@@goki6548yes but you also must determine the length and angle the bungee cord makes with the horizontal to figure out what force balances the 80N, that’s the bit I’m having a bit of trouble with trying to do the problem myself. Update: okay from what I can see it essentially comes down to solving a horrific quadratic equation to determine the new length L+x and there you have the x value but I can’t help but feel like I’m making a mountain from a molehill lol
basically the physics i study. they show you one simple example worked out on a half a4. then finally when you understood it you try to make a similar problem ending up in 4 pages of algebra and calclus and then checking your answer to find later out that you made a mistake midway of the calculation, you go again do everything from the start find another answer which is pretty close to the answer in the back of the book and then questioning if you made a mistake or the author rounded up the answer and finally after everything you get good at it, to realize that the test next week doesnt even include the topic of the problem you solved.
I relate soooooo hard to this, can spend hoooours on questions and often get "the essence correct", like the answer might have correct form but theres always some fucking small calculation mistake
These types of videos are so helpful to students. Showcases the initial thought process, mistakes when simplifying the problem too much and how to re evaluate. 10/10
Rarely do I comment, but this is fucking awesome. keep up the good work. I see a lot of comments being negative about your lengthy approach but that approach is what i like. Talking and thinking through problems really makes one understand what is really going on.
Go on bro! Hate the negative comments! Trying to understand and rejecting the bad approaches to solve one problem is what we miss these days on tutorials! All the best ❤
As someone who is a Jee student i completely disagree with my peers in the comment section calling it a waste of time to solve a question in 2 hours. This is ENGINEERING not jee advanced its not the same at all. His point of taking the video to this length was intentional . So stop acting immature and appreciate him for being so passionate about what he is doing instead of flexing about knowing how to solve it by shortcut method.
I think this is the most complicated explanation you could give to this problem. Many references you make could be taken for granted for a student who is coming to deal with a problem like this.
Very hectic and writing is all twisted around and does not follow any particular sequence, even if you are trying to understand alone writing in sequence helps a lot. But a least you solved it i do think taking this much time is not normal but if this is the first time you are doing this kind of problem then its ok.
I was surprised, too, but after getting into the video, it's like the above comment said. The whole video is essentially walking through thoughts, assumptions, why they're faulty, and how to find truth. I think this is valuable as it shows the reality of the kind of considerations and ways in which people view things. I'd have solved the problem in the most popular way, but after seeing the connection between the spring constant and the deformation equation, I quickly realized its value. Further on, he eventually highlighted something within the book the question came from, which was essentially that if you followed everything in the textbook, you probably still wouldn't have been able to solve the problem perfectly because that particular problem requires a bit of creativity and a deeper understanding/ intuition behind the topic.
@@TheKingFlappyjack I am curious, did you think I was not in university? Also these maths I did back in the first year of college so not even university level. If I was not in uni I would not even understand this shit lol, why would I watch shit I don't even understand. Weird guy you are.
Grade 12 here. None of my classes requires me to do questions like this - not even my calculus class. I already applied for this career. Lord, help me.
Week long problems are the best. When you solve it after having declared defeat and already moved onto the next chapter only to have your subconscious tap you on the shoulder to tell you, "Hey, buddy... it's this."
Thanks! I had to look that one up! It is just confusing when 1 divided by cos is NEVER said to be (cosx)^-1 My Eshbach Handbook of Engineering Fundamentals says, "the symbol sin^-1 x means the angle whose sine is x, and is read "inverse sine of x" "anti-sine of x" or "arc sine x"
it doesn't have to be, but you will end up spending the same amount of time learning about the topics and field regardless. Spending more time on a single problem like this ensures that you develop a deep understanding.
when i first came across this problem, i orignally sovled for the amount of energy being applied to keep the bungie cord stretched like that, turns out that was wrong and had to start over lol
I arrived to the following fourth degree equation for the length of the cord: L^4 - 2 L0 L^3 + (L0^2 - b^2 - P^2/(4k^2)) L^2 + 2 L0 b^2 L - L0^2 b^2 = 0 The acceptable solution is L ≈ 1.066m U = 1/2 k (L - L0)^2 ≈ 6.55J Is there an easier way to do this?
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The official answer for this problem appears to be a little different from the true equilibrium. At equilibrium a virtual displacement of the force P at the point C should equal to the increase in the potential energy stored in the springs as a result of the work-energy theorem. For a string under tension (T) we have: T = k*d where T(Newtons) is the tension, k(Newton/meter) is the spring constant, and d(meters) is the displacement of the spring from the "unstretched" position. We know that the energy in a spring is given by: E=0.5*k*d^2=(T^2)/(2*k). Because the system is symmetric we can just analyze one side (i.e. a half-spring) and double the results to get the energy in the whole-spring. The problem states that k=140 N/m in the whole spring. The spring constant for a half-spring is easily found to be: 2*k. In my full solution which I cannot replicate here it was found that the stretch or displacement in each half-spring is: d=0.13268(meters) at equilibrium and at the vertex where the two springs meet where we apply the force P=80N (i.e. the point C) the angle is theta=43.50(degrees). Given this information lets proceed show that the system is indeed in equilibrium. Assume k=140N/m, b(peg separation)=380mm, L_0=760mm, P=80N is applied as shown in the problem diagram. First calculate the tension in each half-spring (remember that we double the spring constant k of the whole-spring to 2*k). We have T=(2*k)*d=2*140*0.13268 = 37.15 Newtons of tension in the half-string. Now that we have the tension in the half-spring lets show that this equates to an equilibrium position with the external force on the half-spring. The P=80N force is divided between the two half-springs. Therefore we project a force of P/2=40N onto the direction of the half-spring. The angle is theta/2=21.75(degrees) where theta was assumed to be the correct equilibrium angle as stated above. Finally, the projection of the force onto direction of the half-spring is: (P/2)*cos(theta/2) = 40*cos(21.75 degrees) = 37.15N. Notice this equals to the internal spring tension found in the prior paragraph. This proves the system is in equilibrium since the internal spring tension equals the applied external force (i.e. the projection of the force onto the direction of the on the half-spring). Now lets find the energy stored in the system at equilibrium. The energy stored in a half-spring is: 0.5*(2*k)*(d=0.132685)^2 = 2.46456 Joule. The energy stored in the whole spring is simply double that of the half-string from symmetry. Therefore the energy stored in the whole string is: 4.92912 Joules. This is less than the official answer. The displacement of the vertex (i.e. point C where the load is applied) from the unstretched position to the equilibrium position in the direction of the applied force P=80N is 0.143(meters) which is slightly less than the official answer. The total length of the full-spring in equilibrium with applied force (P=80) can easily be found by solving: Total Energy = 0.5*k*(L - L_0)^2 for the variable L. We write 4.92912 = 0.5*(140N/m)*(L - 0.760m)^2 which solving for L gives us L=1.02536 meters as the length of the full-spring when stretched by the applied force of 80N. Alternatively we can find L as follows: L is the length of the whole-spring plus the displacement(d) in both of the half-springs. Therefore L = L_0 + 2*d = 0.76 + 2*0.13268 = 1.02536 meters which agrees with the result above. The full solution involves an application of virtual work and use of the work-energy theorem. To solve for the displacement (d) and the angle between the two springs (theta) it turns out that you can use Excel Solver to solve this implicit equation which with a little effort is easy to derive: 2*d = P/(2*k) * cos(theta/2) WHERE: cos(theta/2) = SQRT(1 - (b/(2*l))^2) AND l=0.5*L_0 + d The end result is that you will find d(the displacement for a single half-spring from the unstretched position) and theta(the angle between the two half-springs.
Thanks for the reply! Double check your multiplication of cos(angle). I think it should be divided by cos(angle) in you excel solver equation and in equilibrium. A simple check is that if 40N is being pulled straight up and needs to be balanced by the cord pulling at some angle. The tension in the cord has to be greater than 40N because only the vertical component of the tension in the cord can withstand that 40N. Analyzing half the cord.
@@holdmybipolar I divided by cos(angle) in my Excel spreadsheet and everything is now fixed. The updated results are: Energy=6.546 Joules, delta_C=168.79mm, angle beween half-cords=41.77 degrees, total length of the whole cord=1.0658 meters. I definitely learned alot by working through this problem. It is always a challenge trying to find your own errors? Thanks for the help.
Mechanics of Materials _ james m gere _ 5th edition _ This is my text book from college but I should buy the new one that colleges are using these days. Most all problems haven't changed tho. Just the numbers.
Hey there, I have a few questions dor you to answer - Where did you find your questions(which page) - Which college major did you pursue - How did you come up with this solution (in a summary)? Anyways, I'm going to pursue civil engineering but little did I know most people have spent hours on this single problem... Edit: I hope that this solution or maybe the summary helps me a lot, or maybe do you have another question to solve?
page 176 Mechanics of Materials - author james m gere. I completed a bs at montana state university in civil engineering/enviromental engineering. I will post tomorrow the short or quick answer to this problem. I like to spend a lot of time on problems because I want to make the problem simpler. That means that I might solve 2 or more problems before getting to the actual problem at hand. That can take more time. Then I like to use numbers that might be easier to recognize. I use the analogy that we for sure know miles per hour in a car and distances (if in america). Then I like to imput the actual numbers. This allows me to have a check on reality that I didn't make anymistakes. Then I like to go back and see what I did to those numbers to come up with generic equations. That is where I then look at those equations piece by piece to see if I understand what the equations mean. Some people can just look at an equation and do the previous steps that I have laid out in their head. But most cannot. So it becomes important to develope that intuition when anaylyzing equations. And sometimes for most physics problems. One equation can be derived with letters of the input values so that you then can change any parameter of the problem. With in reason. This problem requires an itteritive solution or a Solver based solution. So one equation cannot be used to set up the inputs and generate and answer.
never graduated high school lots of meaning behind each letter I want to understand but honestly it would probably a waste of time, especially when I can barely remember or even apply complex information, sometimes even basic information like basic math 😭
I had an amazing mechanics teacher that tortured us with insane homework problems (that if we failed meant we failed the course even if we got 100% on everything else). How many carts on springs can one mind take!
My apologies! I think I am just calling it all mechanical engineering for the algorithm. Calling statics would reach 5 people. Calling it a question in my Mechanics of Materials book would reach 10. But for whatever reason calling something Engineering...
You solved using the equilibrium condition that P=2Tsin(θ) where θ is 60 degrees, but that's never true. θ was 60 degrees before any substantial force was applied; when a force was applied it moves to what you're calling θ₂, and that's what you should be using in your equilibrium equation. I don't see how you can find θ₂ from θ in the way that you have since θ=60 degrees is never of any real consequence, it's not where the equilibrium occurs. You can find θ₂ directly by picturing a triangle with sides L/2 + x, b/2, and sqrt((L/2)^2-(b/2)^2 ) + δ (that's just the length from before the force was applied plus the displacement term δ), using the Pythagorean theorem to solve for δ in terms of x, and using sin(θ₂) = sqrt( (L/2)^2-(b/2)^2 ) + δ ) / (L/2 + x) -- where you're plugging in the value of δ that you've found -- to find the component of the spring's resistive force in the direction opposite to P. (You could use cosine instead of sine, as you did in your solution, but you'll need the length of the side of the triangle with the δ term solved in terms of x either way.) I did this and found **U=6.41J, only a slight difference from what you found. Let me know if you disagree or have any questions. **I found a small mistake in how I typed in my equation when solving that explains the discrepancy.
Hi, I made a video on how you set up the problem. It might have a slight diversion toward the end tho. I still get the exact answer in the book. Let me know if have any extra comments on how you went about it! ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-lP41GAzoCQ0.html
@@holdmybipolar I've found a small error in how I typed in my equation that would explain our different end results. I'll leave my original comment the same and leave a small note indicating the mistake. What I still don't understand is how you used the equilibrium condition P=2Tsin(θ) where θ is 60 degrees in your first solution. The θ we're interested in is never 60 degrees. Maybe I misunderstood your first video? Very cool to you make a video about it. Those are so nasty equations.
Oh, I think to answer your actual question is that I probably did solve it using angle intial is 60 degrees just to see how the answer for energy would differ from the actual answer. In this last video this is the time stamp on how I set up the intial condition and did and iteration to solve for final angle. ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-lP41GAzoCQ0.html I will make a video on Iterative Methods to find solutions as well has showing that Material Mechanics equilibrium is usually always taken before loads are applied BUT that is technically WRONG. But displacements with steel and concrete are so small it won't change the equilibrium much. That is a crazy assumption for most physics problems. And I don't think most Materials students know that isn't the actual equilibrium when loads are applied. So then when a problem using a rubber band has a lot of displacement the final equilibrium is way different than before loads were applied. But all the examples using energy assume intial = final even tho a small displacement occurs for steel structures. I don't like the way the text book doesn't clearly state this assumption so I will make a video on that.
Hi Mate, which chapter does the concept of strain energy usually show up? I’m a Civil Eng. student and I passed Mechanics of Materials, but have not seen any sign of Strain Energy in my courses… kinda bothers me now, just stuck in my head… great video by the way.
I will do more problems on that. Because I hate the concept of how energy is taught and then how simple and useful of a tool its used. It is used mainly or most obviously in the chapter 2 sub chapter on impacts. Because you drop a weighted collar on a steel bar and the coller has an initial velocity of zero but a height of h. Therefore potential energy. Its dropped so right before impact it has a velocity=sqrt(2gh) before impact. But most importantly the coller will come to a dead stop after it elongates the steel bar. That acts as a spring and the energy of either initial potential engery mgh or max kinetic energy sqrt(2gh) has to equal the "energy stored in the steel bar" And basically energy just being equal to the average force over the distance of stoppage. Therefore the average force over the change in length of the steel bar. Since all materials act a spring. The average force is not the max force. Because with zero elongation comes zero force to elongate zero distance. And because springs are linear with force. The average force is always max force divided by 2. Therefore energy is Fmax/2 times change in length. But Fmax for a material is change in length*E*A/L so Max Force comes up twice in the energy equation or you could say change in length comes up twice in the energy equation. Because E*A/L is the spring constant for any material. Sot the energy equation becomes the classic 1/2*k*d^2 where d^2 is the change in length of the steel bar and k is EA/L a material property of a steel bar. This text book proof will be the next video I make. The other part of chapter 2 where Energy comes up is indeterminant structures that have one load on them in one direction. The definition of energy is Load average which will most always be max load or given load P/2 then multiplied by the distance change in length. That simply gives you one extra equation for a problem that you couldn't solve because with inderterminate layups you always have one more unknown than equations. The bummer is that chapter 2 in my book covers multiple special cases that really slow down the learning process of basic material mechanics.
This guy just wanted to make the video long nothing more( The Question is pretty easy . Students here do tougher questions than these in just 3-4 minutes
Thanks man! Your so right! Everything about the subject is just multiplying and dividing numbers. But I did flunk out of this course and had to retake it. That is why I am so focused to explain it so its easy peasy. Thanks again!
