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Bob traverses a chasm by stringing a rope between a tree on one side of the chasm and a tree on the opposite side, 25 m away, Fig. 4-65. Assume the rope can provide...

Question

Bob traverses a chasm by stringing a rope between a tree on one side of the chasm and a tree on the opposite side, 25 m away, Fig. 4-65. Assume the rope can provide a tension force of up to 29 kN before breaking, and use a "safety factor" of 10 (that is, the rope should only be required to undergo a tension force of 2.9 kN). (a) If Bob's mass is 72.0 kg, determine the distance $x$ that the rope must sag at a point halfway across if it is to be within its recommended safety range.

Bob traverses a chasm by stringing a rope between a tree on one side of the chasm and a tree on the opposite side, 25 m away, Fig. 4-65. Assume the rope can provide a tension force of up to 29 kN before breaking, and use a "safety factor" of 10 (that is, the rope should only be required to undergo a tension force of 2.9 kN). (a) If Bob's mass is 72.0 kg, determine the distance $x$ that the rope must sag at a point halfway across if it is to be within its recommended safety range. ($b$) If the rope sags by only one fourth the distance found in ($a$), determine the tension force in the rope. Will the rope break?



Answers

Bob traverses a chasm by stringing a rope between a tree on one side of the chasm and a tree on the opposite side, 25 m away, Fig. 4-65. Assume the rope can provide a tension force of up to 29 kN before breaking, and use a "safety factor" of 10 (that is, the rope should only be required to undergo a tension force of 2.9 kN). (a) If Bob's mass is 72.0 kg, determine the distance $x$ that the rope must sag at a point halfway across if it is to be within its recommended safety range. ($b$) If the rope sags by only one fourth the distance found in ($a$), determine the tension force in the rope. Will the rope break?

So let's draw a free body diagram for this mass or the person we can say. So this point mass essentially would represent the person. And then here, this would be forced tension. Aah! This would be again forced tension. And actually, let's bad placement. My apologies. Right here. We can draw this dashed line and this would be considered Fada. Ah, this again would be considered status of the same angle on either side. We can say that this would be the dash line, uh, going up from the mass or the person would be ax and then we know that this length here would be 12.5 meters. So here, um we also have the weight of the person and this. Yes. So this would be the full force full free body diagram for this person. Now here we need Thio. Ah, right. Noon second law for that, um, small piece of rope So we can say that Sigma fc in the UAE direction would simply be equal to two times force tension times Sign of Fada minus mg And this should be equal zero because the person on the rope is not moving. Therefore, if we wanted to find Seita. Fate Oh would be equal to arc sine of the person's weight, divided by two times the force tension and we can solve. So arc sine of 72.0 kilograms multiplied by 9.80 meters per second squared. This would be divided by two times 2900 Newtons and we find that this is giving us six point nine a degrees. What's I don't want it to be so cramped so we can say again six 0.98 eight degrees. And at this point, we can say that tangent of X in order to find this ex distance here we can say tangent of Fada with equal X divided by 12.5 meters being this distance here. So essentially we're trying to set up a triangle. Um, where this force tension length here would be our iPod news Essentially so we can say that X would be equal to 12.5 meters. Ah, this would be multiplied by tangent of 6.988 degrees and we find that X here is gonna be 1.532 meters and we can round X with equal 1.5 meters. So we have to round 22 significant figures and at this point, we can. For part B, we can use the same equation to solve for the tension force. However, here Ah, this the sag and the rope would only be 1/4 of that of the ex found in part a So X here would be equal to 1/4 of 1.532 meters. This is equaling 0.383 meters and then we're going to say that fade out with the ark tan of 0.383 divided by 12.5. So these two, uh, lengths would stay constant. We find that veda is gonna be 1.755 degrees and then we want to find force tension. So forced tension would then be equal to M G, divided by two times sign of data and so this would be equal to 72.0 kilograms multiplied by 9.80 meters per second squared. This would be divided by two times sign of 1.7 seven 1.755 degrees and we find that the force tension for part B. It would be equal to 11 0.5 times 10 to the third noons and here we know that the rope will not break. Um however, it does exceed ah, the recommend. They recommend attention by a factor of about four so the line won't break. However, it's way past the recommended tension. That's why if the SAG is greater, it's safer for the for the person on the ropes so we can say Ah, rope will break. That is the end of the solution. Thank you for watching.

Right. What's this? Chapter nine Problems 79. Um, if we look here to read the problem, it's a bit of a long one. I'm going to assume you have the book with you on this one in the interest of time Problem nine Problem 79 in chapter nine. I should say, uh, essentially, we have a We have, uh, an adventurer going using a mountain climbing technique. And they need to get past a music, traverse a deep chasm. And essentially, we want to see if the rope that they're going to use to traverse between this chasm you're going to break or not based off of the tension in the rope. This is very similar to two problems before to our problems before Schefter Chapter nine. Problem 75. Uh, in that you can you just need to do the sum of the forces. So we're looking for X here. We want to see how far it's sagging and then based off of how far it sags. We also want to see what the tension in the rope is. And if it's gonna break or not, Right. So let's calculate all these things we can use the mutant. Second long. This will tell us everything we need to know. So first, of course, as we always do with force with forces. Let's drop free body diagram. Let's go look something like this and equip it. Remember our vector signs. So 2025 meters both ways. We have angles there at the top ex, the ex distance. We don't know what Saget is for the person. They're obviously being affected by gravity. They have mass. And there's tension pulling them in both ways as the rope is tied into the trees. Um, yes. So to include the safety factor, the tension must be less than or equal to 2900 Nugent's, that is to say, Yeah, but the tension must be no more than 2900 means. And so we have the exact same equation from problem 75 to times the, uh why component of the tension, which is just ft attention for its multiplied by sine data. It's attracted by energy, is equal to zero. And with that, if we have the maximum force possible there, which was just calling 2900 for now, then say that would be ah, well, you have all these numbers, right? G is 9.1 massive 75 kilograms and the FT max is 2900 news. You can calculate fatal minimum to be seven degrees at most. So this means weaken more or less interchange, sign and tangent as before. Small angle approximation. Um, but also from the triangle that we see on the left side of the road by Bagram. Well, it could be the left or the right. Let's just say neither triangle we can use are triggered. No metric, uh, brain, too. So this problem So you say tangent data. It was attention. Data man is equal to X men over 12.5 meters. We can calculate the distance the minimum distance needed to have for there to be a sag. And as we can see, we are too calculated the angle needed so we can just plug that in, multiply by 12.5 and we find that be sag, you're going to be 1.6 meters. That's to say the minimum distance the rope must sag based off of the maximum force is going to be 2900. It's gonna be 1.6 meters now taking that sag and multiplying it by 1/4. Let's look at if the Tyrolean traverse is set up incorrectly so that the rope sags by only 1/4 the distance found. What's the tension? Then we want to say, OK, let's multiply at 1.6 my fourth, so it's gonna be about 0.4. But let's do it exactly. It'll be 0.3992 meters. We can use this distance to find out what the tension on the rope is by working backwards. So what we can do here We were the next page. So, as I said the most by by 1/4 to get 0.3992 Use that distance. Find the tension. And so essentially, we flip around the changing equation and take the arq Tana both sides to cancel out the changing on the left. To get the angle we plug in our new distance rather than 1.6. Now it will be a 0.3992 meters, uh, calculator angle. Our new angle is going to be 1.8 to 9 degrees, which we can then plug into the sign below the sign of the equation below and mg over to science data will give us the tension with with all of our numbers plugged in should be about 12,000 Newtons more exactly. It'll be 11,512 neuters, um, and with this the rope line on break. But, uh, rather than the safety factor being 10 it's only going to be four. Because if you divide 12,000 by four, you should get about 3000. Which is what our, uh, forces here. Right there's about 5900. Just $100. This is 500 off, but still pretty close to 12. So that's a factor of 10. Would have had a 29 killer Newton attention.

Mhm. Here for the solution for the part A. The answer is option C. Option C. That is Part three is the answer. Now for the explanation. Right. The expression for the vertical force acts on the back that is sigma. If I go to to design tita minus MG and we consider it evacuation one her F. Y. Is the vertical force acts on the back. He is the tension force M. Is the mass of the bag. G. Is the acceleration due to gravity and tita is the angle. Now we rewrite the above expression for the 10s and 4s. So it will be to design tita minus M equal to zero. So T will be equal to MG divided by two scientists to And we consider it as a equation too. Conclusions from the above equation. The tencent force depends on both the fourth separation and say therefore the correct option is part three country. Now for the part B. Right. The expression for the angle of the rope that is tante taker to y divided by X. So from it, he will be quite often universe sign Uh to equal to 10 inwards by by X. And we consider it as a creation three. Here, tita is the angle of the slope X. Is the horizontal distance between the tree and back. And why is the vertical displacement of the rope? So T will be equal to MG by two scientists to and we consider it as a creation. For now. We substitute 0.20 metaphor Y and 5.0 m for X. In equation three to find tita. So here by substituting the value in equation three, we get to take or to 10 0.20 m divided by 5.0 m. And from here we get to take or to 2.29 degree. Now substitute five point europe K G four M 2.29 degree 42 9.8 m per second squared for G. In the equation four. So the tea will be equal to 5.0 program, 9.8 m per second squared, divided by to sign two point to 9° from where we get equal to 6.1 multiplied by 10 to the power to New 10. Therefore the tense and forced. The rope is 6.1 multiplied by 10 to the power to. Okay, so this is a complete solution. Please go through this. Thank you.

For Karen and Jim. We have a mass and then going up is the force tension in the rope. Karen using using her dynamic rope and Jim using his static rope and then going down is the force of gravity or MG. We know that the final velocity is going to be square root of two g h. So this is simply be equal to two times nine point eight times the height of the drop. So two meters, and then this will equal and then this week was six point two six meters per second. We need to find the acceleration, so acceleration is equal to velocity. Final squared, minus velocity initial squared, divided by two times the height. And this will equal We know that the final velocities zero. So we can simply say negative six point two six squared, divided by two times to the height. Or rather, Delta Delta. Why, ah is gonna be one meter rather because this is going to be Ah, the height is essentially the height of the drop. Ah, here it's better to call it actually Delta. Why? Because Delta why specifically is after she hits the ground. So essentially after gym or Karen hit the ground. Their bodies essentially compress one meter S o they they can either be bending their knees or ah or ah bending their knees or rolling or anything in order to increase that delta. Why? Because increasing the delta Why will allow you to have a lower acceleration are a lower deceleration in this case, and then that will in turn reduce the amount of force that is felt by the climate themselves. So this will be equal tio negative nineteen point six meters per second squared. We know that Sigma F why equals m A. Why and making so that this is going to be equal to M G minus force tension. So force of the force tension or force of the rope divided by mg is simply going to be equal to M G minus a y, divided by MG EMS. Cancel out and then this is giving us a nine point eight minus negative. Nineteen point six divided by nine point eight and this is giving us three. So essentially the rope holds three times her weight again. This is for Karen, so the rope can hold three times her weight. However, this factor of three represents essentially the degree in which she's feeling the drop. Essentially. So. The higher the number, the higher the ratio, the more force that is felt by the person the climber themselves. So here, three this is not too bad in the sense that ah one would be able Teo Ah, quota. Handle it. Essentially, I don't The NOAA boat bones wouldn't be broken. Ah, given this ah, given off. Given this force tension, however we have Next. Next. Ah, we have gyms. Rope. Now Jim's rope is much is static and his delta y for the rope is point three zero meters. So here the delta y is one meter, and that was due to the elasticity of the rope. How much the rope was stretching if Karen Inventor Venter knees as she was hitting the ground. A lot of factors can contribute to this Delta y, however, with Jim's static rope that doesn't stretch as much. We have adult a Y of point three zero meters, so here F rope, divided by AMG is going to be again G minus eight. The vibe i G. And then here This is going to equal nine point eight and then plus six point two six squared, divided by ah two times point three all divided by nine point eight. And we're getting Ah, this is equal to seven point seven. So this would be seven times seven point seven times his weight. However, here ah, the greater again the greater the number, the more likely the more Ah, chance for injury. So here Jim is much more likely to be injured. Uhm, force net on his body is ah relatively high. And that would be the end of the solution. Thank you for watching.


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