Quick forces conceptual question

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The discussion revolves around deriving the expression for the acceleration of mass m1 in a double pulley system. It highlights the equations of motion for both m1 and m2, emphasizing the relationship between their accelerations, specifically that a2 = -0.5a1. A key point raised is the tension in the rope, questioning whether T1 can equal F1 on 2 due to the system's acceleration. The consensus is that if the rope and pulleys are massless and frictionless, the tension remains constant throughout the rope. This understanding is crucial for accurately solving the problem.
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Homework Statement



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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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