Atwood Clarification: Tensions & Masses

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In a frictionless Atwood machine with a massless pulley and string, the tensions T1 and T2 in the strings holding masses M1 and m2, respectively, are not equal when M1 is greater than m2. When the system is at rest, T1 equals M1g and T2 equals m2g, leading to the conclusion that T1 must be greater than T2. During acceleration, the center of mass of the system lowers as M1 descends, indicating that the system's dynamics are affected by the differing masses. T3, the tension in the string holding the entire system, is derived from the forces acting on the pulley and cannot be negative, but its relationship to T1 and T2 must be carefully analyzed. The discussion highlights the complexities of tension in non-equal mass systems and the importance of understanding the forces at play.
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Homework Statement


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.


Homework Equations


T1 is tension in string holding M1; T2 is tension in string holding M2.


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 can't find an example of), will it have a negative tension compared with the other tensions?
 
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Vanessa23 said:

Homework Statement


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.


Homework Equations


T1 is tension in string holding M1; T2 is tension in string holding M2.


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

when there is acceleration then the system CM does not change,
what makes you say that?
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?
the CM will lower

also, since T3 is being taken into account (which I can't find an example of), will it have a negative tension compared with the other tensions?
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.
 
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?
 
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