### Video Transcript

In the figure, determine the sum of the moment vectors of the forces 86 and 65 newtons about π in newton-centimeters.

For this question, weβll be calculating the moment vectors of two different forces. For the sake of clarity, lets label this first force of 86 newtons as πΉ one and the second force of 65 newtons as πΉ two. In order to calculate the moment of a force about a point, we can use the formula π equals π« cross π
, where π, π«, and π
are all vector quantities.

Before moving forward, let us briefly remind ourselves what each of these terms represents. π is, of course, our moment. Although we can think of moments as rotational quantities, they can still be represented by vectors. The magnitude of the vector, of course, represents the magnitude of the moment. The direction of the vector gives us the axis of rotation of our moment and also gives us the direction of rotation in 3D space. Next, weβve already seen that πΉ represents our forces. Note, however, that the π
in this formula is a vector quantity, and this will become important in a moment. Finally, π« is the position vector of the point of application of our force πΉ.

To get a clearer understanding of this, letβs use our diagram. The first force, πΉ one, is being applied at point πΆ. Next, we remind ourselves that when working with moments, we calculate them about a certain point. This question asked us to calculate the moments about point π, which weβll consider as the origin of our system. For our first force πΉ one, we will therefore need the position vector of point πΆ, that is, the vector from π to πΆ. We can now mark this on our diagram, and weβll label it as the vector π« one for later. Great, now that weβve recapped what each of our terms means, letβs move on to the calculation steps.

At this stage, we recall that the cross product can be calculated using the determinant of a three-by-three matrix as shown here. We wonβt go too far into detail on this. But as a quick reminder, the elements of the first row are the three basis vectors π’, π£, and π€, the second row are the components of our position vector π«, and the third row are the components of our force π
. And remember that the cross product is not a commutative operation, which means π« cross π
is not the same as π
cross π«. And this means that if we were to accidentally switch row two and three of our determinant, we would get a moment vector which points in the opposite direction from our desired result.

One final side note before we go forward, note that the question has asked us to express the moment in newton-centimeters. Since our forces have been given in newtons and all of the dimensions in our system have been given in centimeters, we wonβt need to worry about any unit conversions during this question. Okay, to calculate the moment of our 86-newton force about the origin, weβll first need the position vector, which weβve called π« one. From our diagram, we can see that point πΆ has a displacement of eight centimeters in the π₯-direction.

But again, weβre ignoring the units of centimeters, so weβll simply represent this as eight π’. Point πΆ does not have a displacement in the π¦-direction. It does, however, have a displacement of six units in the π§-direction. This means that the position vector π« one is eight π’ plus six π€. For this step, we have used the measurements on our cuboid diagram to give us the components of our position vector.

This cuboid can also tell us the components of our force vector π
one. We know from our question that πΉ one has a magnitude of 86 newtons. But from the diagram, we see that this vector is pointing in the π¦-direction, which means it has no π₯- and π§-component at all. The vector π
one can therefore be represented as 86π£. Note that we could also write π« one and π
one in vector form as shown. As we said earlier, weβll be calculating the moment of our first force, which weβll call π one, using the cross product of π« one and π
one. We can do so using the determinant of this three-by-three matrix with the component values of π« one and π
one substituted in.

Evaluating this cross product is somewhat of a lengthy calculation to write out in full. It should be noted that in our particular instance, we have a few helpful elements as zeros which should reduce our calculation. For the π’-component of our moment, we have zero times zero, which is of course zero, minus 86 times six. For the negative of the π£-component, we will have eight times zero minus zero times six. So this term will vanish. And for our π€-component, we will have eight times 86 minus zero times zero. Simplifying all of these terms, we are left with the moment π one of our first force. And this is minus 516π’ plus 688π€.

Great, now all we need to do is find the moment of our second force, which weβll call π two. And weβll be adding this to π one. Letβs clear some space to do so. The force π
two acts at the point πΌ, π« two is the position vector of πΌ, and we should be able to see that this vector is nine units in the π¦-direction and six units in the π§-direction. This means our vector is nine π£ plus six π€. The force vector π
two is slightly less trivial. We have its magnitude as 65 newtons. But since it does not simply point in one of the cardinal directions, weβll need to work out its components.

To do this, weβll again be using our diagram. Letβs consider what we would see if we took a side-on view from this perspective. Here, we form the following diagram. Our force π
two acts in a plane parallel to the π₯π§-plane. It therefore has no component in the π¦-direction, and we can therefore form a right triangle with the π₯- and the π§-components. We should be able to see that this vector has a component in the negative π§-direction and in the positive π₯-direction. In order to find the components exactly, we can use the measurements given on the diagram to give us the ratios of the side lengths. The force π
two connects the diagonals of the rightmost face of our cuboid.

Although our components are not six and eight directly, we can say that this triangle will follow the same proportions. And this means that the side length will be of the ratio six to eight. And we can represent this as six π and eight π if we want to be rigorous. Now, we could proceed to find the ratio of our hypotenuse using the Pythagorean theorem, but thereβs a handy shortcut here. We might recognize that saying something is in the ratio of six to eight is the same as saying itβs in the ratio three to four. This might make it easier to recognize our triangle as a three-four-five Pythagorean triple.

Our final step is then to find the right triangle with the ratios of three-four-five and hypotenuse of magnitude 65. Using our handy constant π, we can say that five π is equal to 65, and π is therefore 13. This means that our π₯- and π§-components are 52 and 39, respectively. Although the shortest length in this triangle is 39 units, it should be worth noting that the component of our force actually points in the negative π§-direction. And so the component is negative 39. Great, using our Pythagorean triple, we have found that the force π
two has components of 52 in the π₯-direction and negative 39 in the π§-direction.

We can now move on to calculating π two, the moment of our second force, using the determinant of a three-by-three matrix as before. Here are the calculation steps we can use to evaluate this determinant. If we simplify this, we are left with the following result for π two. We now have both moment vectors π one and π two, and the only remaining step is to add them together. Letβs clear some space to do so.

We can add these two moments together by summing the individual π’-, π£-, and π€-components of the vectors. Note that π one does not have a π£-component. So we simply take the π£-component of π two. Simplifying this, we are left with negative 867π’ plus 312π£ plus 220π€. This is the answer to our question. We have found sum of the moment vectors of the forces π
one and π
two about the origin π. Remember that our answer is indeed in units of newton-centimeters as requested by the question. This is because in our calculations, we used units of centimeters for our position vector π« and units of newtons for our force πΉ.