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.
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!!
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!
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 ;)
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.
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).
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
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
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.
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).
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.
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 :)
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.
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.
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.
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.
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''.
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).
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!
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"
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.
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)
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.
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.
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.
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.
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.
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.
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.
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.
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.
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
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?
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.
Very clear descriptions here that have really helped the students I teach. Thanks.
A Level Physics Online lol u copy him?
it is rare to see one legend commenting on the video of another
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.
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.
DrPhysicsA you should make a patreon!
DrPhysicsA
DrPhysicsA-S/TG
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!!
ohh really
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Sorry - don't know. My vids are intended to cover the broad A level material of the main A Level courses.
wow, what a teacher. at 63 im still learning stuff. been a welder for many years and this explanation has helped enormously. Many thanks.
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!
It's 11 years later, how are you now
@@waakoshaldon7506 would you believe it, I took a career in teaching myself
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 ;)
thank you very very much,sir.I am the best physic student in my class right now.I'm truly appreciate your work.
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.
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).
your videos are pulling through my a-levels, keep it up!
I cannot thank you enough for this video, you explain the concepts so well.
Yes. The SI units use kg, m and sec. So if a measurement is in mm you need to convert it to m.
Man your amazing at teaching physics, your videos always helped me and friends a lot !!!!
Thanks for your tutorials. You are helping people all over the world.your tutorials are clear and easily to understand.
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!
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
the simplest explanation I have ever see about tensile..... all can understand... you are amazing sir
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
simple,clear amazing videos,very useful for the beginner, thnks you.
this is so convenient! the teaching is good and you can rewind and pause. Its very helpful.
Your videos are brilliant, mate. Thank you!
Thanks.
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.
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).
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.
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 :)
Straight away subscribed you. Great teaching, thanks!
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.
Thanks a million for your amazing videos!!!! :D
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.
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.
This is extremely helpful, thank you
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.
Thank you sir. Brilliant explanation!
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''.
Very Very Interesting fact about Young modulus = Work/volume. I did learn something new. thank you .
Thank you very much! My GCE physics unit two exam is today. I'm feeling more confident on this topic now!
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).
You helped me loads on my way to an A overall in physics and an A* in physics5!!! Got into university :D:D:D
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It's this time of the year again :D
your videos are a great help! thank you, keep it up!
Congratulations. Have a great time at uni.
You're awesome! thank you, helped me a lot
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!
Doing a great job......thank u sir ..
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"
simple but detailed illustrations there.
Thanks!
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.
Incredible explanation. Helps a lot.
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)
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.
I love You so much. This thing help me so much!
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.
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.
knowledgeable video ........thanks
Thank you this was very helpful!
Thanks for the Explanation.
i love you. really brilliant video!
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.
thanks a lot sir........u are just brilliant sir,u r doing a great job for students like us....thanks once again sir
Reasonably comprehensive and comprehendible, but not especially compelling or advanced. Worth watching
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.
Good. Hope the exam went well.
One Word. "Awesome". :D Thanks a bunch! :)
Thank you!!
Stress is proportional to strain.
Thanks, great video.
So helpful. Thanks so much
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MrPhysicsA I LOVE YOU!
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The practical aspects determine whether a material will be malleable and ductile or whether it is brittle or plastic.
Thank You.I really thank you.
QUESTION:
why does the yield point (the point at which the material stretches with constant or reduced load) occur?
Thanks for the help!
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.
thanks u soo much sir this helps me a lot
Great explanation sir!
Thanks for your explanation
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.
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.
thanks it was really really helpful.
Very helpful indeed.. Thank you
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.
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
Really tanks man
It helps me alot
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?
great video!
wow thanks for this very useful for my revision got the exam in one month
excellent presentation .. well done clear and detailed
Very helpful, thank you :)
Thanks.
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.
Well explained, thanks
MRphysicsA thanks alot very good teaching!!!!
Your accent makes it that much better :)
Thank you! :)