange triangle’s larger leg, which is 𝑑. If we cross multiply a bit, we find that 𝑑 equals seven times twenty-four twenty-fifths.

Entering this value in our calculator, we find it’s equal to exactly 6.72. So that’s the perpendicular distance in centimeters between the lines of action of our two 73-newton forces, which means that this distance here, which is what we wanted to solve for all along, must be one-half of that. In other words, the magnitude of the moment of this couple is two times 73 times 6.72 over two. Note also that these forces tend to create a clockwise rotation about our axis. So this moment is a negative moment.

We’re finally ready to add all of our terms together to solve for 𝑀. When we do, we find it’s exactly 395.44. The units of our forces were newtons, and the units of our distances were centimeters. So our final answer is that the total moment created by these three couples is 395.44 newton centimeters.

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