Video: Finding the Resultant Moment Vector of Two Forces about the Origin in Three Dimensions

In the figure, if the forces 𝐅₁ = βˆ’7𝐒 βˆ’ 𝐣 + 3𝐀 and 𝐅₂ = βˆ’7𝐒 + 8𝐣 βˆ’ 6𝐀 are acting on the point 𝐴, where 𝐅₁ and 𝐅₂ are measured in newtons, determine the moment vector of the resultant about the point 𝑂 in newton-centimeters.

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Video Transcript

In the figure, if the forces 𝐅 one equal to negative seven 𝐒 hat minus 𝐣 hat plus three 𝐀 hat and 𝐅 two equal to negative seven 𝐒 hat plus eight 𝐣 hat minus six 𝐀 hat are acting on the point 𝐴, where 𝐅 one and 𝐅 two are measured in newtons, determine the moment vector of the resultant about the point 𝑂 in newton centimeters.

So in our diagram, we have what looks like three pipes connected together. And this object is positioned on a set of 3D axes. We have the π‘₯-axis pointing to the right, the 𝑦-axis pointing up, and the 𝑧-axis pointing out of the screen. We’ve also been given some measurements, which describe the height, width, and depth of the object. And we can also see that two points, 𝑂 and 𝐴, have been labeled. 𝑂 is the origin of our axes. And the question tells us that 𝐴 is the point at which two forces, 𝐅 one and 𝐅 two, are acting.

In this question, we’re being asked to determine the moment vector of the resultant about the point 𝑂. So we can break this question down into two parts. First, we need to find the resultant. And once we’ve done that, we need to determine the moment vector produced by this resultant about point 𝑂. In this question, since there are only two forces acting, 𝐅 one and 𝐅 two, we know that the resultant must refer to the resultant of these forces.

Now, to find the resultant of two forces, we simply need to find their vector sum. In other words, the resultant or total force 𝐅 𝑇 is equal to the sum of 𝐅 one and 𝐅 two. If we start with the π‘₯-components, we can see that 𝐅 one has an π‘₯-component of negative seven 𝐒 hat and 𝐅 two also has an π‘₯-component of negative seven 𝐒 hat. The sum of these is negative 14𝐒 hat. Next, looking at the 𝑦-components of 𝐅 one and 𝐅 two, 𝐅 one has a 𝑦-component of negative 𝐣 hat and 𝐅 two has a 𝑦-component of positive eight 𝐣 hat. The sum of these is positive seven 𝐣 hat. And finally, looking at the 𝑧-components, we have positive three 𝐀 hat and negative six 𝐀 hat, which sum to negative three 𝐀 hat. So this is the resultant force.

We can represent this idea by drawing force vectors on our diagram. Since the two forces 𝐅 one and 𝐅 two are acting at the same point, their overall effect on the system is the same as the effect of their resultant force 𝐅 𝑇. This means that when we determine the moment vector which is produced by this resultant about point 𝑂, we’re actually finding the overall moment which is produced by the two forces together.

To do this, let’s recall that the moment vector 𝐌 produced by a force acting at a point is equal to the vector cross product of the displacement vector 𝐑 and the force vector 𝐅. 𝐑 is the displacement vector of the point at which the force acts, relative to the point that we’re calculating moments about. In this question, we want to calculate the moment about the origin 𝑂 and the forces acting at the point 𝐴. So 𝐑 is the displacement vector that takes us from the origin to point 𝐴. And since we’re looking to find the moment that’s produced by the resultant, the force vector that we’re using is 𝐅 𝑇, which we’ve calculated.

If we want to find the cross product of 𝐑 and 𝐅 𝑇, we first need to find the components of 𝐑. And we can do this by using the measurements given in the diagram. And since we found the resultant force, let’s remove 𝐅 one and 𝐅 two from our diagram to make it a bit clearer.

To find the displacement vector 𝐑, let’s think about how we would get from the origin to point 𝐴. One way we could do this is to imagine traveling along this system of three pipes. So starting from the origin, we could travel up to the top of the first pipe. And we can see from the diagram that this is a displacement of 12 centimeters in the positive 𝑦-direction. And traveling from here to the end of the next pipe would involve a displacement of eight centimeters in the positive 𝑧-direction. And finally, traveling from here to 𝐴 would involve a displacement of nine centimeters in the positive π‘₯-direction. So, in total, the displacement vector 𝐑 has an π‘₯-component of positive nine centimeters, a 𝑦-component of positive 12 centimeters, and a 𝑧-component of positive eight centimeters.

So in vector notation, 𝐑 equals nine 𝐒 hat plus 12𝐣 hat plus eight 𝐀 hat. So we can now find the overall moment vector by finding the cross product of 𝐑 and 𝐅 𝑇. Here, we can note that because 𝐑 is expressed in centimeters and 𝐅 𝑇 is expressed in newtons, when we calculate the cross product of 𝐑 and 𝐅 𝑇, the result will be expressed in newton centimeters, which is the units that we’re asked to use in the question.

To calculate the cross product of these two vectors, we need to find the determinant of a three-by-three matrix, where the elements in the top row are the unit vectors 𝐒 hat, 𝐣 hat, and 𝐀 hat. The elements in the middle row are the π‘₯-, 𝑦-, and 𝑧-components of the vector 𝐑, expressed without their unit vectors. And the elements in the bottom row are the π‘₯-, 𝑦-, and 𝑧-components of the vector 𝐅 𝑇, also expressed without unit vectors.

Note that the order in which these vectors are written down is important. The first vector goes in the middle row, and the second vector goes in the bottom row. This means that the cross product of 𝐑 and 𝐅 𝑇 is not the same as the cross product of 𝐅 𝑇 and 𝐑. So we need to be careful to remember the order of the displacement vector and the force vector whenever we recall this equation.

Going back to our calculation, we can now fill in the components of the two vectors 𝐑 and 𝐅 𝑇. Remember that we write these components without their unit vectors. So 𝐑 has an π‘₯-component of nine, a 𝑦-component of 12, and a 𝑧-component of eight. And 𝐅 𝑇 has an π‘₯-component of negative 14, a 𝑦-component of seven, and a 𝑧-component of negative three.

Calculating this determinant is effectively done in three parts. First, we have the unit vector 𝐒 hat multiplied by 12 times negative three minus eight times seven. Next, we subtract the unit vector 𝐣 hat multiplied by nine times negative three minus eight times negative 14. And finally, we add to this the unit vector 𝐀 hat multiplied by nine times seven minus 12 times negative 14.

We can now simplify each of these terms. 12 times negative three is negative 36. And we’re then subtracting eight times seven, which is 56. And negative 36 minus 56 is negative 92, leaving us with negative 92𝐒 hat. Looking at the next term, nine times negative three is negative 27. And we’re then subtracting eight times negative 14, which is negative 112. Subtracting negative 112 is of course the same as adding 112. And negative 27 plus 112 is 85. So, in total, this term simplifies to negative 85𝐣 hat.

Now, looking at the third and final term, we have nine times seven, which is 63. And we’re subtracting 12 times negative 14, which is negative 168. Once again, we’re subtracting a negative number. So this is the same as adding 168. And 63 plus 168 is 231. So, in total, this final term is 231𝐀 hat. So this is our final answer.

The resultant moment produced about the point 𝑂 is negative 92𝐒 hat minus 85𝐣 hat plus 231𝐀 hat. Since we expressed 𝐑 in centimeters and 𝐅 𝑇 in newtons, then our result, which is the cross product of these two vectors, is expressed in newton centimeters.

It’s worth noting that we could’ve obtained the same answer by calculating the moment produced by the force 𝐅 one and the moment produced by the force 𝐅 two separately and then adding them together to find their resultant moment. However, this would’ve taken longer as it would involve calculating two cross products, one for 𝐅 one and one for 𝐅 two. Since both of these forces were acting at the same point, it’s faster in this case to just add the two forces together and find their resultant force and then just calculate a single cross product. However, in cases where we have forces acting at different points on an object, we wouldn’t be able to use this shortcut. We would instead have to calculate the moment produced by each force and then add those moments together to find their resultant moment.

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