Faulty approach to solving a double Atwood's machine

[Rations]

Hello. I'm wondering why my approach to solving this problem (Morin 3.2) is faulty.

Homework Statement

A double Atwood's machine is shown, with masses $m_1$, $m_2$, and $m_3$. Find the accelerations of the masses.

Homework Equations

$F = ma$

The Attempt at a Solution

Let's say the string over the bottom pulley has tension $T$ and the string over the top pulley has tension $T_2$. Assuming that the pulleys and ropes are massless and defining the upwards direction as positive, I have several equations:

(1) For $m_1$: $T_2 - m_1g = m_1a_1$
(2) For $m_2$: $T - m_2g = m_2a_2$
(3) For $m_3$: $T - m_3g = m_3a_3$
(4) For the lower pulley: $T_2 - 2T = 0$
(5) From conservation of string: $a_2 = -a_3$

If I substitute $-a_3$ into (2) and then solve (2) and (3) for $T$, I find $T = \frac{2m_2m_3g}{m_2 + m_3}$. From (4), $T_2$ is twice this value.

Then, I substitute my expression for $T_2$ into (1) and solve for $a_1$, which yields:
a_1 = g\frac{4m_2m_3 - m_1(m_2 + m_3)}{m_1(m_2 + m_3)}

This answer is close to what Morin writes, but is missing a term in the denominator. After reading his solution, which uses the fact that $a_1 = -\left(\frac{a_2 + a_3}{2}\right)$, I feel that I can understand why his approach works. However, I do not see where I made a mistake.

 

[Qwertywerty]

From conservation of string: a2=−a3a_2 = -a_3

This is where you have made a mistake . Why ?

Conservation of string can be looked at in two ways .
Since you use all accelerations in one direction , this would be an easier way to understand -

Total power by string must be zero ( Why ? ) . Power = F.v , so total sum of of each individual power provided by string must be equal to zero .

So 2T*v₁ + T*v₂ + T*v₃ = 0 , or ,
2v₁ + v₂ + v₃ = 0 .

So what relation do you get for acceleration ?

I hope this helps .

 

[Rations]

Oh, right. I just realized that $a_2 = -a_3$ only holds in the reference frame of the pulley, not the ground. Don't know why that point eluded me for so long. Thanks.

 

[SammyS]

Hello. I'm wondering why my approach to solving this problem (Morin 3.2) is faulty.

Homework Statement

A double Atwood's machine is shown, with masses $m_1$, $m_2$, and $m_3$. Find the accelerations of the masses.
View attachment 86504

Homework Equations

$F = ma$

The Attempt at a Solution

Let's say the string over the bottom pulley has tension $T$ and the string over the top pulley has tension $T_2$. Assuming that the pulleys and ropes are massless and defining the upwards direction as positive, I have several equations:

(1) For $m_1$: $T_2 - m_1g = m_1a_1$
(2) For $m_2$: $T - m_2g = m_2a_2$
(3) For $m_3$: $T - m_3g = m_3a_3$
(4) For the lower pulley: $T_2 - 2T = 0$
(5) From conservation of string: $a_2 = -a_3$

If I substitute $-a_3$ into (2) and then solve (2) and (3) for $T$, I find $T = \frac{2m_2m_3g}{m_2 + m_3}$. From (4), $T_2$ is twice this value.

Then, I substitute my expression for $T_2$ into (1) and solve for $a_1$, which yields:
a_1 = g\frac{4m_2m_3 - m_1(m_2 + m_3)}{m_1(m_2 + m_3)}

This answer is close to what Morin writes, but is missing a term in the denominator. After reading his solution, which uses the fact that $a_1 = -\left(\frac{a_2 + a_3}{2}\right)$, I feel that I can understand why his approach works. However, I do not see where I made a mistake.

Your error is with the equation you get from the "conservation of string", as Qwertywerty points out.

A simple way to see that this is erroneous is: Suppose that m₂ = m₃ . Then we must have a₂ = a₃ , but these won't be zero in general.