More 2D Equilibrium Problems - Video Tutorials & Practice Problems

1

example

Inclined beam against the floor

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15m

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Hey, guys. So in this example, we have a inclined beam that's held against the floor. So you have a force f here, and it's also has the hinge here supporting it. So let's check it out. We have 100 kg um, beam. So massive was 100. It has length of 4 m. And it's how did equilibrium this means all the forces or zero add up to zero and all the torques add up to zero by a hinge down here and by a force that you apply f right here. The beam is held at an angle of 30 degrees above the horizontal. So this angle here is 30. I'm gonna put the 30 in here s so that I can draw the M G in here and your forces directed a degrees above the horizontal. So the distance between your force all the way to the horizontal right here is 50. Now, if this is a 30 this is a 30 as well. So we're gonna split up that 50 into you gotta 30 here. And if the bottom is 13 the whole thing is 50. It means that the top over here is 20. So 20 plus 30 equals 50. Okay, we wanna find F We also want to find So that's part A. We want to find the magnitude and direction of the Net force so this f can get split into FX and f y. Okay. And if the FX is this way, remember, the force have to cancel. There's only one force in the X axis, which is to the right, So the hinge must pull this thing back to the left to hold it in place. So I'm gonna say that the hinge has on a checks. We're going to assume that the the vertical hinge force will be up. So we're gonna assume that I have an h y. That is up. And if this assumption is correct, our total h net will look like this. Okay, so have the neck H Force H nets. And then I have the angle, which is theta of h. So we're looking for F, and we're looking for H Nets hands Feedem hitch. Cool. So let me just highlight the stuff we're looking for. And the way we're going to solve this is by writing that the sum of all forces equals zero. The on the X axis. The sum of all forces equals zero on the Y axis and if necessary, which it will be we're gonna right, um, talk equations or at least one torque equation. Okay, One key difference from how I'm going to solve this versus some of the previous questions we solved, Um is that instead of instead of working with FX and F y in their component forms, I'm actually gonna just work with F on. The reason is, in this particular case, it's going to be simpler. So if you have a question where you have, like, a bar like this held by a rope, um, you had a tension that split into tension y intention X tension X produced no torque torque of tension. Mexico zero. So attention X was kind of useless. It didn't really do much, So it's easier to think of tea since X was useless. It was easier to think of t as just t y and then keep these two separate the x and y instead of working with just the total vector t um so these questions were a little bit simpler, so it was better to do that? Um, here. We got a bunch of angles. They got the third, you got the 50 whatever. So it's gonna be simpler to just work with, um, the complete form of the vector, the vector form and not the components, the individual components. We're gonna work with the entire vector. It doesn't matter. You could have done it the other way, and you would've it would've worked just as well. But you just have one more force because you have FX. And that's why it's two forces as opposed to just having F, which is one. So I'm gonna do that. Um, it would have worked the other way, but that's how we're gonna roll for this one. All right, So, um, the forces in the X axes are a checks and and effects, so I can write that effects in a checks are equal to each other because they cancel each other the forces on the Y axis r f y h y and then mg. So I can write that f y plus h y equals m G. Now notice that I don't know FX. I don't know what checks. I don't know if why I don't know each y I know mg So we don't know. There's a ton of stuff we don't know here. Eso This is not gonna be enough. I'm gonna have to write a third equation, which is gonna be a torque equations. Some of all torques equals zero about some reference axis. Now, remember, the way you want to do this is the reason why we're writing this equation is because we're looking for F. So you want to write your torque equation away from this point? Right? So let's say you got points. 123.3 is the worst point to write the torque equation about because if you were to do this, if you were to write to some of our talks on 0.0.3, um, these guys you're basically treating this is the access, and these forces will not produce a torque. In this case, we're dealing with F Onley F would not produce a talk about 0.3, which means it's not going to show up in the equation. And the whole point of writing an extra equation is so you can solve for F. So you wanted equation where F is on the equation. So to do that, you're gonna write the torque equation about one or two. Those are much better choices. Now, The first rule is to write the torque equation on an access away from your target variable. The second rule is to pick out of the available options if you have multiple, which we do, um, to pick a point with the most forces acting on it so that the most a number of topics will be canceled. So we have the fewest number of terms, basically, wherever ships more complicated. So it's more complicated here, and we're going to write the torque about 0.1. Okay, so let me do that. Talk about 0.1, and this looks like this. Here's one. Let's put a little bar here, and I have m g going down. And I have f going up like this. Okay, this is 30 right here. All right. Um, this here is 30. And this here is 20. Okay, so let's figure out which way these torques, which torques what kind of torture produced here. So if you have the bar like this, mg pushes it down, this is gonna be a clockwise torques. Its negative. So the torque due to M G will be negative and the f it's a little bit harder to see. It's gonna give you a positive torque. One way you could know that is that if if mgs negative, the other one has to cancel. So I have to be positive. Another way you can see this is you can extend the our vector, right. So instead of thinking of it as like this, you can think of it as you have a long a long beam like this, and it's being pushed like this. So basically slide extend your beam and slide the force this way instead from delete this Now it's easier. You gotta force pushing this way. It's gonna cause this thing to go counterclockwise, okay. And really, the reason that happens is because because the force is counterclockwise right away from the extension of your are vectors. If you keep extending our vector, here's our vector. The force happens this way. So the force causes this kind of rotation so the torque of F will be positive. So I have one negative one positive, which means they cancel each other out torque of M g. I could write torque of M G equals torque of F, and then I can expand this equation. So torque of m g is gonna be m g r sine of data and torque of f is gonna be f r sign of data. Let's draw our our vectors. So the our vector for M G looks like this r equals two and the are vector for Let me write someone else R equals two right there. And the are vector for this guy is r equals four. Okay, so the distance, the length of the our vector for M G is too, because it's halfway and f happens all the way at the end of the beam. So it's four. What about the angles? The angle for mg. If this is a 30 we're supposed to use this angle right here between the our and the M G. So it's gonna go up here. Um, so instead of 30 it's going to be 60. Okay, It's the complementary angle on Ben here. The angle between our and F is a little bit more complicated, but we can extend the our vector. Here's f if you extend your are vector right here. It's easy to see that the angle you're supposed to use is not the 30 not the 50. It's the 20. Okay, so the angle we're supposed to use here is 20. All right? So if you plug this in, um, this entire thing here M G is gonna be to sign of 60. This is four F sign of 20 and the whole thing on the left side is going to be equals four points. 342 That's the sign of 20 F. And if you solve for F, you get the F equals F equals So I got the whole left. That's good news. Now check it out. Now that I know F because I know the angle for F. The angle for F is 50 right? I confined FX. And why so from F I confined? This is the path I confined effects and f y. Now notice here that if I have effects, I know HX so it's actually the same number. And if I know f y, um, if I know f y, I confined h y using that equation so I can get. It's not the same number, but I can get to H y. And once I know both of these guys, I can get to H in the angle. Okay, so that's what we're gonna do. So how do I find effects? FX is just f co sign of Fada. So it's 1266 co sign of 50. Because the angle I have to be very careful with the angle for FS 50. And if you do this, I have. FX is e have here 8 8 14 Newton's F Y is sign of 50 which is 9 70. Okay, that's 9 70 Newton's. This is the same as a checks, so I can say HX is 8 14. So I got that. So I know f I know effects f y I know HX. Now I have to get a Chuan I using this equation up here. Okay, so let's get going. So h y plus t y equals m G, which means h y equals m G minus T Y m G is 100 times 10 minus t y, which is 9 70. this means that h y equals 30. Okay. H y equals 30. Now, technically, HX is, uh, negative because it's going to the left, right, Because he has to go away opposite from your force FX. This way. Um, this thing is going to be up. We assumed that it was up on DWI. Got a positive, which means that that assumption was correct. Um, it also has to help out r h wife R h y is 70. It's not enough. It's not enough to holds. It needs a little bit of help. All right, cool. So if you wanna if you wanna draw this real quick what you end up with just to get a visual, what you end up with is a tiny h wine, because this is 30 and a huge. I mean, this isn't even to scale. It's a much bigger difference than this. And a huge paychecks. 8 14 Negative, because it's to the left. So now if you find a church h told is gonna be a check squared, which is 8 14 square. The fact that it's negative doesn't matter, because it's gonna get squared anyway. 30 square square to the whole thing. This is just by factoring theory. Because it's a factor Edition on doesn't give 8 15. It makes sense that you get 8. 15 Newton's because the total vector is like this. This is such a tiny angle right here that you imagine that the blue line and the red line they're almost the same line. So they're almost the same length. You can also find theta, which we expected to be a tiny angle. And it's gonna be the ark Italian of y over X. Why is 30 x is 8 14? Okay, And if you do this, the answer is 2.1 degrees. Now I'm actually gonna plug in. I'm plugging in. Both of these guys is positive when you plug two things is a positive on the arc Tangent equation, your calculator, things that you're talking about, this things you're talking about, a 30 positive and, um, 8 14 positive. So it's actually giving you the angle on the first quadrants. The good thing is that if you are instead on the second quadrants, um, the angle is exactly the same. It's just the image. It's the image about the y axis. So if this is 2.1 here. It means that this English 2.1 there. So what I would do is I would draw. I would draw this here. Okay, Let me just have this here so that you have it as part of that explanation. But your final for your final answer, I would draw eso. You can illustrate where that angle goes. Okay? Yeah, that looks terrible, but we're going to get there, um, 2.1 degrees so you can show exactly where that angle goes. Uh, this is 30. This is 8. 14. This is 8. 15. Okay, so this is the final H vector. If you wanna also show where the angle goes and that's it for this one. Cool. So little tricky, different angles have to figure out which ones to use. Hopeful. This made sense. But let me do you have any questions and let's get going

2

Problem

Problem

A 200 kg, 10 m-long beam is held at equilibrium by a hinge on the floor and a force you apply on it via a light rope connected to its edge, as shown. The beam is held at 53° above the horizontal, and your rope makes an angle of 30° with it. Calculate the angle that the Net Force of the hinge makes with the horizontal (use +/– for above/below +x, and use g=10 m/s^{2}.)

A

θ = −65.9°

B

θ = −21.1°

C

θ = +21.1°

D

θ = +65.9°

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