A description of Hooke's Law, the concepts of stress and strain, Young's Modulus (stress divided by strain) and energy stored in a stretched material
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@DrPhysicsA11 жыл бұрын
Sorry - don't know. My vids are intended to cover the broad A level material of the main A Level courses.
@PhysicsOnline9 жыл бұрын
Very clear descriptions here that have really helped the students I teach. Thanks.
@Sibasish077 жыл бұрын
A Level Physics Online lol u copy him?
@mikeoxlong20773 жыл бұрын
it is rare to see one legend commenting on the video of another
@eyeris2926 жыл бұрын
Probably one of the best videos I've seen regarding the topic of stress-strain, springs, etc. not only from a practical standpoint but from an experimental standpoint as well. DrPhysics, thank you for supplying the community with multi-faceted ways of thinking that is applicable not only to students but also to potential real-world applications in a work environment as well.
@sarkiesarkie49189 жыл бұрын
wow, what a teacher. at 63 im still learning stuff. been a welder for many years and this explanation has helped enormously. Many thanks.
@kenhooke629710 жыл бұрын
Hooke's Law so clearly explained, and the associated physics too. Thank you. Always interested in Hooke's Law. Robert Hooke is part of our family tree!!
@sonamdixit41216 жыл бұрын
ohh really
@jaan7356 жыл бұрын
👌🏻😂😂
@DrPhysicsA11 жыл бұрын
Thanks for kind comment. Young's modulus is defined as stress over strain which is pressure (F/A) divided by strain (extension over original length). So E = F/A / x/l which can be rearranged to E = Fl/Ax
@DrPhysicsA9 жыл бұрын
j lee - these videos are designed for the syllabuses of AQA, OCR, Edexcel and CIE. Not all of them will be relevant for each course.
@mp9239 жыл бұрын
DrPhysicsA you should make a patreon!
@azaneenurarif21027 жыл бұрын
DrPhysicsA
@pushpamahadevan72427 жыл бұрын
DrPhysicsA-S/TG
@DrPhysicsA11 жыл бұрын
Young's Modulus will apply to anything where stress is proportional to strain. So if the proportionate extension is related to the pressure or stress (force over area).
@RyanSeeRolyPoly10 жыл бұрын
You are a ledge, I have been studying this in science for weeks and my teacher does not explain shit all, I've just learned how to do this in a quarter of an hour. Cheers pal, have a nice day ;)
@DrPhysicsA11 жыл бұрын
Not uniquely. The material in the A Level Physics playlist covers the main material in the Edexcel, AQA A/B and OCR A/B courses except for some biophysics which I have not covered.
@yaredaddis4846 жыл бұрын
Thanks for your tutorials. You are helping people all over the world.your tutorials are clear and easily to understand.
@chriswang13819 жыл бұрын
Man your amazing at teaching physics, your videos always helped me and friends a lot !!!!
@Homeworlder11 жыл бұрын
I've spent 6 weeks with my teacher rabbiting on at me about Young's Modulus but she never once said what it actually is. Thanks to this I finally understand how simple it is! This is an excellent video, thank you!
@waakoshaldon75063 ай бұрын
It's 11 years later, how are you now
@Homeworlder3 ай бұрын
@@waakoshaldon7506 would you believe it, I took a career in teaching myself
@trobe521910 жыл бұрын
your videos are pulling through my a-levels, keep it up!
@DrPhysicsA10 жыл бұрын
Yes. The SI units use kg, m and sec. So if a measurement is in mm you need to convert it to m.
@andrepedersen11477 жыл бұрын
Continuously amazed over your brilliance in both explaining and teaching. There come few great teachers these days, but you're surely one of the better ones. Been watching your videoes during my whole bachelor's degree. Few teach subjects as easy and clear as you do. Thanks! Keep up the good work!
@goliath35772 жыл бұрын
I cannot thank you enough for this video, you explain the concepts so well.
@sudheesh8828 жыл бұрын
the simplest explanation I have ever see about tensile..... all can understand... you are amazing sir
@donnertang93195 жыл бұрын
00:00 Hooke's law F=kx 06:23 Stress and Strain Stress(tensile strength)=F/A Strain=x/l W=1/2 Fx= (kx^2)/2 12:53 Young's Modulus E=Stress/Strain=Fx/lA Energy in stressed material = 1/2 (stress)* strain or the area under the stress to strain graph
@DrPhysicsA11 жыл бұрын
In the case of a spring, the extension (x) is the distance between the mean position and the extended position. Force = kx where k is the spring constant. But if you consider a spring oscillating then the force is constantly varying since it is proportional to the extension which itself is constantly varying.
@turicaederynmab534311 жыл бұрын
I love learning more about math and physics! I struggle with other topics so focussing on my passions in my spare time will help me become a better phycisist in future.
@abdoali20888 жыл бұрын
simple,clear amazing videos,very useful for the beginner, thnks you.
@SUONIndustry12 жыл бұрын
thank you very very much,sir.I am the best physic student in my class right now.I'm truly appreciate your work.
@MCmontageX8 жыл бұрын
Hi there, this was a very clear and informative video although i would like to add that you could use searle's apparatus to measure the young modulus of a wire. This involves adding a second wire parallel to the test wire, the second wire acts as a control wire where by any changes in temperature do not affect the end results due to the addition of the second wire. Also a vernier scale could be installed between the two wires which you use to gauge how far the test wire has extended as opposed to the control. Also i think that you need to explain the fact that the young modulus which can be calculate graphically only applies to the straight portion of the stress/strain graph. Young modulus can only be measured within the limits of proportionality. Thanks for making the video, i just wanted to add a little of my knowledge just to clarify a few things.
@DrPhysicsA11 жыл бұрын
Well work done is the area under the curve. If the curve is regular then you might have a formula you can use. Otherwise its a case of adding up the squares (if its plotted on graph paper).
@FatMonkey139511 жыл бұрын
How I wish you were here in January of 2012, but hey, Thank you so much, I'll finally be acing physics2 this time around!
@DrPhysicsA11 жыл бұрын
K is the spring constant such that F = KX, where F is the force and X is the extension. Stress is force over area. Strain is extension over original length. From that you should be able to derive an equation for K.
@simplisticirony58967 жыл бұрын
Your videos are brilliant, mate. Thank you!
@DrPhysicsA7 жыл бұрын
Thanks.
@DrPhysicsA11 жыл бұрын
You find the cross-sectional area by measuring it using a device which measures the circumference accurately. The tension will usually just be the weight applied to the wire which you will usually determined.
@DrPhysicsA11 жыл бұрын
At 13:55, as I indicated in my reply to an earlier comment, I was actually just establishing the dimensionality. E = stress/strain = F/A / x/L = FL/Ax = units of work/energy / units of volume - hence energy per unit volume. The graph at 15:30 better sets out your point. The energy per unit volume stored in a stretched wire is 0.5 x stress x strain = 0.5 (F/A) (x/L).
@DrPhysicsA11 жыл бұрын
It will certainly distort if you crush it. Not sure if that is "crossing the elastic limit" since that term is usually reserved for over-stretching the spring.
@MrHellworth11 жыл бұрын
QUESTION: why does the yield point (the point at which the material stretches with constant or reduced load) occur?
@DrPhysicsA11 жыл бұрын
Well you could watch my 44 A Level Physics revision videos (assuming you are doing A levels or equivalent exams) but they are really only revision videos and can't replace the original tuition. Good luck with the exam.
@abdullahkabbani812310 жыл бұрын
Straight away subscribed you. Great teaching, thanks!
@DrPhysicsA11 жыл бұрын
I guess the point they were making is that if the material returns to its original state then the material was being stretched within its elastic limit. ie it had not gone beyond that point in which case it would not have done so. A spring can be loaded and unloaded and still obey F=kx as long as you always keep within the elastic limit. But if the spring gets deformed with too heavy a load then the F=kx rule will no longer apply.
@umarhakimin7 жыл бұрын
Thank you sir. Brilliant explanation!
@squablywablly11 жыл бұрын
this is so convenient! the teaching is good and you can rewind and pause. Its very helpful.
@DrPhysicsA12 жыл бұрын
I think you've answered your own question. Wk = Fx when the force is constant. If the force varies (as it does with Hooke's law) then you have to integrate each element of F dx to find total work done. So Wk = Integral F dx. In the case of Hooke's law for, say, a spring the force varies linearly with x (since F=kx). So you get a straight line relationship between F and x. The integral in this case is just the area of the triangle under that curve, which is half the base times the height ie Fx/2.
@zahrahf65765 жыл бұрын
Thanks a million for your amazing videos!!!! :D
@beeplorizon007 жыл бұрын
This is extremely helpful, thank you
@DrPhysicsA11 жыл бұрын
The practical aspects determine whether a material will be malleable and ductile or whether it is brittle or plastic.
@johnvalerkossi88096 жыл бұрын
Very Very Interesting fact about Young modulus = Work/volume. I did learn something new. thank you .
@kevinbeckenham387210 жыл бұрын
We need good educational Films on youtube like these films,because it is revision for me. My thanks go's to the lecturer & person who produced the films and also youtube.''Thank's''.
@DrPhysicsA11 жыл бұрын
You are right that the limit of proportionality comes first and is usually closely followed by the elastic limit. Hooke's law still applies at the limit of P, but if you go beyond the elastic limit then the material will be permanently stretched/deformed. There is some material on Work, Energy and Power at the back end of the vid on "Classical Mechanics - A Level Physics"
@NebulaeCat10 жыл бұрын
Wow you've helped me A LOT. My script at university is absolutely terrible comparing to this :) saved me for today's lab, as I was really struggling to get it all :)
@DrPhysicsA11 жыл бұрын
Well I assume that the "bar" you refer to is capable of being stretched - so is in the form of a wire. Measure length of wire and diameter (from which cross sectional area can be calculated). Suspend wire from a suitable fixed point. Hang weights on the wire and measure the extension for each weight (but dont go beyond elastic limit). Plot Force/Area against extension/ original length. The slope is Young's Modulus (ie F/A / x/l)
@DrPhysicsA12 жыл бұрын
Hi. Hooke's Law doesn't apply on an atomic scale because of Heisenberg's uncertainty principle. At the atomic scale all measurements are uncertain. But atomic vibrations can be thought of as similar to the simple harmonic vibrations of a spring as in my videos on SHM.
@teslatesla52853 жыл бұрын
Veljko Milković, an academic and inventor from Novi Sad, has done something great that has not been done by any Serbian inventor before. Milković invention of the mechanical oscillator is widely used worldwide, a testament to the fact that over 500 foreign companies use, sell and manufacture pendulum-based machines used in the heavy industry. The purpose of the two-stage meganic oscillator is multifaceted, because the character of the machine (two-arm lever with pendulum) allows its use as a press, water pumps, compressor, crusher, power generator, mini power plants.
@shekharbhattarai92626 жыл бұрын
Since,the change in length(strain)depends upon the force(stress),wouldn't it be more appropriate to choose stress along x-axis and strain along y-axis?
@DrPhysicsA11 жыл бұрын
I am no expert on this but it is to do with molecular structures. During the elastic stretching the molecular bonds are stretched but the structure remains in tact. The yield point arises when the bonds start to break and the material cannot then return to its original state.
@DrPhysicsA12 жыл бұрын
Stress = f/a will always be true but in the case of a spring it is very complicated and not much use. In the case of a wire hanging vertically with a weight F=mg on the end, then the relevant area is the cross sectional area of the wire. But for a spring the wire is coiled and it would be difficult to assess the cross sectional area to which the force applied.
@hatatske12 жыл бұрын
Thank you very much! My GCE physics unit two exam is today. I'm feeling more confident on this topic now!
@DrPhysicsA10 жыл бұрын
Stress is proportional to strain.
@mehzabinchowdhury90087 жыл бұрын
Doing a great job......thank u sir ..
@Fergorilla12 жыл бұрын
your videos are a great help! thank you, keep it up!
@farris66918 жыл бұрын
You're awesome! thank you, helped me a lot
@donnertang93195 жыл бұрын
Is there is a connection between the elastic potential energy and kinetic energy? I noticed that one is given in (1/2)(kx^2) and the other(1/2)(mv^2), they look similar with m and k both being constants and v with x being variable
@osheensingh60797 жыл бұрын
thanks a lot sir........u are just brilliant sir,u r doing a great job for students like us....thanks once again sir
@DrPhysicsA11 жыл бұрын
My A level playlist covers material for OCR A and B, AQA and Edexcel, with some CIE as well. I can't really tell you how to convert a C to an A other than to go thro the material thoroughly and perhaps practice exam questions, examples of which you can find online. All good wishes for the exam.
@entrop1e79 жыл бұрын
thank you :D you saved my life
@solomont.ayernor741410 жыл бұрын
simple but detailed illustrations there. Thanks!
@lamaalnajjar329210 жыл бұрын
You're just amazing !!! . But I'm wondering about calculating the extension of group of springs in parallel or in series . Could you makeup a video on that ?
@DrPhysicsA11 жыл бұрын
You can't calculate tensile strength from info in this video. Tensile strength is the maximum stress that a material can withstand while being stretched or pulled before deforming. It is usually found by performing a tensile test and recording the stress versus strain. Tensile strength is defined as a stress, which is measured as force per unit area.
@felon55737 жыл бұрын
knowledgeable video ........thanks
@anilhatiboglu453410 жыл бұрын
if i didn't understand wrong, young's modulus is actually the work done to per unit volume which is streching . But if i think of a spring, what is the volume? spring would have a free space inside the helix shape unlike a wire. is it still consistent?
@DrPhysicsA11 жыл бұрын
You may find something helpful in my video on Momentum in 2D - A Level Physics
@DrPhysicsA11 жыл бұрын
Elastic - a stretched material will return to original shape cos atoms can be pulled apart up to a limit and the move back to equilibrium position when load removed. Plastic - stretch leads to permanent deformation - atoms dont return to original position. You may need to look up how atoms are organised in metals, ceramics, polymers and combinations.
@DrPhysicsA11 жыл бұрын
Well for A Level physics its probably sufficient to say that the bonds within the crystal structure are atomic or molecular. But at an engineering level it all gets much more complex. It's not something I've studied at that level.
@xLusbyy9 жыл бұрын
Incredible explanation. Helps a lot.
@kitbattarbee81849 жыл бұрын
Hi, when I learned this I was taught that when you do an experiment to plot a stress-strain graph, the area does change (particularly during plastic deformation when the material starts to neck). I don't really fully understand this so I could be wrong.
@george106189 жыл бұрын
fantastic video as it covers the course perfectly keep it up but can u help me with the part, what would the springs constant be if we used two wires in series and the same for parallel
@DrPhysicsA10 жыл бұрын
Wk is Fx/2 But F is kx by Hookes Law So E = Wk = kx2/2
@DrPhysicsA11 жыл бұрын
Well the modulus of elasticity is usually the same as Young's modulus which is stress/strain. Stress is F/A and strain is x/L. So E = F/A / x/L = FL/xA. So F/A = Ex/L. That means that T in your equation must equate to Stress.
@FatMonkey139511 жыл бұрын
You helped me loads on my way to an A overall in physics and an A* in physics5!!! Got into university :D:D:D
@DrPhysicsA11 жыл бұрын
Strictly it is Energy per unit volume = 1/2 * stress * strain. On your second point I was actually establishing the dimensionality. E = stress/strain = F/A / x/L = FL/Ax = units of work/energy / units of volume - hence energy per unit volume.
@lucaciurares56069 жыл бұрын
I have an elastic tube, and inside the tube i put wather uder diffrent pression to observe deformation of the tube. How can i compute Young modulus, knowing the pressures and the deformation of tube.
@kelvinho24529 жыл бұрын
I love You so much. This thing help me so much!
@DrPhysicsA11 жыл бұрын
Congratulations. Have a great time at uni.
@oscarheath55078 жыл бұрын
Reasonably comprehensive and comprehendible, but not especially compelling or advanced. Worth watching
@popedope48429 жыл бұрын
Great explanation sir!
@ilikegreenthings9611 жыл бұрын
i love you. really brilliant video!
@jayesh59068 жыл бұрын
Thanks for the Explanation.
@mahmoudm4517 жыл бұрын
You said that the young modulus is the work done per stretched volume, and work done is equal to the energy transformed, so basically the energy per stretched volume is equal to the gradient of the stress vs strain graph, so why are we taking the area?
@Onjago8 жыл бұрын
Thank you this was very helpful!
@chocolatehammer68887 жыл бұрын
It's this time of the year again :D
@kefahboumoghdob283411 жыл бұрын
Too much useful session. It is simple and powerful very thank you.
@DrPhysicsA11 жыл бұрын
The units of Young's Modulus are indeed N/m^2 (ie the units of pressure).
@eeesss95937 жыл бұрын
There is unit for strain which is mm/m. that shows change in each unit of length
@EIIjot10 жыл бұрын
Thanks, great video.
@vukbrankovic64955 жыл бұрын
How 1/2 stress × strain can be same as 1/2kx(squared)? Both equations are suposed to be potentional energy but i think second isnt correct? Am i wrong and why?
@Funkymonkfunkymonk8 жыл бұрын
Thank you!!
@rowanmusic155611 жыл бұрын
Hello, firstly can I just thanks for putting up these videos. They've helped me enormously with AS Physics so far. At 4:28 you call the point illustrated with an arrow the elastic limit. Isn't this the limit of proportionality? I thought the elastic limit came after the limit of proportionality. On another note, are you planning to publish any videos on work,energy and power? Cheers.
@omersohail9510 жыл бұрын
One Word. "Awesome". :D Thanks a bunch! :)
@DrPhysicsA12 жыл бұрын
Good. Hope the exam went well.
@rikk93917 жыл бұрын
please please do a detailed video on superconductivity and the property associated with it.
@ihtishammiraj77011 жыл бұрын
Hey DoctorPhysics :) Thannk you soo much for these videos, theyve been a reaaal help Can u tell me though which yeard did a physics practical in stress in springs come out for the cambridge a-level exams? wud be real helpful thanking u in advance
@shenry418510 жыл бұрын
perfect!
@Osama_Alkadomi Жыл бұрын
how to find the cross sectional area of the wire/do we need to know how to do it
@masudurrahman82655 жыл бұрын
If force,F is applied on both end of wire what will be the equation of young's modules?
@Daniel-jd4wz9 жыл бұрын
4:30 - wouldn't it be better to describe it as the limit of proportionality rather than elastic limit? The material still exhibits elastic properties at the limit of proportionality - the only difference being that force exerted doesn't equal the extension. I've always learnt them as two different points, but I might be wrong. Thought I'd make a point of it. Thanks.