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