### Video Transcript

Given that π΄π΅πΆπ· 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 π΅.

Recall that the magnitude of a
moment π of a force π
acting from a point π about a pivot point π is given by
the magnitude of π
multiplied by the perpendicular distance π 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 π΅. 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 π΅: the force of three newtons acting
along the line π΄π΅ and the force of three newtons acting along the line π΅πΆ. Therefore, the perpendicular
distances between the lines of action of these two forces and the pivot point π΅ are
both zero. Therefore, the magnitude of their
moments about the point π΅ 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
π΅. On the diagram, for the force of
two newtons acting along the line π΄π·, this is the distance π one, since this is a
square and the line π΄π· is perpendicular to the line π΄π΅. Likewise, for the force of four
newtons acting along the line π·πΆ, this is the distance π two, since π·πΆ is
likewise perpendicular to πΆπ΅. And finally, for the force of four
root two newtons acting along the line π΄πΆ, this is the distance π 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 π΅. So they will all be positive
moments.

Therefore, the algebraic sum of the
moments π is equal to two times π one plus four times π two plus four root two
times π three. π one and π two are both equal to
the length of the side of the square, so these are both seven centimeters. π 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 π three is seven root two over
two. Therefore, π 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
π΅, 70. And the unit is newton
centimeters.