You are correct. My approach can work, but I left out the "Ma" term. The force down should be mg - ma - Ma or mg + Mg - T1 - T2 = ma + MA. This video is definitely a "redo". Thanks for pointing that out.
If the second pulley has mass, its mass should be considered when you did T1+T2-(mg-ma)=0 at (4m30sec). Instead of being m shouldn't it be m+M ? Besides at your final acceleration formula, at 4m45sec, when you make m=0, a=0, but if the pulley has mass it will still fall down even without the mass m, right ? By the way, great explanation, as always ! :)
@@MichelvanBiezen I believe the answer comes out to: a=(m+M)/(m+3.5M)*g It is basically a free fall problem, but the mass is slowed down by the rotational inertia of the pulleys. If mass m is 50lbs and each pulley is 1lb, the mass will fall at 95% the free fall acceleration. I would have guessed that the pulleys were more negligible than that. But, they would be more negligible in a more typical pulley problem.
@@MichelvanBiezen professor i have an idea for those videos where you have made mistakes . why not change the title and make these videos " misteries " .That is, to let students find the mistakes (because they probably learn more like that ) not exactly learn more but put their learnings in " action " . What do you think ?
Flying butter is asking, system not in equilibrium or constant motion, m has acceleration downward? How sigmaF=0?
You are correct. My approach can work, but I left out the "Ma" term. The force down should be mg - ma - Ma or mg + Mg - T1 - T2 = ma + MA. This video is definitely a "redo". Thanks for pointing that out.
@@MichelvanBiezen all good sir, if theres a mistake from you, its only giving a man hope and dreams
We all make mistakes. That is part of being human. A viewer usually will catch it. 🙂
If the second pulley has mass, its mass should be considered when you did T1+T2-(mg-ma)=0 at (4m30sec). Instead of being m shouldn't it be m+M ? Besides at your final acceleration formula, at 4m45sec, when you make m=0, a=0, but if the pulley has mass it will still fall down even without the mass m, right ? By the way, great explanation, as always ! :)
You are correct, that there are some errors, we'll have to reshoot this video. 🙂
@@MichelvanBiezen
I believe the answer comes out to:
a=(m+M)/(m+3.5M)*g
It is basically a free fall problem, but the mass is slowed down by the rotational inertia of the pulleys. If mass m is 50lbs and each pulley is 1lb, the mass will fall at 95% the free fall acceleration. I would have guessed that the pulleys were more negligible than that. But, they would be more negligible in a more typical pulley problem.
is the acceleration constant?
Thnku
You are welcome. 🙂
Thank sir🙏
Most welcome
Its very interesting video
Yes, but not the error I made. We'll have to reshoot this video.
@@MichelvanBiezen no problem professor . Noone is perfect . We learn from our mistakes .
@@MichelvanBiezen professor i have an idea for those videos where you have made mistakes . why not change the title and make these videos " misteries " .That is, to let students find the mistakes (because they probably learn more like that ) not exactly learn more but put their learnings in " action " . What do you think ?
Yes, we usually leave them in for that purpose and then we'll reshoot the video with the correct solution.
@@MichelvanBiezen very good . 😉😁👍
🙋
🙂
You are wonderful as always
Thank you. 🙂
🙂
Glad you liked it.
@@MichelvanBiezen 👍