# How do I find the reaction between two blocks when only one has friction?

The problem is as follows:

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?.

Hint.

Calling $$\alpha = 37^{\circ}$$

Block $$A$$ (no friction)

$$F\cos\alpha - H = m_a a$$

Block $$B$$ (friction)

$$H-\mu m_b g = m_b a$$

both blocks move at the same acceleration then

$$\frac{1}{m_a}\left(F\cos\alpha-H\right) = \frac{1}{m_b}\left(H-\mu m_b g\right)$$