Hey guys, so in this video, we're going to start solving equilibrium questions that are 2-dimensional, and I'm going to give you an example of a classic problem in equilibrium: the ladder problem. Let's check it out.
So far, we've solved equilibrium problems that were essentially 1-dimensional, meaning all the forces acted in the same axis. Either you had all the forces on the x-axis or all the forces on the y-axis, most of them on the y-axis. And even if you had something at an angle like this, let's say you had something like this, that's still essentially 1-dimensional because the angles were the same when we wrote the torque equation, and they canceled. So in all the problems we've solved so far, sine of theta in the torque equation never really mattered because it either canceled or it was just the sine of 90, which is 1.
Now we're going to finally solve some problems where that's not the case. Right? We're going to actually have to worry about the angle. More advanced problems, as it says here, will include problems in 2 dimensions, in 2 axes. And some of them, in some of these cases, we may need to decompose the forces. Some of them will be decomposed. Remember, however, that torques are scalars. Torques are scalars, so we will never need to decompose them. Okay? So 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.
Alright. So let's check out this problem here. We have a ladder of mass 10 kilograms, with m=10kg, and it's uniformly distributed. That means that the mg of the ladder will act right in the middle. So I'm going to say this is 2 meters, and this here is 2 meters as well. The ladder has a length of 4, that's why I did 2 and 2, and it rests against a vertical wall while making an angle of 53° with the horizontal shown. Since this is 90°, this is 53°, and the other angle is 90° minus 53°, which is 37°. Let's put that there. We're going to calculate several components. These are all the typical questions you might see in a classic ladder problem.
I want to find the normal force at the bottom of the ladder, and a bunch of other things. Before I list them, let's discuss the forces here. There's an mg that pulls downwards. Naturally, there's a normal here that pushes upwards. Because the ladder is resting against the vertical wall, there's also a normal force right here, and the normal is always perpendicular to the surface. So, normal is going to be like this. There are 2 normal forces here: normal bottom and normal top.
Notice that we want this ladder to be in complete equilibrium, so all forces and torques need to cancel. But if you look at what we have now, there's a force going to the left, but no forces going to the right, at least not yet drawn. That means this ladder would not be in equilibrium. There has to be a force going to the right, and that force will be friction over here. And because we want the ladder not to move, this is static friction. There's enough force for everything to cancel.
This is what we have. I can write that the sum of all forces on the x-axis equals 0, meaning the forces in the x-direction cancel each other out. So I am going to have normal at the top equals static friction at the bottom. Sum of all forces on the y-axis equals 0, meaning nbottom=mg, the two cancel. Now I can write more equations. I can write torque equations. Sum of all torques at any point P equals 0. There are 3 points here where I might want to write this: point 1 at the bottom, point 2 in the middle, and point 3 at the top. These are points where forces happen. Remember, you want to write your torque equations around the point where a force occurs, so you have fewer terms in the torque equation.
Here, we want to find the normal force at the bottom of the ladder. This one is the easiest to find. Normal at the bottom is just mg. Here we have mg, so nbottom=10*10N, giving us 100N. That's easy.
Now, what about the normal force at the top of the ladder? To solve for that, I would need to know static friction. I don't have enough information just yet to find static friction, so we're going to need to write a torque equation. If you want to find the normal force at the top, you want to write a torque equation using your axis being somewhere else. The reason is you want your normal top to show up in the equation. If you put the axis of rotation at the bottom, then there will be no torque due to normal at the top, and it won't be part of the equation. So we're going to carefully select our axis to be one of the other two points. I'm going to pick the bottom here, as that is typically the best of the two to pick. You want to write your torque equation about a point with as many forces as possible so that you have as few terms as possible.
Sum of all torques at point 1 equals 0. There are two torques there. There's point 1, here's the ladder. mg is acting here, and then n top is acting here. The r-vector from here to mg looks like this, from the axis to mg looks like this. And then from the axis to ntop looks like this. The bottom one has a length of 2 meters and the top one has a length of 4 meters, half and the total distance. The angle that I'm going to use here is not the 53° right here, but instead, I'm going to use the 37° right here. That's what we're going to use.
We're going to use theta=37. For this one here, I'm supposed to use this angle, and if this angle is 53, then this is 53 as well. They have different r-vectors and different thetas. You have to be very careful here.
Alright. So hopefully, that makes sense. Let's continue. I'm going to say that this torque is going this way, the torque of mg, and the torque of normal is going that way, and I'll show you that just in a second. So imagine we have your ladder looking like this. mg is pushing down, so it's trying to cause a rotation in this direction, and this is clockwise, so
