Ch 15: Rotational EquilibriumWorksheetSee all chapters
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Ch 15: Rotational Equilibrium
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Torque & Equilibrium
Review: Center of Mass
Beam / Shelf Against a Wall
Center of Mass & Simple Balance
Equilibrium with Multiple Objects
More 2D Equilibrium Problems
Equilibrium in 2D - Ladder Problems
Equilibrium with Multiple Supports

Example #1: Inclined beam against the floor


Hey guys! In this example we have an inclined beam that's held against the floor. You have a force F here and it also has the hinge here supporting it. Let's check it out. We have a 100-kilogram beam so mass equals 100. It has length of 4 meters and it's held at equilibrium. This means all the forces 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. This angle here is 30. I'm going to put the 30 in here so that I can draw the mg in here. Your force is directed at 50 degrees above the horizontal. The distance between your force all the way to the horizontal right here is 50. If this is a 30, this is a 30 as well. We're going to split up that 50 into you got a 30 here. If the bottom is 30 and the whole thing is 50, it means that the top over here is 20. 20 + 30 = 50. We want to find F, so that's part A. We want to find the magnitude and direction of the net force. This F can get split into FX and FY. If the FX is this way, remember the forces 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. I'm going to say that the hinge has an HX. We're going to assume that the vertical hinge force will be up, so we're going to assume that I have an HY that is up. If this assumption is correct, our total H net will look like this. I have the net H force H net and then I have the angle which is _ of H. We're looking for F and we're looking for H net and _ H. Let me just highlight this stuff weÕre looking for. The way we're gonna solve this is by writing that the sum of all forces equals zero on the x-axis. The sum of all forces is equal zero on the y-axis. If necessary which it will be, we're going to write torque equations or at least one torque equation. One key difference on how I'm going to solve this versus some of the previous questions we solved is that instead of working with FX and FY in their component forms, I'm actually going to just work with F. The reason is in this particular case, it's going to be simpler. If you had a question where you had like a bar like this held by a rope, you had a tension that split into tension y and tension x. Tension X produce no torque. Torque of tension x equals zero, so tension X was kind of useless. It didn't really do much. It's easier to think of T since X was useless, it was easier to think of T as just Ty and then keep these two separate, the x and y instead of working with just the total vector T. These questions were a little bit simpler so it's better to do that. Here we got a bunch of angles. You got the 30, you got the 50, whatever. It's going to be simpler to just work with the vector form and not the components, the individual components. We're going to work with the entire vector. It doesn't matter. You could have done it the other way and it would have worked just as well. But you just have one more force because you have FX and FY. ThatÕs two forces as opposed to just having F which is one. I'm going to do that. It would have worked the other way, but that's how we're going to roll for this one.

The forces in the x axis are HX and FX so I can write that FX and HX are equal to each other because they cancel each other. The forces in the Y axis r FY HY and then mg. I can write that FY + HY = mg. Notice that I don't know FX, I don't know HX, I don't know FY, I donÕt know HY. I know mg. There's a ton of stuff we donÕt know here. This is not going to be enough. I'm going to have to write a third equation which is going to be a torque equation, sum of all torques equals zero about some reference axis. Remember, the way you want to do this is the reason why we're writing this equation is because we're looking for F. You want to write your torque equation away from this point. Let's say we got points one, two, three. Point three is the worst point to write the torque equation about because if you were to do this, if you were to write the sum of all torques on point three, these guys you're basically treating this as the axis. These forces will not produce a torque. In this case we're dealing with F only. F would not produce a torque about point three which means it's not going to show up in the equation. The whole point of writing an extra equation is so you can solve for F. You want an equation where F is on the equation. To do that, you're going to write the torque equation about point one or two. Those are much better choices. The first rule is to write the torque equation on an axis away from your target variable. The second rule is to pick out of the available options if you have multiple which we do, to pick a point with the most forces acting on it so that the most number of torques will be canceled so you have the fewest number of terms. Basically wherever it's more complicated. It's more complicated here and we're going to write the torque about point one. Let me do that. Torque about point one. This looks like this. Here's point one. Lt's put a little bar here. I have mg going down and I have F going up like this. This is 30 right here. This here is 30 and this here is 20. Let's figure out what kind of torques are produced here. If you have the bar like this and mg pushes it down, this is going to be a clockwise torque so it's negative. The torque due to mg will be negative. The F, it's a little bit harder to see. It's going to give you a positive torque. One way you can know that is that if mg is negative, the other one has to cancel so it has to be positive. Another way you can see this is you can extend the r vector. Instead of thinking of it as like this, you can think of it as you have a long beam like this and it's being pushed like this. Basically, slide, extend your beam and slide the force this way instead. Now it's easier. You got a force pushing this way. It's going to cause this thing to go counterclockwise. Really the reason that happens is because the force is counterclockwise away from the extension of your r vector. If you keep extending your r vector, here's the r vector, the force happens this way so the force causes this kind of rotation. The torque of F will be positive. I have one negative, one positive which means they cancel each other out. I can write torque of mg = torque of F and then I can expand this equation. Torque of mg is going to be mg r sin_ and torque of F is going to be F r sin_. Let's draw our r vectors. The r vector for mg looks like this, r = 2. The r vector for this guy is r = 4. The length of the r vector for mg is 2 because it's halfway and F happens all the way at the end of the beam so it's 4. What about the angles? The angle for mg if this is a 30, we're supposed to use this angle right here between the r and the mg so itÕs the angle up here. Instead of 30, it's going to be 60. It's the complementary angle. Here, the angle between r and F is a little bit more complicated but we can extend the r vector. Here's F. If you extend your r vector right here, it's easy to see that the angle you're supposed to use is not the 30, not the 50, itÕs the 20. The angle we're supposed to use here is 20. If you plug this in, this entire thing here mg is going to be 100(10)(2)sin60. This is 4F sin20. The whole thing on the left side is going to be 1732 = 4 (.342) that's the sine of 20, F. If you solve for F, you get that F = 1266. I got the whole F. That's good news.

Check it out. Now that I know F because I know the angle for F, the angle for F is 50, I can find FX and FY. From F, I can find, this is the path, I can find FX and FY. Notice here that if I have FX, I know HX. It's actually the same number. If I know FY, I can find HY using that equation. It's not the same number but I can get to HY. Once I know both of these guys, I can get to H and the angle. That's what we're going to do. How do I find FX? FX is just F cos_, so it's 1266 cos50, because the angle, you have to be very careful that the angle for F is 50. If you do this, I have FX is 814 N. FY is 1266 sin50, which is 970 N. This is the same as HX so I can say HX is 814, so I got that. I know FX, FY, I know HX. Now I have to get HY using this equation up here. Let's keep going. HY + TY = mg, which means HY = mg Ð TY. Mg is 100*10 Ð TY which is 970. This means that HY = 30. Technically HX is negative because it's going to the left, because it has to go away opposite from your force FX this way. This thing is going to be up. We assumed that it was up and we got a positive which means that that assumption was correct. It also has to help out our HY. If our HY is 970, it's not enough to hold. It needs a little bit of help. If you want to draw this real quick, what you end up with just to get a visual what you end up with is a tiny HY 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 HX 814 negative because it's to the left. If you find H, H total is going to be HX^2 which is 814^2. The fact that it's negative doesn't matter because it's going to get squared anyway. 30^2, square root of the whole thing. This is just by Pythagoream theorem because it's vector addition. This is gonna give 815. It makes sense that you get 815 N because the total vector is like this. This is such a tiny angle right here that you imagine that the blue line in the red line are almost the same line so they're almost the same length. You can also find _ which we expect it to be a tiny angle. It's going to be the arc tangent of Y/ X. Y is 30, X is 814. If you do this, the answer is 2.1 degrees. I'm actually plugging in both of these guys as positive. When you plug two things as a positive on the arctangent equation, your calculator thinks that you're talking about this. It thinks you're talking about a 30 positive and an 814 positive. It's actually giving you the angle on the first quadrant. The good thing is that if you are instead on the second quadrant, the angle is exactly the same. It's just the image. It's an image about the y-axis so if this is 2.1 here, it means that this angle is 2.1 there. What I would do is I would draw this here. Let me just have this here so that you have it as part of that explanation. But for your final answer, I would draw so you can illustrate where that angle goes. That looks terrible but weÕre just gonna leave it there. 2.1 degrees so you can show exactly where that angle goes. This is 30, this is 814, this is 815. This is the final H vector if you want to also show where the angle goes. That's it for this one. A little tricky, different angles, had to figure out which ones to use. Hopefully this made sense but let me don't have any questions and let's keep going.

Practice: 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/s2.)