# Video: Finding the Moment of a Couple Equivalent to a System of Two Forces Acting on a Rhombus

π΄π΅πΆπ· is a rhombus in which its diagonal π΄πΆ = 7 cm and πβ π΄ = 60Β°. Given that two equal forces, each of magnitude 45 N, are acting along π΄π· and πΆπ΅, respectively, find the magnitude of the moment of the couple rounded to two decimal places if necessary.

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### Video Transcript

π΄π΅πΆπ· is a rhombus in which its diagonal π΄πΆ is seven centimeters and the measure of angle π΄ is 60 degrees. Given that two equal forces, each of magnitude 45 newtons, are acting along π΄π· and πΆπ΅, respectively, find the magnitude of the moment of the couple rounded to two decimal places if necessary.

Here, weβve sketched out our rhombus π΄π΅πΆπ·. We can recall that a rhombus is a four-sided shape where all the sides have the same length. Along with this, the opposite interior angles of the shape are the same. In this case then, the measure of angle π΄ is the same as that of angle πΆ. And likewise, the angles at π΅ and π· are the same.

In our problem statement, weβre told that the length of the diagonal line from corner π΄ to corner πΆ is seven centimeters and, along with this, that the measure of angle π΄, this angle here, is 60 degrees. Weβre told further about two forces acting along two of the sides of our rhombus, one from point π΄ to point π· and the other from point πΆ to point π΅. The magnitude of each force is 45 newtons.

Knowing all this, we want to solve for the magnitude of the moment of the couple. This word βcoupleβ refers to the pair of 45-newton forces. In general, if we have two forces which are equal in magnitude but act in opposite directions, and the two forces donβt lie along the same line of action, then we say that they form a couple. And more than that, if we define an axis of rotation between these lines of action, and note that the perpendicular distance between these lines and our axis of rotation is π, then we can say that the moment due to this couple β weβll call it π sub π β equals two times πΉ times π.

Note that often weβll see a subscript that indicates the force and the distance are perpendicular to one another. This then is the relationship thatβll let us solve for the magnitude of the moment of our couple of forces. As weβve drawn our rhombus, the lines of action of these forces run horizontally. And we can mark the perpendicular distance between these lines here and call that distance π. Weβve chosen this particular spot for our vertical line indicating this distance on purpose. It will let us take advantage of the fact that π is the vertical height of this right triangle. Another triangle here is this one in orange. Itβs not a right triangle, but we do know the length of this side of it. And more than that, we can solve for the interior angles of this orange triangle.

To start doing that, letβs clear some space at the top of our screen. And weβll draw an expanded sketch of our rhombus and these two triangles. Considering this view, what weβve said is we can solve for the interior angles of our orange triangle. To start doing that, letβs recall that the entire measure of angle π΄, this angle here, is 60 degrees. Now, this diagonal from π΄ to πΆ cuts that angle exactly in half. In other words, this angle here in our triangle is 30 degrees. And itβs the same thing over here because corner πΆ, like corner π΄, has a complete angle measure of 60 degrees. And so half of that, the interior angle of our orange triangle, is 30 degrees.

Knowing these two angles then and knowing that the sum of all the interior angles of a triangle must add up to 180 degrees, we can say that the measure of our orange triangle here must be 180 degrees minus two times 30 degrees. Thatβs 120 degrees. And now what we want to do is solve for the length of this side here, which we see is one side of our orange triangle. And itβs the hypotenuse of our pink triangle.

We can do this using the law of sines. This law says that, given any triangle with sides π, π, and π and corresponding interior angles πΌ, π½, and πΎ, the ratio of a side length to the sine of its corresponding angle is equal to the ratio of any other side length to the sine of its corresponding angle. In the case of our orange triangle, we know this side length of seven centimeters and this corresponding angle of 120 degrees. And we also know this interior angle, and we want to know this side length here.

If we call that side length β, because as we said it is the hypotenuse of the pink triangle, then leaving out units, we can write that β over the sin of 30 degrees equals seven over the sin of 120 degrees. That implies that β equals seven times this ratio of sines. The sin of 30 degrees is one-half, while the sin of 120 degrees is the square root of three over two.

With these values in place, we see that if we multiply both the denominator and the numerator here by two, then multiplied by those values, the factors of one-half will become one. And so we find that β equals seven over the square root of three. Knowing this, letβs consider our pink triangle. And note that since we know that this angle here is 120 degrees and that angle plus this angle, whatever it is, must equal 180 degrees, that tells us that this interior angle is 60 degrees.

Notice that we now know this side length as well as the corresponding opposite angle. And we want to solve for this side length, while we do know its corresponding opposite angle. Once again then, we can use the law of sines. π over the sin of 60 degrees equals β over the sin of 90 degrees. So the distance π equals β times this ratio of sines. The sin of 60 degrees equals exactly the square root of three over two, while the sin of 90 degrees is one. This entire ratio of sines then equals the square root of three over two. And when we plug in our known value of β, multiplying these values together, the square root of three cancels out. And we find that π equals seven-halves.

Weβre very close now to being able to solve for the magnitude of the moment of our couple. We can recall that the forces in our couple both have a magnitude of 45 newtons and that these lines are perpendicular to our distance π. The one thing we need to be careful of is, in our equation for the moment of a couple, this distance π is one-half the total perpendicular distance between the lines of action of our two forces. The π that we calculated is this total distance. So we could divide this distance in half to use it in this equation for π sub π. Or leaving out units, we could simply write that the magnitude of the moment of our couple equals 45 multiplied by seven over two, where rather than dividing this value by two and then multiplying it by two, as per our equation, weβve left it as it is, seven over two. 45 times seven-halves is exactly 157.5. This result is in units of newtons, the units of our force, multiplied by centimeters, the units of our distance.

We see then that we donβt need to round to two decimal places because this is our answer exactly. The magnitude of the moment of this couple is 157.5 newton centimeters.