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Atwood clarification

  1. Oct 28, 2007 #1
    1. The problem statement, all variables and given/known data
    In a system where the atwood is suspended by a massless string with T3 and M1>m2. M1 is lower than m2. Relate Tensions and masses if it is frictionless, pulley and string are massless.


    2. Relevant equations
    T1 is tension in string holding M1; T2 is tension in string holding M2.


    3. The attempt at a solution
    I believe that when there is no acceleration:
    T3=M1+m2
    T1=M1g
    T2=m2g
    but shouldn't T1=T2?

    when there is acceleration then the system CM does not change, but the CM of M1 and m2 is lowered since M1 falls lower. Or will it stay the same since M1 is already lower than m2?

    also, since T3 is being taken into account (which I cant find an example of), will it have a negative tension compared with the other tensions?
     
  2. jcsd
  3. Oct 28, 2007 #2

    Kurdt

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  4. Oct 28, 2007 #3

    PhanthomJay

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    Yes, good observation. So what does that tell you about the system when M1 and m2 are not equal?

    what makes you say that?
    the CM will lower

    there's no such thing as a negative tension. Once you solve for the string tension around the pulley (where you must note that T1=T2 for strings wrapped around massless frictionless pulleys), you can find T3 with a FBD of the massless, at rest pulley.
     
  5. Oct 28, 2007 #4
    So if it is true that in this situation T3=M1+m2, then why is it that T3 is less than M1+m2+the mass of the pulley (when it isn't massless)?
    and why is T1 greater than T2?
    and T3 greater than T1+T2?
     
    Last edited: Oct 28, 2007
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