Newtonian Mechanics - Pulley block system, finding acceleration [closed]

Question

The question is to find acceleration of the block m₁ in the above system (please look at diagram). So I took tension along String 2 to be T and thus Tension along String 1 would be 2T.

Acceleration of m₁ I took to be 'a' (rightwards), so, as tension on the masses m₂ and m₃ is half that of m₁, I took the acceleration of m₂ and m₃ to be '2a'.

Then I set up laws of motion equations

m₃g - T = 2m₃a for the mass m₃

and, m₂g - T = 2m₂a for the mass m₂

Then dividing each of them throughout by m₃ and m₂ and taking a on the LHS and adding, I got $a = T(1/m2-1/m3)/4$

But I cant eliminate T. Can someone help me on where I went wrong/ suggest amends in the method.

Answer 1

Let a be the acceleration of m1 to the right. This is also the downward acceleration of the pulley supporting m2 and m3. Let $\delta$ be the downward acceleration of m3 relative to this pulley. Then the absolute downward acceleration of m3 is $a+\delta$ and the absolute downward acceleration of m2 is $a-\delta$. So the force balances read: $$m_1a=2T$$ $$m_2(a-\delta)=m_2g-T$$ $$m_3(a+\delta)=m_3g-T$$

The key to doing many pulley problems is to first focus on the kinematics of the motion (as in this problem).