Vertical acceleration of mass 1 in pulley system

[tigerseye]

Now I am supposed to find the magnitude of the vertical acceleration of mass 1, in terms of mass 1, gravity and the tension.
I found that a_1= ((m_1*g)-2T)/2m_1 but that isn't right. Any help would be great

 

[Yegor]

a_1= ((m_1*g)-T)/m_1

a_1= (4/5)*(m_1*g)/(m_1+m_2)

Is it correct?

 

[Yegor]

Sorry. I'd like to say:
a_1= 4*(m_1*g)/(4*m_1+m_2)

If it is correct, i'll try to explain

 

[tigerseye]

Can you explain how you got that?

 

[Yegor]

there is important, that a_2=(a_1)/2 (look carefull at string).
On first mass acts 2 forse: gravity and tension. So:
m_1*a_1= (m_1*g)-T
But on the 2nd mass acts F=2*T
m_2*a_2=2*T
m_2*(a_1)/2=2*T
We have a system:
m_2*(a_1)/2=2*T
m_1*a_1= (m_1*g)-T
We don't know a_1 and T.
If we solve:
T=m_2*(a_1)/4
And a_1= 4*(m_1*g)/(4*m_1+m_2)
Am I right?

 

[tiger_striped_cat]

I didn't check your algebra on your last step but it looks right.

to derive your accelearation relation, you don't have to "look at the strings" and think it through. in these more difficult problems you should derive a constant from the problem. In this case, the length of the string is constant. If I call the length from the wall to the stationary pully L1 the length from the stationary pully to the moving pully x and the length from the stationary pully to m2 y then I can derive the formula:

L1+x+x+y=LT

Where LT is the total length of the string. Takeing time derivatives of both sides twice:

$$2\ddot{x}+\ddot{y}=0$$

then negnative sign you get should be argued away. So then you get the relations for acceleration that you derived.