wedge constraints | constrained motion | nlm
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- เผยแพร่เมื่อ 7 ก.พ. 2025
- motion of connected blocks
motion of connected bodies
motion on the surface of wedges
wedges motion
Newton's laws of motion
wedge constrained motion explained
wedge constrained motion
constrained motion
constraints motion
nlm
NLM
class 11 physics chapter Newton's laws of motion
wedge constrained simplified
class 11 physics
#jee2025 #jeemainphysics #jeeadvanced #physics #class11physics #nlm #constrainedmotion
wonderful explanation sir...
11th me ye concept samajh nhi aaya thaa , but aaj jee mains 2025 se 11 din phle aache se samajh aa gya..❤❤❤
Thank you so much.
Glad ...you understand it 👍
1 saal se mereko wedge problem samjh nhi aa rha tha mgr aap 16 min. Me samjha diye thanks a lot ❤❤❤❤❤❤❤❤
Glad ...I could help you.. thanks for your feedback...it keeps me motivated.
@@doubtdestroyer iam in class 10 and preparing for Olympiad i started my preparation from class 9 sir please aap advanced ke related ach ache illustrations upload kijiye ki jisse hme apna concept building me madad ho .💕💕💕💕
Ye displacement wala method best hai. Badi aasani se ques. Ho jata hai, thanks a lot.
Wow amazing explanation guru ji❤❤❤
Thank you for your appreciation 🙏
Thanks a lot bhaiya ❤
We want more type of this videos
Sure...you will get them...
Nice explanation sir
Thank you Sandeep ji... please share and make sure your friends also subscribe...it is motivation for me.. keep smiling and keep learning..
Sir plz take question no. 64 on cengage on nlm in which a cylinder is present between the wedges moving with different velocity
You will get it soon...
Hey...you can solve it by this very simple approach.
First let's see the situation by sitting on any wedge . Simple one is sit on wedge which is moving with u velocity in backward direction.
To this observer the other wedge will appear to move by 3u velocity. And cylinder will move on inclined plane.
Now since cylinder and wedge are in contact. This means horizontal component of cylinder velocity is 3u. This means v cos30 = 3u
V = 2√3 u.
Now since this observer is also moving with u velocity in backward direction so from ground ( for ground observer)
The net velocity of cylinder should be vector addition of these two velocities.
If you add these two velocities u and 2√3 u by vector addition
You will get net velocity= √7u.
If you still not get it...join telegram channel and ask this question there I will post it with diagrams.
Sir is there is any other approach to solve it
@@Ashu-w8t it is simplest ...you can directly solve by taking ground as reference frame but that will be slightly difficult to imagine so to solve.
❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤
Thank you 🙏