How do I find the maximum value of a force to be applied to a body such as when pulled from another move together?

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?.

• 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).
– amd
Nov 18 '19 at 22:28

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

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