Discover Tension in Systems I and II: A Physics Homework Question

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antigen123
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Homework Statement


PLEASE click on attached picture below:

What can be concluded about the tension in the cords holding the balls stationary in Systems I and II when both systems are at rest?


In both systems the mass of the larger ball M2 is exactly twice the mass of the smaller ball M1.Assume all cord lengths are equal (L1,L2,L3,L4). Assume both systems are at rest. Assume no frictional forces.




Homework Equations


Sum of forces=ma
T-mg=ma

The Attempt at a Solution



At looking at the picture I figured T2>T4 because the mass is twice as big. However what I don't understand is why T1=T3? I am completely lost and need a conceptual understanding of this.

side note: I did not use the Tensions equation because I think its more of a conceptual question however if it can be used please show me how to incorporate. Thank You in advance physics masters!
 

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At the conceptual level, the cord L1 doesn't 'know' about cord L2. As far as it is concerned, there's just some total system below it pulling down. What is the weight of that total system?
For the equations, you should write them out and post them. That's how this forum works.
 
I did write them down. However I am unsure how to use them exactly and need some help in this (T-mg=ma).
 
antigen123 said:

The Attempt at a Solution



At looking at the picture I figured T2>T4 because the mass is twice as big. However what I don't understand is why T1=T3? I am completely lost and need a conceptual understanding of this.

Consider the total weight that's below T1, and compare that with the total weight that's below T3. Are those total weights equal to each other, or is one greater than the other?
 
antigen123 said:
I did write them down. However I am unsure how to use them exactly and need some help in this (T-mg=ma).

If you want to work on this algebraically using symbols, that's certainly possible. Why don't you begin by showing us what you have so far? Label each of the masses and tension forces separately. You can also try drawing appropriate force diagrams.