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

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_ag. 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_ag and we just separate into its components. That's m_ag_y and then we have m_ag_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_bg, 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_bg minus the tension is equal to m_ba. 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_bg 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_ag_x and our friction is going to be negative. So this is going to be m_ag_x minus f_k is equal to mass times acceleration. Remember we can expand both this m_ag_x and this f_k because we have those equations. This is going to be tension minus, this is m_ag 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_ag is the weight This m_ag 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_ag_y, which is equal to m_ag times the cosine of theta. We did this a bunch of times already so this normal force really just becomes m_ag which remember this m_ag 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.

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.

• A. 1.23 N
• B. 2.21 N
• C. 1.67 N
• D. 2.28 N

Blocks on a Wedge

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.

Here’s what students ask on this topic:

How do you determine the type of friction acting on a system of objects on an inclined plane?

To determine the type of friction acting on a system of objects on an inclined plane, you need to analyze the problem text. The problem will usually specify whether the friction is static or kinetic. Static friction occurs when the objects are not moving relative to each other, while kinetic friction occurs when there is relative motion between the objects. For example, if the problem states that a block is moving up an incline, you are dealing with kinetic friction. Identifying the type of friction is crucial for setting up the correct equations to solve the problem.

What are the steps to solve a problem involving connected systems on inclined planes with friction?

To solve a problem involving connected systems on inclined planes with friction, follow these steps: 1) Draw free body diagrams for each object to visualize forces like weight, normal force, tension, and friction. 2) Identify the type of friction (static or kinetic) from the problem text. 3) Use Newton's second law, $F=m\times a$, to set up equations for each object, considering the direction of acceleration. 4) Solve the system of equations to find the unknowns, such as acceleration and tension. This method integrates concepts of force, friction, and motion effectively.

How do you draw free body diagrams for objects on an inclined plane with friction?

To draw free body diagrams for objects on an inclined plane with friction, follow these steps: 1) Identify all forces acting on the object, including weight (mg), normal force (perpendicular to the surface), tension (if connected by a cable), and friction (opposing motion). 2) Represent the weight force as a vector pointing straight down. 3) Draw the normal force perpendicular to the inclined plane. 4) Include the tension force along the direction of the cable. 5) Add the friction force opposing the direction of motion. 6) Tilt the coordinate system to align with the incline, breaking the weight force into components parallel and perpendicular to the plane.

How do you use Newton's second law to solve for acceleration in a system of objects on an inclined plane with friction?

To use Newton's second law to solve for acceleration in a system of objects on an inclined plane with friction, follow these steps: 1) Write the equation $F=m\times a$ for each object. 2) For each object, sum the forces in the direction of motion and set them equal to the mass times acceleration. 3) Include forces such as tension, weight components, normal force, and friction. 4) Solve the equations simultaneously to find the unknowns, such as acceleration and tension. This method allows you to determine the system's acceleration by integrating the effects of all forces involved.

What is the role of the normal force in problems involving inclined planes with friction?

The normal force plays a crucial role in problems involving inclined planes with friction. It acts perpendicular to the surface of the inclined plane and is responsible for balancing the perpendicular component of the weight force. The normal force is also essential for calculating frictional forces, as friction is proportional to the normal force. For kinetic friction, the frictional force is given by $f$_k = μ_k N, where $\mu$_k is the coefficient of kinetic friction and $N$ is the normal force. Understanding the normal force helps in accurately determining the frictional forces and solving the problem.