# Finding acceleration

We have to find the acceleration of the blocks in the above picture considering all the surfaces to be smooth and tension is in string is uniform.

Since, in the horizontal direction, there is no force, that implies that the block of mass $M$ does not have any acceleration.

Therefore $A=0$.

Now, if the blocks of mass $2m$ and $m$ are moving with an acceleration $a_r$ relative to the bigger block (of mass $M$) .

Now, we can write the force equations for the smaller blocks, that is $2mg-T=2ma_r$ and $T-mg=ma_r$.

Solving this, we get $a_r=\dfrac{g}{3}$. However, the answer given to the above problem is $a_r=0$ as well, that is , whole system is at rest. Is my method wrong or the given answer is wrong?

Another similar problem that is bothering me is this:

Now, here similar to the previous problem I have assumed that the bigger block of mass $4m$ moves towards right and the smaller blocks move with an acceleration $a_r$ relative to the bigger bloc towards left.

Again in horizontal direction, net force is $0$ so I wrote $4mA+m(A-a_r)=0 \implies a_r=5A$ but the given answer in this case is $A=\dfrac{g}{23}$ and $a_r=\dfrac{6g}{23}$. Now my very first equation becomes wrong due to this. Where is the fault?

Note by Vilakshan Gupta
6 months, 2 weeks ago

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Sort by:

- 6 months, 2 weeks ago

Acceleration is $\displaystyle a = \frac{g}{3}$ as you claim. Maybe both blocks have equal mass $m$ or the answer given is wrong. :)

- 6 months, 2 weeks ago

No, the blocks aren't of equal mass. I have one more similar problem, and again my answer does not match with the given answer. Give it a try too, I will upload it too.

- 6 months, 2 weeks ago

Ok, I'll give it a try. Do mention me there so that I'll get notified. :)

- 6 months, 2 weeks ago

Wait, are the blocks being moved to right by a force in the question?

- 6 months, 2 weeks ago

No. They are not. The system is released from rest. I have assumed the accelerations.

- 6 months, 2 weeks ago

- 6 months, 2 weeks ago