Mass on a moving wedge

[@NegativeAccelerate]

This guy deserves all the compliments he’s getting. He rly helped me.

[@LT-pp2vr]

you are right!!!!!!!

[@RexGalilae]

Around 8000 engineers and physicists of the future have benefited from this great video! Great job and thanks a lot!!

[@robert.m6755]

Now 44000!

[@AnamKhan-si5py]

More than 100000.... It's 2022,Sep,2

[@Heavenly_Asura]

80000

[@lakshyashukla2740]

Helped a lot... You have generalized everything and that will help me to solve any problem.. Thanks a lot!!

[@Nsut_Aim]

Hello bhaiya You gave JEE 8 years ago How is your lifestyle now? I am going to give it in 2025

[@HarshRajAlwaysfree]

This so helpful I was stuck here for hours to figure out what was happening

[@projectz935]

nice explanation....the problem can also be solved without using s pseudo force and that method gives more insight into the details of the problem....but anyways the ghost force approach (in non inertial frame of wedge) is easier and shorter....

[@jeenius5882]

Sir be like, "Joh yeh tera torture hai ... woh mera warm-up hai"

[@ParthTyagi]

Mujhe lag hi raha tha apna koi a JEE/NEET aspirant mil hi jayega comment section me?

[@mohitforimprovement]

@@ParthTyagi Hehe

[@devmehi]

for such problem, is it possible to solve without using pseudo force (ghost force) by putting observer or frame of reference on ground.?

[@govindankandan4527]

Sir, thank you so much. Your video was immensely helpful.

[@santoshkesavan]

Thank You very much explained a difficult problem very well.

[@MrMariodesouza]

It is possible expressing the motion of m in terms of the relative motion, but there is a lot of algebra.

[@ipsitasingh8339]

What's the professor's name? His explanations are amazing! Thanks a lot for sharing this super useful and insightful content. :-)

[@Besinnung]

He is Alpar Sevgen from Bogazici University

[@laibajadoon7585]

Thnkiuu sir...it helps me alot to solve a problem...a great respect from...PAKISTAN

[@carlkpsplucky5554]

I’m not a engineer or anything, just studying for the ASVAB. This should be a simple question for u guys. “If a wedge Is made longer relative to its height, how does the force increase?”

[@supriya1479]

thank u sooo much sir u helped me a lot to solve many of the tough problems

[@mathOgenius]

can anybody please tell why acceleration of M is not negative because it's going backwards and other is positive..because it's going forward

[@foxtrot8325]

Because he already decided the direction of acceleration, so he only used the magnitudes of the acceleration in the problem. This is also the reason that value of R(Reaction force or Normal force) is a scalar(number) and not a vector in terms of i hat and j hat. Though the problem can be done with vector if you want signs. It won't be tidy though.

[@mathOgenius]

@@foxtrot8325 ohk thanks , It was an old comment , I have done it with vectors too, anyway thanks for explaining

[@Messi-is-goat_10]

​@@mathOgeniusbro!! You're here?!!

[@chups3421]

Why do you not include the portion of the block 'm' (mg cos(Θ)) weight acting on the wedge 'M' in the reaction force(R)? Can sbd explain?

[@jaycho3241]

This helped me a lot. Thanks :)

[@physicsexcellencegroup8945]

Awesome lecture sir, very well explain

[@SanjayKumar-dh1en]

Sir you are awesome ..... Keep posting even more videos...

[@AnishKumar-qx5un]

Hats off to you, sir!

[@mohammedsrivastava5917]

Thank you very much Sir!

[@jaysun1309]

Thank you!

[@ayushijain8671]

Thank you very very much.....You explained really well. May God Bless.....

[@laxmi_reels]

Thanks alot sir 🙏💐

[@krishankantgupta2483]

concept clearing video

[@NegativeAccelerate]

Go to 5:50 to skip warm up

[@ParthTyagi]

22:37- Sir, won't it be mg Sin^2 Q? Isn't the square on sin Q missing? Or maybe I am confused. Can someone please confirm?

[@trivikram4962]

did u figure out what was wrong, am having the same issue

[@zukamot]

Thanks, I have spent many time with this problem, without make a real progress, I understand a little more tanks to you.

[@gamingzeraora443]

just use psuedo force and convert it into a rested block

[@naveens800]

nice video...helped me a lot..

[@sayakghatak3218]

sir please dont feel hesitated while teaching , U are good enough !

[@abhinovenagarajan.s7237]

How can we solve this question without using a pseudo force?

[@shubhronildutta1563]

We can use a Lagrangian to model the problem, and then use Euler Lagrange Equations to solve it without pseudo force.

[@luigiluigi8297]

The best introductory physics course thanks

[@alluarjun3952]

Sir ,you state the block is not jumping from the wedge,why it is so?,how will we decide this...if R makes wedge to move horizontal then why not it makes block jump from wedge and move parabolicaly..

[@zhgshdbssss6536]

Allu Arjun i have the same question. I understand that Rsin(theta) makes the wedge have a horizontal acceleration, but which force (or component of a force) provides the small mass with that horizontal acceleration?

[@milessodejana2754]

That's why it's called a ghost force.

[@RexGalilae]

when the wedge is pushed towards the right, why is the ghost force still acting to the right? shouldn't it be acting to the left now?

[@HarshRajAlwaysfree]

Rex Galilae don't count me in though I'm a future scientist this video is a shit?!!!

[@aryanparekh9314]

There are no forces to acting on the block to let it keep going with the wedge, why would it have any acceleration in the same direction as that of the wedge? all the normal forces act in the opposite direction

[@boranxiii]

you should think of it as one body.Because the direction they are going to is the same,so are the velocity and acceleration vectors.

[@alejandrodeharo9509]

the solutionary of physics sears zemansky book provider has a different answer

[@seanheeran5017]

why is r not mgcos0?

[@RexGalilae]

R should be equal to mgcos0+mAsin0 AFAIK. This is because R must counter both the force due to gravity AND the force due to the block's movement

[@foxtrot8325]

@@RexGalilae Yes, but that's like saying x + 1 = 11 - x instead of 2x + 1 = 11. You are right about the expression for R, but the acceleration A is written in terms of R.

[@RexGalilae]

@@foxtrot8325 My comment is over 4 years old, man. I was probably in my first year of college lmao Right now, I'm 1 year past graduation and in a line of work that has nothing to do with mechanics

[@foxtrot8325]

It is mgcos(θ) in an immovable wedge, do you know why? It is so because there is no motion of block perpendicular to the surface of wedge, so acceleration of block perpendicular to the surface is zero. Sine F=ma, and a=0 so F=0. That means force perpendicular to the surface of wedge is zero, and this means that the components of all forces perpendicular to surface should add up to zero or that opposite forces come out to be equal. Force on block away from wedge is R(the reaction force), and the component of gravity perpendicular to the surface of wedge is mgcos(θ), equating the two gives R = mgcos(θ). The same thing also applies in the problem with movable wedge, since there is no motion of block perpendicular to the surface of wedge(if you observe the block while standing on the wedge - that is in the reference frame of the wedge), you just equate the forces on the block going 'into' the wedge and 'out' of the wedge. But here in addition to the Reaction force(R) and gravity, we also have a pseudo force(due to us observing the block in an accelerating frame of reference). So just equate the forces and you will find that the reaction force(R) does not come out to be mgcos(θ). The interesting part is that in the expression for R, if you set M(mass of wedge) to be infinite, it says that R should then be mgcos(θ). This should be intuitive since an infinitely massive wedge is akin to a non movable wedge and so the expression for R just comes out to be the usual easy case that is R = mgcos(θ).

[@foxtrot8325]

@@RexGalilae lol I expected this. My comment might be useful for someone who stumbles on the same problem.

[@ModernTruthRevelation]

tam bir bakkal hesabı. şaka gibi. neyse gene de benzeri olmayan bir sorunun çözümü o yüzden teşekkürler.

[@HarshRajAlwaysfree]

oh the video with sound was here

[@HarshRajAlwaysfree]

Hello past self glad to see you again

[@barzipori8564]

An amazing job. thanks a lot!

[@arnavsingh5197]

AMAZINGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGG!!!!!!!!!!!!!!!!!1

[@ishaankulkarni1396]

This was great. Watched it for JEE advance revision

[@superdoer]

thanks!

[@batuhansaroglu7434]

acil İngilizce öğrenmem lazım

[@lakshgoyal1324]

He did not give clear understanding

[@debjitgangopadhyay7713]

Lol

[@rmsvideos1335]

Where the heck does mAcos theta come from around 10 min

[@oxygen2671]

In 6:19 Why he didn’t make the force on the opposite direction ?

[@lakshgoyal1324]

😡😡😡😡😡

[@sameeraggarwal782]

G is down....i don't care!! Damn😂

[@lakshgoyal1324]

He wasted my time

[@lakshgoyal1324]

poor explanation

[@deltaagartala6530]

total waste of 26 minute from my course.