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A traffic light is suspended by three cables. If angle 1 is 38degrees, the mass of the light is 16 kg , and the magnitude of T1is 93 N, what is the magnitude of T2 ...

Question

A traffic light is suspended by three cables. If angle 1 is 38degrees, the mass of the light is 16 kg , and the magnitude of T1is 93 N, what is the magnitude of T2 ?

A traffic light is suspended by three cables. If angle 1 is 38 degrees, the mass of the light is 16 kg , and the magnitude of T1 is 93 N, what is the magnitude of T2 ?



Answers

A traffic light is suspended by three cables. If angle 1 is 38 degrees, the mass of the light is 16 kg , and the magnitude of T1 is 93 N, what is the magnitude of T2 ?

All right, so let's get right into this problem. So how can we do? Jan Kal is 9.1 point 19. So this is read to the problem of Quicken. Write down some stuff we needed and continue. So the traffic light hangs from a pole, as shown in the figure below the uniform aluminum pole, a. D a. 7.2 meters long and has a massive 12 kilograms. The mass of the traffic light is 21.5 kilograms. Determined a attention in be horizontal massless cable CD and B the vertical and horizontal components of the force exerted by the pivot A on the aluminum pole. Okay, so a few things we should write down, right? Let's let's take on the length of the aluminum pole and some juice on the side. That's how far can I go, baby. So just on a side here, that's right. No, no. Think a little bit more to the left. Okay, so the first quantity we're given the length of the uniform. A limit impulse was right that down his l 7.2 meters and it has a mass. Let's give the pole a mass of little M. Who's the smaller Mass? 12 point? Okay, geez, kilograms massive. The traffic light. Let's go Big M comes bigger 21.5 kilograms and that's mostly what's given to us. We look at the picture, the angle between the massless cable and then move in and pull us 37 degrees, and that will be handy later. So let's write that down or list. Yes, I won't write his name, but it would be important to remember them as well. So h just a Samos height footage from the pivot to the top of the math lis cable is going to be 3.8 meters, and I think that's all that's all of our for reasons that are going to be important. So I guess the next few wanted up how you approach the southern one. That's one thing that's very important when you're talking about forces is noon. Second says that horse, equal to the sum of the forces of a system, is equal to the mass of that system, multiplied by the acceleration of that system. And so if you have some forces, maybe you want to find out force one or something, we have to forces acting on some mass, and you want to find out the acceleration. Then you could do something like this. Or if you want to find out one of the forces, you know the other one. And you could say that if the system wasn't moving, you could say that one forces the opposite of the other and things like that. So it's very important to use this kind of formalism to talk about forces. At the same time, you can talk about rotational force, which would just called tour. The Secret I Alfa is adjust the rotational and logs moment of inertia and angular acceleration. And so if both of these are equal to zero, which is what happens whenever a systems in equilibrium, then you can use these equations to help find certain quantities and talk. I didn't write, but it's also equal to it was in the vector form R across F. This means just distance crossed with a force. And so if you don't and if you don't know what a cross product is that it'll be good to look up, and I understand. But it's not. It's it's not essential to the problem. We can just talk about the magnitude and we'll be fine. So the magnitude of the torque is just going to be a magnet. Don't be our vector. The distance vector times magnitude of the force Specter times a man times a sine of the angle between them. So we have something like, you know, fulcrum, ena, seesaw that and a force. Let's say somebody sitting on this side kid is sitting here and it's going to be a force due to gravity based off that kid, right? And it's going to be acting at some distance, are so the angle between them Is this a 90 degree angle here? But if he was, I don't know, for some reason hanging off this I differently. That's it. We would rather you drive like this, and maybe he's going at an angle now. Then there would be that angle between memory, and so the force of the wind, but different anyway, this will be I'll be very useful to solve this problem so and I'll keep everything up here in the non Scott of the next page. I have this inspector or SPECTRE, and they're in the angle between. So with all these things in mind. How do we solve this problem? Essentially, when you're dealing with forces, you want to use it. We called FBT. There's a free body diagrams. So you have like a mass. If you have a mass on a table or something and you want to find out so the forces on it, then you drop as everybody. Begum will be taking this and drawing it as a dot and then saying OK, well, what forces are happening on this thing? Even upward force. Of course, the normal force do. Teo is why the box is not falling into the earth and then the earth is exerting some force on it, right in some gravity here forced you to gravity. And since the box is not moving, we could say that the son of the force is zero. So in minus mg which is what after you would be people to zero. And so we see that the normal force is just the positive of the the force of gravity. So this could work for a lot of different situations, and we're going to use that here so that the pole is in equilibrium so that the Network zero. But also it's since it's also, since it's in equilibrium, the net force is also equal to zero. Swinging in both of these equations is going to be very nice for us. So from the free body diagram, which will draw later, we can calculate the network about the lower end of the pole with counterclockwise talks as positive. What this means is you have to kind of choose a direction when you're talking about some of the forces is kind of easier than that you don't have toe. You do have to set up a coordinate system, but it's not. It is not as important as me talking about torque. We're talking about torque. You want to choose either counterclockwise or clockwise as your direction. You talked about the forces so that one of the horses, if we have to people and say now so one of the horses will be positive and b negative, depending on where you place the people and all. And so let's go ahead and do that. Let's just draw everybody diagram on the next page here of our system and you know it. So let's look at the we'll get the horses drop. Hold here. This is at a length L Right then we know we have this angle of the top. Someone's gonna draw a lot of the situation here. I'm going to call this data for now. But that way, do you remember this is 37 degrees, right? So let's see the pivot at the bottom is going to act. You know, it's going to push No, in the same direction as the pole, right, P, But we want the vertical and horizontal components. We only watch this piece and this piece, Right, Pete? Wine at p X. Yeah. Okay. Ah, And geometry tells us that if you have the line like this, right and another lines and learn to parallel lines than the angles, the opposite English should be the same. They're so they're they're both going to be safe. You gotta try a little bit closer. So it's not matching with force. Good. Don't look so good, baby. When you get the idea, that's that's your generation. Make it look a little better. He's back. You know, just this here. Now it looks worse, but as you can see the data so we still have this fourth here in the diagonal direction F p. And we're going to find when I look for the excellent. Why later? So that's that'll be something will do for the second part of the problem. Write the first part. We're still talking about this, some of the torque. So the pushing one asked, right is what's going to cause a tour? Uh, here. So, like I said, when you're dealing with a fulcrum, it depends on where you place this. And I didn't. When you have a Caesar on a fulcrum, it depends on where the people are sitting more where the forces are on on the line. And so I'll say, towards the middle of this, we'll know at the exact millwright the center of mass of the pole itself is going to feel a force due to gravity. So we're going to have mg here, and the traffic light is going to feel something similar, right? So the traffic lights the very, very top here, but now has a different mass, right? So these are just supporting. This is just F G for the big masses, Just effigy for the little mass rates. Traffic light will feel big mg and the the only hole itself will feel little mg. I can't. And then there's one on the force I'm forgetting right is that I have this upward force to do the in between the center of the earth, the pole and the traffic light. You have the cable, right? You have thiss massless cable sitting here. So basically for that for that, I'm gonna have this force due to the cable right here, because that's pulling it horizontally, right? So Beijing, right, So we can do the sum of the talks now basically, So some of the talks is going to be zero mean. Like we said, this is the system, is you. The system is in equilibrium. And then let's talk about which which forces are 1,000 torques? We'll definitely the cable forces going to if if the pool was free to move around, some open the cable force of definitely going to cause a tour. And since since it's on the top side and we're taking kind of clockwise to be putting counterclockwise to be our positive direction, that would mean that clockwise is going to be negative then That means that since the cables aero is above the line. It's going to be positive here, so forced from the cable, which is going to be the tension. Basically, that we're searching for is going to be multiplied by the distance that it that it's act. And so if we remember from the problem this distance, write it in green. This is from the cable to the bottom of the hole just going to be a church. So that means the torque due to the cable, it's just going to be FC times H and then the other forces are going to be the other Forces are going to be negative because they're on the underside of the of the pole, so we won't have minus since we sort of basically we're dealing with these three, these three forces causing torques the fourth to the force of the cable, the force due to gravity of the traffic light on the horse to the gravity of the of the old self. So those three forces our cousin tourists in the hole. So second must do the torque due to the pole itself. And so we have to think. Okay, well, what's the distance that this What's the distance that this force act and I forgot to do it earlier. But let's let's figure a general way to write Thie X during but the X distance here. So l is the normal distance. That's a full, you know, 7.2 meters. But more generally if if, well, if we allowed the data to change even though we know it's 37 and we could write it more generally as ill coastline data, right. So if it was L close and data just happen horizontally, then this is the center of mass is going to be at the center, which is halfway in between, right, So it's just going to be This is that this force of gravity acts on is going to be in the middle. So just halfway in between the leaders, the force of gravity mg times L over to our distance force, times distance, and then this is just going to be a number, right? With his co sign of 37 and then something similar will happen for the traffic light. But as you could guess, since the distance is full here, we're going to have to include the full El no sign 37. So you could focus on 37. Of course, on a day that it would be the same thing since we define favor here, you know? And then we would just meet to do algebra afterwards, right? Pretty easy problem Afterwards. Let me just start from the top Gear's on the toxic zero legal too. F c h minus M g l over to coast data My big m G l coast. All right, so now you seem to do algebra. We want to find this right. FC FC is what we're looking for this, right? So let's go and just move on. These guys can't really see that one. Let's move these guys to the left side. Right, So they just become positive. We have mg l over too post time data minus big mg l coast and data. And it's pretty simple from here. I wouldn't want to divide H on both sides is just about the whole thing by H. And so we have this horse from the cable. Otherwise, not us. Attention is going to be well and we can simplify this top. Actually, we look at a cold of the heart, right? FC is going to be with the one over. H, I'm just taking this out in the front so I don't have to write. Write it. But looking at this top expression, there's a common factor, right? Definitely coz I'm is there else they're on both sides in G. Stay right. So we can actually pull this out towards G L Corps sign painter and then multiply that itself. Bye within the factored mass, right? So what's left? I'm pulling a G Yoko's and favor here. I'll just have little em over too, right? It's a little m over two plus Big M. Oh, this should be a plus two. I'm sorry if these became positive as they moved to the left side. Right. Then we have the next line. Here s so, uh, yeah, back to this. So I factored rgl coast on beta, and I just have some of the masses here multiplied by some other stuff. And and it's not actually over. No. Trying to erase this just doesn't need to be so large just right. You can just write people here, remember? H I'm so so this will be our attention basically. So it's gonna be one over too. Too far down. This coach of the other side. H was what through Puneet. So you're that one over three meters. I was just keeping units out. But remember that it is very important to keep up with your units. So one over 3.8 meters. I was 9.1. I'm 7.2. I was co sign of 37. And that's it, Sung Eun. All of that multiplied by six. Because the mass of the aluminum post 12 and one divided by two, plus 21.5. And this should give you the correct number. So I'll let you guys work that out. And you can No, that's fine, because this is just plugging in numbers, right? Just molded by these three. That should be its own recognition. To put his son Prince. He's sorry, but you have these four quantities inside 27.5 coast under 77.29 21. Multiply all those and divide by 3.8. You should get some number of multiple, and then you can have the unit correctly. His mutants, right. Ah. Then for the last part, we want to talk about the force is redrawing our worst agony of it. You want to talk about this force, right? FT. Why, if X is not, does not rewards until, actually that's better F p X and there's later in between here that we cash in here data. So there's this force from the pivot, right? I was what we wanted to do, so the same without using this. It's a very useful tool to save. My used to. Some of the talks is zero. In this situation, we can say the son of the forces of zero, so we don't have to think about rotational stuff. And the Net force on the pole is also zero, since it's in equilibrium, which is what allows him to do this. So essentially just one right in second law. Which is this people zero. If it's not moving in equals that made this moving so it could be zero here that still a fine state. We're going to write me and second line both excellent. Why directions to solve for the forces at each of those points? So let's see what we can find. And of course, this is with all the other all yet Maybe I should You should. You should look to this. We should look to this diagram right when we're talking about these forces. Because we have this force is the force of the cable is acting in the negative X direction. Right. And you also have these forces acting on the negative. What direction? The force due to gravity. Basically, it'll be very important to remember those, even though I didn't draw them in the last, uh, everybody diagram. So keeping that in mind, uh, you have some of the forces in the ex direction. Of course, this is equal to zero, right? But of course, this is positive. This forces, this one is pointing in the right direction, which we're going. Let's just say that positive. We're defining it of the positive direction, so f p X is going to be positive. And like I said before the cable news draught in here, I guess f c. It's going to be negative since he's coming minus FC. So essentially this Just tell us that the force ah, the force in the ex direction due to the pivot is just people to the force of the cable. So whatever number you found earlier from plugging in these stuff on pace. For whatever number this is, it's just going to go in here. And that's going to be the FC and Newton's right, some number and Newton's. And so yeah, essentially. But that people forces just the same as the cable force, since since it's an equilibrium thing, we could do a similar thing for the Y direction, right? So some of the forces in the UAE. So he asked, What's positive? What's negative? And as you can see here, this is positive. So f p. Y going to start us off. Then we had little MG and big mg here, right? So that means these are both going to be negative, both reporting downwards. I won't have little mg, big G and, of course, the sum of the forces of zero. Like we said before, because the system's in new glittering. Both of these are true. So then if we just movie's over the two masses like we did before on the force, and due to the pivot in the white direction is just the sum of the masses, and so that should conclude the problem. Uh, I don't think there's anything else I want to clarify. Yeah, thanks for listening.

Here for the solution first we had the expression for the force acting in vertical direction. To obtain the tension that is to to sign 15 degree equal to MG. Or by rearranging this, we get equal to MG by sign 15 degree and we consider it as the equation. One. Here I am is the mask, he is the attention and he is the acceleration due to gravity. Now we write the expression for the Seer modelers. It is as equal to stress divided by a strength stress divided by strength, or we can regulate the by a, divided by their little by a lot. And we consider it as a question too. Here, delta alibi L. Not is the fractional change in the land? A. Is the area. Now we re arrange equation to to find the expression for the frictional change in land. That is Delta L. Biological duty by asked by I square and we continue it as aggression. For now we substitute 45 kg four M 9.8 m per second. Described for G. In the equation one to find attention. So here by substituting the values, we get equal to 45 kg, multiplied by 9.8 m per second square, divided by to sign 15 degree. And from here we get equal to 85 to N. Now we substitute 852 N. 40 20 multiply by 10 to the power 10 and permit meters go forward as 50.50 multiplied by 10 to departments to M police car. For a an equation for to find the share modelers. So here by substituting the value, we get delta held by L not equal to 852 and divided by pi 0.50 m hole square 20. Multiply by 10 to the power 10 and per m square. And from here we get it 5.4 multiplied by 10 to the power minus five. Therefore, the fractional increase in the land due to weight of the land is 5.4. Multiplied by 10 to the power minus five. So this is a complete solution. Please go through this. Thank you.

All right. This question wants us to consider traffic Lee suspended by two cables and find the tension in each case cause they're both hanging from the saying same angle of 20 degrees. So what we need to dio is draw free body diagram to find the forces. So this is air traffic, Lee, and we have its weight pulling down on it and it says the traffic light weighs £12. Then we have one tension force pulling it that way and another attention force pulling it that way. And it said that both of these have an angle of 20 degrees. So if the traffic light isn't moving than F net in the extraction as to be equal to zero and F net in the wind direction has to equal zero. So if we look at our force diagram again, we see that in the wind direction we have the way plus the tension in the wider action from each pole and in the ex direction. We just have the tension in the ex direction from each so since each pole is at the same angle, we know that all these forces air the same, which means the why component of one is equal to the white component of two and same thing for the x component. So why we have the weight plus two times the tension in the wider action than X, we have two times the tension in the extraction. We're just gonna call this tension for us. T so will solve for y first. So we know that the F net in the wider action equals 12. And that's a minus 12 cause weight's pulling down and then plus two times our attention force. So we know that the net forest has to be equal to zero so we can solve. So the tension in the wider action is £6. So now we can update our diagram. So this is 12 this six and this is also six. So now we need to solve for the tensions in the ex direction. So we do that by using our newly found tension in the why along with our angle of 20 degrees. So we see that tangent of 20 is equal to opposite over adjacent. So our attention in the ex direction is equal to six, divided by the tangent of 20 and I plug that into my calculator here and I got 16.485 pounds. So now here is our complete free body diagram. This is 12 down, then thes air, both six and the ex components or both. 16.485 So now we can solve for the magnitude of our attention force. So magnitude is just the square root of the components squared. And when I did this, I got £17.54. So again, just to run through how he did this. So first we set up a free body diagram. Then I realized that the net force has to be equals zero. So he summoned the forces, realized that since they're hung at the same angle, the tensions and each wires the same, then we use that to solve for the tension in the wider action for each cable. Then, since we now have a side of a triangle and the ankle we were able to solve for the X component and then when we had both components, we're able to solve for the magnitude of the overall tension

Have a diagram of the system. We know that, um the system has static equilibrium and as well as translational equilibrium in the X and Y directions. So we can say that the sum of torques we know equals zero. And this would equal forced tension. H minus m g times. I'll over too Times co sign of Fada minus m g l Co sign If ada and then we know that the force tension then must equal em over two plus m. This would be multiplied by G. L Co sign of Fada, divided by H and so we can solve force. Tension would be equaling 6.0 kilograms plus 21.5 kilograms multiplied by 9.80 meters per second squared multiplied by 7.20 meters, multiplied by co sign of 37 degrees, all divided by age of 3.80 meters. And we find that the force tension is equal in approximately 400 and eight Newton's. So this would be your answer for the force tension. We're going to solve the sum of forces in the ex direction. This is equaling zero, and this would equal force Tibbett at the act of the force of the pivot in the Ex direction minus force tension. And so the forced force of the pivot The X component of the force of the pivot is equaling force tension and this would be equal to again 408 Newtons. So this would be your answer for the X component of the force of the pivot and then using the sum of forces in the wind direction, this would be the equal to the force of the pivot. Some why minus m g minus mg equals zero. And we find that the force of the pivot in the wind direction will be equaling M plus and times G. This is equaling 33.55 kilograms, rather multiplied by 9.80 meters per second squared. This is equaling approximately 328 Newtons. This would be your answer for the why component of the force of the pivot. That is the end of the solution. Thank you for


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