Forces & Load: Forces Terms & Math

Okay, let's talk about force components qualitatively. This is the ARE, the Architect Registration Exam, not the PE, the Professional Engineering Exam. It's not. So we need to know these things conceptually, big picture, and we should be able to get the answer on a multiple choice qualitatively without crunching numbers, even though the question may be about numbers. Okay. So, some terms and definitions first regarding forces. The arrowhead is significant because if the force is pushing on the point of application, then that is compression. Versus if it were pulling on the point of application, then we have tension. So, a pull is tension, pushing is compression. The inclination or the angle which the force makes with the horizontal, the vertical, or the rise-run slope is critical to understand that as long as the force has an angle less than 45 degrees, then its run is greater than its rise. And if the angle is greater than 45 degrees, then obviously the rise is greater than the run. And just to dispel some myth here, the rise is equivalent of the sine of an angle versus the run is equivalent of the cosine. And the slope is equivalent to the tangent of an angle. So this is the engineer's language, and this is the architect's language. They are one and the same. So let's not worry about the sine, cosine, tangent. Instead, let's think of rise, run, slope, the language we're used to seeing. So something like this, an angle like this, would have much more run than rise. Run greater than rise. Which means whatever this angle is, its cosine is larger than its sine. And if the angle is doing that, whatever that angle is, its rise is greater than its run. And therefore, the sine is greater than the cosine. So, these terms are exchangeable. They correspond to each other. So let's look at this diagram here, and let's see that force number 1, 2, 3 are less than 45 degrees. Even this force, which is maybe 44 degrees, they all have a larger run than rise. Therefore, their cosine is going to be larger than their sine. And sine and cosine are fractions, because they're parts of that much. They are less than. The hypotenuse is larger than the vertical and larger than the horizontal. Therefore, sine and cosine are fractions. They're not greater than 1. Excellent. So for forces 1, 2, 3, the run is greater than the rise, the cosine is greater than the sine. Once you pass 45 degrees, this is forces 4 and 5, then all of a sudden your rise is greater than your run, and your rise is the sine of the angle would be greater than the cosine. Very good. That applies in all quadrants. Clearly, this angle is going to have more run than rise, and this angle, oops, this force here is going to have more rise than run. Excellent. Direction is critical. This one is going up and to the right. so its vertical is positive as long as that is the positive y-axis and that's the positive x-axis. And this force number 1 is going to the right, therefore its horizontal is positive, versus force number 2 is going up and to the left, therefore its horizontal is negative. And force number 3 is going to the left and down. Down and to the left. Down is negative. That's its vertical. And to the left is negative. That's its horizontal. And finally, force number 4 is going down and to the right. And down is negative. To the right is positive. It's important to keep these things in mind because I need to know if the component is positive or negative based on the diagram. Now, transmissibility is a huge word, but the concept is very straightforward and simple. If I have a force doing that, then I can slide it anywhere on its line of action without changing its magnitude or arrowhead, And I can remove this force and say that this force is the same as the previous one. As long as the magnitude and arrowhead are not changed, then sometimes it's more advantageous to see the force in this direction. Likewise, this force could translate into that one. Okay, transmissibility. You can move a force anywhere on its line of action if you don't change the magnitude or the arrowhead. So now let's talk about components. I have this one force here, and I project on the horizontal and on the vertical. And these are called the components of the force. So the vertical component is this much, and the horizontal component is this much. Very good. Those are the components of the force. Let's please keep in mind that this force specifically is shallower than 45 degrees, which means the horizontal is larger than the vertical. Let me correct that. So I'm going to make sure that this one is smaller than the horizontal. quantitatively. The closer you get to 45, the closer the vertical and horizontal components are to each other. At 45, the vertical equal to the horizontal, versus at this angle, the horizontal is much greater than the vertical. Okay. If the angle is steep, if the inclination of the force is steep, then its vertical component is larger than its horizontal component. So here's the vertical component. This one is going up and to the right. So its vertical component is up. Its horizontal component is to the right. And H is much smaller than V because it's pretty steep. So you can express this as any pair in this parallelogram. Okay, so V is way greater than H versus, in this example, H was greater than V. Okay, so that concludes the introduction to forces. What we're going to do next is take a look at the math. So now to get into the math, let's remember that we need to approach this qualitatively instead of quantitatively. Yes, I'm going to be doing math, but I don't want to do sine, cosine, tangent, opposite over hypotenuse, adjacent over hypotenuse, opposite over adjacent, all that stuff from physics. I want to approach it as rise-run slope in an architect's manner. Just like the slope of a roof being 3 and 12, that has meaning. Okay, so looking at these two forces, A and B, each is 1,000 pounds. A is at 30 degrees with the horizontal versus B is at 70 degrees with the horizontal. Excellent. So what we do here is, without thinking too much, I just go to a calculator and I get two fractions, the sine and cosine of 30 degrees. So from a calculator, this is 0.5, this is 0.866. Very good. I get these two fractions and I don't do opposite adjacent blah, blah, blah. Instead, I just break down the force. You gave me 1,000 pounds. I'm going to break it down into two components, and one component is 500. The other component is 866 pounds. Now, I just look at the diagram, and I remember what we said earlier. Are you less than 45 or more than 45? This angle is less than 45 means its run is larger than its rise. Well then, I have two numbers, 500 and 866, and I know that the horizontal is larger than the vertical. Must be, you're the horizontal, you're the smaller of the two, you're the vertical. Done. Those are the two components of force A. That was for force A. Now let's look at force B. Force B is at a different angle. It's at 70 degrees. Same thing, I go to a calculator, I get two fractions. the sine and cosine of 70 degrees. I've looked these up. The sine is 0.94. The cosine is 0.342. And I multiply by the force, and I get two components. I'm using 1,000 pounds because it simplifies the math instead of all the number crunching. Of course, this applies whatever that force is, 1562.9, same concept. So one component is 940 pounds. The other component is 342 pounds. Very good. Let's look at the diagram again, and let's see that this force is pretty steep. It's more than 45 degrees. Must be, this is the vertical component. It's larger than the run. So, in this case, we go back and we say, okay, which is larger, 940 or 342? 342. 940 is larger IC. You must be the vertical. You're the smaller of the two, the horizontal. So that's how we do this. It's a lot easier just blindly go get the sine and cosine as fractions, break down the force, then look at the diagram. Is the force inclination more than 45 degrees or less? That's how you determine which is the larger of the two. Instead of sine opposite over hypotenuse, cosine adjacent over hypotenuse, let's not go there, because it will get ugly if I give an angle with a vertical. That will not work. You'll get confused. Okay, so let's take a look at what I just said, which is let's give an angle with the horizontal, the other one with the vertical. So now for force A, again, it's 30 degrees. And here's my 45. Let me remember where my 45 is. Somewhere here. And force A has an inclination less than 45. Force B is steeper. It has an inclination greater than 45 degrees. Okay, so back to force A. It looks like the sine and cosine of 30 degrees. We get those fractions again. And they are 0.866 and 0.5. And I multiply by 1,000 pounds. And I get the same two components as before because I'm using the same exact numbers. So 500 and 866. Excellent. Now, which is which? That's the critical part. We just look at the diagram, and we see that this is less than 45, and the run is greater than the rise, and so must be horizontal is larger than vertical. Okay, looking at these two numbers, 500, 866, the larger is horizontal, the smaller is vertical. Now, this force was going up and to the left, up and to the left. Up means the vertical component is positive. To the left means the horizontal component is negative. Who was horizontal? You were. And this one is positive. Excellent. I'm done with force A. Now, let me look at force B. Force B has a sine of 20 degrees and a cosine of 20 degrees. which means that this is 0.342 and this is 0.94. Same as before, I'm just going to multiply by the 1,000 pounds, and I will end up with two components, same as before, 342 pounds and 940 pounds. Now, the angle is given with the vertical. So this might be a little confusing, but if you look at the diagram, you will not get it wrong. I don't care that the angle is given with the horizontal or the vertical. I'm looking at this, and I'm seeing that the vertical is much less, sorry, much greater than the horizontal. So it must be the larger of the two is the vertical, the smaller of the two is the horizontal. Okay? So please do not do opposite over hypotenuse. you might make a mistake. I would rather look at the diagram. It's clear the vertical is much larger than the horizontal, so is 940 greater than 342. This was going up and to the left, so up is positive, to the left is negative. So minus 342 on the horizontal, and the positive V is 940 pounds. Excellent. So let's take a look at this example now, and let's see what happens with the 1,000 pounds. It's clear we've done this before. We ended up with two components. It's still 1,000 pounds, so we ended up with two components, and those components were 500 and 866. Now, looking at this, I tried to find my 45. It's over here. This angle is shallower than 45. Well, then, must be that the run is greater than the rise. This is the run. This is the rise. Horizontal is greater than vertical. This was going down and to the left. Down and to the left. So those are my two components. Very good. Now, looking at the second force, B, let me do this, please. Let me get it out of my way and move it up to here. Good. So I have some room to write. Now, as far as B goes, whenever you are given a rise-run slope, it's like the slope of a roof or a truss is usually dimensioned. So we have its rise, we have its run for each panel. So, to find the components of this force, we must have this hypotenuse of the dimension triangle. 12, 16, A square plus B square equals C square gives me 20. If you're quick with math, this is 3 times 4, this is 4 times 4, must be this is 5 times 4, Because in a right-angled triangle, if the numbers are nice, then this relationship should work. Now, what happened here? Okay, so if the numbers are nice, it must be 3, 4, 5. Okay, so bottom line is I got this dimension to be 20. So now, what happens is two similar triangles. One is 12, 16, and a hypotenuse of 20. The other one, the other triangle, is this one that has, on the hypotenuse, it has 1,000. It has a vertical component, and it has a larger horizontal, because 16 is greater than 12. That's the proportion of run to rise, the same as h to v. Very good. So for this force, a quick shortcut to find the components is one component, the horizontal, is 12 over 20, and the other component is 16 over 20. So let's do that, and let's come up with the components of this force given its rise and its run, 12 over 20, 12 to the hypotenuse, and 16 to the hypotenuse of 1,000 pounds. So this gives 600 pounds, and this one gives 800 pounds. So those are the components of the 1,000 pounds, given a rise of 12, a run of 16, and a slope of 20. So looking at the diagram, it looks like the horizontal is larger than the vertical. 16 is more than 12. So it must be you're the horizontal, you're the vertical in that proportion. And this force was going down and to the right. So horizontal is positive, vertical is down. So those are the components of this force. Please be careful when given 16 to 12. You don't take the ratio of 12 to 16. That means nothing. It's the tangent of the angle. I don't want the tangent of the angle. So instead, I always have to find this dimension of the proportion triangle. In this case, it's 20. Excellent. So that's it for this example. Let's transition to this example where A is still the same. It's 1,000 pounds at 30 degrees with the horizontal, although its arrowhead is flipped. No big deal. B, on the other hand, in the previous example, the run was 16, now the rise is 16. Earlier, the vertical was 12, now the horizontal is 12. Very good. So for A, I'm not going to change anything. A, still two components, 500 and 866. And it's shallower than 45 degrees. So the run is bigger than the rise. So the run is bigger than the rise. So the run, this is H, and this is the smaller of the two. This one is going up and to the right, positive and positive, versus the second force. The second force this time is doing 1,000 pounds, but its proportion triangle is 16 rise and 12 run. I still need that 20, which is 12 squared plus 16 squared square root, the Pythagoras theorem. And now I'm going to set up my proportion 12 over 20 and 16 over 20 of the force of the 1,000 pounds.