Video: Finding the Moment Vector of an Inclined Force about the Origin in Three Dimensions

In the figure shown, a force of magnitude 23โˆš2 newtons acts at a point ๐ด. Determine the moment vector of the force about the origin ๐‘‚ in N โ‹… m.

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Video Transcript

In the figure shown, a force of magnitude 23 root two newtons acts at a point ๐ด. 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 ๐ด to a point ๐ท, which looks like itโ€™s on the ground. The tension in this wire exerts a force at point ๐ด with a magnitude of 23 root two newtons. And this whole system is positioned on a set of 3D axes, with the ๐‘ฅ-axis pointing out of the screen, the ๐‘ฆ-axis pointing to the right, and the ๐‘ง-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 ๐ด. To do this, we can use this equation. This tells us that the moment vector ๐Œ, 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, thatโ€™s the 23 root two newton force that acts at ๐ด. The vector ๐‘ is the displacement vector of the point at which the force acts, relative to the point about which weโ€™re calculating moments.

So in this question, since the force is acting at ๐ด and we want to calculate moments about the origin ๐‘‚, the vector ๐‘ that weโ€™re interested in is the vector that takes us from ๐‘‚ to ๐ด. So using vector notation, we can say that the vector ๐‘ is equal to the vector ๐Ž๐šจ.

Now, in order to calculate the cross product of ๐‘ and ๐…, we need to know their ๐‘ฅ-, ๐‘ฆ-, and ๐‘ง-components. And in order to work these out, we can use the information thatโ€™s provided in the diagram. So letโ€™s start by finding the components of the vector ๐‘.

We previously noted that ๐‘ is equal to the vector ๐Ž๐šจ. And weโ€™ve drawn this on our diagram in orange. We can see that this vector points vertically upwards, in other words in the positive ๐‘ง-direction. And we can also see that itโ€™s five meters long. In other words, this vector has an ๐‘ฅ-component of zero, a ๐‘ฆ-component of zero, and a ๐‘ง-component of positive five. So in unit vector notation, we can write this vector as zero ๐ข plus zero ๐ฃ plus five ๐ค. And since both of these terms are zero, we can just write ๐‘ as five ๐ค.

Next, we need to find the components of the vector ๐…, which weโ€™ve drawn in blue on our diagram. Working out the components of ๐… is a little bit trickier than working out the components of ๐‘. Letโ€™s start by writing down what we know about ๐…. Well, firstly, weโ€™ve 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 ๐ด and ๐ท, we know that the vector ๐… acts towards ๐ท from ๐ด. In other words, we could say that the vector ๐… is parallel to the vector ๐šจ๐ƒ.

At this point, itโ€™s important to note that although ๐… acts along ๐šจ๐ƒ, itโ€™s not equal to ๐šจ๐ƒ. ๐… is a force vector, and ๐šจ๐ƒ is a displacement vector. So the magnitude of ๐… has nothing to do with the magnitude of ๐šจ๐ƒ.

Now, the two statements weโ€™ve 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 itโ€™s parallel to ๐šจ๐ƒ. So letโ€™s start by finding the components of ๐šจ๐ƒ.

Now, the vector ๐šจ๐ƒ describes the displacement that would need to take place to take us from ๐ด to ๐ท. In order to make this journey, weโ€™d need to travel five meters in the negative ๐‘ง-direction, three meters in the positive ๐‘ฅ-direction, and four meters in the positive ๐‘ฆ-direction. This means that the vector ๐šจ๐ƒ has a ๐‘ง-component of negative five, an ๐‘ฅ-component of three, and a ๐‘ฆ-component of four. So, using unit vector notation, we can write the vector ๐šจ๐ƒ as three ๐ข plus four ๐ฃ minus five ๐ค. 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 ๐šจ๐ƒ, that actually means we can obtain ๐… by scaling ๐šจ๐ƒ. In other words, since the two vectors are pointing in the same direction, then if we were to scale or stretch ๐šจ๐ƒ by the right amount, then it would actually be equal to ๐…. Mathematically, we do this by multiplying the vector ๐šจ๐ƒ by some scalar constant, which we could call ๐พ. In other words, we just multiply it by a number.

So we can obtain ๐… by just multiplying this vector ๐šจ๐ƒ by some number such that its magnitude becomes 23 root two. But what number is this? Whatโ€™s the scalar constant that we need to multiply ๐šจ๐ƒ 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 ๐šจ๐ƒ. 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. 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 ๐šจ๐ƒ is relatively straightforward. We do this by dividing ๐šจ๐ƒ by its own magnitude. So we can say that the unit vector ๐ฎ is equal to ๐šจ๐ƒ divided by the magnitude of ๐šจ๐ƒ.

We can find the magnitude of ๐šจ๐ƒ using the three-dimensional form of Pythagorasโ€™s theorem. This magnitude is 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, three squared is nine, four squared is 16, and negative five squared is 25. And nine plus 16 plus 25 is 50.

Now, 50 isnโ€™t a square number. So root 50 doesnโ€™t 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 weโ€™ve shown that the unit vector ๐ฎ, which is the unit vector pointing in the direction of the vector ๐šจ๐ƒ, is equal to ๐šจ๐ƒ divided by five root two, where five root two is the magnitude of ๐šจ๐ƒ. And since weโ€™ve shown that ๐… is equal to 23 root two times ๐ฎ, this means that ๐… is equal to 23 root two times ๐šจ๐ƒ over five root two.

We can simplify this by writing it as 23 root two over five root two times ๐šจ๐ƒ and then canceling a factor of root two in the numerator and denominator, leaving us with ๐… equals 23 over five times ๐šจ๐ƒ. So weโ€™ve shown that multiplying the vector ๐šจ๐ƒ by this scalar constant, 23 over five, gives us ๐….

So letโ€™s now calculate the components of ๐… by multiplying each component of ๐šจ๐ƒ by 23 over five. This gives us an ๐‘ฅ-component of three ๐ข times 23 over five, a ๐‘ฆ-component of four ๐ฃ times 23 over five, and a ๐‘ง-component of negative five ๐ค times 23 over five. Simplifying each of these terms, three ๐ข times 23 over five is 13.8๐ข. Four ๐ฃ times 23 over five is 18.4๐ฃ. And negative five ๐ค times 23 over five is negative 23๐ค. So weโ€™ve 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 ๐šจ๐ƒ. So by finding the unit vector that points along ๐šจ๐ƒ and multiplying that by 23 root two, we obtained ๐… equals 13.8๐ข plus 18.4๐ฃ minus 23๐ค.

So now that weโ€™ve 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Œ. 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 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 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 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….

Letโ€™s 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 ๐‘ฅ-component of zero, a ๐‘ฆ-component of zero, and a ๐‘ง-component of five. Next, looking at the vector ๐…, this has an ๐‘ฅ-component of 13.8, a ๐‘ฆ-component of 18.4, and a ๐‘ง-component of negative 23.

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

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

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 thereโ€™s our final answer. The moment vector about the origin produced by the force shown in the diagram is equal to negative 92๐ข plus 69๐ฃ.

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