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# Equilibrium in 2D - Ladder Problems

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Sections
Center of Mass & Simple Balance
Equilibrium with Multiple Objects
More 2D Equilibrium Problems
Equilibrium in 2D - Ladder Problems
Equilibrium with Multiple Supports
Torque & Equilibrium
Review: Center of Mass
Beam / Shelf Against a Wall

Concept #1: Equilibrium in 2D - Ladder Problems

Transcript

Hey guys! In this video, we're going to start solving equilibrium questions that are two-dimensional and IÕm going to give an example of a classic problem in equilibrium that is the latter problem. Let's check it out. So far we've solved equilibrium problems that were essentially one-dimensional, meaning all the forces acted in the same axis, either you had all the forces in the x-axis or all the forces in the y-axis, most of them in the y-axis. Even if you had something at an angle like this, let's say you had something like this, that's still essentially one-dimensional because the angles were the same when we wrote the torque equation and they cancel. In all the problems weÕve solved so far, sin_ in the torque equation never really mattered because it either canceled or it was just the sin90 which is 1. Now we're going to finally solve some problems where that's not the case. We're going to actually have to worry about the angle. More advanced problems as it says here will include problems in two dimensions in two axes. Some of them in some of these cases, we may need to decompose the forces. Some of them will be decomposed. Remember however the torques are scalars so we will never need to decompose them. We're going to decompose forces in problems, but we don't have to decompose torques because torques are scalar. They may be positive or negative but they are scalars. They don't have a vector direction. Let's check out this problem here.

We have a ladder of mass 10 kilograms, so m = 10 and it's uniformly distributed. What that means is that the mg of the ladder will act right at the middle so I'm going to do this here and say this is 2 meters and this here is 2 meters as well. The ladder has length 4, that's why I did 2 and 2. It rests against a vertical wall while making an angle of 53 with the horizontal shown. This is 53. This by the way, because this is 90, if this is 53, this is 90 Ð 53 which is 37. Let's just put that there. When I calculate a bunch of stuff, these are all the things you might see in a classic ladder question. I want to find the normal force at the bottom of the ladder and a bunch of other stuff. But before I read the list, let's talk about what forces we have here. You have an mg that pulls you down. Obviously there's a normal here that pushes you up because the latter is resting against the vertical wall. There's also normal right here and normal is always perpendicular to the surface, so normal is going to be like this. There are two normal forces here. There's normal bottom and thereÕs normal at the top. Notice that obviously we want this ladder to be in complete equilibrium so all forces cancel and all torques cancel. But if you look at what we have right now, there's a force going to the left but there's no forces going to the right. At least I haven't drawn them yet. That means this ladder would not be an equilibrium. There has to be a force going to the right and that force will be friction over here. Because we want the ladder not to move, this is static friction. There's enough forces that everything cancel. These are all the forces you're going to have. I can write that the sum of all forces in the x-axis equals zero. What this means is that the forces in the x cancel each other, so I'm going to have normal top equals friction static. That's friction at the bottom. Sum of all forces on the y-axis equals zero, so this means that N bottom equals mg. The two cancel. N bottom = mg. I can write more equations. Now I can write torque equations. Sum of all torques at any point p equals zero. There are three points here what I might want to write this. There's the point one here at the bottom, two at the middle and three at the top. These are points where forces happen. Remember, you want to write your torque equations about the point where a force happens so you have fewer terms when you write out the torque equation. Let's see.