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# More 2D Equilibrium Problems

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Sections
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

Transcript

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.