### Video Transcript

In this video, we’re looking at
moments in 3D. Specifically, we’ll look at how we
can represent a moment using a vector and how we can calculate the size and
direction of moments using cross product multiplication.

Let’s start by reminding ourselves
of what moments are. Generally, we can think of a moment
as a rotational or twisting force. For example, if we used a wrench to
turn a bolt, we could say that we exert a moment on the bolt. Now, moments are always defined
relative to some point in space. And we say that a moment acts about
this point. For example, if we apply some force
𝐹 to the end of the wrench, then it would be possible to calculate the moment
exerted about the center of the bolt. And the size of this moment
effectively describes how hard we’re turning the bolt.

Now, for two-dimensional systems
like this, we would typically calculate the moment by multiplying force and
distance. Specifically, we multiply the
magnitude of the applied force, in this case 𝐹, by the perpendicular distance
between the line of action of the force and the point that we’re calculating the
moment about. So in this example, if we want to
calculate the moment about the center of the bolt and the force is acting along this
dashed line, then the distance we use in our calculation is the perpendicular
distance between this dashed line and the center of the bolt. In this case, that happens to be
the same as the distance between the center of the bolt and the point at which the
force acts.

Multiplying these two quantities
together gives us the magnitude of the moment about the center of the bolt. And because the force is acting to
rotate the bolt in the counterclockwise direction, we’d also want to note down that
this moment is acting counterclockwise. Now, this works well for
two-dimensional problems. However, when we deal with
three-dimensional problems, words like clockwise and counterclockwise aren’t very
helpful. Instead, we need to be careful to
exactly define the direction of a moment. And the way that we do this is by
representing moments using vectors.

To see how we can do this, let’s
consider the same example of a wrench turning a bolt. But this time, rather than having
everything in the plane of the screen, we’ll consider how the system looks in 3D
space.

Okay, so here we have the head of
the bolt located at the origin of a set of 3D axes. We have the 𝑥-axis pointing to the
right of the screen, the 𝑦-axis pointing up, and the 𝑧-axis pointing out of the
screen. The handle of the wrench is aligned
along the 𝑥-axis. And let’s say that the force vector
which is applied to the handle points into the screen. So it’s perpendicular to the
𝑥-axis and parallel to the 𝑧-axis. Let’s consider the moment about the
center of the head of the bolt that this force produces.

At this point, we can note that if
we view the system from above, so we’re looking in the negative 𝑦-direction, we
would say that the force applies a counterclockwise moment. But if we view the system from
below, so we’re looking in the positive 𝑦-direction, we would see that the force
applies a clockwise moment. This shows us that descriptive
terms like clockwise and counterclockwise have limited use in 3D systems. Instead, we can define the exact
orientation of the moment produced by representing it with a vector.

In this case, the moment produced
by the force 𝐅 about the center of the bolt would actually be represented by a
vector pointing in the positive 𝑦-direction. Since this vector represents a
moment, let’s call it 𝐌. Now as we would expect, the
magnitude of this vector corresponds to the magnitude of the moment. So the larger the moment exerted
about the center of the bolt, the bigger this arrow would be. However, the direction of this
vector has a slightly less obvious meaning. After all, we think of a moment as
a rotational quantity. And in this case, the rotation that
occurs is in the plane of the 𝑥- and 𝑧-axes. So it might seem strange that the
moment vector is pointing vertically up.

It’s important to note that the
direction in which a moment vector points is not actually the direction in which the
moment physically acts. Instead, a moment vector points in
the same direction as the axis of rotation. A useful way of visualizing the
relationship between the direction of rotation and the direction that the moment
vector points is to use the right-hand rule. If we make a fist with your right
hand and extend our thumb in the direction of a moment vector, then the curl of our
fingers will show us the direction of rotation that that moment causes.

In our example, the direction of
the moment vector 𝐌 signifies that the force would cause the wrench to rotate in
this direction. So this is how we can interpret a
moment vector. But how do we actually calculate
it? Well, in two-dimensional problems,
we often use the equation 𝑀 equals 𝐹 times 𝑑. That is, the size of a moment is
equal to the magnitude of a force multiplied by the perpendicular distance between
the point we’re calculating moments about and the line of action of the force.

For 3D problems, we essentially use
a vector version of this equation. So instead of calculating just the
magnitude of a moment 𝑀, we now want to calculate a moment vector. And similarly, instead of using the
magnitude of a force in our calculation, we now use a force vector. Now, as we’ve said, 𝑑 represents
the perpendicular distance between the line of action of the force and the point
that we’re calculating moments about. We can think of the vector version
of this quantity as being the displacement vector between the point that we’re
calculating moments about, let’s call this 𝐴, and the point at which the force
acts, which we can call 𝐵. The vector that takes us from 𝐴 to
𝐵 we can just call 𝐑.

And multiplying the force vector by
this vector 𝐑 gives us the moment vector 𝐌. So we can see that this 3D vector
equation is similar to the equation that we would’ve used for two-dimensional
problems. But it’s important to remember at
this point that multiplying vectors is more complicated than multiplying
numbers. Specifically, to find the moment
vector 𝐌, we need to find the vector cross product of the displacement vector 𝐑
and the force vector 𝐅. Here, it’s useful to note that
cross multiplying 𝐑 by 𝐅, like this, is not the same as cross multiplying 𝐅 by
𝐑. Mixing up the order of these
vectors actually changes the result. So it’s important that we always
multiply these two vectors in the right order.

The cross product of 𝐑 and 𝐅 is
given by the determinant of this three-by-three matrix. The elements in the top row of this
matrix are 𝐢 hat, 𝐣 hat, and 𝐤 hat. These are the unit vectors in the
𝑥-direction, 𝑦-direction, and 𝑧-direction, respectively. In the middle row of this matrix,
we have 𝐑 𝑥, 𝐑 𝑦, and 𝐑 𝑧, which are the 𝑥-, 𝑦-, and 𝑧-components of the
displacement vector 𝐑. And finally, in the bottom row of
this matrix, we have 𝐅 𝑥, 𝐅 𝑦, and 𝐅 𝑧, which are the 𝑥-, 𝑦-, and
𝑧-components, respectively, of the force vector 𝐅.

We’ll look more closely at exactly
how to calculate the determinant of a three-by-three matrix like this in the
following example. But for now, let’s just note that
calculating this moment gives us a three-dimensional moment vector. This consists of three components:
𝐌 𝑥 times 𝐢 hat, 𝐌 𝑦 times 𝐣 hat, and 𝐌 𝑧 times 𝐤 hat. So 𝐌 𝑥 tells us the magnitude of
the vector that points in the 𝑥-direction, 𝐌 𝑦 tells us the magnitude of the
vector that points in the 𝑦-direction, and 𝐌 𝑧 tells us the magnitude of the
vector that points in the 𝑧-direction. The sum of all these components is
equal to the vector 𝐌.

When we’re considering the
different components of a moment vector, it can be useful to think of each component
as the amount of rotational force about an axis pointing in a certain direction. For example, the 𝑥-component of
our moment vector describes the component of the moment which would rotate an object
around an axis pointing in the 𝑥-direction. Looking at our example diagram
again, we can see that the moment vector produced in this case points entirely in
the 𝑦-direction. This tells us that the axis of
rotation produced by this moment points in the 𝑦-direction.

We can also see that while the
force vector 𝐅 acts in the 𝑧-direction and the displacement vector 𝐑 acts in the
𝑥-direction, both the 𝑥- and 𝑧-components of the moment vector in this case are
zero. This shows us how the cross product
of two vectors is always perpendicular to those vectors.

Okay, now that we’ve talked about
how to interpret and calculate moment vectors, let’s take a look at an example
problem.

If a force 𝐅 equal to six 𝐢 minus
seven 𝐣 minus eight 𝐤 is acting at a point 𝐴 five, negative eight, 11, find the
magnitude of the component of the moment of 𝐅 about the 𝑦-axis.

We can start by drawing a diagram
of the situation. We’re told that point 𝐴 has an
𝑥-coordinate of five, a 𝑦-coordinate of negative eight, and a 𝑧-coordinate of
11. That puts it about here. We’re told that a force vector 𝐅
equal to six 𝐢 minus seven 𝐣 minus eight 𝐤 acts at this point. So we can draw this force vector as
an arrow on our diagram like this. Now, we’re being asked to find the
magnitude of the component of the moment of 𝐅 about the 𝑦-axis. Let’s break this down step by step
so we can see exactly what the question is asking here.

Firstly, we can recall that a
moment, that is, a rotational force, can be represented by a vector. Generally, we can call this vector
𝐌. Now, we can calculate this vector
𝐌 by finding the vector cross product of the position vector at which the force
acts relative to the point we’re calculating moments about and the force vector
itself. So calculating this cross product
tells us the moment that’s produced by the force 𝐅. In other words, it tells us the
moment of 𝐅.

Now, when we calculate the cross
product of two three-dimensional vectors, like 𝐑 and 𝐅, the result is a vector
with the same number of dimensions. That means that when we calculate
the moment vector 𝐌, we obtain a three-dimensional vector. And we can write the components as
𝐌 𝑥 times 𝐢 hat plus 𝐌 𝑦 times 𝐣 hat plus 𝐌 𝑧 times 𝐤 hat. Each of these terms is one of the
components of the moment.

We can see that the question asks
us for the magnitude of one of the components. Specifically, it’s after the
magnitude of the component about the 𝑦-axis. In order to understand what this
means, let’s recall that the direction of a moment vector is parallel to the axis of
rotation of the moment. This means that the 𝑥-component of
a moment vector describes the component of that moment that rotates about an axis
pointing in the 𝑥-direction. The 𝑦-component of the moment
vector describes the component of that moment that rotates about an axis pointing in
the 𝑦-direction. And the 𝑧-component of that moment
vector describes the component of the moment that rotates about an axis pointing in
the 𝑧-direction.

What this means is that if we
calculate a moment about the origin, then, for example, the 𝑥-component of the
moment vector describes the component of the moment which causes rotation about the
𝑥-axis itself. Similarly, the 𝑦-component of the
moment vector describes the component of the moment which causes rotation about the
𝑦-axis, which is what the question is asking us about.

Finally, the fact that the question
just asks for the magnitude of this component just means that we need to give the
size of this vector component without writing it as a vector. This bit is fairly
straightforward. If the 𝑦-component is 𝐌 𝑦 times
𝐣 hat, then the unit vector 𝐣 hat just tells us the direction of this vector. It points in the 𝑦-direction. And the magnitude of vector is just
given by 𝐌 𝑦. So in order to find what this
question is looking for, we need to calculate the moment produced by the force 𝐅
about the origin. And in order to do this, we need to
find the cross product of these two vectors.

Remember that the vector 𝐑 is the
position vector of the point at which the force acts, in this case 𝐴, relative to
the point that we’re calculating moments about. In this case, that’s the
origin. So the vector 𝐑 is the position
vector going from the origin to the point 𝐴. That means that 𝐑, written as the
sum of its components, is effectively given by the coordinates of the point 𝐴
multiplied by the respective unit vectors. So we can say 𝐑 is equal to five
times 𝐢 hat minus eight times 𝐣 hat plus 11𝐤 hat.

To calculate the cross product of
𝐑 and 𝐅, we need to find the determinant of this three-by-three matrix, where the
elements in the top row of the matrix are the three basis vectors that we’re
using. The elements in the middle row of
the matrix are the three components of the first vector that we’re cross
multiplying. In this case, that’s 𝐑. And the elements in the bottom row
are the components of the second vector that we’re cross multiplying, in this case
𝐅. Notice that the order of 𝐑 and 𝐅
is important here. The cross product of 𝐑 and 𝐅 is
not the same as the cross product of 𝐅 and 𝐑.

There are effectively three parts
to calculating the determinant of this matrix. Firstly, we have the unit vector 𝐢
hat multiplied by 𝐑 𝑦 times 𝐅 𝑧 minus 𝐑 𝑧 times 𝐅 𝑦. Next, we subtract the unit vector
𝐣 hat multiplied by 𝐑 𝑥 times 𝐅 𝑧 minus 𝐑 𝑧 times 𝐅 𝑥. And finally, we have the unit
vector 𝐤 hat multiplied by 𝐑 𝑥 times 𝐅 𝑦 minus 𝐑 𝑦 times 𝐅 𝑥. These three terms are the 𝑥-, 𝑦-,
and 𝑧-components of the moment vector 𝐌. However, because we’re only
interested in the magnitude of the component of this moment vector that goes about
the 𝑦-axis, that means we’re only interested in this middle term. The magnitude of this component is
given by negative 𝐑 𝑥 times 𝐅 𝑧 minus 𝐑 𝑧 times 𝐅 𝑥.

𝐑 𝑥 is the magnitude of the
𝑥-component of the displacement vector 𝐑. In this case, that’s five. And 𝐅 𝑧 is the magnitude of the
𝑧-component of the force vector 𝐅. That’s negative eight. 𝐑 𝑧 is the magnitude of the
𝑧-component of 𝐑, which is 11. And 𝐅 𝑥 is the magnitude of the
𝑥-component of 𝐅, which is six. Altogether, that gives us negative
five times negative eight minus 11 times six. Five times negative eight is
negative 40, and 11 times six is 66. Negative 40 minus 66 is negative
106. And since there’s a negative sign
before the parentheses, this gives us a final answer of 106. And since the question doesn’t
specify any particular units of force or displacement, we can just say that our
final answer is 106 units of moment.

Let’s now recap the key points that
we’ve looked at in this video. Firstly, we’ve seen that a moment
can be represented by a three-dimensional vector 𝐌 which points in the direction of
the axis of rotation of the moment. Mathematically, the moment about a
point 𝐴 that’s produced by a force 𝐅 acting at a point 𝐵 is given by the vector
cross product of 𝐑 and 𝐅, where 𝐑 is the position vector of point 𝐵 relative to
point 𝐴. We can also use the right-hand rule
to remember the relationship between the directions of these vectors. If we make a fist with our right
hand and extend our thumb in the direction of a moment vector, then the curl of our
fingers shows us the direction of the rotation produced by this moment.