### Video Transcript

In the figure shown, a force of magnitude 23 root two newtons acts at a point ๐ด. Determine the moment vector of the force about the origin ๐ in newton meters.

Looking at the diagram provided, we can see we have an object consisting of a base and two columns on top of one another. And a wire connects the top of this object at a point marked ๐ด to a point ๐ท, which looks like itโs on the ground. The tension in this wire exerts a force at point ๐ด with a magnitude of 23 root two newtons. And this whole system is positioned on a set of 3D axes, with the ๐ฅ-axis pointing out of the screen, the ๐ฆ-axis pointing to the right, and the ๐ง-axis pointing up.

Our job is to find the moment vector of the force about the origin ๐. In other words, we want to find the vector that describes the moment which is produced by the force acting at ๐ด. To do this, we can use this equation. This tells us that the moment vector ๐, which is produced by a force acting at a distance from some point, is equal to the vector cross product of a displacement vector ๐ and force vector ๐
.

Now, the force vector ๐
is simply the force vector that produces the moment. So in this question, thatโs the 23 root two newton force that acts at ๐ด. The vector ๐ is the displacement vector of the point at which the force acts, relative to the point about which weโre calculating moments.

So in this question, since the force is acting at ๐ด and we want to calculate moments about the origin ๐, the vector ๐ that weโre interested in is the vector that takes us from ๐ to ๐ด. So using vector notation, we can say that the vector ๐ is equal to the vector ๐๐จ.

Now, in order to calculate the cross product of ๐ and ๐
, we need to know their ๐ฅ-, ๐ฆ-, and ๐ง-components. And in order to work these out, we can use the information thatโs provided in the diagram. So letโs start by finding the components of the vector ๐.

We previously noted that ๐ is equal to the vector ๐๐จ. And weโve drawn this on our diagram in orange. We can see that this vector points vertically upwards, in other words in the positive ๐ง-direction. And we can also see that itโs five meters long. In other words, this vector has an ๐ฅ-component of zero, a ๐ฆ-component of zero, and a ๐ง-component of positive five. So in unit vector notation, we can write this vector as zero ๐ข plus zero ๐ฃ plus five ๐ค. And since both of these terms are zero, we can just write ๐ as five ๐ค.

Next, we need to find the components of the vector ๐
, which weโve drawn in blue on our diagram. Working out the components of ๐
is a little bit trickier than working out the components of ๐. Letโs start by writing down what we know about ๐
. Well, firstly, weโve been told that its magnitude is 23 root two newtons. And because this force vector arises from tension in the wire that connects the points ๐ด and ๐ท, we know that the vector ๐
acts towards ๐ท from ๐ด. In other words, we could say that the vector ๐
is parallel to the vector ๐จ๐.

At this point, itโs important to note that although ๐
acts along ๐จ๐, itโs not equal to ๐จ๐. ๐
is a force vector, and ๐จ๐ is a displacement vector. So the magnitude of ๐
has nothing to do with the magnitude of ๐จ๐.

Now, the two statements weโve just written down describe the magnitude and the direction of ๐
. But in order to calculate the components of ๐
, we need to mathematically describe the direction that ๐
acts in. We know that itโs parallel to ๐จ๐. So letโs start by finding the components of ๐จ๐.

Now, the vector ๐จ๐ describes the displacement that would need to take place to take us from ๐ด to ๐ท. In order to make this journey, weโd need to travel five meters in the negative ๐ง-direction, three meters in the positive ๐ฅ-direction, and four meters in the positive ๐ฆ-direction. This means that the vector ๐จ๐ has a ๐ง-component of negative five, an ๐ฅ-component of three, and a ๐ฆ-component of four. So, using unit vector notation, we can write the vector ๐จ๐ as three ๐ข plus four ๐ฃ minus five ๐ค. So we know that ๐
points in the direction of this vector, but it has a magnitude of 23 root two.

At this point, it can be useful to think about what it means for two vectors to be parallel. If ๐
is parallel to ๐จ๐, that actually means we can obtain ๐
by scaling ๐จ๐. In other words, since the two vectors are pointing in the same direction, then if we were to scale or stretch ๐จ๐ by the right amount, then it would actually be equal to ๐
. Mathematically, we do this by multiplying the vector ๐จ๐ by some scalar constant, which we could call ๐พ. In other words, we just multiply it by a number.

So we can obtain ๐
by just multiplying this vector ๐จ๐ by some number such that its magnitude becomes 23 root two. But what number is this? Whatโs the scalar constant that we need to multiply ๐จ๐ by in order to get ๐
? One way we can figure this out is to find the unit vector which points in the direction of ๐จ๐. Letโs call this unit vector ๐ฎ, denoted with a hat symbol to show that itโs a unit vector.

Remember that a unit vector has a magnitude of one. So if we can find this unit vector and then multiply it by 23 root two, this will give us the vector that points in the same direction but that has a magnitude of 23 root two. In other words, it will give us ๐
. Fortunately, finding the unit vector that points in the direction of ๐จ๐ is relatively straightforward. We do this by dividing ๐จ๐ by its own magnitude. So we can say that the unit vector ๐ฎ is equal to ๐จ๐ divided by the magnitude of ๐จ๐.

We can find the magnitude of ๐จ๐ using the three-dimensional form of Pythagorasโs theorem. This magnitude is given by the square root of the ๐ฅ-component squared plus the ๐ฆ-component squared plus the ๐ง-component squared. Simplifying the denominator of this expression, three squared is nine, four squared is 16, and negative five squared is 25. And nine plus 16 plus 25 is 50.

Now, 50 isnโt a square number. So root 50 doesnโt have an integer answer. However, it is possible to further simplify the denominator. The square root of 50 can be written as the square root of 25 times two. And this can be written as the square root of 25 multiplied by the square root of two. The square root of 25 is five. So weโve shown that the unit vector ๐ฎ, which is the unit vector pointing in the direction of the vector ๐จ๐, is equal to ๐จ๐ divided by five root two, where five root two is the magnitude of ๐จ๐. And since weโve shown that ๐
is equal to 23 root two times ๐ฎ, this means that ๐
is equal to 23 root two times ๐จ๐ over five root two.

We can simplify this by writing it as 23 root two over five root two times ๐จ๐ and then canceling a factor of root two in the numerator and denominator, leaving us with ๐
equals 23 over five times ๐จ๐. So weโve shown that multiplying the vector ๐จ๐ by this scalar constant, 23 over five, gives us ๐
.

So letโs now calculate the components of ๐
by multiplying each component of ๐จ๐ by 23 over five. This gives us an ๐ฅ-component of three ๐ข times 23 over five, a ๐ฆ-component of four ๐ฃ times 23 over five, and a ๐ง-component of negative five ๐ค times 23 over five. Simplifying each of these terms, three ๐ข times 23 over five is 13.8๐ข. Four ๐ฃ times 23 over five is 18.4๐ฃ. And negative five ๐ค times 23 over five is negative 23๐ค. So weโve now found the components of ๐
.

We knew that ๐
had a magnitude of 23 root two and that it pointed in the direction of the vector ๐จ๐. So by finding the unit vector that points along ๐จ๐ and multiplying that by 23 root two, we obtained ๐
equals 13.8๐ข plus 18.4๐ฃ minus 23๐ค.

So now that weโve found the components of the vectors ๐ and ๐
, all we need to do is calculate that cross product. And this will tell us the moment vector ๐. The cross product of ๐ and ๐
is given by this three-by-three determinant, where the elements in the top row are the unit vectors ๐ข, ๐ฃ, and ๐ค. The elements in the middle row are the ๐ฅ-, ๐ฆ-, and ๐ง-components of the displacement vector ๐, written without unit vectors. And the elements in the bottom row are the ๐ฅ-, ๐ฆ-, and ๐ง-components of the force vector ๐
, also written without unit vectors.

Note that when we write the cross product, the order in which the two vectors are written affects the position in this determinant. So the first written vector ๐ goes in the middle row, and the second vector ๐
goes in the bottom row. This means that the cross product of ๐ and ๐
is not the same as the cross product of ๐
and ๐. So whenever we recall this formula, we need to be careful to remember the order of ๐ and ๐
.

Letโs now fill in the numerical values of the elements in this determinant. Starting with the components of ๐, we can see that it has an ๐ฅ-component of zero, a ๐ฆ-component of zero, and a ๐ง-component of five. Next, looking at the vector ๐
, this has an ๐ฅ-component of 13.8, a ๐ฆ-component of 18.4, and a ๐ง-component of negative 23.

This determinant is effectively calculated in three parts. First, we have the unit vector ๐ข multiplied by zero times negative 23 minus five times 18.4. Next, we subtract the unit vector ๐ฃ multiplied by zero times negative 23 minus five times 13.8. And finally, we add on the unit vector ๐ค multiplied by zero times 18.4 minus zero times 13.8.

Now, we just need to simplify each term. Starting with the ๐ข-term, we have zero times negative 23, which is zero. And weโre subtracting five times 18.4, which is equal to 92. Zero minus 92 is negative 92. So this term simplifies to negative 92๐ข. Next, looking at the ๐ฃ-term, we have negative ๐ฃ multiplied by zero times negative 23 โ thatโs zero โ minus five times 13.8, which is 69. Zero minus 69 is negative 69, and negative ๐ฃ times negative 69 is positive 69๐ฃ. Finally, looking at the ๐ค-term, we have ๐ค multiplied by zero times 18.4, which is zero, minus zero times 13.8, which is also zero. Zero minus zero is zero, and ๐ค times zero is zero. So this term disappears. So weโve now found the components of our moment vector ๐.

We can also note that because we found the moment vector ๐ in meters and the force vector ๐
in newtons, then our answer, which we obtained by finding the cross product of ๐ and ๐
, is expressed in newton meters as specified in the question.

So thereโs our final answer. The moment vector about the origin produced by the force shown in the diagram is equal to negative 92๐ข plus 69๐ฃ.