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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. ($g=10\,\frac{m}{s^2}$)

Sketch of the problem

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?.

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1 Answer 1

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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

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

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