Video Transcript
As shown in the figure, 𝐴𝐵𝐶𝐷 is a square of side length one meter. Two forces, each of magnitude 16 kilogram-weight, are acting along the lines 𝐴𝐵 and 𝐶𝐷. And another two forces, each of magnitude 𝐹, are acting at 𝐵 and 𝐷, where one of them makes an angle of 15 degrees with line 𝐵𝐶 and the other makes an angle of 15 degrees with line 𝐷𝐴. If the couple formed by the first two forces is equivalent to that formed by the other two, determine the magnitude of 𝐹 and round to two decimal places.
So in this question, we have two couples. The first is the couple formed by the two 16-newton forces. And the second is the couple formed by the two forces of magnitude 𝐹. Now, of course, the question tells us that these are forces. But a couple consists of two equal noncolinear parallel forces acting in opposite directions. So we could’ve deduced this from the diagram itself. Now, whilst the resultant force of a couple is zero, the resultant itself isn’t; it’s a pure moment. Now, in fact, we’re told that the couple formed by the first two forces is equivalent to that formed by the other two. And so we can calculate the sum of the moments of each couple and set them equal to one another and solve for 𝐹.
Now, of course, a moment is found by multiplying the magnitude of the force by the perpendicular distance 𝑑 from the pivot. And when working with couples, this is independent of the reference location. So for ease, we’re going to construct the diagonals of this square and take moments about the center, the points where they intersect. Before we do though, let’s calculate the length of the diagonal. This will be important later on. We’re told that the side length of our square is one meter. And since the diagonal divides the square into two congruent right triangles, we can use the Pythagorean theorem to work out the length of that diagonal. Let’s define that to be 𝑥. By the Pythagorean theorem, 𝑥 squared equals one squared plus one squared or 𝑥 squared equals two.
We take the positive square root of two, and we find that 𝑥 is equal to root two. And so the length of our diagonal is root two meters. We’re going to take moments about the midpoint of this square, so that’s halfway along this line. Let’s call that point 𝑂. And we’ll begin by taking moments of our 16-newton forces. Now, remember, we said a moment is the product of the force and the perpendicular distance from the pivot. This tells us that the force and the distance themselves must be perpendicular to one another. So what is the component of our 16-newton force that is perpendicular to the line segment 𝐴𝐶?
Well, since the diagonal of the square bisects the angle at each vertex, angle 𝐵𝐴𝐶 shown is 45 degrees. This means the angle outside the square must also be 45 degrees since 45 plus 45 is 90. Then, we can construct a right triangle as shown. The hypotenuse of this is that 16-newton force. And we want the component of this force that’s adjacent to the angle of 45 degrees. And so since we have the adjacent and the hypotenuse, we can use the cosine ratio. This is 16 cos of 45. The same can be said at the other side for this 16-newton force acting at point 𝐶. So let’s take moments about 𝑂 of the 16-newton forces and we’re going to work in a clockwise direction.
The distance that these points are acting from point 𝑂 is root two over two. So each force has a moment of 16 cos of 45 times root two over two. These are both acting in the same direction. That’s the clockwise direction. And so their moments are going to be positive. So the sum of our moments is 16 cos 45 times root two over two plus 16 cos 45 times root two over two. Now, cos of 45 is indeed root two over two. So we can rewrite each of our moments as 16 times root two over two times root two over two. And this simplifies then to eight plus eight, which is of course 16, and that’s newton meters.
So let’s clear some space and take moments of our 𝐹 newton force. Now, once again, each of these forces acts at a distance of root two over two meters from point 𝑂. But we need that distance and that force to be perpendicular to one another. Since the angle given is 15 degrees and we know that the diagonal bisects the angle at the vertex, so it’s 45 degrees, we know that the angle here must be 30. 30 plus 15 plus 45 gives us 90 degrees. Once again, we need to calculate the component of the force that’s perpendicular to line segment 𝑂𝐵. So adding the right triangle shown, let’s find the adjacent side in this triangle. We know the hypotenuse. This means we can use the cosine ratio once again. Since the hypotenuse is 𝐹 newtons, the adjacent will be 𝐹 cos 30. And the same holds down at point 𝐷.
So this time, we’ll take moments about point 𝑂 of our 𝐹 newton force and we’re going to treat the clockwise direction as positive once again. At point 𝐵, we have 𝐹 cos 30 acting at a distance of root two over two meters from point 𝑂. And the same happens over at point 𝐷. We find the sum of these and they’re both positive because they’re trying to rotate the shape in a clockwise direction. cos of 30 is root three over two. So we get two lots of 𝐹 times root three over two times root two over two. And that simplifies to root six over two 𝐹.
Once again, we’re working in newton meters. And we were told that the couple formed by the first two forces is equivalent to that formed by the other two. So we can say that 16 must be equal to root six over two 𝐹. The equation then is root six over two 𝐹 equals 16. And we’ll solve for 𝐹 by dividing both sides by root six over two. 16 divided by root six over two is 13.0639 and so on. Correct to two decimal places, that’s 13.06. And of course, our forces are in kilogram weight. So the magnitude of 𝐹 correct to two decimal places is 13.06 kilogram weight.
