Atwood Clarification: Tensions & Masses

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In summary, in a system where the atwood is suspended by a massless string with T3 and M1>m2. M1 is lower than m2. T1 is greater than T2 due to the lack of friction and the system remains in equilibrium. However, when there is acceleration, the system CM changes and the CM of M1 and m2 is lowered since M1 falls lower. T3 is greater than T1+T2 due to the added mass of the pulley.
  • #1
Vanessa23
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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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  • #3
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.
 
  • #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:

1. What is Atwood Clarification?

Atwood Clarification is a scientific experiment that involves two masses connected by a string over a pulley, and is used to study the concept of tension in a system.

2. How is tension measured in an Atwood Clarification experiment?

Tension is measured by the difference in the masses on either side of the pulley, as well as the force exerted by the string connecting the masses.

3. What is the purpose of an Atwood Clarification experiment?

The purpose of an Atwood Clarification experiment is to study the relationship between tension and masses, and how changes in one affect the other.

4. What are the factors that affect tension in an Atwood Clarification experiment?

The factors that affect tension include the masses of the objects, the length of the string, and the gravitational force acting on the masses.

5. How is Atwood Clarification used in real-life applications?

Atwood Clarification is used in real-life applications such as weight measurement systems, elevator systems, and construction of bridges and buildings.

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