You have no idea how much this has helped me. I've been having problems with multiple pulleys like those in problem 5 & 6 and I have national exams in a week! While watching this video, something clicked and I was able to calculate everything myself perfectly without looking at your calculations. Thank you sir, I really appreciate this.
Your video made these pulley problems so much more understandable and simpler! Thank you so much for explaining the solutions clearly and in an organized way.
Being brave enough to be flexible with your coordinate axes is a crucial skill to have in mathematics and physics :) It oftentimes reduces problem's difficulty enormously.
You can find acceleration easily by this trick Here : Net Force on the system / Total Mass of the system Tho you can't find the tension. Btw great video 👍
Thank you so much teach, this really helped me to comprehend pulley problems way better now. Especially how the use of a pulley can help with lifting big weights. Amazing video and explanation!
Not to be dramatic, but I have been crying for the past 24 hours, petrified of failing my physics exam because none of my study material was making sense to me till i found this video. I cant thank you enough!!!!
I liked your channel to make the total number of likes divisible by 3 made with three consecutive natural numbers including the smallest of the set. I liked then when this happened. For thinkers the number is 123 which I liked and made for his video. I am an IIT JEE aspirant and these problems I did in my first basic paper solving of nlm for pulley mass systems during my preparation. Edited to say thank you for helping me refresh and revise a bit. I am still preparing and hope to get selected the next time. Please pray for me😊
Thanks a lot man, even if I am from totally different field and trying to get another job which involves physics, I am getting the hang of these pulleys, slopes etc. thanks to you. Appreciate it
I really enjoy physics but the time that it was taking to figure out every single concept was concerning as my teacher is a bit... unclear? Finding your channel has given me hope in not dropping the entire subject! :) Thank you!!!
Hi, I appreciate your ways of solving those problems. However, I think, the question a) of Example 5 is a bit imprecise: You ask: what is the force for lowering the weight with constant speed of 2 m/s and your answer is m/2*g. The answer is principally correct because a=0 and v=const. But, if you apply this force, nothing would move, if I assume that tha weight was at rest. So, you first have to decrease this force to a value lower than m/2*g to produce an acceleration. Then the weight is accelerated from v=0 to a velocity different from zero. If the speed of the weight has reached v = 2 m/s due to this acceleration, then you can change the force to m/2*g and now v=const=2 m/s. In numbers: If you apply a force of m/4*g then the acceleration is -5 m/s^2 and it takes 0.4 s to reach v=2 m/s. If you apply 80% of m/2*g= 0.4*m*g, then a = - 2 m/s^2 and it takes 1 second to reach v=2 m/s.
after struggling from 2 months to understand this concept it is finally clear. as a student preparing for neet this is an amazing vid to clear your basics and are there faster methods to calculate the acceleration cause i gotta solve it under a min anyways thank you for the lovely vid
T1 not= T2, because in the past we've been assuming that the tension in the rope is uniform everywhere. In this problem the rope is not slipping and the pulley is not massless, the tension is not constant in the rope. The tension depend on the mass. If Mass1 > mass2, tension 1 > tension 2 in the rope.
Awesome video I have a question. Regarding question #1, if the bigger block is rendered immovable, is the tension on the rope the same as if it were moving?
Thank you so much for this video sir. I was struggling with these types of questions for a long time, I’m solving ib past papers that have to do with pulleys and I’m finally getting it.
Can you help me with this problem:In a container of negligible mass, there are 100 grams of water at a temperature of 80 degrees Celsius, 150 grams of water at a temperature of 10 degrees Celsius are added. After the thermal equilibrium is established, a piece of ice at a temperature of -25 degrees Celsius is added. Find the temperature of the water before adding the ice. , the heat needed to melt the ice, the final temperature of the water in the glass.
Hello! Thank you for making this video. But I have a question. In problem 6, letter a indicate that we should lower the mass with a constant speed of 2.0 m/s, hence a=0 m/s^2 Then the net force should be Fnet =ma =0 Fnet = 0 Mg - 6F =0 since the mass is lowered... Mg = 6f
In problem 5 & 6, what is the meaning of speed 2.0 m/s, any relation to the Tension force ?? Would it be the Tension force is the same when speed is 1.0, 2.0, 3.0 etc. ?? Does it mean the block can move for a long time if friction force in this force system can be ignored ??
Alternatively, you could treat it as one big object and directly add up/subtract the masses, ignoring tension for these problems. Lil shortcut to get to the same equation for a. Then pick a block to solve for T
I don't think the two M's will move, 'cause they are right next to each other and (even though the big M is heavier) they will be balanced, so the same forces acting on them. 6:59
4:20 why did we Substracted the weight from the tension how did we know that the weight is bigger because it might be the other way around because the inertia of the first object is bigger then the second object so the tension force would be bigger then the weight
He defined the downward direction as positive, and since the mg vector is downwards (positive) and the T vector is upward (negative), the equation is mg -T = ma (which can also be written as -T + mg = ma).
Five years later and you're still helping struggling students like me! Thank you so much for your work.
Please don't delete this video. It'll help a lot of people. Your teaching style is awesome!
Literally the best video ive seen on pulleys.
Yeah, it's really good :)
Scare to think that there are people out there who live for TH-cam vids re pulleys.
You can scrub 'literally' - it isn't doing anything.
Please kick me. All best.
@@notreallydavid, shut up, your being rude! @#$&!
I agree
You just cured my life time phobia of pulleys and FBDs. Thank you.
You must be adored by your community, need more ppl like you in this world
Love the way you explain and speak . These little things help ALOT
honestly a lifesaver man, i look at problems like these and think "theres no way i can solve this". Thanks for the video!
Undoubtedly a masterclass and literally the best video for pulleys on the internet. I can't find words on how to thank you mate
You have no idea how much this has helped me. I've been having problems with multiple pulleys like those in problem 5 & 6 and I have national exams in a week! While watching this video, something clicked and I was able to calculate everything myself perfectly without looking at your calculations. Thank you sir, I really appreciate this.
This video actually helped me understand pulleys better than my own class. Thank you so much!
Your video made these pulley problems so much more understandable and simpler! Thank you so much for explaining the solutions clearly and in an organized way.
Why the best people are so under-rated
Respect for you sir❤❤
You are amazing. I was strugling to understand this concept and you just explained it so well I didn't have to watch the full video. Thank you man.
Thank you so much!! This video has finally helped me understand pulleys after months of hitting dead ends while studying for the MCAT
i've been looking for a video like this for hours you're a lifesaver
Thank you very much. This is such a systematic and organized way of thinking and solving.
Thank you for the kind words! I greatly appreciate comments like yours
I have an AP physics exam in 30 hours, 20 MCQs and 3 FRQs. I feel invisible to any pulley problem.
you have no idea how much you saved my grade. tysm!
This is such a good explanation that you now have a new sub.
@Titus Solomon these sites don't work
you save me from fail physics exam fail thank you so much !
THIS VIDEO HELPED ME A LOT , COULDNT UNDERSTAND WHERE EACH FORCE COMPONENT CAME FROM BUT NOW I KNOW , THX SO MUCH
Bros a saviour to the society kudos to you
Great method. I used to be afraid of choosing downward as positive y-axis. Now I am not. Great job!
Being brave enough to be flexible with your coordinate axes is a crucial skill to have in mathematics and physics :) It oftentimes reduces problem's difficulty enormously.
@@PenandPaperScience you good
@@numberone6955 :D
@@PenandPaperScienceikr it would feel like a silly fear but as a beginner we all have been through these phases
You can find acceleration easily by this trick
Here : Net Force on the system / Total Mass of the system
Tho you can't find the tension.
Btw great video 👍
pully problems were my kryptonite in physics class but with your videos, this helps resolves it keep up the great content
Thank you so much teach, this really helped me to comprehend pulley problems way better now. Especially how the use of a pulley can help with lifting big weights. Amazing video and explanation!
this is soo good it makes me wanna cry.....helped a lot and i have a test tomorrow tnxxx
Not to be dramatic, but I have been crying for the past 24 hours, petrified of failing my physics exam because none of my study material was making sense to me till i found this video. I cant thank you enough!!!!
You got this!
honestly , the best pulley video on the platform ...
mister it will more useful if you have explained the x y direction and cos thing more in depth
👍
Just wonderful. Easiest way for solving the pulley problems..watching from BANGLADESH
how can you not like this video, this man is awesome💙
Can't understand why you have only 17.9k subs only
I think that quite a lot for a physics channel. Thanks for the the support. I'll get there.....eventually.
I liked your channel to make the total number of likes divisible by 3 made with three consecutive natural numbers including the smallest of the set. I liked then when this happened. For thinkers the number is 123 which I liked and made for his video. I am an IIT JEE aspirant and these problems I did in my first basic paper solving of nlm for pulley mass systems during my preparation. Edited to say thank you for helping me refresh and revise a bit. I am still preparing and hope to get selected the next time. Please pray for me😊
Bhai aapka JEE ka exam ho gya??
I like the animation.
the lecturers intonation is not sleepy to hear.
Very well explained how and why
May god bless you physics ninja. this had helped me a lot in solving past papers.
Of what???
@@adityajha4936 physics
@@fatimaarif8392 Hmmm
The best video for pulley questions 💯💯
you are a lifesaver, idk why my lecturer won't explain the steps but now I know ty king
You got this!
Thank God for you. My teacher hasn't explained any of this.
Thank you for this video, you explain this concept so clearly and simply!
Glad it was helpful!
U ve got the art of teaching man...❤
This channelll is underrated ✨ wish a few billion subscribers ahead! 😌
Thanks a lot man, even if I am from totally different field and trying to get another job which involves physics, I am getting the hang of these pulleys, slopes etc. thanks to you. Appreciate it
You can do it!
I really enjoy physics but the time that it was taking to figure out every single concept was concerning as my teacher is a bit... unclear? Finding your channel has given me hope in not dropping the entire subject! :) Thank you!!!
That was the best explanation!
Glad you think so!
@PhysicsNinja I have a question..in two pulley prblm ..if we pull the string with force F then wouldn't the tension forces be in opposite directions?
Thankyou for the short and sweet explanation !
Do you have a full explanation of this Topic?
The best explanation I have ever seen, thank you bro
Thank you so much!
Thanks you man you are best from all our indian teachers
You explained it very easily
So nice of you
Hi, I appreciate your ways of solving those problems. However, I think, the question a) of Example 5 is a bit imprecise: You ask: what is the force for lowering the weight with constant speed of 2 m/s and your answer is m/2*g. The answer is principally correct because a=0 and v=const. But, if you apply this force, nothing would move, if I assume that tha weight was at rest. So, you first have to decrease this force to a value lower than m/2*g to produce an acceleration. Then the weight is accelerated from v=0 to a velocity different from zero. If the speed of the weight has reached v = 2 m/s due to this acceleration, then you can change the force to m/2*g and now v=const=2 m/s. In numbers: If you apply a force of m/4*g then the acceleration is -5 m/s^2 and it takes 0.4 s to reach v=2 m/s. If you apply 80% of m/2*g= 0.4*m*g, then a = - 2 m/s^2 and it takes 1 second to reach v=2 m/s.
please dont delete this shit, still strong in 2024
Please upload more videos for also other topic it very helpful sir liked it
The best video i have ever seen on pulleys, appreciate from brazil
Man you are a life saver. I have been acing calc 2, but I was struggling so much with pulleys lol
You got this!
You can also include the infinite pulley problems and the concept of equivalent mass.
after struggling from 2 months to understand this concept it is finally clear. as a student preparing for neet this is an amazing vid to clear your basics and are there faster methods to calculate the acceleration cause i gotta solve it under a min anyways thank you for the lovely vid
thank you soo much for this explanation, I couldn't understand my teachers explanations at all, but thanks to this I have been enlighten.
who would not like it. you are very helpful. Thanks alot
You're welcome!
Wow all pulleys in one place helped a lot
I have a physics olympiad in 2 days and this really helped understand pulleys :)
Good luck! You’ve got this
Is it good for International Olympiads???
Is it enough for physics olympiad??
Superb Explanation !!!
Thank you for such a great explanation!!
omg thank you for explaining is so simply and not skipping your train of thought!
Glad it was helpful!
Sir your voice is so cool! I can hear it all day🛐
You can feel the Tension between the blocks in his vohhice
great video.
T1 not= T2, because in the past we've been assuming that the tension in the rope is uniform everywhere. In this problem the rope is not slipping and the pulley is not massless, the tension is not constant in the rope. The tension depend on the mass. If Mass1 > mass2, tension 1 > tension 2 in the rope.
Awesome video
I have a question. Regarding question #1, if the bigger block is rendered immovable, is the tension on the rope the same as if it were moving?
This is soooo helpul cause its free
Thank you so much, my school uses these models all the time and this will help massively
Thanks for the video it really helps me understand p6
Glad it helped
Thank you so much for this video sir. I was struggling with these types of questions for a long time, I’m solving ib past papers that have to do with pulleys and I’m finally getting it.
now this is a masterpiece 🌟
Amazing videos. Really helps to recall the concepts. Thanks a lot. In problem 5a, isn't it better to state constant velocity?
One of the best tutors 🔥 🔥
good explanations, just wish you added the force of frictions as well.
You're a genius mannn.. ❤❤
Can you help me with this problem:In a container of negligible mass, there are 100 grams of water at a temperature of 80 degrees Celsius, 150 grams of water at a temperature of 10 degrees Celsius are added. After the thermal equilibrium is established, a piece of ice at a temperature of -25 degrees Celsius is added. Find the temperature of the water before adding the ice. , the heat needed to melt the ice, the final temperature of the water in the glass.
Hello! Thank you for making this video. But I have a question. In problem 6, letter a indicate that we should lower the mass with a constant speed of 2.0 m/s, hence a=0 m/s^2
Then the net force should be
Fnet =ma =0
Fnet = 0
Mg - 6F =0 since the mass is lowered...
Mg = 6f
Just like in problem 5
In problem 5 & 6, what is the meaning of speed 2.0 m/s, any relation to the Tension force ??
Would it be the Tension force is the same when speed is 1.0, 2.0, 3.0 etc. ??
Does it mean the block can move for a long time if friction force in this force system can be ignored ??
Hii I have a question why we don't consider the rotational motion of pulley in this video???
May Allah bless you and your family this is sooooo helpful
Sir, your are great
In my class room ,I am responding in good way why because
I am listening your class
Thank you sir
Thank you so very much may God bless you sir
Alternatively, you could treat it as one big object and directly add up/subtract the masses, ignoring tension for these problems. Lil shortcut to get to the same equation for a. Then pick a block to solve for T
❤️ from india 👍 fabulous video it clear all my doubt
in question 6a, would there be a difference if i wanted to lower it a constant speed of 8m/s instead of 2m/s?
Exactly what I was wondering, same with the question before it, like the speed value was never used in the problem.
Awesome video. Thank you!
for problem 6, what if two of the pullies are NOT connected to the block itself? Would it only be 4*Tension applied on the block then?
Thank you very much
Great explanation 🤝
understood all the concepts clearly thank u very much 😄
May please also calculate the time
Thank you so much I had difficulty understanding tension and this video cleared it up
I don't think the two M's will move, 'cause they are right next to each other and (even though the big M is heavier) they will be balanced, so the same forces acting on them. 6:59
Which software do you use??
thank you!!!!! these questions were the exact ones in my paper!!
Please share video on blocks and springs
Thank you so much, this video is very helpful!
On the first example, why is the second tension negative if it’s pointing up?
wooooow !! thanks p6ninja
4:20 why did we
Substracted the weight from the tension how did we know that the weight is bigger because it might be the other way around because the inertia of the first object is bigger then the second object so the tension force would be bigger then the weight
He defined the downward direction as positive, and since the mg vector is downwards (positive) and the T vector is upward (negative), the equation is mg -T = ma (which can also be written as -T + mg = ma).
you're a saint