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.

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?. :help: