Forces & Load: Force Addition Algebraic Math 3

So let us conclude the algebraic method, or rather the resultant, the equilibrium, the components, the force addition, all in one problem. I have two forces, 100 pounds, given an angle of 20 degrees with the vertical. The other force is 260 pounds, and it's at a slope of rise 5, run 12. Please remember, this is not likely on the ARE. It's just a good review of what we've covered. Individual pieces of this question might be on the ARE. Very good. So thinking graphically, I'm going to draw a parallel to the first force. I'm going to draw a parallel to the second force. And I'm going to remember that my resultant needs to be here. Looks like the answer is going to be more than 260, more than 100. looks like the angle is somewhere between these two forces. Okay, so that's the graphic method. Let's go ahead and get started on the algebraic method. And to do that, I must resolve each force into its components. So let me start with the 260-pound force, and let's remember how we did this one. When you have a rise-run slope, we're going to be seeing this with trusses. So I think that's pretty important. What I have here is 260 pounds, and it's a 5 in 12. And we said we must find that hypotenuse of the proportion triangle. Now 5 square plus 12 square root gives me 13. So this hypotenuse of the proportion triangle is 13. And to break down the 260 pounds into vertical and horizontal components, we take 5 over 13, we take 12 over 13, 5 over 13, and 12 over 13. That's the way I break down a force when I have the force, sorry, the proportion triangle instead of the angle. So I'm going to take 5 over 13 times 260 pounds, and I'm going to take 12 over 13 times the 260 pounds to end up with two components. Looks like the first one is 100 pounds. Looks like the second one is 240 pounds. Now let me look at my diagram and recognize that this is in the proportion of 12 run, 5 rise. Looks like the horizontal is larger, so must be that 240 is the horizontal, and the smaller of the two is the vertical. 240 pounds over here, and 100 pounds over here. This was up and to the right, therefore both are positive. By the way, 240 divided by 100 is the same as 12 divided by 5. That's the whole idea. Okay, so that's the first one. The second one says 20 degrees with the vertical. It's 100 pounds. We've done this one multiple times so far. Sine of 20 degrees is 0.342. Cosine of 20 degrees is 0.94. We multiply by the force of 100. And we end up with two components. One of them is 34.2, and the other one is 94. And we look at the diagram again. we see that the horizontal is the smaller of the two, and the vertical is the larger of the two. So must be, this is the vertical, this is the smaller of the two. With the 260, the larger was the horizontal, the smaller was the vertical. Very good. So let me set up my table of horizontal and vertical components so that I solve this problem using the algebraic method. And I am trying to break down two forces and find their resultant. The first one is 260 pounds. And the second force is 100 pounds. The 260 pounds broke down into 240 on the horizontal. And on the vertical, it was plus upwards of 100 pounds. And the 100 pound force at 20 degrees with the vertical broke down into two components, 94 and 34.2 with the larger being the vertical. So this is the 94 point, sorry, 94 and this is the 34.2 pounds. So the horizontal is to the right plus 34.2 pounds and the vertical is upward worth 94 pounds. So adding these up, I end up with 274.2 on the horizontal and on the vertical 194 pounds. So the resultant of these two forces will have a horizontal of 274. Let me go back and do my graphic, this one. So the resultant is going to be 274 on the horizontal, 0.2, and on the vertical, it's going to be 194. Very good. So the resultant itself is the square root of 274.2 square plus 194 square. And that turns out to be how much? 335.89 or 84. I can't read my own handwriting. Whatever. 84, let's say, pounds. So that's the resultant. 335.84. Very good. We still need to define this angle. How much is this angle? This angle has a rise of 194. Let me redraw this. Here's what we're looking at. A rise of 194, a run of 274, and a hypotenuse of 335.84. So this angle theta, the tangent of theta is rise over run, or 194 divided by 274.2, which turns out to be 0.7075. So I need an angle whose tangent is 0.7075, So I'll go to my calculator and hit 10 inverse, which is usually on your calculator, shift or second. The key above the tangent is usually tangent inverse. So tangent inverse of 0.7075 is a certain angle, and that angle is 35.28 degrees. Very good. So let me conclude with this drawing that says, here's my axes. and here's my resultant. It is 335.84 pounds and the angle it subtends is 35.28 with the horizontal and if you want to have the equilibrium, the equilibrium is just basically equal and opposite to that one so the equilibrium is also 335.84 but its angle is this much, which makes it 180 plus 35, which is 215.28 degrees. Okay.