Very clear descriptions here that have really helped the students I teach. Thanks.
@Sibasish07
7 жыл бұрын
A Level Physics Online lol u copy him?
@mikeoxlong2077
3 жыл бұрын
it is rare to see one legend commenting on the video of another
@eyeris292
6 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.
@kenhooke6297
10 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!!
@sonamdixit4121
6 жыл бұрын
ohh really
@jaan735
6 жыл бұрын
👌🏻😂😂
@Homeworlder
11 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!
@Shalie7506
4 ай бұрын
It's 11 years later, how are you now
@Homeworlder
4 ай бұрын
@@Shalie7506 would you believe it, I took a career in teaching myself
@sarkiesarkie4918
9 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.
@DrPhysicsA
11 жыл бұрын
Sorry - don't know. My vids are intended to cover the broad A level material of the main A Level courses.
@DrPhysicsA
11 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
@DrPhysicsA
9 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.
@mp923
9 жыл бұрын
DrPhysicsA you should make a patreon!
@azaneenurarif2102
7 жыл бұрын
DrPhysicsA
@pushpamahadevan7242
7 жыл бұрын
DrPhysicsA-S/TG
@SUONIndustry
12 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.
@turicaederynmab5343
11 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.
@DrPhysicsA
11 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).
@RyanSeeRolyPoly
10 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 ;)
@FatMonkey1395
11 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!
@DrPhysicsA
11 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.
@yaredaddis484
6 жыл бұрын
Thanks for your tutorials. You are helping people all over the world.your tutorials are clear and easily to understand.
@MCmontageX
8 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.
@andrepedersen1147
7 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!
@DrPhysicsA
11 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.
@donnertang9319
5 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
@sudheesh882
9 жыл бұрын
the simplest explanation I have ever see about tensile..... all can understand... you are amazing sir
@DrPhysicsA
11 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.
@squablywablly
11 жыл бұрын
this is so convenient! the teaching is good and you can rewind and pause. Its very helpful.
@chriswang1381
9 жыл бұрын
Man your amazing at teaching physics, your videos always helped me and friends a lot !!!!
@DrPhysicsA
11 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).
@teslatesla5285
3 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.
@trobe5219
10 жыл бұрын
your videos are pulling through my a-levels, keep it up!
@DrPhysicsA
11 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.
@NebulaeCat
11 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 :)
@kevinbeckenham3872
10 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''.
@goliath3577
2 жыл бұрын
I cannot thank you enough for this video, you explain the concepts so well.
@FatMonkey1395
11 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
@DrPhysicsA
11 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.
@DrPhysicsA
11 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"
@DrPhysicsA
11 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.
@MrHellworth
11 жыл бұрын
QUESTION: why does the yield point (the point at which the material stretches with constant or reduced load) occur?
@johnvalerkossi8809
6 жыл бұрын
Very Very Interesting fact about Young modulus = Work/volume. I did learn something new. thank you .
@DrPhysicsA
11 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).
@DrPhysicsA
12 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.
@abdoali2088
8 жыл бұрын
simple,clear amazing videos,very useful for the beginner, thnks you.
@DrPhysicsA
11 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.
@DrPhysicsA
11 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.
@osheensingh6079
8 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
@DrPhysicsA
11 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.
@DrPhysicsA
11 жыл бұрын
The practical aspects determine whether a material will be malleable and ductile or whether it is brittle or plastic.
@DrPhysicsA
12 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.
@DrPhysicsA
12 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.
@oscarheath5507
9 жыл бұрын
Reasonably comprehensive and comprehendible, but not especially compelling or advanced. Worth watching
@chocolatehammer6888
7 жыл бұрын
It's this time of the year again :D
@DrPhysicsA
11 жыл бұрын
Congratulations. Have a great time at uni.
@DWolf16
3 ай бұрын
Hi, I understand completly these operations and concepts. However, what I seem not to find/understand is how can I calculate the Young's Modulus if my input data has many times given Tensile Strength values and Tensile Elongation (in %) Would these "Strangth and Elongation" be considered "Stress" and "Strain" respectively? I am confused and don't know if i am doing it right.
@DrPhysicsA
11 жыл бұрын
Stress is proportional to strain.
@DrPhysicsA
12 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.
@DrPhysicsA
11 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.
@alvinsu8740
11 жыл бұрын
@Dr.PhysicsA,Do have a video for the Analysing Forces in Equilibrium? I need you teach how me how to form a diagram to make it easier so that I can know which SIN,COS or TAN I need to applied.
@shekharbhattarai9262
7 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?
@DrPhysicsA
11 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.
@DrPhysicsA
11 жыл бұрын
Wk is Fx/2 But F is kx by Hookes Law So E = Wk = kx2/2
@DrPhysicsA
12 жыл бұрын
Good. Hope the exam went well.
@Mech.Masters
9 жыл бұрын
How do we come to know the exact value of yield stress of a material??? Because it is difficult to find the actual point when the yielding starts???
@DrPhysicsA
11 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.
@DrPhysicsA
11 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.
@DrPhysicsA
11 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.
@mahmoudm451
7 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?
@masudkamali4109
11 жыл бұрын
Dr.PhysicsA can you tell me how i can find out the cross sectional area and tension in the wire in young modules. if you have a video can you please put the link on the message board.
@shehreyaraijaz4090
7 жыл бұрын
I didn't get the end where it says kx^2/2 is basically the area under a triangle as its compared to 1/2 *stress*strain (which definitely is the area of a triangle) however kx^2/2 isn't the area of a triangle, as it contains square, it should be 1/2*k*x. I am confused, I know I'm wrong I just need an explanation. Thanks.
@DrPhysicsA
11 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)
@lucaciurares5606
9 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.
@anilhatiboglu4534
10 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?
@simplisticirony5896
8 жыл бұрын
Your videos are brilliant, mate. Thank you!
@DrPhysicsA
7 жыл бұрын
Thanks.
@hatatske
12 жыл бұрын
Thank you very much! My GCE physics unit two exam is today. I'm feeling more confident on this topic now!
@donnertang9319
5 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
@ali-bb7so
8 жыл бұрын
Could you tell me the difference between elastic limit and yield point
@DrPhysicsA
8 жыл бұрын
Elastic limit is when the material stops behaving elastically and begins behaving plastically - from here onwards it would no longer return to it's original length. Yield point is when the material suddenly starts to stretch without any extra load.
@Osama_Alkadomi
Жыл бұрын
how to find the cross sectional area of the wire/do we need to know how to do it
@erinwilliams3965
11 жыл бұрын
I may not fail Physics thank you so much!
@rowanmusic1556
11 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.
@VukBrankovicvvv
6 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?
@johnbyrne7083
6 жыл бұрын
Hi. Lots of great stuff here. Forgive me if someone has already mentioned. You make a quick reference to the Youngs Modulus equaling Work per unit Volume. I am not sure if this is correct. Is not the area under a stress-strain graph (not the gradient) equal to the Strain energy per unit volume? I think your annotation mentions this, but doesn't correct the final part of the formula above. Thanks for any clarification.
@aidanmarshall558
11 жыл бұрын
@DrPhysicsA i am doing kevlar for a physics project and was wandering how to explain the fact the kevlar has a youngs modulus yet its is flexible. I thought that this would be contradictory due to youngs modulus being a measure of stiffness. Any help would be greatly appreciated thanks.
@gagangahir5976
6 жыл бұрын
does the spring constant K equals to the gradient on the force and extension graph??? @drphysicsA
@kitbattarbee8184
10 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.
@mahmoudm451
7 жыл бұрын
I didn't understand why we take the average force?
@chriswinnard7739
7 жыл бұрын
Mahmoud Matar force*distance = work done in general when moving something, or in this case stretching/compressing. Here force isn't constant so less work will be done each moment when you start to stretch something compared to when its almost fully stretched. the power (work done per second) changes so average amount of energy transfer is needed to give an overall value. that probably doesn't help but its kind of hard to explain
@mahmoudm451
7 жыл бұрын
Shon Wuls Oh I think I got what you mean, if the force is constant, at that time work=force*distance, but if the force is varying throughout the stretch/compression, then we take the average force isn't it?
@chriswinnard7739
7 жыл бұрын
Mahmoud Matar yeah, but force will always be varying because more force is required each moment when extension is longer as F=kx
@mahmoudm451
7 жыл бұрын
Shon Wuls that means that the more it strerches, the more force you need to apply to stretch it more, and that's principally the reason why the force varies?
@chriswinnard7739
7 жыл бұрын
Mahmoud Matar exactly that
@masudurrahman8265
6 жыл бұрын
If force,F is applied on both end of wire what will be the equation of young's modules?
@xLusbyy
9 жыл бұрын
Incredible explanation. Helps a lot.
@robertsoto9004
10 жыл бұрын
I am having trouble understanding the difference of Flexural Modulus @ 1% Deformation versus Flexural Modulus @ 1% Secant.
@toyfactory6684
11 жыл бұрын
Hi. In my Mechanics 3 (Edexcel) textbook it says that Hookes law is: T = (Lambda / L)x, where Lambda is the modulus of elasticity, L is original length and x is the extension. The book describes Lambda as the modulus of elasticity. Does this mean that lambda is Young's Modulus E? Or is lambda EA, where A is the cross sectional area? I am quite confused because wikipedia says that the modulus of elasticity is stress/strain, which is E, however I think it's wrong. Thanks in advance
@WildHorsesRear
11 жыл бұрын
So, is Hooke's law a law stating that the stress applied to a material is proportional to the strain on the material, or is it a law stating that the extension of the spring is proportional to the force stretching it (provided the elastic limit of the spring is not exceeded) or both?
@jlee9727
9 жыл бұрын
Is this for any specific exam board or is it universally applicable ?
@badshabz1
11 жыл бұрын
When you talking about bonds being brocken when the matial is streched beyond limit of proportionality are you refering to intermolecular bonding?
@johnbenjamin7047
6 жыл бұрын
the books i am using for studies are stating that 1Gpa= 1x10^3 N/mm^2 i am confused
@howtomakeanything9031
6 жыл бұрын
Isn't it possible to extend something to infinity and it aquaere it's original position
@AbdulKaderAbdeenAgha
7 жыл бұрын
So cool, really good video, I think you missed 2 points: 1- you explained stress and strain and have not include them in the video title 2- you missed using the symbols for both stress and strain which are sigma and epsilon respectively. Thanks
@anotherdoseyt
6 жыл бұрын
Abd Alkader Abdeen Agha f that’s not really necessary in understanding the concept the only symbols truly needed is the SI units
@umarhakimin
7 жыл бұрын
Thank you sir. Brilliant explanation!
@0113slimshady
11 жыл бұрын
How would you calculate the work done from a Force vs. Extension graph if the graph was curved.
@emanaref8170
6 жыл бұрын
Is there any other example for young's modules ?
@Daniel-jd4wz
9 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.
@mahmoudm451
7 жыл бұрын
does that mean that the area under the F against x graph, will give us the work done?
@feitan5052
7 жыл бұрын
Yeah, you could say the elastic potential energy stored in the spring.
@abdullahkabbani8123
10 жыл бұрын
Straight away subscribed you. Great teaching, thanks!
@JarredDavidson
3 жыл бұрын
So helpful. Thanks so much
@beoptimistic5853
3 жыл бұрын
kzitem.info/news/bejne/13x6s5V-imacim0 ..👍
@lisapersson5818
12 жыл бұрын
Your accent makes it that much better :)
@fenerbahce262
11 жыл бұрын
@Doctorphysics is this OCR A? And could you tell me how to get an A in physics, my course grade is a C what do I need to do to get A in AS physics(OCR A) please respond asap!
Пікірлер: 304