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Question Video: Finding the Moment of a Force in Three Dimensions Mathematics

In the figure, ๐ด๐ต is a rod fixed to a vertical wall at end ๐ด. The other end ๐ต is connected to a wire ๐ต๐ถ, where ๐ถ is fixed to a different point on the same vertical wall. If the tension in the wire equals 65 N, calculate the moment of the tension about point ๐ด in newton-meters.

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Video Transcript

In the figure, ๐šจ๐šฉ is a rod fixed to a vertical wall at end ๐ด. The other end ๐ต is connected to a wire ๐šฉ๐‚, where ๐ถ is fixed to a different point on the same vertical wall. If the tension in the wire equals 65 newtons, calculate the moment of the tension about point ๐ด in newton meters.

Weโ€™ve been provided with a diagram of the system on a set of 3D axes. The ๐‘ฅ-axis is pointing out of the screen, the ๐‘ฆ-axis is pointing to the right, and the ๐‘ง-axis is pointing up. Here, we can see the rod with one end attached to the wall as described. And we can see that this wire connects the other end of the rod to the wall as well. So, the diagram only shows us what appears to be two pieces of wall floating in space. But the question makes it clear that both the wire and the rod are connected to the same vertical wall. So, the two walls shown in the diagram are actually parts of the same wall.

There are three specific points that are labeled in the diagram and also mentioned in the question: ๐ด, ๐ต, and ๐ถ. ๐ด is the point where one end of the rod is connected to the wall, and this is also the origin about 3D axes. ๐ต is where the other end of the rod is connected to the wire and ๐ถ is where the other end of the wire is connected to the wall.

The question asks us to calculate the moment of the tension about point ๐ด in newton meters. In other words, we want to calculate the moment about point ๐ด, which is produced by the tension in the wire. And to do this, we need to use this equation. This equation tells us that the moment vector ๐Œ is equal to the cross product of a displacement vector ๐‘ and a force vector ๐…. Specifically, ๐… is just the force vector that produces the moment. So, in this question, thatโ€™s the force vector acting at ๐ต due to the tension in the wire. And that ๐‘ is the displacement vector of the point at which the force acts relative to the point that we want to calculate the moment about.

So in this question, because the force is acting at point ๐ต and we want to calculate the moment about point ๐ด, the vector ๐‘ is the displacement vector that would take us from ๐ด to ๐ต. In other words, itโ€™s the vector ๐šจ๐šฉ. In this question, we havenโ€™t been given the components of ๐‘ or ๐…. So in order to work out their cross product, we need to find out their components from the information given in the question. Letโ€™s start by finding the components of the vector ๐‘.

Now, we previously noticed that ๐‘ is equal to the vector ๐šจ๐šฉ, where ๐ด is located at the origin and ๐ต is located 12 meters away in the positive ๐‘ฆ-direction. So, the vector ๐šจ๐šฉ has a magnitude of 12 meters and points in the positive ๐‘ฆ-direction. Equivalently, we could say that the components of this vector in the ๐‘ฅ- and ๐‘ง-direction is zero and the component in the ๐‘ฆ-direction is 12. Written as a vector in meters, this gives us zero ๐ข plus 12๐ฃ plus zero ๐ค, and in its simplest form, thatโ€™s just 12๐ฃ.

Okay, so next, we just need to find the components of the force vector ๐…. Then we can calculate the cross product of ๐‘ and ๐…, and this will give us the moment vector ๐Œ. Unfortunately, finding the vector ๐… is a bit trickier than finding ๐‘. Now, ๐… describes the force vector which is acting at ๐ต due to the tension in the wire. This means that we know two things about the vector ๐….

Firstly, the magnitude of ๐… is 65 newtons, as weโ€™re told that thereโ€™s 65 newtons of tension in the wire. And secondly, because the wire goes from ๐ต to ๐ถ, we know that the force acting at ๐ต must go along the line ๐šฉ๐‚. In other words, the vector ๐… is parallel to ๐šฉ๐‚. Itโ€™s useful to remind ourselves at this point that even though ๐… acts along ๐šฉ๐‚, itโ€™s not equal to ๐šฉ๐‚. ๐… is a force vector, and ๐šฉ๐‚ is a displacement vector. So, the magnitude of ๐… has nothing to do with the magnitude of ๐šฉ๐‚.

We can make this a bit clearer in our diagram by drawing the vector ๐… in a different color to the wire. Now, these two facts actually describe ๐… completely. After all, we have its magnitude and its direction. However, we need to calculate the actual components of ๐… in the ๐‘ฅ-, ๐‘ฆ-, and ๐‘ง-direction so that we can calculate ๐Œ. In order to find these components, we first need to better describe the direction that ๐… points in. We said that itโ€™s parallel to ๐šฉ๐‚, but what are the components of ๐šฉ๐‚? We can figure these out by looking at the measurements in the diagram.

To get from point ๐ต to point ๐ถ, weโ€™d need to travel 12 meters in the negative ๐‘ฆ-direction, three meters in the positive ๐‘ง-direction, and then four meters in the negative ๐‘ฅ-direction. So expressed in meters, the vector ๐šฉ๐‚ has an ๐‘ฅ-component of negative four ๐ข, a ๐‘ฆ-component of negative 12๐ฃ, and a ๐‘ง-component of three ๐ค. So, ๐… is parallel to this vector, but we know it has a magnitude of 65 newtons.

At this point, itโ€™s useful to remind ourselves that if ๐… is parallel to ๐šฉ๐‚, that means we can obtain ๐… by scaling ๐šฉ๐‚. In other words, since the two vectors point in the same direction, then if we stretch the vector ๐šฉ๐‚ or scale it by the right amount, then it will be equal to ๐…. Mathematically, we do this by multiplying the vector ๐šฉ๐‚ by some scalar constant ๐ค, in other words a number. So, we can obtain ๐… by multiplying the vector ๐šฉ๐‚ by a number so that its magnitude becomes 65. But what number is this? Whatโ€™s the scalar constant that we need to multiply ๐šฉ๐‚ by to give us ๐…?

One way we can solve this problem is to find the unit vector that points in the direction of ๐šฉ๐‚. In this case, letโ€™s call this unit vector ๐ฎ denoted with a hat symbol to show that itโ€™s a unit vector. Remember that a unit vector has a magnitude of one. If we can find this unit vector, then multiplying it by 65 will tell us the vector that points in the same direction and has a magnitude of 65. In other words, it will tell us ๐…. Now, fortunately, finding the unit vector that points in the direction of some vector is relatively straightforward. Itโ€™s done by simply dividing that vector by its own magnitude. That means ๐ฎ, the unit vector that points in the direction of ๐šฉ๐‚, is given by dividing the vector ๐šฉ๐‚ by the magnitude of ๐šฉ๐‚.

We can calculate the magnitude of ๐šฉ๐‚ by using the three-dimensional form of Pythagorasโ€™s theorem. Itโ€™s given by the square root of the ๐‘ฅ-component squared plus the ๐‘ฆ-component squared plus the ๐‘ง-component squared. Simplifying the denominator of this expression, negative four squared is 16, negative 12 squared is 144, and three squared is nine. 16 plus 144 plus nine is 169, and the square root of 169 is 13.

So, weโ€™ve shown that because ๐šฉ๐‚ has a magnitude of 13, dividing it by 13 gives us the unit vector that points in the direction of ๐šฉ๐‚. Since weโ€™ve shown that ๐… is equal to 65 times this unit vector ๐ฎ, this means that ๐… is equal to 65 times ๐šฉ๐‚ over 13, which we can equivalently write as 65 over 13 times ๐šฉ๐‚. In other words, scaling ๐šฉ๐‚ by a factor of 65 over 13 gives us ๐….

Note that weโ€™re also technically changing the units from meters to newtons. When we multiply the vector ๐šฉ๐‚ by a factor of 65 over 13, we effectively multiply each of the components of ๐šฉ๐‚ by 65 over 13. This gives us an ๐‘ฅ-component of negative four ๐ข times 65 over 13, a ๐‘ฆ-component of negative 12๐ฃ times 65 over 13, and a ๐‘ง-component of three ๐ค times 65 over 13. Note that 65 over 13 simplifies to just five. So, weโ€™re really multiplying each of these components by five.

We can now simplify this one term at a time. Negative four times 65 over 13 or negative four times five is negative 20, leaving us with an ๐‘ฅ-component of negative 20๐ข. Looking at the next term, we have negative 12๐ฃ times 65 over 13. This is equal to negative 12๐ฃ times five, which is negative 60๐ฃ. And finally, three ๐ค times 65 over 13 is three ๐ค times five, which is equal to 15๐ค. So, there we have it. We knew that ๐… had a magnitude of 65 newtons and pointed in the same direction as ๐šฉ๐‚. So, we found ๐… by calculating the unit vector in the direction of ๐šฉ๐‚ and then multiplying this by 65.

Now that we found the components of the displacement vector ๐‘ and the force vector ๐…, we can calculate the moment vector thatโ€™s produced by ๐… by working out the cross product of ๐‘ and ๐…. To do this, we calculate this three-by-three determinant where the elements in the top row are the unit vectors ๐ข, ๐ฃ, and ๐ค. The elements in the middle row are the ๐‘ฅ-, ๐‘ฆ-, and ๐‘ง-components of the displacement vector ๐‘, written without their unit vectors. And the elements in the bottom row are the ๐‘ฅ-, ๐‘ฆ-, and ๐‘ง-components of the force vector ๐…, also written without their unit vectors.

๐‘ has an ๐‘ฅ-component of zero, a ๐‘ฆ-component of 12, and a ๐‘ง-component of zero. And ๐… has an ๐‘ฅ-component of negative 20, a ๐‘ฆ-component of negative 60, and a ๐‘ง-component of 15. This determinant is then calculated effectively in three parts. First, we have the unit vector ๐ข multiplied by 12 times 15 minus zero times negative 60. Next, we subtract the unit vector ๐ฃ multiplied by zero times 15 minus zero times negative 20. And finally, we add the unit vector ๐ค multiplied by zero times negative 60 minus 12 times negative 20.

Okay, now, we just need to simplify each term. Looking at the ๐ข-term, we have 12 times 15 which is 180. And weโ€™re subtracting zero times negative 60, which is of course zero. So the ๐ข-term simplifies to 180๐ข. Looking at the ๐ฃ-term, we have zero times 15, which is zero. And weโ€™re subtracting zero times negative 20, which is also zero. This means that this term simplifies to negative ๐ฃ times zero, which is of course just zero. So, we donโ€™t need to write this term down. Finally, looking at the ๐ค-term over here, we have zero times negative 60 which is zero. And then, weโ€™re subtracting 12 times negative 20. 12 times negative 20 is negative 240. So, we have zero minus negative 240, which is just 240. So, this term simplifies to 240๐ค.

So, because weโ€™ve now found the cross product of ๐‘ and ๐…, that means that this is equal to the moment vector ๐Œ. We can also note that because we expressed the displacement vector ๐‘ in meters and the force vector ๐… in newtons, then weโ€™ve calculated a moment vector in newton meters as specified in the question. So, this is our final answer. The moment about point ๐ด which is produced by the tension shown in the diagram expressed in newton meters is 180๐ข plus 240๐ค.

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