Question Video: Finding the Magnitude of Two Forces in a Coplanar Couple Mathematics

Video Transcript

𝐴𝐵𝐶𝐷 is a square. The forces 𝐹, two newtons, 𝐹, and two newtons act along the directions 𝐴𝐵, 𝐶𝐵, 𝐶𝐷, and 𝐴𝐷, respectively. Find the value of 𝐹 for the system to be in an equilibrium state.

We begin by sketching the square 𝐴𝐵𝐶𝐷 as shown. There are four forces 𝐹, two newtons, 𝐹, and two newtons acting along the four sides of the square. These forces form two couples, where a couple is a pair of parallel, but not coincident, forces of equal magnitudes and opposite directions. We have been asked to find the value of 𝐹 for the system to be in an equilibrium state. And we know for this to be true, the sum of the moments must equal zero. And the moment of a couple is equal to the magnitude of the force multiplied by the distance between the lines of action.

In this question, both couples have forces acting at points 𝐴 and 𝐶. This means that the distance between the lines of action will be the same for both couples. Convention dictates that moments acting in a counterclockwise direction are positive. In this question, the two-newton couple is acting in a counterclockwise direction. So the moment of this couple will be equal to two multiplied by 𝑑.

As the couple of force 𝐹 is acting in a clockwise direction, this will have a negative moment equal to negative 𝐹 multiplied by 𝑑. We know that the sum of these two moments equals zero. Simplifying the left-hand side, we have two 𝑑 minus 𝐹𝑑 is equal to zero. Since the distance 𝑑 cannot equal zero, we can divide through by this. We can then add 𝐹 to both sides such that 𝐹 is equal to two. And we can therefore conclude that the value of 𝐹 for the system to be in an equilibrium state is two newtons.