Clarification about the acceleration problem

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The discussion revolves around solving an acceleration problem involving multiple masses and forces. The user constructs a free body diagram for mass M, identifying forces including gravity, normal reaction, and tensions from connected masses. They argue that their approach, which considers the interactions between the masses and the pulley, provides better insight into the system's dynamics. Despite differing from the book's solution, the user confirms that their reasoning led to the correct answer. Overall, the logic applied in analyzing the forces appears sound and is validated by peer feedback.
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Hi

I have solved this problem but I have some doubts about the logic I have used at one point.
I have attached the problem in 1.jpg. Now I am trying to construct a free body diagram
for the mass M. First, there is force \inline{F} in the right direction. And as usual, there is force of gravity \inline{Mg} downward and the normal reaction
\inline{N} upward. Also there would be a force exerted by the mass \inline{m_2} towards the left, let's call it \inline{F_{21}}. From top, there would be a force exerted by the mass \inline{m_1} on \inline{M}.
Now let's say that there is a tension \inline{T} in the string when the system is going to the right. Since pulley is frictionless, the tension in both sides would be same and since the pulley is attached to the mass \inline{M} any force exerted on the pulley would be exerted on the mass \inline{M}. Since force \inline{T} is exerted
on the mass \inline{m_1} ,by Newton's third law, the force \inline{T} would be exerted by the m1 on the pulley, which means on the mass
\inline{M} , towards left.. With similar reasoning, we can say that a downward force
\inline{T} is exerted on the mass \inline{M}.

Do you think that was the valid reasoning. I got the right answer. The book's solution
manual does it differently but doing my way gives more insight into the forces
between the masses m1,m2 and \inline{M}.

thanks
 

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Your reasoning sounds good to me.
 
Thank You, Doc..
 
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