Quick forces conceptual question

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SUMMARY

The discussion focuses on deriving the expression for the acceleration of mass m1 in a double pulley system involving two masses, m1 and m2. The key equations used are the sum of forces for m1, represented as Fx = F2 on 1 = m1a1, and for m2 as Fy = F1 on 2 + T1 - m2g = m2a2. The relationship a2 = -0.5a1 is established, and it is confirmed that tension T1 equals F1 on 2 due to the assumption of a massless and frictionless rope and pulley system, ensuring uniform tension throughout the rope.

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  • Understanding of Newton's second law of motion
  • Familiarity with pulley systems and tension forces
  • Basic knowledge of kinematics and acceleration
  • Concept of massless ropes and frictionless pulleys
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Homework Statement



aknightfigure.jpg

Find an expression for the acceleration of m1.

Homework Equations





The Attempt at a Solution


I did this earlier, and looking back on it I have a question:

Sum of forces for m1:
Fx= F2 on 1= m1a1

Sum of forces for m2:
Fy = F1 on 2 + T1 - m2g = m2a2

I got the expression after substituting them into each other and saying a2 = -.5a1, I'm just wondering why you can say T1= F1 on 2 , because since there is acceleration, can you consider the tension throughout the rope to be the same?
 
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It's a double pulley problem with m1 sitting on a table, attached to a rope that goes around a pulley at the edge of the table. Then the string passes through another pulley attached to the top of a suspended m2. Then the rope goes up and is attached to a ceiling.
 
bcjochim07 said:
Sum of forces for m1:
Fx= F2 on 1= m1a1
The only horizontal force on m1 is the rope tension T.

Sum of forces for m2:
Fy = F1 on 2 + T1 - m2g = m2a2
Careful. The tension force pulls up twice on m2 (via the pulley).

I got the expression after substituting them into each other and saying a2 = -.5a1, I'm just wondering why you can say T1= F1 on 2 , because since there is acceleration, can you consider the tension throughout the rope to be the same?
As long as the rope is massless and the pulley is massless and frictionless, then the tension is the same throughout the rope.
 

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