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Connected Objects On Inclined Planes With Friction

Patrick Ford
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Hey guys. 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 gonna 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 incline plane. We've got the weights and the masses of both boxes, and the coefficient of kinetic friction, we're also told is we're also told that Block B is moving up the degree in clients. We know which kind of friction is going to be acting. It's going to be kinetic, and so we want to do. Is find the acceleration of the system. So let's get started. We want to draw a free body diagrams for both objects. Let's go ahead and start with Block A block is the one that's on the incline plane. We've done this a bunch of times before. We've got the weight force that acts straight down MSG we've got the normal force that is perpendicular to the surface. Then 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 moves up the 30 degree incline. So that means that friction has to oppose that. That's gonna be our kinetic friction. Alright, so that's your free body diagram. Now we do. Is we just tilt our coordinate system? We get rid of this m a G and we just separated into its components. That's m 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 forest which acts straight down MBG. And then we got the tension. All right, so now we move onto 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 three, which is writing f equals M A, and we're gonna go ahead and start with the simplest object, which is Object B. So we're gonna write some of all forces in the Y axis equals M B times A. We've got two 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 is moving up the incline. So basically, anything that goes up, around and over is going to be your direction of positive. So for be anything that points down is going to be positive for a anything that points up the ramp is going to be positive. All right, so we expand our forces we've got R M B G minus. The tension is equal to M B A and we actually have the masses in the weight so we can go ahead and simplify. Remember that the weights of Block B is already given to us. This is 100 Newtons, 900 kg. So that means that this MBG is already 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, um the hanging block. Right? So we've got these two 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 direction Z equals mass times acceleration and whoops. We've got mass A times a All right, So anything that points up the ramp is going to be positive, our tension force and then anything that points down the ramp like RMG X and our friction is gonna 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 MG X and this f k because we have those equations. This is gonna 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. All right, now we're gonna go ahead and start replacing the values that we know. We've got tension minus and then we've got to 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 are 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 mg y, which is equal to mg times the coastline of Fada. With this bunch of times already. So this normal force really just becomes mg which remember this MG is 40 times the co sign of 30 degrees. So you don't have to add a 98 there because you've already taken care of it with the weight force So this is equal to the mass, which is 41 times the acceleration. Alright, so basically, you're just gonna replace all these? Uh, you're just gonna plug all this stuff into your calculator or you're gonna get is 20 for this guy, and then you're gonna get five points to 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 gonna get negative 25 points to because they're both negative. So this is gonna be 4.1 a. And so these are your two equations, right? So we've got acceleration intention 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 gonna solve this system of equations here. All right, so we're gonna use equation addition to solve. These are basically going to stack these two things on top of each other. So if the equation number one is tension minus 25 points, two equals 4.1 a and then equation number two is gonna be 100 minus. Tension equals, um, 10 points, two times a. All right, so we've got is we're going to eliminate these tensions and then basically, you're just going to add straight down 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 square. 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.