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The problem is as follows:

The system shown in the figure from below is at rest. The masses for both objects are given as follows, $m_{B}= 3 m_{A}=6\,kg$, the coefficient of friction between the block $A$ and $B$ is $0.5$ and the friction is negligible between block $B$ and the floor. Using the provided information. Find the maximum value of $F$ such as the blocks will move together.

Sketch of the problem

The alternatives given on my book are as follows:

$\begin{array}{ll} 1.&21\,N\\ 2.&23\,N\\ 3.&18\,N\\ 4.&20\,N\\ 5.&2.2\,N\\ \end{array}$

In this problem I'm totally lost at. Can somebody help me with the FBD?. The only thing which I could come up with was that:

$F=\left(m_{A}+m_{B}\right)a$

Then:

$F=\left(6+2\right)a = 8a$

Then for the block $A$

$F-f_{s}=0$

$F=f_{s}$

But here's where I'm confused at as:

$F= \mu m_{A}g = \frac{1}{2} \left( 2 \right) \times 10$

$F= \mu m_{A}g = \frac{1}{2} \left( 2 \right) \times 10 = 10$

I can't relate exactly how to use this information. A FBD would greatly help me. Can somebody help me with this?.

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    $\begingroup$ You seem to be having a lot of trouble with this subject. You’ve asked a good half dozen questions about these exercises in a short span (not to mention your other questions from previous days). $\endgroup$
    – amd
    Nov 18 '19 at 22:28
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Hint.

For body $A$

$$ \mu_s m_a g = m_a\alpha $$

for body $B$

$$ F-\mu_sm_a g = m_b\alpha $$

(Both bodies have the same acceleration)

then

$$ \frac{1}{m_a}\mu_2m_ag = \frac{1}{m_b}\left(F-\mu_s m_a g\right) $$

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