Question Video: Finding the Moment of a Couple Equivalent to a System of Two Couples Acting on the Sides of a Parallelogram | Nagwa Question Video: Finding the Moment of a Couple Equivalent to a System of Two Couples Acting on the Sides of a Parallelogram | Nagwa

# Question Video: Finding the Moment of a Couple Equivalent to a System of Two Couples Acting on the Sides of a Parallelogram Mathematics • Third Year of Secondary School

## Join Nagwa Classes

π΄π΅πΆπ· is a parallelogram, in which π΄π΅ = 10 cm, π΅πΆ = 10 cm, and πβ π΄π΅πΆ = 150Β°. Forces of magnitudes 3, 6, 3, and 6 newtons are acting along π΄π΅, πΆπ΅, πΆπ·, and π΄π·, respectively. Find the moment of the resulting couple.

07:17

### Video Transcript

π΄π΅πΆπ· is a parallelogram, in which π΄π΅ equals 10 centimeters, π΅πΆ equals 10 centimeters, and the measure of angle π΄π΅πΆ is 150 degrees. Forces of magnitudes three, six, three, and six newtons are acting along π΄π΅, πΆπ΅, πΆπ·, and π΄π·, respectively. Find the moment of the resulting couple.

In our figure, we see this parallelogram with one force acting on each side. On the top and bottom sides, we could call them, are opposing forces of six newtons. On the other two sides are forces of three newtons, which also act in opposite directions. Our question asks us to solve for the moment of the resulting couple. Now we could think of these four forces as one combined couple, that is, two equal and opposite forces that donβt lie along the same line of action. Or we could equivalently think of these four forces as two couples. Either approach will get us the same result for the overall moment created.

As background, if we have two equal and opposite forces, here weβve called them πΉ and negative πΉ, whose lines of action are separated by a perpendicular distance π from an axis of rotation here, then we say that these forces form a couple. And this couple creates a moment, weβve called it π sub π, about this axis. The magnitude of π sub π is two times πΉ times π. Sometimes in this equation, weβll see this symbol included, which tells us that the force is perpendicular to the distance π. Note then that, in a couple, each force contributes equally to the moment of that couple. Effectively, we take the moment due to one of those two forces and multiply it by two to find the moment created by the couple.

Regarding the two couples of forces that act on the sides of our parallelogram, if we can solve for the perpendicular distances between the lines of action of these couples, here weβve called these distances π six and π three, respectively, then we can use that information to solve for the overall moment of these two couples. To start doing that, letβs clear some space on screen. And we can begin by writing a governing equation for the overall moment π that we want to calculate. That moment is equal to the sum of the moments created by our six-newton force couple and our three-newton force couple.

If we think first of the moment created by our six-newton force couple, when we recall the expression for the moment due to a couple, we can recognize that this moment will equal two times the magnitude of the force six newtons multiplied by half of the perpendicular distance between the lines of action of the two forces in the couple. We can note that two multiplied by one-half is one. So the magnitude of this moment simplifies to six times π sub six. What then is this distance π sub six? To solve for it, letβs consider this right triangle, where π sub six is the side length. This triangle, by the way, is similar to the triangle with π sub six that we see in our figure, which means that this angle right here is identical to this angle in our parallelogram.

We can now note that the angle at corner π΅ of our parallelogram is 150 degrees. And because this is a parallelogram, that means the angle at corner π·, here, has the same measure. Now, for a four-sided shape, like we have here, the sum of the interior angles of all four corners is 360 degrees. As we look to solve for this angle, which we donβt yet know, we can recognize first that itβs the same as the angle measure at corner π΄. And therefore, if we call this angle πΌ, we can write that 360 degrees equals two times 150 degrees, those are the angle measures at corners π΅ and π·, plus two times our unknown angle πΌ. Two times 150 degrees is 300 degrees. And if we subtract 300 degrees from both sides of this equation, we find that 60 degrees equals two times πΌ. Therefore, πΌ equals 30 degrees.

Returning to our right triangle with π six as its vertical side, we can say that the topmost corner in this triangle is point π· in our parallelogram, and the left corner is point πΆ. Thinking back to our problem statement, we were told that the length of the line segment from π΄ to π΅ is 10 centimeters, as is the length of the segment from π΅ to πΆ. And if π΄ to π΅ is 10 centimeters, because we have a parallelogram, so is πΆ to π·. So then, we can say the hypotenuse of our right triangle is 10 centimeters.

At this point, letβs recall the law of sines. This law tells us that, for a right triangle, the ratio of any given side length to the sine of the angle opposite that side is equal to that same ratio for any other side length in the triangle, which means, coming back to this right triangle, that π six over the sin of 30 degrees equals 10 over the sin of 90 degrees. Since the sin of 90 degrees is one and the sin of 30 degrees is one-half, we can multiply both sides of this equation by one-half, and we find that π six equals five. This number has units of centimeters which for now weβll leave off.

So then, we have a value for π six to substitute in to our equation for π sub six π. Once weβve done that, we can now consider whether this moment will be positive or negative according to our sign convention. Because these six-newton forces tend to create a counterclockwise rotation about our axis, the moment created by this couple will be positive. Knowing all this, we can now move on to solve for π sub three π. This is the moment created by our three-newton force couple. According to our equation, we would say that the magnitude of this moment equals two times three newtons multiplied by the distance π three over two. Once again, two multiplied by one-half is equal to one. So the magnitude of this moment equals the force magnitude multiplied by π three.

Once again, we can create a right triangle similar to the one in our original sketch. The far right corner of our new triangle is point π΄, and the far left corner is point π·. Note that when it comes to this interior angle, itβs the same as the angle here in our parallelogram. In other words, itβs equal to πΌ. πΌ, we recall, is 30 degrees. In our problem statement, weβre told that the side length π΅πΆ in our parallelogram is 10 centimeters long. That means that side π΄π· shares the same length, and so the hypotenuse of our right triangle here is 10 centimeters long.

Once again, we can make use of the law of sines, now to solve for π three. π three over the sin of 30 degrees equals 10 over the sin of 90 degrees. Once again, the sin of 30 degrees is one-half, and the sin of 90 degrees is one. So if we multiply both sides of this equation by one-half, we find that π three, like π six, is five centimeters. And now we take this value and we substitute it in for π three here.

Now letβs consider whether this moment is positive or negative. We see that these three-newton forces tend to create a clockwise rotation about our axis. By our sign convention then, this moment is negative. Here that is what we can say about the overall moment created by these two couples. Itβs equal to five times six minus three or 15. And we recall that the units of our forces are newtons and the units of our distances are centimeters. Our final answer then is that the moment created by these four forces, whether we think of them as two separate couples or one combined couple, is 15 newton centimeters.

## Join Nagwa Classes

Attend live sessions on Nagwa Classes to boost your learning with guidance and advice from an expert teacher!

• Interactive Sessions
• Chat & Messaging
• Realistic Exam Questions