### Video Transcript

A force of magnitude six newtons is
acting on πΆ and is represented by a vector in a plane perpendicular to the π¦-axis
as shown in the figure. Determine its moment vector about
π΄ in newton-centimeters.

Looking at the diagram, we can see
that we have a force vector with a magnitude of six newtons which is acting on an
object positioned on a set of 3D axes. We can see that the π₯-axis is
pointing out of the screen, the π¦-axis is pointing to the right, and the π§-axis is
pointing up. Three specific points have been
labeled on the object. These are π΄, which is the point
about which we want to find a moment vector; π΅, which is located at the origin of
our axes; and πΆ, which is the point at which the force vector is acting.

The question asked us to calculate
the moment vector about π΄ thatβs produced by the force acting at πΆ. To calculate the moment vector, we
can use the following equation. This tells us that a moment vector
π is given by the cross product of a displacement vector π and a force vector
π
. The displacement vector π is
specifically the displacement vector at which the force acts relative to the point
that weβre calculating a moment about. In this case, weβre calculating
moments about π΄ and the forces acting at πΆ. This means that in this question
the displacement vector π that weβre looking for is the vector that takes us from
π΄ to πΆ.

And using vector notation, we can
say that the vector π is equal to the vector ππ. We can now see both the vectors π
and π in our diagram. The challenge in this question is
that in order to calculate the cross product of these vectors, we need to find out
what their components are. We can do this by using the
measurements in the diagram. Letβs begin by finding the
displacement vector π.

We can do this by thinking about
how we would get from π΄ to πΆ. Starting at π΄, we can move some
distance in the positive π§-direction and then some distance in the positive
π¦-direction. And this will take us to point
πΆ. Using the measurements on the
diagram, we can see that the distance we would have to move in the π§-direction is
16 centimeters plus eight centimeters, which gives us 24 centimeters. We then see that we would have to
move 16 centimeters in the π¦-direction.

The displacement vector π has an
π₯-component of zero centimeters, a π¦-component of 16 centimeters, and a
π§-component of 24 centimeters. We can write this as zero π’ plus
16π£ plus 24π€, where π’, π£, and π€ are unit vectors in the π₯-, π¦-, and
π§-directions, respectively. We will also note at this point
that we are measuring this vector in centimeters.

Our next step is to find the
components of the force vector π
. We can see from the diagram that
the line of vector π
is 60 degrees away from a line parallel to the π₯-axis and 30
degrees away from a line parallel to the π§-axis. We can therefore conclude that
vector π
lies in the π₯π§-plane and is therefore perpendicular to the π¦-axis as
stated in the question. This means that the π¦-component of
the force vector is zero.

If we now imagine looking at this
vector from a different perspective, so weβre looking in the negative π¦-direction,
weβd see something like the diagram shown, where this is the π§-component of the
force vector, which we could call π
sub π§, and this is the π₯-component of the
force vector π
sub π₯. Since the vector π
and its two
components π
sub π§ and π
sub π₯ form a right triangle, we can use trigonometry to
work out the magnitudes of the two components.

The magnitude of vector π
is six
newtons. So, we can think of the length of
the hypotenuse of the triangle as six. This means that the π§-component
has a magnitude of six sin 60 degrees, and the magnitude of the π₯-component is six
cos 60 degrees. Itβs important to note that these
are just the magnitudes of the components. Since π
sub π§ points in the
negative π§-direction, this means that the π§-component of π
is negative six sin 60
π€. And since π
sub π₯ points in the
positive π₯-direction, the π₯-component of π
is six cos 60 π’. We know that sin of 60 degrees is
equal to root three over two and cos of 60 degrees is one-half. This means that the π§- and
π₯-components simplify to negative three root three π€ and three π’,
respectively.

Recalling that the π¦-component is
equal to zero, we have π
is equal to three π’ plus zero π£ minus three root three
π€. This time, we note that this vector
is measured in newtons. We are now in a position to
calculate the cross product of the two vectors. The vector cross product of π and
π
is given by the determinant of this three-by-three matrix, where the elements in
the top row are the unit factors, π’, π£, and π€. The elements in the middle row are
the π₯-, π¦-, and π§-components of the displacement vector π written without unit
vectors. And the elements in the bottom row
are the components of the vector π
, also written without unit vectors.

It is important to note that the
order in which these vectors are written down affects their position in the matrix,
which in turn will affect the result we get. In other words, the cross product
of π and π
is not equal to the cross product of π
and π. This means we need to be careful to
keep the right order when we calculate the cross product. Since weβve calculated all the
components of π and π
, we can now fill these into our matrix. The π₯-component of π is zero, the
π¦-component is 16, and the π§-component is 24. In vector π
, the π₯-component is
three, the π¦-component is zero, and the π§-component is negative three root
three.

The determinant of any
three-by-three matrix is effectively found in three parts. Firstly, we have the unit vector π’
multiplied by 16 multiplied by negative three root three minus 24 multiplied by
zero. The expression inside the
parentheses simplifies to negative 48 root three minus zero. So, our first term is negative 48
root three π’.

Next, we subtract the unit vector
π£ multiplied by zero multiplied by negative three root three minus 24 multiplied by
three. This time, our parenthesis
simplifies to zero minus 72. As this is equal to negative 72,
our second term becomes positive 72π£. Finally, we have the unit vector π€
multiplied by zero multiplied by zero minus 16 multiplied by three. The expression inside the
parentheses is equal to negative 48. So, the third term in our vector is
negative 48π€.

We now have the final answer to the
question. The moment vector of the force
acting at πΆ about point π΄ is negative 48 root three π’ plus 72π£ minus 48π€. And in this case, since our
displacement vector was measured in centimeters and our force vector was measured in
newtons, the moment vector is measured in newton-centimeters.