Guys, let's check this problem out. So we've got these two blocks that are connected to each other on a wedge shape. So basically, we've got two different angles to consider. It's like you get two different rough inclined planes that are kind of sandwiched together. Right? But basically, we want to figure out what the magnitude of the system's acceleration once we release it. We know there's going to be some friction involved. And so, really, we're just going to treat this like any other problem. Let's go ahead and draw the free-body diagrams for A and B. Alright. So for A, basically, what we've got is a free-body diagram. We know we have our weight force. This is going to be mag. Our tension force is going to point up the incline, that's T, and our normal force points perpendicular to the surface. Now we know we're also going to have some friction. There's some coefficients. But remember, we have to figure out what direction that friction is going to go. Is it going to point up the ramp or down the ramp? We actually don't know. All we're told is that the system is going to start moving. So remember, whenever you don't know the direction of the friction, you're going to have to figure out what the system would do if there weren't friction. So without friction, where would the acceleration point? So without friction here, what happens? Well, basically, I've got these two blocks. This one is 5 kilograms. This one is 2 kilograms. The 5 kilograms is on the steeper incline. Right? It's more inclined like that. So pretend there's no friction for a second. You have a heavier object that is on a steeper incline, a lighter object that is on a shallower incline. So without friction, if you were to let the system go, basically, the acceleration point this direction. It would point in the direction of the heavier object. So because of that, this acceleration, we know that the friction force has to point down the ramp for object A. So this is going to be some friction. We're also told that the system begins moving once you release it. It begins moving. So basically, we know that F is equal to F_k, and that actually takes care of step 2 for us. We know what kind of friction we're dealing with. This is a coefficient of kinetic. So now we just basically separate our mg into its components. So this is going to be m_Ag_y, and this is m_Ag_x. Alright? Now we're going to do something similar for object B except it's going to be on a different inclined plane. So here, we've got our free-body diagram for B. So we've got the weight force, m_Bg. And so now what happens is my tension points in this direction. This is my tension. So here's my normal force. It's kind of like flipped from object A. And so now what happens is we're also going to have some friction. So again, without friction, the acceleration is going to be up over to the right. So for object B, object B wants to slide down this way. So we know that this friction here is actually going to point up the surface. That's going to be the F_k. And so now we just split up our mg. So this is going to be m_Bg and then m_Bg_x. And then we just don't really need these anymore. So those are our free-body diagrams. We also figured out what type of friction we're dealing with. And now we just go ahead and get into our F equals m A. We want to figure out the acceleration, so we're going to use F equals m A. Right? So we've got the sum of all forces in the x-axis equals mass times acceleration. Now just pick the direction of positive. Basically, if our acceleration is going to point up or over to the right, then that means that for object A, the positive direction is this way. And then for B, it's that way. Alright? So I've got my direction of positive here and here. Alright. So when we expand our forces I've got my tension that points up and then my F_k points actually I've got my m_Ag_x So this is m_Ag_x - friction is going to be m_Aa. This is our target variable. Now just basically expand out all of these terms. We have T - m_Ag times the sine of, this is going to be theta A. Remember that there are two different angles to consider. There's this one and this one. So I've got this one is going to be μ_k, the coefficient of friction. And then remember, this is going to be μ_k times the normal force, and the normal force is going to be equal to m_Ag times the cosine of theta so we have m_Agcosine of theta A and that equals m_Aa. Alright? So basically, what happens is these are all just a bunch of numbers when you plug them in, and so we're going to simplify. So this m_Ag_x actually just becomes 5.07. And then this μ_k times m_Agcosine theta really just becomes 3.79 So this equals the mass of A which is 2 times a. So we can simplify once more and this is just going to be T - 8.86 equals 2a. We can't go any further because now we have two unknowns. This is going to be our first equation here. Let's go to the other F equals m A. So now we have the sum of all forces equals mass B times a. We know that they're going to have the same acceleration. So now we're going to use the downward direction like this. And so we have our m_Bg_x - our tension, - F_k, is equal to m_Ba. Now let's do the same exact thing. We're going to expand all these terms here. So this is going to be m_Bg sine theta B. This is just the other angle. Right? Just keep track of your variables. And then this is just going to be the μ_k times the normal. So this is going to be μ_k times m_Bg times the cosine of theta B, and this is equal to m_Ba. Alright? So just like before, these are really just a bunch of numbers when you plug them into your calculator. You have m's, g's, thetas, and all the coefficients. So you can just go ahead and solve for this. Right? This is going to be 24.5 - tension - 8.49 equals, and then this is going to equal 5a. So when you simplify this, this is 16.01 - tension equals 5a. Alright, so now we have our two equations. So this is going to be equation number 1, equation number 2. And so now we just use equation substitution to solve for this. So bring these equations down here. I've got T - 8.86 equals 2a. Then I've got 16.01 - tension equals 5a. So you add these straight down, your tensions cancel, and then basically you end up with 8 points. Actually, you know, you end up with 7.15 equals 7a. And so therefore your acceleration is going to be 1.02 meters per second squared. Now the problem asks for the magnitude of acceleration, so we don't have to worry about the signs or anything like that. But the positive just means it's going to accelerate in the direction that we thought it would. So we go back to our answer choices, and our answer choice is B. So let me know if you guys have any questions, and that's it for this one.