Sorry about the re upload folks! Thank you so much for spotting a "deliberate mistake" in the first version of the video where I missed adding the mass of the bar when balancing forces. Thanks for spotting it! You are great!: )
Sir when solving this problem , we can easily take the moments about the point where the 5 kg mass is suspended. Considering clockwise moments as positive , d*(F1) - (1-d)*(F2) =0 d*(k1*x) = (1-d)*(k2*x) (since the extension is same in the two springs.) d*(10) = (1-d)*(15) 2*d = (1-d)*3 5*d = 3 therefore , d =0.6 m Hope this approach is better. 😊 Anyway , your video is a great one !!! ❤
hey sir, please make a workbook in equilibrium moments and resolving forces. these questions really need loads of practice, and ive almost run out of most AS questions in this topic
in the case of springs, we can use directly extension=force/spring constant. We tend to need the original length if we had to work out something such as strain. Hope this helps!
interesting question, when an unusual question appears I think the best is to focus on the fundamentals. I.e. the trick to this one is realising the bar will be horizontal only if the extensions are the same. The rest involves having a great knowledge of the syllabus and problem solving and solve multiple 2 markers to reach the final answer. I'd seek all the longest calculations questions you can find and have a go at them : ) Good luck!!
This is very good question actually these kind of questions i do for practice before solving IIT JEE question ""Please explain one of the jee advanced phisics problem "" Thankyou sir Lots of love from India 🇮🇳❤
Sorry about the re upload folks! Thank you so much for spotting a "deliberate mistake" in the first version of the video where I missed adding the mass of the bar when balancing forces. Thanks for spotting it! You are great!: )
Thank you so much. You’re one of the best teachers and you keep uploading! Don’t stop!
thank you so much for the very kind comment, will do!
This question is very interesting combining springs in parallel with moments is something i have never seen before. Good thing i got it right 😅
well done, thank you for the comment!
Sir when solving this problem , we can easily take the moments about the point where the 5 kg mass is suspended. Considering clockwise moments as positive ,
d*(F1) - (1-d)*(F2) =0
d*(k1*x) = (1-d)*(k2*x) (since the extension is same in the two springs.)
d*(10) = (1-d)*(15)
2*d = (1-d)*3
5*d = 3 therefore , d =0.6 m
Hope this approach is better. 😊 Anyway , your video is a great one !!! ❤
i also came to the same conclusion wonder if it is right or not
@tanvirfarhan5585 this approach is right . Also, it is easy
did it by finding the net torque by taking the hanging point as hinge and equating to zero, nice problem
well done!
hey sir, please make a workbook in equilibrium moments and resolving forces. these questions really need loads of practice, and ive almost run out of most AS questions in this topic
great idea! It will take some time but I will have it ready this year. Thank you for the suggestion
Is extension of springs the same even if we don't know the original lengths of them?
in the case of springs, we can use directly extension=force/spring constant. We tend to need the original length if we had to work out something such as strain. Hope this helps!
Hi sir, do you have any advice on what to do when you're doing a complex question like this but doesn't understand the mark scheme either?
interesting question, when an unusual question appears I think the best is to focus on the fundamentals. I.e. the trick to this one is realising the bar will be horizontal only if the extensions are the same. The rest involves having a great knowledge of the syllabus and problem solving and solve multiple 2 markers to reach the final answer.
I'd seek all the longest calculations questions you can find and have a go at them : ) Good luck!!
@@zhelyo_physics Thank you very much for answering my question! all the best!
thanks
anytime!
can you do 1.5/2.5 (force/total force as a ratio) x 1.0m to get 0.60m? or is the goal to hit all the marks on mark scheme
What point are you taking moments with respect to? I have a feeling yes, but I'd need to write it out to check.
This is very good question actually these kind of questions i do for practice before solving IIT JEE question
""Please explain one of the jee advanced phisics problem "" Thankyou sir Lots of love from India 🇮🇳❤
thank you very much for the kind comment!
I have a random suggestion. Do/go over some math olympiad/exam problems. I kinda wanna see what that will look like. Thank you!
Oh, very interesting...: ) Blackpenredpen has some amazing Math Olympiad Problems I'd highly recommend. Thank you for the suggestion.
Great video! But must ask are you going to continiue solving different physics olympiad problems?
I absolutely will, sorry I haven't been able to post recently on this topic. Life has been very busy, but shall get back onto it!
@@zhelyo_physics No problem, I was just curious
Dumb question, but could you please explain why the extensions must be the same? Other than that, I understood the calculation.
not a dumb question! If the beam is balanced it is horizontal meaning that each spring extension would be the same. Hope this helps!
Any possibility that you could do another ESAT video, considering the ESAT is in at least 9 days time. Thanks
working on a maths video as we speak for it : ) Stay tuned this week...I will try to do it ASAP.