7. Friction, Inclines, Systems

Inclined Planes

# Connected Block on a Ramp

Patrick Ford

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Hey, guys, let's get started here. We've got these two connected objects that are on a frictionless incline. So this actually is going to be a connected systems of objects on an inclined planes. Really? Just gonna combine a lot of steps that we know about connected systems and also inclined plane. So let's check this out. Right. Who? We've got these two objects. Let's go ahead and start with the free body diagram for object. A. So this is the one that's on the inclined plane. So remember, we're going to have the weight force that acts straight down, So this is going to be m a G. Now we have a normal force. That normal force doesn't point upwards. It acts perpendicular to the surface. So the normal force for a actually is going to look like this. This is gonna be my normal. There's also a cable that's attached this box here. So we know we're gonna have some tension for us at points this way. But that's it. There's no friction. There's no applied forces, anything like that. Let's look at object B. This is a little bit simpler. So we have these just, uh, two forces. We really just have the weight force, which is MBG that act straight down. And then we have the tension force and these tensions are actually gonna be the same because they're part of the same cable that goes over the pulley. So that's the first step. However, one of these objects is on an inclined plane. So we have to do first is we actually have to tilt the coordinate system. So we want the new Why access to be basically in the direction of that normal force, and then we have a choice. We can either choose down the incline to be positive or up the incline to be positive. So if you think about this, what's going to happen is that when you release this system, object B is going to go down like this, right? It's gonna accelerate downwards. And so because of that, this object is going to go up the ramp. So I'm gonna choose this direction to be my positive X. All right, so that's what I've got here, which means that I'm gonna have to decompose this mg into its components. So I've got my m g y. That's going to be down here. This is M A G Y. And this component here is M A G X, and that completes our free body diagram. Now, we're just gonna choose the direction of positive. But we already just talked about that. Basically, anything that goes up and down like this, like over and down, is going to be the direction of positive. All right, so that's that. Now we want to write the F equals M A. Starting with the simplest object. The simplest object is the one with the fewest forces in that object B. So we have here is F equals M B times A. Except we're not looking at the X Forces were looking at the white forces right now, remember, for object, be anything that points downwards is going to be positive. So when it comes to our f y, we expand, this are MBG is positive and our tension is negative and this equals m B times A. So we can figure this out real quickly. We know that the mass is equal to five for this B block, so this mg is actually just gonna be 49 once you multiply five by 9.8. So you get this 49 minus T equals five A. I want to figure out the acceleration, but I can't because I have this tension force. It's unknown, so I'm just going to go to the other object now. Object? A So this is the sum of all forces in the X axis. I'm now looking up the incline like this, and this is gonna be mass eight times acceleration. Remember the object a is going to accelerate this way. Right? Okay, So all your forces in the X axis are gonna be tension and tension is along the direction of positive. So this is our T then we have minus M. A G X is equal to L a times a. Now remember, this mg X component of gravity is not mg. It's actually mg times the sine of theta. So this is equal to mass times acceleration, so we can go ahead and figure this out. Mass is two g is 9.8 and now we're going to use the sine of the angle that was given to us. This is 53 degrees right here. So we have the sign of 53 this is equal to mass, which is two times acceleration. So I'm just gonna try to scoop this over. All right, So we get here once you plug this all in is you're gonna get 15.7, so you get 15.7 here is equal to two times a, so we have the acceleration. And again, we have this tension force that's unknown here. So I'm gonna have these two equations with two unknowns. I'm going to label them one and two. And here what I'm gonna do is now, I'm gonna have to solve this by using equation Substitution. We do. That equals May. Now, we're gonna solve this by using equation addition. So this is my first equation here. Remember, this is gonna be tension minus 15.7 equals to a now, for the second equation would have lined up so that the non target variable goes away. So this is going to be 49 minus T and then I'll line up the equation. Signs like this equals five a day. So this is my equation Number two. You're gonna add them straight down like this so the tensions will cancel. You're going to get 49 minus 15.7 is equal to seven A. So when you go and actually saw for this, you're gonna get it as an acceleration that is four points, 8 m per second squared. All right, so that's it for this one, guys. And that basically means that our equation or our answer is answer choice A. That's the magnitude of the acceleration.

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