In the given diagram, the block A and B are place in contact. The contact between the blocks is rough while the contacts at wall and ground are smooth. The two blocks remain at rest.
The contact force acting between the blocks is R. The minimum required co – efficient of friction between the blocks for them to stay at rest is $\mu .$ Find the value of $\frac{2}{5}\times R\times \mu$ in the units of Newton   

Figure shows a free body diagram of lower block:
 
Under the condition of block A to be just likely to slip, $f_{\max }=\mu N$    
Using Newton’s laws: $N\sin {37}^o-\mu N\cos {37}^o=0$ and $N\cos {37}^o+\mu N\sin {37}^o=50N$  
Thus, the required coefficient of friction is: $\mu =0.75$ 
If R is contact force with $R_{1}and{R}_2$ as their components along normal and tangent to the contact of the block
 
$R_1\sin {37}^o-R_2\cos {37}^o=0$ and $R_1\cos {37}^o+R_2\sin {37}^o=50N$
$R_1\left(\frac{3}{4}\right)=R_{2}and{R}_1\frac{4}{5}+\left(\frac{3}{4}{R}_1\right)\frac{3}{5}=50N$ 
$R_1=40Nand{R}_2=30NmakingR=50N$ 
$\frac{2}{5}\times R\times \mu =\frac{2}{5}\times 50\times 0.75=15N$