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

A force of magnitude 6 N is acting on 𝐢 and is represented by a vector in a plane perpendicular to the 𝑦-axis as shown in the figure. Determine its moment vector about 𝐴 in newton-centimeters.

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Video Transcript

A force of magnitude six newtons is acting on 𝐢 and is represented by a vector in a plane perpendicular to the 𝑦-axis as shown in the figure. Determine its moment vector about 𝐴 in newton-centimeters.

Looking at the diagram, we can see that 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 π‘₯-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; 𝐡, which is located at the origin of our axes; and 𝐢, which is the point at which the force vector is acting.

The question asked us to calculate the moment vector about 𝐴 that’s produced by the force acting at 𝐢. To calculate the moment vector, we can use the following 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 forces acting at 𝐢. This means that in this question the displacement vector 𝐑 that we’re looking for is the vector that takes us from 𝐴 to 𝐢.

And using vector notation, we can say that the vector 𝐑 is equal to the vector 𝐀𝐂. We can now 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 find out what their components are. We can do this by using the measurements in the diagram. Let’s begin by finding the displacement vector 𝐑.

We can do this by thinking about how we would get from 𝐴 to 𝐢. 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 point 𝐢. Using the measurements on the diagram, we can see that the distance we would have to move in the 𝑧-direction is 16 centimeters plus eight centimeters, which gives us 24 centimeters. We then see that we would have to move 16 centimeters in the 𝑦-direction.

The displacement vector 𝐑 has an π‘₯-component of zero centimeters, a 𝑦-component of 16 centimeters, and a 𝑧-component of 24 centimeters. We can write this as zero 𝐒 plus 16𝐣 plus 24𝐀, where 𝐒, 𝐣, and 𝐀 are unit vectors in the π‘₯-, 𝑦-, and 𝑧-directions, respectively. We will also note at this point that we are measuring this vector in centimeters.

Our next step is 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 a line parallel to the π‘₯-axis and 30 degrees away from a line parallel to the 𝑧-axis. We can therefore conclude that vector 𝐅 lies in the π‘₯𝑧-plane and is therefore perpendicular to the 𝑦-axis as stated in the question. This means that the 𝑦-component of the force vector is zero.

If we now imagine looking at this vector from a different perspective, so we’re looking in the negative 𝑦-direction, we’d see something like the diagram shown, where this is the 𝑧-component of the force vector, which we could call 𝐅 sub 𝑧, and this is the π‘₯-component of the force vector 𝐅 sub π‘₯. Since the vector 𝐅 and its two components 𝐅 sub 𝑧 and 𝐅 sub π‘₯ form a right triangle, we can use trigonometry to work out the magnitudes of the two components.

The magnitude of vector 𝐅 is six newtons. So, we can think of the length of the hypotenuse of the triangle as six. This means that the 𝑧-component has a magnitude of six sin 60 degrees, and the magnitude of the π‘₯-component is six cos 60 degrees. It’s important to note that these are just the magnitudes of the components. Since 𝐅 sub 𝑧 points in the negative 𝑧-direction, this means that the 𝑧-component of 𝐅 is negative six sin 60 𝐀. And since 𝐅 sub π‘₯ points in the positive π‘₯-direction, the π‘₯-component of 𝐅 is six cos 60 𝐒. We know that sin of 60 degrees is equal to root three over two and cos of 60 degrees is one-half. This means that the 𝑧- and π‘₯-components simplify to negative three root three 𝐀 and three 𝐒, respectively.

Recalling that the 𝑦-component is equal to zero, we have 𝐅 is equal to three 𝐒 plus zero 𝐣 minus three root three 𝐀. This time, we note that this vector is measured in newtons. We are now in a position to calculate the cross product of the two vectors. The vector cross product of 𝐑 and 𝐅 is given by the determinant of this three-by-three matrix, where the elements in the top row are the unit factors, 𝐒, 𝐣, and 𝐀. The elements in the middle row are the π‘₯-, 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.

It is important to note that the order in which these vectors are written down affects their position in the 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 right order when we calculate the cross product. Since we’ve calculated all the components of 𝐑 and 𝐅, we can now fill these into our matrix. The π‘₯-component of 𝐑 is zero, the 𝑦-component is 16, and the 𝑧-component is 24. In vector 𝐅, the π‘₯-component is three, the 𝑦-component is zero, and the 𝑧-component is negative three root three.

The determinant of any three-by-three matrix is effectively found in three parts. Firstly, we have the unit vector 𝐒 multiplied by 16 multiplied by negative three root three minus 24 multiplied by zero. The expression inside the parentheses simplifies to negative 48 root three minus zero. So, our first term is negative 48 root three 𝐒.

Next, we subtract the unit vector 𝐣 multiplied by zero multiplied by negative three root three minus 24 multiplied by three. This time, our parenthesis simplifies to zero minus 72. As this is equal to negative 72, our second term becomes positive 72𝐣. Finally, we have the unit vector 𝐀 multiplied by zero multiplied by zero minus 16 multiplied by three. The expression inside the parentheses is equal to negative 48. So, the third term in our vector is negative 48𝐀.

We now have the final answer to the question. The moment vector of the force acting at 𝐢 about point 𝐴 is negative 48 root three 𝐒 plus 72𝐣 minus 48𝐀. And in this case, since our displacement vector was measured in centimeters and our force vector was measured in newtons, the moment vector is measured in newton-centimeters.

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