Your teaching method is so great. If you are making any mistakes pls edit that portion before uploading so that it will give more confidence to student community.
I realy found your tutorial videos very helpfull,more appreciation to your effort in explaining this topic,thanks .am final year student of mechanical engineering
In this problem where is mg component ? because that is also a force and is to be applied on both the blocks M1 and M2. That particular mass multiplied by acceleration due to gravity (i.e g = 9.81 m/s^2) has to be applied on both respective blocks in the free body diagram of both the blocks. My question is why is it not included in free body digram as well as rest of the problem. Thank you.
This acceleration due to gravity is still an acceleration which is the second derivative of displacement, which is there. It's just that in this case you know the value of the displacement
In that case we have to consider mass between damper and spring as zero so that now we have two mass bodies one with zero mass and another with given mass and rest of the process I same hope u understand
i think gravitational force is already acting on the system in the condition when we dont apply any force. so we just have to take the force applied and reaction due to that force .
In this problem, the force is applied to a system at equilibrium condition which means the gravity is already applied and the system reaches its equilibrium condition.
hi madam, indeed you have a call and a willing heart to teach, i really appreciate you madam, thank you so much
Mam you such a great teacher you always help me to understand the topic easily suxh a great skill of teaching 💕
you have helped me alot today. i am having a control sytem exam now and youre my saving grace
Your teaching method is so great. If you are making any mistakes pls edit that portion before uploading so that it will give more confidence to student community.
Yes you are right
She corrected her mistake.
I think its better when she doesn't edit, it also makes students familiar where they can do such mistakes
@@DishanksahuYes, You are right
@@UltimateStudy ruclips.net/video/XS9uSSq_npM/видео.html
You are the best teacher i ever know.
I realy found your tutorial videos very helpfull,more appreciation to your effort in explaining this topic,thanks .am final year student of mechanical engineering
rip
@@aboodatiyat6255 ruclips.net/video/XS9uSSq_npM/видео.html
Your way of teaching if wery 👍good, and I know nothing about Control systems after classes. But Your videos helped me a lot
ruclips.net/video/XS9uSSq_npM/видео.html
Not gonna lie, this is quality content
Madam is beginner but hats off to her for her great efforts.
She's not a beginner sir - probably this is the first lecture you've watched - She's good
MAM
K2 Is between y1 and y2 so we need to take k2(y2- y1)
Which u mentioned in previous video
Mistake yeah she had to take
Abe sale pura video to dekh le pehle
You are excellent in your teaching, you just save by day!
your teaching method is so great.
I really appreciate your work and lecture delivering method
Mam for m2 the force of k2 is common for m1 and m2 so it must come as k2(y2-y1)
Ya correct
Testing your IQ😂
ruclips.net/video/XS9uSSq_npM/видео.html
Actually for m1 we should take y2-y1
@@nani_alpha ruclips.net/video/TTVrswZNor4/видео.html
Ma'am your are a best teacher
May Allah All teacher give lecture as ma'am
wow what a great skill to educate your so unique mam God bless love from Ethiopia long live our teacher
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You're great!. Thank you very much, you explain very well. Greetings from Chile.
I'm taking this class in about 2 months. I'm going to try to be ahead of my class
Amazing teacher she is...🥰
Just great, no body is perfect...
J'apprécie beaucoup votre effort and committment.. thanks a million
Mam .... I like ur teaching soò much 😘😘😍
Very well explained. More examples please.
madam wahan p k2(y2-y1) aaega..... koi baat nhi, mistake sabse hoti h..
Thank you for the explanation. I really needed help
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Thank you for the clear explanation. Was helpful to me.
Mrs. Swarna i hope you have a beautiful life, god bless you thank you
Thank you so much ma'am, the video helped a lot.
(Y1-Y2 )*K2 is for first body . The reaction is on second body or negate the. - (Y1-Y2 )*K2 so it becomes (Y2-Y1 )*K2
Madam u r teaching so super 👌 Madam I need cs exam important questions for exam plz 🙏
Super explanation Sister.....👌👌👌👌
In this problem where is mg component ? because that is also a force and is to be applied on both the blocks M1 and M2. That particular mass multiplied by acceleration due to gravity (i.e g = 9.81 m/s^2) has to be applied on both respective blocks in the free body diagram of both the blocks. My question is why is it not included in free body digram as well as rest of the problem. Thank you.
This acceleration due to gravity is still an acceleration which is the second derivative of displacement, which is there. It's just that in this case you know the value of the displacement
And also there is no Laplace transform of g😌
mg component is ma where a is acceleration due to gravity.
That mg component is applied in downward direction which should be there in this case
@@MShazarul ruclips.net/video/XS9uSSq_npM/видео.html
very good teaching.
In a free body diagram of M2, the Fk2= K2(y2-y1) not K2Y2
She got mentioned at the end
You are great ❤
Madam you are my life saver madam
Dear madam, just want to know why we do not consider the gravity force here? Thanks
We must..
If zero position is taken to be that at which the system is at equilibrium with gravity, then gravity can be ignored in the analysis.
applied force = opposing force
not according to newton law, but by ambert's principle
Rest in peace....
is there a mistake in second FBD? you must clear it k2(y2-y1)
Same doubt...bro
Same
Watch the video till the end. It was a mistake and she corrected it later
@@muhammedshafikp9964 ruclips.net/video/XS9uSSq_npM/видео.html
Explain me this
perfectly explained thank you!
Thanks mam for this this video☺😌
Thank u so much mam🙏
Superb class mam
Love you so much mam.... ❤️
Thank you ❤
Laplace transform of second order differential equation ......... don't you think it is wrong
It is d²y/dt²=s²y(s)-sy(0)-y'(0)
Great teaching madame
in the second body digram its should be K2 + ( y2 - y1) right?
Great respect for you
Thank you madam.
thank you so much. it saved my project :D
confusion may occur , because of the way you indicate direction of forces, others vertical others horizontal
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What will happen if frictionless surface is given ?
Will the Damper element become =0 ?
Y2(s)/F(s) is just one of the possible system transfer functions.
Thank u very helpful
Very good sister
What will we do when we have both spring and damper in series connected to mass.
exactly what i wanted to know!
me2
In that case we have to consider mass between damper and spring as zero so that now we have two mass bodies one with zero mass and another with given mass and rest of the process I same hope u understand
Hello,does the force fk2 pulling away M1 or pushing in? Am confused 😕
Thanks for the correction
Mam can you please clarify why you taken fk2=k2(y1 -y2) in FBD1 if one end is fixed?
I am wondering the same
But madam for k2 its not fixed should be k2(y2-y1)
You are good, make some videos on grounded chair representation
Your lecture is very much helping but why dont we take mg as gravitational force into consideration??
She established as a framework that the initial strains o the springs happen when both masses gravitational forces are applied.
Thank you ma
it's great
How are the directions of Fk1 & Fk2 same???
Yea, they should be opposite!
Why did she used Y1 - Y2 instead of Y2 - Y1
For M2 we are not having fixed end ,but why you are taking just k2*X2?
Wo + k2(y1-y2) hoga , unhone sirf y2 likha hai
M2 ke liye Kiya ja Raha hai isliye K2(Y2-Y1) hoga
Thnks miss
ruclips.net/video/XS9uSSq_npM/видео.html
Thank you mam
Thank you
why don/t we consider gravitational force??
i think gravitational force is already acting on the system in the condition when we dont apply any force. so we just have to take the force applied and reaction due to that force .
In this problem, the force is applied to a system at equilibrium condition which means the gravity is already applied and the system reaches its equilibrium condition.
the gravitational force between these two closer bodies m1 and m2 is very very small. G=6.67*10^-11...thats why we dont consider it
@E lemme guess
u read that book on vibrations and came here only to drop some knowledge?
lol
Here don't we want to consider gravitational force ( Mg)
Hi madam, im your student in youtube, in the last part 18mins. I cant understand how did simplify your answer.
What is the name of the book or module is that??
Plz take power system analysis of single line diagram
thank you so much love you
Awesome 👍👍👍🙆
Actually for m1 we should take y2-y1
nice representation ........
Mam agar lecture Hindi me hota to or bhi achha lagta
Btw great teacher
thank u mam......
I think she is not tired that's why referring voltage to velocity and f2 k2* x2
why is total force on m2 = 0?
Coz there no applied force
Tqq u Mam.
iyi ki varsın ablammmm
WHAT IF WE DOT APPLY ANY EXTERNAL FORCE. WHAT WILL BE THE TRANSFER FUNCTION
plz give solution of linear system book by D K Cheng of analogous system
Thankuuuu
Tq madam
3:40
❤️❤️❤️❤️❤️
Please suggest a book
How to find out only y1 and y2
ruclips.net/video/XS9uSSq_npM/видео.html
should have done FBD longitudinally
3 nodes dhi chepandi medam
Can anyone please explain to me how fm1 is acting on the side of mass 1?
♥️🙏
To teaching mam
with 1.01M subscribers avoid silly mistakes