The system is moving over a rough surface. It is known that only block $A$ is fritionless. Find the modulus of the reaction, in $N$ between the blocks $A$ and $B$. Consider that the coeficcient of kinetic friction between the floor and block $B$ is $0.5$ and the mass of $A$ is $3\,kg$ and mass of $B$ is $2\,kg$, respectively. ($g=10\,\frac{m}{s^2}$)

The alternatives given in my book are:

$\begin{array}{ll}

1.&40\,N\\

2.&30\,N\\

3.&22\,N\\

4.&12\,N\\

5.&25\,N\\

\end{array}$

What I attempted to do was to find the acceleration for block $A$ and assume that would be the acceleration for the system:

$a=\frac{F\cos37^{\circ}}{m_a}$

Then:

$a=\frac{F\cos37^{\circ}}{m_a}=\frac{50\left(\frac{4}{5}\right)}{3}=\frac{40}{3}$

Then for block $B$.

$F\cos37^{\circ}-R-f_k=m_b a$

$R=F\cos37^{\circ}-f_k-m_b \left(\frac{F}{m_a}\right)$

This would become into

$R=F\cos37^{\circ}-f_k-m_b \left(\frac{F}{m_a}\right)$

$R=F\cos37^{\circ}-\mu_k \left(m_b g + F\sin 37^{\circ}\right)-m_b \left(\frac{F\cos37^{\circ}}{m_a}\right)$

Therefore pluggin the information given would become into:

$R=50\cos37^{\circ}-0.5 \left(2\times 10 + 50 \sin 37^{\circ}\right)-2 \left(\frac{50\cos37^{\circ}}{3}\right)$

$R=40-25-2 \left(\frac{40}{3}\right)$

But as it can be seen I'm no closer to the supposed answer which is $22$

What can It be wrong with my method?. Can somebody help me here? Can somebody help me with the right FBD for this as well?. :help: