1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Deriving Expression
  1. Derive an Expression (Replies: 1)

Loading...