𝐴𝐵𝐶𝐷 is a square, where the five forces, measured in newtons, are acting on it as shown in the figure. If the system of forces is equivalent to a couple, determine 𝐹 sub one and 𝐹 sub two.
We see in the diagram that there are five total forces — one of 20 newtons, one of 13, one of nine square root of two, 𝐹 one, and 𝐹 two — acting on the corners of the square. In particular, the forces originate from two of the corners of the square: corner 𝐴, where three of the forces originate, and corner 𝐶, where the other two do.
We’re told that if we consider all five of the forces together that they’re equivalent to a force couple. And given that information, we want to solve for 𝐹 sub one and 𝐹 sub two. To begin solving for 𝐹 sub one and 𝐹 sub two, we can recall the definition of a force couple, that it’s a pair of parallel forces with equal magnitude and opposite direction, which do not lie on the same line of action.
This means that the overall forces that originate at point 𝐴 and point 𝐶 should be parallel. They should have equal magnitude and opposite direction. Let’s consider what those forces are based on the information given.
At each of these two corners, we’ll endeavor to solve for the vertical as well as the horizontal components of the net force from each corner. Starting with corner 𝐴, we can see as we consider the horizontal component of this force that it has two contributors. There’s the force of 13 newtons acting completely horizontally. And there’s the horizontal component of the force of nine times the square root of two newtons.
Since the direction of that force bisects a 90-degree angle, we know that the horizontal component is equal to the magnitude of the force overall, nine times the square root of two, multiplied by the cos of 45 degrees. And since the cos of 45 degrees is equal to the square root of two over two, our expression simplifies to 13 newtons plus nine newtons, or 22 newtons. That’s the horizontal component of the force originating at point 𝐴.
When we consider the vertical component of that force, we see that it consists of the vertical component of nine times the square root of two, which is nine root two times the sin of 45 degrees, plus the unknown force 𝐹 sub two.
Just as with the horizontal component, we can write the vertical component of the force bisecting our 90-degree angle as nine times the square root of two times the square root of two over two, or simply nine newtons. So the vertical component of the total force acting from point 𝐴 is nine newtons plus 𝐹 sub two.
Now we move on to consider the forces originating at corner 𝐶. When we look at the vertical component of this force, we see that it’s equal seemingly to 20 newtons. And the horizontal component is equal to 𝐹 sub one. Because these two net corner forces form a force couple, we can generate two equations from the information we figured out so far.
First, we know that the horizontal components of the two forces, 22 newtons in the one case and 𝐹 sub one in the other, are equal. That is, 𝐹 sub one is equal to 22 newtons. And further, the vertical components of these two forces, nine newtons plus 𝐹 sub two and 20 newtons, are also equal to one another. This implies that 𝐹 sub two equals 20 newtons minus nine newtons, or 11 newtons.
So we’ve solved for 𝐹 sub one and 𝐹 sub two, the force magnitudes which make the overall forces in this diagram equal a force couple.