Mass on a moving wedge
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- เผยแพร่เมื่อ 26 ธ.ค. 2024
- This video contains an excellent analysis to a rather hard problem: A mass resting on a wedge, which itself is free to slide on a horizontal plane. The problem is complicated because (a) there are two accelerations to keep track of, and (b) the kinematic constraint is not-so-easy to apply. Special attention to the careful use of a ghost force (which is fine!) in the accelerating frame.
Quiz 1: webtest.boun.ed...
Quiz 2: webtest.boun.ed...
Quiz 3: webtest.boun.ed...
Homework: webtest.boun.ed...
( The full playlist "Introductory Physics" is at
• Introductory Physics )
This guy deserves all the compliments he’s getting. He rly helped me.
you are right!!!!!!!
Around 8000 engineers and physicists of the future have benefited from this great video!
Great job and thanks a lot!!
Now 44000!
More than 100000....
It's 2022,Sep,2
80000
This so helpful I was stuck here for hours to figure out what was happening
Helped a lot... You have generalized everything and that will help me to solve any problem.. Thanks a lot!!
Hello bhaiya
You gave JEE 8 years ago
How is your lifestyle now?
I am going to give it in 2025
Sir be like, "Joh yeh tera torture hai ... woh mera warm-up hai"
Mujhe lag hi raha tha apna koi a JEE/NEET aspirant mil hi jayega comment section me?
@@ParthTyagi Hehe
Sir, thank you so much. Your video was immensely helpful.
Thank You very much explained a difficult problem very well.
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....
for such problem, is it possible to solve without using pseudo force (ghost force) by putting observer or frame of reference on ground.?
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
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.
@@autumn_auburn ohk thanks , It was an old comment , I have done it with vectors too, anyway thanks for explaining
@@mathOgeniusbro!! You're here?!!
thank u sooo much sir u helped me a lot to solve many of the tough problems
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?
did u figure out what was wrong, am having the same issue
Thnkiuu sir...it helps me alot to solve a problem...a great respect from...PAKISTAN
What's the professor's name? His explanations are amazing! Thanks a lot for sharing this super useful and insightful content. :-)
He is Alpar Sevgen from Bogazici University
Thanks, I have spent many time with this problem, without make a real progress, I understand a little more tanks to you.
This helped me a lot. Thanks :)
I had the same problem on a problem set, but I did it such that theta was the wedge's angle of inclination, and the "true course" of the mass sliding down would have a steeper angle, alpha. Is it necessary that alpha+theta=90 deg? It almost feels intuitively correct, but I'm having difficulty proving it... and I lost a few marks because of this, but I don't understand how it's wrong. Could you please clarify. Thx
sir please dont feel hesitated while teaching , U are good enough !
It is possible expressing the motion of m in terms of the relative motion, but there is a lot of algebra.
The best introductory physics course thanks
Sir why do we consider that acceleration of wedge is constant, when block is sliding over it. Can we prove it.
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?”
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?
Sir you are awesome ..... Keep posting even more videos...
Go to 5:50 to skip warm up
Awesome lecture sir, very well explain
Hats off to you, sir!
Thank you very very much.....You explained really well. May God Bless.....
In 6:19 Why he didn’t make the force on the opposite direction ?
How can we solve this question without using a pseudo force?
We can use a Lagrangian to model the problem, and then use Euler Lagrange Equations to solve it without pseudo force.
concept clearing video
why is r not mgcos0?
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
@@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.
@@autumn_auburn
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
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).
@@RexGalilae lol I expected this. My comment might be useful for someone who stumbles on the same problem.
Thank you very much Sir!
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?
Rex Galilae don't count me in though I'm a future scientist this video is a shit?!!!
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..
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?
That's why it's called a ghost force.
nice video...helped me a lot..
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
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.
@@boranxiii They are going at the opposite directions. One is moving left and the other is moving right.
Thanks alot sir 🙏💐
This was great. Watched it for JEE advance revision
Thank you!
just use psuedo force and convert it into a rested block
An amazing job. thanks a lot!
the solutionary of physics sears zemansky book provider has a different answer
tam bir bakkal hesabı. şaka gibi. neyse gene de benzeri olmayan bir sorunun çözümü o yüzden teşekkürler.
oh the video with sound was here
Hello past self glad to see you again
Where the heck does mAcos theta come from around 10 min
thanks!
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acil İngilizce öğrenmem lazım
Lol
He did not give clear understanding
G is down....i don't care!! Damn😂
😡😡😡😡😡
He wasted my time
poor explanation
total waste of 26 minute from my course.