Two masses and two pulleys problem

[Mr Davis 97]

All right, I think I get it now. Back to the original problem.

So I have four pieces of information:

$T_1 - W_1 = m_1 a_1$
$T_2 - W_2 = m_2 a_2$
$T_1 = 2T_2$
$a_2 = 2 a_1$

When I combine all of this, I get the equation $\displaystyle a_1 = \frac{g(2 m_2 - m_1)}{m_1 - 4 m_2}$, which is not correct because I need to have that if the masses are equal, then $\displaystyle a_1 = \frac{g}{5}$. What am I doing wrong?

 

[ehild]

All right, I think I get it now. Back to the original problem.

So I have four pieces of information:

$T_1 - W_1 = m_1 a_1$
$T_2 - W_2 = m_2 a_2$
$T_1 = 2T_2$
$a_2 = 2 a_1$

When I combine all of this, I get the equation $\displaystyle a_1 = \frac{g(2 m_2 - m_1)}{m_1 - 4 m_2}$, which is not correct because I need to have that if the masses are equal, then $\displaystyle a_1 = \frac{g}{5}$. What am I doing wrong?

You have to take the the sign of accelerations into account. m₁ and m₂ accelerate in opposite directions. If m₁ moves upward, the hanging pulley moves downward, and so does m_2. So a₂=2a₁ is not true.

 

[fiziksisphun]

so a(2)=-2a(1)?

 

[kuruman]

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