A force $\overrightarrow{F}=3 \hat{i}+4\widehat{j}$ acts on block as shown. The force of friction acting on the block is (g = 10 ms^–2)

$\overrightarrow{\mathrm{F}}=3\widehat{\mathrm{i}}+4\widehat{\mathrm{j}}$

$\begin{array}{l}\mathrm{Fcos}\mathrm{\theta }=1\mathrm{N}\\ \mathrm{Fsin}\mathrm{\theta }=4\mathrm{N}\\ \text{Limiting Friction }=\mathrm{\mu }(\mathrm{Mg}-\mathrm{Fsin}\mathrm{\theta })\\ =0.3(10-4)\\ =1.8\mathrm{N}\end{array}$

$\text{ Friction }=-1.8\widehat{\mathrm{i}}\mathrm{N}$ maximum friction but applied force is 3 N so frictional force is  1.8 N but in negative X direction