# Homework Help: Deriving Expression

1. Oct 17, 2006

So the problem I am working on is: In the figure, find an expression for the acceleration of (assume that the table is frictionless). I have attached the figure of reference.

I have so far tried to solve by first looking at m2 to get:
2T-m_2*g=m_2*a so T= (m_2*(a+g))/2

Then by looking at m1 I got:
T=m_1*a so a=(m_2*g)/(2*m_1-m_2)
This is supposidely wrong.
Can anyone see what I did wrong and how to fix it?

Should it be a=(m_2*g)/(2*m_1+m_2) ?

#### Attached Files:

• ###### expimage.doc
File size:
51 KB
Views:
89
2. Oct 17, 2006

### gunblaze

wheres the figure ?

3. Oct 17, 2006

sorry...forgot it at first

4. Oct 17, 2006

here is the image

#### Attached Files:

• ###### expimage.doc
File size:
51 KB
Views:
73
5. Oct 17, 2006

Any thoughts?

6. Oct 18, 2006

any smart person have any clue about where i went wrong?

7. Oct 18, 2006

how does a=(m_2*g)/(2*m_1+m_2) sound?

8. Oct 18, 2006

okay...when I assumed a1 not equal to a2 which i wasnt doing originally, i got 2T-m_2*g=-m_2*a_2 and T=m_1*a_1
sbsitiuting and shifting things around I got:

a_1=(m_2*g-m_2*a_2)/(2*m_1)

does thi seem right?

should my anyswer depend on a_2 or are they assumed to be the same, therefor giving me a=(m_2*g)/(2*m_1+m_2) ?

Can someone work the problem and tell me if they are getting what I have of something different?

9. Oct 18, 2006

### tim_lou

notice that if there is no slipping, a_2 should be one half of a_1 because of the string.

10. Oct 17, 2007

### mantillab

Did anyone solve this? I am working on the same problem and am completely stumped.