Three mountain climbers who are roped together in a line are a...

Question

Three mountain climbers who are roped together in a line are ascending an icefield inclined at 29° to the horizontal (Fig. 4–67). The last climber slips, pulling the second climber off his feet. The first climber is able to hold them both. If each climber has a mass of 75 kg, calculate the tension in each of the two sections of rope between the three climbers. Ignore friction between the ice and the fallen climbers.
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Accepted Answer

Welcome back. Everyone in this problem. During a rescue mission, three firefighters are linked together by a rope as they ascend a steep slope. The slope is inclined at 35 degrees to the horizontal. As shown in the diagram unexpectedly, the last firefighter loses footing, leading the second firefighter to also lose balance. Fortunately, the first firefighter successfully supports both of them with each firefighter weighing 70 kg, determining the tension in each of the two segments of the rope connecting the three firefighters assume negligible friction between the slipping firefighters and the surface. For our answer choices. A says that it's 3.9 multiplied by 10 squared newtons. B 4.78 multiplied by 10 squared newtons C 7.9 multiplied by 10 squared newtons and D 9.5 multiplied by 10 squared newtons. No, in this problem, if we're going to determine the tension in each of the two segments of the rope, then we have to think about the forces that are acting in our diagram for our firefighters. OK. So let's consider a free body diagram. Let's try to draw a free body diagram here where for our free body diagram or Y and X axis are relative to our planes. Now, first of all, starting with our first firefighter, we know that his weight acting downwards. As a matter of fact, let me put, put our forces in blue here, his weight acting downwards is MG. OK? And his normal force is acting relative to the plane as N, OK. We also know that probably I should draw this a bit better. OK. We also know that his, the, the normal OK, is inclined at the degrees to the horizontal in this case or 35 degrees. But for now, let's just say that theta equals 35 degrees to make it easier to represent on our diagram. And we know for our firefighter the tension in the first rope, let's call that F one, OK? Is acting to the right while F two, the tension in the second rope here, sorry, not F two, the tension or the force acting opposite to F one, the tension in the first part of the group would be the force of our firefighter because remember our first firefighter is supporting everyone else now, for a second firefighter, assuming he's at rest, he also has a normal force and a weight that's acting downwards. OK? And you know what, let me put theta equals 35 degrees somewhere else to make space. OK? I think I can put it, let me put it on the other diagram, say that it's 35 degrees Now, remember again, he'll also have who is acting this way theta. Ok. So we have our first firefighter, our second firefighter and our third firefighter. Ok. And for our second firefighter, tension in the group F one will be acting this way while our tension in the second segment F two will be acting that way. Ok. Finally, for our third firefighter, this is the one who initially lost his balance. Then he also has a normal force acting upwards with MG acting downwards an anger theater two or vertical weight acting and attention F two that is acting towards uh our second firefighter. OK? Because he's pulling him here. Now notice that each firefighter weighs 70 kg. So the mass is the same for all three and we can take G or acceleration due to gravity to be 9.8 m per second squared. Now, when we think about the weight of all our firefighters, they're going to have a X component and a Y component or X component, notice that our X component is opposite to our anger theta. So that means the X component of our weight. Let's make a note here. OK. The X component of our weight is going to be equal to MG sign theater. OK? And the Y component of our weight is adjacent to our angle theta. So it's going to be equal to MG cos theta. OK. Now, since we have these for the components of our weight, we can start to resolve our forces and by resolving eventually, we will be able to figure out the tension between the segments in each of the two segments of the rope, which would have been F one. Ok. Now, what do we know, let's try to resolve the forces in the X direction for the lowest firefighter. So let's start with the third firefighter, firefighter three. Ok. Now, at three, we know that assuming he is at rest, the sum of forces in the X direction are going to be equal to zero. If we look at him, we know that his weight will be acting in the opposite direction to F two. OK. So that means negative ft sorry, negative F two multiplied by sine, sorry, negative F two. My bad plus MG sign theater is going to be equal to zero. OK? Because again, the tension in the second row F two is acting to the left while the weight, the X component of the weight of a firefighter which is MG sign theta is acting to the right. Now. If that's the case, this tells us then that F two is going to be equal to MG sine theta which when we substitute our values, our mass is 70 kg or acceleration due to gravity 9.8 m per second squared. And our theater is 35 degrees. Therefore, if two is going to be equal to that value, which when we put that into our calculator, we should get 393.4734. Newton. OK. So that's the value of F two. Now, if we want to figure out F one, the tension in the first, be between the 1st and 2nd firefighter, then we can also write Newton's second law. But in this case, in the X direction for the lowest fire for the middle firefighter also assuming that he is at risk. So at two, OK? Since he is at rest, then the sum of forces in the X direction are going to be equal to zero. And if we think about those forces, if we go back to, I probably should have written that in blue. But if we go back to our second firefighter, notice that F one is acting to the left F two is acting to the right and the weight of a second firefighter is also acting to the right. Therefore, that means negative F one acting to the left plus F two, acting to the right plus the X component of his weight, MG Sine theta also acting to the right are going to be equal to zero. Now, this is pretty good because we can use this to solve for F one. Because if we add F one to both sides, it tells us that F one is going to be equal to F two plus MG sine Theta. Again, we know that MG Sine theta is F two earlier, we had found that out. OK? We have found out F two equals MG sine theta. So we can represent this then to say that F one is going to be equal to F two, F two or two F two. OK? Since that's the case, that means F one tension in that first segment is gonna be equal to 309 sorry two multiplied by F two which is 393.4734 newtons. And when we do that, we find that it's equal to 786.9468 newtons, which if we write it in units of 10 squared, that's going to be approximately 7.9 multiplied by 10 squared newtons, writing it, that is writing it in standard form. Rather, therefore, this will be our attention F one. And thus, if we look back on our answer choices that tells us C is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

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