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

Jun 2017
218
6
Lima, Peru
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?. :help:
 

skeeter

Math Team
Jul 2011
3,129
1,695
Texas
forces on block A

$\displaystyle \sum F_x = F\cos(37) - N_x = 40 - N_x$

$\displaystyle \sum F_y = F\sin(37) + m_A g - N_{Ay} = 30 + 3g - N_{Ay} = 0$

forces on block B

$\displaystyle \sum F_x = N_x - f_x = N_x - 0.5(2g)$

$\displaystyle \sum F_y = m_B g - N_{By} = 2g - N_{By} = 0$


$40 - N_x = N_x - 10 \implies N_x = 25 \text{ N}$