# Video: Determining the Moment of a Force about a Point in Three Dimensions

Given that a force of magnitude 6 N is acting on 𝐶 as in the diagram, determine its moment vector about 𝐴 in newton-centimeters.

08:27

### Video Transcript

Given that a force of magnitude six newtons is acting on 𝐶 as in the diagram, determine its moment vector about 𝐴 in newton-centimeters.

Looking at the diagram, we can see that, indeed, we have a force vector with a magnitude of six newtons, which is acting on an object positioned on a set of 3D axes. We can see that the 𝑥-axis is pointing out of the screen, the 𝑦-axis is pointing to the right, and the 𝑧-axis is pointing up. Three specific points have been labeled on the object. These are 𝐴, which is the point about which we want to find a moment vector; 𝐵, which is located at the origin of our axes; and 𝐶, which is the point at which the force vector is acting.

The question asks us to calculate the moment vector about 𝐴 that’s produced by the force acting at 𝐶. To calculate the moment vector, we can use this equation. This tells us that a moment vector 𝐌 is given by the cross product of a displacement vector 𝐑 and a force vector 𝐅. The displacement vector 𝐑 is specifically the displacement vector at which the force acts relative to the point that we’re calculating a moment about. In this case, we’re calculating moments about 𝐴 and the force is acting at 𝐶. This means that in this question, the displacement vector 𝐑 that we’re looking for is the vector that takes us from 𝐴 to 𝐶. And using vector notation, we can say that the vector 𝐑 is equal to the vector 𝐀𝐂.

So now we can see both the vectors 𝐅 and 𝐑 in our diagram. The challenge in this question is that in order to calculate the cross product of these vectors, we need to figure out what their components are. And we can do this by using the measurements in the diagram. So, let’s start by finding the displacement vector 𝐑. We can do this by thinking about how we would get from 𝐴 to 𝐶. Starting at 𝐴, we can move some distance in the positive 𝑧-direction and then some distance in the positive 𝑦-direction, and this will take us to 𝐶. Using these measurements, we can see that the distance we would have to move in the 𝑧-direction is 16 plus eight centimeters, which gives us 24 centimeters.

And this measurement shows us that we would then have to move 16 centimeters in the 𝑦-direction. This means the displacement vector 𝐑 has an 𝑥-component of zero centimeters, a 𝑦-component of 16 centimeters, and a 𝑧-component of 24 centimeters. We can write this as zero 𝐢 hat plus 16𝐣 hat plus 24𝐤 hat. And we’ll make a note that we’re measuring this vector in centimeters.

So next we need to find the components of the force vector 𝐅. We can see from the diagram that the line of vector 𝐅 is 60 degrees away from this line, which is parallel to the 𝑥-axis, and 30 degrees away from this line, which is parallel to the 𝑧-axis. From this, we can conclude that the vector 𝐅 lies in the 𝑧𝑥-plane. In other words, its 𝑦-component is zero.

If we imagine looking at this vector from a different perspective, so we’re looking in the negative 𝑦-direction, we’d see something like this, where this is the 𝑧-component of the force vector, which we could call 𝐅 𝑧, and this is the 𝑥-component of the force vector, 𝐅 𝑥. Since the vector 𝐅 and its two components 𝐅 𝑧 and 𝐅 𝑥 form a right-angle triangle, we can use trigonometry to work out the magnitudes of the two components. Since the magnitude of the vector 𝐅 is six newtons, we can think of the length of the hypotenuse of this triangle as six. This means that the 𝑧-component has a magnitude of six sin 60 degrees and the magnitude of the 𝑥-component is six cos 60 degrees. It’s important to remember that these are just the magnitudes of these components.

Since 𝐅 𝑧 points in the negatives 𝑧-direction, this means the 𝑧-component of 𝐅 is negative six sin 60𝐤 hat. And since 𝐅 𝑥 points in the positive 𝑥-direction, the 𝑥-component of 𝐅 is six cos 60𝐢 hat. Since sin 60 has a value of root three over two, we can express the 𝑧-component as negative six times root three over two 𝐤 hat, which is negative three root three 𝐤 hat. And since the value of cos 60 is a half, we can express the 𝑥-component of 𝐅 as six times a half 𝐢 hat, which is, of course, three 𝐢 hat. Remembering that the 𝑦-component of 𝐅 is zero, we can express the vector 𝐅 as three 𝐢 hat plus zero 𝐣 hat minus three root three 𝐤 hat, making a note that this vector is measured in newtons.

Now, in the vector 𝐑 since the 𝑥-component is zero, we don’t actually need to write this term down. And likewise, the 𝑦-component of 𝐅 is zero, so we don’t need to write this term down. However, here we’ve left them on the screen so that we can see exactly how these components are used when we calculate this cross product, which is the next thing we need to do. The vector cross product of 𝐑 and 𝐅 is given by the determinant of this three-by-three matrix. The elements in the top row are the basis vectors 𝐢 hat, 𝐣 hat, and 𝐤 hat. The elements in the middle row are the 𝑥-, 𝑦-, and 𝑧-components of the displacement vector 𝐑 written without unit vectors. And the elements in the bottom row are the components of the vector 𝐅 also written without unit vectors.

We can notice that the order in which these vectors are written down affects their position in this matrix, which in turn will affect the result we get. In other words, the cross product of 𝐑 and 𝐅 is not equal to the cross product of 𝐅 and 𝐑. This means we need to be careful to keep the order right when we calculate a cross product. Since we’ve calculated all the components of 𝐑 and 𝐅, we can fill these into our matrix. The 𝑥-component of 𝐑 is zero, the 𝑦-component of 𝐑 is 16, and the 𝑧-component is 24. Now, looking at the vector 𝐅, the 𝑥-component is three, the 𝑦-component is zero, and the 𝑧-component is negative three root three.

The determinant of this matrix is effectively found in three parts. First, we have the unit vector 𝐢 hat multiplied by 16 times negative three root three minus 24 times zero. Next, we subtract the unit vector 𝐣 hat multiplied by zero times negative three root three minus 24 times three. And finally, we have the unit vector 𝐤 hat multiplied by zero times zero minus 16 times three.

Now, at this point, we can notice that all of the multiplications we’re performing involve multiplying a component of 𝐑 with a component of 𝐅. Since 𝐑 is measured in centimeters and 𝐅 is measured in newtons, this means that each of these multiplied quantities represents some quantity in newton-centimeters, which is the unit that we’ve been asked to use in the question. So we can see that when we calculate the cross product of two vectors, the units in which the result is expressed are the product of the units of the two vectors that we’re multiplying.

Getting back to our calculation, let’s now work out the values of all of these products inside the parentheses. 16 times negative three root three is negative 48 root three, and 24 times zero is zero. Zero times negative three root three is, of course, zero. And we’re subtracting 24 times three, which is 72. Next, we have zero times zero which is zero. And we’re subtracting 16 times three, which is 48. Now, let’s clear some space on the screen and move this line of working up here. And we can then simplify each of the terms. Negative 48 root three minus zero is, of course, just negative 48 root three. And this is being multiplied by 𝐢 hat. Next, we have zero minus 72, which is negative 72. And this is being multiplied by negative 𝐣 hat. Negative 72 times negative 𝐣 hat gives us positive 72𝐣 hat.

Finally, for the last term, we have zero minus 48, which is negative 48 times 𝐤 hat. And this is the final answer to our question. To find the moment vector about a point 𝐴 that’s produced by a force acting at point 𝐶, we need to find the cross product of the displacement vector from 𝐴 to 𝐶 and the force vector. This gives us a result of negative 48 root three 𝐢 hat plus 72𝐣 hat minus 48𝐤 hat. And in this case, since our displacement vector was measured in centimeters and our force vector was measured in newtons, this means that our result is measured in newton-centimeters.