Blocks in a room which is being accelerated

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The discussion centers on the equations of motion for two masses in an accelerating room, focusing on the forces acting on each mass, including pseudo-forces and friction. The derived friction coefficient, μ, was calculated as 0.6, but there are concerns about the correctness of the equations and the sign of the friction. An alternative calculation suggests a different value for μ, specifically 2/3, based on the forces acting on the masses. The choice of vertical direction for the x-axis is noted as unconventional, and there is a suggestion to include intermediate steps for clarity. Overall, the conversation highlights discrepancies in calculations and the need for clearer presentation of the problem-solving process.
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Homework Statement
The two bodies shown in the picture have the same mass and are inside a room which is accelerating to the right with ##2 \frac{m}{s^2}##. The coefficient of static friction is the same for both bodies. Determine the minimum value of the coefficient of static friction such that the blocks don't slide with respect to the room and the tension.
Relevant Equations
Newton's equations
I wrote the equations for ##m_2## with respect to the room, which are:
##x) T-Fr_1-f_1 *=0##
##y) N_1-P_1=0##

For ##m_1## we hace:
##x) N_2=f_2 *##
##y) T+Fr_2 -P_2=0##

Where ##f*_1## and ##f*_2## are the pseudo-forces and ##Fr_1## and ##Fr_2## are the friction forces that contain ##\mu##

Then I solved for ##\mu## and got ##\mu =0.6##

Are my ideas correct? Because I wasn't sure about the sign of the friction.
 

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x is in vertical direction? An unconventional choice.
You could have shown the intermediate steps. I get a different answer.

Working purely in the single degree of freedom: m*g is a force towards "m2 goes down", m*a is a force against it (where a=2 m/s^(2)), m*g*μ is against it (friction from m1), m*a*μ is against it (friction from m2).
m(g-a) = mμ(g+a), μ = 8/12 = 2/3.
 
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