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.

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestWhy do you think it is the same system?

Log in to reply

Why is the acceleration of the 2nd pulley wrt 2nd block \(a\)?

Log in to reply

@jatin yadav. @Anish Puthuraya @David Mattingly. Please help.

Log in to reply

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.

Log in to reply

You misunderstood me.

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

Log in to reply

See the mistake which you did while solving this senario....

https://brilliant.org/community-problem/infinite-atwood-machine/

Log in to reply

There is also another such question which i am resharing so that you may answer it.

Log in to reply

s in place of 10m/ssLog in to reply

I have no problem with 10 & 9.8

Log in to reply