### Video Transcript

The forces of magnitudes πΉ sub one
equals five root 673 newtons and πΉ sub two equals 16 root 569 newtons act along
vectors ππ and ππ, respectively, as shown in the figure. Given that vectors π’, π£, and π€
are a right system of unit vectors in the directions of π₯, π¦, and π§,
respectively, find the sum of the moments of the forces about point π in
newton-meters.

Weβre going to begin by reminding
ourselves how we find the moment of a force about a given point. Since weβre working with a system
of vectors in three dimensions, weβre going to use the cross product method. This says that the moment vector of
a force about a point is equal to the cross product of the vector π from that point
to anywhere on the line of action of the force and the vector force itself. Now, in this question, weβre
actually given the magnitude of the forces, rather than their vector form. So, letβs begin by finding the
vectors that represent forces πΉ sub one and πΉ sub two.

Now, since these forces act along
the vectors ππ and ππ, respectively, we know force πΉ sub one must be some
multiple of the vector ππ, whilst force πΉ sub two must be some multiple of the
vector ππ. So, letβs find these two
vectors. Letβs begin by identifying the
vector of point π΄ on our diagram, taking π at the base of this unit to be the
origin. Since point π΄ lies on the π§-axis
, as shown on the diagram, at a height of 16 meters above point π, in coordinate
form, we can say point π΄ has coordinates zero, zero, 16.

Weβll now repeat this process for
point π΅. This lies in the π₯π¦-plane, and so
its π§-coordinate is going to be zero. It lies at a perpendicular distance
of 4.75 units from π¦ and 10 units from π₯, meaning its coordinate is 4.75, 10,
zero. One last time letβs repeat this for
point πΆ. Again, this lies in the π₯π¦-plane,
so it will have a π§-coordinate of zero. It lies at a perpendicular distance
of 6.25 meters from the π¦-axis, and it lies at a perpendicular distance of five
meters from the π₯-axis, but in the negative π¦-direction. And so, its coordinates are 6.25,
negative five, zero.

This means we can now calculate the
vector ππ by considering the vector ππ and the vector ππ. If ππ is the vector that takes us
from the origin to point π΅ and ππ takes us from the origin to point π΄, then the
vector ππ is the vector ππ minus the vector ππ. So, that is 4.75, 10, zero minus
the vector zero, zero, 16. And of course, we can subtract
vectors by simply subtracting their components. So, we see that vector ππ is
negative 16.

In a similar way, we can calculate
the direction vector ππ by subtracting the vector ππ from the vector ππ. Thatβs 6.25, negative five, zero
minus zero, zero, 16. And so, we have vector ππ to be
6.25, negative five, negative 16.

We said that force πΉ sub one is
going to be some multiple of this vector ππ, where force πΉ sub two is going to be
some multiple of the vector ππ. Letβs clear some space and consider
this in a little more detail. Another way to think about this is
that the ratio of the vector force πΉ sub one with its magnitude must be equivalent
to the ratio of the vector ππ with its magnitude. This allows us to form an equation
for force πΉ sub one. We multiply through by the
magnitude of πΉ sub one in the equation that shows the ratios. And we see that the vector π
sub
one must be equal to the magnitude of πΉ sub one times the vector ππ divided by
the magnitude of ππ.

Then, since the magnitude of the
vector is found by calculating the positive square root of the sum of the squares of
its components, we see that force πΉ sub one is five times the square root of 673
times the vector ππ divided by its magnitude, which is the square root of 4.75
squared plus 10 squared plus negative 16 squared. Now, evaluating the denominator of
this fraction and we actually get three times the square root of 673 over four.

This allows us to simplify by
dividing through by a factor of root 673. And then five divided by
three-quarters is equivalent to five times four-thirds; itβs twenty-thirds. So, our vector force π
sub one is
twenty-thirds times the vector ππ, which is 4.75, 10, negative 16. We can, of course, multiply a
vector by a scalar by multiplying the individual components. And that gives us that vector force
π
sub one is the vector 95 over three, 200 over three, negative 320 over three.

Weβre now going to clear some space
and repeat this process for the vector force π
sub two. This time, we can say that the
ratio of the force πΉ sub two with its magnitude must be equal to the ratio of the
vector ππ with its magnitude. This time, we get 16 times root 569
times the vector ππ divided by its magnitude. Now, if we take the positive square
root of the sum of the squares of its components, we get three root 569 over
four. Once again, we notice that we can
divide through by that constant factor of root 569. Then, 16 divided by three-quarters
is 64 over three. So, we need to multiply vector ππ
by sixty-four thirds. And so, we see that the force
vector π
sub two is four hundred thirds, negative three hundred and twenty thirds,
negative one thousand and twenty-four thirds.

And thatβs great because we now
have enough information to work out the moment of each force and hence their
sum. Since both forces are acting at
point π΄ and weβre calculating the moments about point π, the sum of the moments is
the vector ππ crossed with the vector π
one plus the cross product of ππ and π
two. Using the definition of the cross
product as a determinant, we find the sum of the moments is equal to the determinant
of the matrix π’, π£, π€, zero, zero, 16, 95 over three, 200 over three, negative
320 over three and the determinant of π’, π£, π€, zero, zero, 16, 400 over three,
minus 320 over three, negative 1024 over three.

We know then that to calculate the
determinant of a three-by-three matrix, we multiply each element in the top row by
the determinant of the two-by-two matrix that remains when we eliminate that row and
that column. So, itβs π’ times the determinant
of the two-by-two matrix zero, 16, 200 over three, negative 320 over three. And this determinant, of course, is
zero times negative 320 over three minus 16 times 200 over three.

Next, we subtract π£ times the
determinant of the two-by-two matrix with elements zero, 16, 95 over three, negative
320 over three. We then add π€ multiplied by the
third two-by-two determinant. But actually, this determinant is
zero. Finally, letβs repeat this for our
second determinant, and the sum of the moments is as shown. All thatβs left to do is to add the
π’- and π£-components, noting that the π€-components sum to zero. And when we do, we get 640π’ plus
2640π£.

And so, by calculating the actual
vector forces πΉ sub one and πΉ sub two and then using the cross product, weβve
calculated the sum of the moments of our two forces about point π in newton-meters
to be 640π’ plus 2640π£.