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Homework Help: Deriving Expression

  1. Oct 17, 2006 #1
    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:

  2. jcsd
  3. Oct 17, 2006 #2
    wheres the figure ?
     
  4. Oct 17, 2006 #3
    sorry...forgot it at first
     
  5. Oct 17, 2006 #4
    here is the image
     

    Attached Files:

  6. Oct 17, 2006 #5
    Any thoughts?
     
  7. Oct 18, 2006 #6
    any smart person have any clue about where i went wrong?
     
  8. Oct 18, 2006 #7
    how does a=(m_2*g)/(2*m_1+m_2) sound?
     
  9. Oct 18, 2006 #8
    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?
     
  10. Oct 18, 2006 #9
    notice that if there is no slipping, a_2 should be one half of a_1 because of the string.
     
  11. Oct 17, 2007 #10
    Did anyone solve this? I am working on the same problem and am completely stumped.
     
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