Hey guys, let's check out this example where we are asked to find the torque produced by a force with 2 different angles. So here we have a fish catching your bait. Your fishing pole is at an angle of 50 degrees above the x-axis. So, let's say you are like this, holding onto your fishing pole that is 3 meters long, weighs 2 kilograms, and it makes an angle of 50 degrees right there. We want to calculate the torque that's produced on your fishing pole about an axis of rotation in your hands. In other words, think that the fishing pole rotates about the hands right there. So that's the axis of rotation, right here. If the fish pulls on it with 40 newtons, that's a force directed at 20 degrees below the x-axis. The line is here. You're pulling from here. The fish is pulling at an angle direct at 20 degrees below the x-axis. So here's the positive x-axis. 20 degrees below the x-axis looks like this. I'm going to do negative 20 over here. And the fish is pulling with a force of 40 newtons. We want to know how much of a torque this force produces on this axis right here. So what we're going to do is to write the torque equation, torque equals \(fr \sin(\theta)\). Now remember the steps. The first thing we're going to do is to write the, draw our \(r\) vector. Then we're going to figure out what \(\theta\) is, and then we're going to plug it into the equation. Okay? So what is our \(r\) vector? \(R\) vector is an arrow. It's a vector from the axis of rotation to the point where the force happens, where the force is applied on the object. The axis of rotation is here. The force pulls right there. We draw an arrow. This is our \(r\) vector. Now, the length of the \(r\) vector is 3 meters. So that's what I'm going to put here. \(40 \times 3 \times \sin(\theta)\). Now, drawing the \(r\) vector is important, actually not so much so that you can figure out how long it is because you could have just looked at this pole and said the pole is 3 meters long. That's the answer. What's really important about drawing the \(r\) vector is so that you can figure out what angle to use, which is the hardest part. You've got to make sure you're using the right angle. So here, should we use 50? Should we use 20? Should we use negative 20? It actually turns out that in this problem, it's none of these options. It's a different angle. It's a combination of the 2. I want to remind you that \(\theta\) is the angle between \(f\) and \(r\). And to figure out which angle goes between them, the technique I like to use is to try to get \(f\) and \(r\) to be pointing from the same common point, something like this. And then it's just a matter of finding this angle right here. To do this, I'm going to shift \(f\) around or shift \(r\) around so that they start from the same point. So what I'm going to do here is I'm going to push, I'm going to shift my \(r\) up so that they both start from this point. So let me draw this again. I have 50. Let me draw this a little lower. I have 50 over here, and then I have 20 here. What I'm going to do is I'm going to get the \(r\) vector and push it over here so that I have \(r\) and \(f\). And then the angle I need is the angle between these two guys. Now, you see here how there is a 50 between the \(r\) and the x-axis, right? Right here, there's a 50 gets transferred over here. And now I hope you see that the total angle between \(r\) and \(f\) is actually 70 degrees. So you're supposed to add up those 2 guys. This the entire purpose of this question was to look into how to solve questions with nontrivial, \(\theta\) values, angle values, and how to figure out the correct angle to use. If you multiply all of this, you get a 113 newtons-meter, and that's your final answer. Cool. That's it for this one. Let me know if you have any questions. Let's keep going.