A force $F=\stackrel{\wedge }{i}+4\stackrel{\wedge }{j}$  acts on a block as shown. Their

Here $N+4=mg=10$

$\therefore N=6N$

$\therefore f_{\max }=0.4\times 6=2.4N$

The horizontal applied force is $F_H=1N$

$\therefore F_H<f_{\max }$ , so static friction comes into play

$\therefore F_H=f$  and acceleration of the block is zero

$\therefore f=1N$

$\therefore f=-\left(1N\right)\stackrel{\wedge }{i}$