Systems of Objects on Inclined Planes with Friction

7. Friction, Inclines, Systems

Systems of Objects on Inclined Planes with Friction - Video Tutorials & Practice Problems

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1

concept

Connected Objects On Inclined Planes With Friction

Video duration:

6m

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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.

2

Problem

Problem

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

3

example

Blocks on a Wedge

Video duration:

6m

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Hey, guys, let's check this problem now. So we've got these two blocks that are connected to each other on a wedge shape. So basically, this we've got two different angles to consider is like you have two different rough inclined planes that are kind of sandwiched together, right? But basically, we want to figure out what's the magnitude of the systems acceleration. Once we release it, we know there's gonna be some friction involved. And so really, we're just gonna treat this like any other problem. Let's go ahead and draw the free body diagrams for A and B. All right, so for a basically we've got is we've got a free body diagram. We know we have our weight force. This is gonna be m A g r. Tension forces 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 going to point up the ramp or down the ramp. We actually don't know. What 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 gonna 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 had these two blocks. This one is 5 kg. This one is 2 kg. The 5 kg 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've pointed the direction of the heavier object. So because of that, this acceleration, uh, we know that the Friction Force has two point down the ramp for object. A. So this is gonna be some friction. We're also told that the system begins moving. Once you release, it begins moving. So basically we know that F is equal to F K, and that actually takes care of step two for us. We know what kind of friction we're dealing with this is coefficient of kinetic. So now we just basically separate RMG into its components. So this is going to be m A g y. And this is m A G X. All right, now we're gonna do something similar for Object B, except it's gonna be on a different inclined plane. So here we've got our free body diagram for Be So you've got the Weight Force MBG. 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 into the rights. So for object B object to be wants to slide down this way. So we know that this friction here is actually gonna point up the surface That's going to be the F. K. So now we just split up RMG. So this is going to be M B g y and then M B g X and then we just don't really need these anymore. So those are 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 over into the rights, then that means that for object a the deposited directions this way and therefore be it's that way. All right, so I've got my direction of positive here and here. All right, so when we expanded our forces, I've got my tension that points up and then my f k points. Actually, I've got my mg x. So this is m A G X minus. Friction is going to be m A. So this is our free, our target variable. So now we just basically expanded all of these terms. We have t minus m a g times the sine this is going to be fatal. A. Remember that. There's two different angles to consider. There is this one and this one So I got this one is going to be, um u K coefficient of friction. And then remember, this is going to be, um u K times the normal force and the normal force is going to be equal to mg times the cosine of theta. So we have m a g co sine of theta A and that equals M A. Alright, so basically, what happens is these are just a bunch of numbers when you plug them in. Uh, and so we're going to simplify. So this m a G X actually just becomes 55 point oh seven and then this mu k times m a g cosign theta really just becomes 3.79. So this equals the mass of A, which is two times a so you can simplify once more, and this is just gonna be t minus 8. equals to a We can't go any further because now we have two unknown. So this is gonna be our first equation here. Let's go to the other. The other, um, f equals m A. So now we have some of our forces equals mass B times A. We know that they're going to have the same acceleration. So now we're gonna use the downward direction like this. And so we have our M B g X minus. Our tension minus F K is equal to M B A. We now just do the same exact thing. We can expand all these terms here, so this is gonna be mbg sine theta B. This is just the other angle, right? Just keep track of your variables, and then this is just gonna be the mu k times the normal. So this is gonna be the UK Times m b g times the co sign of to be and this is equal to M B. Hey, All right, so just like before, these are really just a bunch of numbers. When you plug them into your calculator, you have mgs, datas and all the coefficients, so you can just go ahead and sulfur this right, this is gonna be 24.5, minus tension minus 8.49 equals, and this is going to equal five A. So when you simplify this, this is 16 point a one minus tension equals, uh, five. This is five a. All right, so Now we have our two equations. So this is the equation. Number one equation number two. And so now I just use equation substitution to sulfur This. So bring these equations down here. I've got t minus 8.86 equals to a then I've got 16 point. A one minus. Tension equals five A. So you add these straight down your tensions cancel and basically you end up with eight points. Actually, do you end up with seven 0.15 equals seven A. And so therefore your acceleration is going to be 1. 1.2 m per second square. Now the problem masses 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 answer choices and our answer choice is be so let me know if you guys have any questions and that's it for this one

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