If 𝐅 equal to negative 19𝐢 plus 𝐿𝐣 plus two 𝐤 acts at a point 𝐴 with coordinates negative three, five, negative three and the moment of 𝐅 about the origin point is equal to four 𝐢 plus 63𝐣 plus 101𝐤, find the value of 𝐿.
So here we’re thinking about force 𝐅 which acts at a point 𝐴. And we’re told that this force produces a moment about the origin that can be represented by this vector. We can notice that we have numerical values for the 𝑥- and 𝑧-components of 𝐅, but the 𝑦-component is represented by this constant 𝐿. And it’s our job to find the value of 𝐿. Because this question talks about the moment vector that’s produced by a three-dimensional force acting at a certain point in space, let’s start by recalling the equation 𝐌 equals 𝐑 cross 𝐅.
This equation tells us that we can calculate a moment vector by taking the cross product of a displacement vector called 𝐑 and a force vector 𝐅. In this equation, 𝐅 is simply the force that produces the moment. So this corresponds to the vector called 𝐅 that we’ve been given in the question. 𝐑 is the displacement vector of the point at which the force acts relative to the point about which we’re calculating moments.
In this question, we’re told that the force acts at point 𝐴 with coordinates negative three, five, negative three. And the question talks about a moment that’s calculated about the origin point. This means that the vector 𝐑 we’ll use in our calculation is equal to the displacement vector that goes from the origin to point 𝐴. This is the same as saying it’s the displacement vector of 𝐴 relative to the origin. This vector is easy to find. We’re told in the question that the coordinates of 𝐴 are negative three, five, negative three. In other words, the point 𝐴 lies negative three units in the 𝑥-direction, five units in the 𝑦-direction, and negative three units in the 𝑧-direction from the origin. This means that the displacement vector that takes us from the origin to 𝐴 is given by negative three 𝐢 plus five 𝐣 minus three 𝐤.
We can also note that the moment vector 𝐌 is given in the question. It’s equal to four 𝐢 plus 63𝐣 plus 101𝐤. So looking at this equation, it seems like we know a lot about the vectors used. We’re told 𝐌 in the question, we can easily figure out the components of 𝐑, and we know two of the three components of 𝐅. The only unknown quantity in this equation is the 𝑦-component of 𝐅.
Now, given that we want to find this unknown component of 𝐅, it’s tempting to try to rearrange this equation to make 𝐅 the subject. For example, we might try to divide both sides of the equation by the vector 𝐑, as this looks like it might tell us the vector 𝐅, including its unknown component. Unfortunately, this idea isn’t correct. Firstly, it isn’t possible to divide a vector by another vector. And secondly, this equation uses a cross product. This isn’t the same as general multiplication. So even if we could divide both sides of the equation by the vector 𝐑, this wouldn’t actually help us rearrange the equation. In fact, the cross product operation doesn’t actually have an inverse, which means there’s no way of rearranging this equation to make 𝐅 the subject.
So, how do we find this unknown component of 𝐅? Well, here’s the idea. First, we obtain an expression for the moment vector 𝐌 by taking the vector 𝐅 as it’s given in this question and substituting it into this equation along with 𝐑. Since 𝐅 is expressed in terms of this unknown quantity 𝐿, the expression we end up with for 𝐌 will also be in terms of 𝐿. Next, we compare this expression for 𝐌, which is in terms of 𝐿, to the actual components of 𝐌, which are given in the question. By comparing the components from our expression to the components given in the question, we’ll be able to form equations that we can solve to find the value of 𝐿. So let’s start by calculating an expression for 𝐌.
We know that 𝐌 is equal to the cross product of 𝐑, which we found, and 𝐅, which is given in the question. This cross product is given by this three-by-three determinant, where the elements in the top row are the unit vectors 𝐢, 𝐣, and 𝐤. The elements in the middle row are the 𝑥-, 𝑦-, and 𝑧-components of the displacement vector 𝐑, written without their unit vectors. And the elements in the bottom row are the components of 𝐅, also written without unit vectors. Let’s start by substituting in the components of 𝐑. The 𝑥-component of 𝐑 is negative three 𝐢. So we write negative three here in our determinant. The 𝑦-component of 𝐑 is five 𝐣. So we write five into this spot. And the 𝑧-component is negative three 𝐤. So we write negative three here.
Next, we do the same thing for the components of the vector 𝐅. We know that 𝐅 is equal to negative 19𝐢 plus 𝐿𝐣 plus two 𝐤. So the elements in the bottom row of this determinant are negative 19, 𝐿, and two. Next, we need to compute this determinant. This is done in three parts. First, we have the unit vector 𝐢 multiplied by five times two minus negative three times 𝐿. Next, we subtract the unit vector 𝐣 hat multiplied by negative three times two minus negative three times negative 19. And finally, we add the unit vector 𝐤 multiplied by negative three times 𝐿 minus five times negative 19.
We can now simplify each of these terms. Looking at the first term, we have five times two, which is 10. And we’re subtracting negative three 𝐿, which is the same as adding three 𝐿. Looking at the next term, negative three times two is negative six. And we’re then subtracting negative three times negative 19. Well, negative three times negative 19 is positive 57. So we’re left with negative six minus 57, which is negative 63. Because we have negative 63 times negative 𝐣, in total, this term becomes positive 63𝐣.
Now, looking at the final term, we can see that inside the parentheses we have negative three 𝐿 minus five times negative 19. Five times negative 19 is negative 95. And since we’re subtracting this, this is equivalent to adding 95. And just to make things a bit more consistent, let’s swap around the negative three 𝐿 and positive 95 to give us 95 minus three 𝐿. And finally, we’ll move the unit vectors 𝐢 and 𝐤 to the end of those terms.
Okay, so now we’ve completed the first thing on our list. We obtained an expression for the moment vector 𝐌 in terms of this unknown quantity 𝐿.
We can now move on to the second part of our solution, comparing our expression for 𝐌 to the actual components of 𝐌 that are given in the question. The idea behind doing this is that the expression we’ve found describes exactly the same moment as this vector given in the question. In fact, we can see that the 𝑦-component of our expression is the same as the 𝑦-component given in the question, which is a good sign that we’re on the right track. The fact that these two vectors express the same quantity means that the 𝑥-components and 𝑧-components of these vectors must also be the same.
This enables us to come up with two equations. Firstly, because the 𝑥-component of our expression must be equal to the 𝑥-component of this vector, we can say that 10 plus three 𝐿 must be equal to four. And because the 𝑧-component of our expression must be equal to the 𝑧-component given in the question, we can also say that 95 minus three 𝐿 is equal to 101. Both of these equations are linear equations with a single unknown quantity, 𝐿. This means that solving either one of these equations should tell us the value of 𝐿. If we start with the equation on the left, we can see that subtracting 10 from both sides gives us three 𝐿 equals negative six. And then dividing both sides by three tells us that 𝐿 is equal to negative two.
Just to be safe, let’s solve this equation too and see if it gives us the same answer. Subtracting 95 from both sides of the equation gives us negative three 𝐿 equals six. And then dividing both sides of the equation by negative three also gives us 𝐿 equals negative two. So there’s our final answer. By coming up with our own expression for 𝐌 using the information given in the question and comparing this to the numerical values of the components of 𝐌 that we’re given in the question as well, we were able to find the unknown component of the force.
So, when a force equal to negative 19𝐢 plus 𝐿𝐣 plus two 𝐤 acts at the point 𝐴 with coordinates negative three, five, negative three and we know that the moment produced about the origin point is equal to four 𝐢 plus 63𝐣 plus 101𝐤, the value of 𝐿 must be negative two.