Question Video: Finding the Sum of Moments of Five Forces Acting on a Square Mathematics • Third Year of Secondary School

Video Transcript

Given that 𝐴𝐵𝐶𝐷 is a square with side length seven centimeters and forces acting on it as shown in the figure, calculate the algebraic sum of the moments about vertex 𝐵.

Recall that the magnitude of a moment 𝐌 of a force 𝐅 acting from a point 𝑃 about a pivot point 𝑂 is given by the magnitude of 𝐅 multiplied by the perpendicular distance 𝑑 between the pivot point and the line of action of the force.

The question asks us to find the algebraic sum of the moments about the vertex 𝐵. So this is our pivot point. We have five forces acting on the square. Two of them have a line of action that passes directly through the pivot point 𝐵: the force of three newtons acting along the line 𝐴𝐵 and the force of three newtons acting along the line 𝐵𝐶. Therefore, the perpendicular distances between the lines of action of these two forces and the pivot point 𝐵 are both zero. Therefore, the magnitude of their moments about the point 𝐵 are also both zero, so we can ignore them.

For the remaining forces, we need to calculate the perpendicular distance between their lines of action and the point 𝐵. On the diagram, for the force of two newtons acting along the line 𝐴𝐷, this is the distance 𝑑 one, since this is a square and the line 𝐴𝐷 is perpendicular to the line 𝐴𝐵. Likewise, for the force of four newtons acting along the line 𝐷𝐶, this is the distance 𝑑 two, since 𝐷𝐶 is likewise perpendicular to 𝐶𝐵. And finally, for the force of four root two newtons acting along the line 𝐴𝐶, this is the distance 𝑑 three.

We also need to determine the direction, positive or negative, of each of the moments of each of the forces. The question tells us to use the convention that counterclockwise moments are positive. From the diagram, we can see that all three of the forces have a line of action in the counterclockwise direction around the point 𝐵. So they will all be positive moments.

Therefore, the algebraic sum of the moments 𝐌 is equal to two times 𝑑 one plus four times 𝑑 two plus four root two times 𝑑 three. 𝑑 one and 𝑑 two are both equal to the length of the side of the square, so these are both seven centimeters. 𝑑 three is equal to half the length of the diagonal of the square. The diagonal of any square is root two times its side length. So 𝑑 three is seven root two over two. Therefore, 𝐌 is equal to two times seven plus four times seven plus four root two times seven root two over two.

The root twos here will multiply to give two, canceling with the two on the denominator, leaving just four times seven. Performing this calculation and simplifying gives us the magnitude of the algebraic sum of the moments about vertex 𝐵, 70. And the unit is newton centimeters.