### Video Transcript

In this video, weβre looking at moments in 3D. Specifically, weβll 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.

Letβs 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 πΉ 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 weβre 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 πΉ, by the perpendicular distance between the line of action of the force and the point that weβre 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, weβd 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 arenβt 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, letβs consider the same example of a wrench turning a bolt. But this time, rather than having everything in the plane of the screen, weβll 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 π₯-axis pointing to the right of the screen, the π¦-axis pointing up, and the π§-axis pointing out of the screen. The handle of the wrench is aligned along the π₯-axis. And letβs say that the force vector which is applied to the handle points into the screen. So itβs perpendicular to the π₯-axis and parallel to the π§-axis. Letβs 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 weβre looking in the negative π¦-direction, we would say that the force applies a counterclockwise moment. But if we view the system from below, so weβre looking in the positive π¦-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 π
about the center of the bolt would actually be represented by a vector pointing in the positive π¦-direction. Since this vector represents a moment, letβs call it π. 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 π₯- and π§-axes. So it might seem strange that the moment vector is pointing vertically up.

Itβs 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 π 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 π equals πΉ times π. That is, the size of a moment is equal to the magnitude of a force multiplied by the perpendicular distance between the point weβre 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 π, 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 weβve said, π represents the perpendicular distance between the line of action of the force and the point that weβre calculating moments about. We can think of the vector version of this quantity as being the displacement vector between the point that weβre calculating moments about, letβs call this π΄, and the point at which the force acts, which we can call π΅. The vector that takes us from π΄ to π΅ we can just call π.

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

The cross product of π and π
is given by the determinant of this three-by-three matrix. The elements in the top row of this matrix are π’ hat, π£ hat, and π€ hat. These are the unit vectors in the π₯-direction, π¦-direction, and π§-direction, respectively. In the middle row of this matrix, we have π π₯, π π¦, and π π§, which are the π₯-, π¦-, and π§-components of the displacement vector π. And finally, in the bottom row of this matrix, we have π
π₯, π
π¦, and π
π§, which are the π₯-, π¦-, and π§-components, respectively, of the force vector π
.

Weβll 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, letβs just note that calculating this moment gives us a three-dimensional moment vector. This consists of three components: π π₯ times π’ hat, π π¦ times π£ hat, and π π§ times π€ hat. So π π₯ tells us the magnitude of the vector that points in the π₯-direction, π π¦ tells us the magnitude of the vector that points in the π¦-direction, and π π§ tells us the magnitude of the vector that points in the π§-direction. The sum of all these components is equal to the vector π.

When weβre 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 π₯-component of our moment vector describes the component of the moment which would rotate an object around an axis pointing in the π₯-direction. Looking at our example diagram again, we can see that the moment vector produced in this case points entirely in the π¦-direction. This tells us that the axis of rotation produced by this moment points in the π¦-direction.

We can also see that while the force vector π
acts in the π§-direction and the displacement vector π acts in the π₯-direction, both the π₯- and π§-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 weβve talked about how to interpret and calculate moment vectors, letβs take a look at an example problem.

If a force π
equal to six π’ minus seven π£ minus eight π€ is acting at a point π΄ five, negative eight, 11, find the magnitude of the component of the moment of π
about the π¦-axis.

We can start by drawing a diagram of the situation. Weβre told that point π΄ has an π₯-coordinate of five, a π¦-coordinate of negative eight, and a π§-coordinate of 11. That puts it about here. Weβre told that a force vector π
equal to six π’ minus seven π£ minus eight π€ acts at this point. So we can draw this force vector as an arrow on our diagram like this. Now, weβre being asked to find the magnitude of the component of the moment of π
about the π¦-axis. Letβs 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 π. Now, we can calculate this vector π by finding the vector cross product of the position vector at which the force acts relative to the point weβre calculating moments about and the force vector itself. So calculating this cross product tells us the moment thatβs produced by the force π
. In other words, it tells us the moment of π
.

Now, when we calculate the cross product of two three-dimensional vectors, like π and π
, the result is a vector with the same number of dimensions. That means that when we calculate the moment vector π, we obtain a three-dimensional vector. And we can write the components as π π₯ times π’ hat plus π π¦ times π£ hat plus π π§ times π€ 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, itβs after the magnitude of the component about the π¦-axis. In order to understand what this means, letβs recall that the direction of a moment vector is parallel to the axis of rotation of the moment. This means that the π₯-component of a moment vector describes the component of that moment that rotates about an axis pointing in the π₯-direction. The π¦-component of the moment vector describes the component of that moment that rotates about an axis pointing in the π¦-direction. And the π§-component of that moment vector describes the component of the moment that rotates about an axis pointing in the π§-direction.

What this means is that if we calculate a moment about the origin, then, for example, the π₯-component of the moment vector describes the component of the moment which causes rotation about the π₯-axis itself. Similarly, the π¦-component of the moment vector describes the component of the moment which causes rotation about the π¦-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 π¦-component is π π¦ times π£ hat, then the unit vector π£ hat just tells us the direction of this vector. It points in the π¦-direction. And the magnitude of vector is just given by π π¦. So in order to find what this question is looking for, we need to calculate the moment produced by the force π
about the origin. And in order to do this, we need to find the cross product of these two vectors.

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

To calculate the cross product of π and π
, 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 weβre using. The elements in the middle row of the matrix are the three components of the first vector that weβre cross multiplying. In this case, thatβs π. And the elements in the bottom row are the components of the second vector that weβre cross multiplying, in this case π
. Notice that the order of π and π
is important here. The cross product of π and π
is not the same as the cross product of π
and π.

There are effectively three parts to calculating the determinant of this matrix. Firstly, we have the unit vector π’ hat multiplied by π π¦ times π
π§ minus π π§ times π
π¦. Next, we subtract the unit vector π£ hat multiplied by π π₯ times π
π§ minus π π§ times π
π₯. And finally, we have the unit vector π€ hat multiplied by π π₯ times π
π¦ minus π π¦ times π
π₯. These three terms are the π₯-, π¦-, and π§-components of the moment vector π. However, because weβre only interested in the magnitude of the component of this moment vector that goes about the π¦-axis, that means weβre only interested in this middle term. The magnitude of this component is given by negative π π₯ times π
π§ minus π π§ times π
π₯.

π π₯ is the magnitude of the π₯-component of the displacement vector π. In this case, thatβs five. And π
π§ is the magnitude of the π§-component of the force vector π
. Thatβs negative eight. π π§ is the magnitude of the π§-component of π, which is 11. And π
π₯ is the magnitude of the π₯-component of π
, 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 thereβs a negative sign before the parentheses, this gives us a final answer of 106. And since the question doesnβt specify any particular units of force or displacement, we can just say that our final answer is 106 units of moment.

Letβs now recap the key points that weβve looked at in this video. Firstly, weβve seen that a moment can be represented by a three-dimensional vector π which points in the direction of the axis of rotation of the moment. Mathematically, the moment about a point π΄ thatβs produced by a force π
acting at a point π΅ is given by the vector cross product of π and π
, where π is the position vector of point π΅ relative to point π΄. 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.