Question Video: Finding the Magnitude of the Resultant of Two Inclined Forces Acting on a Vertical Rod about a Point in Three Dimensions Mathematics • Third Year of Secondary School

Video Transcript

The forces of magnitudes 𝐹 sub one equals five root 673 newtons and 𝐹 sub two equals 16 root 569 newtons act along vectors 𝐀𝐁 and 𝐀𝐂, respectively, as shown in the figure. Given that vectors 𝐢, 𝐣, and 𝐤 are a right system of unit vectors in the directions of 𝑥, 𝑦, and 𝑧, respectively, find the sum of the moments of the forces about point 𝑂 in newton-meters.

We’re going to begin by reminding ourselves how we find the moment of a force about a given point. Since we’re working with a system of vectors in three dimensions, we’re going to use the cross product method. This says that the moment vector of a force about a point is equal to the cross product of the vector 𝑟 from that point to anywhere on the line of action of the force and the vector force itself. Now, in this question, we’re actually given the magnitude of the forces, rather than their vector form. So, let’s begin by finding the vectors that represent forces 𝐹 sub one and 𝐹 sub two.

Now, since these forces act along the vectors 𝐀𝐁 and 𝐀𝐂, respectively, we know force 𝐹 sub one must be some multiple of the vector 𝐀𝐁, whilst force 𝐹 sub two must be some multiple of the vector 𝐀𝐂. So, let’s find these two vectors. Let’s begin by identifying the vector of point 𝐴 on our diagram, taking 𝑂 at the base of this unit to be the origin. Since point 𝐴 lies on the 𝑧-axis , as shown on the diagram, at a height of 16 meters above point 𝑂, in coordinate form, we can say point 𝐴 has coordinates zero, zero, 16.

We’ll now repeat this process for point 𝐵. This lies in the 𝑥𝑦-plane, and so its 𝑧-coordinate is going to be zero. It lies at a perpendicular distance of 4.75 units from 𝑦 and 10 units from 𝑥, meaning its coordinate is 4.75, 10, zero. One last time let’s repeat this for point 𝐶. Again, this lies in the 𝑥𝑦-plane, so it will have a 𝑧-coordinate of zero. It lies at a perpendicular distance of 6.25 meters from the 𝑦-axis, and it lies at a perpendicular distance of five meters from the 𝑥-axis, but in the negative 𝑦-direction. And so, its coordinates are 6.25, negative five, zero.

This means we can now calculate the vector 𝐀𝐁 by considering the vector 𝐎𝐀 and the vector 𝐎𝐁. If 𝐎𝐁 is the vector that takes us from the origin to point 𝐵 and 𝐎𝐀 takes us from the origin to point 𝐴, then the vector 𝐀𝐁 is the vector 𝐎𝐁 minus the vector 𝐎𝐀. So, that is 4.75, 10, zero minus the vector zero, zero, 16. And of course, we can subtract vectors by simply subtracting their components. So, we see that vector 𝐀𝐁 is negative 16.

In a similar way, we can calculate the direction vector 𝐀𝐂 by subtracting the vector 𝐎𝐀 from the vector 𝐎𝐂. That’s 6.25, negative five, zero minus zero, zero, 16. And so, we have vector 𝐀𝐂 to be 6.25, negative five, negative 16.

We said that force 𝐹 sub one is going to be some multiple of this vector 𝐀𝐁, where force 𝐹 sub two is going to be some multiple of the vector 𝐀𝐂. Let’s clear some space and consider this in a little more detail. Another way to think about this is that the ratio of the vector force 𝐹 sub one with its magnitude must be equivalent to the ratio of the vector 𝐀𝐁 with its magnitude. This allows us to form an equation for force 𝐹 sub one. We multiply through by the magnitude of 𝐹 sub one in the equation that shows the ratios. And we see that the vector 𝐅 sub one must be equal to the magnitude of 𝐹 sub one times the vector 𝐀𝐁 divided by the magnitude of 𝐀𝐁.

Then, since the magnitude of the vector is found by calculating the positive square root of the sum of the squares of its components, we see that force 𝐹 sub one is five times the square root of 673 times the vector 𝐀𝐁 divided by its magnitude, which is the square root of 4.75 squared plus 10 squared plus negative 16 squared. Now, evaluating the denominator of this fraction and we actually get three times the square root of 673 over four.

This allows us to simplify by dividing through by a factor of root 673. And then five divided by three-quarters is equivalent to five times four-thirds; it’s twenty-thirds. So, our vector force 𝐅 sub one is twenty-thirds times the vector 𝐀𝐁, which is 4.75, 10, negative 16. We can, of course, multiply a vector by a scalar by multiplying the individual components. And that gives us that vector force 𝐅 sub one is the vector 95 over three, 200 over three, negative 320 over three.

We’re now going to clear some space and repeat this process for the vector force 𝐅 sub two. This time, we can say that the ratio of the force 𝐹 sub two with its magnitude must be equal to the ratio of the vector 𝐀𝐂 with its magnitude. This time, we get 16 times root 569 times the vector 𝐀𝐂 divided by its magnitude. Now, if we take the positive square root of the sum of the squares of its components, we get three root 569 over four. Once again, we notice that we can divide through by that constant factor of root 569. Then, 16 divided by three-quarters is 64 over three. So, we need to multiply vector 𝐀𝐂 by sixty-four thirds. And so, we see that the force vector 𝐅 sub two is four hundred thirds, negative three hundred and twenty thirds, negative one thousand and twenty-four thirds.

And that’s great because we now have enough information to work out the moment of each force and hence their sum. Since both forces are acting at point 𝐴 and we’re calculating the moments about point 𝑂, the sum of the moments is the vector 𝐎𝐀 crossed with the vector 𝐅 one plus the cross product of 𝐎𝐀 and 𝐅 two. Using the definition of the cross product as a determinant, we find the sum of the moments is equal to the determinant of the matrix 𝐢, 𝐣, 𝐤, zero, zero, 16, 95 over three, 200 over three, negative 320 over three and the determinant of 𝐢, 𝐣, 𝐤, zero, zero, 16, 400 over three, minus 320 over three, negative 1024 over three.

We know then that to calculate the determinant of a three-by-three matrix, we multiply each element in the top row by the determinant of the two-by-two matrix that remains when we eliminate that row and that column. So, it’s 𝐢 times the determinant of the two-by-two matrix zero, 16, 200 over three, negative 320 over three. And this determinant, of course, is zero times negative 320 over three minus 16 times 200 over three.

Next, we subtract 𝐣 times the determinant of the two-by-two matrix with elements zero, 16, 95 over three, negative 320 over three. We then add 𝐤 multiplied by the third two-by-two determinant. But actually, this determinant is zero. Finally, let’s repeat this for our second determinant, and the sum of the moments is as shown. All that’s left to do is to add the 𝐢- and 𝐣-components, noting that the 𝐤-components sum to zero. And when we do, we get 640𝐢 plus 2640𝐣.

And so, by calculating the actual vector forces 𝐹 sub one and 𝐹 sub two and then using the cross product, we’ve calculated the sum of the moments of our two forces about point 𝑂 in newton-meters to be 640𝐢 plus 2640𝐣.