solve-the-given-problems-for-the-spring-mechanism-in-exercise-31-of-section-104-find-d-b-d-a-note-th

Question

Solve the given problems.
For the spring mechanism in Exercise 31 of Section $10.4,$ find $d b / d A .$ (Note that $a$ and angle $B$ are constants.)

Ren Jie Tuieng · Accepted Answer

in his question, given the seat on the way object that has just issue a que IHS oscillating you with a spring My success story spring in his oscillating on a frictionless surface. Now the frequency of this constellation is given to be tree hurts and the amplitude off this oscillation is shops. Yeah, reach indicated a figure No Saudi cute or the object right now is at its, uh, equilibrium point right x zero and oscillates left and right to de maximum point, which is the amplitude off Joe CIA on both sides for spot. We are looking at finding what it sees spring constant. So the spring constant we can equate it. We&#x27;ve or station or the angler frequency off our system. And graphic unsee is actually close to squaring off the spring constant throughout a variety mess rearranging this equation. So now we have what we need. We have yet the frequency is treated right. So we got a converted into angry frequency, which is a question to pie f. Then we will have to pie square type I and on a mess off the subject, it&#x27;s also given to be 170 grams, which is 0.17 kilograms Print in S I units. So substituting the very scene two pi f f is trees hurts M e 0.17 that we have 60 new tents per meter. From the next part, we don&#x27;t find any creature that described this position off were given a starts off at the amplitude where x this shelf. So assuming the X positive X is in the right direction, so be we&#x27;ll meet X any given point in time, the question would be starting grief. Shoaf celery juice. What you played by Ida called ST or ST? Yes, that is how the simple honey harmonic motion is described us, which is Sinus oId Oh, so you can either be signed. All co sign multiplied by oh, sign and co sign off Omega T Oh, my eyes d in graphic. Now we have to decide where a sign of course. Sinus more suitable. And the way to do that is by looking at I was starting point because I was starting point. You keep step X off. Zero must be close to shops here and we know that. Sign off when 20 0 these bracket will be zero so sign off. Zero. We know it&#x27;s zero. So you cannot be signed money out there for course saying assange zero is one and just nice were more bellicose I into this job we get Jawf. So our final answer should be that XTC ghost zero times Go sign Omega T. Do you know what my guys to pie times if tree huts times t so it is so final question.

Henry York · Answer

problem, 76 asks for the X, Y and Z components of the £40 resultant force on the plate at a In part of the problem had to start off with F of X. It will be equal to the result in force of negative. £40 since is in the negative X direction times a sign of 30 degrees to get it into the xz plane and then times the co sign of 35 degrees to get it along the X axis. This is equal to £16.4 now for the force in the Y direction. It&#x27;s also and the negative y direction we take the result in force and multiply it simply times a co sign of 30 degrees. To get that force along the Y axis. The answer is negative £34.6 and lastly, we&#x27;re looking for the force in the Z direction. Take the result in force, which is going in the positive Z direction and multiply it by the sign of 30 degrees to get it along the XY plane. Then we&#x27;ll multiply it by the sign of the 35 degree angle to get it along the Z axis. This gives us an answer of £11.5 and part B. This problem asks for X, y and z components of the angle that the result in force acts. We can use the equation that the co sign of the angles in the X direction Is he gold too, the force in that direction over the result in force. And since we just found the X component of the results and force, we can plug that in along with the resultant force given in the problem and solve for data to get data in the X direction is 114 degrees. We can use the same equation to find the winds e components of data as well. We get that data in the Y direction comes out to be 150 degrees. We get the data in. The Z direction comes out to be 73.3 degrees

Dading Chen · Answer

Okay, So you should take it with a graph so you can tell that Empty three people, 0.8 year. Okay, so for the first part is going to appoint a zero a meter and four questions he was asking after the bank of the velocity. So, Omega, can we go to power Batik, which to our time here. And, um if you take a look at ah, grab as you can tell the time Here it will be 4.1 2nd Okay, because it takes about four seconds to play toe to complete one cycle. So therefore, the angler velocity should 1/2 pie, which is about 1.6 radian per second and four questions. See, while we know the Anguilla velocity zero squared OK, over him. Case the spring constant. An embassy at Mass. So therefore, the spring constant case, um, times, it&#x27;ll make us work, and we know the masses, given Israel point Israel killed. So they&#x27;re trying to determine the spring constant so they spring constant should be equal to zero point a zero kilograms times 1.6 radium for second and a square. And this will give us the spring constant is about two point. Oh, new temple meter and question being was asking us was body. So the velocity Ah, um What? What? What? Italians. People. The one second. Well, what time is it for the ones happen that these placement become the amplitude. So we know musical Omega Times Square to a score of minus pack square. So at this point, ah, acts become a which is amplitude So therefore we&#x27;ll have me is equal to Omega square Oh, a square minus a square, a square minus is what you call zero. So there will. The speed at one point all of seconds should even zero meters per second as this one&#x27;s you guys remember. So the speed at a contemplative points is always zero okay, And the acceleration of the point is the greatest, Which is the next question here. So at that point, since the last question was asking us the man until the acceleration at time, one second, which is when the when emotions at the end of two point. So therefore the acceleration is the greatest. So a mask is you go to any Omega Square acts which is equal to negative 1.6 radio for a second, um, to a barber to entice the displacement at a point, which is the amplitude, which is your point. Throw point. Uh, I&#x27;m sorry. 0.0 a meter and this will give us the maximum acceleration. Uh, is about 0.2 meter for second square and these are the answers. I&#x27;m sorry. Eso nurse will be negative here. Sorry about that. So it&#x27;s negative. Zero point two meter for a second square. And remember, the question was asking us the amplitude. So mpasu here should be just positive. 0.2 meters per second square. Okay, so these are the answers. What is?

Brian Ketelobeter · Answer

What? Ah, good day, ladies and gentlemen. Uh, again, we&#x27;re looking at problem number 33 here. And in this problem again, we&#x27;re not really emphasizing the solving method so much, um, or more, uh, setting up a problem and then looking at what we get, um, we get it. So a lot of ah, this is actually using things from 32. I didn&#x27;t technically need to do I&#x27;m here, so I&#x27;m just sort of grabbing, um, information from 32. So if any of this stuff you&#x27;re not aware of, it&#x27;s because you haven&#x27;t looked at problem number 32 or you don&#x27;t know how to solve, um, previous problem. So I&#x27;m sort of just summarizing pretty much what you didn&#x27;t regard. Um, so in the first case, in the first part of it, when you set up the ordinary differential equation, um, in the way it is away, how can I put this? So the 10 here, which is as ah, mass of the object. Um, this is tthe e m. Why? Double prime term? In other words, this is the force term. Um, this here. Ah. See, remember, correctly, um is is this the k or you. I&#x27;m getting now. Which terminus is, um So once the K and one&#x27;s the beach? No, I This script, I can&#x27;t remember what it is. Um, was variable that I should have it here. Okay, so, uh, this is really the I think that. See you. Yeah. So the 60 here. Sorry. This is the B y prime. And it&#x27;s instant, um, I guess there&#x27;s accent. It&#x27;s, uh it refers to the the damping term, um, which which comes from friction and then the stiffness term. McKay, uh, that comes from this spring. So this is sort of the spring constant. So you just I mean, the center of the problem is really to just input those terms into the equation. And the previous one, um, has no, um is un down. So this is a case for you have damping. The other one is undammed, and you&#x27;re sort of twisted pair between the two of them. What happens, eh? So the first case. So now if we look at what the difference equation is, um yeah. Ah, Uh, uh. Okay. So what I want to do is just mention here, um, that when you go through the boy solving and everything you&#x27;re going to see. Um, this is your equation here. Um, and then we&#x27;ll me, we want to talk about setting it to zero. Okay, um, we just sat y equals zero solve. And what you get is this tangent 42 b negative three halves and then that tells you that t is equal. Thio this. But you&#x27;re, um the, uh This is not four divided by pi, by the way. It&#x27;s for pie. It was supposed to be four pi. This number is negative. Um, which doesn&#x27;t mean anything, but you can add four pi to it. You can ask basically, by adding numbers a pie multiples of pie to this to this initial value of tea I was looking for the first time is that, um, this would be positive, so I can&#x27;t remember what the number wasit was. Something like negative 14 or something. Um, which was, like, negative 14 seconds or something. And so, by adding four pi to it, I get back. Um, the 1.6 tree, which is the first time it adds to its positive. So this tells me basically that in 1.63 seconds, it gets back to zero. Okay. And, um so that&#x27;s the first time it it passes. Zero. Um, and then the frequency here, as defined lips in the previous problem, the frequency comes from the, um hola. Say this the B term. So are the beta term. So remember, when you solve this ordinary difference equation, you&#x27;re going to get a e l tha t sometimes, you know, co sign bait. Ah, t a plus sign beta t. So, um, the frequency, as defined in the previous problem, is dis beta term. And in that case, it is actually beta divided by two pi, which, in this case, just to over pi. Um and it makes sense in the other problem, because it&#x27;s just a It&#x27;s just a non oscillation population. Eri um how would you say that? Like a looping spring. So spring all the springs really doing is going, you know, something like lumps, uh, Springs. Just going back and forth. Boy is terrible. Uh, yeah. Okay. It&#x27;s not gonna do what I wanted it to do. Essentially, what I want to do is this. So spring starting year going over here, you know, sort of going back and forth. So the oscillation here or the frequency basically is looking at How long is it, Uh, how long is it going going to take to get back to here? In that case, it makes perfect sense. Is has pie. Um, in this case, um, in the damp term, because what&#x27;s happening is the spring will start out sort of going wide, and then it will sort of damp and so sort of decrease in some sort of go eggs. This basically sort of like that s o the the, uh, what you call it. It&#x27;s not measuring how long it&#x27;s taking to get back at this point, but how long it&#x27;s taking to go from its furthest 0.2 between its two furthest points of oscillation. In this case, it&#x27;s, um, you&#x27;ll notice that in the undead apt frequency or in the undammed, um, case that has a higher frequency habits. Because this guy&#x27;s losing energy with time. Um, then in the damp cave. So So this case is losing energy, going to the damping. So, um, and that&#x27;s that&#x27;s that&#x27;s really it is that the damping term basically causes the system to lose energy because it&#x27;s official. Um and that causes it to lose energy, which causes the motion to just sort of fall Go to zero. Okay, so that&#x27;s really it for here. Um, you know, I I had to read, um, about this and section, um, 4.1 and then a little bit more. 4.3. So, um, you know, I it&#x27;s probably useful to actually read about how what these varying terms mean and so on and so forth. It&#x27;s probably helpful pouring a little bit about it, but, I mean, it&#x27;s basically what it tell it gives you is a little idea of what to expect your solutions toe look like, even if you know, you don&#x27;t necessarily care. I got but, uh okay, thank you very much.

Eric Mockensturm · Answer

So in this problem way have a a spring that is attached to two points. And for this problem, we&#x27;re looking at the spring from A to B. So we have a post here, and that&#x27;s gonna be our y axis. And we have our X axis, um, and then rz in this direction. And so then we have the spring attached here, and then it&#x27;s attached to some point in the X Z play and the angle if we project that down on their this angle here is 35 degrees and we are told that a C this&#x27;ll angle here is 30 degrees. And then the tension I&#x27;m assuming is going to be acting is in this direction. We&#x27;re just from from being a is, uh let&#x27;s see here that they gave us this £50. So the magnitude is £50. So to get the components, let&#x27;s see here to get the y component, we just need to take fifth magnitude times cosine of 30 degrees on. That gives us £43.3 to get the X component. We need to take the this vector and then find this this factor in here and then projected onto here, so that means we projected with sign of 30 degrees and then co sign of 35 degrees. And that gives us minus £20.5. And now we need to find to find this one way project down here with sign of 30 degrees and then project over here with signs of 35 degrees, and that gives us minus £14.3. Now, to find the the angles between this factor, I guess we actually haven&#x27;t, you know, here and each of these axes we take the component in that direction developed, provided by the magnitude on take the universe co sign, and that gives us between the X and the X axis. And this factor is 114 degrees. So if you made a plane that passed through this factor and then it was tilted so class through the X axis and it was tilted so paid, um, passed through this factor. So if you had this plane and you&#x27;re just looking at it like this on this is the X axis. And now this is the vector because the plane is tilted Well, so this is the X axis on, then, because the plane is tilted, the vector is in this plane. So that&#x27;s this vector. And then this would be 114 degrees. So if you basically make a plane that goes through this factor and this axes on, then look at the angle between two factors in that plane. This is what you get. And so we can see just immediately that that should be 30 degrees with the Y axis, since that was actually a given. So if you looked at the plane that went through the vector here on the Y axis on Ben, if we look at the Z component, the Z angle, we take the universe co sign of the Z component divided by the magnitude, and that gives us 101 107 degrees. So again, you can kind of see how that would sweeps out here a little over 90 degrees

Problem

For the cantilever column in Exercise 32 of Secti…

Problem 48 Easy Difficulty

Solve the given problems.
For the spring mechanism in Exercise 31 of Section $10.4,$ find $d b / d A .$ (Note that $a$ and angle $B$ are constants.)

Answer

Related Courses

Basic Technical Mathematics with Calculus

Related Topics

Discussion

Top Calculus 1 / AB Educators

Heather Z.

Samuel H.

Michael J.

Joseph L.

Calculus 1 / AB Courses

Precalculus Review - Intro

Functions on the Real Line - Overview

Recommended Videos

Watch More Solved Questions in Chapter 27

Video Transcript

Allyn J. Washington, Richard S. Evans

Basic Technical Mathematics with Calculus

Heather Z.

Samuel H.

Michael J.

Joseph L.

Precalculus Review - Intro

Functions on the Real Line - Overview

What do you need help with?

Textbooks

Ask a question

Courses

AI Tutor

Basic Technical Mathematics with Calculus

Heather Z.

Samuel H.

Michael J.

Joseph L.

Precalculus Review - Intro

Functions on the Real Line - Overview

Allyn J. Washington, Richard S. EvansBasic Technical Mathematics with Calculus

Heather Z.

Samuel H.

Michael J.

Joseph L.

Allyn J. Washington, Richard S. Evans

Basic Technical Mathematics with Calculus