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.
