Pearson+ LogoPearson+ Logo
Start typing, then use the up and down arrows to select an option from the list.

Shortcut for Solving Connected Systems of Objects

Patrick Ford
Was this helpful?
Hey guys. So we've already seen how to solve these kinds of problems where you have connected systems of objects. In fact, we've already solved these exact problems before in previous videos, which is why I have them written out already. So what I wanna do in this video, I want to show you a very powerful shortcut that you can use to solve the accelerations in these kind of system problems. Let's go ahead and check this out. So we've already seen this problem again, we had a three and a five kg block being pulled by a force and to solve this, we had to write F equals that. We had to draw free body diagrams. We had to write F equals M. A. For both. Come up with two equations. Then we had to use equation addition or substitution. And then you finally could figure out that the answer was 3.75. I'm gonna show you how to do this in 20 seconds. And basically the shortcut is that we're going to combine all of these smaller masses like M A M. B. And we're basically just gonna imagine there's a single large object and combine all the masses together. For example, we have these 23 and five kg blocks, but I'm just gonna imagine now there's a big block that's eight. And what happens is when you imagine this, there's only one force that's pulling on this block, which is F equals 30. So you might be wondering what happens to these tension forces that were connecting the two objects and basically they go away. So when you do this, you're gonna ignore any tensions or normals between the two objects or three objects, whatever There's So basically any connecting forces. So these two tensions basically is that they don't exist anymore. And the only force that is external, the one that's not between is the 30. So now what we do is write F. Equals M. A. But now we're gonna right F. Equals big M. Times A. So the only force that's acting now is the 30 and this equals A. And so A. Equals 3.75 m per second squared, which is exactly the same answer that we got when we did it the long way. So super powerful shortcut. When you're solving for the acceleration, let's check out the other example here again, we've already done this one before. We have this four kg block the two kg books hanging. And then we want to calculate the acceleration. So again, we had to draw the free body diagrams for both. Come up with the F. Equals M. A. For both through the equation addition for both. And then finally you figure out the answer. So instead what you can do is you can just lump these both together into a single object. This is four. This is two, which means the large object is actually gonna have a massive six. And so now what happens is again the tension forces between those two objects are going to go away. So what's the only external force that's really acting on this? It's actually just the weight force that was on object B. Remember that the weight force of A is canceled out by this normal force. So basically as if you have one block that's being pulled down by a weight force of MBG. So even though we combine the masses together into a single six kg block, it's as if the weight force is really only still from the two kg block, that's acting all right. So then we do our F equals M. A. This is F equals big M times A. So remember this is going to be M. B. G. Now, just like we did before, remember how we considered anything to the right and down as positive. It's the exact same thing here. Anything to the right and down is gonna be positive. So that means our MbG is positive here, and this equals big M times A. So this is gonna be two times 9.8 equals six. A. This is 19.6 equals six A. So we get A. Is equal to 3.27. And that is exactly the same answer that we got before. Let's do these next two examples really quickly. So I can show you again how this works. So now we're gonna calculate this acceleration of this three block system. So we wanted to do this, we have to draw three free body diagrams. We have to write F equals M A. Three times. We have to get three equations do equation edition and all that stuff. And instead what you can do is you can just lump these to get these together into a single box of 10, Right? So it's two plus three plus five is 10. And so now it's as if the only force that's acting is this 40. All the tensions that are between these blocks, right? There's tensions between the two and the three, the three and the five. All of those go away. So now what happens is we just write our F equals big M times A. There's only just the 40 equals 10 A. And so your acceleration is four. That's the answer. Now, this last one here, remember this is like an Atwood kind of machine here. And what happens here is that you have to wait forces. Remember this is gonna be mbG really? This is going to be like six times 9.8. And then you also have this one which is M A G, which is four times 9.8. And then you have the two tension forces. There was attention up and attention up like this. Well, again, if you just combine them into a single object, it's as if you have one block that's actually 10. And what happens is the tension forces will go away. And the only to external forces are really just these weight forces. Remember that the rule that we use is that anything up to the right and down was going to be positive. So you actually have these two weight forces that are kind of like pulling against each other. It's the exact same thing here. Anything up to the right and down is going to be positive except when you replace this into one big object, it's kind of as if you have the weight force which is MbG which we know is six times 9.8 is 58.8, which is pulling it down. But then on the other side you have this upward force which is M. A. G. Which is really just 39.2. That's pulling it kind of up. Right? Again, we're just kind of just imagine this is a single object, that's kind of what it works out to. So now when you use your F equals big M times A. Now you have to external forces, remember anything that's too, that's downwards is going to be positive, anything upwards is gonna be negative. So what happens here is that you have 58.8 minus 39.2 equals 10 A 19. equals 10 A. And you get a equals 1.96 which is exactly what we got when we work this out the long way, That's it for this one guys, let me know if you have any questions