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In Situation 2nd , answer would remain same !! Cause ,with any reference frame , it's weight will comes to be same ...am I right !! Respected Sir nd pg team , a lot of questions in my mind ,how can I ask them ?...I never get any reply on my commented questions !!...
Vro sal bhar mai galt teacher se fas gya like nkc sir uncedmy now for adv I start these concepts lec with adv question practice chapter wise adv tak complete hojaega mughe barosa hai mera accha ho jaega kyuki dropper bhi hu aur fast capture kar parha pahle se kafi portion kiya hai thankyou sir my first comment ❤️ sir half hour me hi itna portion bata dete jitna boht teacher 1.30 min me explain k KArte usme me kuch skip bhi kar.dete sirs lecture is so beneficial
I want some idea of first question... Tell me where I am wrong if person has acceleration "a" then that rope will have acceleration "2a". Then if me make equation by supposing that whole system is going upward. 2T-60g=60(2a) Now we just need to calculate "a" to calculate weight by N-40g=40a And by N we can find weight but for "a" we need another equation to find T and I am not getting that second equation please help urgent...
Sir Can we also explain the upward acceleration of elevator situation as when lift is accelerating upward, every instant the inertia of body is trying to keep it in its previous position or previous speed, so that leads to an increased weight on the weighing machine, and vice versa for downward?? Kindly reply and tell if this explanation is correct.
Just resolve the components mgsin(x) = Nsin(x)cos(x) along the incline and you will get the normal force Or you can apply newton’s 2nd law along the line parallel to ground : after applying equation will be Nsin(x) = ma And since mg is perpendicular, therefore its component along the line would be 0. Hope it helps!
@J Swati First take only the man system individually and write apply the NLMs u will get the eqn as T+N=600, then u can consider the whole sytem and apply NLMs u will get an eqn like:- T=200+N. Then u can solve and get N=200, thus reading=N/g=200/10=20 I considered g=10m/s^2
simple. only difference is that for ground reference you resolve block's forces in FBD along and perpendicular ground horizontal. in pseudoforce we resolve along and perpendicular incline surface. so for ground frame we get mg=n cos theta means n=mg/cos theta=15g/(root3/2)=30g/root3 for weighning machine reading divide by g and answer is 10 root 3
1 boy will apply a force of 30g downward to keep the lift at rest T+T=40g+20g,T=30g Therefore,boy applies 30g downward,and also 40g downward due to gravity 30g+40g=N Therefore,N/g=70
@@s_shiva30g will be upwards for fbd of boy because in fbd t is used and force applied by boy ,answer is coming 10 for me because 30g - 40g - N =0 (equilibrium) N = 10 g
boy will apply a force of 30g downward to keep the lift at rest T+T=40g+20g,T=30g Therefore,boy applies 30g downward,and also 40g downward due to gravity 30g+40g=N Therefore,N/g=70
boy will apply a force of 30g downward to keep the lift at rest T+T=40g+20g,T=30g Therefore,boy applies 30g downward,and also 40g downward due to gravity 30g+40g=N Therefore,N/g=70
boy will apply a force of 30g downward to keep the lift at rest T+T=40g+20g,T=30g Therefore,boy applies 30g downward,and also 40g downward due to gravity 30g+40g=N Therefore,N/g=70
Because if it is winding continuously at the rate of 10m/s2 then how to find weight of man on weighing machine because he pulled up by string if it is not in equilibrium
@J Swati boy will apply a force of 30g downward to keep the lift at rest T+T=40g+20g,T=30g Therefore,boy applies 30g downward,and also 40g downward due to gravity 30g+40g=N Therefore,N/g=70
eqn for lift : T-(N+10g) = 10(winding acceleration/2) -(i) eqn for boy : T+N-30g = 30(winding acceleration/2) -(ii) Solving above eqns, we get: N=150, so reading =15kg. Hope this helps.
@@nobarakugisaki_._ bro see: The winding machine is itself a part of the system which doesn't has it's mass itself. Think as if there is no winding machine but the length of the string is getting short at the rate of 10m/s². Then by conventional method, if platform has to go by x distance up then string will be reduced by 2x length. Now if you look at the problem then, don't you think the platform from left side would try to fall due to gravity while that of right side would be attached. So, One assumption is there that the boy's feet is sticked to the weighing machine and weighing machine is also sticked to the platform. By this logic there would be a kind of cohesion and adhesion force in reality. But we ignore them here for simplicity. By this way, only one tension would act on the platform on right side due to winding machine.
boy will apply a force of 30g downward to keep the lift at rest T+T=40g+20g,T=30g Therefore,boy applies 30g downward,and also 40g downward due to gravity 30g+40g=N Therefore,N/g=70
boy will apply a force of 30g downward to keep the lift at rest T+T=40g+20g,T=30g Therefore,boy applies 30g downward,and also 40g downward due to gravity 30g+40g=N Therefore,N/g=70
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Sir I'm confused. I should learn physics from my coaching or from onion physics?
Sir am not getting your DPP yet
Sir plzz comment the answers of homework illustration 1,2
In Situation 2nd , answer would remain same !! Cause ,with any reference frame , it's weight will comes to be same ...am I right !!
Respected Sir nd pg team , a lot of questions in my mind ,how can I ask them ?...I never get any reply on my commented questions !!...
hats off for ashish sir for saying thank you to poji every time it erases the board
I am regretting why I got know this channel late.but also happy that I haVe some time. This is pure gold for jee aspirants. Believe me
It's not I'm regretting, it's I regret
If I had known about this channel earlier I wouldn't have got problem in solving mehchanics questions. Thanks sir.
Vro sal bhar mai galt teacher se fas gya like nkc sir uncedmy now for adv I start these concepts lec with adv question practice chapter wise adv tak complete hojaega mughe barosa hai mera accha ho jaega kyuki dropper bhi hu aur fast capture kar parha pahle se kafi portion kiya hai thankyou sir my first comment ❤️ sir half hour me hi itna portion bata dete jitna boht teacher 1.30 min me explain k
KArte usme me kuch skip bhi kar.dete sirs lecture is so beneficial
@@harsh_katiyr 1st attempt kaisa hua bhai
10:28 Ashish sir being ashish sir😂, this is one fine situation based example
😂😂😂😂
HW1- 10 kg
HW2- 15kg
Ya got same
Hey,can u say me how did u get the answers??
HWIL - 1, reading = 10 kg and HWIL - 2, reading = 20 kg
i have got the same answers as well!
@@SamayKrishna-y7m hi, which standard are you in
@@siddharthm7384 XI
can u tell idea for 1st question?
I want some idea of first question...
Tell me where I am wrong if person has acceleration "a" then that rope will have acceleration "2a". Then if me make equation by supposing that whole system is going upward. 2T-60g=60(2a)
Now we just need to calculate "a" to calculate weight by
N-40g=40a
And by N we can find weight but for "a" we need another equation to find T and I am not getting that second equation please help urgent...
Thank you very much sir In leacher se meri physics ke sare concept strong ho rhe h again thank you
Refer pathfinder
Sir, I promise you that I will get 100 percentile in physics in JEE Mains and Advanced. Please bless me to achieve my goal.
All the best Bhai
Kis class me ho?
@@takkutiwari1 Bhai class 11 me hu
did u get big brother?
Sir Can we also explain the upward acceleration of elevator situation as when lift is accelerating upward, every instant the inertia of body is trying to keep it in its previous position or previous speed, so that leads to an increased weight on the weighing machine, and vice versa for downward??
Kindly reply and tell if this explanation is correct.
Nice method to get more score
Sir plzz Homework IL 2 ka solution bta do plzz!!!! Confirm kr do
sir pls tell should we make notes in between of lectures or after full lectures
can be easily made in between and sir also prefers making notes simultaneously as it saves our time .
@@divyanshuverma552 thanks
Sir ji says 13:53 ground se bath ke solve karna hai.. Mean while I solve that problem on ground of my park
Abhi bhi baithe ho kya?
@@Kajal-l2c9k aapke liye to hum khade hai ;)
Thankyou pongeee is constant 😂😂
Sir how can i get previous lecture. Pls help me.
Thanks
15:16 sir I solved this without pseudo but
Kgf is coming m/cos^2 theta
Is this series stopped?
15:25 koi isse trad. Method se kaise solve hoga bta do 🙏🏻🙏🏻
Just resolve the components mgsin(x) = Nsin(x)cos(x) along the incline and you will get the normal force
Or you can apply newton’s 2nd law along the line parallel to ground : after applying equation will be Nsin(x) = ma
And since mg is perpendicular, therefore its component along the line would be 0.
Hope it helps!
Last question answer - 10 and 25
HWIL-1-10KG
HWIL-2-20KG
@J Swati First take only the man system individually and write apply the NLMs u will get the eqn as T+N=600, then u can consider the whole sytem and apply NLMs u will get an eqn like:- T=200+N. Then u can solve and get N=200, thus reading=N/g=200/10=20
I considered g=10m/s^2
Like for same answer
HW IL-1 reading =5kg
HWIL -2 reading =20kg
Is it OK
HW ka solution de dijiyega plzz
thanks sir
Kisi ki bhi dpp download hogyi plz send me
This is my gmail --- niveshkr149@gmail.com
Send karo yrrr
13:55 how do you solve such questions with ground reference
simple. only difference is that for ground reference you resolve block's forces in FBD along and perpendicular ground horizontal.
in pseudoforce we resolve along and perpendicular incline surface.
so for ground frame we get mg=n cos theta
means n=mg/cos theta=15g/(root3/2)=30g/root3
for weighning machine reading divide by g and answer is 10 root 3
@@TMT05 thank you
10:18 😂😂😂😂😂😂
(T+N)-30g=30a
(40+N)-T=10a
1
boy will apply a force of 30g downward to keep the lift at rest
T+T=40g+20g,T=30g
Therefore,boy applies 30g downward,and also 40g downward due to gravity
30g+40g=N
Therefore,N/g=70
@@s_shiva30g will be upwards for fbd of boy because in fbd t is used and force applied by boy ,answer is coming 10 for me because
30g - 40g - N =0 (equilibrium)
N = 10 g
Sir pls tell me kis sequence me apke videos ko dekna hai
Any sequence.... Every lecture is independent
Sir homework ka answer bta do plss!???
22:17 [20kg]
mera bhi yahi ans aaya par sahi he ya galat hai ye pata nhi chal rha hai.....
the eq'ns i got are
N+T=40g
T=20g
BY SOLVING I GOT READING=20.4kg
Bro you didn't consider normal force exerted by weighing machine on system (cabin)
HWIL-10Kg HWIL-15Kg
10:29 😂
HW2 25 kg
sir dpp is not download
hw1 : 70 kg
Hw2: Boy will lose contact with the surface the the string is winding upon the machine. Hence N = 0
Can anyone tell me why Sir keeps the onion lectures unlisted on RUclips ??
Bcz this series is for dedicated student not for utube roamers
@@vinay3726 ohh fine 👍🏻👍🏻
HW IL 1). 10 kg
HW IL 2). 15 kg
Is it right?
boy will apply a force of 30g downward to keep the lift at rest
T+T=40g+20g,T=30g
Therefore,boy applies 30g downward,and also 40g downward due to gravity
30g+40g=N
Therefore,N/g=70
@@s_shiva in FBD we consider forces applied on body, not the forces applied by body.
Same
HWIL 1 - 10Kg
HWIL 2 - 20Kg
Same bro
Par confirm nahi ho pata ki kiska sahi hai
Glt h 15 kg h 2 ka
Correct
@@vinay3726 tere baap ne bola tha
HWIL1:10kg
Hey tumhara selection hua remember you from pahul sir's session on vedantu
Check out 23:00 HWIL -2 ANS IS 25KGF
Sir can you please make some revisions checklist for jee advanced plz sir it's our humble request 🙏🙏🙏
Done with pseudo force
SIR PLEASE DO SOLVE OR ATLEAST TELL THE ANSWERS OTHERWISE IT WILL NOT BE BENEFICIAL IF WE GET WRONG ANSWERS ,PLEASE DO TELL!!!!!
Sir thoda aur samjha kar padhaya kigeye please sir utna jaldi hum chijo ko catch nahi kar pate hain
pleace says
1-20kg
Ist answer is 80/3 and 2 and is 80
Hwil 1 -10 kg
Hwil 2- 20 kg
boy will apply a force of 30g downward to keep the lift at rest
T+T=40g+20g,T=30g
Therefore,boy applies 30g downward,and also 40g downward due to gravity
30g+40g=N
Therefore,N/g=70
mujhe to sir ka board h nhi smjh aata h smart h ya marker wala h😂😂🤔
Hw1 :10kg
Hw2:20kg
boy will apply a force of 30g downward to keep the lift at rest
T+T=40g+20g,T=30g
Therefore,boy applies 30g downward,and also 40g downward due to gravity
30g+40g=N
Therefore,N/g=70
HW L2 reading = 25 kg (approx)
Pls confirm
Can you pls tell me how you got it?
Bro is illustration second hw is in equlibrium state
Because if it is winding continuously at the rate of 10m/s2 then how to find weight of man on weighing machine because he pulled up by string if it is not in equilibrium
@@Siddharth2407 yup it is not in EQUILIBRIUM
HW IL 1=10kg
HW IL 2=25kg
Kisi ki bhi dpp download hogyi plz send me
This is my gmail --- niveshkr149@gmail.com
Send karo yrrr
How HWIL - 2 ans is 25
HW IL1 Reading =10 kg
Kisi ki bhi dpp download hogyi plz send me
This is my gmail --- niveshkr149@gmail.com
Send karo yrrr
@J Swati boy will apply a force of 30g downward to keep the lift at rest
T+T=40g+20g,T=30g
Therefore,boy applies 30g downward,and also 40g downward due to gravity
30g+40g=N
Therefore,N/g=70
@J Swati i also got the same answer
HWIL-1=10KG
HWIL-2=15KG
How did you get 15 for 2nd question?
How 15kg.?
eqn for lift : T-(N+10g) = 10(winding acceleration/2) -(i)
eqn for boy : T+N-30g = 30(winding acceleration/2) -(ii)
Solving above eqns, we get: N=150, so reading =15kg.
Hope this helps.
How u wrote eqn for boy ?
Why the acceleration of both will be half anyway? By constraint this is still the same..
@@nobarakugisaki_._ bro see:
The winding machine is itself a part of the system which doesn't has it's mass itself. Think as if there is no winding machine but the length of the string is getting short at the rate of 10m/s². Then by conventional method, if platform has to go by x distance up then string will be reduced by 2x length.
Now if you look at the problem then, don't you think the platform from left side would try to fall due to gravity while that of right side would be attached. So,
One assumption is there that the boy's feet is sticked to the weighing machine and weighing machine is also sticked to the platform.
By this logic there would be a kind of cohesion and adhesion force in reality. But we ignore them here for simplicity.
By this way, only one tension would act on the platform on right side due to winding machine.
@@RahulKumar-jl9ki your query is answered in last part of my reply to @Nezuko.
@@nobarakugisaki_._ also try to get acceleration of both by unconventional method that is work energy method. It's quite easy. You'll get it.
hwil 2 33.34 kg
ans of 1 hw question is 10 kg
HW IL 1 = 10 kg
HW IL 2 = 35 kg
60kg
10kg
Answers:
HWIL-1: reading= 60kg
HWIL-2: reading= 60kg.
boy will apply a force of 30g downward to keep the lift at rest
T+T=40g+20g,T=30g
Therefore,boy applies 30g downward,and also 40g downward due to gravity
30g+40g=N
Therefore,N/g=70
Daaru peeke solve kiya hai kya?
@@arramareddynihithreddy1932 🤣🤣
Bro reading is 10 kg
@@s_shiva bhai thoda sa confidence mujhe bhi dede, solving k liye 0 marks but confidence ke liye full!!!
1)HWIL=60Kg
2)HWIL=60Kg
boy will apply a force of 30g downward to keep the lift at rest
T+T=40g+20g,T=30g
Therefore,boy applies 30g downward,and also 40g downward due to gravity
30g+40g=N
Therefore,N/g=70