# Still stuck on two blocks and pulleys

1. Nov 14, 2004

### 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

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2. Nov 14, 2004

### Yegor

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

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

Is it correct?

3. Nov 14, 2004

### 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

Last edited: Nov 14, 2004
4. Nov 14, 2004

### tigerseye

Can you explain how you got that?

5. Nov 15, 2004

### 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?

6. Nov 15, 2004

### 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 lenght 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.