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Spider webAn orb weaver spider with mass of 0.28 of 4.7 x 109 N/m? and grams hangs vertically by one of its threads. The thread has = Youngs modulus radius of 9.8 x...

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Spider webAn orb weaver spider with mass of 0.28 of 4.7 x 109 N/m? and grams hangs vertically by one of its threads. The thread has = Youngs modulus radius of 9.8 x 10"What the fractional Increase the thread s length caused by the spider?10 3 Eubmit2) Suppose 78 kg person hangs vertically from nylon rope . What radlus must the rope have If Its fractional increase In length Is to be the same as that of the spiders thread? (The Youngs modulus for nylon 3.7 * 109 Nlm )Submi

Spider web An orb weaver spider with mass of 0.28 of 4.7 x 109 N/m? and grams hangs vertically by one of its threads. The thread has = Youngs modulus radius of 9.8 x 10" What the fractional Increase the thread s length caused by the spider? 10 3 Eubmit 2) Suppose 78 kg person hangs vertically from nylon rope . What radlus must the rope have If Its fractional increase In length Is to be the same as that of the spiders thread? (The Youngs modulus for nylon 3.7 * 109 Nlm ) Submi



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BIO Spiderweb An orb weaver spider with a mass of 0.26 g hangs vertically by one of its threads. The thread has a Young's modulus of $4.7 \times 10^{9} \mathrm{N} / \mathrm{m}^{2}$ and a radius of $9.8 \times 10^{-6} \mathrm{m}$ . (a) What is the fractional increase in the thread's length caused by the spider?
(b) Suppose a 76 -kg person hangs vertically from a nylon rope. What radius must the rope have if its fractional increase in length is to be the same as that of the spider's thread?

In the party we have to find what is the fractional increase in the threats length caused by the spider? That means we have to find Delta lbl. We know that young models is equal to F into l divided by a into the l From here we can ride. There's the l by l is a call to F by a into the young one. Let's discuss that formula. Now we have to find the force of pension in the threat, which is a call to Mars in two great personal acceleration. So this is a cult OMG So the mass is equal to 0.26 g. That means 0.26 to 10 to the power minus three multi private G, which is 9.8. So this force is equal to 2.5 ft eight in 2 10 to the power minus three newton. This is the value of force. If we talk about that area, area is a call to pie are square area is a call to buy our square Suffian to radius various is 9.8 into 10 to the power minus 6 m. This is the radius square So if we solve, the area is equal to 3.16 312016 and to 10 to the power. My understand, we did a square. This is the area. This is the area. No, we have to find Delta l Y l. So this is a call to F A phone by into a delta. And by l is a call to f upon A in Dubai. So this is a call to militar. Lbl will be called UFO aid. Why not f? This is why I will be in the denominator. Yeah, so this is equal to have upon way into a right. So we will put the Value Forces 2.54 into 10 to the power minus three. They went by young wordless, which is 4.7 to 10 to the power nine. The right way area, which is 3.1 16 to 10 to the power. I understand. So Bill to n by n is a cold Oh, 1.8 into 10 to the power minus 31 point into 10 to the power minus 31.8 to 10 to the power minus three. This is the fractional change in the length. This is an answer for party Now, for part, we we have to find the massive person is 76 kg so young were less of nylon is 0.370 point 37 into 10 to the power 10 Newtons per meter square And this time force will be equal to M into G now masses 76 kg So this is 76 into 9.8. That means it is a call to 74 4744.8 Newton This is the value of force So we have to find that areas So we know that again we will use that formula That area is a call to half upon why into delta while mhm so here. Now we have to find the radius so we know that radius area is equal to Pi r square This is f often buy into delta l Y l. So from here we can say the areas will be equal to f a phone buy into why into delta l Y l in Underwood So we will put the values value of forces 7 44 0.8, divided by five into young models is 0.37 into 10 to the power 10 into Delta, via which is 1.8 into 10 to the power minus three. So this is equal to radius is a call to 5.967 into 10 to the power minus three into 10 to the power minus 3 m. Okay, or we can write it is a call to 0.60 centimeters.

Solving party. We know that f by a physical toe. Why dll by l here. Why is Young's model ISS and Dell is equal toe f by a in tow l by y is equal toe this so on Going forward Dellal is equal. Toe this into this into 0.1 toe by pie 0.75 into £10 minus three square in two point toe 10 power 10. So dll is coming as this so dll can be written as 0.166 Now sold in part B. Maximum weight is equal toe 10 Sile force in tow area. So putting the value of 10 Sile force as well as area Here we get the maximum weight as 17.67 Newton.

All right. Questions. 30 ah ha sido the spider and her web on orb Weaver. Spider sits in her web that is supported by three radio strands. These real strands of these red lines attached to three different points on I Believe a wall. I assume that the radio strands contribute to supporting the weight of the spider. If the mass of the spider is 5.2 tents and the negative four kilograms and the radio strands were all under the same tension, find the manager of the tension T. This is, ah, representation of the diagram. Perhaps the angles aren't necessary to scale, but the scenario still holds true, so you're given each of these angles. So, for instance, for theater, one of the top left, this art line is what's given to be data one. This art line top two stated to which I'm con calling the state a 123 Just thanks. It'll be easier. Um, with each of their degrees, let me see that the system is in equilibrium. There's no force acting upward or downward or left the rate. So first thing we need to do because we recognize this is a static problem. We'll need to look at the different forces in each direction. So we're leaving my axes ax. And why? So this new we know that if the gravity So the force acting on the spider will be played down again, that's given in the question. So the weight of the spider is the negative direction. I'm gonna look first to see the forces in the wind direction, and from there come well, Look, if we look at the forces in the wider and the extraction, right. So if you look at actually we'll get you to these different points connecting the spider in the wall. So this is our top left here. This is a sinner. You won or throwing top left. So one is top left to his top rate and three is bottom. Okay, so Syria one, our attention acts here, we'll get our parts Are vertical component attention here is going positive. Hurry up! Has scientists. So here it contributes a positive T sign theater one Syria to our top, right. Our attention is here therefore, that this is their angle Then we're contributing are forcing her positive. Why? Direction to be tied? Attention time, science theater, too. And senior three in the bottom. We have a negative component. So is re Streeter. This is theater, and the vertical component of this is a negative. Attention. T sign data three, General. The vertical force, components of tension, vertical force, components of the Web. And each of these different, um, anchor points along the walls through our net force of why acting on the system as a whole, you can be a positive T scientist. A one positive t signed dated, too. A negative attention component to the angles. Dated three. So think of T sign data three. And, of course, the spiders on the web as well. So she contributes to this forest. And of course, it is zero because of the static. So looking at the situation, we're giving mass and were given all of these angles, so we don't need to look at any other component. Um, so we look at our forces, we can rearrange this so that we have tea and itself for attention. So somebody removed the left side of the quick the mg to the right side of the equation. Honest. L's right up. Well, I'm going out for simplicity. so we can collect our sign term attention Term scientist A one plus sign dated too minus sign theater three. All this equals mg mg. Therefore we can solve her attention to be saying that one post signed dated Thio might a scientist a three, all that is under the MG again him was given All the angles were given. So are resulting tension acting and each of these components Reporting to two significant figures is 28 0 0.23 mutants very small compared to other problems of the sea, which is other 20 on the order of tens of Newton's or sometimes even hundreds of Newton's. But we expect very, very, very lightweight spider that she wouldn't be able to generate something with a huge amount of tension. So, yeah, there's your final answer.

Hi there. So for this problem we are told that the youngs model is for a spider silk that we are going to Dennett Tate with Why is equal to four jigga paschal. Which can also be written as four times 10 to the nine past school and can with town and stressed up We're going to call that this dress and that is equal to one point for giggle paschal that we can write us 1.4 times 10 to the nine paschal. And a single web is drawn has a cross sectional area That is given and that is equal to one times 10 To the -11 meters to the square. Yeah. And A web is made up of 50 radio strands. So the the picture effort is problem is the following mm this one right here and as you can see there is a bach that lands in the center of an horizontal web so that the web stretches that war. So for part a of this problem, we need to fine the ball you of the max if the maximum stress, It's absurd ID on each strand. We need to calculate the angle tita that the way makes with the horizontal. So we need to find that angle tita, that one that is in here represented it. So in that sense, um what we are going to use is first, we're gonna a student the trick the geometry of the web. We are going to have the test in here, we have this situation, We have that this is a distance theme. Of course this is 90°. This is the angle that we want to obtain. This is the distance, the radial distance. And this in here is the radial distance. Plus that change in the length. Now we know that we can obtain that by the equation of the maximum stress. That is equal to john's modelers. Times change in the length over the initial length. So and here we can obtain this ratio between the change in length and initial length because we can pass this to the other side. So we will have the stress over the youngs modelers. And we know these two values. We know that the stress is 1.4 times 10 to deny paschal. And the youngs model is for this um for the A spider silk is equal to four times 10 to the nine paschal. So from this we obtain that this ratio Is equal to 0.35. Now, since in this case L is represented by the radial distance, then we will have that delta L'd over art is equal to 0.35. Now, with this information we can use um trigonometry and to obtain the body of tita because as you can see from the figure, we can use the consign of teeth. Huh? Because that is the act isn't side, which is art over the hypothalamus, which is our plus delta our delta L. That change in the length. So we can substitute that in here. So we have art over R plus delta ELT. And we're substituting here this expression right here that we know is this. So we will have that this is equal to simplifying this one. Plus delta Over art. And we substitute about you that we obtained before the 0.35 in this. And we will have to obtain the angle we apply the concern of minus one to both sides of this equation. So we will have that. The angle tita is equal to take a sign of minus want of 1/1 plus 0.35. So from this we obtain an angle dita of 42.2°. So that's the solution for part a of this problem. Now for part B we are asked what does the mass of a back have to be in order to exert this maximum stress on the web. So to obtain that we apply Newton's second law in this case the situation we are going to apply um we're gonna see all of the forces that are acting on this block. Now we have the weight of the buck. That is the mass times acceleration due to gravity and the tension of the silk. This tension and because this is the angle in here when we some all of the forces in the white component, we will half that that is the tension sign of tita minus the weight because it is pointing downwards and this is equal to zero because we are in that condition. So solving for detention we will have that that is the mass times acceleration due to gravity over this sign of tita. No. To determine the mass of the bug. We know that the force over the area and the only force that is being applied to this is the tension. So we have attention over the area is equal to the mass times acceleration due to gravity. The the conceptual art at times the sine of tita. And we know that this is this is the maximum stress that corresponds to the maximum stress and this is 1.4 times 10 to the nine newtons per meter squared. So solving for the mass we will have that the mass is equal to the process optional area. Remember that we need to multiply this by 50 because they are 50 strands. So we have 50 Times the process optional area which is one times 10 To the -11 m to the square. And these times the sine of the angle that we obtain in part A which is 42.2°. And these times the maximum stress. And there is divided by the acceleration due to gravity. So from this we obtained a mass of 48g. So that is the maximum mass that this silk can. That this web can with them. Now for parts scene, What we need to obtain is if the wet 0.10 m in radios. So we are given the value for the radio, zero 10 m then we need to obtain how far does the wet extend. So we need to find that value. D. This value right here. They spelled you. So in order to do that um Mhm. We know that the downward exemption of the web is the lack of the right triangle opposite to tita to the angle. So the hypothesis we know again that is are over are the radial distance plus the difference in the length. So we can use the sign of data because the sign of Dita in this case is going to be the distance the overt. There's some of these two expressions and solving for D. We will have that that is equal to this product. And we know that we can take out our from this so we will have this sign of T turn. We know that this in here is 0.35 and this one in here is given is 0.1 m. So from here we're going to obtain that this is 0.10m times 1.35 times sine of 42.2°. So from this we obtain the distance d. is equal to 9.1 cm. So this is a solution for this problem. Things


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