So we've seen how to solve connected systems of objects problems, inclined planes, and also friction. And basically, we're just going to combine all three of those together in this video because sometimes you're going to have problems that will combine all of these things together. The key difference though is that unlike previous videos where you don't know what kind of friction is acting, you almost always know which kind of friction is acting in these problems. You'll know from the problem text whether you're dealing with static or kinetic friction. Let me go ahead and show you. Basically, we're going to use all the same problem-solving steps for all three of these ideas. It's really all the same steps. So let's go ahead and check out the problem here. We've got these two blocks on this 30 degree inclined plane. We've got the weights and the masses of both boxes and the coefficient of kinetic friction. What we're also told is we're also told that block b is moving up the 30 degree incline, so we know which kind of friction is going to be acting. It's going to be kinetic. And so what we want to do is find the acceleration of the system. So let's get started. We want to draw free body diagrams for both objects. Let's go ahead and start with block a. Block a is the one that's on the inclined plane. We've done this a bunch of times before. We've got the weight force that acts straight down, m_{a}g. We've got the normal force that is perpendicular to the surface, and we've got some tension that acts because of the cable, and then we also have some friction. We have some coefficient of kinetic friction here. And we're told that block a is being pulled up the ramp. It moves up the 30 degree incline. So that means that friction has to oppose that. That's going to be our kinetic friction. Alright? So that's your free body diagram. Now what we do is we just tilt our coordinate system. We get rid of this m_{a}g and we just separate into its components. That's m_{a}g_{y} and then we have m_{a}g_{x} that points down the incline. Okay? So now block b, we've also done this one. The hanging block is basically just going to be the weight force which acts straight down, m_{b}g, and then we've got the tension. Alright. So now we move on to the second step, which actually would be determining the type of friction. But remember the problem text just told us that we're going to deal with kinetic friction so we're already done with that. Now we move on to step 3 which is writing f = m_{a} and we're going to go ahead and start with the simplest object which is object b. So we're going to write the sum of all forces in the y-axis equals m_{b} times a. We've got 2 forces, but actually first I forgot. Let me back up for a second. We have to choose the direction of positive. We've also done this a bunch of times before. It's basically going to be the direction of the acceleration of the system. We're told block a is moving up the incline, so basically anything that goes up around and over is going to be your direction of positive. So for b, anything that points down is going to be positive for a anything that points up the ramp is going to be positive. Alright? So we expand our forces. We've got our m_{b}g minus the tension is equal to m_{b}a. We actually have the masses and the weights, so we can go ahead and simplify. Remember that the weight of block B is already given to us. This is a 100 Newtons, not a 100 kilograms. So that means that this m_{b}g is already a 100. Minus tension and then we've got the mass which is 10.2. That's what you get when you divide by 9.8. So basically, this is your equation for the hanging block. Right? So we've got these 2 unknowns and so we have to go to the other object to figure out another equation. So we do this for block a, we've got the sum of all forces in the x directions is equal to mass times acceleration and, whoops. We've got, mass a times a. Alright? So anything that points up the ramp is going to be positive, our tension force. And then anything that points down the ramp like our m_{a}g_{x} and our friction is going to be negative. So this is going to be m_{a}g_{x} minus f_{k} is equal to mass times acceleration. Remember we can expand both this m_{a}g_{x} and this f_{k} because we have those equations. This is going to be tension minus, this is m_{a}g times the sine of theta minus and then friction, kinetic friction, remember is mu_{k} times the normal. So this is mass times acceleration. Alright. Now we're going to go ahead and start replacing the values that we know. We've got tension minus and then we've got remember this m_{a}g is the weight This m_{a}g is actually equal to 40 newtons This is 40 times the sine of 30 and then for our kinetic friction, our coefficient is 0.15 so we've got that and now we multiply by the normal. Remember that on inclined planes, the normal is just equal to your m_{a}g_{y}, which is equal to m_{a}g times the cosine of theta. We did this a bunch of times already so this normal force really just becomes m_{a}g which remember this m_{a}g is 40 times the cosine of 30 degrees. So you don't have to add a 9.8 there because you've already taken care of it with the weight force. So this is equal to the mass, which is 4.1 times the acceleration. Alright. So, basically, you're just going to replace all these, you're just going to plug all the stuff into your calculator. What you're going to get is 20 for this guy, and then you're going to get 5.2 for this guy over here. So, basically, what we have is tension. And then be careful. We have minus 20 and then minus 5.2. So when you combine those things together, you're actually going to get negative 25.2 because they're both negative. So this is going to be 4.1a. And so these are your 2 equations. Right? So we've got acceleration and tension that are unknown. So this is basically my second unknown equation. Now I'm going to move on and solve whoops. Now I'm going to move on and I'm going to solve this system of equations here. Alright. So I'm going to use equation addition to solve these. We're basically going to stack these two things on top of each other. So equation number 1 is tension minus 25.2 equals 4.1a, and then equation number 2 is going to be 100 minus tension equals 10.2 times a. Alright? So we've got is we're going to eliminate these tensions, and then basically you're just going to add straight down. And what you end up getting is you end up getting 74.8 equals 14.3 a, and so your acceleration is going to equal 5.2 meters per second squared, and that's the answer. So, right, there's nothing new here. Just going to combine all the different steps that we've seen so far. So let me know if you guys have any questions.

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# Systems of Objects on Inclined Planes with Friction - Online Tutor, Practice Problems & Exam Prep

In problems involving connected systems, inclined planes, and friction, it's essential to identify the type of friction—static or kinetic. For example, when analyzing two blocks on a 30-degree incline, draw free body diagrams to visualize forces like weight, normal force, tension, and kinetic friction. Use Newton's second law, F = m × a, to set up equations for each block, considering the direction of acceleration. Solving these equations reveals the system's acceleration, integrating concepts of force, friction, and motion effectively.

### Connected Objects On Inclined Planes With Friction

#### Video transcript

Two blocks made of different materials, connected by a string, slide down a 30° inclined plane. Block A has mass 8kg, and the coefficient of kinetic friction between Block A and the incline is 0.35. Block B has mass 4kg, and the coefficient of friction between block B and the plane is 0.25. After the blocks are released, find the tension in the cord.

1.23 N

2.21 N

1.67 N

2.28 N

### Blocks on a Wedge

#### Video transcript

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_{A}g_{y}, and this is m_{A}g_{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_{B}g. 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_{B}g and then m_{B}g_{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_{A}g_{x} So this is m_{A}g_{x} - friction is going to be m_{A}a. This is our target variable. Now just basically expand out all of these terms. We have T - m_{A}g 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_{A}g times the cosine of theta so we have m_{A}gcosine of theta A and that equals m_{A}a. 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_{A}g_{x} actually just becomes 5.07. And then this μ_{k} times m_{A}gcosine 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_{B}g_{x} - our tension, - F_{k}, is equal to m_{B}a. Now let's do the same exact thing. We're going to expand all these terms here. So this is going to be m_{B}g 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_{B}g times the cosine of theta B, and this is equal to m_{B}a. 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.

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