A solid block of mass 5kg is placed on a rough horizontal surface, in x – y plane as shown:
The friction coefficient between the surface and the cube is 0.4. An external force is applied on the cube given as: $\overrightarrow{F}=(6\widehat{i}+8\widehat{j}+20\widehat{k})N.$ Assuming $g=10m/s^2,$ the frictional force acting on the block is $\overrightarrow{f}=-(x\widehat{i}+y\widehat{j})N$ Find the value of (x + y) 

If the normal force acting on the block is N: 50N – $N$ – 20 N = 0 $\Rightarrow N$= 30 N
The limiting friction at the contact becomes: $f_{\max }=0.4\left(30N\right)=12N$  
So, the block will slide only if horizontal component of applied force exceeds a magnitude of 10N, thus, the block will not slide with horizontal component of force of $(6\widehat{i}+8\widehat{j})N.$
Thus, the frictional force developed should be - $(6\widehat{i}+8\widehat{j})N$