 In this example problem, we are going to analyze the effect of a 100 kilonewton force acting on the horizontal stabilizer of a commercial jet aircraft. But before we jump into calculations, let's try and understand what the horizontal stabilizer does and what we could possibly expect it to do to the aircraft. If you think about the function of a horizontal stabilizer, it is a control surface that generates forces to provide pitch control and stability. It allows the aircraft to pitch up, like this, or down, like this. And that motion is a rotation about a specific point, its center of gravity to be precise. And notice it is reasonable to expect that the forces on this surface will generate a moment about a center of gravity because a moment is what causes rotation. So what can we say about the moment this force generates about the center of gravity? If we want to analyze what this moment would be, we need to gather some information. First, we need to establish a reference coordinate system. The aircraft body makes a convenient reference line, so let's define this as the x-axis. I will then define the y-axis as positive upwards and center this coordinate system on our point of interest, 0.0. Next, we need to obtain some more information about the orientation and location of the forces on the horizontal stabilizer. For this problem, our force is oriented 60 degrees counterclockwise from the x-axis. The linear distance from 0.0 to the location of the force acting on the stabilizer is 13.2 meters, measured at an angle of 20 degrees. With this information established, we are now ready to answer the question, what is the moment generated at 0.0 by the 100 kilonewton force? There are multiple ways to solve this problem, but we will start with the method of vector decomposition. We can now begin to build our model of the problem. We will start by representing the entire aircraft as a single point, its center of gravity, or 0.0. We can then put our force on the stabilizer in this model, which we will denote as vector F. And from our problem description, we know that F has a magnitude of 100 kilonewtons, and it's oriented at 60 degrees. Next, we can define our position vector from 0.0 to this force as position vector R. From our problem, we also know that this vector has a magnitude of 13.2 meters and is oriented 20 degrees. Now, we want to decompose these vectors into X and Y components. Starting with the position vector, it will decompose into component RX and component Ry. Similarly, vector F will decompose into FX and FY. The reason we decompose the vectors into components is that all of our decomposed distances are either parallel or perpendicular to our decomposed force components. This allows us to determine the moment by scalar multiplication, as the moment of a force about a given point is simply the force multiplied by the perpendicular distance from that point to the force. Knowing this, we can easily see that a moment is generated by our force component FY with its moment arm RX. This component of our moment will tend to cause a counterclockwise rotation as shown here. And in order to see that this counterclockwise rotation is positive, we have to look at our coordinate system. We established X as positive to the right, Y as positive upwards, which would mean that Z is going to be positive out of the screen. If we use the right hand rule and align our right hand with our thumb pointing in the positive Z direction, our fingers will wrap around our palm in a counterclockwise direction. This defines our positive moment direction. Looking at force component FX, we see that FX is perpendicular to Ry, so we will get a moment that is Ry times FX. This will tend to cause a clockwise rotation, which will be thus negative according to our coordinate system and right hand rule. The total moment at point O will just be the summation of these two moment components. So we will get that our moment is equal to RX times FY minus Ry times FX. Now we need to calculate all of the various components. Our X will simply be R times cosine of 20 degrees, which will give us a value of 12.4 meters if we substitute the 13.2 meters into this equation. Similarly, Ry is going to be R, the magnitude of R, which is 13.2 meters times sine of 20 degrees, which will give us 4.51 meters. For our forces we have to use our angle of 60 degrees. So FX will be equal to F cosine 60 degrees and FY will be F sine 60 degrees, giving us 50 kilonewtons and 86.6 kilonewtons respectively. Now we can see, do a little bit of a sanity check to whether these numbers make sense, because it's very easy to make errors with swapping sine and cosine. And there's various rules, you might know the acronym SOCA TOA to figure out the relevant components that would come up with your sine and cosine. But we can always look back at the problem and see if it makes sense. Here for our position vector, we see that we calculated that Rx is significantly larger than Ry. This makes sense for our angle of 20 degrees, because we can see that Rx from this angle should be significantly larger than Ry. So we can be pretty confident that we're correct there. Similarly with our force components, FX is smaller than FY and for that 60 degree and this orientation we expect this component FX to be smaller than FY. We can now substitute all of these values into this equation and get that our moment is equal to 12.4 meters times 86.6 kilonewtons minus 4.51 meters times 50 kilonewtons, giving us a total moment of 848 kilonewtons meters acting in a counterclockwise or positive direction. Earlier I mentioned that there are multiple ways this problem could be solved. So let's take a look at another method using the vector cross product. The moment of a force is defined vectorally as R cross F, which can be shown here with both letters as the vector or with the vector notation over top of the variable. Now it's very important to remember that the order is important for a cross product. So this is R cross F and R cross F does not equal F cross R. You have to be very careful with that. So in order to evaluate this cross product we need to know vector R and vector F in vector notation. Now luckily for us we already calculated the Rx and Ry components in the previous calculation. 12.4 and 4.51 and Rz has to be equal to 0 because our vector R is contained completely in the xy plane. Similarly for our force vector we calculated Fx and Fy already and Fz is 0 because our force is completely contained within the xy plane. Now that we have these vectors we can solve the vector cross product. And I'm going to use the Amsterdam method for solving for this cross product. And if you're unfamiliar with the Amsterdam method there will be a link in the video description to another video describing how to apply this method. But you may have learned another method for doing cross product that's also okay to apply. So in the Amsterdam method what we do is we take our position vector and our force vector because it's R cross F. So we have R and we have F and we stack it on top of a copy of itself. Okay so we have position vector R here as well as a copy of position vector R, force vector F and a copy of force vector F. What we then do is cross out the top and bottom row of this stack. And now what we can do to calculate the components of our moment is actually do a cross multiplication of this array. So what we do is take the diagonal products so we get 4.51 multiplied by 0 and we subtract 0 times 86.6. So in this method if it's a downwards diagonal multiplication it's positive if it's an upwards diagonal multiplication. Like here it's downwards from 4.5 point to 0 and upwards from 0 to 86.6 we get positive and negative. Now these products are all multiplied by 0 so the final result is 0. But now we can move down one row and we get 0 times 50 minus 12.4 times 0 which indeed gives us 0 again. And the final row we get 12.4 times 86.6 minus 4.51 times 50 which will be equal to 848. And this gives us our x, y and z components of our moment so we can write that in vector notation as shown here. These are the two main approaches for obtaining the moment of a force F about 0.0 for the problem. But there are ways to set the problem up in an even smarter way to minimize the number of calculations you need to do, particularly for the vector decomposition approach. Can you figure out a more efficient way to set up this problem?