Blocks in a room which is being accelerated

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SUMMARY

The discussion centers on the analysis of forces acting on two masses, ##m_1## and ##m_2##, in an accelerating room. The equations presented include the balance of forces involving pseudo-forces and friction, leading to a calculated coefficient of friction, ##\mu = 0.6##. However, an alternative calculation suggests a different coefficient of friction, ##\mu = 2/3##, based on the forces acting on the masses. The discrepancy arises from differing interpretations of the forces and their directions.

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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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