Video Transcript
The forces of magnitudes 𝐹 sub one
equals five root 673 newtons and 𝐹 sub two equals 16 root 569 newtons act along
vectors 𝐀𝐁 and 𝐀𝐂, respectively, as shown in the figure. Given that vectors 𝐢, 𝐣, and 𝐤
are a right system of unit vectors in the directions of 𝑥, 𝑦, and 𝑧,
respectively, find the sum of the moments of the forces about point 𝑂 in
newton-meters.
We’re going to begin by reminding
ourselves how we find the moment of a force about a given point. Since we’re working with a system
of vectors in three dimensions, we’re 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 𝑟 from that point
to anywhere on the line of action of the force and the vector force itself. Now, in this question, we’re
actually given the magnitude of the forces, rather than their vector form. So, let’s begin by finding the
vectors that represent forces 𝐹 sub one and 𝐹 sub two.
Now, since these forces act along
the vectors 𝐀𝐁 and 𝐀𝐂, respectively, we know force 𝐹 sub one must be some
multiple of the vector 𝐀𝐁, whilst force 𝐹 sub two must be some multiple of the
vector 𝐀𝐂. So, let’s find these two
vectors. Let’s begin by identifying the
vector of point 𝐴 on our diagram, taking 𝑂 at the base of this unit to be the
origin. Since point 𝐴 lies on the 𝑧-axis
, as shown on the diagram, at a height of 16 meters above point 𝑂, in coordinate
form, we can say point 𝐴 has coordinates zero, zero, 16.
We’ll now repeat this process for
point 𝐵. This lies in the 𝑥𝑦-plane, and so
its 𝑧-coordinate is going to be zero. It lies at a perpendicular distance
of 4.75 units from 𝑦 and 10 units from 𝑥, meaning its coordinate is 4.75, 10,
zero. One last time let’s repeat this for
point 𝐶. Again, this lies in the 𝑥𝑦-plane,
so it will have a 𝑧-coordinate of zero. It lies at a perpendicular distance
of 6.25 meters from the 𝑦-axis, and it lies at a perpendicular distance of five
meters from the 𝑥-axis, but in the negative 𝑦-direction. And so, its coordinates are 6.25,
negative five, zero.
This means we can now calculate the
vector 𝐀𝐁 by considering the vector 𝐎𝐀 and the vector 𝐎𝐁. If 𝐎𝐁 is the vector that takes us
from the origin to point 𝐵 and 𝐎𝐀 takes us from the origin to point 𝐴, then the
vector 𝐀𝐁 is the vector 𝐎𝐁 minus the vector 𝐎𝐀. 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 𝐀𝐁 is
negative 16.
In a similar way, we can calculate
the direction vector 𝐀𝐂 by subtracting the vector 𝐎𝐀 from the vector 𝐎𝐂. That’s 6.25, negative five, zero
minus zero, zero, 16. And so, we have vector 𝐀𝐂 to be
6.25, negative five, negative 16.
We said that force 𝐹 sub one is
going to be some multiple of this vector 𝐀𝐁, where force 𝐹 sub two is going to be
some multiple of the vector 𝐀𝐂. Let’s 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 𝐹 sub one with its magnitude must be equivalent
to the ratio of the vector 𝐀𝐁 with its magnitude. This allows us to form an equation
for force 𝐹 sub one. We multiply through by the
magnitude of 𝐹 sub one in the equation that shows the ratios. And we see that the vector 𝐅 sub
one must be equal to the magnitude of 𝐹 sub one times the vector 𝐀𝐁 divided by
the magnitude of 𝐀𝐁.
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 𝐹 sub one is five times the square root of 673
times the vector 𝐀𝐁 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; it’s twenty-thirds. So, our vector force 𝐅 sub one is
twenty-thirds times the vector 𝐀𝐁, 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
𝐅 sub one is the vector 95 over three, 200 over three, negative 320 over three.
We’re now going to clear some space
and repeat this process for the vector force 𝐅 sub two. This time, we can say that the
ratio of the force 𝐹 sub two with its magnitude must be equal to the ratio of the
vector 𝐀𝐂 with its magnitude. This time, we get 16 times root 569
times the vector 𝐀𝐂 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 𝐀𝐂
by sixty-four thirds. And so, we see that the force
vector 𝐅 sub two is four hundred thirds, negative three hundred and twenty thirds,
negative one thousand and twenty-four thirds.
And that’s 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 𝐴 and we’re calculating the moments about point 𝑂, the sum of the moments is
the vector 𝐎𝐀 crossed with the vector 𝐅 one plus the cross product of 𝐎𝐀 and 𝐅
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 𝐢, 𝐣, 𝐤, zero, zero, 16, 95 over three, 200 over three, negative
320 over three and the determinant of 𝐢, 𝐣, 𝐤, 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, it’s 𝐢 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 𝐣 times the
determinant of the two-by-two matrix with elements zero, 16, 95 over three, negative
320 over three. We then add 𝐤 multiplied by the
third two-by-two determinant. But actually, this determinant is
zero. Finally, let’s repeat this for our
second determinant, and the sum of the moments is as shown. All that’s left to do is to add the
𝐢- and 𝐣-components, noting that the 𝐤-components sum to zero. And when we do, we get 640𝐢 plus
2640𝐣.
And so, by calculating the actual
vector forces 𝐹 sub one and 𝐹 sub two and then using the cross product, we’ve
calculated the sum of the moments of our two forces about point 𝑂 in newton-meters
to be 640𝐢 plus 2640𝐣.
