Pseudo Force & its applications | NLM for JEE & NEET | Class 11

Topics covered in Lecture 22 of Pseudo Force & its applications in chapter Newton's Laws of Motion for JEE & NEET in Onion Physics are listed below with timestamps:
Timestamps:
00:00 Pseudo force
03:00 Pseudo force explained on a boy on accelerating truck
03:56 Inertial and non inertial reference frame
05:31 Pseudo Force on block as seen by boy on truck
08:21 Why pseudo force is important
09:13 Important points about Pseudo force
10:21 Pseudo force on a block over a triangular wedge
13:57 Acceleration of wedge for block to remain at rest on wedge
16:36 Normal reaction by wedge on block
18:31 Force on wedge to keep block at rest over it using pseudo force
21:01 Pseudo force for pulley block system on a wedge
25:47 Free Body Diagram and Pseudo force on a pulley block system on wedge
Link for PIM videos on Pseudo force:
Pseudo Force in Newton's Laws for JEE & NEET (explained): • Pseudo Force in Newton...
Pseudo Force Application & Reference Frame for JEE & NEET: • Pseudo Force Applicati...
JEE and NEET aspirants must crucially work along the concept of Pseudo force (Class 11 Physics, Chapter 5) as it cuts out many problem solving steps while working on problems. Pseudo force will help JEE and NEET aspirants come a step ahead of the competition by providing an alternative faster method to solve numericals at a fast pace.
Onion Physics is a FREE You Tube series for High Quality Online Physics Classes for JEE & NEET preparation. In this series, complete physics is taught through Illustrations based on different physical situations. JEE Main and NEET are competitive exams in which about 50 percent of questions are based on applications of physical situations and the rest are based on memory and fundamental analytical concepts. In JEE Advanced all questions asked are purely on applications of concepts in simple as well as complex experiments and physical situations. Onion Physics is focused on learning through examples and different physical situations which makes it the best online jee coaching available in Physics.
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Class 11 Revision Checklist for JEE Main & NEET : • Motion in One Dimensio...
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#rigidbodydynamics #rotationalmotion #jeemain #jeeadvanced #neet #jeepreparation #neetpreparation #onionphysics

[@physicsgalaxyworld]

Catch the perfect explanation of Pseudo force in 3min: kzbin.info/www/bejne/i3XNi2xpq86rmKs Watch Pseudo Force Application & Reference Frame for JEE & NEET: kzbin.info/www/bejne/poDYaYenndloitk For views, queries and suggestions about upcoming videos on our channel, viewers can reply to this comment. All replies to this comment will be answered soon !

[@PrinceGupta-kr6qz]

Sirrr please increase the frequency to 6 in a week plsss ssss pls sss sirrrrrrr

[@vinay3726]

Sir plzz increase the video frequency , 1 lecture everyday will be very helpful for 2021 aspirant or u can plzz cover the more topics in 1 video its a request sir 🙏🙏

[@madhavdhage1643]

28:09 there should be a Tcos(θ) rightwards↦ in calculation of a wrt wedge right???!

[@vishalpandey5326]

It's more interesting when you know the concepts..

[@soumodeepchatterjee3248]

Yes sir that's correct. And I think that's why sir tells to watch the concept videos already before watching the onion physics lecture although I just read the theory part from book and even that helped to improve my understanding.

[@bodiminds6571]

Agreed , used abj sirs lectures , solved a few problems in competishun sheet and now watching this gold

[@varunbaggan1912]

Every Time Same answer comes both with pseudo force & Eq. of Motion method 👍 Thnks Sir ❤️

[@krishnajadhav9527]

Thanks sir ...your this free teaching like at home is going to a help a lot for those students who can't afford coaching like me!

[@RajKumar-rz5ww]

Tu toh bhikhari nikla

[@akashkumarsaha5191]

Haa bhai,ma bhi.❣️👍

[@carryminnnnati]

bro what doing now ??

[@undisputedrush8463]

He is not only an teacher 😃 but also a mentor 😃😃.

[@loser3357]

Huehue 😋😋😋🤩🥳😘

[@hayatkhan4700]

Sir please explain, in the last case, why in horizontal direction tere was no relative acc of block of mass m wrt mass 2m?

[@jaishreeram5808]

THANKS AND LOVE YOU SO MUCH SIR...MJA AAGYA ✍🏻✍🏻✍🏻✍🏻✍️

[@TanushKarkade]

Yes sir got the answer by traditional method too Thank you sir

[@7_thala_for_a_reason]

yes I have got the same answer using traditional method pulley system question

[@244maneerafsan2]

sir onion physics is amazing

[@YdvSyAero]

Amazing amazing and amazing

[@RajnishKumar-jh9ql]

23:46 yes sir i am getting the answer in ground frame

[@sahilshrivastava5398]

24:40 yes sir, i have got the answer from traditional method too🙂🙂🙂

[@rishabhpalsingh2644]

Hello brother can you tell how to do it by traditional method B ki fbd smjh nhi aarhi

[@granth6665]

how

[@i.anupamalok]

Yes ,I have got the same result, sir

[@vishalgoswami1785]

S3: Let b be the acceleration of block and acceleration of system be a F=7ma For blocks: mb=T+ma 2mb=2mg-T Adding them: 3mb=2mg+ma Now they are rest realative to C, b=0 2g=a Hence F=14mg

[@chiragbirla5606]

Best level

[@abhinavarya7752]

S3 ko traditional method se kaise solve karey?

[@poojasonawane7181]

24:00 done by traditional method

[@advaitk624]

same answer with traditional method too sir :) 25:00

[@himanshuacharya215]

Thanks sir

[@bijaykryadav7906]

How we will get acc of both block as we can't use a and 2a as acc of block as sir said it is with reference to each other

[@sujaldebnath01]

Sir why you write 2mg in downward direction?

[@krishnajadhav9527]

Thanks sir!

[@aakankshajain8988]

Thank you sir Please sir thoda adv 2021 le baare mai bhi btaye

[@rashimamittal2281]

mujhe sir se zyada acha sir ka poogy laga

[@sujaykarmakar7349]

23:07 sir why normal force don't pass through the centre of the block... pls explain sir

[@mdsohelraja285]

It's perpendicular to the centre of block. Sir has just drawn the arrow sign a little downward so as to keep the diagram neet and clean.

[@sujaykarmakar7349]

@@mdsohelraja285 okay

[@sujaykarmakar7349]

@@parthpariwandh okay thnx

[@julichaudhary8491]

Thanks sir 👍

[@user-zt6co9hk5z]

error at 21.00 ans is 60/root3

[@user-zd6yi6rw2i]

Yes yes i have got the answer s3

[@nikitasharma7209]

21:24 same quest come in jee mains 22

[@PrinceKumar-le8nd]

Yes sir I got the same result 😁😁

[@yogijnuite]

H.w IL2 answer =15kg

[@Abhishek-vu4yh]

sir also start for class 12

[@yogijnuite]

H.w IL1 answer =20kg

[@Physicslover596]

30:58 Sir ground frame se bhi kia but 4 eqns and 4variable arahi hai and use solve krne me thodi dikaat hui!

[@shivannagk7537]

Eqns type karke bhejdo please 🥺

[@vaishnavisingh6654]

Got it by traditional method

[@borntobehuge]

30:00 anyone got the same equation of block by traditional method ?

[@miwi7502]

I got the equation for wedge but I’m unable to get the right one for the small mass m. What about you?

[@vishnuksharma]

13:30 what if force is more than (M+m)gtan thetha?

[@TheAditya-zt9pb]

Yes i am getting same result with traditional method

[@avneeshyadav6814]

I got same result

[@granth6665]

21:56 cant get ans by normal method

[@xc8684]

sir apne to bola tha pseudo force is the force on A wrt B. So agar block wedge pe h or usko wedge ke respect mai rest mai rhna h then. block ka net acceleration wedge ke respect mai 0 hoga to pseudo force to 0 aega.

[@karankumarsharma_4410]

Same doubt

[@TMT05]

being in rest means net force is 0. pseudo force is one force that acts on block when seen by viewer on wedge. other force that acts on block is gravity. components of gravity and pseudo force along incline cancel each other and thus net force and net acceleration on block=0. as seen by viewer. this is where condition comes that magnitude of acceleration of wedge must be certain value for pseudoforce to = gravity component

[@RohitRajChaudary]

Sir but in non inertial frame third law is valid 4:56

[@AyushYadav-tc4xt]

24:40 sir done using traditional method !

[@subhamsouravbehera4091]

Hey bro tumne traditional method se kese solve Kiya can you please 🙏🏻🙏🏻 help me ??? I am unable to solve all problems in traditional method please bro !!!!

[@insanefacts8190]

sir without pseudo ho gaya

[@anayraj3315]

sir i m jain how can i consume ur content

[@Akash-wb9mq]

28.18 how the TT got cancelled????

[@dbdoxbxb]

Opposite directions and acting on same body

[@Mohit-eq3hz]

I have solved it by traditional method, I noticed the total force on A in right direction it is (T+ma) and the total acting on B in downward direction it is (2mg-T) these must be equal So, ma = 2mg Hence a = 2g and putting it in F = 7ma we got the answer 😊😊😊😊

[@dhruvgupta8146]

Kya koi batayega ki relative acceleration wale question mein kyun aisa koi pseudo ka concept nhi hai

[@subhamsah1601]

Sir non lnertial frame me Newton's third valid hota he

[@Alphamatics1234]

Non inertia frame ma Newtonian laws kam nahi karta

[@rohangupta3363]

sir who is pogg1? app kise thanks bolte hai bar bar? btw very poerful lecture

[@armanagrawal5548]

jo board clear kar rha hai

[@ngxftbl]

Pseudo force fayda: 4to8 marks in jee main and adv.😅😂

[@Vidy884]

Ye pogi kon h yr bich me ud ke Jo aata h

[@LakshmiDeepakChalla]

Poji is a fictional character in Ashish sir class

[@yasharora6772]

Jee main 2022 que 21:28 thanks sir

[@hyp-en]

24:43 able to get the answer by traditional method ❤

[@user-cv9hq2un5x]

29:39 virtual work se answer nahi ha raha 👎

[@xyz...8475]

Yes bro I use work tension method to find a and I get different acceleration

[@abhinavarya7752]

No one has dare to solve S3 by traditional method. If anybody has dare then write.

[@prince.j.]

Very true😂

[@abhinavarya7752]

@technicaldreams2509 you too need to worry about your future bro. So you too kindly solve.

[@abhinavarya7752]

@technicaldreams2509 Mee too bro. Look when I commented!

[@user-wq3lk3pu7x]

It is easy to solve with traditional method you just need to take T = 2mg from block B as there is no acceleration in vertical direction then fron block A ( T = 2MG ) ma = 2mg as there is no force acting on block A other then T so then if we solve both equations we get ( a= 2g ) and then put a and m of whole system in eq. F = MA And you have your answer buddy ..

[@solenoid6545]

Bsdk apne kaam se kaam rakh na. Sabko challenge kyu kar Raha h. Sab teri tarah nahi ki ho jaae to comment karke sabke saamne hero ban ne ki koshish kare.

[@rudranshsharma6913]

a=2gsin(theta) /7 2a=4gsin(theta) /7 N=mg/7[cos(theta) (7+2cos(theta)) -2] T=[ 3mgsin(theta) +mg sin2(theta) ]/7 The answers to the last question....

[@dbdoxbxb]

2gsintheta/3 hoga shayad

[@TMT05]

hi. can u revisit this problem please lol when i solve with pseudoforce i get a=2gsin(theta) /(7-4cos theta) but with conventional method I get a=2gsin(theta) /(7-2cos theta)

[@ShauryaAggarwal24]

@@TMT05 what process did you use for the convential method? beacuse i got the same answer as you using psuedoforce

[@loser3357]

😋😋😋😋🥰😍🤩🥳😘

[@ramtokas0001]

Sir ji kbhi has bhi liya kro 😂

[@sandeepshah101]

Sir poji kon... 🧐😐

[@manthansagar200]

Bhoot

[@ambujyadav5978]

sir ka sex slave

[@Freezeali]

Sir very punctual about ur time

[@aryangupta904]

sir pls itna clear krdo ki If a block is kept on a moving wedge (with accelration a) then if we have to draw the Free body diagram of block i) with respect to ground will we consider pseudo force? ii) if ques asks to draw the free body diagram of block with respect to wedge(moving with acc a) will we consider pseudo force now? sir in dono ka ans dedo pls

[@pallavigupta4408]

1st case me no 2nd case me yes