FIGURE EX7.17 shows two 1.0 kg blocks connected by a rope. A second rope hangs beneath the lower block. Both ropes have a mass of 250 g. The entire assembly is accelerated upward at 3.0 m/s^2 by force F. (b) What is the tension at the top end of rope 1?

Video transcript

Mhm Hey everyone in this problem, three metallic blocks of mass 2.5 kg and three metallic chains, each of mass 500 g are assembled. As shown, the chains are welded onto the metal blocks. The third chain is welded below the lowest block. An external force F E X T accelerates the fabrication at 2.5 m per second squared. We're asked to determine the tension at the top of chain one. Now we're given this diagram of the system and we have this external force acting upwards on block one and all of the blocks are arranged vertically. So we have block one at the top again with this external force acting upwards off of it at the bottom of block one connecting it to block two, we have chain one connecting block two to block three, we have chain two and at the bottom of block three hanging from it, we have chain three. Now we're giving six answer choices in this problem. Option A 6.2 Newtons option B 80 noons, option C 110 newtons, option D 74 Newtons, option E 47 Newtons and option F 31 newtons. Now I'm gonna go back up so we can see this full diagram and we're looking for the tension at the top of chain one. I'm gonna draw a little red star at the top of chain one. OK. That's right below block one. And that is where we're looking for the tension. Now, the easiest way to solve this problem is to draw a boundary. And what we're gonna do is we're gonna draw a boundary separating what's inside of our system and what's outside of our system. So we're gonna draw our boundary at that point where we're looking for the tension and inside of our system inside of here, we have chain one, chain two and block three. All of these internal forces we're going to add to zero. What do I mean by that? Well, you can imagine if we look at chain two, OK? When we're looking at a free body diagram of block two, OK? We have the tension from chain to acting downwards. And then if we go and we look at the everybody diagram of block three, we have that exact same tension T two. But now it's acting upwards and those two kind of cancel each other out when we're looking at the entire system. So all of these internal forces add to zero. That means that we only need to consider the external forces. OK. So if we take a free body diagram of our system, we're gonna draw our system as the dog. OK? And that system contains all of those internal components and where all of those internal forces add to zero and externally, what do we have? Well, we have this force acting upwards and that's gonna be the force block one and chain one. We're at the very top of chain one. So we have this force acting between this block one and this chain one. It's gonna be acting upwards and acting downwards. We're gonna have gravity and this is gonna be gravity of the entire system. So we have F G of this system. OK? Because hanging from that point with that tension is gonna be all of those things. Chain one, chain, two, chain three, block two and block three. So we need to consider the gravity, uh the gravitational force from all of those. So this is the free body diagram of our system and we're gonna take up to be positive. Now, we're called Newton's law. Newton's law tells us that the, some of the forces and this is Newton's second law that we're referring to some of the forces. And in this case, we're talking about the forces of the system, it's gonna be equal to the mass of the system multiplied by the acceleration of the system. Now, for some of the forces, we're talking about the Y direction because that's, we only have forces in the Y component or the vertical component. OK. So in the positive Y direction, we have this force from block one chain one. And in the negative direction, we have the force of gravity of the system. So we have F B one C one minus F G. This is equal to the mass of the system multiplied by the acceleration of the system. Let's keep going, give ourselves some more room to work here. Now, this force of gravity, what is that equal to? Well, we have F B one C one, the force of gravity is gonna be equal to the mass. OK. In this case, the mass of the system multiplied by the gravitational acceleration G, this is gonna equal the mass of the system multiplied by the acceleration of the system. Now this force F B one C one, this is what we're trying to find. OK. This is that tension at the top of chain one, the force acting upwards on chain one. OK. So that's what we're looking for. So let's go ahead and isolate for that value F B one C one, it's going to be equal to, we can add MS I multiplied by G to move it to the right hand side. And then we can factor out that mass of the system. So we get the mass of the system multiplied by G plus the acceleration of the system. Now, what is the mass of our system? Can you recall we have to consider everything inside of that system. And if we look back at our diagram, we, where we drew that boundary, everything below that boundary is inside of our system. So we have all three chains, chain, one, chain two and chain three as well as block two and block three. So when we're talking about the mass of our system, we need to add the mass of all five of those components. So we get MC one four M B two, what MC Two plus M B three plus MC three. OK. Where the CS indicate chain, the fees indicate block and then the number just corresponds to what number we are looking at. OK. And we're multiplying that by G plus the acceleration of the system. All right. So we've got this expression all broken out as much as we can. Now, we need to substitute in these values. Now, we're told that the mass of each chain is 500 g. We want to convert this to our standard unit of kilograms. OK? And also the blocks were given the mass in kilograms. So we wanna be consistent there as well. And so the mass of a chain is equal to 500 g. We're gonna multiply that and in each kilogram, we have 1000 g. So we're gonna multiply by one kg divided by 1000 g, the unit of gram will divide out. And so what we're essentially doing is dividing by 1000 and we get 0.5 kg. So now for each chain, our mass is 0.5 kg. For each block we're told the mass is 2.5 kg. So we have 0.5 kg plus 2.5 kg plus 0.5 kg plus 2.5 kg plus 0.5 kg. And we're gonna multiply that by G which is 9. m per second squared plus the acceleration of the system. And we're told that this external force accelerates the fabrication at 2.5 m per second squared. And so the acceleration of the system is 2.5 m per second squared. All right, we're almost there. We just need to add everything up, multiply it. So if we do that, we're gonna get 6.5 kg multiplied by 12.3 m per second squared. We multiply the two together we have kilogram meter per second squared. That's gonna give a unit which is equivalent to a Newton. OK. So we get a Newton, which is what we want for a force and this is gonna be equal to 79.95 newtons. Now, our answer choices are rounded to the nearest Newton. So we round to the nearest Newton and we get that the force, the tension at the top of chain one is approximately 80 newtons which corresponds with answer choice. B thanks everyone for watching. I hope this video helped see you in the next one.