7. Friction, Inclines, Systems

Systems of Objects on Inclined Planes with Friction

# Blocks on a Wedge

Patrick Ford

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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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