Pully System on an Inclined Plane: Understanding Direction of Acceleration

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The discussion centers on understanding the direction of acceleration in a pulley system on an inclined plane, particularly the confusion surrounding the inclusion of a vertical force (m2g) in a horizontal force equation. Participants express frustration over the teacher's vague explanation regarding why acceleration is considered positive and how it relates to forces in different directions. Clarification is sought on the relationship between m2g and other forces in the equation, with suggestions that additional pulleys may be necessary for accurate modeling. The need for a clearer problem statement and diagram is emphasized, as well as the importance of following guidelines when posting physics problems. Overall, the conversation highlights the complexities of analyzing forces in multi-directional systems.
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Homework Statement



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So I am pretty okay with most of this problem. And I asked my teacher for clarification but he just sort of said "because it's positive" and I get that's it's positive. Apparently "everyone's" struggling with that, and therefore we are going over this problem to cement the fact that we consider the direction of acceleration to be positive...but that doesn't help me. I get, clearly, how why whatever that we consider m2g positive in this problem. What I don't get is why we can put m2g, a y direction force, into an other wise x direction equation...the fact that it's positive...I don't see how that negates or affects the fact that it's a y direction force in this problem

Homework Equations


See first picture
ΣF=ma

The Attempt at a Solution


As I was writing this I had a thought:

Do we use m2g because m2g is equivalent to Ft2...and Ft2 is equal to Ft1+Ft3...and ft1 and ft3 are x direction forces?? Is that why?
 

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Please start your explanation at the beginning.
Yellowkies_3275 said:
I get that's it's positive
. What is positive?
The photo that you posted does not show a lot of data.
There is no problem statement. What are you trying to calculate?
Please follow the guidelines when you post a problem.
 
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I suspect the diagram is wrong. It appears to show that a rope goes straight across between two pulleys, with a second rope tied at the middle. The instantaneous acceleration of m2 would be g, since it would not immediately cause any significant acceleration of the other two masses.

To arrive at the equation shown, there would need to be two more pulleys in the middle. One rope would go over one of these to connect m1 to m2, and another rope over the other to connect m3 to m2.
If that is correct, you can arrive at the equation using virtual work. If m2 descends a small distance dy then each of the other masses moves up slope by dy. Write out the energy balance for that and cancel the dy's.
 
Is the equation in the image part of the solution provided by your teacher or your attempt at the solution?
 
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