Each block has mass \(m\). We are finding acceleration of topmost block.

Assume the tension \(T\) acting on topmost block and acceleration \(a\) in upward direction.

Then tension on second block is \(\frac{T}{2}\). If we see the block w.r.t the pulley that it is connected to, then apply psuedo force $ma$ in upward direction (as the pulley it is connected to is moving in downward with $a$), then the system observed is same as the above, I.e acceleration will be same.

Hence \[T-mg=ma\\ \frac{T}{2}+ma\text{ (psuedo force) }-mg=ma \]

We obtain \(a=g\) but the answer should be ...(this is a question on brilliant and I can't share answer)

Please help pinpoint the error.

## Comments

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TopNewestWhy do you think it is the same system? – Pratik Shastri · 2 years, 4 months ago

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Why is the acceleration of the 2nd pulley wrt 2nd block \(a\)? – Pratik Shastri · 2 years, 4 months ago

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@jatin yadav. @Anish Puthuraya @David Mattingly. Please help. – Megh Parikh · 2 years, 4 months ago

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All the blocks will be relative to each other as all the blocks are not stable, and thus the acceleration of each block will be different and you took them as same. just find the relation between the accelerations of different blocks and find a sequence and similarly you will find that acceleration of topmost block is g/2. ,try by taking a suitable observable position. – Ronak Pawar · 2 years, 4 months ago

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I am seeing block2 wrt pulley2. Then pulley2 is at rest and it also has a system similar to pulley one. The acceleration of block2 is \(a\) wrt pulley2 – Megh Parikh · 2 years, 4 months ago

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https://brilliant.org/community-problem/infinite-atwood-machine/ – Ronak Pawar · 2 years, 4 months ago

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There is also another such question which i am resharing so that you may answer it. – Megh Parikh · 2 years, 4 months ago

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s in place of 10m/ss – Ronak Pawar · 2 years, 4 months agoLog in to reply

I have no problem with 10 & 9.8 – Megh Parikh · 2 years, 4 months ago

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